NLM
1.
Which of Newton's laws of motion is involved in rocket propulsion ? Ans. Newton's third law of motion.
2.
Action and reaction are equal in magnitude and opposite in direction. Then, why do they not balance each other ? Ans. Action and reaction act on different bodies.
3.
A body is acted upon by a number of external forces. Can it remain at rest ? Ans. If the vector sum of all the external forces is zero, then the body will remain at rest.
4.
Identify the contact surface :
(a)
(b)
(c)
(d)
(e)
//////////////////////////////////////////////////////////////
(f)
(g) 5.
6.
7.
///////////////////////////////////////////////////
Find the velocity and acceleration of block A.
Ans. VA = 2 m/s (), aA = 4 m/s2 () Find the velocity of block B.
Ans. VB = 5 m/s ( )
Find all the normal reactions and the accelerations : (a) Ans.
(b) (a) N = 100 N, a = 2 m/s2 (c) N = 20 N, a = 2 m/s2|
RESONANCE
(c)
(d)
(b) N = 50 N, 150 N, a = 0 (d) N = 50 N, 150 N, a = 2 m/s2
Page # 1
8.
9.
Draw the FBD for the following systems : (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
A body of mass m is suspended by two strings making angles and with the horizontal. Find the tension in the strings. See figure
(A) T1
mg cos T2 sin( )
(C*) T1 10.
mg cos mg cos , T2 sin( ) sin( )
12.
13. 14.
mg sin sin( )
(D) none of these
A small block B is placed on block A of mass 1 kg and length 20 cm. If initially the block is placed at the right end of block A. A constant horizontal force of 10 N is applied on the block A. All the surfaces are assumed frictionless. Find the time in which B separates from A. [RRD PEARSON\Q.17/Pg-78]
(A*) 0.2 s 11.
(B) T1 T2
(B) 0.32 s
(C) 0.39 s
(D) 0.45 s
A toy train consists of three identical compartment A, B and C. It is being pulled by a constant force F along C. The ratio of the tensions in the string connecting AB and BC is (A) 2 : 1 (B) 1 : 3 (C) 1 : 1 (D*) 1 : 2 [RRD PEARSON\Q.1/Pg-84] Two blocks of mass 4 kg and 6 kg are placed in contact with each other on a frictionless horizontal surface. If we apply a push of 5N on the heavier mass, the force on the lighter mass will be (A*) 2N (B) 4 N (C) 5 N (D) None of these [RRD PEARSON\Q.4/Pg-85] Two masses m1 and m2 are attached to a string which pass over a frictionless fixed pulley. Given that m1 = 10 kg and m2 = 6 kg and g = 10 ms–2, What is the acceleration of the masses? (A*) 2.5 ms–2 (B) 5 ms–2 (C) 20 ms–2 (D) 40 ms–2 Two masses m1 and m2 are attached to the ends of a string which passes over a pulley attached to the top of an inclined plane. The angle of inclination of the plane in . Take g = 10 ms–2
RESONANCE
Page # 2
If m1 = 10kg, m2 = 5kg, = 30º, what is the acceleration of mass m2? [RRD PEARSON\Q.30/Pg-86] (A*) zero (B) (2/3) ms–2 (C) 5 ms–2 (D) 10 ms–2 15.
A body is placed over an inclined plane of angle . The angle between normal reaction and the weight of the body is (A)
16.
(B)
– 2
(C*) –
Find out the mass of block B to keep the system at rest :
(a)
(D) 0
[RRD PEARSON\Q.35/Pg-87] [BUT DPP#13/Q.11]
(b)
Ans. 10 kg 17.
18.
Find out the accelerations of the blocks and tensions in the strings.
[BUT DPP#13/Q.13]
Ans. zero
Find the velocity of A.
[Q.12/Build-up-DPP-9] Ans.. 19.
VA = 24 m/s ()
Find the acceleration of B.
[Q.13/Build-up-DPP-9] Ans.
aB = 2m/s2 (
RESONANCE
)
Page # 3
20.
Find the velocity of A.
[Q.15/Build-up-DPP-9] Ans. 21.
VA = V cos ()
Find the acceleration of wedge A
[Q.2/Build-up-DPP-10]
22.
Ans. aA = b tan
Find the acceleration of wedge A.
[Q.3/Build-up-DPP-10]
23.
Find the acceleration of B.
[Q.4/Build-up-DPP-10]
a cos 1 Ans. aB = cos 2
For the following questions assume. 24.
Find the acceleration of A w.r.t. ground.
[Q.8/Build-up-DPP-10] Ans. – b ˆi – 4b ˆj 25.
Find the acceleration of C w.r.t. ground
RESONANCE
Page # 4
[Q.9/Build-up-DPP-10]
26.
Ans. a ˆi – 2 (a + b) ˆj
Find the velocity of B w.r.t. ground.
[Q.10/Build-up-DPP-10]
Ans. – 4 ˆi + 8 ˆj 27.
Find the acceleration of B w.r.t ground.
[Q.6/Build-up-DPP-10] Ans. a (1 – cos) ˆi + a sin ˆj 28.
Sol.
29.
Sol.
Inside a horizontally moving box, an experimenter finds that when an object is placed on a smooth horizontal table and is released, it moves with an acceleration of 10 m/s2. In this box if 1 kg body is suspended with a light string, the tension in the string in equilibrium position. (w.r.t. experimenter) will be. (Take g = 10 m/s2) [Made PKS, 2005] (A) 10 N (B*) 10 2 N (C) 20 N (D) zero Acceleration of box = 10 m/s 2 Inside the box forces acting on bob are shown in the figure
T=
(mg)2 (ma)2 = 10 2 N
The block of mass ‘m’ is being pulled by a horizontal force F = 2 mg applied to a string as shown in figure (where ‘g’ is acceleration due to gravity). The pulley is massless and is fixed at the edge of immovable table. The force exerted by supporting table to the pulley is [Made MPS-2005]
(A) mg
(B) 2mg
The F.B.D. of pulley is as shown
(C)
3 mg
(D*) 2 2 mg
Let T1 , T2 and FS be the forces exerted by the horizontal string, vertical string and the support on the
RESONANCE
Page # 5
massless pulley respectively. Then T1 + T2 + FS = 0 or | FS | = | T1 + T2 | = 2 2 mg ( Tension in each string is | T1 | = | T2 | = 2 mg) 30.
If the resultant of three forces F1 = pi + 3j - k, F2 = -5i + j + 2k and F3 = 6i - k acting on a particle has magnitude equal to 5 units, then the value(s) of p is (are) (A) -6 (B*) -4 (C*) 2 (D) 4 [1994]
31.
Select the incorrect statement : (A)
(B*)
(C*) Sol.
(D)
A powerful man pushes a wall and the wall gets deformed. The magnitude of force exerted by
the wall on man will be equal to the magnitude of force exerted by man on the wall.
Ten person are pulling horizontally an object on a smooth horizontal surface in different
directions. The resultant acceleration of the object is zero. Then the pull of each man is equal to the pull of the other nine men.
A massless object cannot exert force on any other object.
All contact forces are electromagnetic in nature.
(B) & (C)
The direction of pull of each man is in opposite direction to that of net pull by nine other people. A massless body can exert force on other bodies if other bodies exert force on it
32.
Sol.
B & C are correct choices. marks shall be awarded for either correct choice.
Two springs are in a series combination and are attached to a block of mass ‘m’ which is in equilibrium. The spring constants and the extensions in the springs are as shown in the figure. Then the force exerted by the spring on the block is :
k1 k 2 (A*) k k (x 1 + x 2) 1 2
(B) k 1x 1 + k 2x 2
(C*) k 1x1
(A, C) Tension in both springs are same i.e. k 1 x 2 = k 2 x 2 = force exerted by lower spring on the block.
RESONANCE
(D) None of these
Page # 6
Comprehensions # 1 Two blocks (as shown in figure) are connected by a heavy uniform rope of mass 4 kg. An upward force of 300 N is applied on upper block. (g = 10 m/sec2) [D TEST 2007]
33.
The acceleration of the system is equal to ; (A) 20 m/sec2 upward (B*) 10 m/sec2 upward (C) 10 m/sec 2, downward (D) 30 m/sec 2, downward
34.
The tension at the top of the heavy rope is equal to (A) 90 N (B) 60 N (C*) 180 N
(D) 240 N
35.
The tension at the mid-point of the rope is equal to (A) 240 N (B) 180 N (C) 70 N
(D*) 140 N
Sol.
T 2 – (5 + 2)g = (5 + 2) a T 2 = 140 N
Match the following : 36. (p) force between the earth and the falling stone (i) Gravitational force (q) The pressing force between one block and another block (ii) Electromagnetic force (r) The stretching force developed in a spring. (iii) Strong nuclear force (s) The force between the proton and the neutron in a nuclear (iv) Weak nuclear force Ans. (p-1), (q-ii), (r-iii), (s-iii) 37.
The expression given below have their usual meaning match them with column on right side u v t 1. S = 2
2. Internal force of a system always add upto zero 3. To stop a car in minimum distance break is applied such that car does not slip Ans. 1-b, 2-d, 3-e, 4-g
RESONANCE
(a) variable acceleration
(b) Constant acceleration (c) Newton’s second law (d) Newton’s third law (e) Bodies which can deform (f) Bonding between the molecules (g) s > k
Page # 7
38.
Assertion : The pressing force between two blocks is an example of electromagnetic interaction force. Reason : At microscopic level, all bodies are made of charged constituents (Nuclei and electron.) so any mechanical contact causes mutual forces between there charges. [Made A.K. 2007] (A*) If both Assertion and Reason are true and the Reason is correct explanation of Assertion. (B) If both Assertion and Reason and true but Reason is not a correct explanation of Assertion (C) If Assertion is true but Reason is false. (D) If both Assertion and Reason are false.
39.
A monkey of mass 20 kg is holding a vertical rope. The rope can break when a mass of 25 kg is suspended from it. What is the maximum acceleration with which the monkey can climb up along the rope? [Pearson/Page No. 85/Q.No. 8] (A) 7 ms–2 (B) 10 ms –2 (C) 5 ms –2 (D*) 2.5 ms –2
40.
A cricket player catches a ball of mass 100g moving with a velocity of 25 m/s. If the ball comes to rest in 0.1s after being caught, then the force of the blow exerted on the hand of the player is : [Pearson/Page No. 87/Q.No. 41] (A) 4N (B) 40 N (C*) 25 N (D) 250 N
41.
Find the extension in the spring if : (i) a = g/2 upwards (ii) a = g/2 downwards
Ans. (i) 42.
(iii) a = 0
3 mg 1 mg mg (ii) (iii) 2 k 2 k k
Find out the reading of the weighing machine in the following cases.
Ans. 10 3 N or
3 kg
43.
A block of metal is lying on the floor of a bus. The maximum acceleration which can be given to the bus so that the block may remain at rest, will be (A) g2 (B) 2 g (C*) g (D) /g [Q.14/Objective Physics/Pg.-85]
44.
A block of mass 4kg is suspended through two light spring balances A and B. Then A and B will read respectively : [GRB/obj.physics/pg.148/Q.251]
(A*) The reading of A is 4kg (C*) The reading of B is 4kg
RESONANCE
(B) The reading of A is 0 kg (D) The reading of B is 2kg
Page # 8
Comprehension 45. Mr. Abdulla wishes to move a heavy block of mass 10 kg by means of an engine through a string such that reading of the spring balance is 5 kg. The acceleration of the block will be :
(A*) 4m/s2 46.
(B) 2m/s 2
(C) 2.6m/s2
(D) none of these
If there is a weighing machine on which the 10 kg block is placed and two birds 1 and 2 as shown in figure, the weighing machine reads 7.5 kg and the friction coefficient between the block A and weighing machine is = 0.6. Mr. Abdulla himself applies a force of 5N as shown. Then the acceleration of the block will be : (Bird '2' is of mass 2 kg)
(A) 3.5 m/s2
(B) 4m/s 2
(C*) 0 m/s 2
(D) 1 m/s 2
47.
What will be the acceleration of the block, if Abdulla makes the bird 1 flew away ? (A) 0 m/s2 (B) 1 m/s2 (C) 0.5 m/s2 (D*) 0.3 m/s 2
48.
What will be the mass of the bird 1 ? (A) 600 gm (B) 1 kg
49.
Find the acceleration of B.
(C*) 500 gm
(D*) can not be determined
[Q.11/Build-up-DPP-9] Ans. 50.
aB = 2 m/s2 ()
If block A is released from rest, then find the extension in spring and accelerations of the blocks of mass m just after release.
(a) Ans. 51.
(b) (a) aA = g ; aB = 0 ; x = mg/K ; (b) aA = g ; x = 0
Find out the accelerations of the block B in the following systems :
[Q.9/Build-up-DPP-13]
RESONANCE
Page # 9
Ans. aB = 2g/3 52.
A bob is suspended with the help of a thread whose breaking load is twice the weight of the bob. Taking g = 10 ms–2, what is the minimum time in which the bob can be raised by 10 m? (A*) 2 s (B) 2 2 s (C) 1/ 2 s (D) 1 s
53.
Find out the time taken by the block A of mass ‘m’ to reach pt. B from pt. A. The length of inclined plane is .
Ans. 54.
t=
8 3g
System is shown in the figure and man is pulling the rope from both sides with constant speed ' u'. Then the velocity of the block will be: [ Made 2004]
(A*)
3u 4
(B)
3u 2
(C)
u 4
(D) none of these
55.
The velocity time graph of a lift moving downwards is a straight line inclined to the time axis at 45°. If mass of the lift is m kg. What is the effective weight (in newton) of the lift? Take g = 10 ms –2 (A) m (B*) 9m (C) 10 m (D) None of these
56.
Find out all the normal reactions acting in the system.
Ans.
N1 = m (g cos + a sin)
Ng = N1 cos + Mg
57.
A book is lying on the table. What is the angle between the action of the book on the table and the reaction of the table on the book? [Q.20/Objective Physics/Pg. 86] (A*) 180° (B) 90° (C) 45° (D) 0°
58.
When we kick a stone, we get hurt. Due to which one of the following properties does it happens? (A) velocity (B) momentum (C) inertia (D*) reaction [Q.40/Objective Physics/Pg. 87]
59.
Find the velocity of A with respect to pulley P.
Ans.
VAP = 8 m/s ()
RESONANCE
Page # 10
60.
In the figure M = 2m. The coefficient of friction at all surfaces is . String is light and inextensible and all the pulleys are smooth. D.C. Pandey_pg.220_30
M
61.
m
The free body diagram for block M will be as follows (True/False).
T
T N1
N1
Mg
T
a
Ans.
False (normal reaction and friction due to the small block is not shown).
62.
The free body diagram for block m will be as follows (True/False).
T
N2
a
N2 mg
Ans.
False (acceleration will be 2a).
63.
Figure shows a small block A of mass m kept at the left end of a plank B of mass M = 2m and length . The system can slide on a horizontal road. The system is started towards right with the initial velocity v. The friction coefficients between the road and the plank is 1/2 and that between the plank and the block is 1/4. Find D.C. Pandey_pg.219_29
A
B
(a) the time elapsed before the block separate from the plank. (b) displacement of block and plank relative to ground till that moment.
Ans. (a) t = 4 3g 64.
2 5 (b) SA = 4 V 3g 3 , SB = 4 V 3g 3 ,
The diagram shows a cord supporting a picture. Which of the graphs shown in figure correctly represents the relationship between the tension T in the cord and the angle ? [JP DPP-24 _Q.2 ]
RESONANCE
Page # 11
(A)
65.
(C)
(D)
In the figure is shown the top view of two horizontal forces pulling a box along the floor. (a) (b)
66.
(B)
A batch DPP 46_19.7.05
How much work does each force do as the box is displaced 70 cm along the vertical line? What is the total work done by the two forces in pulling the box this distance?
[ Ans.: (a) w 1 = 29.7 J, w 2 = 51.89 J (b) 81.2 J]
The force required to stretch a spring varies with the distance as shown in the figure. If the experiment is performed with the above spring of half length, the line OA will [3 marks] USS_Pg.296_22
F
(A*) Shift towards F-axis (C) Remain as it is
O
A x
(B) Shift towards X-axis (D) Become double in length
67.
In the system of pulleys , the value of M such that 14 kg block remains at rest is :
68.
(A) 28 kg (B) 35 kg (C*) 24 kg (D) 42 kg Two blocks of masses m 1 = 5 kg and m 2 = 2 kg hang on either side of a frictionless cylinder as shown in the figure . If the system starts at rest , what is the speed of m 1 after it has fallen 40 cm ?
DPP 47_ACJ_05-06
DPP 48_ACJ_05-06
[ Ans. :
24 m/s ] 7
RESONANCE
Page # 12
69.
A 1 kg block ‘B’ rests as shown on a bracket ‘A’ of same mass . Constant forces F1 = 20 N and F2 = 8
N start to act at time t = 0 when the distance of block B from pulley is 50 cm . Time when block B reaches the pulley is _______ .
DPP 48_ACJ_05-06
[ Ans. : 0.5 sec ] 70.
71.
A ladder that is 3.0 m long and weighs 200 N has its center of gravity 120 cm from the foot. At its top end is 50 N weight. Compute the work required to raise the ladder from a horizontal position on the ground to a vertical position. DPP 49_ACJ_05-06 [ Ans : 390 J ]
Two blocks with masses m 1 = 4 kg and m 2 = 5 kg are connected by a light rope and slide on a frictionless wedge as shown in the figure . Given that it starts at rest , what is the speed of m 2 after it has moved 40 cm along the incline ? DPP 49_ACJ_05-06
[ Ans. : 1.19 = 8/3 5 ] 72.
(i)
(ii)
(iii)
(iv)
(v)
In all the given cases blocks are at rest, are in contact and the forces are applied as shown. All the surfaces are smooth. Then in which of the following cases, normal reaction between the two blocks is zero : (A) (i) , (iv) (B) (ii) , (iii) (C) (iii) (D*) (v) 73.
A block of 6 kg is put between two smooth walls. If F = 50 2 is also applied as shown in figure, then [5 Marks]
[Made M. Pathak]
P&Q_DPP 44_8
(A*) Interaction force on the block due to walls = 50 ˆi 110 ˆj (B) Interaction force on the walls due to the block = 50 ˆi 110 ˆj (C) If F were reversed, now interactron force on the block due to wall = 50 ˆi 110 ˆj
50 ˆ i m/sec 2. (D*) If F were reversed, now the acceleration of the block = 6
RESONANCE
Page # 13
Sol. Ny = 110 Nx = 50 Nblock 50 ˆi 110ˆj So A is correct N wall will be opposite of Nblock = 50 ˆi 110 ˆj So B is incorrct. If F were reversed
it will loose contact with verticle wall and Ny = 10 ˆj So option C is incorrect.
Net force = 50 ˆi 6 a So D is correct. 74.
50 ˆ a i 6
A block of mass m is placed on a wedge. The wedge can be accelerated in four manners marked as (1), (2), (3) and (4) as shown. If the normal reactions in situation (1), (2), (3) and (4) are N1, N2, N3 and N4 respectively and acceleration with which the block slides on the wedge in situations are b1, b2, b3 and b4 respectively then : [5 Marks] P&Q_DPP 44_9
(A*) N3 > N1 > N2 > N4 (B) N4 > N3 > N1 > N2
Sol.
(C*) b2 > b3 > b4 > b1
(D) b2 > b3 > b1 > b4
(1) Balancing forces perpendicular to incline N = mg cos37° + ma sin37° N1 =
RESONANCE
4 3 mg + ma 5 5
Page # 14
and along incline mg sin37° – ma cos37° = mb1
b1 =
3 4 g– a 5 5
(2) Similarly for this case get N2 = N2 =
4 3 mg – ma 5 5
4 3 mg – ma 5 5
and b2 =
3 4 g+ a 5 5
(3)
Similarly for this case get N3 =
4 4 3 3 mg + ma and b3 = g + a 5 5 5 5
Similarly for this case get N 4 =
4 4 mg – ma 5 5
(d)
and b4 = 75.
3 3 g– a 5 5
Find the tensions in the strings (1), (2) and (3) and the acceleration of the mass ‘m’ just after : [S-05-06/DPP 39/Q.10]
(a) string (1) is cut Ans.
(b) string (2) is cut
(a) T1 = 0 ; T2 = mg ; T3 = 2 mg ; a = g (b) T1 = mg ; T2 = 0 ; a = 0 (c) T1 = mg ; T2 = 0 ; T3 = 0 ; a = 0
RESONANCE
(c) string (3) is cut
Page # 15
76.
A body of mass 10 kg is on a rough inclined plane of inclination = sin 1 (3/5) with the horizontal. When a force of 30 N is applied on the block parallel to and upward the plane, the total reaction by the plane on the block is nearly along : [S-05-06/DPP 44/Q.5]
(A*) OA 77.
(B) OB
(C) OC
(D) OD
Three equal balls 1, 2, 3 are suspended on springs on below the other as shown in the figure. OA is a weightless thread. (a) If the string is cut, the system starts falling. Find the acceleration of all the balls at the initial instant.
(b) Find the initial accelerations of all the balls if we cut the spring BC which is supporting ball 3 instead of cutting the thread. [Bank/Physics_Bank/Jaipur/DPP(A)/Q.10] Ans. [(a) 3g, 0, 0, (b) 0, g, g] 78.
System is shown in the figure. Assume that cylinder remains in contact with the two wedges.The velocity of cylinder is A-Batch_DPP-58_05-06_3
(A)
19 4 3
u m/s 2
(B)
13 u m/s 2
(C)
3 u m/s
(D*)
7 u m/s
79.
In the figure shown the velocity of different blocks is shown. The velocity of C is : (A) 6 m/s JP DPP_51_9 (B*) 4 m/s (C) 0 m/s (D) none of these
80.
Two blocks A (5 kg) and B (3 kg) resting on a smooth horizontal plane are connected by a spring of stiffness 294 N/m. A horizontal force F of magnitude 3 × 9.8 N acts on A as shown. At the instant B has an acceleration of 4.9 m/s2 towards left : JP DPP_51_10
(A) the acceleration of A is 3 × 1.96 m/s2 (C) the extension in the spring is 0.1 m
RESONANCE
(B*) the acceleration of A is 3 × 0.98 m/s2 (D) the extension in the spring is 0.2 m
Page # 16
81.
82.
83.
A ball of mass m is attached to the lower end of a light vertical spring of force constant k. The upper end of the spring is fixed. The ball is released from rest with the spring at its normal length, and comes to rest again after descending through a distance x : JP DPP_52_1 (A) x = mg/k (B*) x = 2mg/k (C*) The ball will have no acceleration at the position where it has descended through x/2 (D*) The ball will have an upward acceleration equal to g at its lowermost position. Consider the situation shown in figure. Initially the spring is unstretched when the system is released from rest. Assuming no friction in the pulley, find the maximum elongation of the spring. JP DPP_52_2 Ans. [2 mg/k]
In the figure shown neglecting friction and mass of pulleys, what is the acceleration of mass B? fp=kesan'kkZ,vuqlkj?k"kZ.k,oaiqyhdsnzO;ekudksux.;ekusa]nzO;ekuBdkRoj.kD;kgksxk? [ 3.46_NLM ]
(A) 84.
g 3
(B)
5g 2
(C)
2g 2
(D*)
2g 5
Two blocks A and B of equal mass are connected by a light inextensible string passing over two light smooth pulleys fixed to the blocks. The parts of the string not in contact with the pulleys are horizontal. A horizontal force F is applied to the block A as shown. Then: M.Bank_NLM_3.57 nksxqVdsArFkkBftudsnzO;ekulekugS],dgYdhvrU;jLlh}kjk¼jLlhtksxqVdksalsca/ksnksfpduhiqyhesa lsxqtjrhgSA½tqM+sgq,gSaAjLlhdkogHkkxtksiqyhlstqM+kughagS]{kSfrtgSA,d{kSfrtcyFxqVdsAijfp=kkuqlkj yxk;ktkrkgS]rks-
(A) the acceleration of A will be equal to that of B (B*) the acceleration of A will be greater than that of B (C) the acceleration of A will be less than that of B (D) none of the above is correct. (A)Adk Roj.k] B ds Roj.k ds cjkcj gksxk (B*)Adk Roj.k] B ds Roj.k ls vf/kd gksxk (C)Adk Roj.k] B ds Roj.k ls de gksxk (D)buesalsdksbZugha 85.
A weightless inextensible rope on a stationary wedge forming angle with the horizontal. One end of the rope is fixed to the wall at point A. A small load is attached to the rope at point B. The wedge starts moving to the right with a constant acceleration. Determine the acceleration a1 of the load when it is still on the wedge. M.Bank_NLM_8.12 ,dgYdhvrU;jLlhfLFkjostij{kSfrtlsdks.kcukrsgq,j[khgSAjLlhdk,dfljkfcUnqAijfLFkjgSAfcUnq B ij,dNksVklknzO;ekutksM+j[kkgSAostnk;havksjfu;rRoj.klsxfrdjuk'kq:djnsrkgSAB ijtqM+snzO;ekudk
RESONANCE
Page # 17
Roj.k a1 Kkr dhft,] tcfd ;gvHkhHkhost ijgks-
M.Bank_NLM_8.12
[ Ans: 2 a sin (/2) ] 86.
In the figure shown the blocks A & C are pulled down with constant velocities u . Acceleration of block B is : [ Made 2003] fn[kk;sx;sfp=kesaArFkkC CykWdksadksfu;rosxuls[khapktkrkgSACykWdBdkRoj.kgksxk : [ Made 2003] M.Bank_NLM_8.13
(A)
87.
(C)
u2 tan2 sec b u2 sec 2 tan b
(D) zero ‘'kwU;
(B*) m g
(C) 2 m g
(D) none of these
In the figure shown, find out the value of [ assume string to be tight ] [ Made 2004] fn[kk;sx;sfp=kesadkekucrkb;s[jLlhdksdlkgqvkekus] [ Made 2004] M.Bank_NLM_8.25
(A*) tan1 89.
u2 tan3 b
In the figure shown the car moves down with a constant acceleration g. A bob of mass m is attached to a string whose other end is tied to the ceiling of the car. If the bob remains stationary relative to the car. Then tension in the string is: M.Bank_NLM_5.20
(A) m g/2 88.
(B*)
3 4
(B) tan1
4 3
(C) tan1
3 8
(D)noneofthesebuesalsdksbZugha
A block of weight 9.8N is placed on a table. The table surface exerts an upward force of 10 N on the block. Assume g = 9.8 m/s 2. [M.Bank_NLM_4.3] 9.8N Hkkj okyk ,d xqVdk est ij j[kk gqvk gSA est dh lrg xqVds ij Åij dh vksj 10 N dk cy yxkrh gSA ekfu, g = 9.8 m/s2 (A*) The block exerts a force of 10N on the table xqVdk est ij 10N dk cy yxkrk gSA (B) The block exerts a force of 19.8N on the table xqVdk est ij19.8N dk cy yxkrk gSA (C) The block exerts a force of 9.8N on the table xqVdk est ij 9.8N dk cy yxkrk gSA (D*) The block has an upward acceleration. xqVdsdkÅijdhvksjRoj.kgksxkA
RESONANCE
Page # 18
90.
91. 92.
Two weights W 1 & W 2 in equillibrium and at rest, are suspended as shown in figure. Then the ratio W 1/ W 2 is about : [M.Bank_NLM_3.57] nksHkkjW1 rFkkW2 fp=kkuqlkjyVdsgq,gSrFkklkE;koLFkkvkSjfojkeesagSrksW1/W2 dkvuqikrgksxk:
(A*) 5/4
(B) 4/5
(C) 8/5
(D) none of these buesalsdksbZugha
In the previous question the tension in the string attached to the wall is:
iwoZç'uesanhokjlstqM+hjLlhesarukogksxk& (A*) 0.6 W 1
(B) 0.8 W 1
(C) W 1
[M.Bank_NLM_3.58]
(D) none of these buesalsdksbZugha
Two smooth spheres each of radius 5 cm and weight W rest one on the other inside a fixed smooth cylinder of radius 8 cm. The reactions between the spheres and the vertical side of the cylinder are : [M. Ban k (0 7-
08)_NLM_4.2]
8 lseh0f=kT;kdsfpdusfLFkjcsyudsvUnjnksfpdusxksysçR;sdf=kT;k5lseh0rFkkW HkkjokysfLFkjj[ksgSA
xksyksadhm/oZnhokjksadslkFkçfrfØ;k,saKkrdjks\
93.
(A) W/4 & 3W/4
(B) W/4 & W/4
(C*) 3W/4 & 3W/4
(D) W & W
Two blocks A and B of masses 4 kg and 12 kg respectively are placed on a smooth plane surface. A force F of 16 N is applied on A as shown. The force of contact between A & B is: 4 fdxzk0 rFkk 12 fdxzk0 nzO;eku dsnksCykWdØe'k% ArFkkB fpdus {kSfrtlrgij j[ks gq, gSaA ,d 16 N dkcy F, Aij fp=kkuqlkj yxk;k tkrk gSAArFkk B ds e/; yxus okys lEidZ cy dk eku gksxk & [M.Bank(07-08)_N.LM_4.7]
(A) 4 N 94.
(B) 8 N
(C*) 12 N
(D) 16 N
The block of mass ‘m’ equal to 100 kg is being pulled by a horizontal force F = 2 mg applied to a string as shown in figure (Take g = 10 m/s2). The pulley is massless and is fixed at the edge of an immovable table. What is the value of force exerted by the supporting table on the pulley (in Newton) ‘m’xqVdsdknzO;eku100kggSvkSjbls{kSfrtcyF=2mgtksfdfp=kkuqlkjyxrkgS]ls[khapktkrkgSA(g=10m/ s2ysa)AiqyhnzO;ekughugSrFkkfLFkjestdsfdukjsijc¡/khgqbZgSAest}kjkiqyhijyxk;sx,cydkeku(U;wVuesa)
gksxkA
Sol.
Made_MPS_2005
[M.Bank(07-08)_NLM_3.56]
The F.B.D. of pulley is as shown
RESONANCE
Page # 19
Let T1 , T2 and FS be the forces exerted by the horizontal string, vertical string and the support on the massless pulley respectively. Then T1 + T2 + FS = 0 or | FS | = | T1 + T2 | = 2 2 mg ( Tension in each string is | T1 | = | T2 | = 2 mg) Ans. 2828 95.
A horizontal force of magnitude F is applied on a body of mass m as shown in figure. The magnitude of net normal reaction on the block is : ,dm nzO;ekudhoLrqijF cyfp=kkuqlkjyxk;ktkrkgSAxqVdsijifj.kkehvfHkyEcçfrfØ;kcydkifjek.kgksxk : [M.Bank(07-08)_NLM_4.9]
96.
(A*) 10 2 N
(B)
10
2
N
(C) 10 N
(D) none of these buesalsdksbZugha
Consider the system as shown in the figure. The pulley and the string are light and all the surfaces are frictionless. The tension in the string is (g = 10 m/s 2).
fp=kesan'kkZ;svuqlkjfLFkfrdksekfu,AiqyhrFkkjLlhnzO;ekughugSrFkklHkhlrg?k"kZ.kghugSAjLlhesarukogksxkA
97.
(g = 10 m/s2).
[M.Bank(07-08)_NLM_3.30]
(A) 0 N
(B) 1 N
(C) 2 N
(D*) 5 N
A painter is applying force himself to raise him and the box with an acceleration of 5 m/s2 by a massless rope and pulley arrangement as shown in figure. Mass of painter is 100 kg and that of box is 50 kg. If g = 10 m/s 2, then: [M.Bank(07-08)_NLM_3.45] 2 (fp=kkuqlkj,disUVjLo;acyyxkdjviusvkidksrFkkck¡Dldks5 m/s dsRoj.kls,dnzO;ekughujLlhoiqyhO;oLFkk }kjkfp=kkuqlkjÅijmBkrkgSA;fnisUVjdknzO;eku100 kg rFkkck¡Dldk50 kg gSrFkkjLlhdknzO;ekuux.;gS (;fn g = 10 m/s2 ) rks
tension in the rope is 1125 N (jLlh esa ruko 1125 N) tension in the rope is 2250 N (jLlh esa ruko 2250 N) force of contact between the painter and the floor is 375 N (isUVj rFkk Q'kZ ds chp lEidZ cy 375 N gS) (D) none of these (buesalsdksbZugha) For painter ; R + T – mg = ma R + T = m(g + a) ............(1) For the system ; (A*) (B) (C*)
Sol.
RESONANCE
Page # 20
2T – (m + M)g = (m + M)a 2T = (m + M) (g + a) where ; m = 100 kg M = 50 kg a = 5 m/sec2
and ; 98.
150 15 2 R = 375 N. T=
=
..............(2)
1125 N
Two blocks ‘A’ and ‘B’ each of mass ‘m’ are placed on a smooth horizontal surface. Two horizontal force F and 2F are applied on the 2blocks ‘A’ and ‘B’ respectively as shown in figure. The block A does not slide on block B. Then the normal reaction acting between the two blocks is : (A and B are smooth) nks xqVds ‘A’o ‘B’, izR;sd dk nzO;eku‘m’,d fpduh {kSfrt lrg ij j[ks gSaA nks {kSfrt cyF o 2F Øe'k% xqVdsAo ‘B’ijfp=kkuqlkjvkjksfirgSAxqVdkA, xqVdsB ijughafQlyrkgSrksnksxqVdksadse/;dk;ZjrvfHkyEcizfrfØ;kcy
gSA ¼A rFkkB dh laifdZr lrg ?k"kZ.kghu gS½
(A) F Sol.
(B) F/2
nksnzO;ekufudk;dkRoj.kgSa=
99.
(C)
Acceleration of two mass system is a =
FBD of block A
[Q.33/RK_BM/Constrained Motion] [M.Bank(07-08)_NLM_8.29]
F 2m
F
(D*) 3F
3
F leftward 2m
ck;havksj
N F
xqVds Adk eqDr oLrq js[kkfp=k
60°
30°
mF N cos 60° – F = ma = solving N = 3 F 2m In the shown mass pulley system, pulleys and string are massless. The one end of the string is pulled by the force F = mg. The acceleration of the block will be [M.Bank(07-08)_NLM_3.51] fp=kesan'kkZ;sf?kjuhnzO;ekufudk;esaf?kjuhrFkkjLlhnzO;ekughugSAjLlhdk,dfljkcyF=mg}kjk[khapk
tkrkgSAxqVdsdkRoj.kgksxk&
(A) g/2 100.
Sol.
(B) 0
(C*) g
(D) 3g
The system starts from rest and A attains a velocity of 5 m/s after it has moved 5 m towards right. Assuming the arrangement to be frictionless every where and pulley & strings to be light, the value of the constant force F applied on A is : [Made 2006, VSS, GRSTUX] [M.Bank(07-08)_NLM_3.54] fudk;fojkelsizkjEHkgksrkgSrFkkAnk;havksj5m pyusdsi'pkr~5m/s dkosxizkIrdjrkgSAO;oLFkkdkslHkhtxg ?k"kZ.kjfgrof?kjfu;ksaoMksfj;ksadksgYdhekursgq,Aijvkjksfirfu;rcyFdkekugS-
(A) 50 N (B)
RESONANCE
(B*) 75 N
(C) 100 N
(D) 96 N Page # 21
a=
v 2 25 = 2.5 m/s2 2s 10
For 6 kg : – F – 2T = 6a For 2 kg : – T – 2g = 2 (2 a) From (1) & (2) F = 75 N 101.
In the figure shown, find out the value of [ assume string to be tight ] fn[kk;sx;sfp=kesadkekucrkb;s[jLlhdksdlkgqvkekus] [ Made 2004]
(A*) tan1 102.
ugha
3 4
(B) tan1
4 3
(C) tan1
3 8
[M.Bank(07-08)_NLM_8.23]
(D) none of these buesa ls dksbZ
A system is as shown in the figure. All speeds shown are with respect to ground. Then the speed of Block B with respect to ground is : [M.Bank(07-08)_NLM_8.27] fp=kesa,dfudk;iznf'kZrgSAlHkhpky/kjkrydslkis{kgSaArks/kjkrydslkis{kCykWdBdhpkygksxhA
[Q.31/RK_BM/Constrained Motion]
Sol.
(A) 5 m/s (B)
(C) 15 m/s
(D) 7.5 m/s
1 + 2 2 = constant
d 1 2 d 2 + =0 dt dt ( 5 + 5) + 2 (5 + vB) = 0 or
103.
(B*) 10 m/s
vB = 10 m/s
Three blocks are connected by strings as shown in figure and pulled by a force F = 60 N. If m A = 10 kg, m B = 20 kg and m C = 30 kg, then : rhu CykWdksa dks fp=kkuqlkj jLlh ls tksM+dj F = 60 N cy ls [khapk tkrk gSA ;fn mA = 10 kg, mB = 20 kg rFkk mC = 30 kg gks rks: [M.Bank_NLM_3.34]
RESONANCE
Page # 22
(A) acceleration of the system is 2 m/s 2 (B*) T 1 = 10 N (C*) T 2 = 30 N 104.
fudk;dkRoj.k2m/s2gksxkA
(D) T 1 = 20 N & T 2 = 40 N
A cart of mass 0.5 kg is placed on a smooth surface and is connected by a string to a block of mass 0.2 kg. At the initial moment the cart moves to the left along a horizontal plane at a speed of 7 m/s. (Use g = 9.8 m/ s2) nzO;eku0.5fdxzk-dh,dxkM+h]0.2fdxzk-nzO;ekudsfi.MlsjLlh}kjktqM+hgS]izkjEHkesaxkM+hfpdus{kSfrtlery
ijck;ahavksj7eh-/ls-. dhpkylsxfrekugS(g=9.8eh-/ls2 ysa) [M.Bank_NLM_3.47]
0.5 kg
[Modified AKS_2007]
0.2 kg 2g towards right. 7 (B*) The cart comes to momentary rest after 2.5 s. (C*) The distance travelled by the cart in the first 5s is 17.5 m. (D) The velocity of the cart after 5s will be same as initial velocity. (A*) The acceleration of the cart is
(A*) xkM+hdkRoj.k
2g nk;harjQgksxkA 7
(B*)xkM+h2.5lsd.M i'pkr~{kf.kd:dsxhA
(C*)igys5 lsd.MesaxkM+h}kjkr;dhxbZnwjh17.5eh-gSA (D)5lsd.Mi'pkr~xkM+hdkosxizkjfEHkdosxdslekugksxkA Sol.
0.2 g = 0.7 a a =
2g m/s 2 7
For the case, it comes to rest when V = 0
ml fLFkfr ds fy, tc ;g fLFkj gksrk gS V = 0 2g t 0 = 7 + 7
49 t = 2g = 2.5 s
Distance travelled till it comes to rest
7 m/s
a
T T
T = 0.5 a
fLFkjvoLFkkvkusrdpyhxbZnwjh
a 0.2g
0.2 - T = 0.2 a
2g s 0 = 72 + 2 7
S = 8.75 m So in next 2.5s, it covers 8.75 m towards right. Total distance = 2 x 8.75 = 17.5 m After 5s, it speed will be same as that of initial (7 m/s) but direction will be reversed.
105.
vr%vxys2.5 ls-esa;gnk;havksj8.75 eh-nwjhr;djsxkA dqy nwjh = 2 x 8.75 = 17.5 m 5 ls-i'pkr~bldhpky]izkjfEHkdpky(7 eh-/ls-)dscjkcjgksxhijUrqfn'kkfoijhrgksxhA
A 1 kg block ‘B’ rests as shown on a bracket ‘A’ of same mass. Constant forces F 1 = 20 N and F 2 = 8 N start to act at time t = 0 when the distance of block B from pulley is 50 cm.Time when block B reaches the pulley is _______. (Assume that friction is absent every where. Pulley and string are light.
,d 1 fdxzk- nzO;eku dk CykWd ‘B’ leku nzO;eku ds CykWd ‘A’ ij fLFkj j[kk gS t = 0 ij cy F1 = 20 N rFkk
RESONANCE
Page # 23
F2 = 8 N yxk;stkrsgSblle;CykWdB dhiqyhlsnwjh50 cm gSrksCykWdB fdrusle;esaiqyhrdigqapsxk_____(;g ekfu,fdlHkhtxg?k"kZ.kvuqifLFkrgSA¼iqyhrFkkjLlhgYdhgS) [M.Bank(07-08)_NLM_3.2]
[ Ans: 0.5 sec. ] 106.
In the figure if blocks A and B will move with same acceleration due to external agent, there is no friction between A and B, then the magnitude of interaction force between the two blocks will be () : fp=kesa;fnCykWdArFkkB fdlhckácydsdkj.k,dlekuRoj.klsxfrdjsarFkkArFkkB dse/;dksbZ?k"kZ.kcy
ughagksrksnksuksaCykWdksdse/;vfHkyEcçfrfØ;kcydkekuD;kgksxkA [M.Bank_NLM_4.10]
Sol.
(A*) 2 mg/cos N cos
(B) 2 mg cos (C) mg cos N = 2 mg/cos . Ncos
N a
Nsin
(D) none of these
2m A
a
2mg
107. A 2 kg toy car can move along an x axis. Graph shows force F x, acting on the car which begins at rest at time t = 0. The velocity of the particle at t = 10 s is :
,d2 kg dh f[kykSuk dkj x v{k ds vuqfn'k xfr dj ldrh gSA xzkQ cy Fx, dks çnf'kZr djrk gS] tks le; t = 0 ij fojke ij dkj ij yxuk çkjEk gksrk gSA t = 10 s ij d.k dk osx gS % [Made A.K.S. sir] [M.Bank_NLM_7.35]
Fx(N) 4
0 -2
Sol.
(A) – i m/s
dp
= pf – pi
=
F dt
pi = 0 Net Area = 16 – 2 – 1
4
(B) – 1.5 i m/s
8
9 10
= Area under the curve.
t(s)
11
(C*) 6.5 i m/s
(D) 13 i m/s
= 13 N-s
13 = 6.5 i m/s 2 [As momentum is positive, particle is moving along positive x axis.] = Vf =
RESONANCE
Page # 24
108.
Sol.
109.
Sol.
110.
Three rigid rods are joined to form an equilateral triangle ABC of side 1m. Three particles carrying charges 20 C each are attached to the vertices of the triangle. The whole system is at rest always in an inertial frame.The resultant force on the charged particle at A has the magnitude. rhulqn`<+NM+s1eh-Hkqtkdsleckgqf=kHkqtABCds:iesatksM+hx;hgSArhud.kftuesaizR;sdij20 Cvkos'k gS]f=kHkqtds'kh"kksZijtksM+sx;sgSAlEiw.kZfudk;,dtM+Roh;funsZ'kkrU=kesafojkekoLFkkesagSAAijfLFkrvkosf'kr d.k ij ifj.kkeh cy dk ifjek.k gS & [M.Bank_HCV_Ch.5_Ob.1_8] (A*) zero
(B) 3.6 N
Fnet m a
(C) 3.6 3 N
(D) 7.2 N
a = acceleration of charge of particle at A = 0
Fnet = 0. Two springs are in a series combination and are attached to a block of mass ‘m’ which is in equilibrium. The spring constants and the extensions in the springs are as shown in the figure. Then the force exerted by the spring on the block is : [M.Bank_NLM_1.16] nksfLizaxJS.khla;kstuesagSrFkk‘m’nzO;ekudsxqVdslstqM+sgSatkslkE;koLFkkesagS]rksfLizax}kjkxqVdsijvkjksfircy gS-
k1 k 2 (A*) k k (x 1 + x 2) (B) k 1x 1 + k 2x 2 (C*) k 1x1 (D) None of these mijksDresadksbZugha 1 2 (A C) Tension in both springs are same nksuksafLiazxksaesarukolekugS i.e. vFkkZr~ k1 x1 = k2 x2 = mg = force exerted by lower spring on the block. xqVds ij fupys fLizax }kjk vkjksfir
cy
Mass m shown in figure is in equilibrium. If it is displaced further by x and released find its acceleration just after it is released. Take pulleys to be light & smooth and strings light. fp=kesafn[kk,vuqlkjnzO;ekumlkE;koLFkkesagSAvxjblsxnwjhrdvkSj[khapktkrkgSvkSjfQjNksM+fn;ktkrk
gSrksNksM+usdsBhdcknbldkRoj.kD;kgksxk\iqyhrFkkjLlhgYdsgSvkSjiqyhfpduhgSA
[M.Bank(07-08)_NLM_1.18]
[Made 2005, PKS Sir] 4kx 2kx 4kx (B) (C*) 5m 5m m Sol.(C) Initially the block is at rest under action of force 2T upward and mg downwards. When the block is pulled downwards by x, the spring extends by 2x. Hence tension T increases by 2kx. Thus the net unbalanced force on block of mass m is 4kx.
(A)
111.
acceleration of the block is =
(D) none of these
4kx m
Figure shows a man of mass 50 kg standing on a light weighing machine kept in a box of mass 30 kg. The box is hanging from a pulley fixed to the ceiling through a light rope, the other end of which is held by the man himself. If the man manages to keep the box at rest, the weight shown by the machine is _______ N.
RESONANCE
Page # 25
fp=kkuqlkj50kg dk,dO;fDrgYdhHkkjekiue'khuij[kM+kgqvkgStks30kgdsckWDlesj[khgqbZgSAckWDldksjLlh ds,dfljslsiqyh}kjkVkaxktkrkgSrFkknwljkfljkO;fDrds}kjkidM+kgqvkgSA;fnO;fDrckWDldksfLFkjj[krkgS rksHkkjekiudkikB;kad_______ U;wVuesagksxkA [M.Bank(07-08)_NLM_2.1]
[ Ans: 100 N ] T
N
Sol.
N
30g
T = N + 30 g T + N = 50 g N = 100 Nt. 112.
Sol.
T
........(i) ........(ii)
Inside a horizontally moving box, an experimenter (who is stationary relative to box) finds that when an object is placed on a smooth horizontal table and is released, it moves with an acceleration of 10 m/s2. In this box if 1 kg body is suspended with a light string, the tension in the string in equilibrium position. (w.r.t. experimenter) will be. (Take g = 10 m/s2) [Made PKS 2005_GRSTU] [M.Bank_NLM_5.11] ,dxfreku{kSfrtcDlsesa],diz;ksxdehZikrkgSfdtcoLrqdks{kSfrtfpduhestijj[krsgS]rks;g10eh-/ls-2 ds Roj.klsxfrekugksrhgSAblcDlsesatc1fdxzk-dhoLrqdksgYds/kkxslsyVdkrsgSa]rks/kkxsesalkE;oLFkkesaiz;ksxdehZ ds lkis{k ruko gksxk &(fn;k gS g=10m/s2) (A) 10 N (B*) 10 2 N (C) 20 N (D) zero 2 Acceleration of box = 10 m/s Inside the box forces acting on bob are shown in the figure cDlsdkRoj.k=10eh-/ls-2
cDlsdsvUnjoLrqijdk;Zjrcyfp=kesan'kkZ;sax;sgSA
T= 113.
(mg)2 (ma)2 = 10 2 N
In the figure shown, A & B are free to move. All the surfaces are smooth. (0 < < 90º) fp=k esa n'kkZ, vuqlkjArFkkB xfr djus ds fy, LorU=k gSA lHkh lrg fpduh gSA (0 < < 90º) [M.Bank_NLM_6.5] (A*) the acceleration of A will be more than g sin A dk Roj.k g sin ls T;knk gksxkA (B) the acceleration of A will be less than g sin A dk Roj.k g sin ls de gksxkA (C) normal force on A due to B will be more than mg cos B ds dkj.k Aij vfHkyEc cy mg cos ls vf/kd gksxkA (D*) normal force on A due to B will be less than mg cos A ij B ds dkj.k vfHkyEc cy mg cos ls de gksxkA
RESONANCE
Page # 26
N
Sol.
ma0 (Pseudo)
A
a
B
a0
mg
ma0 sin + N = mg cos N < mg cos Hence, (D) is true. ma0 cos + mg sin = ma a = g sin + a0 cos A
a
N = mgcos – ma0sin
a0cos a0
a0sin
Hence acceleration of A = 114.
mg cos
(a a 0 cos )2 (a 0 sin )2 g sin .
Find the acceleration of the 6 Kg block in the figure. All the surfaces and pulleys are smooth. Also the strings are inextensible and light. [Take g = 10 m/s2] fp=kesa6KgCykWddkRoj.kD;kgksxklHkhlrgrFkkf?kjuh?k"kZ.kjfgrgSrFkkjLlhvfoLrkfjrgSA[Takeg=10m/s2] [Made_RA_2006_RSTG] [M.Bank_NLM_6.6]
[2] Sol.
All the blocks will be having the same acceleration along the length of the string. So, Applying Newtons law along the string on A,B & C. 6g – 2g sin300 – 2g = (6 + 2 + 2)a or
115.
3g = 10a
a = 3 m/s2
3g 10 Ans. a = 3 m/s2
a=
System is shown in the figure. Assume that cylinder remains in contact with the two wedges. The velocity of cylinder is [M.Bank(07-08)_NLM_8.33] fp=kesa,dfudk;iznf'kZrgSekfu;sfdcsyunksostksdse/;lEidZesajgrk gSAcsyudkosxgScsy u
u m/s
Sol.
(A)
19 4 3
u m/s 2
(B)
13 u m/s 2
(D) Method - I fof/k- As cylinder will remains in contact with wedge A
RESONANCE
30°
(C)
3 u m/s
30°
2u m/s
(D*)
[Made BKM, 2005]
7 u m/s
D;ksafd csyu ost A ds laidZ esa gS
Page # 27
Vx = 2u
As it also remain in contact with wedge B ;gostBdslkFkHkhlaidZesajgrkgSA u sin 30° = Vy cos30° – Vx sin30° sin 30 U sin 30 + cos 30 cos 30 Vy = Vx tan30° + u tan 30°
Vy = Vx
V=
Vy = 3u tan30° =
Vx2 Vy2 =
Method - II In the frame of A 116.
Sol. 117.
and
3u
7 u Ans.
fof/k- II fof/kAdsra=kesa
3u sin 30º = Vycos30º Vy = 3u tan 30º =
o Vx = 2u
3u V=
Vx2 Vy2 =
7 u Ans.
A lift is falling with an acceleration 2 m/s2. A ball of mass 100 gm is attached at one end of the string and the other end is fixed to the ceiling of the lift. The ball remains at rest relative to lift. The tension in the string is: (g = 10 m/s2) [M.Bank_NLM_5.9] ,dfy¶V2m/s2 dsRoj.klsfxjjghgSAjLlhds,ddksusls100gmdhxsantqM+hgS,oanwljkfljkNrlscU/kkgqvk
gSAxsanfy¶Vdslkis{kfojkeesajgrhgSAjLlhesarukogS& (A) 1.2 N (B*) 0.8 N T = ma (g – a) = 0.1 (10 – 2) = 0.8 N
(C) 10 N
(D) 0.2 N
Three equal balls 1,2,3 are suspended on springs one below the other as shown in the figure. OA is a weightless thread. The balls are in equilibrium [M.Bank(07-08)_NLM_1.2]
fp=kesan'kkZ,vuqlkjrhu,dtSlhxsan1,2,3 fLizaxdhlgk;rklsfp=kkuqlkj,dnwljslstqM+hgqbZgSAOA,d nzO;ekughujLlhgSAxSansalke;koLFkkesagSA
(a) If the thread is cut, the system starts falling. Find the acceleration of all the balls at the initial instant
vxjjLlhdksdkVfn;ktkrkgSrksfudk;fxjuk'kq:djnsrkgSAizkjfEHkd{k.kijlHkhxsanksadkRoj.kKkrdhft,A
(b) Find the initial accelerations of all the balls if we cut the spring BC, which is supporting ball 3,
RESONANCE
Page # 28
instead of cutting the thread. Sol.
vxjgejLlhdhtxgxsan3dkstksM+usokyhfLizaxdksdkVnsrsgSrkslHkhxsanksadkizkjfEHkdRoj.kKkrdhft,A [ Ans: (a) 3 g 0, 0, (b) 0, g g ] T
(a) For A, For B,
T AB = 2mg, T BC = mg 2mg + mg = maA T AB – mg – T BC = maB 2mg – mg – mg = maB T BC – mg = mac
(b)
T AB = 2mg T AB – mg = maB 2mg – mg = maB aB = g () aA = 0 & aC = g().
118.
TBC
AB
aA = 3g
A m
maB = aB = 0 ac = 0.
TAB
mg
B m mg
TBC
Cm mg
TAB m B m
aB
mg
A bob is hanging over a pulley inside a car through a string . The second end of the string is in the hand of a person standing in the car . The car is moving with constant acceleration 'a' directed horizontally as shown in figure . Other end of the string is pulled with constant acceleration ' a ' (relative to car) vertically. The tension in the string is equal to [ Made 2003 ]
,dckscdkjdsvUnj,dMksjhds}kjkf?kjuhlsyVdjgkgSA,dMksjhls,dckWcyVdkgSAMksjhdknwljkfljkdkj esa[kM+s,dO;fDrdsgkFkesagSAdkj{kSfrtfn'kkesafu;rRoj.k afp=kkuqlkjxfr'khygSAMksjhdknwljkfljkfu;rRoj.k a ¼dkjdslkis{k½ lsÅ/okZ/kjuhps[khapktkrkgSAMksjhesarukocjkcjgS– [M.Bank(07-08)_NLM_5.10] (A) m g2 a 2
Sol.
(B) m g2 a 2 – ma
(C*) m g2 a 2 + ma
(D) m(g + a)
(C)
(Force diagram in the frame of the car) ¼dkj ds rU=k esa cy fp=k½ Applying Newton’s law perpendicular to string mg sin = ma cos a tan = g
Mksjh ds yEcor~ U;wVu ds fu;e yxkus ij &
Applying Newton’s law along string T – m g2 a 2 = ma
T = m g 2 a 2 + ma Ans.
Mksjh ds yEcor~ U;wVu ds fu;e yxkus ij T – m g2 a 2 = ma T = m 119.
g 2 a 2 + ma Ans.
Shown in the figure is a system of three particles of mass 1 kg, 2 kg and 4 kg connected by two springs. The acceleration of A, B and C at an instant are 1 m/sec2, 2 m/sec2 and 1/2 m/sec2 respectively directed as shown in the figure external force acting on the system is : fp=kkuqlkjnwljkfudk;es 1kg,2kgrFkk4kgdksnksfLçaxlstksM+ktkrkgSAfdlh{k.kA,BrFkkCdkRoj.k1m/ sec2,2 m/sec2 rFkk1/2m/sec2 fp=kkuqlkj fn'kk esa gSA fudk; ij yxus okyk ckº; cy gksxkA [M.Bank(0708)_NLM_1.10]
RESONANCE
Page # 29
(A) 1 N 120.
Sol.
121.
Sol.
(B) 7 N
(C*) 3 N
(D) 2N
Figure shows a 5 kg ladder hanging from a string that is connected with a ceiling and is having a spring balance connected in between. A boy of mass 25 kg is climbing up the ladder at acceleration 1 m/s2. Assuming the spring balance and the string to be massless and the spring to show a constant reading, the reading of the spring balance is : (Take g = 10 m/s2) [Made AA_2006_GRSTU] ,d 5fdxzk0 dhlh<+h ,d jLlh }kjk Nr lsyVdk;hx;h gS buds e/; ,d dekuhnkj rqyk tqMh gSA 25fdxzk0nzO;eku dk,dO;fDr1eh0/lS02 dsRoj.klslh<+hijp<+jgkgSA;gekuysfddekuhnkjrqykrFkkjLlhHkkjghugS]dekuhnkj rqyk dk ikB~;kad gksxk& (fn;kgSg=10m/s2) [M.Bank_NLM_1.20]
(A) 30 kg (B*) 32.5 kg (C) 35 kg (D) 37.5 kg If reading of spring balance is T, then applying NLM on (man + ladder) system ;fndekuhrqykdkikB~;kadTgS]rks¼yM+dk$lh<+h½fudk;ijU;wVudkfu;eyxkusij T – (25 + 5)g = 25 a T – 30g = 25 a T – 300 = 25(1) T = 325 N = 32.5 kg. A body of mass 32 kg is suspended by a spring balance from the roof of a verticallyoperating lift and going downward from rest. At the instants the lift has covered 20 m and 50 m, the spring balance showed 30 kg & 36 kg respectively. The velocity of the lift is: [M.Bank(07-08)_NLM_1.11] fLFkjkoLFkklsÅ/okZ/kjuhpsdhvksjtkjgh,dfy¶VdhNrls,dfLçaax}kjk32kgnzO;ekudhoLrqyVdhgq;hgSA fy¶V }kjk 20m o50m,nwjh r;djus okys {k.kksa ij fLçax dkikB;kad Øe'k%30kgo36kggksrk gS fy¶V dk osx % (A) decreasing at 20 m & increasing at 50 m (B*) increasing at 20 m & decreasing at 50 m (C) continuously decreasing at a constant rate throught the journey (D) continuously increasing at constant rate throughout the journey (E) remaining constant throughout the journey. (A)20 m ij ?kV jgk gS rFkk 50m ij c<+ jgk gSA (B*)20 m ij c<+ jgk gS rFkk50m ij ?kV jgk gSA (C);k=kkdsnkSjkuyxkrkj,dfu;rnjls?kVjgkgSA (D);k=kkdsnkSjkuyxkrkj,dfu;rnjlsc<+jgkgSA (E);k=kkdsnkSjkufu;rjgrkgSA N = m (g – a) , N < mg if a () and N > mg if a ( ) Reading of spring balance is less than m if a () and reading of spring balance is greater than m if a ( )
RESONANCE
Page # 30
122.
In the figure shown all contact surfaces are smooth. Acceleration of B block will be: fp=kesafn[kkbZfLFkfresalHkhlEidZlrgfpduhgSAxqVdsBdkRoj.kgksxk:
[M.Bank(07-08)_NLM_3.39]
Sol.
(A*) 1 m/s 2 (B) 2 m/s2 1 (C) 3 m/s2 (D) none of these buesa ls dksbZ ugha Let aB = a then aA = 3a [by string constraint] T = maA T = 3ma .........(i) mg – 3 T = m aB mg – 3 T = ma .........(iI)
123.
mg – 9 ma = ma a = a = 1 m/s 2
g 10
In the figure a block ‘A’ of mass ‘m’ is attached at one end of a light spring and the other end of the spring is connected to another block ‘B’ of mass 2m through a light string. ‘A’ is held and B is in static equilibrium. Now A is released. The acceleration of A just after that instant is ‘a’. In the next case, B is held and A is in static equilibrium. Now when B is released, its acceleration immediately after the release is 'b'. The value of a/b is : (Pulley, string and the spring are massless) [M.Bank(07-08)_NLM_1.17] fp=kesanzO;eku'm'dkCykWd‘A’gYdhfLizaxds,dfljslstqM+kgSrFkkfLizaxdknwljkfljkgYdhjLlh}kjknzO;eku 2mdsnwljsCykWd‘B’lstqM+kgSA‘A’dksfdlhusidM+kgSrFkk'B'fLFkjlkE;oLFkkesagSAvcAdksNksM+rsgSaAml{k.k dsBhdcknCykWd'A'dkRoj.k'a'gSAnwljhfLFkfresa‘B’dksidM+rsgSrFkkA fLFkjlkE;oLFkkesagSAvctcBdksNksM+rs gS]rksNksM+usdjusdsrqjUrcknbldkRoj.kbgSA
[Made RKV, AS 2005]
(B) undefined vLi"V
(A) 0 Sol.
(C*) 2
For first case tension in spring will be Ts = 2mg just after 'A' is released. 2mg – mg = ma a = g
(D)
1 2
In second case Ts = mg
gy.
2mg – mg = 2mb b = g/2 a/b = 2
izFkefLFkfresafLizaxesarukogksxk Ts = 2mg 'A' dks eqDr djus ds rqjUr ckn 2mg – mg = ma a = g Ts = mg 2mg – mg = 2mb b = g/2 a/b = 2
f}rh;fLFkfresa
124.
In the figure shown all the surface are smooth. All the blocks A, B and C are movable, x-axis is horizontal and y-axis vertical as shown. Just after the system is released from the position as shown. [M.Bank_N.LM_4.11] n'kkZ;safp=kesalHkhlrgfpduhgSAlHkhfi.MA,BrFkkCxfrdjldrsgSAx-v{k{kSfrtrFkky-v{kn'kkZ;svuqlkjm/
okZ/kjgSAfudk;dksfn[kkbZxbZfLFkfrlseqDrdjrsgSrksbldsrqjUrckn&
RESONANCE
Page # 31
y
A B
Sol.
x
C
[Made RKV_2007]
Horizontal Surface
(A*) Acceleration of 'A' relative to ground is in negative y-direction (B*) Acceleration of 'A' relative to B is in positive x-direction (C*) The horizontal acceleration of 'B' relative to ground is in negative x-direction. (D*) The acceleration of 'B' relative to ground directed along the inclined surface of 'C' is greater than g sin . (A*)tehudslkis{k'A'dkRoj.k_.kkRedy-fn'kkesagksxkA (B*)'B'dslkis{k'A'dkRoj.k/kukRedx-fn'kkesagksxkA (C*)tehudslkis{k'B'dk{kSfrtRoj.k_.kkRedx-fn'kkesagksxkA (D*) tehu ds lkis{k 'B' dk Roj.k urry 'C' ds vuqfn'k gsin ls T;knk gksxkA (Tough) There is no horizontal force on block A, therefore it does not move in x-direction, whereas there is net downward force (mg – N) is acting on it, making its acceleration along negative y-direction. Block B moves downward as well as in negative x-direction. Downward acceleration of A and B will be equal due to constrain, thus w.r.t. B, A moves in positive x-direction. fi.MijdksbZ{kSfrtcyughagS]vr%;gx-fn'kkesaughapyrkgS]tcfdblijuhpsdhvksjifj.kkehcy, (mg–N)yxrk gS]tksfdbldsRoj.kdks_.kkRedy-fn'kkesansrkgSA fi.MBuhpsdslkFk&lkFk_.kkRedx-fn'kkesaxfrekugksrkgSAArFkkBdkuhpsdhvksjRoj.kc)rklscjkcjgksrkgS] vr%Bdslkis{k]A/kukRedx-fn'kkesaxfrekugSA
B
B
fr fØ; k C } kj k v fHky Ec i z Normal reaction due to C Due to the component of normal exterted by C on B, it moves in negative x-direction. CijB}kjk vfHkyEc ds ?kVd ds dkj.k ;g _.kkRed,x-fn'kk esa xfreku gSA NA
B
NC
Mg
The force acting vertically downward on block B are mg and NA(normal reaction due to block A). Hence the component of net force on block B along the inclined surface of B is greater than mg sin. Therefore the acceleration of 'B' relative to ground directed along the inclined surface of 'C' is greater than g sin CykWd Bij Å/okZ/kj uhpsdhvksj cy mg+NA(CykWd Ads dkj.k vfHkyEc izfrfØ;k)gSA blfy;s CykWd Bij dk;Zjr
usV cy dk Bdh ur lrg ds vuqfn'k ?kVd mgsinls vf/kd gSA blfy;s Hkwfe ds lkis{k Bdk Roj.k] ftldh fn'kk C dh ur lrg ds vuqfn'k gS] gsin ls vf/kd gSA 125.
Initially the spring is undeformed. Now the force 'F' is applied to 'B' as shown. When the displacement of 'B' w.r.t. 'A' is 'x' towards right in some time then the relative acceleration of 'B' w.r.t. 'A' at that moment is: izkjEHkesafLizaxvladqfprgSAvcBijn'kkZ;svuqlkjcy'F' yxk;ktkrkgSAtcdqNle;ckn'B' dk'A' dslkis{knk;ha vksjfoLFkkiu'x' gSrksml{k.k'B'dk 'A' dslkis{kRoj.kgksxk& [M.Bank_N.L.M._1.24]
(A)
F 2m
RESONANCE
(B)
F kx m
(C*)
F 2 kx m
(D) none of these
Page # 32
Sol.
F - Kx = mb and kx = ma Hence m (b – a) = F – 2kx Ans. (C).
126.
In the figure shown, a person wants to raise a block lying on the ground to a height h. In both the cases if time required is same then in which case he has to exert more force. Assume pulleys and strings light. fp=kesan'kkZ,vuqlkj],dvknehxqVdsdkstehulshÅpk¡bZrdmBkukpkgrkgSAnksauksfLFkfr;ksaesavxjle;
cjkcj yxrk gS rks dkSulh fLFkfr esa vko';d cy dk eku vf/kd gksxkA ;g ekfu, fd iqyh rFkk jLlh gYds gSA [M.Bank(07-08)_N.L.M._3.26]
[Made 2006, RKV, GRSTU] [3.35 NLM]
Sol.
(A*) (i) (B) (ii) (C) same in both nksuksa esa leku (D) Cannot be determined fu/kkZj.k ugha dj ldrs 1 2 Since, h = at a should be same in both cases, because h and t are same in both cases as 2 given.
127.
In (i) F 1 – mg = ma. F1 = mg + ma.
mg ma F1 > F2 . 2 A weight W is supported by two strings inclined at 60º and 30º to the vertical. The tensions in the strings are T 1 & T 2 as shown. If these tensions are to be determined in terms of W using a triangle of forces, which of these triangles should you draw? (block is in equilibrium) ,dHkkjW nksjfLl;ksatksÅ/oZls60ºrFkk30ºij>qdhgSjLlh;ksaesarukofp=kkuqlkjT1rFkkT2gSA;fnburukoksadks W ds:iesacyksadkf=kHkqtcukdjKkrdjusgSrksfuEuesalsdkSulkf=kHkqt[khapsaxs? ¼HkkjlkE;koLFkkesagS½ In (ii) 2F2 – mg = ma
[M.Bank_N.L.M._
3.13]
(A)
(B)
RESONANCE
F2 =
(C)
(D)
(E*)
Page # 33
Sol.
= Ans. (E) 128.
,
+
+
=
=0
,
=
System is shown in the figure and man is pulling the rope from both sides with constant speed ' u'. Then the speed of the block will be: [Made 2004] M.Bank_NLM_8.12
fp=kesaiznf'kZrfudk;esa,dvknehfu;rpky'u'lsjLlhdksnksuksfljksals[khapjgkgSACykWddhpkygksxhA
(A*) Sol.
u=
3u 4
(B)
(C)
0 v1 v1 v 2 –v2 u , =v, =v 2 2 2
Hence v = velocity of M = NLM 129.
3u 2
u 4
(D)noneofthesebuesalsdksbZugha
3u 4
System is shown in the figure. Velocity of sphere A is 9 m/s. Then speed of sphere B will be : fp=kesaiznf'kZrfudk;esa;fnxksysAdkosx9m/sgksrksxksysBdhpkygksxhA [ Made 2004] [M.Bank_NLM_8.14]
(A) 9 m/s Sol.
(B*) 12 m/s
(C) 9
9 cos = v sin
19 – R = tan 12
RESONANCE
5 m/s 4
(D)none of thesebuesa ls dksbZ ugha
(i) (ii) Page # 34
130.
Sol.
(R + 5)2 = (12) 2 + (19 – R)2 R = 10 Hence from (i) and (ii) v = 12 m/s2
A block of mass m is placed on a wedge. The wedge can be accelerated in four manners marked as (1), (2), (3) and (4) as shown. If the normal reactions in situation (1), (2), (3) and (4) are N1, N2, N3 and N4 respectively and acceleration with which the block slides down on the wedge in situations are b1, b2, b3 and b4 respectively then : [Made M. Pathak] [M.Bank_NLM_5.18] m nzO;eku dk xqVdk ,d ost ij j[kk gSA n'kkZ;s vuqlkj ost pkj izdkj ls Rofjr gksrk gS (1), (2), (3) o (4)A ;fn fLFkfr;ksa(1), (2), (3)o(4) esavfHkyEcizfrfØ;kØe'k%N1,N2,N3 oN4 gSa,oaRoj.kftllsxqVdkostijbufLFkfr;ksa esafQlyrkgS]osb1,b2, b3 vkSjb4gSarks:
(A*) N3 > N1 > N2 > N4 (B) N4 > N3 > N1 > N2 (A), (C)
(C*) b2 > b3 > b4 > b1
a
(1)
mgsin37°
N
(D) b2 > b3 > b1 > b4 macos37° ma (t
M+Roh; cy )
masin37°
mg mgcos37°
Balancing forces perpendicular to incline N = mg cos37° + ma sin37° ur ry ds yEcor~ cyksa dks larqfyr djus ij N = mg cos37° + ma sin37° N1 =
4 3 mg + ma 5 5
b1 =
3 4 g– a 5 5
and along incline mg sin 37° – ma cos 37° = mb1 ,oa ur ry ds vuqfn'k mg sin 37° – ma cos 37° = mb1
(2)
(t
masin37° N
M+Roh; cy ) ma
macos37° mgsin37°
a mg mgcos37°
Similarly for this case get N 2 =
4 3 mg – ma 5 5
4 5
3 5
blh izdkj bl fLFkfr ds fy, ge ikrs gSa N2 = mg – ma and
vkSj b2 = N2 =
RESONANCE
3 4 g+ a 5 5
4 3 mg – ma 5 5 Page # 35
(3)
Similarly for this case get N3 =
4 4 mg + ma 5 5 4 5
4 5
blh izdkj bl fLFkfr ds fy, ge ikrs gSaN3 = mg + ma and vkSj b3 =
(d)
3 3 g+ a 5 5
Similarly for this case get N4 =
4 4 mg – ma 5 5 4 5
4 5
blh izdkj bl fLFkfr ds fy, ge ikrs gSa N4 = mg – ma and vkSj b4 = 131.
3 3 g– a 5 5
The force acting on a body moving along x axis varies with the position of the particle as shown in the fig. The body is in stable equilibrium at Bank new _WPE_34 [M.Bank(07-08)_WPE_6.3] x v{kdsvuqfn'kxfr'khy,doLrqijdk;Zjrcyd.kdhfLFkfrdslkFkfp=kkuqlkjifjofrZrgksrkgSAoLrqfdlfLFkfr
ijLFkk;hlkE;koLFkkesagSA
Sol.
132.
Sol.
(A) x = x1 (B*) x = x 2 (C) both x 1 and x 2 (D) neither x 1 nor x2 (A) x = x1 (B*) x = x 2 (C)x1 rFkkx2nksauks (D)x1 rFkkx2nksauksugha Body will be in equilibrium at both x1 & x2 as at these points force will be zero. At x2 on increasing x force becomes -ve & on decreasing x force becomes + ve. so force & displacement have opposite signs. so it a pt. of stable eq. S1 S2 S3 S4
: : : :
S1 S2 S3 S4
: : : :
Newton's third law depends on Newton's second law. Newton's first law can be derived from Newton's second law. All the three Newton's laws are independent of each other. A stationary body is kept stationary on ground. Then the gravitational force exerted by earth on block and normal reaction exerted by block on earth is an example of action reaction pair illustrating Newton's third law.
U;wVu dk rhljk fu;e] U;wVu ds f}rh; fu;e ij fuHkZj djrk gSA U;wVu ds f}rh; fu;e ls] U;wVu ds izFke fu;e dks O;qRiUu fd;k tk ldrk gSA U;wVudslHkhrhuksafu;e,dnwljslsLorU=kgSA ,doLrqdkstehuijfojkeesaj[kktkrkgSrcCykWdiji`Foh}kjkyxk;kx;kxq:Rokd"kZ.kcyrFkkCykWd }kjk i`Foh ijyxk;kx;k vfHkyEccy]U;wVu ds rhljs fu;e esafØ;k o izfrfØ;k ;qXe dks crkrs gSA
(A*) F T F F (B) F F T T (C) T T T F S1 : Newton’s third law is independent of first law. Hence S1 is false.
(D) T F F F
S2 : Newton’s first law is a special case of second law when a = 0. S2 is true. S3 : Newton’s first law is derived from second law. S3 is false. S4 : The normal reaction by earth on block and by block on earth form action and reaction pair. S4 is false. S1 : U;wVu dk rhljk fu;e] igys fu;e ls Lora=k gSA vr% S1 vlR; gSA S2 :U;wVu dk izFke fu;e]U;wVu dsnwljs fu;edk ghfo'ks"kfLFkfr gSA tc a=0.S2 lR; gSA S3 : U;wVu dk izFke fu;e dks U;wVu ds f}rh; fu;e ls O;qRiUu fd;k tk ldrk gSA vr% S3 vlR; gSA S4 :CykWd ij i`Foh }kjk yxk;k x;k vfHkyEc cy o CykWd }kjk i`Foh ij yxk;k x;k vfHkyEc cy fØ;k izfrfØ;k ;qXe cukrsgSaAvr%S4 vlR;gSA
RESONANCE
Page # 36
133.
A particle is projected with a velocity 20 m/s from bottom of inclined smooth, fixed plane of inclination 30º and height 10 m. Find out the maximum height (from ground) reached by the particle. (g = 10 m/s2) ,dCykWd dks = 30° dsfpdusurrydsfuEurefcUnqls20 m/s dsosx lsiz{ksfir fd;k tkrk gSA urry dh Å¡pkbZ 10 m gSA CykWd urry ij vf/kdre fdruh Å¡pkbZ ¼tehu ls½ ij igq¡psxkA (g = 10 m/s2)[M.Bank(07-08)_WPE_5.42]
[Ans.: 12.5 m] 134.
A bead of mass m is located on a parabolic wire with its axis vertical and vertex at the origin as shown in figure and whose equation is x2 = 4ay. The wire frame is fixed and the bead can slide on it without friction. The bead is released from the point y = 4a on the wire frame from rest. The tangential acceleration of the bead when it reaches the position given by y = a is : [M.Bank(07-08)_CM_1.8]
fp=kkuqlkjijoy;dhlehdj.kx2=4aygSAbldhÅ/okZ/kjv{krFkk'kh"kZewyfcUnqijgS]blijoy;rkjijmnzO;eku dh eudkfLFkrgSA rkj dk Ýse fLFkj gS rFkkeudk ¼eudk½fcuk ?k"kZ.k ds ijoy; ij fQly ldrk gSA rkj Ýse ij y =4afcUnqlseudkfLFkjkoLFkklsNksM+ktkrkgSAtceudky=afLFkfrijigq¡prkgSrksbldkLi'kZjs[kh;Roj.kgS: BM_CM_161
[MB_Q. 1.8]
[Q.161/RK_BM/Circular Motion] [Made MPS, 2005]
Sol.
g 3g (B) 2 2 x2 = 4ay Differentiating w.r.t. y, we get yds lkis{k vodyu (A)
(C*)
g
(D)
2
g
5
dy x = dx 2a
dy =1 hence vr% = 45° dx the component of weight along tangential direction is mg sin . HkkjdkLi'kZjs[kh;fn'kkesa?kVd mgsin.
At (2a, a),
hence tangential acceleration is g sin =
vr% Li'kZ js[kh; Roj.k gsin= 135.
g
g
2
2
A rod AB is shown in figure. End A of the rod is fixed on the ground. Block is moving with velocity
3 m/s towards right. The velocity of end B of rod when rod makes an angle of 60º with the ground is: [ Made 2004]
[M.Bank_NLM_8.16]
,dNM+ABfp=kesaiznf'kZrgSANM+dkAfljkHkwfeijfLFkj¼c¡/kk½gSrFkkCykWdnk¡;hvkSj 3 m/sosxlsxfrdjjgk gSAtcNM+/kjkryls60º dkdks.kcuk;srksNM+dsfljsBdkosxgksxk&
RESONANCE
Page # 37
[Made 2004]
Sol.
(A) 3 m/s (B*) 2 m/s Let AB = , B = (x , y) v B = vx ˆi + vy ˆj vB =
3 ˆi + v y ˆj
x2 + y2 = 2
136.
Sol.
2x vx = 2y vy = 0
(D) 3 m/s
(i)
3 +
3 + (tan600) vy = 0 vy = – 1 Hence from (i) v B = 3 ˆi – ˆj Hence vB = 2 m/s
y v =0 x y
A body with mass 2kg moves in x direction in the presence of a force which is described by the potential energy-displacement graph. If the body is released from rest at x=2m, then its speed when it crosses x = 5m is : ,d2kgnzO;ekudkfi.M,dcytksfdfLFkfrtÅtkZ&foLFkkiuoØ}kjkçnf'kZr gS fd mifLFkfresaxfn'kk esaxfr dj jgk gSA ;fn bls fcUnq x=2m,ls NksM+k tk;s rks x=5mfcUnq dks ikj djrs le; bldh pky gS& [M_Bank(07-08)_WPE_7.7]
(A) zero (B) 1 ms1 (C*) 2 ms1 Loss in potential energy = gain in kinetic energy 6–2=
Alternate : Sol.
(C) 2 3 m/s
(D) 3 ms 1
1 . 2 . (v2) 2 v = 2 m/s.
du =–F dx Since, Slope is constant and negative from x = 1 to x = 3.5 m, the force is accelerating and constant (constant acceleration case)
Slope of U – X graph is
Acceleration (x = 1 to 3.5) = F/m = tan/m = Using ; v2 = u2 + 2aS from x = 2 to x = 3.5 m
(10 / 2.5) = 2m/s2 2
3 v2 = 2(2) =6 2 From x = 3.5 to x = 4.5 :
&
(2 / 1) = – 1m/s2 2 v22 = v12 – 2aS2 a2 =
RESONANCE
Page # 38
= v22 = 4 v2 = 2m/s After which it will move with constant speed. U – X xzkQ
dk
du =–F dx
Ans.
pqfd
x = 3.5 ls x = 4.5 rd: & = 138.
(10 / 2.5) = 2m/s2 2
iz;ksx ls
=6
(2 / 1) = – 1m/s2 2 v22 = v12 – 2aS2 v22 = 4 v2 = 2m/s a2 =
blds i'pkr ;g fu;r pky ls xfr djsxk
Ans.
A lift is moving upwards with a constant speed of 5 m/s. The speed of one of the pulleys is 5 m/s as shown. Then the speed of second pulley is : ,d fy¶V 5m/sdh fu;r pky ls Åij dh vksj xfreku gSA n'kkZ;s vuqlkj rFkk ,d f?kjuh dk osx 5m/sgS rks nwljh f?kjuhdhpkygksxh% [M.Bank(07-08)_NLM_8.32]
[Made BKM 2005]
Sol. 139.
(A*) 5m/s (B) zero (C) 7.5 m/s (D) 10 m/s (A*) 5m/s (B)'kwU; (C) 7.5 m/s (D) 10 m/s23. In the frame of the lift first pulley will be stationary so velocity of second pulley will be zero & so in the frame of ground it moves with 5 m/s A man is standing on a cart of mass double the mass of the man. Initially cart is at rest on the smooth ground. Now man jumps with relative velocity 'v' horizontally towards right with respect to cart. Find the work done by man during the process of jumping. [Made 2004] [M.Bank(07-08)_COM._3.30]
,dvknehydM+hdhxkM+hij[kM+kgSAxkM+hdknzO;ekuvknehdsnzO;ekulsnqxqukgSAizkjEHkesaxkM+hfpduhlrgij :dhgqbZgSAvc;gvknehxkM+hdslkis{knkfguhvksjlkis{kosx'v'ls{kSfrtesadwnrkgSAdwnusdhizfØ;kesavkneh}kjk m fd;kx;kdk;Zgksxk&
/////////////////////////////////////////////
Sol.
Let the velocity of man after jumping be ‘u’ towards right. Then speed of cart is v-u towards left. From conservation of momentum mu = 2m(v – u) m
u
v-u
RESONANCE
/////////////////////////////////////////////
Page # 39
2v 3 hence work done by man = change in K.E. of system
u=
2v v 1 1 1 1 = mu2 + 2m (v – u)2 = m + 2m 2 2 2 2 3 3 2
Ans. 140.
Sol.
1 mv 2 3
2
=
mv 2 3
In the figure shown the blocks A & C are pulled down with constant velocities u . Acceleration of block B is : fn[kk;sx;sfp=kesaArFkkC CykWdksadksfu;rosxuls[khapktkrkgSACykWdBdkRoj.kgksxk : [ Made 2003] [M.Bank_NLM_8.11]
(A)
u2 tan2 sec b
(B*)
Let vB = v()
u2 tan3 b
(C)
u2 sec 2 tan b
(D) zero ‘'kwU;
u = vcos
du dv d = cos – v sin ...........(1) dt dt dt d d v (b cot ) = – v = sin2 dt dt b
Using ...... (1)
v sin2 0 = aB cos – (v sin ) b u v 2 sin3 aB = = b cos cos aB = 141.
u 2 tan 3 Ans. (B) b
2
sin3 b cos
STATEMENT-1 : A man standing in a lift which is moving upward, will feel his weight to be greater than when the lift was at rest. (F_only) STATEMENT-2 : If the acceleration of the lift is ‘a’ upward, then the man of mass m shall feel his weight to be equal to normal reaction (N) exerted by the lift given by N = m(g+a) (where g is acceleration due to gravity) [Test_PQF combined_(CT-4_19-08-07)_Paper-1_Q.12] oDrO;-1: ,d vkneh fy¶V esa [kM+k gS rFkk fy¶V Åij dh rjQ py jgh gS rks og O;fDr bl ckj fy¶V ds fojke esa
gksusdhrqyukesages'kkvf/kdHkkjeglwldjsxkA oDrO;-2: ;fn fy¶V dkÅijdhrjQRoj.k'a'gSrksmnzO;ekudk O;fDr }kjk eglwlfd;k x;k Hkkj fy¶V }kjk vkneh ijyxk;sx;svfHkyEccyNdscjkcjgksxktgkaN =m(g+a)(;gkagxq:Rokd"kZ.kdsdkj.kRoj.kgS) (A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D*) Statement-1 is False, Statement-2 is True (A) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA (B) oDrO;-1 lR; gS] oDrO;-2 lR; gS ; oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k ugha gSA
RESONANCE
Page # 40
Sol.
142.
(C) oDrO;-1 lR; gS] oDrO;-2 vlR; gS ; (D*) oDrO;-1 vlR; gS] oDrO;-2 lR; gS [Easy] If the lift is retarding while it moves upward, the man shall feel lesser weight as compared to when lift was at rest. Hence statement1 is false and statement 2 is true.
;fnÅijdhrjQtkrhfy¶VeafnrgksrhgSrksvkneh}kjk eglwlfd;kx;k Hkkj blckjokLrfodHkkjlsdegksxkA vr% oDrO;-1 vlR; gS] oDrO;-2 lR; gS
Two blocks, of masses M and 2M, are connected to a light spring of spring constant K that has one end fixed, as shown in figure. The horizontal surface and the pulley are frictionless. The blocks are released from rest when the spring is non deformed. The string is light. [Made A.K.S. sir] [M.Bank(07-08)_NLM_1.26]
M o2MnzO;ekudsnksxqVdsfLizaxfu;rkadKds,dgYdsfLizaxlstqM+sgSaftldk,dfljkfp=kkuqlkjfLFkjgSA
{kSfrtlrg,oaiqyh?k"kZ.kjfgrgSaAxqVdsfojkelsNksM+stkrsgSatcfLizaxvfod`r¼fcukf[kapk½gSAMksjhgYdhgSA K
M
2M
(A*) Maximum extension in the spring is
fLizaxesavf/kdrefoLrkj
4 Mg gSA K
4 Mg . K
(B*) Maximum kinetic energy of the system is
2 M2 g2 K
2 M2 g2 gS K (C*) Maximum energy stored in the spring is four times that of maximum kinetic energy of the system.
fudk; dh vf/kdre xfrt ÅtkZ
fLizax esa laxzfgr vf/kdre ÅtkZ fudk; dh vf/kdre xfrt ÅtkZ dh pkj xquh gSA
(D) When kinetic energy of the system is maximum, energy stored in the spring is
Sol.
4 M2 g 2 K
4 M2 g 2 gSA K Maximum extension will be at the moment when both masses stop momentarily after going down. Applying W-E theorem from starting to that instant. k f – k i = W gr. + W sp + W ten. 1 2 0 – 0 = 2 M.g.x + Kx + 0 2
tc fudk; dh xfrt ÅtkZ vf/kdre gS] rks fLizax esa lxzfgr ÅtkZ
4 Mg K System will have maximum KE when net force on the system becomes zero. Therefore x=
2 Mg = T and T = kx
x=
2 Mg K
Hence KE will be maximum when 2M mass has gone down by Applying W/E theorem k f – 0 = 2Mg. kf =
2 M2 g2 K2
RESONANCE
2 Mg . K
1 4 M2 g 2 2 Mg – K. 2 K K2 Page # 41
1 4 Mg 8 M2 g2 = Maximum energy of spring = K . 2 K K Therefore Maximum spring energy = 4 × maximum K.E. 2
When K.E. is maximum Spring energy =
i.e. (D) is wrong. 143.
1 4 M2 g 2 .K . 2 K2
=
2 Mg . K
2 M2 g 2 K2
Two blocks of masses m 1 and m 2 are connected as shown in the figure. The acceleration of the block m 2 is: m1 rFkkm2 nzO;ekudsnksCykWddksfp=kkuqlkjjLlhlstksM+ktkrkgSAm2 nzO;ekudsCykWddkRoj.kD;kgksxk: [M.Bank(07-08)_NLM_3.42]
(A*) Sol.
x=
Let
m2 g m1 m2
(B)
m1 g m1 m2
a = accn of m1 accn of pulley =
If acceleration of m2 = b
(C)
a0 a = 2 2
4 m2 g m1 g m1 m2
(D)
m2 g m1 4 m2
b a = 2 2 Hence a = b T = m 1 a , m 2g – T = m 2 a 144.
Then
0+
m2g a = m m 1 2
A force F is applied on block A of mass M so that the tension in light string also becomes F when block B of mass m acquires an equilibrium state with respect to block A. Find the force F. Give your answer in terms of m, M and g. M nzO;eku ds CykWd Aij Fcy bl çdkj yxk;k tkrk gS fd jLlh esa ruko Fgks tc CykWd B(nzO;eku m)Ads lkis{k lkE;koLFkk fLFkfr çkIr dj ysaA Fdk eku Kkr djksA (viuk mÙkj m,MrFkk gds :i esa nksA) [Made PKS 2005] [8] [ M . B a n k ( 0 7 08)_NLM_5.12]
Sol. Applying Newton's law on the system in horizontal direction F = (M + m) a. Now consider the equilibrium of block B w.r.t. block M
F F = (mg) + (ma) = (mg) + m m M 2
RESONANCE
2
2
2
2
Page # 42
145.
F = 2
1
m 2 g2
m2
m M
2
;
F=
mg
m 1 m M
2
System shown in figure is in equilibrium. The magnitude of change in tension in the string just before and just after, when one of the spring is cut. Mass of both the blocks is same and equal to m and spring constant of both springs is k. (Neglect any effect of rotation)
fn,x,fp=kesafudk;lkE;koLFkkesagSAtc,dfLçaxdksdkVfn;ktkrkgSrksjLlhesaruko esadqyifjorZu fLçaxdsdkVusdsBhdigysvkSjBhdcknesaD;kgksxknksuksaxqVdksadknzO;eku,dtSlkgSvkSjm dscjkcjgS vkSjnksuksafLçaxdkfLçaxfu;rkadKgSA¼?kw.kZudsizHkkodksux.;ekusa½ M.Bank_NLM_1.14 [Made 2004]
(A*) 146.
mg 2
(B)
mg 4
(C)
3m g 4
(D)
3m g 2
A boy and a block, both of same mass, are suspended at the same horizontal level, from each end of a light string that moves over a frictionless pulley. The boy starts moving upwards with an acceleration 2.5 m/s2 relative to the rope. If the block is to travel a total distance 10 m before reaching at the pulley, the time taken by the block in doing so is equal to : [Made AKS_2007_GRSTU] [M.Bank_NLM_3.46]
/k"kZ.kjfgrf?kjuhlsikfjr,dgYdhjLlhdsizR;sdfljsij]leku{kSfrtLrjij],dyM+dk,oa,dfi.Mnksuksaleku nzO;ekudsyVdsgSAyM+dkÅijdhvksjjLlhdslkis{k2.5eh-/ls-2 dsRoj.klspyukizkjEHkdjrkgSA;fnfi.M]f?kjuh rdigqapusesadqynwjh10eh-r;djrkgS]rksfi.Mdks,slkdjusesayxkgqvkle;cjkcjgSA
10 m
m
(A) Ans. Sol.
(C)
10
2
s
(D) 8s
(B) Acceleration of boy and pulley will be same equal to 1.25 m/s2 w.r.t. ground. Hence yM+ds,oaf?kjuhnksuksadktehudslkis{kRoj.kcjkcj,oa1.25eh-/ls-2 dscjkcjgksxkAvr% 10 =
147.
(B*) 4s
8s
m
1 (1.25) t2 t = 4 sec. 2
In the position shown collar B moves to the left with a velocity of 150 mm/s. Determine: fn[kk;sx;sfp=kesadkWyjB ck¡;hvkSj150 mm/s dsosxlsxfrdjrkgS rkscrkb;sA [M.Bank_NLM_8.25] [Made 2004]
RESONANCE
Page # 43
the velocity of collar A dkWyjAdkosx the velocity of portion C of the cable rkjdsC Hkkxdkosx the relative velocity of portion C of the cable with respect to collar B. dkWyjBdslkis{krkjdsC Hkkxdkvkisf{kdosx [ Ans.: VA = 300 right, VC = 600 left, VCB = 450 left ] (a) (b) (c)
148.
A rigid rod leans against a vertical wall at a point A. The other end of the rod is on the horizontal floor. Somebody is pushing point A of the rod downwards with constant velocity. Will B move with constant velocity ? What is the path of the centre of the rod. M.Bank_NLM_8.35 ,dn`<+NM+fdlhÅ/oZnhokjdsfcUnqAij>qdhgS]NM+dknwljkfljk{kSfrtQ'kZijgS]dksbZO;fDrNM+dksfcUnqA ijfu;rosxlsuhpsdhvksj/kdsyjgkgSrksD;kfljsBdkosxfu;rgksxk\NM+dsdsUnzdkfcUnqiFkD;kgksxk\
[ Ans. No, Circular] 149.
A wedge of height h is released from rest with a light particle P placed on it as shown . The wedge slides down an incline which makes an angle with the horizontal . All the surfaces are smooth , P will reach the surface of the incline in time : h ÅWPkkbZdkosturftlij,dgYdkd.kP j[kkgSdksfojkelsfp=kkuqlkjNksMktkrkgSAost dks.kdsurryij fQlyrk gSA lHkh lrg fpduh gSA d.k P ur ry dh lrg ij fdrus le; ds ckn igqpsxkA
(A*) 150.
2h
(B)
gsin 2
2h gsincos
(C)
2h gtan
(D)
2h
gcos 2
A lift of total mass M kg is raised by cables from rest to rest through a height h . The greatest tension which the cables can safely bear is nM kg wt . Show that the shortest interval of time in which the
2nh ascent can be made is , (n1)g
1/ 2
.
M nzO;eku dh ,d fy¶V] rkjksa (cables) ds }kjk fojke ls fojke rd h ÅpkbZ rd mBkbZ tkrh gSA rkjksa dk vf/kdre lqjf{kr ruko ftruk rkj lgu dj ldrk gS] nM fdxzkµHkkj gSA çnf'kZr djks fd U;wure le;kUrjky ftles fy¶V
2nh
RofjrdjldrsgSa] (n1)g
Sol.
1 v T= h 2 0
v0 = 0 + (h – 1) gt1 0 = v0 – g (T – t1) v0 = gT – gt1 . 151.
1/ 2
dscjkcjgSA 2h T= v 0
Two blocks A and B of mass 1 kg & 2 kg respectively are connected by a string, passing over a light frictionless pulley. Both the blocks are resting on a horizontal floor and the pulley is held such that string remains just taut. At moment t = 0, a force F = 20 t newton starts acting on the pulley along vertically upward direction, as in Figure. Calculate : [M.Bank(07-08)_NLM_4.13] 1kgrFkk2kgnzO;ekudsnksCykWdksadksjLlh}kjkf?kjuhlstksM+ktkrkgSAf?kjuh?k"kZ.kjfgrrFkkgYdhgSAnksuksaCykWd {kSfrtlrgijfojkeesagSrFkkf?kjuhblrjggSfdjLlhruhgqbZgSAt = 0 ijF = 20 t U;wVudkcyf?kjuhijÅij dhvksjfp=kkuqlkjyxk;ktkrkgSAx.kukdjks: RESONANCE Page # 44
Sol.
(a) velocity of A when B loses contact with the floor (b) height raised by the pulley upto that instant (c) work done by the force F upto that instant. (a) CykWdAdkosxtc CykWdB /kjkrylslaidZNwVtkrkgSA (b) bl{k.krdf?kjuh}kjkr;Å¡pkbZ (c) bl {k.k rd cy F }kjk fd;k x;k dk;Z [ Ans: 5 m/s, 5/6 m, 175/6 J ] A looser contact at t = 1 sec and B at t = 2 sec. Between t = 1 to 2 sec. , aA = T – 1 g = 10 t – 10
dv = 10 t – 10 dt
t
dv 10 ( t – 1) dt 1
( t – 1)2 = 5 (t – 1)2 2 At t = 2 , V = 5 (2 – 1)2 = 5 mgs (b) height raised by A = S ? V = 10
sol.
ds (1) = 5 (t – 1)2 dt
s =
(1)
Ans. of (a)
5 3
t 2
5( t – 1)3 ds = 3 t 1
Height raised by pulley between t = 1 to t = 2 sec. = (c)
W F + W G = K W F – 1 g (5/3) =
WF = 152.
50 175 25 + = J 3 6 2
1 (1) (5)2 2
5 5 = m 6 2
Ans.
A block of mass m 1 is placed on a wedge of an angle , as shown. The block is moving over the inclined surface of the wedge. Friction coefficient between the block and the wedge is µ1, where as it is µ2 between the wedge and the horizontal surface. If µ1 =
1 , = 45°, m 1 = 4 kg, m 2 = 5kg and g = 10 m/s 2, 2
find minimum value of µ 2 so that the wedge remains stationary on the surface. Express your answer in multiple of 10–3. nzO;ekum1 dk,dfi.Mdks.kokysostijfp=kkuqlkjj[kkgSafi.M]ostdhur&lrgijÅijxfrekugSA?k"kZ.kxq.kkad 1 2
fi.M rFkk ost ds chp µ1 gS tcfd ;g ost rFkk {kSfrt lrg ds chp µ2 gSA ;fn µ1 = , = 45°, m1 = 4 kg, m2 = 5kg rFkkg = 10 eh-/ls-2 gksrksµ2 dkU;wureekuKkrdjks]ftllsost]lrgijfLFkjjgsAviusmRrjdks10–3 dsxq.kd esaO;DrdjksA [M_Bank_Friction_Q. 2.72] µ1
m1
m2
Sol.
µ2
[Made RKV_2007]
Taking block + wedge as system and applying NLM in horizontal direction
fi.M$ostdksfudk;ysdjrFkk{kSfrtfn'kkesaU;wVudkfu;eyxkusij f2
= m 1a cos
RESONANCE
Page # 45
= m 1 [g(sin – µ1 cos )] cos Again applying NLM in vertical direction
........... (1)
iqu%U;wVudkfu;em/okZ/kjfn'kkesayxkusij
(m 1 + m 2)g – N2 = m 1 a sin N2 = (m 1 + m 2)g – m 1 sin (g sin – µ1g cos ) f For limiting condition lhekUr fLFkfr ds fy, f2 = µ2N2 ........... (2) 2 From (1) and (2) (1) rFkk(2) ls
µ1
m1
a
m2 µ2
N2
m1 cos (g sin 1g cos ) µ2 = (m m )g m sin (gsn g cos ) 2 2 1 1 Using values ekuksadkiz;ksxdjusij 1 = 125 x 10–3 8 Ans. 125
µ2 =
N2
s co
g
f2
m1
N = m1g cos
µ1
Alt.
Applying NLM in vertical direction N2 = m 2g + m 1g cos 2 + µ1m 1g sin Applying NLM in horizontal direction f 2 = µ2N2 = m 1g cos sin – µ 1m 1g cos2 On Solving µ2 =
.......... (1)
1 Ans. 8
153.
In the figure shown 'P' is a plate on which a wedge B is placed and on B a block A is placed. The plate is suddenly removed and system of B and A is allowed to fall under gravity. Neglecting any force due to air on A and B, prove mathematically that the normal force on A due to B is zero. fp=kkuqlkjIysV'P'ijCykWdArFkkBj[kstkrsgSaAIysVPvpkudgVkyhtkrhgSrFkkfudk;(BrFkkA)dksxq:Rods izHkkoesafxjusfn;ktkrk gSAgokds}kjkArFkkBijcyux.;gSAxf.krh; :i ls fl)dhft, fdA ijB ds}kjk vfHkyEc izfrfØ;k cy 'kwU; gSA [ M_Bank_NLM_Q. 4.16(a)]
Sol.
F.B.D. of block A
Applying newtons second law for block A in vertical direction mAg – N cos = mA g ; where is the angle of the wedge N cos = 0 as < 90° N=0 154.
In the figure PQRS is a frictionless horizontal plane on which a particle A of mass m moves in a circle of radius r with an angular velocity such that 2 r = g/3. Another particle of mass m is tied to A through an inextensible massless string. O is the hole through which string passes down to B. B can move only vertically. The tension in the string at this instant will be:
RESONANCE
Page # 46
fp=kesaPQRS ,d?k"kZ.kjfgr{kSfrtrygSftlijm nzO;ekudk,dd.kA,r f=kT;kdso`Ùkesadks.kh;osxls bl çdkj xfr djrk gS fd 2 r = g/3Am nzO;eku dk ,d nwljk d.k ,d nzO;ekughu vforkU; Mksjh }kjk Als ca/ kkgqvkgSAO ogfNnzgSftlesagksdjMksjhBrdtkrhgSABdsoym/okZ/kjxfrdjldrkgSAbl{k.kMksjhesaaruko gksxk:
(A) mg/3
(B*) 2 mg/3
(C) mg/6
T
Sol. m
T
a
m a
g T – m = ma 3
(1)
mg – T = ma
(2)
mg = mg – T 2T = 4 mg/ 3 3
T = 2 mg /3 155.
m w2r (C.F.F)
mg
T – m w2r = ma
T–
Ans. (B)
A bus is moving with a constant acceleration a = 3g/4 towards right. In the bus, a ball is tied with a rope and is rotated in vertical circle as shown. The tension in the rope will be minimum, when the rope makes an angle = _____ . [Made M. Pathak Sir] M.Bank_Circu_M_3.57 ,dcla=3g/4fu;rRoj.klsnka;hrjQxfrdjjghgSAclesa,dxsantksjLlhlstqM+hgSAm/okZ/kjryesafp=kkuqlkj o`Ùkh;xfrdjjghgSAjLlhesarukoU;wuregksxktcjLlh dks.kcukrhgSArks = _____ gksxk
(A*) 53°
Sol.
(B) 37°
a=3g/4
(C) 180 – 53°
(D) 180 + 37°
B Tmin
inertial force
53
0
M(3g/4)
530 A
Fnet
156.
(D) none
Tmax
mg
Fnet is shown in the figure. So, tension will be max. at point A and will be min. at point B.
In the figure shown C is a fixed wedge on horizontal surface. Blocks A and B are of masses m and 2m respectively are kept as shown in figure. They can slide along the inclined plane smoothly. The pulley and string are massless. Take = 30° and g = 10m/s2. The inclined planes are very long. A and B are released from rest. 2 seconds after the release, 'B' is caught for a moment and released again. Find out the speed of 'A' just before the instant when the string becomes tight again.
RESONANCE
Page # 47
fp=kesafLFkjostC{kSfrtlrgijj[kkgSAmo2mnzO;ekudsCykWdArFkkBfp=kkuqlkjj[kstkrsgSACykWdurry ds vuqfn'k fQly ldrs gSA f?kjuh rFkk jLlh nzO;eku jfgr gSA =30°rFkk g=10m/s2gSA ur ry cgqr yEck gSA A rFkkBdksfojkelsNksM+ktkrkgSANksM+usds 2lsd.Mi'pkr~'B'dksidM+ktkrkgSrFkkfQjNksM+ktkrkgSrksCykWd'A' dhbl{k.klsigyspkyD;kgksxhtcjLlhnwljhckjruhgksxhA [4]
Sol.
[M.Bank(07-08)_NLM_4.16] [Made_RKV_2006_RSTG] Initially acceleration of B and A is same along the string which is given by
2 mg sin 30 mg sin 30 g = 3m 6 After 2 second, speed of both A and B is a=
VB = VA =
g 10 ×2= m/s 6 3
Now, When B is caught for a moment and released again, speed of B becomes zero,
10 m/s up the inclined due to which string will become slack. But A will 3 deccelerate and B will accelerate. Because of this the string will become tight again after A and B travel the same distance Let this time interval be t
while A is still having a speed
10 1 1 ×t– g sin 30° × t2 = g sin 30° × t2 3 2 2
2 sec 3 At this time speed of A is given by
10 g = t ; as t 0 3 2
t=
V= 157.
10 10 2 g 10 – ×t = – × =0 3 3 3 2 2
Ans. (b) speed of A = 0
A system is shown in the figure. A man standing on the block is pulling the rope. Velocity of the point of string in contact with the hand of the man is 2 m/s downwards. The velocity of the block will be: [ assume that the block does not rotate ] [ Made 2004] [M.Bank(07-08)_NLM_8.13]
fp=kesa,dfudk;iznf'kZrgSACykWdij[kM+kO;fDrjLlhdks[khapjgkgSaAO;fDrdsgkFklslEidZokyhjLlhdkosx 2m/suhpsdhvksjgSACykWddkosxgksxk[ekfu;sfdCykWd?kw.kZuughadjrkgSA] [ Made 2004]
Sol.
(A) 3 m/s (B)
4
3
(B*) 2 m/s
2
1
(C) 1/2 m/s
(D) 1 m/s
v
2 m/s
1 + 2 + 3 + 4 = C
RESONANCE
Page # 48
d 3 d 1 d 2 d 4 + + + =0 dt dt dt dt –v – v + 0 + v + 2 = 0 v = 2m/s
158.
A ring of radius R lies in vertical plane. A bead of mass ‘m’ can move along the ring without friction. Initially the bead is at rest at the bottom most point on ring. The minimum constant horizontal speed v with which the ring must be pulled such that the bead completes the vertical circle [M.Bank(07-08)_Circular Motion_3.9] Rf=kT;kdkoy;m/okZ/kjryesafLFkrgSA‘m’nzO;ekudk,deudkoy;dsvuqfn'kfcuk?k"kZ.kdsxfrdjldrkgSA
izkjEHkesaeudkoy;dsU;wurefcUnqijfojkeesagSAoy;dksfdlU;wurefu;r{kSfrtpkyls[khapukpkfg,ftlls eudk o`Ùkh; xfr dj ldsA
[Made MPS-2005]
Sol.
(A)
(B*)
3gR
(C)
4gR
5gR
(D)
5 . 5 gR
In the frame of ring (inertial w.r.t. earth), the initial velocity of the bead is v at the lowest position.
The condition for bead to complete the vertical circle is, its speed at top position vtop 0 From conservation of energy
or 159.
1 1 m v 2top + mg (2R) = mv2 2 2
v=
4 gR
A person wants to slide down a rope whose breaking strength is safely.
[NLM]
3 th of his weight. He can come down 4
,d O;fDr jLlh ds }kjk uhps vkus dh dksf'k'k dj jgk gSA jLlh esa vf/kdre lguh; ruko (breakingstrength) Sol. 160.
3 4
mldsHkkjdk HkkxgSAD;koglqjf{kruhpsvkik;sxkA True
System is shown in figure and wedge is moving towards left with speed 2 m/s. Then velocity of the block B will be: [ Made 2004] fp=kesaiznf'kZrfudk;esaxqVdkosx2m/spkylsck¡;hvksjxfrdjjgkgS]rksCykWdBdkosxgksxkA M.Bank_NLM_8.21
Sol.
(A) 3 m/s (B) 1 m/s Assume that angle of indination = = 60º
RESONANCE
(C*) 2 m/s
(D) 0 m/s
Page # 49
VB = 161.
Sol.
162.
2
120º
2
=
2 2 2 2 2(2)(2) cos120 º = 2 m/s
Ans. (C)
Column I describes the motion of the object and one or more of the entries of column II may be the cause of motions described in column I. Match the entries of column I with the entries of column II. dkWye I esa fudk; dh xfr dks n'kkZ;k x;k gSA tcfd dkWye II ,d ;k ,d ls T;knk fudk; dh xfr ds dkj.k dks n'kkZ;kx;k gSAdkWyeIls dkWyeIIdkfeyki dhft,A [Arihant_Pg.24_Q.51] Column I Column II (A) An object is moving towards east. (P) Net force acting on the object must be towards east. (B) An object is moving towards east with (Q) At least one force must act towards east. constant acceleration. (C) An object is moving towards east with (R) No force may act towards east. varying acceleration (D) An object is moving towards east with (S) No force may act on the object. constant velocity. LrEHkI LrEHkII (A)fudk;iwoZfn'kkesaxfrdjjgkgSA (P)fudk; ijifj.kkeh cyiwoZdh vksj yxuk pkfg,A (B)fudk;iwoZesafu;rRoj.klsxfrdjjgkgSA (Q) de ls de ,d cy iwoZ dh fn'kk esa yxkuk pkfg,A (C) fudk;iwoZdhvksjifjofrZrRoj.klsxfrdjjgkgSA(R)iwoZdhvksjdksbZcyughayxldrkA (D)fudk;iwoZdhvksjfu;rosxlsxfrdjjgkgSA (S) fudk; ij cy ugha yx ldrkA Ans. A – R,S ; B – P,Q ; C – P,Q ; D – R,S For ‘A’ : It is not mentioned whether the object is accelerated or moving with constant velocity. So nothing can be predicted with surity. If not net force is acting along east, then also it can move with constant velocity, and if no force is acting at all, then also it can move with constant velocity. For ‘B’ and ‘C’ : As the object is accelerated (whether uniform or non-uniform) a force must act on the object in such a manner that a component or whole of the force whould be along east, and also the net force must be towards east. For ‘D’ : It is moving with constant velocity, so net force must be zero that implies no force may act on the object.
fudk;fp=kkuqlkjçkjEHkesalkE;koLFkkdhfLFkfresa
The system shown below is initially in equilibrium :
Sg
Spring1
C
B
D Spring2
A
Column I [NLM] (A) Just after the spring 2 is cut, the block D (B) Just after the spring 2 is cut, the block C (C) Just after the spring 2 is cut, the block A (D) Just after the string connecting A and B is cut, the block D LrEHkI (A) fLçax 2ds dkVus ds Bhd ckn CykWd D (B)faLçax 2ds dkVus ds Bhd ckn CykWd C (C) fLçax 2ds dkVus ds Bhd ckn CykWdA
RESONANCE
E
Column II (P) (Q) (R) (S)
has acceleration up has acceleration down has acceleration zero has acceleration g upward
(P) (Q) (R)
LrEHkII ÅijdhvksjRofjrgksxkA uhpsdhvksjRofjrgksxkA Roj.k'kwU;gksxkA Page # 50
Sol.
163.
(D)ArFkk Bds chp dh fLçax dkVus ds Bhd ckn CykWd D (S) gRoj.k ls Åijdhvksj Rofjr gksxkA Ans. A–P;B–P;C–R;D–R Just after Spring 2 is cut. The net force acting on block D changes and it is acting in upwards direction and hence D accelerates. As there is no change in elongation of Spring 1 th equilibrium of A and B would’t be disturbed. Similarly, we can find reasons for ‘D’ . A hinged construction consists of three rhombus with the ratio of sides (5 : 3 : 2). Vertex A3 moves in the horizontal direction with velocity V. Velocity of A2 will be : fp=kkuqlkj,ddhyfdrfudk;rhuleprqHkqZtftudhHkqtkvksaesa5:3:2dkvuqikrgS]lscukgS;fnA3 fljsdk{kSfrt fn'kkesaosxVgSArksA2fljsdkosxgksxkA [M.Bank(07-08)_NLM_8.36]
(A) 2.5 V 164.
(B) 1.5V
(C) (2/3)V
(D*) 0.8V
A rod of length 2 is moving such that its ends A and B move in contact with the horizontal floor and vertical wall respectively as shown in figure. O is the intersection point of the vertical wall and horizontal floor. The velocity vector of the centre of rod C is always directed along tangent drawn at C to the fp=kkuqlkj,d2yEckbZdhNM+blizdkjxfrdjrhgSAfdbldsfljsArFkkB{kSfrt/kjkryrFkkm/oZnhokjdslEidZ esaxfrdjrsgSA'O'fcUnqm/okZ/kjnhokj,oa{kSfrtQ'kZdkdVkufcUnqgSANM+dsdsUnzCdkosxlfn'kges'kkmlLi'kZ js[kkdsvuqfn'kgksxktksfddsfcUnqCijMkyhtkrhgS& [M.Bank(07-08)_NLM._8.34] [Made 2005, MPS] (A) circle of radius
whose centre lies at O 2 (C) circle of radius 2whose centre lies at O
(B*) circle of radius whose centre lies at O (D) None of these
f=kT;kdso`ÙkftldkdsUnzOijfLFkrgSA (B*)f=kT;kdso`ÙkftldkdsUnzO ijfLFkrgSA 2 (C)2f=kT;kdso`ÙkftldkdsUnzO ijfLFkrgSA (D)buesalsdksbZughA At any instant of time the rod makes an angle with horizontal, the x & y coordinates of centroid of rod are (A)
Sol.
x = cos y = sin x2 + y2 = 2 Hence the centre C moves along a circle of radius with centre at O. velocity vector of C is always directed along the tangent drawn at C to the circle of radius whose centre lies at O. 165. In the pulley system shown in figure, block C is going up at 2 m/s and block B is going up at 4 m/s, then the velocity of block A on the string shown in figure, is equal to : fp=kkuqlkj ,d f?kjuh fudk; ds fy, CykWd C, 2 m/s ds osx ls Åij dh vksj rFkk CykWd B, 4 m/s, ds osx ls Åij dh vksjxfrdjjgkgSrksjLlhijfLFkrCykWdAdkosxgksxkA [Made Ashish Arora 2007] P & F [M.Bank(07-08)_NLM_8.39]
RESONANCE
Page # 51
(A) 2 m/s
(B*) 4 m/s
(C) 6 m/s
(D) 8 m/s
Sol. 1 + 2 + 3 = constant
0 1 2 3 (V – 4) + (V – 2) + (–2) = 0 2V = 8
166.
V = 4 m/s
Figure shows two blocks A and B connected to an ideal pulley string system. In this system when bodies are released then : (neglect friction and take g = 10 m/s2) fp=kesan'kkZ;svuqlkjnksfi.MArFkkB,dvkn'kZf?kjuh&jLlhfudk;lstqM+sgSAblfudk;esatcfi.MksadkseqDrdjrs
gSarks%¼?k"kZ.kdksux.;ekusarFkkg=10eh-@ls-2 ysa½
[M.Bank(07-08)_NLM_3.48]
[Made AA_2007]
(A) Acceleration of block A is 1 m/s2
Afi.M dkRoj.k 1eh-/ls-2gSA
(C) Tension in string connected to block B is 40 N
BlstqM+hjLlhesa ruko40NgSA
(B*) Acceleration of block A is 2 m/s2
Sol.
(D*) Tension in string connected to block B is 80 N Applying NLM on 40 kg block 40 fdxzk fi.M ij U;wVu ds fu;e yxkus ij 400 – 4T = 40 a For 10 kg block T = 10.4 a Solving a = 2m/s2 10 fdxzk- fi.M ds fy,T = 10(4 a)
gy djus ij a = 2m/s2 T = 80 N
RESONANCE
Afi.M dk Roj.k2eh-/ls-2gSA
BlstqM+hjLlhesa ruko80NgSA T
2T
4a
10kg
2T
4T 40kg
a
Page # 52
167.
Two blocks A (5kg) and B(5kg) attached to the ends of a spring constant 1000 N/m are placed on a smooth horizontal plane with the spring undeformed. Simultaneously velocities of 10m/s and 4 m/s along the line of the spring in the same direction are imparted to A and B then nksxqVdsA(5kg)oB(5kg)fLizaxfu;rkad1000N/mds,dfLizaxdsfljksaijtqM+sgq,,dfpdus{kSfrtryijj[ksgSaA fLizaxvfoÑrgSAAoBdks,dlkFklekufn'kkesafLizaxdhjs[kkdsvuqfn'kØe'k%10m/so4m/sdsosxfn;stkrs gSarks[Test(20-05-07)_PQ_PT-2_Q.12] 10
k = 1000 N/m
5kg
4
5kg
(A*) when the extension of the spring is maximum the velocities of A and B are same. (B*) the maximum extension of the spring is 30cm. (C) the first maximum compression occurs /56 seconds after start. (D*) maximum compression and maximum extension occur alternately. (A*)tc fLizax dk foLrkjvf/kdre gS rks AoBds osxleku gSA (B*) fLizaxdkvf/kdrefoLrkj30cm gSA (C)izFkevf/kdrelEihMu]izkjEHkgksusds/56lsd.MckngksrkgSA (D*)vf/kdre lEihMu o vf/kdre foLrkj ,d ds ckn ,d gksrs jgrs gSaA Sol.
1 m1m 2 1 2 2 2 m1 m 2 (V1 – V2) = 2 kx
1 ( 5 ) (5 ) 1 (10 – 4)2 = × 1000 x2 2 55 2 2.5 (36) = (1000) x2
(25) (36 ) 10 1 = x2 1000 ( 25 ) (36 ) = x2 10000
(5 ) ( 6 ) = x2 10 x = 0.30 m
Also
=
k
= 20 sec. T=
1000 (5 ) (5 ) 55
2 20 10
The first maximum compression occurs
xfr 'kq: gksus ds mijkUr
T sec. after start. 4 40
T sec.ckn igyh ckj vf/kdre laihMu gksxkA 4 40
Comprehension (7 - 10 ) vuqPNsn Following are three equations of motion
[M.Bank-08-9)_NLM_comprehension]
xfr dh rhu lehdj.ksa nh xbZ gS &
1 2 at v(s) = u 2 2 a s v(t) = u + at 2 Where ; S, u, t, a, v are respectively the displacement (dependent variable), initial velocity (constant), time taken (independent variable), acceleration (constant) and final velocity (dependent variable) of the particle after time t. tgk¡S, u, t, a, v Øe'k% foLFkkiu ¼ijra=k pj½] çkjfEHkd osx (fu;r), le; ¼Lora=k pj½] Roj.k (fu;r) rFkk t le; S(t) = ut +
168.
i'pkr~d.kdkvfUre osx ¼ijra=kpj½gSA
Find displacement of a particle after 10 seconds starting from rest with an uniform acceleration of 2m/
RESONANCE
Page # 53
s 2. Sol. 169. Sol. 170. Sol. 171.
Sol.
172.
,d d.k fojkekoLFkk ls,d leku Roj.k 2 m/s2 ls xfr djrk gS rks 10 lSd.M i'pkr~ d.k dk foLFkkiu gksxkA (A) 10 m S = ut + at2
(B*) 100 m
1 x2(10)2 = 100 m 2 Find the velocity of the particle after 100 m – 100 m foLFkkiu ds i'pkr~ d.k dk osx gksxk & (A) 10 m/s (B*) 20 m/s v = u + at v = 0 + 2 x 10 = 20 m/s.
(C) 50 m
(D) 200 m
(C) 30 m/s
(D) 0 m/s
S=0+
Find the velocity of the particle after 10 seconds if its acceleration is zero in interval (0 to 10 s) – ;fn d.k dk Roj.k (0 to 10 s) varjky esa 'kwU; gks rks 10 lSd.M i'pkr~ bldk osx gksxk & (A) 10 m/s (B) 20 m/s (C) 30 m/s (D*) 0 m/s v=u v = o m/s
Find the displacement of the particle when its velocity becomes 10 m/s if acceleration is 5 m/s 2 all through– tc d.k dk osx 10 m/s gks rks ml le; d.k dk foLFkkiu gksxkA (d.k dk ROj.k lnSo 5 m/s2 ) gSA (A) 50 m (B) 200 m (C*) 10 m (D) 100 m v2 = u2 + 2ar (10)2 = 0 + 2 × 5 × 5 5 = 10 m
In the figure shown, a person wants to raise a block lying on the ground to a height h. In both the cases if time required is same then in which case he has to exert more force. Assume pulleys and strings light. fp=kesan'kkZ,vuqlkj],dvknehxqVdsdkstehulshÅpk¡bZrdmBkukpkgrkgSAnksauksfLFkfr;ksaesavxjle;
cjkcj yxrk gS rks dkSulh fLFkfr esa vko';d cy dk eku vf/kd gksxkA ;g ekfu, fd iqyh rFkk jLlh gYds gSA [M.Bank(07-08)_NLM_3.26]
[Made 2006, RKV, GRSTU] [3.35 NLM]
Sol.
(A*) (i) (B) (ii) (C) same in both nksuksa esa leku (D) Cannot be determined fu/kkZj.k ugha dj ldrs 1 2 Since, h = at a shoule be same in both cases, because h and t are same in both cases as 2 given.
In (i) F 1 – mg = ma. F1 = mg + ma. In (ii) 2F2 – mg = ma
RESONANCE
F2 =
mg ma 2
F1 > F2 . Page # 54
173.
A cart of mass 0.5 kg is placed on a smooth surface and is connected by a string to a block of mass 0.2 kg. At the initial moment the cart moves to the left along a horizontal plane at a speed of 7 m/s. (Use g = 9.8 m/ s2) ?k"kZ.kjfgrlrgijj[khnzO;eku0.5fdxzk-dh,dxkM+h]0.2fdxzk-nzO;ekudsfi.MlsjLlh}kjktqM+hgSAizkjEHkesaxkM+h
{kSfrtryijck;ahvksj7eh-/ls-. dhpkyls xfrekugS(g=9.8eh-/ls2 ysa)
[M.Bank(07-08)_NLM_3.47]
0.5 kg
[Modified AKS_2007] 0.2 kg
2g towards right. 7 (B*) The cart comes to momentary rest after 2.5 s. (C*) The distance travelled by the cart in the first 5s is 17.5 m. (D) The velocity of the cart after 5s will be same as initial velocity. (A*) The acceleration of the cart is
(A*) xkM+hdkRoj.k
2g nk;harjQgksxkA 7
(B*)xkM+h2.5lsd.M i'pkr~{kf.kd:dsxhA
(C*)igys5 lsd.MesaxkM+h}kjkr;dhxbZnwjh17.5eh-gSA (D)5lsd.Mi'pkr~xkM+hdkosxizkjfEHkdosxdslekugksxkA Sol.
0.2 g = 0.7 a a =
2g m/s 2 7
For the case, it comes to rest when V = 0
ml fLFkfr ds fy, tc ;g fLFkj gksrk gS V = 0 2g t 0 = 7 + 7
49 t = 2g = 2.5 s
Distance travelled till it comes to rest
7 m/s
a
T T
T = 0.5 a
fLFkjvoLFkkvkusrdpyhxbZnwjh
a 0.2g
0.2 - T = 0.2 a
2g s 0 = 72 + 2 7
S = 8.75 m So in next 2.5s, it covers 8.75 m towards right. Total distance = 2 x 8.75 = 17.5 m After 5s, it speed will be same as that of initial (7 m/s) but direction will be reversed.
vr%vxys2.5 ls-esa;gnk;havksj8.75 eh-nwjhr;djsxkA dqy nwjh = 2 x 8.75 = 17.5 m 5 ls-i'pkr~bldhpky]izkjfEHkdpky(7 eh-/ls-)dscjkcjgksxhijUrqfn'kkfoijhrgksxhA 174.
Sol.
175.
A uniform thick string of length 5 m is resting on a horizontal frictionless surface. It is pulled by a horizontal force of 5 N from one end. The tension in the string at 1m from the force applied is: ,dlekueksVhjLlhdhyEckbZ5m gStks?k"kZ.kjfgrlrgijj[khgSA;g5N ds{kSfrtcyls,dfljsls[khaphtkrh gSrkscyyxkusokysfcUnqls1m dhnwjhijjLlhesarukogksxk: [ M . B a n k ( 0 7 08)_NLM_3.11] (A) zero (B) 5 N (C*) 4 N (D) 1 N a = 5/M a T M M 5 5N .a = . 5–T= 5 5 M 4M/5 M/5 T = 4 N. In the system shown in the figure the friction and mass of rope is negligible, then acceleration of 2 m is: Page # 55
RESONANCE
uhpsfn[kk;sx;sfudk;esajLlhnzO;ekujfgrrFkk?k"kZ.kjfgrgSArks2m nzO;ekudkRoj.kgksxk&
[M.Bank(07-08)_NLM_3.36]
(A)
g 5
(B)
2g 5
(C*) 0
(D)
5g M 5 . 2 5 M
P1 P2
Sol.
a 2m aP2 =
am 0 2
P2 mg – T = ma
2T
T
a 2 From eq. (i) & (ii) a=0 Ans. 2T – 2mg = 2m.
176.
T
m
T
a
mg
2m
2T
a/2
2mg
...........(i) .........(ii)
A weight W is supported by two strings inclined at 60º and 30º to the vertical. The tensions in the strings are T 1 & T 2 as shown. If these tensions are to be determined in terms of W using a triangle of forces, which of these triangles should you draw ? ,dHkkjW nksjfLl;ksatksÅ/oZls60ºrFkk30ºij>qdhgSijlarqfyrgSAjLlh;ksaesarukofp=kkuqlkjT1 rFkkT2 gSA;s rukoksaW ds:iesacyksadkf=kHkqtcukdjKkrdjusgSArks rqefuEuesalsdkSulkf=kHkqt[khapkxks?[M.Bank(0708)_NLM_3.13]
(A)
177.
(B)
(C)
(D)
(E*)
Figure shows two pulley arrangements for lifting a mass m. In figure (a) the mass is lifted by attaching a mass 2 m while in figure (b) the mass is lifted by pulling the other end with a downward force F=2 mg, If f a & f b are the accelerations of the masses in figures (a) and (b) then (Assume string is massless and pulley is ideal)
RESONANCE
Page # 56
fp=kesanksiqyhO;oLFkkn'kkZ;hx;hgStksm nzO;ekumBkrhgSA(a)esanzO;ekudks2m nzO;ekudklgk;rklsmBkrs gStcfd(b) esam nzO;ekudksnwljsfljsijF=2mg cyuhpsdhvksjyxkdjmBkrsgSA;fnfa ofb nksuksanzO;ekuksads Roj.kgSarks¼ekuyksjLlhnzO;ekujfgrgS,oaf?kjuhvkn'kZgSA½ [M_Bank(06-07)\Q.1.10\NLM]
(A) f a = f b Sol.
2m m g g = fa = 3 2m m
(B) f a = f b/2
(a)
(b)
(C*) f a = f b/3
(D) f a = 2 f b
2mg mg =g m So, f a = f b/3. fb =
180.
A painter is applying force himself to raise him and the box with an acceleration of 5 m/s2 by a massless rope and pulley arrangement as shown in figure. Mass of painter is 100 kg and that of box is 50 kg. If g = 10 m/s 2, then: [M.Bank(07-08)_NLM_3.45] 2 (fp=kkuqlkj,disUVjLo;acyyxkdjviusvkidksrFkkck¡Dldks5 m/s dsRoj.kls,dnzO;ekughujLlhoiqyhO;oLFkk }kjkfp=kkuqlkjÅijmBkrkgSA;fnisUVjdknzO;eku100 kg rFkkck¡Dldk50 kg gSrFkkjLlhdknzO;ekuux.;gS (;fn g = 10 m/s2 ) rks
tension in the rope is 1125 N (jLlh esa ruko 1125 N) tension in the rope is 2250 N (jLlh esa ruko 2250 N) force of contact between the painter and the floor is 375 N (isUVj rFkk Q'kZ ds chp lEidZ cy 375 N gS) (D) none of these (buesalsdksbZugha) For painter ; R + T – mg = ma R + T = m(g + a) ............(1) For the system ; 2T – (m + M)g = (m + M)a 2T = (m + M) (g + a) ..............(2) where ; m = 100 kg M = 50 kg a = 5 m/sec2 (A*) (B) (C*)
Sol.
and ; 181.
150 15 2 R = 375 N. T=
=
1125 N
In the figure shown the acceleration of A is, a
contact with B)
A
= 15ˆi 15ˆj then the acceleration of B is: (A remains in [M.Bank(07-08)_NLM_1.8]
n'kkZ;sx,fp=kesaAdkRoj.kgS, aA = 15ˆi 15ˆj rksB dkRoj.kgksxk: (A, B lEidZesajgrkgSA)
RESONANCE
Page # 57
Sol.
(A) 6 ˆi (B) 15 ˆi (D) From wedge constraint
(C) 10 ˆi
(D*) 5 ˆi
aAY
ostca/kuls
(a A ) = (a B ) a aAX cos 53° – aAY cos 37° a = aB cos 53º 37° aB = – 5 m/s aB 5 ˆi A rod AB is shown in figure. End A of the rod is fixed on the ground. Block is moving with velocity 37°
AX
B
182.
3 m/s towards right. The velocity of end B of rod when rod makes an angle of 60º with the ground is:
,dNM+ABfp=kesaiznf'kZrgSANM+dkAfljkHkwfeijtM+or~(fixed) gSrFkkCykWdnk¡;hvkSj 3 m/sosxlsxfrdj jgk gSA tc NM+ /kjkry ls 60º dk dks.k cuk;s rks NM+ ds fljs B dk osx gksxk& [ Made 2004] [M.Bank(0708)_NLM_8.16]
(A) 3 m/s 60º
Sol.
(D) 3 m/s
V
For regular contact V sin60º =
V = 2 m/s. 183.
(C) 2 3 m/s
(B*) 2 m/s
3
Ans.
In the Figure, the blocks are of equal mass. The pulley is fixed & massless. In the position shown, A is given a speed u and vB= the speed of B. (< 90°) [M.Bank(07-08)_NLM_8.7]
fp=kkuqlkjnksuksaxqVdslekunzO;ekudsgSAiqyhfu;rrFkknzO;ekughugSAfp=kesafn[kkbZfLFkfresaAdkspkyu nh tkrh gSA vB= B dh pkyA (< 90°) P
B
/// /
/// /
/// //
A
u
(A*) B will never lose contact with the ground B dHkhHkhtehulslEidZughaNksM+sxkA (B) The downward acceleration of A is equal in magnitude to the horizontal acceleration of B. A dkuhpsdhvkSjRoj.k]B ds{kSfrtRoj.kdsifjek.kdscjkcjgksxkA (C) vB = u cos (D*) vB = u/cos
RESONANCE
Page # 58
P
Sol.
B
/// /
/// /
/// //
A
u
From string constrained motion VBcos = u
u cos For angle < 90º Vertical component of tension will be less than mg. B will never lose contact with the ground.
VB =
184.
In the Figure, the pulley P moves to the right with a constant speed u. The downward speed of A is vA, and the speed of B to the right is vB. fn, x, fp=k esa f?kjuh P nka;h rjQ u fu;r osx ls xfreku gSA Adh uhps dh rjQ pky vA, gS rFkk B dh nka;h rjQ pkyvBgSrks [M.Bank(07-08)_NLM_8.9]
(A) (C) (D*) Sol.
185.
vB = vA (B*) vB = u + vA vB + u = vA the two blocks have accelerations of the same magnitude
nksauksCykWdksadsRoj.kdkifjek.klekugSA
BP + AP = constant
d BP d AP =0 dt dt u – VB + V A = 0 VB = u + VA Ans. dv B d(u VA ) = dt dt aB = 0 + aA aB = aA Ans.
The pulley moves up with a velocity of 10 m/sec. Two blocks are tied by a string which passes over a pulley. The velocity V will be _________. Given: vB = 5 m/s iqyh 10 m/sec osxlsÅijdhvksjxfr'khygSAnksxqVdksadksjLlhtks iqyh ds Åij ls xqtjrh gS dhlgk;rk ls ck¡/kktkrk gSA osxV _________ gksxkAfn;kgSvB = 5 m/s [Modified_NLM_8.6] [M.Bank(07-08)_NLM_8.5]
[ Ans: 25 m/s Sol.
AP + BP = constant
d AP d BP 0 dt dt 10 – V + 10 + 5 = 0 V = 25 m/s .
RESONANCE
]
P Ans.
V
Page # 59
186.
Sol.
Two blocks ‘A’ and ‘B’ each of mass ‘m’ are placed on a smooth horizontal surface. Two horizontal force F and 2F are applied on the 2blocks ‘A’ and ‘B’ respectively as shown in figure. The block A does not slide on block B. Then the normal reaction acting between the two blocks is : nks xqVds ‘A’o ‘B’, izR;sd dk nzO;eku‘m’,d fpduh {kSfrt lrg ij j[ks gSaA nks {kSfrt cyF o 2F Øe'k% xqVdsAo ‘B’ijfp=kkuqlkjvkjksfirgSAxqVdkA, xqVdsB ijughafQlyrkgSrksnksxqVdksadse/;dk;ZjrvfHkyEc izfrfØ;kcygSASA [Q.33/RK_BM/Constrained Motion] [M.Bank(07-08)_NLM_8.29]
(A) F
(B) F/2
(D)
(C)
Acceleration of two mass system is a =
nksnzO;ekufudk;dkRoj.kgSa= FBD of block A
F 2m
F leftward 2m
N
60°
30°
mF solving N = 3 F 2m
In the figure shown the velocity of lift is 2 m/s while string is winding on the motor shaft with velocity 2 m/s and block A is moving downwards with a velocity of 2 m/s, then find out the velocity of block B. [ Made 2004] M.Bank_NLM_8.26 fn[kk;s x;s fp=kesa fy¶Vdkosx2m/sgS tcfd jLlh2m/sds osx lseksVj 'kk¡¶V ij fyiV jgh gS rFkk CykWdA, 2m/s osxlsuhpsxfrdjjgkgsS]rksCykWdBdkosxKkrdjks& [S-(06-07)_BEK_DPP-23_Q.5]
(A) 2 m/s
ugha
VP =
(B) 2 m/s
V1 V2 2
4 VB 2 VB = 8 m/s . 2=
188.
(D*) 3F
xqVds Adk eqDr oLrq js[kkfp=k
N cos 60° – F = ma =
Sol.
3
ck;havksj
F
187.
F
(C) 4 m/s
(D*) none of these buesa ls dksbZ
2m/s
VB
B
A 2m/s
Two blocks A & B with mass 4 kg and 6 kg respectively are connected by a stretched spring of negligible mass as in figure. When the two blocks are released simultaneously the initial acceleration of B is 1.5 m/s 2 westward. The acceleration of A is: M.Bank_NLM_1.9
nksfi.MAoBftudsnzO;ekuØe'k%4fdxzko6fdxzkgS,dnzO;ekujfgr[khaphagqbZfLizaxlstqM+sgq,gSAtcnksuksa fi.Mksadks,dlkFkNksM+ktkrkgSrksBdkRoj.kif'pedhrjQ1.5m/s2 gSAAdkRoj.kgSA [S-(06-07)_BEK_DPP-23_Q.6]
RESONANCE
Page # 60
Sol.
(A) 1 m/s2 westward (C) 1 m/s 2 eastward (A) 1 m/s2 if'pe dh rjQ (C) 1 m/s2 iwoZ dh rjQ T = MBaB T = 6 × 1.5 = 9N T = MAaA
9 aA = = 2.25 m/s2 4 189.
Ans.
(B*) 2.25 m/s2 eastward (D) 2.75 m/s 2 westward (B*) 2.25 m/s2 iwoZ dh rjQ (D) 2.75 if'pe dh rjQ a
A
T
T
T
1.5m/s
T
2
B
Match the following : [Made RA sir Batch A 2007] [M.Bank(07-08)_NLM_6.18] Three blocks of masses m 1, m 2 and M are arranged as shown in figure. All the surfaces are frictionless and string is inextensible. Pulleys are light. A constant force F is applied on block of mass m 1. Pulleys and string are light. Part of the string connecting both pulleys is vertical and part of the strings connecting pulleys with masses m 1 and m 2 are horizontal. nzO;ekum1,m2 rFkkMdsrhufi.Mfp=kesan'kkZ;svuqlkjla;ksftrgSaAlHkhlrg?k"kZ.kjfgrrFkkjLlhvfoLrj.kh;
gSAf?kjfu;k¡ojLlhgYdhgSaAnzO;ekum1 dsfi.Mij,dfu;rcyF vkjksfirdjrsgSaAnksuksf?kjuhdkstksM+us okyhjLlhdkHkkxm/oZgSvkSjnzO;ekum1rFkkm2 dksf?kjuh;ksalstksM+usokyhjLlhdsHkkx{kSfrtgSa&
(A) Acceleration of mass m 1 (B) Acceleration of mass m 2
F (Q) m m 1 2
(D) Tension in the string
m 2F (S) m m 1 2
(C) Acceleration of mass M
(A) nzO;eku m1 dk Roj.k (B) nzO;eku m2 dk Roj.k (C) nzO;eku M dk Roj.k Sol.
F (P) m 1
(D) jLlhesaruko
(A) Q (b) Q (C) R (D) S
(R) zero
F (P) m 1
F (Q) m m 1 2
(R)'kwU;
m 2F (S) m m 1 2
FBD’s
T = m 2a. F – T = m 1a
RESONANCE
Page # 61
F = (m 1 + m 2)a T = m 2a 190.
m 2F T = m m . 1 2
F a = m m 1 2
F x = 0, aM = 0
Three identical balls 1,2,3 are suspended on springs one below the other as shown in the figure. OA is a weightless thread. M.Bank_NLM_1.2
fp=kesan'kkZ,vuqlkjrhu,dtSlhxsan1,2,3 fLizaxdhlgk;rklsfp=kkuqlkj,dnwljslstqM+hgqbZgSAOA,d nzO;ekughujLlhgSA
(a) If the thread is cut, the system starts falling. Find the acceleration of all the balls at the initial instant
vxjjLlhdksdkVfn;ktkrkgSrksfudk;fxjuk'kq:djnsrkgSAizkjfEHkd{k.kijlHkhxsanksadkRoj.kKkrdhft,A
(b) Find the initial accelerations of all the balls if we cut the spring BC which is supporting ball 3 instead of cutting the thread. Sol.
vxjgejLlhdhtxgxsan3dkstksM+usokyhfLizaxdksdkVnsrsgSrkslHkhxsanksadkizkjfEHkdRoj.kKkrdhft,A [ Ans: (a) 3 g 0, 0, (b) 0, g g ]
(a)
T AB = 2mg, T BC = mg TAB A m TAB
mg
B m mg
TBC
TBC Cm mg
For A, 2mg + mg = maA For B, T AB – mg – T BC = maB 2mg – mg – mg = maB T BC – mg = mac (b)
aA = 3g maB = 0 ac = 0.
aB = 0
T AB = 2mg TAB m B m
aB
mg
T AB – mg = maB 2mg – mg = maB aB = g () aA = 0 & aC = g().
RESONANCE
Page # 62
191.
A body of mass 32 kg is suspended by a spring balance from the roof of a verticallyoperating lift and going continuously downward from rest. At the instants the lift has covered 20 m and 50 m, the spring balance showed 30 kg & 36 kg respectively. The velocity of the lift is: [ M . B a n k ( 0 7 08)_NLM_1.11] fLFkjkoLFkklsyxkrkjÅ/okZ/kjuhpsdhvksjtkjgh,dfy¶VdhNrls,dfLçaax}kjk32kgnzO;ekudhoLrqyVdh gq;hgSAfy¶V}kjk20mo50m,nwjhr;djusokys{k.kksaijfLçaxdkikB;kadØe'k%30kgo36kggksrkgSfy¶Vdk
osx%
(A) decreasing at 20 m & increasing at 50 m 20m ij ?kV jgk gS rFkk 50m ij c<+ jgk gSA (B*) increasing at 20 m & decreasing at 50 m 20m ij c<+ jgk gS rFkk50m ij ?kV jgk gSA (C) continuously decreasing at a constant rate throught the journey
;k=kkdsnkSjkuyxkrkj,dfu;rnjls?kVjgkgSA
(D) continuously increasing at constant rate throughout the journey
;k=kkdsnkSjkuyxkrkj,dfu;rnjlsc<+jgkgSA (E) remaining constant throughout the journey.
;k=kkdsnkSjkufu;rjgrkgSA Sol.
Ans. 192.
N = m (g – a) , N < mg if a () and N > mg if a ( ) Reading of spring balance is less than m if a () and reading of spring balance is greater than m if a ( ) (B) In the figure shown all contact surfaces are smooth. Acceleration of B block will be: fp=kesafn[kkbZfLFkfresalHkhlEidZlrgfpduhgSAxqVdsBdkRoj.kgksxk:
[M.Bank(07-08)_NLM_3.39]
Sol.
(A*) 1 m/s 2 (B) 2 m/s2 Let aB = a then aA = 3a [by string constraint] T = maA mg – 3 T = m aB
193.
(C) 3 m/s 2
T = 3ma mg – 3 T = ma
mg – 9 ma = ma a =
g 10
(D) none of these buesa ls dksbZ ugha
(i)
a = 1 m/s 2
In the figure a block ‘A’ of mass ‘m’ is attached at one end of a light spring and the other end of the spring is connected to another block ‘B’ of mass 2m through a light string. ‘A’ is held and B is in static equilibrium. Now A is released. The acceleration of A just after that instant is ‘a’. In the next case, B is held and A is in static equilibrium. Now when B is released, its acceleration immediately after the release is 'b'. The value of a/b is : (Pulley, string and the spring are massless) [M.Bank(07-08)_NLM_1.17] fp=kesanzO;eku'm'dkCykWd‘A’gYdhfLizaxds,dfljslstqM+kgSrFkkfLizaxdknwljkfljkgYdhjLlh}kjknzO;eku 2mdsnwljsCykWd‘B’lstqM+kgSA‘A’dksfdlhusidM+kgSrFkk'B'fLFkjlkE;oLFkkesagSAvcAdksNksM+rsgSaAml{k.k dsBhdcknCykWd'A'dkRoj.k'a'gSAnwljhfLFkfresa‘B’dksidM+rsgSrFkkA fLFkjlkE;oLFkkesagSAvctcBdksNksM+rs gS]rksNksM+usdjusdsrqjUrcknbldkRoj.kbgSA
RESONANCE
Page # 63
[Made RKV, AS 2005]
Sol.
(A) 0 (B) undefined vLi"V For first case tension in spring will be Ts = 2mg
(C*) 2
(D) 1/2
just after 'A' is released.
2mg – mg = ma a = g In second case Ts = mg
gy.
2mg – mg = 2mb b = g/2 a/b = 2
izFkefLFkfresafLizaxesarukogksxk
Ts = 2mg 'A' dks eqDr djus ds rqjUr ckn 2mg – mg = ma a = g
f}rh;fLFkfresa
Ts = mg 2mg – mg = 2mb b = g/2 a/b = 2
194.
System is shown in the figure and man is pulling the rope from both sides with constant speed ' u'. Then the speed of the block will be: [Made 2004] M.Bank_NLM_8.14
fp=kesaiznf'kZrfudk;esa,dvknehfu;rpky'u'lsjLlhdksnksuksfljksals[khapjgkgSACykWddhpkygksxhA
Sol.
(A*)
3u 4
8 + 9 = c2
RESONANCE
(B)
3u 2
(C)
u 4
(D)noneofthesebuesalsdksbZugha
Page # 64
7
d 8 d 9 0 dt dt –V+u=0 u = V. 1 + 2 + 3 + 4 + 5 + 7 = 0
9
8 2
1
4 3
5
6
d 1 d 2 d 3 d 4 d 5 d 6 d 7 0 dt dt dt dt dt dt dt V + (V – V1) + (–V1) + 0 + (–V1) + (V1) + u = 0 2V – 4V1 + u = 0 2u + u = 4V1
V1 = 195.
3u . 4
In the figure shown all the surface are smooth. All the blocks A, B and C are movable, x-axis is horizontal and y-axis vertical as shown. Just after the system is released from the position as shown. [M.Bank(0708)_N.LM_4.11]
n'kkZ;safp=kesalHkhlrgfpduhgSAlHkhfi.MA,BrFkkCxfrdjldrsgSAx-v{k{kSfrtrFkky-v{kn'kkZ;svuqlkjm/ okZ/kjgSAfudk;dksfn[kkbZxbZfLFkfrlseqDrdjrsgSrksbldsrqjUrckn& (P Batch) y
A B
Sol.
x
C
[Made RKV_2007]
Horizontal Surface
(A*) Acceleration of 'A' relative to ground is in negative y-direction (B*) Acceleration of 'A' relative to B is in positive x-direction (C*) The horizontal acceleration of 'B' relative to ground is in negative x-direction. (D*) The acceleration of 'B' relative to ground directed along the inclined surface of 'C' is greater than g sin . (A*)tehudslkis{k'A'dkRoj.k_.kkRedy-fn'kkesagksxkA (B*)'B'dslkis{k'A'dkRoj.k/kukRedx-fn'kkesagksxkA (C*)tehudslkis{k'B'dk{kSfrtRoj.k_.kkRedx-fn'kkesagksxkA (D*) tehu ds lkis{k 'B' dk Roj.k urry 'C' ds vuqfn'k gsin ls T;knk gksxkA (Tough) There is no horizontal force on block A, therefore it does not move in x-direction, whereas there is net downward force (mg – N) is acting on it, making its acceleration along negative y-direction. Block B moves downward as well as in negative x-direction. Downward acceleration of A and B will be equal due to constrain, thus w.r.t. B, A moves in positive x-direction. fi.MijdksbZ{kSfrtcyughagS]vr%;gx-fn'kkesaughapyrkgS]tcfdblijuhpsdhvksjifj.kkehcy, (mg–N)yxrk gS]tksfdbldsRoj.kdks_.kkRedy-fn'kkesansrkgSA fi.MBuhpsdslkFk&lkFk_.kkRedx-fn'kkesaxfrekugksrkgSAArFkkBdkuhpsdhvksjRoj.kc)rklscjkcjgksrkgS] vr%Bdslkis{k]A/kukRedx-fn'kkesaxfrekugSA
B
B
fr fØ; k C } kj k v fHky Ec i z Normal reaction due to C Due to the component of normal exterted by C on B, it moves in negative x-direction. CijB}kjk vfHkyEc ds ?kVd ds dkj.k ;g _.kkRed,x-fn'kk esa xfreku gSA NA
B
RESONANCE
Mg
NC
Page # 65
The force acting vertically downward on block B are mg and NA(normal reaction due to block A). Hence the component of net force on block B along the inclined surface of B is greater than mg sin. Therefore the acceleration of 'B' relative to ground directed along the inclined surface of 'C' is greater than g sin CykWd Bij Å/okZ/kj uhpsdhvksj cy mg+NA(CykWd Ads dkj.k vfHkyEc izfrfØ;k)gSA blfy;s CykWd Bij dk;Zjr 196.
usV cy dk Bdh ur lrg ds vuqfn'k ?kVd mgsinls vf/kd gSA blfy;s Hkwfe ds lkis{k Bdk Roj.k] ftldh fn'kk C dh ur lrg ds vuqfn'k gS] gsin ls vf/kd gSA System is shown in the figure. Velocity of sphere A is 9 m/s. Then speed of sphere B will be: fp=kesaiznf'kZrfudk;esa;fnxksysAdkosx9m/sgksrksxksysBdhpkygksxhA [ Made 2004] [M.Bank(07-08)_NLM_8.14]
rB = 5m 9m /s
(A) 9 m/s
(B*) 12 m/s
(C) 9
5 m/s 4
(D)none of thesebuesa ls dksbZ ugha
Sol.
9 cos = v sin
197.
(i)
19 – R = tan (ii) 12 (R + 5)2 = (12) 2 + (19 – R)2 R = 10 Hence from (i) and (ii) v = 12 m/s2 For the following system shown assume that pulley is frictionless, string is massless (m remains on M) : [M.Bank(07-08)_NLM_6.16] n'kkZ;sx;sfudk;dsfy,ekukf?kjuh?k"kZ.kjfgrgSoMksjhnzO;ekujfgrgS(m,Mijj[kkjgrkgS)&
M.Bank_NLM_6.16 Find Kkr dhft, & (a) the acceleration of the block A. (b) Normal reaction on m is (force on C due to B) (c) the force on the ceiling (a)xqVds AdkRoj.kgS% (b) m ij vfHkyEc çfrfØ;k gS (B ds dkj.k C ij cy) (c) Nr ij cy gS % [RK_TOPIC/NL/Q.No.7 made subjective]
RESONANCE
[4]
Page # 66
Sol. By newtons law on system of (A, B, C) (a) (M + M – m) g = (2M + m) a
mg 2M m (b) free body diagram ‘C’ block
a=
mg – N = ma
N=m N=
2M gm 2M m
(c) T – Mg = M
gm g 2M m
mg for A block 2M m
T = Mg +
for pulley
Mmg 2M m
P = 2T + Mg = 2Mg + =
2Mmg + Mg 2M m
6M 3m 2m Mg 2M m
6M 5m P = 2 M m Mg Ans. 198.
mg 2Mmg (6M 5m) Mg (b) 2M m 2M m (c) 2M m
(a)
System is shown in the figure. Assume that cylinder remains in contact with the two wedges. The velocity of cylinder is [M.Bank(07-08)_NLM_8.33] fp=kesa,dfudk;iznf'kZrgSekfu;sfdcsyunksostksdse/;lEidZesajgrkgSAcsyudkosxgScsy u
u m/s
(A)
19 4 3
u m/s 2
RESONANCE
(B)
13 u m/s 2
30°
(C)
3 u m/s
30°
2u m/s
(D*)
[Made BKM, 2005]
7 u m/s Page # 67
Sol.
(D) Method - I fof/k- As cylinder will remain in contact with wedge A Vx = 2u
D;ksafd csyu ost A ds laidZ esa gS
As it also remain in contact with wedge B ;gostBdslkFkHkhlaidZesajgrkgSA u sin 30° = Vy cos30° – Vx sin30° sin 30 U sin 30 + cos 30 cos 30 Vy = Vx tan30° + u tan 30°
Vy = Vx
V=
Vy = 3u tan30° =
Vx2 Vy2 =
Method - II In the frame of A
and 199.
3u
7 u Ans.
fof/k- II fof/kAdsra=kesa
3u sin 30º = Vycos30º Vy = 3u tan 30º =
o Vx = 2u
3u
V=
Vx2 Vy2
=
7 u Ans.
A light spring is compressed and placed horizontally between a vertical fixed wall and a block free to slide over a smooth horizontal table top as shown in the figure. The system is released from rest. The graph which represents the relation between the magnitude of acceleration ‘ a ‘ of the block and the distance ‘ x ‘ travelled by it (as long as the spring is compressed) is M.Bank_N.L.M._1.23
,d gYdh fLizax dks laihfMr fd;k tkrk gS vkSj bls fpduh {kSfrt est ds Åij xfr ds fy, LorU=k ,d CykWd vkSj ,dÅ/okZ/kjn`<+nhokjdschpj[kktkrkgStSlkfdfp=kesafn[kk;kx;kgSAblO;oLFkkdksfojkelsNksM+ktkrk gSAogxzkQtksCykWddsRoj.kadsifjek.kvkSjblds}kjk pyhxbZnwjh xdschplEcU/kdksn'kkZrkgS(tcrd fLizaxlaihfMrdhtkrhgS)
(A) Sol.
(B)
(C*)
(D)
Let the initial compression of spring be . Then the acceleration after the block travels a distance x is (Moderate) ekuk fLizax dk izkjfEHkd laihMu gSA rksCykWd }kjk x nwjh pyus dsckn Roj.k gSA (Moderate)
a=
The graph of a vs x is a rFkk x dk xzkQ gS
RESONANCE
k ( – x) m
Page # 68
200.
Sol.
A hinged construction consists of three rhombus with the ratio of sides (5 : 3 : 2). Vertex A3 moves in the horizontal direction with velocity V. Velocity of A2 will be : fp=kkuqlkj,ddhyfdrfudk;rhuleprqHkqZtftudhHkqtkvksaesa5:3:2dkvuqikrgS]lscukgS;fnA3 fljsdk{kSfrt fn'kkesaosxVgSArksA2fljsdkosxgksxkA [M.Bank(07-08)_NLM_8.36]
(A) 2.5 V (B) 1.5V OA3 = 20 cos , OA2 = 16 cos
(C) (2/3)V
dOA 3 d = –20sin dt dt
V = –20sin
d dt
d V = dt 20 sin velocity of OA2 VA 2
dOA 2 d = –16sin dt dt
–V 16 VA 2 16 sin = V = 0.8 V 20 20sin
201.
(D*) 0.8V
Ans.
System shown in figure is in equilibrium. The magnitude of net change in tension in the string just before and just after, when one of the spring is cut. Mass of both the blocks is same and equal to m and spring constant of both springs is k. (Neglect any effect of rotation)
fn,x,fp=kesafudk;lkE;koLFkkesagSAtc,dfLçaxdksdkVfn;ktkrkgSrksjLlhesaruko esadqyifjorZu fLçaxdsdkVusdsBhdigysvkSjBhdcknesaD;kgksxknksuksaxqVdksadknzO;eku,dtSlkgSvkSjm dscjkcjgS vkSjnksuksafLçaxdkfLçaxfu;rkadKgSA¼?kw.kZudsizHkkodksux.;ekusa½ M.Bank_NLM_1.14 [Made 2004]
Sol.
(A*)
mg 2
(B)
mg 4
Initially tension in string = mg Initially tension in each spring = mg After cut the spring
(C)
3m g 4
(D)
3m g 2
mg
mg
T1
a
ma = mg + T 1 – mg ma = T 1
RESONANCE
Page # 69
T1
a
mg ma = mg – T 1 mg T1 = 2
Change in tension = T – T 1 = 202.
mg 2
A boy and a block, both of same mass, are suspended at the same horizontal level, from each end of a light string that moves over a frictionless pulley. The boy starts moving upwards with an acceleration 2.5 m/s2 relative to the rope. If the block is to travel a total distance 10 m before reaching at the pulley, the time taken by the block in doing so is equal to : [Made AKS_2007_GRSTU] [M.Bank_NLM_3.46]
/k"kZ.kjfgrf?kjuhlsikfjr,dgYdhjLlhdsizR;sdfljsij]leku{kSfrtLrjij],dyM+dk,oa,dfi.Mnksuksaleku nzO;ekudsyVdsgSAyM+dkÅijdhvksjjLlhdslkis{k2.5eh-/ls-2 dsRoj.klspyukizkjEHkdjrkgSA;fnfi.M]f?kjuh rdigqapusesadqynwjh10eh-r;djrkgS]rksfi.Mdks,slkdjusesayxkgqvkle;cjkcjgSA
10 m
m
Ans. Sol.
(A)
8s
(B*) 4s
m
(C)
10
2
s
(B) For block T – mg = ma .............(i) For boy T – mg = m(2.5 – a) .............(ii) From (i) & (ii) 2ma = m(2.5) a = 1.25 m/sec2 Acceleration of boy and pulley will be same equal to 1.25 m/s2 w.r.t. ground. Hence yM+ds,oaf?kjuhnksuksadktehudslkis{kRoj.kcjkcj,oa1.25eh-/ls-2 dscjkcjgksxkAvr%
(D) 8s
1 (1.25) t2 t = 4 sec. 2 Comprehension : (6 - 8) Two blocks A and B of mass 1 kg & 2 kg respectively are connected by a string, passing over a light frictionless pulley. Both the blocks are resting on a horizontal floor and the pulley is held such that string remains just taut. At moment t = 0, a force F = 20 t newton starts acting on the pulley along vertically upward direction, as in Figure. Calculate : [M.Bank(07-08)_NLM_4.13] 1kgrFkk2kgnzO;ekudsnksCykWdksadksjLlh}kjkf?kjuhlstksM+ktkrkgSAf?kjuh?k"kZ.kjfgrrFkkgYdhgSAnksuksaCykWd {kSfrtlrgijfojkeesagSrFkkf?kjuhblrjggSfdjLlhruhgqbZgSAt = 0 ijF = 20 t U;wVudkcyf?kjuhijÅij dhvksjfp=kkuqlkjyxk;ktkrkgSAx.kukdjks: 10 =
203.
Velocity of A when B loses contact with the floor : CykWdAdkosxtcCykWdB /kjkrylslaidZNwV tkrkgSA
RESONANCE
Page # 70
(A*) 5 m/s 204.
(B) 10 m/s
(D) zero
(C) 10/3 m
(D) zero
(C) 30 J
(D) zero
Height raised by the pulley upto that instant :
bl{k.krdf?kjuh}kjkr;Å¡pkbZ (A) 5/3 m
205.
(C) 15 m/s
(B*) 5/6 m
Work done by the force F upto that instant. bl {k.k rd cy F }kjk fd;k x;k dk;Z
175 50 J (B) J 6 3 [ Ans: 5 m/s, 5/6 m, 175/6 J ] Sol. (6 to 8) Block A will loose contact at 10t = 1g t = 1 sec. Block B will loose contact at 10t = 2g t = 2sec. Acceleration of A a = 10t – 1g for 1 < t < 2
(A*)
(a)
dv = 10t – 10 dt Velocity when block block constant
10t
v
0
10t
2
dV (10 t 10 ) dt 1
t2 V = 10 2 10 t 1
2
(b)
= [5 × 4 – 10 × 2] – [5(1)2 – (10)] = 5 m/s Displacement of pulley at t = 2 10t 10t 10 20 1g 1g
dV = 10t – 10 dt V = 5t2 – 10t + C V = 5t2 – 10t + C At t = 1 V=0 C = 5. V = 5t2 – 10t + 5 dx = 5t2 – 10t + 5 dt dx = (5t2 – 10t + 5)dt
RESONANCE
Page # 71
t3 2 x = 5 3 10 t 5t 1
2
Displacement of pulley = (c)
= =
5 x = m. 6 2
Work done by force = mgh + 1 × 10 × 50 25 3 2
5 1 + ×1×5×5 3 2
100 75 6
=
5 m. 3
=
1 mV2 2
175 J. 6
Comprehension : (5 – 7) In the position shown collar B moves to the left with a velocity of 150 mm/s. Determine: fn[kk;sx;sfp=kesadkWyjB ck¡;hvkSj150 mm/s dsosxlsxfrdjrkgS rkscrkb;sA [M.Bank_NLM_8.25] [Made 2004] A
C
B
206.
The velocity of collar A dkWyjAdkosx (A) 300 mm/sec. left (B) 600 mm/sec. left (C*) 300 mm/sec. right (D) 600 mm/sec. right
207.
the velocity of portion C of the cable rkjdsC Hkkxdkosx (A) 150 mm/sec. left (B) 300 mm/sec. right (C) 600 mm/sec. right (D*) 600 mm/sec. left
208.
the relative velocity of portion C of the cable with respect to collar B. dkWyjBdslkis{krkjdsC Hkkxdkvkisf{kdosx (A) 300 mm/sec. left (B) 300 mm/sec.right (C) 450 mm/sec. right (D*) 450 mm/sec. left [ Ans.: VA = 300 right, VB = 600 left, VCB = 450 left ] 1 + 2 + 3 + 4 = 0 1' + 2' + 3' + 4' = 0 (VB – VA) + VB + VA + VA = 0
Sol.
1
VA = – 2VB VA = – 2VB = – 2 × 150 m/sec. = – 300 mm/sec, = 300 mm/sec right VB =
B
A
C
4
3
2
VA VC 2
300 VC 2 VC = 600 mm/sec. 600 mm/sec. left VCB = VC – VB = 600 – 150 = 450 mm/sec. left. 150 =
RESONANCE
Page # 72
209.
Consider the system as shown in the figure. The pulley and the string are light and all the surfaces are frictionless. The tension in the string is (g = 10 m/s 2). M.Bank_NLM_3.30 [GRSTU CT-2(23-09)_Paper2_Q.6]
fp=kesafn[kk,fudk;dksfyft,AiwyhrFkkjLlhgYdhgSrFkklHkhlrg?k"kZ.kghugSArcjLlhesarukogksxk (g = 10 m/s2).
Sol.
(A) 0 N (B) 1 N (C) 2 N (Easy) The acceleration of block of mass m 1 is (Easy) m1 CykWd dk Roj.k
a=
210.
(D*) 5 N
m1 g = 5 m/s 2 m1 m 2
From FBD of block m 1 m1 CykWd dk FBD m 1g – T = m 1a T = m 1 (g – a) = 1 × 5 = 5N
For the following system shown assume that pulley is frictionless, string is massless (m remains on M) : [M.Bank(07-08)_NLM_6.16] n'kkZ;sx;sfudk;dsfy,ekukf?kjuh?k"kZ.kjfgrgSoMksjhnzO;ekujfgrgS(m,Mijj[kkjgrkgS)&
M.Bank_NLM_6.16 Find Kkr dhft, & (a) the acceleration of the block A. (b) Normal reaction on m is (force on C due to B) (c) the force on the ceiling (a)xqVds AdkRoj.kgS% (b) m ij vfHkyEc çfrfØ;k gS (B ds dkj.k C ij cy) (c) Nr ij cy gS % [RK_TOPIC/NL/Q.No.7 made subjective]
[4]
Sol. By newtons law on system of (A, B, C) (a) (M + m – M) g = (2M + m) a
mg 2M m (b) free body diagram ‘C’ block
a=
RESONANCE
Page # 73
mg – N = ma
N=m N=
2M gm 2M m
(c) T – Mg = M
gm g 2 M m
mg for A block 2M m
T = Mg +
for pulley
Mmg 2M m
P = 2T + mg = 2mg + =
2Mmg + mg 2M m
6M 3m 2m mg 2M m
6M 5m P = 2 M m mg Ans. 211.
(a)
mg 2Mmg (6M 5m) Mg 2M m (b) 2M m (c) 2M m
A rod of length 2 is moving such that its ends A and B move in contact with the horizontal floor and vertical wall respectively as shown in figure. O is the intersection point of the vertical wall and horizontal floor. The velocity vector of the centre of rod C is always directed along tangent drawn at C to the fp=kkuqlkj,d2yEckbZdhNM+blizdkjxfrdjrhgSAfdbldsfljsArFkkB{kSfrt/kjkryrFkkm/oZnhokjdslEidZ esaxfrdjrsgSA'O'fcUnqm/okZ/kjnhokj,oa{kSfrtQ'kZdkdVkufcUnqgSANM+dsdsUnzCdkosxlfn'kges'kkmlLi'kZ js[kkdsvuqfn'kgksxktksfddsfcUnqCijMkyhtkrhgS& [M.Bank(07-08)_NLM._8.34] [Made 2005, MPS] (A) circle of radius
whose centre lies at O 2 (C) circle of radius 2whose centre lies at O
(B*) circle of radius whose centre lies at O (D) None of these
f=kT;kdso`ÙkftldkdsUnzOijfLFkrgSA (B*)f=kT;kdso`ÙkftldkdsUnzO ijfLFkrgSA 2 (C)2f=kT;kdso`ÙkftldkdsUnzO ijfLFkrgSA (D)buesalsdksbZughA At any instant of time the rod makes an angle with horizontal, the x & y coordinates of centroid of rod are (A)
Sol.
RESONANCE
Page # 74
x = cos y = sin x2 + y2 = 2 Hence the centre C moves along a circle of radius with centre at O. velocity vector of C is always directed along the tangent drawn at C to the circle of radius whose centre lies at O. 212.
Which of the following statement is not true?
fuEuesalslghrFkkughagS\ (A) the
[Made 2003] [M_Bank_WPE_3.8] Work done by conservative force on an object depends only on the initial and final states and not on path taken.
lajf{krcy}kjkfdlhoLrqijfd;kx;kdk;ZçkjfEHkdrFkkvfUrefLFkfrijfuHkZjdjrkgSAiFkijughaA
(B) The change in the potential energy of a system corresponding to conservative internal forces is equal to negative of the work done by these forces. (C) the
If some of the internal forces within a system are non-conservative, then the mechanical energy of system is not constant.
(D*)
If the internal forces are conservative, the work done by the internal forces is equal to the change in mechanical energy.
gSA 213.
fLFkfrtÅtkZesaifjorZuvkUrfjdlajf{krcy}kjkfd;sx;sdk;Zdk_.kkRedgksrkgSA
;fndqNvkUrfjdcyfdlhfudk;dsvUrjvlajf{krgksrks;kaf=kdÅtkZlajf{krughajgsxhA
;fnvkUrfjdcylajf{krgksrks]vkUrfjdcyksads}kjkfd;kx;kdk;Z;kaf=kdÅtkZesaifjorZudsrqY;gksrk
Two uniform rods of equal length but different masses are rigidly joined to form an Lshaped body, which is then smoothly pivoted about O as shown. If in equilibrium the body is in the shown configuration, ratio M/m will be : [M.Bank(07-08)_Rotation_2.10] lekuyEckbZrFkkfHkUu&fHkUunzO;ekudhnksle:iNM+sLvkdkjesafp=kkuqlkjn`<+:ilstqM+hgqbZgSAbudksO fcUnqlsfp=kkuqlkjyVdk;kx;kgSA;fnlkE;oLFkkesaoLrqiznf'kZrO;oLFkkesagksrksM/m dkvuqikrgksxkA m
90°
Sol.
O 30°
M
(A) 2 (B) 3 (C) 2 (D*) 3 In equilibrium, torques of forces mg and Mg about an axis passing through O balance each other. lkE;oLFkkesa]mgrFkkMgcyksadkcyk?kw.kZO lsikfjrv{kdslkis{k,dnwljsdkslUrqfyrdjnsrkgSvr% mg.
L L cos30° = Mg cos60° 2 2
M 3 m
214. In the pulley system shown in figure, block C is going up at 2 m/s and block B is going up at 4 m/s, then the velocity of block A on the string shown in figure, is equal to : fp=kkuqlkj ,d f?kjuh fudk; ds fy, CykWd C, 2 m/s ds osx ls Åij dh vksj rFkk CykWd B, 4 m/s, ds osx ls Åij dh vksjxfrdjjgkgSrksjLlhijfLFkrCykWdAdkosxgksxkA [Made Ashish Arora 2007] P & F [M.Bank(07-08)_NLM_8.39]
RESONANCE
Page # 75
(A) 2 m/s
(B*) 4 m/s
(C) 6 m/s
(D) 8 m/s
Sol. 1 + 2 + 3 = constant
0 1 2 3 (V – 4) + (V – 2) + (–2) = 0 2V = 8 215.
V = 4 m/s
A heavy spherical ball is constrained in a frame as in figure. The inclined surface is smooth. The maximum acceleration with which the frame can move without causing the ball to leave the frame:
,dHkkjhxsanfp=kesan'kkZ,vuqlkjfLFkrgSAurryfpdukgSAblurrydksvf/kdrefdlRoj.klsysdjtkldrs gSarkfd;gxsanurryijxfrughadjsaA M.Bank_NLM_5.7
216.
Sol.
(A)
g 2
(B) g 3
g 3
(C*)
(D)
g 2
A monkey pulls along the ground mid point of a 10 m long light inextensible string connecting two identical objects A & B each of mass 0.3 kg continuously along the perpendicular bisector of line joining the masses. The masses are found to approach each other at a relative acceleration of 5 m/s2 when they are 6 m apart. The constant force applied by monkey is : M.Bank_NLM_7.29 ,d10 m yEch gYdh vrU; jLlh ls nks ,d leku oLrq, ArFkkB izR;sd nzO;eku 0.3 kg ls cU/ks gSA ,d cUnj tehudhfn'kkesaArFkkBdkstksM+usokyhjs[kkdsyEcor~fn'kkesabljLlhdks,dfu;rcylsyxkrkj[khap jgk gSa ;s nzO;eku ,d nwljs dh vkSj lkis{k Roj.k 5 m/s2 ls xfr dj jgs gS tc muds chp dh nwjh 6m gSA cUnj
}kjk yxk;k x;k fu;r cy dk eku gSA (A) 4 N arel =
(B*) 2 N
5m/sec 2
5 aA = aB = = 2.5 m/sec. 2
.3 kg A
3 = .3 × 2.5 5
3
.3 kg B
T
(D) none
Tsin
F
T=
5 N 4
Force applied by monkey = 2T sin = 2 ×
RESONANCE
Tcos 5
3 4 cos = sin = 5 5 T cos = m(2.5)
T×
(C) 3 N
5 4 × = 2N 4 5 Page # 76
217.
A block of mass 10 kg is lying on the ground. The force exerted by the block on the earth is: [Neglect the effect of rotation of the earth about its own axis and its revolution around the sun] M.Bank_NLM_7.40
,d 10kgnzO;eku dk CykWd /kjkry ij j[kk gSA CykWd ij i`Foh }kjk yxk;k x;k cy gS [i`Foh dk viuh v{k ds lkis{krFkklw;Zdspkjksavksj?kw.kZudksux.;ekusa] [ g = 10 N/m2 ] (A) 100 N 218.
219.
(B) 10 N
(C) greater than 100 N
(D*) none of these
Two blocks A (5 kg) & B (3 kg) resting on a smooth horizontal plane are connected by a spring of stiffness 294 N/m. A horizontal force F of magnitude 3 × 9.8 N acts on A as shown. At the instant B has an acceleration of 4.9 m/s2 towards left : M.Bank_NLM_1.14 (nksfi.MA(5kg)rFkkB(3kg),dfpduh{kSfrtlrgijj[ksagSrFkkfp=kkuqlkj,dfLizaxftldkfLizaxfu;rkad294 N/m gS] ls tqM+h gSA ,d {kSfrt cy F = 3 ×9.8 N fi.M ‘A’ij dk;Zjr gSA ftl le; ‘B’dk Roj.k 4.9 m/s2 cka;h rjQ gSrc)
(A) the acceleration of A is 3 × 1.96 m/s2 (A dk Roj.k 3 × 1.96 m/s2 gS) (B*) the acceleration of A is 3 × 0.98 m/s2 (A dk Roj.k 3 × 0.98 m/s2 gS) (C) the extension in the spring is 0.1 m (fLizax dk foLrkj 0.1 m gS) (D) the extension in the spring is 0.2 m. (fLizax dk foLrkj 0.2 m gS)
A rigid rod leans against a vertical wall at a point A. The other end of the rod is on the horizontal floor. Somebody is pushing point A of the rod downwards with constant velocity. Will B move with constant velocity ? What is the path of the centre of the rod. M.Bank_NLM_8.37 ,dn`<+NM+fdlhÅ/oZnhokjdsfcUnqAij>qdhgS]NM+dknwljkfljk{kSfrtQ'kZijgS]dksbZO;fDrNM+dksfcUnqA ijfu;rosxlsuhpsdhvksj/kdsyjgkgSrksD;kfljsBdkosxfu;rgksxk\NM+dsdsUnzdkfcUnqiFkD;kgksxk\
[ Ans. No, Circular] 220.
Figure shows two pulley arrangements for lifting a mass m. In (a) the mass is lifted by attaching a mass 2 m while in (b) the mass is lifted by pulling the other end with a downward force F=2 mg, If f a & f b are the accelerations of the two masses then fp=kesanksiqyhO;oLFkkn'kkZ;hx;hgStksm nzO;ekumBkrhgSA(a)esanzO;ekudks2m nzO;ekudklgk;rklsmBkrs gStcfd(b) esam nzO;ekudksnwljsfljsijF=2mg cyuhpsdhvkSjyxkdjmBkrsgSA;fnfa ofb nksuksanzO;ekuksadk
Roj.kgSrks
221.
(A) f a = f b
M.Bank_NLM_1.10
(B) f a = f b/2
(C*) f a = f b/3
(D) f a = 2 f b
The pulley moves up with a velocity of 10 m/sec. Two blocks are tied by a string which passes over a pulley. The velocity V will be _________. Given: vB = 5 m/s iqyh10 m/sec osxlsÅijdhvksjxfr'khygSAnksCykWddksjLlhdhlgk;rklstksiqyhdsÅijesaxqtjrhgSck¡/ kk tkrk gS osx V _________ gksxk fn;k gSvB = 5 m/s [Modified_NLM_8.6]
RESONANCE
Page # 77
[ Ans: 25 m/s 222.
]
Two blocks A and B of masses 4 kg and 12 kg respectively are placed on a smooth plane surface. A force F of 16 N is applied on A as shown. The force of contact between A & B is: 4 fdxzk0 rFkk 12 fdxzk0 nzO;eku ds nks CykWd Øe'k% ArFkkB fpdus {kSfrt lrg ij j[ks gq, gSA ,d 16 N dk cy F, Aij fp=kkuqlkj yxk;k tkrk gSAArFkk B ds e/; yxus okyk lEidZ cy dk eku gksxk & M.Bank_NLM_4.8
(A) 4 N 223.
(B) 8 N
Match the following :
(C*) 12 N
[Test ACJ/18.6.06/Q.33] [NLM]
I (a) when a bus starts suddenly, the passengers fall backward (b) When a bullet is fired from a gun, it gives a backward jerk called recoil to the gun man. (c) Rockets move forward by ejecting gases in the backward direction.
fuEu dk feyku dhft, %
(D) 16 N
I
(a) tc,dclvpkud'kq:gksrhgSrks;k=khihNs
dhvksjfxjrsgSA (b) tc,dxksyhcUnwdlspykbZtkrhgS]rks ;gcUnwdpykusokysdksihNs>Vdknsrh gSA (c)jkWdsVxSldksihNsdhvksjNksM+rsgq,]vkxs dhvksjc<+rkgSA
II (P) Newton’s 3 law rd
(Q) Newton’s 2nd law (R) Newton’s 1st law
II
(P) U;wVu dk r`rh; fu;e (Q) U;wVu dk f}rh; fu;e (R) U;wVu dk çFke fu;e
Ans.
(a) R, (b) P, (c) P
224.
Three weights are hanging over a smooth fixed pulley as shown in the Figure. What is the tension in the string connecting weights B and C? M.Bank_NLM_3.14 ,dfpduhn`<+f?kjuhijrhunzO;ekufp=kkuqlkjyVdjgsgSABrFkkCdkstksM+usokyhjLlhesarukofdrukgksxk
\
(A) g
225.
(B) g/9
(C) 8g/9
(D*) 10g/9
Two blocks A & B with mass 4 kg and 6 kg respectively are connected by a stretched spring of negligible mass as in figure. When the two blocks are released simultaneously the initial acceleration of B is 1.5 m/s 2 westward. The acceleration of A is: nksxqVdsArFkkBftudknzO;ekuØe'k%4kgvkSj6kg gSAmUgsa,d[khaphgqbZfLçaxlsftldknzO;ekuux.;gSrkstc
RESONANCE
Page # 78
nksuksaxqVdksadks,dlkFkNksM+ktkrkgSrksBdkRoj.kif'pedhvksj1.5m/s2 gSAAdkRoj.kgksxk: [M. Bank_NLM_1.9]
(A) 1 m/s2 westward (C) 1 m/s 2 eastward 226.
(B*) 2.25 m/s2 eastward (D) 2.75 m/s 2 westward
Initially the spring is undeformed. Now the force 'F' is applied to 'B' as shown. When the displacement of 'B' w.r.t. 'A' is 'x' in some time then the relative acceleration of 'B' w.r.t. 'A' at that moment is : izkjEHkesafLizaxvladqfprgSAvcB ijn'kkZ;svuqlkjcy'F' yxk;ktkrkgSAtcdqN le;ckn'B' dk'A' dslkis{k foLFkkiu'x' gSrksml{k.k'B' dk 'A' dslkis{kRoj.kgksxk& M.Bank_NLM_1.42
(A) 227.
F 2m
(B)
F kx m
(C*)
F 2 kx m
(D) none of these
Two smooth spheres each of radius 5 cm and weight W rest one on the other inside a fixed smooth cylinder of radius 8 cm. The reactions between the spheres and the vertical side of the cylinder are: M.Bank_NLM_4.2
8 lseh0f=kT;kdsfpdusfLFkjcsyudsvUnjnksfpdusxksysçR;sdf=kT;k5cm rFkkW HkkjokysxksysfLFkjj[ks
gSAxksysadschpçfrfØ;krFkkxksysrFkkÅ/oZnhokjdschpçfrfØ;kKkrdjks\
(A) W/4 & 3W/4 228.
229.
Ans.
(B) W/4 & W/4
(C*) 3W/4 & 3W/4
(D) W & W
A block of weight 9.8N is placed on a table. The table surface exerts an upward force of 10 N on the block. Assume g = 9.8 m/s 2. M.Bank_NLM_4.3 9.8N Hkkj okyk ,d xqVdk est ij j[kk gqvk gSA est dh lrg xqVds ij Åij dh vksj 10 N dk cy yxkrh gSA ekfu, g = 9.8 m/s2 (A*) The block exerts a force of 10N on the table xqVdk est ij 10N dk cy yxkrk gSA (B) The block exerts a force of 19.8N on the table xqVdk est ij19.8N dk cy yxkrk gSA (C) The block exerts a force of 9.8N on the table xqVdk est ij 9.8N dk cy yxkrk gSA (D*) The block has an upward acceleration. xqVdsdkÅijdhvksjRoj.kgksxkA
A small block B is placed on another block A of mass 5 kg and length 20 cm. Intially the block B is near the right end of block A (figure). A constant horizontal force of 10 N is applied to the block A. All the surfaces are assumed frictionless. Find the time elapsed before the block B separates from A . HCV_Ch-5_Ex._11[NLM] ,dxqVds Bdks5kgnzO;ekurFkk20cmyEckbZokysnwljsxqVdsAijj[kktkrkgSAçkjEHHkesaxqVdkB,xqVdsAds nk;havksjvfUrefljsijj[kkgSA(fp=kkuqlkj)AxqVdsAij10Ndk,d{kSfrtfu;rcyyxk;ktkrkgSAlHkhlrgksa dks?k"kZ.kghuekusaAxqVdsBds,xqVdsAlsvyxgksuslsiwoZyxsle;dhx.kukdhft,A
0.45 s
RESONANCE
Page # 79
230.
A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in figure. Show that the force (assumed equal for both the friends) exerted by each friend on the rope increases as the man moves up. Find the force when the man is at a depth. ;gvknehdpkSM+kbZdh,d[kkbZesafxjtkrkgSvkSjmldsnksnksLrmls/khjs&/khjsgYdhjLlh}kjkÅijdhvksj[khap jgsgSAfl)dhft,fdgjnksLr}kjkyxk;kx;kcy( nksuksanksLrksadsfy,lekuekfu,)dkekutSls&tSlsvknehÅij tkrkgS] c<+rktkrkgSAcy dkekucrkb;s tcoghxgjkbZij gSA HCV_Ch-5_Ex._12
[Ans : 231.
mg 2 d 4h 2 ] 4h
Figure shows a uniform rod of length 30 cm having a mass of 3.0 kg. The strings shown in the figure are pulled by constant forces of 20 N and 32 N. Find the force exerted by the 20 cm part of the rod on the 10 cm part. All the surfaces are smooth and the strings and the pulleys are light. HCV_Ch-5_Ex._24 30cm yEckbZ okyh ,d leku NM+ dk nzO;eku 3.0kggSA fp=k esa fn[kkbZ xbZ jLlh dks Øe'k%20NrFkk32Nds cy ls [khapktkrkgSA20cmokysHkkx}kjk10cm okysHkkxijyxk,x,cydkekuKkrdhft,AlHkhlrgdksfpdukekusa
rFkkjLlhrFkkiqyhdksgYdkekusaA
[Ans : 24 N ] 232. The monkey B shown in figure is holding on to the tail of the monkey A which is climbing on a rope. The masses of the monkeys A and B are 5 kg and 2 kg respectively. If A can tolerate a tension of 30 N in its tail, what force should it apply on the rope in order to carry the monkey B with it ? Take g = 10 m/s 2. HCV_Ch-5_Ex._38 fp=kesafn[kk,x,cUnjB usacUnjAdhiwaNdksidM+j[kkgStksfdÅijdhvksjp<+jgkgSAArFkkB dsnzO;ekuØe'k%
5 kg rFkk2 kg gSA vxj A viuh iwaN esa 30 N rd dk ruko lgu dj ldrk gS rks cUnj jLlh ij D;k cy yxk, rkfd og
cUnj B dks vius lkFk ys tk ldsa ? g = 10 m/s2 ysaA
[Ans : Between 70 N and 105 N] 233.
Find the acceleration of the block of mass M in the situation shown in figure. All the surfaces are frictionless and the pulleys and the string are light. HCV_Ch-5_Ex._32 fp=kesafn[kk,xbZfLFkfresaMnzO;ekuokysxqVdsdkRoj.kcrkb;sAlHkhlrg?k"kZ.kghugSvkSjiqyhrFkkjLlh
nzO;ekughugSA
RESONANCE
Page # 80
M
30º
2M
[Ans : g /3 up the plane ] 234.
Let m1 = 1 kg, m2 = 2 kg and m3 = 3 kg in figure shown. Find the acceleration of m1, m2 and m3.The string from the upper pulley to m1 is 20 cm when the system is released from rest. How long will it take before m1 strikes the pulley ? HCV_Ch-5_Ex._28
fp=k esa fn[kk, x, m1 =1kg,m2 =2 kgrFkkm3 =3 kgekfu,A m1,m2 rFkkm3 dk Roj.k crkb;sA Åij okyh iqyh ls m1 ds chp dh nwjh 20cmgS tc fudk; dks fojke ls NksM+k tkrk gSA m1 iqyh ls Vdjkus esa fdruk le; yxsxk ?
[Ans : 235.
19 17 g (up), g (down), 21 g (down), 0.25 s ] 29 29 29
In the figure the reading of the spring balance will be : (Assuming that the extension of the spring is not changing) and the inclined plane is smooth. Also assume that the pulley, string, spring and the box of the spring are light. (g = 10m/s2) M.Bank_NLM_1.12 fp=kesfLçaaxdkikB;kadgksxk:(;gekfu,fdfLçaxdksçlkjughacnyjgkgSvkSjurryfpdukgSA;gHkhekfu;sfd iqyh]jLlh]fLçaxrFkkfLçaxdkckWDlgYdsgS) (g=10m/s2) 2m/s
2
kg 10
(A*) 6 kg f 236.
(B) 5 kg f
30°
Fixed
(C*) 60 N
5kg
(D) 60 kg f
A 1 kg block ‘B’ rests as shown on a bracket ‘A’ of same mass. Constant forces F 1 = 20 N and F 2 = 8 N start to act at time t = 0 when the distance of block B from pulley is 50 cm.Time when block B reaches the pulley is _______. (Assume that friction is absent every where. Pulley and string light. ,d 1 fdxzk- nzO;eku dk CykWd ‘B’ leku nzO;eku ds CykWd ‘A’ ij fLFkj j[kk gS t = 0 ij cy F1 = 20 N rFkkF2 = 8 N yxk;s tkrs gS bl le; CykWd B dh iqyh ls nwjh 50 cm gS rks CykWd B fdrus le; esaiqyh rd igqapsxk _____(;g ekfu,fdlHkhtxg?k"kZ.kvuqifLFkrgSAiqyhrFkkjLlhgYdhgS) M.Bank_NLM_3.2
RESONANCE
Page # 81
[ Ans: 0.5 sec. ] 237.
A person is standing on a weighing machine placed on the floor of an elevator. The elevator stars going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and the minimum weights recorded are 72 kg and 60 kg. Assuming that the magnitudes of the acceleration and the deceleration are the same, find (a) the true weight of the person and (b) the magnitude of the acceleration. Take g = 9.9 m/s2.
fdlhfy¶VdhQ'kZijj[khrkSye'khuij,dO;fDr[kM+kgSAfy¶VdqNRoj.klsÅijtkukçkjEHkdjrhgSrFkkdqN le; ,d leku osx ls pyrh gSA vUr esa eafnr gksdj :d tkrh gSA e'khu dk vf/kdre o U;wure ikB;kad72kgrFkk 60kggSARoj.kvkSjeanudkifjek.k,dlekuekursgq,KkrdjksA(a)vknehdklghHkkjA(b)Roj.kdkifjek.kATake 238.
g = 9.9 m/s2. HCV_Ch-5_Ex_15 [Ans : 66 kg and 0.9 m/s|] A bob is hanging over a pulley inside a car through a string . The second end of the string is in the hand of a person standing in the car . The car is moving with constant acceleration 'a' directed horizontally as shown in figure . Find out the constant angle made by the string with vertical when the other end of the string is pulled with constant acceleration ' a ' vertically . Solve this problem in the frame of car . Determine also the tension in the string . ,dm nzO;ekudkckWc(xksyk) dkjdsvUnjjLlhlsyVdkgqvkgSAjLlhdknwljkfljkdkjes[kM+s,dvknehdsgkFk esagSAdkj{kSfrtesafu;rRoj.k'a' lsfp=kkuqlkjxfrdjjghgSAtcjLlhdsnwljsfljsdksfu;rRoj.k'a'lsfp=kkuqlkj xfrdjjghgSAtcjLlhdsnwljsfljsdksfu;rRoj.k'a' lsÅ/oZuhpsdhvksj[khapktkrkgSrksjLlh}kjkcuk,x,
fu;rdks.kdkekucrkb,Ablç'udksdkjdsfunsZ'krU=kesagydhft,AjLlhesarukoKkrdjksA [Made 2003]
M.Bank_NLM_5.12
[Q.247/RK_BM/N.L.]
Sol.
(Force diagram in the frame of the car) Applying Newton’s law perpendicular to string mg sin = ma cos Applying Newton’s law along string
239.
T – m g2 a 2 = ma
a tan = g
T = m g 2 a 2 + ma
Ans.
Two blocks A and B of masses m & 2m respectively are held at rest such that the spring is in natural length. Find out the accelerations of both the blocks just after release. M.Bank_NLM_1.26 ¼nksCykWdArFkkBftudsnzO;ekuØe'k%mo2mgSdksfojkekoLFkkesablrjgj[kktkrkgSftllsdekuhdhyEckbZ lkekU;(natural)jgsrksCykWdksadsNksM+usdsrqjUrcknmudkRoj.kD;kgksxk½
RESONANCE
Page # 82
(A*) g , g 240.
(B)
g g 3 3
(D) g 0
(C) 0, 0
Find the mass M of the hanging block in figure which will prevent the smaller block from slipping over the triangular block. All the surface are frictionless and the spring and the pulleys are light. fp=kesafn[kk,MnzO;ekuokysxqVdsdkekucrkb;srkfd;gNksVsxqVdsdksf=kdks.kh;xqVdsijfQlyuslsjksdldsA
lHkhlrg?k"kZ.kghurFkkdekuhrFkkiqyhgYdsgSA
241.
HCV_Ch-5_Ex._33
M'm ] cot 1 System shown in figure is in equilibrium. The magnitude of net change in tension in the string just before and just after, when one of the spring is cut. Mass of both the blocks is same and equal to m and spring constant of both springs is k. (Neglect any effect of rotation)
[Ans :
fn,x,fp=k esafudk;lkE;oLFkk esa gSA tc,d dekuhdks dkV fn;k tkr gS rks jLlh esa ruko esa dqy ifjorZu dekuhdsdkVusdsBhdigysvkSjBhdcknesaD;kgksxknksuksaxqVdksadknzO;eku,dtSlkgSvkSjm dscjkcjgS vkSjnksuksadekuhdkdekuhfu;rkadKgSA¼?kw.kZudsizHkkodksux.;ekusa½ M.Bank_NLM_1.28
(A*) 242.
mg 2
(B)
mg 4
[Made 2004]
(C)
3m g 4
(D)
3m g 2
A car is speeding up on a horizontal roadwith a constant acceleration a. Calculate in the following situations in the car. (i) A ball is suspended from the ceiling. Find the angle made by the string if the block & string remain static w.r. to car. (ii) A block is kept on a smooth fixed incline and does not slip on the fixed incline with the horizontal. HCV_Ch-5_Ex_41 ,ddkj{kSfrtlM+dijfu;rRoj.kalsc<+rhgqbZpkyalsxfrdj jghgSA dkjesa fuEu fLFkfr dh dYiuk dhft,A (i),d xsan jLlh }kjkNr ls yVd jgh gSA jLlh}kjk cuk,x, dks.k dk eku crkb, vxjxsan rFkkjLlh dkj dslkis{k
RESONANCE
Page # 83
fLFkjjgsA (ii) ,d fpdus fLFkj urry ij ,d xqVdk j[kk gqvk gS vkSj ;g urry ij {kSfrt esa ugha fQly jgk gSA [Ans : tan–1(a/g) in each case]
Comprehension # 3 [Made Rahul Shukla Sir] BCK_20.8.06_Com3 [NLM] Figure shows a weighing machine kept in a lift. Lift is moving upwards with acceleration of 5 m/s 2. A block is kept on the weighing machine. Upper surface of block is attached with a spring balance. Reading shown by weighing machine and spring balance is 15 kg and 45 kg respectively. Answer the following questions. Asume that the weighing machine can measure weight by having negligible deformation due to block, while the spring balance requires larger expansion :(take g = 10 m/s 2) vuqPNsn# 3 [Made Rahul Shukla Sir] fp=kesafy¶Vesaj[kh,dHkkje'khufn[kkbZxbZgSAfy¶VÅijdhvksj5 eh0/lS02 lsRofjr gSAHkkje'khu ij,d
fi.Mj[kkx;kgSAfi.MdhÅijhlrgdks,dfLçaxrqyklstksM+kx;kgSAHkkje'khu,oafLçaxrqyk}kjkn'kkZ;s ikB;kadØe'k%15 fdxzk0,oa45 fdxzk0gSAfuEu ç'uksadkmÙkjnks&
M
Weighing Machine
243. Mass of the object in kg is :
oLrqdknzO;ekufdxzk0esagS& (A) 60 kg
Sol.
(B*) 40 kg
(C) 80 kg
FBD of Block in ground frame : applying N.L. 150 + 450 – 10 M = 5M
15 M = 600 M =
M = 40 Kg Ans.
150 N
(D) 10 kg
450 N
600 15
5 m/s2
Mg = 10 M 244. In the above situation normal acting on the block as seen by an observer in the lift.
mijksDrifjfLFkfresa]fy¶Vesaçs{kd}kjkns[kusijfi.MijvfHkfØ;kcygksxk&
(A) 450 N (B*) 150 N (C) 400 N Sol. Normal on block is the reading of weighing machine i.e. 150 N.
(D) zero 'kwU;
245. If lift is stopped and equilibrium is reached. Reading of weighing machine will be :
;fnfy¶V:dtk;srFkklkE;koLFkkçkIrgks]rksHkkje'khudkikB;kadgksxk&
Sol.
(A) 40 kg (B) 10 kg If lift is stopped & equilibrium is reached then T = 450 N
(C) 20 kg
(D*) zero
'kwU;
450 + N = 400 N = – 50 So block will lose the contact with weighing machine thus reading of weighing machine will be zero.
N
Mg = 400 M
246. If lift is stopped and equilibrium is reached. Reading of spring balance will be :
;fnfy¶V:dtk;srFkklkE;koLFkkçkIrgks ]rksfLçaxrqykdkikB;kadgksxk& (A) zero
(B) 20 kg
T
Sol.
40 g
T = 40 g
RESONANCE
(C) 10 kg
(D*) 40 kg
So reading of spring balance will be 40 Kg. Page # 84
247. Find the acceleration of the lift such that the weighing machine shows its true weight.
fy¶V dk Roj.k Kkr djks ftlls Hkkj e'khu dk ikB;kad lgh otu n'kkZ;s & (A*)
45 m/s 2 4
(B)
T = 450 N
Sol.
40 Kg
85 m/s2 4
(C)
N = 400 N a
22 m/s2 4
(D)
a=
950 400 40
a=
450 45 = m/s 2 Ans. 40 4
60 m/s2 4
Mg = 400 N
248.
A particle is moving with initial velocity u 1ˆi 1ˆj 1kˆ . From the following what should be its acceleration so that it can remain moving in the same straight line ? [Made M.Pathak] [NLM] BEK_30.7.06_20
,dd.kdksizkjfEHkdosx u 1ˆi 1ˆj 1kˆ lsiz{ksfirdjrsgSaAbldkRoj.kD;kgksukpkfg,ftlls;slekuljy js[kkesaxfrekujglds\ Sol.
249.
(A*) a 2 ˆi 2 ˆj 2 kˆ
(B*) a 2 ˆi 2 ˆj 2 kˆ (C) a 3 ˆi 3 ˆj 3 kˆ (D) a 1ˆi 1ˆj To move is a straight line, a must be in the same direction of or directly opposite to the velocity i.e. a 2u A monkey of mass 40 kg climbs on a rope as shown in figure which can bear a maximum tension of 600 N. In which of the following cases will the rope break : The monkey ,d40 fdxzk0 nzO;eku dk cUnj fp=kkuqlkj ,d jLlh ij p<+rk gSA jLlh vf/kdre 600 N dk ruko lgu dj ldrh
gSAbuesa lsfdlfLFkfresajLlhVwVtk,xhAtccUnj (a) climbs up with an acceleration of 6 m s–2 (c) climbs up with a uniform speed of 5 m s–1 (a) 6 m s–2 ds Roj.k ls Åij p<+rk gSA (c) 5 m s–1 dh ,d leku pky ls Åij p<+rk gSA
gSA
Ans. 250.
(Ignore the mass of the rope) (a) T – 400 = 240, T = 640 N. (c) T = 400 N
NCERT XI_5.36 [NLM]
(b) climbs down with an acceleration of 4 m s –2 (d) falls down the rope nearly freely under gravity ? (b) 4 m s–2 ds Roj.k ls uhps mrjrk gSA (d) yxHkx eqDr :i ls xq:Roh; çHkko ls jLlh ls uhps fxjrk
(b) 400 – T = 160, T = 240 N (d) T = 0
A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/ s2.Find the displacement of the block during the first 0.2s after the start. Take g = 10 m/s2. ,d xqVds dksfy¶V ds lery ijfojkeesa j[kktkrk gSAfy¶V 12m/s2 ds Roj.k ls uhps fxjuk 'kq: gksrh gSA xfr 'kq: gksus ds çFke 0.2sesa xqVds dk foLFkkiu fdruk gksxkA (g=10eh0/lSd02 yhft,) [Ans : 20 cm]
RESONANCE
HCV_Ch-5_Ex._42
Page # 85
251.
The rear side of a truck is open and a box of 40 kg mass is placed 5 m away from the open end as shown in Fig. The co-efficient of friction between the box and the truck below it is 0.15. On a straight road, the truck starts from rest and accelerates with a constant acceleation of 2 m s–2. At what distance from the starting point does the box fall off the truck ? (Ignore the size of the box). ,dVªddsiyM+slHkhrjQls[kqysgq,gSvkSjblij[kqysgq,fljsls5mnwjhijfp=kkuqlkj,d40kgnzO;ekudkckWDl j[kk gSA ckWDl o blds uhps dh lrg ds e/; ?k"kZ.k xq.kkad 0.15 gSA ,d lh/kh lM+d ij Vªd fLFkjkoLFkk ls 2ms–2 dsRoj.klspyukçkjEHkdjrkgSrksVªddhçkjfEHkdfLFkfrlsfdruhnwjhijck¡DlVªdlsfxjtk,xkA(cDls dhvkd`fÙkux.;gSA) NCERT XI_5.39
Ans.
Acceleration of the box due to friction = g = 0.15 × 10 = 1.5 m s–2. But the acceleration of the box relative to the truck is 0.5 M s –2 towards the rear end. The time take for the box to fall off the truck = 25 20 s . During this time, the truck covers a distance = 1/2 × 2 × 20 = 20 m. 0 .5
252.
Find the reading of the spring balance as shown in figure (5-E6). The elevator is going up with an acceleration of g/10, the pulley and the string are light and the pulley is smooth. fp=kesan'kkZ,vuqlkjdekuh&rqykdkikB;kad crkb,Afy¶VÅijdhvksj g/10,Roj.klstkjghgSAiqyhrFkkjLlh
gYdhgSrFkkiqyhfpduhgSA
HCV_Ch-5_Ex_16
g/10 1.5 kg [Ans : 4.4 kg] 253.
3.0 kg
Neglecting friction every where, find the acceleration of M. Assume m > m . lHkh txg ?k"kZ.k dks ux.; ekurs gq, M dk Roj.k dk eku Kkr dhft,A ;g ekfu, fdm > m A
M.Bank_NLM_6.5
[ Ans: a = 254.
(m m' )g ] 2M 3m 3m'
System is shown in the figure and man is pulling the rope from both sides with constant speed ' u'. Then the speed of the block will be: [Made 2004] M.Bank_NLM_8.14 fp=kesaiznf'kZrfudk;esa,dvknehfu;rpky'u'lsjLlhdksnksuksfljksals[khapjgkgSACykWddhpkygksxkA
RESONANCE
Page # 86
(A*) 255.
3u 4
(B)
(C)
u 4
(D)noneofthesebuesalsdksbZugha
A hinged construction consists of three rhombus with the ratio of sides (5 : 3 : 2). Vertex A3 moves in the horizontal direction with velocity V. Velocity of A2 will be : M.Bank_NLM_8.38 fp=kkuqlkj,ddhyfdrfudk;rhuleyEcprqHkZHkqtftudhHkqtkvksaesa5:3:2dkvuqikrgS]lscukgS;fnA3 fljsdk {kSfrtfn'kkesaosxVgSArksA2 fljsdkosxgksxkA
(A) 2.5 V 256.
3u 2
(B) 1.5V
(C) (2/3)V
(D*) 0.8V
A boy standing on a stationary lift (open from above) throws a ball upwards with the maximum initial speed he can, equal to 49 ms–1. How much time does the ball take to return to his hands ? If the lift starts moving up with a uniform speed of 5 ms–1 and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands ? If the lift starts moving up with a uniform accleration of 5 ms–2 and the boy again throws the ball up with the maximum speed he can, how long does the ball take to return to his hands ? NCERT XI_3.24
,d ckyd ,dfLFkjfy¶V¼Åijls [kqyh gqbZ½ij [kM+k gqvk] vf/kdre izkjfEHkd pky49 ms–1 ls Åijdh vksj xsan dks Qadrk gSA xsan mlds gkFk esa vkus ls fdruk le; ysxh \ vc vxj fy¶V ,d fu;r pky 5 ms–1 ls Åij dh vksj tkuk'kq: dj nsrh gS vkSjfQj ckydxsan dksmlh vf/kdre pky ls Åij Qsadrk gS rks xsan dks okfil mlds gkFk esa vkus esa fdruk le; yxsxk \ vxj fy¶V Åij dh vksj ,d leku Roj.k 5 ms–2 ls Åij dh vkSj xfr djuk 'kq:djrhgSrFkkckydfQjlsxsandksÅijdhvkSjmlhvf/kdrepkylsQsadrkgSrksxsandksokfilmldsgkFk esavkusesafdrukle;yxsxk\ NCERT XI_3.24 Ans.
10 s, 10 s,
25 s 3
Comprehension # vuqPNsn [Test A batch 2007] ACJ_28.5.06._Comp#2 [NLM] Read the following write-up and answer question no. ... to ... based on that Two smooth blocks are placed at a smooth corner as shown. Both the blocks are having mass m. We apply a force F on the small block m. Block A presses the block B in the normal direction, due to which pressing force on vertical wall will increase, and pressing force on the horizontal wall decrease, as we increase F. ( = 37° with horizontal). As soon as the pressing force on the horizontal wall by block B becomes zero, it will loose the contact with the ground. If the value of F is further increased, the block B will accelerate in upward direction and simultaneously the block A will move toward right. Read the following write-up and answer question no. ... to ... based on that nksfpdusfi.M,dfpdusdksusesafp=kkuqlkjj[ksx;sgSAnksuksafi.MksadknzO;ekumgSAgefi.MAijcyFyxkrsgSA fi.M A,fi.MBdksyEcor~fn'kkesanckrkgSAFc<+kusdsdkj.knhokjijnckocyc<+rkgSvkSj{kSfrtnhokjijncko cy?kVrkgSA({kSfrtls=37°)tSlsgh{kSfrtnhokjij]nckocy'kwU;gksrkgSbldktehulslEidZgVtkrkgSA;fn FdkekuvkSjc<+k;ktk;srksxqVdkBÅijdhfn'kkesaRofjrgksrktkrkgSlkFkghlkFkfi.MAnka;hrjQxfrekugksuk
'kq:gkstkrkgSA
257. What is minimum value of F, to lift block B from ground : fi.MB dks tehu ls Åij mBkus ds fy, F,dk U;wure eku gksxk &
25 mg 12 Sol. For equl. f block (A) (A)
RESONANCE
(B)
5 mg 4
(C*)
3 mg 4
(D)
4 mg 3
Page # 87
F = N sin N = F/sin
To lift block B from ground N cos > mg F cos > mg sin
3 F > mg tan = mg 4
So Fmn =
3 mg 4
258. If both the blocks are stationary, the force exerted by ground on block A is : ;fnnksuksafi.MfLFkjgksrksfi.MAijtehu}kjkyxk;kx;kcygksxk:
3F 3F (B) mg – 4 4 Sol. If both the blocks are stationary, (A) mg +
(C*) mg +
4F 3
(D) mg –
4F 3
Balancing forces along x-direction F = N sin N = F/sin Balancing forces along y-direction Ny = mg + N cos
F = mg + cos = mg + F cot sin 4F My = mg + 3
259. If acceleration of block A is a rightward, then acceleration of block B will be : ;fnfi.MAdkRoj.k'a'nka;hrjQgSrksfi.MBdk Roj.kgksxk& (A*)
3a upwards 4
(B)
4a upwards 3
4a 3a Åij dh vksj (B) Åij dh vksj 3 4 Sol. To keep regular contact a sin = b cos
(A*)
RESONANCE
(C) (C)
3a upwards 5
3a Åij dh vksj 5
(D) (D)
4a upwards 5
4a Åij dh vksj 5
Page # 88
b = a tan = 260.
3 g 4
Find the acceleration of the block of mass M in the situation of figure. The cofficient of friction between the two blocks is 1 and that between the bigger block and the ground is 2 .
fp=kesafn[kk,vuqlkjxqVdsMdkRoj.kKkrdhft,AnksxqVdksadse/;?k"kZ.kxq.kkad 1 vkSjcM+sxqVdsrFkktehuds e/;?k"kZ.k 2 gSA HCV_Ch-6_Ex_28
261.
[2m 2 (M m)g] [Ans : M m[5 2( )] ] 1 2
In a simple Atwood machine , two unequal masses m1 and m2 are connected by a string going over a clamped light smooth pulley. In a typical arrangement (figure ) m1 = 300 g and m2 = 600g. The system is
released from rest. (a) Find the distance travelled by the first block in the first two seconds. (b) Find the tension in the string . (c) Find the force exerted by the clamp on the pulley. ,dlkekU;,VoqMHkkje'khuesa]nksvlekunzO;ekum1 om2 fp=kkuqlkjdlhgqbZf?kjuhlsikfjrjLlhlstqM+sgq, gSA;fnm1 =300 g om2 = 600g gksrFkkfudk;dksfLFkjkoLFkklsNksM+ktk;rks(a) igysCykWd}kjkizFke2lSd.M
esar;nwjhKkrdjks\ (b) jLlh esa ruko Kkr djks (c) [kwaVh ¼dysEi½ }kjk f?kjuh ij yxk;k x;k cy Kkr djks \
HCV_Ch-5_Ex_22
///////////////////////////////////
m1 [Ans : (a) 6.5 m 262.
(b) 3.9 N
(c) 7.8 N]
m2
Consider the Atwood machine of the previous problem. The larger mass is stopped for a moment of 2.0 s after the system is set into motion. Find the time elapsed before the string is tight again. fiNysç'uesa,VoqMe'khudksekfu,AHkkjhokysnzO;ekudksxfrçkjEHkgksusds2.0sckn,d{k.kdsfy,jksdktkrk
gSA ml le; dk eku crkb, tc jLlh esa nqckjk ruko vk tkrk gSA Ans : 2 / 3 s 263.
HCV_Ch-5_Ex_23
A smooth wedge A is fitted in a chamber hanging from a fixed ceiling near the earth’s surface. A block B placed at the top of the wedge takes a time T to slidedown the length. If the block is placed at the top of the wedge and the cables supporting the chamber start accelerating it upward with an acceleration of ‘g’, at the same instant, the block will. (A) take a time longer than T to slide down the wedge. (B*) take a time shorter than T to slide down the wedge. (C) remain at the top of the wedge. (D) jump off the wedge. [HCV_Ch_5-obj-I_11] ,d fpduk ost A,d pSEcj¼d{k½ esafLFkj ¼fix½ djds j[kk tkrk gS]psEcj i`Foh ds utnhd fdlh fLFkj Nr ls yVdk gqvkgSA,dxqVdkBurrydsÅijh¼'kh"kZ½fcUnqlsuhpsiwjhyEckbZrdfQlyusesale;TysrkgSAvxjxqVdk]
RESONANCE
Page # 89
ostdsÅijh¼'kh"kZ½fcUnqijj[kktkrkgSvkSjmlh{k.kjLlhftllspSEcjcU/kkgqvkgS]ÅijdhvksjgRoj.kls Rofjrgksuk'kq:gkstkrhgSArksblckjxqVdkB(A) ost ij uhps fQlyus esa T ls T;knk le; yxsxkA (B*) ost ij uhps fQlyus esa T ls de le; yxsxkA (C) Åij¼'kh"kZ½fcUnqijghcukjgsxkA (D) ost ls mNy tk,xkA [HCV_Ch_5-obj-I_11]
Sol. When chamber starts moving up by acceleration ‘g’, pseudo force mg acts downward on block. Driving force is increased from mg sin to 2 mg sin hence acceleration is increased. 264.
Consider the situation shown in figure. The coefficient of friction between the blocks is . Find the minimum and the maximum force F that can be applied in order to keep the smaller blocks at rest with respect to the bigger block. HCV_Ch-6_WOE_9 fp=kesafn[kkbZxbZfLFkfrdksekfu,AxqVdksadse/;?k"kZ.kxq.kkadµgSAcyFdkU;wureovf/kdreekuKkrdhft;s
rkfdNksVsxqVds]cM+sxqVdsdslkis{kfLFkjjgldsA
Sol.
If no force is applied , the block A will slip on C towards right and the block B will move downward. Suppose
the minimum force needed to prevent slipping is F. Taking A + B + C as the system, the only external horizontal force on the system is F. Hence, the acceleration of the system is
F M 2m Now take the block A as the system. The force on A are (figure), a=
....(i)
(i) tension T by the string towards right , (ii) friction f by the block C towards left, (iii) weigth mg downward and (iv) normal force upward
For vertical equilibrium = mg As the minimum force needed to prevent slipping is applied , the friction is limiting . Thus, f = = mg As the block moves towards right with an acceleration a, T–f = ma or, T – mg = ma Now take the block B as the system. The forces are (figure 6-W14) (i) tension T upward, (ii) weight mg downward, (iii) normal force towards right with an acceleration a, (iv) friction f’ upward
RESONANCE
Page # 90
As the block moves towards right with an acceleration a, = ma As the friction is limiting , f’ = = ma For vertical equilibrium T + f’ = mg or, T + ma = mg Eliminating T from (ii) and (iii)
....(iii)
1 g. 1 When a large force is applied the block A slips on C towards left and the block B slips on C in the upward direction . The friction on A is toward right and that on B is downwards. Solving as above , the acceleration in this case is amin =
amax = Thus , a lies between
1 g 1
1 1 g and g 1 1
From (i) the force F should be between
1 1 (M + 2m) g and ( M + 2m )g. 1 1
265.
The force exerted by the floor of an elevator on the foot of a person standing there is more than the weight of the person if the elevator is
fy¶V ds Q'kZ ij [kM+s vkneh ij Q'kZ }kjk yxk;s x;s cy dk eku vkneh ds Hkkj ls T;knk gksxk vxj fy¶V HCV_Ch-5_Obj.II_6
267.
(A*) going up and speeding up (B) going up and speeding down (C) going down and speeding up (D*) going down and speeding down (A*)Åij tk jgh gS vkSj pky c<+rh tk jgh gSA (B) Åij tk jgh gS vkSj pky ?kVrh tk jgh gSA (C)uhps tk jgh gS vkSj pky c<+rh tk jgh gSA (D*)uhps tk jgh gS vkSj pky ?kVrh tk jgh gSA Figure shows two blocks connected by a light string placed on the two inclined parts of a triangular structure. The cofficients of static and kinetic friction are 0.28 and 0.25 respectively at each of the surfaces . (a) Find the minimum and maximum values of m for which the system ramains at rest.(b) Find the acceleration of either block if m is given the minimum value calculate in the first part (a) and is gently pushed up the incline for a short while.
f=kdks.kh;
2 kg
m
Sol.
45º
45º
(a) Take the 2 kg block as the system. The forces on this block are shown in figure with M = 2 kg. It is assumed that m has its minimum value so that the 2 kg block has a tendency to slip down. As the block is in equilibrium , the resultant force should be zero.
T
N
f
f' Mg
Taking components to the incline
RESONANCE
T
N'
mg
Page # 91
= Mg cos 45º = Mg / Taking components | | to the incline
T + f = Mg sin45º = Mg /
or, T = Mg Mg / 2 – f As its is a case of limiting equilibrium,
2 2
= s
f
Mg Mg – s = (1 – s ) ...(i) 2 2 2 Now consider the other blocks as the system. The forcees acting on this block are shown in figure (6-W-17), Taking components to the incline , or,
T =
Mg
= mg cos45º =
Mg
2 Taking components || to the incline, T = mg sin45º + f’ =
Mg
2 As it is the case of limiting equilibrium Mg = s 2
f’ = s
Thus ,
T=
From (i) and (ii)
Mg 2
+ f’
(1 + s )
....(ii)
m (1 – s ) = M (1 – s )
or,
(1 s ) 1 0.28 m = (1 ) M = × 2 kg 1 0.28 s
=
.....(iii)
9 kg. 8
When maximum possible value of m is supplied , the directions of friction are reserved because m has the tendency to slip down and 2 kg block to slip down and 2 kg block to slip up. Thus , the maximum value of m can be obtained from (iii) by putting s = – 0.28. Thus , the maximum value of m is m= = (b) If m =
1 0.28 × 2 kg 1 0.28 32 kg. 9
9 kg and the sustem is gently pushed , kinetic friction will operate. Thus , 8
Mg f = k . 2
and
f’ =
k mg 2
,
where k = 0.25. If the accleration is a , Newton’s second law for M gives (figure 6-W16). Mg sin45º – T – f = Ma
mg –T– k = Ma 2 2 Applying Newton’s second law m figure (figure 6-W17), T – mg sin45º = f’= ma or,
or,
Mg
T–
Adding (iv) and (v)
RESONANCE
Mg 2
–
k mg 2
=ma
.....(iv)
.......(v)
Page # 92
Mg 2
(1 – s ) –
Mg 2
or,
(1 + s ) = (M +m) a a = a =
M (1 k ) m (1 k ) 2 (M m)
2 0.75 – 9 / 8 1.25 2 ( 2 9 / 8)
g
g
= 0.31 m/s2. 268.
In the arrangement shown in Fig. the mass of the rod M exceeds the mass m of the ball. The ball has an opening permitting fp=kesafn[kkbZfLFkfresaNM+dknzO;ekuM,xsandsnzO;ekumlsvf/kdgSAxsandsfljsblrjg[kqysgS Irodov_1.74
it to slide along the thread with some friction. The mass of the pulley and the friction in its axle are negligible. At the initial moment the ball was located opposite the lower end of the rod. When set free, both bodies began moving with constant accelerations. Find the friction force between the ball and the thread if t seconds after the beginning of motion the ball got opposite the upper end of the rod. The rod length equals .
rkfd;gjLlhdsvuq:idqN?k"kZ.kdslkFkfQlyldsAiqyhdknzO;ekuvkSjiqyhesa?k"kZ.kux.;ekusaAçkjfEHkd {k.kijxsanNM+dsnwljsfljsijmifLFkrgSAtcmUgsaLorU=k:ilsNksM+fn;ktkrkgSrksnksuksaoLrq,afu;rRoj.k ls xfr djuk 'kq: dj nsrh gSA jLlh rFkk xsan ds e/; dk ?k"kZ.k cy dk eku crkb, vxjt lSd.M ds ckn xsan NM+ dsnwljsfljsijigq¡ptkrhgSANM+dhyEckbZ gSA [Ans : Ffr = 2 mM / (M – m)t2 ] 269.
In the Figure, the blocks are of equal mass. The pulley is fixed. In the position shown, A is given a speed u and vB= the speed of B. M.Bank_NLM_8.9 (Only F_13.08.06_17)
fp=kkuqlkjnksuksaxqVdslekunzO;ekudsgSAiqyhfu;rgSAfp=kesafn[kkbZfLFkfresaAdkspkyunhtkrhgSAvB= B dh pkyA
(A*) B will never lose contact with the ground B dHkhHkhtehulslEidZughaNksM+sxkA (B) The downward acceleration of A is equal in magnitude to the horizontal acceleration of B. A dkuhpsdhvkSjRoj.k]B ds{kSfrtRoj.kdscjkcjgksxkA (C) vB = u cos (D*) vB = u/cos 270.
A lift is falling with an acceleration 2 m/s2. A ball of mass 100 gm is attached at one end of the string and the other end is fixed to the ceiling of the lift. The ball remains at rest relative to lift. The tension in the string is: (g = 10 m/s2) M.Bank_NLM_5.11 ,dfy¶V2m/s2 dsRoj.klsfxjjghgSAjLlhds,ddksusls100gmdhxsantqM+hgS,oanwljkfljkNrlscU/kkgqvk
gSAxsanfy¶Vdslkis{kfojkeesajgrhgSAjLlhesarukogS&
RESONANCE
Page # 93
Sol. 271.
(A) 1.2 N (B*) 0.8 N T = ma (g – a) = 0.1 (10 – 2) = 0.8 N
(C) 10 N
(D) 0.2 N
In the system shown in the figure m 1 > m 2. System is held at rest by thread BC. Just after the thread BC is burnt: fp=kesaiznf'kZrfudk;esam1 >m2 gSAfudk;dksjLlhBC}kjklkE;oLFkkesaj[kkx;kgSAjLlhBCdstyusds
rqjUri'pkr~&
M.Bank_NLM_1.41 (A*) (B)
(C*) (D) 272.
acceleration of m 2 will be upwards m2 dk Roj.k Åij dh vkSj gksxkA
m m2 g magnitude of acceleration of both blocks will be equal to 1 m1 m2
1 m2 gdscjkcjgksxkA nksuksaCykWdksadsRoj.kksdkifjek.k m m m
1
2
acceleration of m 1 will be equal to zero m1 dkRoj.k'kwU;dscjkcjgksxkA magnitude of acceleration of two blocks will be nonzero and unequal.
nksuksaCykdksdsRoj.kksdkifjek.kv'kwU;rFkkvlekugksxkA
In the arrangement shown in the figure pulleys A & B are massless and the thread is inextensible. Mass of pulley C is equal to m. If friction in all the pulleys is negligible, then: fp=kesafn[kkbZfLFkfresaf?kjuhArFkkBnzO;ekujfgrgSrFkkjLlhvrU;gSACdknzO;ekum gSAvxjf?kjuhesa
?k"kZ.kux.;gSrks
M.Bank_NLM_3.44
273.
(A) tension in thread is equal to (1/2) mg jLlh esa ruko(1/2) mg gSA (B) acceleration of pulley C is equal to g/2 (downward) f?kjuhC dk Roj.k g/2 (uhps dh vksj) (C) acceleration of pulley A is equal to g (upward) f?kjuhAdk Roj.k g (Åij dh vksj) (D*) acceleration of pulley A is equal to 2 g (upward) f?kjuhAdk Roj.k 2g (Åij dh vksj)
For the following system shown assume that pulley is frictionless, string is massless (m remains on M) :
n'kkZ;sx;sfudk;dsfy,ekukf?kjuh?k"kZ.kjfgrgSoMksjhnzO;ekujfgrgS&
M.Bank_NLM_7.35 Find Kkr dhft, & (a) the acceleration of the block A.
RESONANCE
Page # 94
(b) Normal reaction on m B force on C due to B. (c) the force on the ceiling (a) CykWdAdk Roj.k (b) B ds dkj.k C ij vfHkyEc cy (c) Nr ij cy [RK_TOPIC/NL/Q.No.7 made subjective]
[4]
Sol. By newtons law on system of (A, B, C) (a) (M + m – m) g = (2M + m) a
mg 2M m (b) free body diagram ‘C’ block
a=
mg – N = ma
N=m N=
2M gm 2M m
(c) T – mg = M
gm g 2M m
mg for A block 2M m
T = Mg +
for pulley
Mmg 2M m
P = 2T + mg = 2mg + =
2Mmg + mg 2M m
6M 3m 2m mg 2M m
6M 5m P = 2 M m mg Ans. 274.
(a)
mg 2Mmg (6M 5m) Mg (b) 2M m 2M m (c) 2M m
Three equal balls 1,2,3 are suspended on springs one below the other as shown in the figure. OA is a weightless thread.
fp=kesan'kkZ,vuqlkjrhu,dtSlhxsan1,2,3 fLizaxdhlgk;rklsfp=kkuqlkj,dnwljslstqM+hgqbZgSAOA,d nzO;ekughujLlhgSA
RESONANCE
Page # 95
[M.Bank_NLM_1.3]
(a) If the thread is cut, the system starts falling. Find the acceleration of all the balls at the initial instant
vxjjLlhdksdkVfn;ktkrkgSrksfudk;fxjuk'kq:djnsrkgSAizkjfEHkd{k.kijlHkhxsanksadkRoj.kKkrdhft,A
(b) Find the initial accelerations of all the balls if we cut the spring BC which is supporting ball 3 instead of cutting the thread.
vxjgejLlhdhtxgxsan3dkstksM+usokyhfLizaxdksdkVnsrsgSrkslHkhxsanksadkizkjfEHkdRoj.kKkrdhft,A [ Ans: (a) 3 g 0, 0, (b) 0, g g ] 275.
In the figure shown, A & B are free to move. All the surfaces are smooth. (0 < < 90º) fp=k esa n'kkZ, vuqlkjArFkkB xfr djus ds fy, LorU=k gSA lHkh lrg fpduh gSA (0 < < 90º) M.Bank_NLM_6.8 (A*) the acceleration of A will be more than g sin A dk Roj.k g sin ls T;knk gksxkA (B) the acceleration of A will be less than g sin A dk Roj.k g sin ls de gksxkA (C) normal force on A due to B will be more than mg cos
B ds dkj.k Aij yEcor~ cy mg cos ls vf/kd gksxkA
276.
(D*) normal force on A due to B will be less than mg cos A ij B ds dkj.k yEcor~ cy mg cos ls de gksxkA
Three forces are applied to a square plate as shown in Fig.. Find the modulus, direction, and the point of application of the resultant force, if this point is taken on the side BC.
fp=kesan'kkZ,vuqlkjrhucyoxkZdkjIysVijyxk,tkrsgSAifj.kkehcydkfØ;kfcUnq]ifjek.krFkkfn'kkdkeku crkb,AvxjblfcUnqdks BCHkqtkijfy;ktkrkgSA M.Bank_NLM_3.23
[irodov _1.237-]
277.
[4 marks, 6 min.]
[Ans : Fres = 2F. This force is parallel to the diagonal AC and is applied at the midpoint of the side BC.]
In the figure shown, a person wants to raise a block lying on the ground to a height h. In both the cases if time required is same then in which case he has to exert more force. Assume pulleys and strings light. fp=kesan'kkZ,vuqlkj],dvknehxqVdsdkstehulshÅpk¡bZrdmBkukpkgrkgSAnksauksfLFkfr;ksaesavxjle;
cjkcj yxrk gS rks dkSulh fLFkfr esa vko';d cy dk eku vf/kd gksxkA ;g ekfu, fd iqyh rFkk jLlh gYds gSA
RESONANCE
Page # 96
[Made 2006, RKV, GRSTU] [3.35 NLM] (A*) (i) Sol.
(B) (ii)
Since, h = given.
(D) Cannot be determined
1 2 at a shoule be same in both cases, because h and t are same in both cases as 2
In (i) F 1 – mg = ma. F1 = mg + ma. In (ii) 2F2 – mg = ma
278.
(C) same in both
F2 =
mg ma 2
F1 > F2 .
In the figure shown all contact surfaces are smooth. Acceleration of B block will be: fp=kesafn[kkbZfLFkfresalHkhlEidZlrgfpduhgSxqVdsBdkRoj.kgksxk:
[3.52 NLM]
(A*) 1 m/s 2 279.
(B) 2 m/s2
(C) 3 m/s 2
(D) none of these buesa ls dksbZ ugha
In the shown mass pulley system, pulleys and string are massless. The one end of the string is pulled by the force F = mg. The acceleration of the block will be M.Bank_NLM_7.17 fp=kesan'kkZ;sf?kjuhnzO;ekufudk;esaf?kjuhrFkkjLlhnzO;ekughugSAjLlhdk,dfljkcyF=mg}kjk[khapk
tkrkgSAxqVdsdkRoj.kgksxk&
(A) g/2 280.
(B) 0
(C*) g
(D) 3g
A monkey pulls along the ground mid point of a 10 m long light inextensible string connecting two identical objects A & B each of mass 0.3 kg continuously along the perpendicular bisector of line joining the masses. The masses are found to approach each other at a relative acceleration of 5 m/s2 when they are 6 m apart. The constant force applied by monkey is : M.Bank_NLM_7.29 ,d10 m yEch gYdh vrU; jLlh ls nks ,d leku oLrq, ArFkkB izR;sd nzO;eku 0.3 kg ds cU/ks gSA ,d cUnj tehudhfn'kkesaArFkkBdkstksM+usokyhjs[kkdsyEcor~fn'kkesabljLlhdks,dfu;rcylsyxkrkj[khap
RESONANCE
Page # 97
jgk gSa ;s nzO;eku ,d nwljs dh vkSj lkis{k Roj.k 5 m/s2 ls xfr dj jgs gS tc muds chp dh nwjh 6m gSA cUnj }kjk yxk;k x;k fu;r cy dk eku gSA (A) 4 N 281.
(B*) 2 N
(C) 3 N
(D) none
A 15 kg block B is suspended from an inextensible cord attached to a 20 kg cart A. Neglect friction.
[5.16_NLM] (i) (ii)
Draw FBD of cart A & block B indicating all forces acting on them. Determine the acceleration of the cart & block immediately after the system is released from rest. Mass m shown in figure is in equilibrium. If it is stretched further by x and released find its acceleration just after it is released. Take pulleys to be light & smooth and strings light. fp=kesafn[kk,vuqlkjnzO;ekuMlkE;koLFkkesagSAvxjblsxnwjhrdvkSj[khapktkrkgSvkSjfQjNksM+fn;ktkrk
282.
gSrksNksM+usdsBhdcknbldkRoj.kD;kgksxk\iqyhrFkkjLlhgYdsgSvkSjiqyhfpduhgSA
[Q.1.34_NLM] [Made 2005, PKS Sir]
(A)
4kx 5m
(B)
2kx 5m
(C*)
4kx m
(D) none of these
Sol.(C)
Initially the block is at rest under action of force 2T upward and mg downwards. When the block is pulled downwards by x, the spring extends by 2x. Hence tension T increases by 2kx. Thus the net unbalanced force on block of mass m is 4kx. 283.
4kx m Three blocks A, B and C are suspended as shown in the figure. Mass of each blocks A and C is m. If system is in equilibrium and mass of B is M, then: fp=kesan'kkZ,vuqlkjrhuxqVds A, B vkSjC fp=kkuqlkjyVdsgq,gSAArFkkC çR;sddknzO;ekum gSAvxjfudk; lkE;koLFkkesagSvkSjBdknzO;ekuMgSrks: [3.43.NLM]
acceleration of the block is =
(A) M = 2 m 284.
(B*) M < 2 m
(C) M > 2 m
(D) M 2 m
There is an inclined surface of inclination = 30º. A smooth groove is cut into it forming angle with AB. A steel ball is free to slide along the groove, if the ball is released from the point O. The speed when it comes to A is _____. [ g = 10 m/s 2 ] [7.5 _ NLM] urry dk >qdko = 30º gSA fp=kkuqlkj ABHkqtk ls dks.k ij ,d fpduk [kk¡pk cuk;k tkrk gSA ,d xsan [kk¡ps dsvuqfn'kxfrdjusdsfy,LorU=kgSAxSandksO fcUnqlsNksM+ktkrkgSArksxsandhpkyD;kgksxhtc;gfcUnq A ij igq¡psxh &[g = 10 m/s2 ]
RESONANCE
Page # 98
O
A [ Ans: 40 m / s ] 285.
287.
3m
4m
q
B
In the figure shown neglecting friction and mass of pulleys, what is the acceleration of mass B? fp=kesan'kkZ,vuqlkj?k"kZ.k,oaiqyhdsnzO;ekudksux.;ekusa]nzO;ekuBdkRoj.kD;kgksxk? [ 3.46_NLM ]
(A) 286.
a
g 3
(B)
5g 2
(C)
2g 2
(D*)
2g 5
Two blocks A and B of equal mass are connected by a light inextensible string passing over two light smooth pulleys fixed to the blocks. The parts of the string not in contact with the pulleys are horizontal. A horizontal force F is applied to the block A as shown. Then: M.Bank_NLM_3.57 nksxqVdsArFkkBftudsnzO;ekulekugS],dgYdhvrU;jLlh}kjk¼jLlhtksxqVdksalsca/ksnksfpduhiqyhesa lsxqtjrhgSA½tqM+sgq,gSaAjLlhdkogHkkxtksiqyhlstqM+kughagS]{kSfrtgSA,d{kSfrtcyFxqVdsAijfp=kkuqlkj yxk;ktkrkgS]rks-
(A) the acceleration of A will be equal to that of B (B*) the acceleration of A will be greater than that of B (C) the acceleration of A will be less than that of B (D) none of the above is correct. (A)Adk Roj.k] B ds Roj.k ds cjkcj gksxk (B*)Adk Roj.k] B ds Roj.k ls vf/kd gksxk (C)Adk Roj.k] B ds Roj.k ls de gksxk (D)buesalsdksbZugha
A system in set up as shown in the Figure. At time t = 0, a force P is applied to the block at a constant angle of 370. Magnitude of this force is gradually increased and it is found that the block lifts off from the floor when the extension in the spring is 25cm. The acceleration of the block at this instant is _________. [Friction is absent and the spring was initially in nondeformed state] ,dfudk;fp=kkuqlkjj[kk x;kgSAle;t = 0, ij xqVdsijfu;r dks.k370 ij cyP yxk;k tkrk gSAbl cy dk ifjek.k /khjs&/khjsc<+k;ktkrkgSvkSj;gik;ktkrkgSfdtcfLizaxesaf[kpkao25cm gSxqVdktehulsÅijmBtkrkgSAbl {k.kijxqVdsdkRoj.kgS_________. [?k"kZ.kvuqifLFkrgSrFkkfLizsxizkjEHkesavldqafprvoLFkkesagSA] [1.1_NLM]
RESONANCE
Page # 99
[ Ans: 5 m/sec2 ] 288.
A weightless inextensible rope on a stationary wedge forming angle with the horizontal. One end of the rope is fixed to the wall at point A. A small load is attached to the rope at point B. The wedge starts moving to the right with a constant acceleration. Determine the acceleration a1 of the load when it is still on the wedge. M.Bank_NLM_8.12 ,dgYdhvrU;jLlhfLFkjostij{kSfrtlsdks.kcukrsgq,j[khgSAjLlhdk,dfljkfcUnqAijfLFkjgSAfcUnq B ij,dNksVklknzO;ekutksM+j[kkgSAostnk;havksjfu;rRoj.klsxfrdjuk'kq:djnsrkgSAB ijtqM+snzO;ekudk Roj.k a1 Kkr dhft,] tcfd ;gvHkhHkhost ijgksM.Bank_NLM_8.12
[ Ans: 2 a sin (/2) ] 289.
In the figure shown the blocks A & C are pulled down with constant velocities u . Acceleration of block B is : [ Made 2003] fn[kk;sx;sfp=kesaArFkkC CykWdksadksfu;rosxuls[khapktkrkgSACykWdBdkRoj.kgksxk : [ Made 2003] M.Bank_NLM_8.13
(A) (C) 290.
u2 tan2 sec b
(B*)
u2 sec 2 tan b
u2 tan3 b
(D) zero ‘'kwU;
In the figure shown the car moves down with a constant acceleration g. A bob of mass m is attached to a string whose other end is tied to the ceiling of the car. If the bob remains stationary relative to the car. Then tension in the string is: fn,x,fp=kesadkjfu;rRoj.kg lsuhpsdhvksjxfrdjjghgSA,dm nzO;ekudkxqVdk(bob) jLlhlstqM+kgSftldk
nwljkfljkdkjdhNrlstqM+kgqvkgSAvxjxqVdkdkjdslkis{kfLFkjgS]rksjLlhesarukogS& M.Bank_NLM_5.20
(A) m g/2
RESONANCE
(B*) m g
(C) 2 m g
(D) none of these buesalsdksbZugha
Page # 100
291.
In the figure shown, find out the value of [ assume string to be tight ] [ Made 2004] fn[kk;sx;sfp=kesadkekucrkb;s[jLlhdksdlkgqvkekus] [ Made 2004] M.Bank_NLM_8.25
(A*) tan1 292.
3 4
(B) tan1
4 3
(C) tan1
3 8
(D)noneofthesebuesalsdksbZugha
A heavy spherical ball is constrained in a frame as in figure. The inclined surface is smooth. The maximum acceleration with which the frame can move without causing the ball to leave the frame:
,dHkkjhxsanfp=kesan'kkZ,vuqlkjfLFkrgSAurryfpdukgSAblurrydksvf/kdrefdlRoj.klsysdjtkldrs gSarkfd;gxsanurryijxfrughadjsaA M.Bank_NLM_5.7
(A) 293.
g 2
(B) g 3
(C*)
g 3
(D)
g 2
A block B of mass 0.6 kg slides down the smooth face PR of a wedge A of mass 1.7 kg which can move freely on a smooth horizontal surface. The inclination of the face PR to the horizontal is 45º. Then : ,d xqVdk B ftldk nzO;eku 0.6 kg gS fpduh ost A¼ftldk nzO;eku1.7 kg gS½ ij fQly jgk gSA ost fpduh {kSfrt lrg ij Lora=k :i ls xfr dj ldrk gSA PR lrg dk {kSfrt ls >qdko 45º gS rks : M.Bank_NLM_6.11 (A*) the acceleration of A is 3 g/20 (B*) the vertical component of the acceleration of B is 23 g/40 (C*) the horizontal component of the acceleration of B is 17 g/40 (D) none of these (A*) A dk Roj.k3 g/20 gSA (B*) B ds Roj.k dk Å/oZ ?kVd23g/40 gSA (C*) B ds Roj.k dk {kSfrt ?kVd 17g/40 gSA (D)buesalsdksbZughaA
RESONANCE
Page # 101
