Class 11 Diwali Assignment Physics Solution

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Subject : PHYSICS

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Wishing You & Your Family A Very Happy Ha ppy & Prosperous Deepawali

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2

dx

dy

=c= . Where a, b and partiicle cle move movess in in spac spacee al along the the path path z = ax + by in such a way that mQ.1 A part dt dt o c c are constants. The acceleration of the particle is . g ˆ (B) (2ax2 + 6by2) k ˆ ˆ (D) (bc2x + 2by) k ˆ a (A*) (6ac2x + 2bc2) k (C) (4bc2x + 6ac2) k h u dx dy dz e S[Sol. z = ax3 + by2, = 3ax2 + 2by g y dt dt dt a p B s 2 dx dy dz d z h . t = 3acx2 + 2bcy, = 6acx + 2bc 1 2 a dt dt dt 8 dt 8 M 8 . 5 2 d z w 0 2 2 3 2 = (6ac x + 2bc ) w 9 dt 8 w 9  2x + 2bc 2) ˆ ] 0 & = (6ac a k , m 9 7 oQ.2 A stone stone is proj projecte ected d from a horizon horizontal tal pla plane ne.. It attains attains maxi maxim mum hei heigh ghtt 'H' & 7 c 7 . strikes a stationary smooth wall & falls on the ground vertically below the 3 s 0 e 9 maximum maximum height. height. Assume the collis collision ion to be elastic the height of the t he point on the t he s 3 s wall where ball will strike is: 0 9 a l (A) H/2 (B) H/4 0 C : (C*) 3H/4 (D) none of t hese o e n k[Sol. Becaus ausee hori horizon zontal tal veloci velocity ty is constan constantt so o e[Sol. Bec h T P . 2 u sin θ l w T= a g p w o h w B u 2 sin 2 θ , : ) r given H = , usinθ = 2gH i e t 2g S i . s K b . 2 2gH R e . T= at the t he time of hitting hitting the wall w S g ( a m y i o 3R 3R r r a f The horizontal distance covered is , so time taken to cover c over horizontal distance distance K 4 4 . e R g g a H H 3T 1 a H k 3 h , h = 2gH × 3 – ×g×a× T' = = c u 2g 4 2 2g 2g S a : P s 3 H h y t h = ] a d 4 M u t , s S e r ising vertically with an accelerat ac celeration ion of 4.9 m/s 2 releases a ball 2 seconds after the s dQ.311/12kin A man in a balloon rising s a a l balloon is let go from the ground. The greatest height above the ground reached by the ball is o C l (g = 9.8 m/s2) o n k e (A*) 14.7 m (B) 19.6 m (C) 9.8 m (D) 24.5 m w T o[Sol. v = 0 + 4.9 × 2 = 9.8 9.8 m/s D 1 E h = × 4.9 × 4 = 9.8 m E 1 2 R 0 = v2 – 2gh2 F 2

Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective Ineffective People don't.

Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com m o c . g a h uQ.4 S y B s h t a[Sol. M . w w w & m o c . s e s s a l C o k e T . w w wQ.5 : e t i s b e w [Sol. m o r f e g a k c a P y d u t S dQ.6 a o l n w o D E[Sol. E R F

h2 =

v2 2g

=

9.8 × 9.8 2 × 9.8

= 4.9

H = h1 + h2 = 9.8 + 4.9 = 14.7 m

]

A particle is projected at an angle of 45° from a point lying 2 m from the foot of a wall. It just touches the top of the wall and falls on the ground 4 m from it. The height of the wall is (A) 3/4 m (B) 2/3 m (C*) 4/3 m (D) 1/3 m u 2 sin 2θ

3

e g a p

. 1 8 g 8 8 5 60 = u2, u = 60 0 3 9 8 ucosθ = 30 , usinθ = 30 9 0 2 , 9 t= , 7 30 7 7 3 1 4 2 0 h = 30 × – × 10 × 9 2 30 30 3 0 9 2 4 0 : h = 2 – = m, 3 3 e n o h 4 P h= m ] l 3 a p o h The velocity at the maximum height of a projectile is half its initial velocity of projection. Its range on the B , ) horizontal plane is r i S . 2 2 2 2 u 3u 3u 3u K . (A*) (B) (C) (D) R 2g 2g g 2g . S ( Given:— ucosθ = u/2 a y cosθ = 1/2, θ = 60° i r a K 3 1 . 2u 2 R 2u 2 sin θ cos θ 2 2 g = R= a g h g u S : 3u 2 s h ⇒ R= ] t a 2g M , s e s Two particles instantaneously atA & B respectively 4.5 meters apart are moving with uniform s B' a l velocities as shown in the figure. The former towards B at 1.5 m/sec and the latter perpendicular C to AB at 1.125 m/sec. The instant when they are nearest is: o k A' A 23 B e T 4.5 m

R=6=

, θ = 45°

×

(A) 2 sec 

V1 = 1.5î ,

(B) 3 sec



×

(C) 4 sec

(D*)1

25

sec

V2 = 1.125 jˆ



V21 = 1.125 jˆ − 1.5î Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

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m o c . g a h uQ.7 S y B s h t a[Sol. M . w w w & m o c . s e s s a l C o k e T . wQ.8 w w : e t i [Sol. s b e w m o r f e g aQ.9 k c a P y d u t S d a o[Sol. l n w o D E E R F

(1.125) 2 + (1.5) 2 = 1.875 m/s

| V21 | = t=

3.6 1.875

= 1.92 sec = 1

23

]

25

A river is flowing with a speed of 1 km/hr. Aswimmer wants to go to point 'C' starting from 'A' . He swims with a speed of 5 km/hr, at an angle θ w.r.t. the river. If AB = BC = 400 m . Then the value of θ is: (A) 37º (B) 30º (C*) 53º (D) 45º Condition for reaching the point C

4

e g a p

. 1 8 8 8 5 vy 0 3 tan 45° = , vy = vx 9 vx 8 9 0 (vR + vMcosθ) = vMsinθ , 1 + 5cosθ = 5sinθ 9 7 7 2 7 1 + 5cosθ = 5 1 − cos θ 3 0 On squaring, 9 3 1 + 25cos2θ + 10cosθ = 25 – 25cos2θ 0 2 9 50cos θ + 10cosθ – 24 = 0 0 : On solving, e ⇒ ] θ = 53° n o h P A boat is moving towards east with velocity 4 m/s with respect to still water and river is flowing towards l a north with velocity 2 m/s and the wind is blowing towards north with velocity 6 m/s. The direction of the p o h flag blown over by the wind hoisted on the boat is: B , ) (A*) north-west (B) south-east (C) tan–1 (1/2) with east (D) north r i    S v Bg v BR v Rg = 4î 2 jˆ . K .  R v wg = 6î . S (    a v wB = v w − v B = 6î − 4î − 2 jˆ = 2î − 2 jˆ y i r a Direction will be north-west ] K . R g A girl is riding on a flat car travelling with a constant velocity 10 ms-1 as a h shown in the fig. She wishes to throw a ball through a stationary hoop in u S such a manner that the ball will move horizontally as it passes through : s h the hoop. She throws the ball with an initial speed 136 ms-1 with respect t a to car. The horizontal distance in front of the hoop at which ball has to M , s be thrown is e s (A) 1m (B) 2m (C) 4m (D*) 16m s a l C x = (10 + 136 cos )t ...(1) o k 2 e v y = 0 = 136 sin2 – 2 × 10 × 5 ...(2) T

=

+

+

θ

θ

sinθ =

5=

5 34

, cosθ =

136 ×

5

3 34

t – 5t 2

...(3)

34 Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com t 2 –2t + 1 = 0 ⇒ t=1 So, x = 16 m ] m o c . gQ.10 A particle is projected with a speed V from a point O making an angle of 30° with the vertical. At the a same instant, a second particle is thrown vertically upward from a point A with speed v. The two particle h u reach H, the highest point on the parabolic path of the first particle simultaneously, then the ratio V v e S g y a 2 3 p B (A) 3 2 (B) 2 3 (C*) (D) s 3 2 h . t 1 a 2 2 8 V sin 60° 2gH × 4 8gH 8 M 8 , = V = . [Sol. H = 5 2 g 3 3 w 0 3 w 9 2 8 8gH / 3 V w 9 v = 2gH , = = 0 & 3 v 2gH , m 9 7 o 2 V 7 c 7 = ] . 3 v 3 s 0 e 9 s sQ.11 A particle is projected with a certain velocity at an angle θ above the horizontal from the foot of a given 3 0 9 a l 0 θ plane inclined at an angle of 45° to the horizontal. If the particle strike the plane normally then equals C : − − − − 1 1 1 1 o (A) tan (1/3) (B) tan (1/2) (C) tan (1/ √2) (D*) tan 3 e n k o e h g T P t . [Sol. vx = ux – l 2 w a p w o g h w t 0 = ucos( θ – 45°) – B , : 2 ) r i e t S i . s 2u cos(θ − 45°) K b . t= R g e . w S ( 1 g 2 a m t ,y=0 y = uyt – y i o r 2 2 r a f K . e R 2 2u y g 2 2 u sin(θ − 45°) g a = t = a k g g h c u S a : P 2 2 u sin(θ − 45°) 2u cos(θ − 45°) s = h y t g g a d M u t , 1 s S e = tan(θ – 45°) s d 2 s a a l o C l tan θ − 1 1 o n ⇒ ⇒ θ = tan–1(3) = tan θ = 3 k + θ 1 tan 2 e w T o DQ.12 Velocity time graph of a particle is in shape of a semicircle of radius R as E shown in figure. Its average acceleration from T = 0 to T = R is: E (A) 0 m/s2 (B*) 1 m/s2 R F (C) R m/s2 (D) 2R m/sec2 5

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Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com [Sol.

m o c . g aQ.13 h u S y B s h t a M . w[Sol. w w & m o c . s e s s a l C o k e T . Q.14 w w w : e t i s b e w m o r f e g a k[Sol. c a P y d u t S d a o l n w o D E E R F

u = 0, V = R (at T = R) T = R, V = u + at R=0+a×R a = 1 m/s2 ] A car is moving with uniform acceleration along a straight line between two stops X and Y. Its speed at X and Y are 2m/s and 14m/s. Then 6 e (A) Its speed at mid point of XY is 15m/s g a p (B) Its speed at a point A such that XA : AY = 1 : 3 is 5m/s (C*) The time to go form X to the mid point of XY is double of that to go from mid point to Y. . (D) The distance travel in first half of the total time is half of the distance travelled in the second half of 1 8 8 the time. 8 5 (A) Not possible if acceleration is const. 0 (B) Velocity at mid-point

v2 = u + 2 × a ×



4

106 = v2 + 2 × a × a= (C) Possible (D) Not possible

96 

,v=

3 4

52

]

3 9 8 9 0 , 9 7 7 7 3 0 9 3 0 9 0 : e n o h P is a positive l a p o h B , ) r i S . K . R . S ( a y i r a K . R g a h u S : s h t a M , s e s s a l C o k e T

A particle having a velocity v = v 0 at t = 0 is decelerated at the rate |a| = α v , where α constant. (A*) The particle comes to rest at t =

2 v0

α (B) The particle will come to rest at infinity. (C) The distance travelled by the particle is

2v 30 / 2

α

.

3 / 2

(D*) The distance travelled by the particle is (A)

a = –α√v dv dt 0

= –αv1/2 t

dv

∫ v1 / 2 = ∫0 − αdt

v0

[2v1 / 2 ] 0v

0

= –αt

2[–v01/2] = –αt t=

2 v0 3

α

.

2 v0

α (D) Velocity at any time t is


Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com m o c . g a h u S y B s h t a M . w w w & m o c . s e s s a l C o k e T . w w w : eQ.15 t i s b e w m o r f e g[Sol. a k c a P y d u t S d a o l n w o D E E R F

v

t

dv

∫ v1 / 2 = – ∫0 α dt

v0

v

[2 v1 / 2 ]v = –αt 0

1 / 2

2[ v

7

e g a p

− v10 / 2 ] = –αt

αt    v =  v0 − 2  

2

. 1 8 8 8 5 2 dα 0 t   3  v= =  v 0 9 2  dt  8 9 0 2 α α t   , dα =  v 0  dt 9 7 2  7 0  0 7 3 3 0 αt 2 v 0 t 9 x = v0t + – 3 12 0 2 9 0 : 2 v0 e at t = n o h P 3 / 2 l v 2 0 a x= ] p o 3 h B , ) r i Two towns A and B are connected by a regular bus service with a bus leaving in either direction every S . T minutes. A man cycling with speed of 20km/h in the direction A to B, notices that a bus goes past him K . every t1 = 18 minutes in the direction of motion, and every t 2 = 6 minutes in the opposite direction. What R . is the period T of the bus service? Assume that velocity of cyclist is less than velocity of bus S ( (A) 4.5 minutes (B) 24 minutes (C*) 9 minutes (D) 12 minutes a y i r a K . R g a h u S : 18 s (v – 20) = d = vt h 60 t a M 6 , s (v + 20) = d = vt e 60 s s a 3v – 60 = v + 20 l C v = 40 kmph o k e T

−

∫

∫

−

α

α

α

α

α

(40 + 20) × 6 = 40 T

6 = 40 × T 60 ⇒ T = 6/40 hr = 9 min

]


Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com Q.16 kin A body starts from rest with uniform acceleration. Its velocity after 2n second is v0. The displacement of the body in last n seconds is m

o c . g a h[Sol. u S y B s h t a M . w w w & m o c . s eQ.17 s s a l C o k e[Sol. T . w w w : e t i s b e w m o r f e g a kQ.18 c a P y d u t S d a o l n w o[Sol. D E E R F

(A)

v 0 ( 2n − 3)

(B)

6 v0 = 0 + a × 2n S2n = Sn =

1 2 1

v0 4n

(2n–1)

(C*)

3v 0 n

(D)

4

3v 0 n 2

...(1)

a × (2n)

8

e g a p

2

a × (n)2

2 -----------------S2n – Sn = =

3 2 3 2

an2 ×

v0 2n

...(2) × n2 =

3v 0 n 4

]

An airplane pilot wants to fly from cityA to city B which is 1000 km due north of city A. The speed of the plane in still air is 500 km/hr. The pilot neglects the effect of the wind and directs his plane due north and 2 hours later find himself 300km due north-east of city B. The wind velocity is (A*) 150km/hr at 45°N of E (B) 106km/hr at 45°N of E (C) 150 km/he at 45°N of W (D) 106 km/hr at 45°N of W V = 500 kmph jˆ p/w

Vw/g = v x î + v y jˆ Vp/g = v x î + ( v y + 500) jˆ kmph In two hours Sp/g = 150 2 î + 1150 2 jˆ km = 2 v x î + ( 2 v y + 1000) jˆ

⇒

vx = 75 2 vy = 75 2 Vw/g = 75 2 î + 75 2 jˆ = 150 kmph at 45° N of E

]

An arrangement of the masses and pulleys is shown in the figure. Strings connecting masses Aand B with pulleys are horizontal and all pulleys and strings are light. Friction coefficient between the surface and the block B is 0.2 and between blocks A and B is 0.7. The system is released from rest (use g = 10 m/s2). (A) The magnitude of acceleration of the system is 2 m/s 2 and there is no slipping between block A and block B (B) The magnitude of friction force between block A and block B is 42 N (C) Acceleration of block C is 1 m/s 2 downwards (D*) Tension in the string connecting block B and block D is 12 N (A) 6g – 18 – 1g = 16a a = 2 m/s 2 C moving down (B) f s – 18 – 10 = 4 × 2 f s = 36 (C) ac = 2 m/s2 downward


. 1 8 8 8 5 0 3 9 8 9 0 , 9 7 7 7 3 0 9 3 0 9 0 : e n o h P l a p o h B , ) r i S . K . R . S ( a y i r a K . R g a h u S : s h t a M , s e s s a l C o k e T

Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com m o c . g a h u SQ.19 y B s h t a[Sol. M . w w wQ.20 & m o c . s e s s a l C[Sol. o k e T . wQ.21 w w : e[Sol. t i s b e w m o r f e g a k c a P y dQ.22 u t S d a o l n w[Sol. o D E E R F

(D)

T – 10 = 1 × 2

⇒

T = 12 N

]

9

e

A body of mass 2 kg is placed on a horizontal surface having kinetic friction 0.4 and static friction 0.5. g a If the force applied on the body is 2.5 N, the frictional force acting on the body will be (g = 10 m/s 2) p (A) 8 N (B) 10 N (C) 20 N (D*) 2.5 N . 1 f l· s = µsmg = 0.5 × 2g = 10 N 8 8 Block is stationary P < f l·s 8 5 ⇒ friction force f = P = 2.5 N ] 0

3 9 8 In the arrangement shown in figure, there is friction between the blocks of 9 0 masses m and 2m which are in contact. The ground is smooth. The mass of , the suspended block is m. The block of mass m which is kept on mass 2m 9 7 is stationary with respect to block of mass 2 m. The force of friction between 7 7 m and 2m is (pulleys and strings are light and frictionless) : 3 0 mg mg mg 9 mg (A) (B) (C*) (D) 3 2 0 2 4 3 9 mg = 4ma 0 : a = g/4 e n o f s = ma = mg/4 ] h P l a The maximum value of m(in kg) so that the arrangement shown in the figure is in p o equilibrium is given by h B (A) 2 (B*) 2.5 (C) 3 (D) 3.5 , ) r i Bigger block is not moving S . T = mg ...(1) K . 2T = 0.4 N ...(2) R . T + 10 g = N ...(3) S ( String a y i r 2T a K = 0.4 . T 100 R g 1.6 T = 40 a h T = 25 u S ⇒ m = 2.5 kg ] : s h t Two blocks, A and B, of same masses, are resting in equilibrium on an inclined plane having inclination a M , with horizontal = (>0). The blocks are touching each other with block B higher than A. Coefficient of s e static friction of A with incline = 1.2 and of B = 0.8. If motion is not imminent, s s a (A) < 30° (B*) (Friction)A > (Friction)B l C (C*) < 45° (D) (Friction)A = (Friction)B o k e T

+

α

α α

2mgsinα < 2mgcosα ⇒ tanα < 1

α < 45°

f A = mgsinα + N f B = mgsinα – N


Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com To show N ≠ 0

m f b = 0.8 mgcosα o max c . ⇒ N = mg[sinα – 0.8cosα] g N = mgsecα[tanα – 0.8] a h For α > tan–1 (0.8) u f A > f B for α ≤ tan –1(0.8) f A = f B ] e S g y a p B sQ.2313nl A rope of length L and mass M is being puled on a rough horizontal floor by a constant horizontal force h t . F = Mg. The force is acting at one end of the rope in the same direction as the length of the rope. The 1 a coefficient of kinetic friction between rope and floor is 1/2. Then, the tension at the 8 8 M 8 . 5 midpoint of the rope is w 0 (A) Mg / 4 (B) 2Mg / 5 (C) Mg / 8 (D*) Mg / 2 3 w 9 w 8 9 & 0 , m 9 o 7 7 c . [Sol. a = F − µN = Mg − 0.5Mg = g/2 7 s 3 M M 0 e 9 s 3 s 0 a 9 l T – µMg/2 = Ma/2 0 C : T – Mg/4 = Mg/4 o e k n T = Mg/2 ] o e h T . P wQ.24 A plank of mass 2kg and length 1 m is placed on a horizontal floor. A small block of mass 1 kg is placed l a 13nl p w o on top of the plank, at its right extreme end. The coefficient of friction between plank and floor is 0.5 and w h that between plank and block is 0.2. If a horizontal force = 30 N starts acting on the plank to the right, B , : ) 2 r e the time after which the block will fall off the plank is (g = 10 m/s ) i t i S . (A*) (2/3) s (B) 1.5 s (C) 0.75 s (D) (4/3) s s K b[Sol. a = 2 m/s2 . 1/g e R . w 13 30 − 2 − 15 S ( 2 a2/g = = = 6.5 m/s m a 2 2 y i o r r a2/1 = 4.5 a f K e . 1 1 g 2 R 2 s2/1 = a 2 / 1t 1 = × 4.5 × t a g 2 2 a k h c u a S 2 4 : P t= = sec ] s 3 9 y h t d a u M t , s SQ.2513nl Two wedges, each of mass m, are placed next to each other on a flat floor. A e s d cube of mass M is balanced on the wedges as shown. Assume no friction between s a a l the cube and the wedges, but a coefficient of static friction µ < 1 between the o C l wedges and the floor. What is the largest M that can be balanced as shown o n k w e without motion of the wedges? T o D 2µm m µm (A) (B) (C*) (D) All M will balance E 1− µ 2 2 E R[Sol. 2Ncos45 = Mg ...(1) F 0 1


Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com N

+ mg = N1 ...(2) m 2 o c . N g = µN1 ...(3) a 2 h u e S N N  g y µ + mg a =   p B 2 2   s h t .  2µ  1 a 8 ⇒ N =  1 − µ mg 8 M 8 .   5 w 0 3 w 2µ 9 w 8 ∴ √2 N = 1 − µ mg = Mg 9 & 0 , m 2µm 9 o 7 M= ] 7 c 1− µ . 7 s 3 0 e sQ.26 A force F acting on a particle of mass 5 kg placed on a smooth horizontal surface. F = 40 N remains 9 nl 3 s 0 constant but its vector rotates in a vertical plane at an angular speed 2 rad/sec. If a t = 0, vector F is a 9 l 0 C horizontal, find the velocity of block at t = π 4ω sec. : o e k n o e (A) 1 m/s (B) 2 m/s (C) 2 m/s (D*) 2 2 m/s h T . P l w a p w o w h B At any time t , : [Sol. ) r e i t i S . s K b . e R . 40 cos ωt w S a= = 8cos2t ( 5 m a y i o r r π / 8 a f   8 π   π / 8 K dv = 8 ∫ cos 2t dt = [sin 2t ] 0 = 4 sin  − 0 = 2 2 m/s ] e . 2 4  g  R   0 a g a k h c u aQ.2713nl A sphere of mass m is kept between two inclined walls, as shown in the figure. If S : P the coefficient of friction between each wall and the sphere is zero, then the ratio of s y h t normal reaction (N1 /N2) offered by the walls 1 and 2 on the sphere will be d a u M t (A) tanθ (B) tan2θ , s S (C*) 2cosθ (D) cos2θ e s d s a a l o C l o n k w e T o[Sol. D E E ...(1) Ν1cosθ = mg + N2cos2θ R F N sinθ = N sin2θ ...(2) 1 1

1

2


Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com By eq (2) m o c N1 . ] g N 2 = 2cosθ a h u SQ.28 A particle is projected horizontally from the top of a tower with a velocity v0. If v be its velocity e 11/12/13wpe g y at any instant, then the radius of curvature o f the path of the particle at the point (where the particle is at a p B s that instant) is directly proportional to: h t . (A*) v3 (B) v2 (C) v (D) 1/v 1 a 8   8 M ˆ ˆ 8 v = v 0 i − gtj and a = – g jˆ . [Sol. 5 w 0   3 w  a · v    9   w 8 Component of a ⊥ to v = a −  2  v ∴ 9 v   & 0 2 1

, m   9 v g o 0  7  (gtî + v jˆ) 7 c a ⊥ =  − 2 i.e. 0 2 2  . 7 + v g t  0  s 3 0 e 9 s v 0g 3 s  0 a 9 | a ⊥ | = l ∴ 2 2 + v gt 0 0 C : o e k  2 n 2 2 2 3 / 2 3 o ( v 0 + g t ) e |v| v h T = Also, r = = . P | a | v 0g v g l ⊥ 0 w a p w r ∝ v3 ∴ o w h Option (A) is correct ] ∴ B , : ) r e i t i Q.29 S There are two massless springs A and B of spring constant K and K respectively and K > . 11/12/13wpe A B A s K b KB. If WA and WB be denoted as work done on A and work done on B respectively, then . e R (A*) If they are compressed to same distance, WA > WB . w S ( (B*) If they are compressed by same force (upto equilibrium state) WA < WB m a y (C) If they are compressed by same distance, WA = WB i o r r a f (D) If they are compressed by same force (upto equilibrium state) W > W K e[Sol. For same compression x (say) . g R 0 a g a 1 1 k 2 2 h c WA = k A x 0 & WB = k B x 0 u 2 2 a S : P ⇒ WA > WB [∴ k A > k B] s y h t for same force at equilibrium force = F0 d a u M t , F0 F0 s S xA = , x = e s d k A B k B s a a l o C l 2 F o n 1 0 2 k k x ∴ WA = = w e 2k A T 2 A A o D 2 E F0 E Similarly, WB = 2k B R F ⇒ WB > WA A

B


Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com ∴

(A) & (B) are correct options

]

m oQ.30 13wpe A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging c . vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the g a hanging part on to the table is h u mgL mgL mgL e S (A) mgL (B) (C) (D*) g y 3 9 18 a p B [Sol. If hanging part of chain doesn't get any velocity then (D) is correct option. Also minimum work done is s h given by option (D) and its equal to change in gravitational potential energy of chain. t . 1 a 8 M L 8 M 8 ×g× . = as CM moves by a distance L/6 ∆W = ∆PE 5 3 6 w 0 3 w 9 MgL w 8 = 9 18 & 0 (D) is correct option ] , m 9 o 7 7 c . Q.31 2 7 Power delivered to a body varies as P = 3t . Find out the change in kinetic energy of the body from s 13wpe 3 0 e t = 2 to t = 4 sec. 9 s 3 s (A) 12 J (B*) 56 J (C) 24 J (D) 36 J 0 a 9 l [Sol. Here power delivered is 0 C 2 : P = 3t o e k n If this power results into only kinetic energy change then o e h T 4 . P 4 4 3 l t w 2 a 3 3 P dt 3 t dt ∆KE = ∫ = ∫ = 3 = (4 – 2 ) J = 56 J p w 3 o 2 2 2 w h B , : Power delivered will cause this maximum change in K.E. ) r e i t (B) is correct option ] i S . s K b . eQ.3213wpe A block ‘A’ of mass 45 kg is placed on a block ‘B’ of mass 123 kg. Now block R . w S ‘B’ is displaced by external agent by 50 cm horizontally towards right. During the ( m a same time block ‘A’ just reaches to the left end of block B. Initial & final position y i o r r are shown in figure. Refer to the figure & find the workdone by frictional force on a f K e block Ain ground frame during above time. . g R (A) – 18 Nm (B*) 18 Nm (C) 36 Nm (D) – 36 Nm a g a k h c u a[Sol. Here blocks are moving w.r.t. each other, hence friction force = 0.2 × 45 × 10 = 90 N S : P Given block 'B" moves 50 cm s y h t Also given that block A moves (40 – 10) cm back w.r.t. block 'B' d a u M t ∴ Forward movement of block Ain ground frame = 50 – 30 cm = 20 cm , s S ∴ Work done by friction force = 90 × 0.2 J = 18 J e s d s Work done is positive a a l o C ∴ Option (B) is correct] l o n k w e oQ.3313wpe Aspring of force constant k is cut in two part at its one third length. when both the parts are stretched T D by same amount. The work done in the two parts, will be E (A) equal in both (B) greater for the longer part E (C*) greater for the shorter part (D) data insufficient. R F[Sol. When a spring is cut into two parts each part has spring constant more than that of original spring. If 3 1


Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com k = spring constant & 0 = natural length, then for cut parts

m o c . g a h u If they are stretched by same amount then work done in shorter part will be double than that in the case e S g of longer part. y a p B Option (C) is correct ] ∴ s h t . aQ.3413wpe The horsepower of a pump of efficiency 80%, which sucks up water from 10 m below ground and 1 8 8 2 M ejects it through a pipe opening at ground level of area 2 cm with a velocity of 10 m/s, is about 8 . 5 w (A) 1.0 hp (B*) 0.5 hp (C) 0.75 hp (D) 4.5 hp 0 3 w 9 [Sol. Here, w 8 2 = 2 × 10 –4 m2 9 area = 2 cm & 0 velocity = 10 m/s , m 9 ∴ Volume flow rate = 2 × 10 –3 m3s–1 = vρgh o 7 7 c ∴ Energy required per second = 100 × 10 3 × 2 × 10–3 J = 2 × 100 J = 200 J . 7 s 3 Efficiency is 80% ∵ 0 e 9 s Power of pump = 250 W ∴ 3 s 0 a Hence (B) is correct option ] 9 l 0 C : oQ.35 e k 13wpe Potential energy and position for a conservative force are plotted in graph n o e shown. Then force position graph can be h T . P l w a p w o (A) (B) (C) (D*) w h B , : ) r e i t i [Sol. Here by the graph we can say that S . s U = U0cosr K b . e R dU . w F= – = –U0(–sinr) ∴ S ( dr m a y F = U0sinr i o r r a f Hence correct option is (D) ] K e . g R a g a kQ.3613wpe A constant force produces maximum velocity V on the block connected to the spring of force constant h c K as shown in the fig. When the force constant of spring becomes 4K, the maximum velocity of the u a S block is : P s y h V V t d (A) (B) 2V (C*) (D) V a u 4 2 M t , s S[Sol. Block will gain maximum velocity at the point of equilibrium e s d s F a a l In first case equilibrium elongation = o C l k o n k w e 2 2 T F 1  F  1 o F F · – k   = mV2 ⇒ V= ∴ D k 2  k  2 mk E E R F 4 1


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In second case equilibrium elongation = m 4 k o c . 2 F 1 1 V g  F  F2 2  = mV' F· – ×4k  V' = = ⇒ a 4 k 2 2 2 4mk h  4k  u (C) is correct option ] ∴ S y B sQ.3713wpe A bead of mass 5kg is free to slide on the horizontal rod AB. They are h t connected to two identical springs of natural length h ms. as shown. If initially a bead was at O & M is vertically below L then, velocity of bead at point N will M . be w (A*) 5h m/s (B) 40h/3 m/s w (C) 8h m/s (D) none of these w &[Sol. Natural length of each spring is h m h h o –h= elongation in each spring = ∴ c cos 37° 4 . s & applying work-energy theorem e s 2 s 1 1  h  2 a mv = 2 × k   l 2 2  4  C o v = 5h m/s k e ] ∴ Option (A) is correct T . wQ.38 13wpe Block A in the figure is released from rest when the extension in the spring is w x0. The maximum downwards displacement of the block is w : Mg Mg 2Mg 2Mg x x x − + − + x0 e (A*) (B) (C) (D) 0 0 0 t i 2K 2K K K s b[Sol. Let the block move 'x' downward then elongation in spring is '2x' e 1 1 w k(x0 + 2x)2 – k x 02 = Mgx ∴ 2 2 m o r f ⇒ k x 02 + 4kxx0 + 4kx2 – k x 02 = 2Mgx e g Mg a x 0 x + x = ∵ ≠ ⇒ 0 k 2k c a Mg P − x0 x= ∴ y 2k d Option (A) is correct ] ∴ u t S dQ.3913wpe A smooth semicircular tube AB of radius r is fixed in a vertical plane a and contains a heavy flexible chain of length π r and weight W= π r as o l shown. Assuming a slight disturbance to start the chain in motion, the n w velocity v with which it will emerge from the open end B of the tube is o D  2 π  4gr 2gr  2  E (A) (B) (C) 2gr + π  (D*) 2gr  +  π π  π   π 2  E R[Sol. Initial CM position Final CM position F

5 1

e g a p

K

A


M


Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com m o c . g 2r πr a x = x = 1 h π 2 u ∴ ∆h for CM = x + x1 e S g y a 1 W 2 p B U s W∆h = ∆PE = ∆KE ⇒ 2 g h t . 1 a 8 2 π  8  M 8 . U2 = 2gr  +  5 w  π 2  0 3 w 9 w 8 π 2   9 & U = 2gr  +  0  π 2  , m 9 o Option (D) is correct ] ∴ 7 7 c . 7 sQ.4013wpe A heavy particle hanging from a string of length l is projected horizontally with speed gl . The speed 3 0 e 9 s of the particle at the point where the tension in the string equals weight of the particle is: 3 s 0 a 9 l (A) 2gl (B) 3gl (C) gl / 2 (D*) gl / 3 0 C : o[Sol. Speed at bottom = e g < 2g  k n o e h T 1 1 . P mg (1 – cos ) = mg – mv2 ...(1) θ   l w a 2 2 p w o w h 2 mv B , : Also, T – mgcosθ = ) r e  i t i S . But T = mg s K b . e R mv 2 . w = mg – mgcosθ ∴ S (  m a y i o r 1 mg  r a f i.e. mv2 = (1 – cosθ) K 2 2 e . g R a g 1 1 a n k ∴ eq (1) ⇒ mg(1 – cosθ) = mg – mg(1–cosθ) h c 2 2 u a S : P 2 1 s y h 1 – cosθ = cos θ = ⇒ t d 3 3 a u M t , v = g / 3 ∴ s S e s d Option (D) is correct ] s ∴ a a l o C l o nQ.41 A skier plans to ski a smoo th fixed hemisphere of radius R. He starts from k 13wpe w e rest from a curved smoo th surface of height (R/4). The angle θ at which he T o D leaves the hemisphere is E (A) cos–1 (2/3) (B) cos–1 (5/ 3 ) E R (C*) cos–1 (5/6) (D) cos–1 (5/ 2 3 ) F 6 1


Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com R

m[Sol. ∆h = 4 + R(1 – cosθ) o c . 1 mgR g 2 = mg h = mv {1 + 4(1 – cosθ)} ∆ a 2 4 h u mg mv 2 e S (5 − 4 cos θ) = ∴ g y a 2 R p B s h mv 2 t . 1 mgcosθ – N = a 8 R 8 M 8 . 5 mg w 0 (5 − 4 cos θ) mgcosθ = 3 w 2 9 w 8 9 cos = 5/6 θ & 0 Option (C) is correct ] ∴ , m 9 o 7 c . Q.42wpe A simple pendulum swings with angular amplitude θ. The tension in the string when it is vertical is twice 7 7 s 3 the tension in its extreme position. Then, cos θ is equal to 0 e 9 s (A) 1 / 3 (B) 1 / 2 (C) 2 / 3 (D*) 3 / 4 3 s 0 a [Sol. At extreme v = 0 At vertical position 9 l 0 C : o e k n o e h T . P l w a p w o w h B , : ) r e i t i S . Given T2 = 2T1 i.e. mg(3 – 2cosθ) = 2mgcosθ s K b . 3 – 2 cosθ = 2cosθ cos θ = 3/4 ∴ ⇒ e R . (D) is correct option ] ∴ w S ( m a y i o Q.43 The inclined plane OA rotates in vertical plane about a horizontal axis through r r 13wpe a f K O with a constant counter clockwise velocity ω = 3 rad/sec. As it passes the e . g R position θ = 0, a small mass m = 1 kg is placed upon it at a radial distance r = a g 0.5 m. If the mass is observed to be at rest with respect to inclined plane. The a k h c u value of static friction force at θ = 37° between the mass & the incline plane. a S : (A*) 1.5 N (B) 3.5 N (C) 2.4 N (D) none P s y[Sol. Drawing the FBD in rotating frame we get h t d a u M t , s S e s d s a a l o C l o n k w e T o D As the block is at rest, hence E mgsinθ = f + mω2r E f = mgsinθ – mω2r = 1 × 10 × (3/5) – 1 × 9 × 0.5 = 6 – 4.5 = 1. 5 ∴ R Therefore, force of friction (static) = 1.5 N F 7 1

∴

(A) option is correct

]


Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com Q.44wpe On a particle moving on a circular path with a constant speed v, light is thrown m from a projectors placed at the centre of the circular path. The shadow of the o particle is formed on the wall. the velocity of shadow up the wall is c . g (A*) v sec2φ (B) v cos2φ (C) (A) v cos φ (D) none a h[Sol. Method I

u e S v' cos θ v g y a = R sec θ R p B s h t v' = v sec2 θ . 1 a 8 Method II 8 M 8 . y = R tan θ 5 w 0 3 w dy dv 9 2 w 8 = R sec θ = 9 dt dt & 0 2 V = R sec ( ) θ ω , y m 9 o 7 7 c v   . 7 2  2  Vy = R sec θ = v sec θ] s 3  R  0 e 9 s 3 sQ.4513wpe Two cars A and B start racing at the same time on a flat race µ = 0.1 0 a µ = 0.2 9 l track which consists of two straight sections each of length 100 π r r = 100m 0 r C r = 200m : and one circular section as in Fig. The rule of the race is that each o e k n car must travel at constant speed at all times without ever skidding o e h T (A) car A completes its journey before car B . L = 100 π P l w (B) both cars complete their journey in same time a p w (C) velocity of car A is greater than that of car B o w h (D*) car B completes its journey before car A. B , : ) r e i t i [Sol. v ≤ µrg S . s K b vA ≤ 0.1×100 ×10 = 10 m/s . e R . w vB ≤ 0.2 × 200 × 10 = 20 m/s S ( m a y 200 π + π(100) m i o r r a f t A = = 30 π sec. 10 m / s K e . g R a 200π + π( 200) g a k t B = = 20 π sec.] h c 20 m / s u a S PQ.4613wpe A horizontal curve on a racing track is banked at a 45° angle. When a vehicle goes around this curve : s y h at the curve’s safe speed (no friction needed to stay on the track), what is its centripetal acceleration? t d a u (A*) g (B) 2g (C) 0.5g (D) none M t , s S 2 e s d[Sol. tan θ = v s a a l Rg o C l o n 2 k v w e T o tan 45° = R D g E E v2 R = ac = g ] F 8 1

A B

B

A

A

B

B

A

R


Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com Q.4711/12/13mom When a block is placed on a wedge as shown in figure, the block starts sliding down and the wedge also start sliding on ground. All surfaces are rough. The centre of mass of (wedge + block) m o system will move c . (A) leftward and downward (B*) rightward and downward g a (C) leftward and upward (D) only downward h u[Sol. (B) Wedge

9 1

e S g y a p B s h t . 1 a 8 8 M 8 . 5 w 0 3 w 9 w 8 9 System & 0 , m 9 o 7 7 c . 7 s 3 0 e 9 s 3 s 0 a 9 l 0 external force are gravity friction towards right so com shifts right + downward ] C : o e k n o e Question No. 48 to 50 (3 questions) h T . Two smooth balls A and B, each of mass m and radius R, have their centres at (0,0,R) and at (5R,–R,R) P l w a respectively, in a coordinate system as shown. Ball A, moving along positive x axis, collides with ball B. p w o Just before the collision, speed of ball A is 4 m/s and ball B is stationary. The collision between the balls w h B is elastic. , : ) r e i t i S . s K b . e R . w S ( m a y i o r r Q.48 Velocity of the ball A just after the collision is a f 11/12/13mom K e . (A*) (i + 3 j) m/s (B) (i − 3 j) m/s (C) (2i + 3 j) m/s (D) (2i + 2j) m/s g R a g Before collision After collision a k[Sol. (A) h c u a S : P s y h t d a u M t , s S e s d s a a l o C l o n k w e T o ˆ ˆ vA = 4 sin 30° [cos 60 i + sin 60 j] D E vA = î + 3 jˆ ] E R F


Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com Q.4911/12/13mom Impulse of the force exerted by A on B during the collision, is equal to

m 3 o mi − 3mj) kg-m/s (A) ( 3mi + 3mj) kg-m/s (B) ( c . 2 g a (C*) (3mi − 3mj) kg-m/s (D) ( 2 3mi + 3mj) kg-m/s h u e S[Sol. J m VB − VB g A on B = y f i a p B s ˆ ˆ = m [ 4 cos 30°(cos 30°i − sin 30° j ) − 0] h t . 1 a = (3mi − 3mj) kg-m/s ] 8 8 M 8 . 5 w 0 wQ.5011/12/13mom Coefficient of restitution during the collision is changed to 1/2, keeping all other parameters 3 9 unchanged. What is the velocity of the ball B after the collision? w 8 9 & 0 1 1 + − ( 3 3 i 9 j ) ( 9 i 3 3 j ) (A) m/s (B*) m/s (C) (6i + 3 3 j) m/s (D) (6i − 3 3 ) m/s , m 9 2 4 o 7 7 c After collision . [Sol. Before collision 7 s 3 0 e 9 s 3 s 0 a 9 l 0 C : o e k n o e h T . P l w a p w o w h B , : − − ( V V ) ) 1 r 2 1 e i = t (1) i S − ° 2 ( 0 4 cos 30 ) . s K b . e R V2 – V1 = 3 . w S ( 4 3 m a y (2) m = mV1 + mV2 i o r r 2 a f K e . V2 + V1 = 2 3 g R a g a k 3 3 h c u V2 = m/s a S 2 : P s y h t  3 3 d a V2 = [cos 30° î + sin 30° (− jˆ)] u M t 2 , s S e s d 9ˆ 3 4 ˆ s a a j ] = i− l o 4 4 C l o n k w e T oQ.5111/12/13mom A particle of mass m = 0.1 kg is released from rest from a point A of D wedge of mass M = 2.4 kg free to slide on a frictionless horizontal plane. The E particle slides down the smooth face AB of the wedge. When the velocity of the E wedge is 0.2 m/s the velocity of the particle in m/s relative to the wedge is: R F 10 0 2

(A) 4.8

(B*)

(C) 7.5

(D) 10

Successful People Replace the words 3 like; "wish", "try" & "should" with "I Will". Ineffective People don't.

Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com [Sol.

(1)

O = 0.1 (v1 cos 30°– v0) – 2.4 v0 v1 cos30° = 25 v0

m o c 25(0.2) 5 10 . (2) v = = = m/s ] g 1 3 / 2 3 / 2 3 a hQ.52 A bullet of mass 0.01 kg and travelling at a speed of 500 m/s strike a block of mass of 2 kg which 12/13mom u is suspended by a string of length 5 m. The centre of gravity of the block is found to rise a vertical e S g y a distance of 0.1 m. What is the speed of the bullet after it merge from the block: p B s (A) 780 m/s (B*) 220 m/s (C) 1.4 m/s (D) 7.8 m/s h t [Sol. For block . 1 a 8 V'2 = 2gh 8 M 8 . 2 V' = 2×10×0.1 5 w 0 3 w V' = 2 m/s 9 w 8 Just before and after bullet strikes, momentum conserved 9 & 0 , m 9 0.01 × 500 = 2V' + 0.01 V o 7 7 c . 7 5 – = 0.01 V 2 2 s 3 0 e 9 5 − 2.828 s 3 V= = 217.2 m/s] s 0 0 . 01 a 9 l 0 C : oQ.53 e A ball is dropped from a height h. As it bounces off the floor, its speed is 80 p ercent o f what it was 13mom k n o e just before it hit the floor. The ball will then rise to a height of most nearly h T . P (A) 0.80 h (B) 0.75 h (C*) 0.64 h (D) 0.50 h l w a p w o w h B , : [Sol. ) r e i t i S . s K b . e (0.8 2gh ) R . w h’ = S 2g ( m a y i = 0.64 h ] o r r a f K e . gQ.5413mom In a one-dimensional collision, a particle of mass 2m collides with a particle of mass m at rest. If the R a particles stick together after the collision, what fraction of the initial kinetic energy is lost in the collision? g a k h c u 1 1 a 1 S (A) (B*) (C) (D) none : P 2 3 s 4 y h t d a u [Sol. M t , s S e s d s a a l o C l 2 o n k 2mv + 0 = 3mv’ v’ = v ⇒ w e 3 T o D 2 1  2  E 3m  v  K f E ∆K 2  3  R = 1– K = 1– Ki 1 2 F i 2m v 1 2

2


Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com 3

4

=1– × m 2 5 o c . 1 g = ] a 3 h uQ.5513mom A particle is projected from ground towards a vertical wall 80m away at an angle of 37º with S horizontal with initial velocity of 50m/s. After its collision with wall & then once with ground find at what e g y a distance from wall will it strike the ground again if coefficient of restitution for both collisions is equal to p B s 1/2. h t . 1 (A) 70 m (B) 120 m (C*) 140 m (D) none a 8 8 [Sol. After first collision M 8 . 5 w 0  80  3 w ˆ   Vy = 30 – gt = 30 – 10 = (10m/s) j 9 w 8  40  9 & 0 1 , m Vx = − (40) = − 20 î 9 2 o 7 7 c . 7 80 s 3 t 1 = = 2 sec 0 e 9 40 s 3 s 0 a × 2 30 9 l t = T – t = –2 = 4 sec 0 2 1 C : 10 o e k n Before second collision o e h T Vx = – 20 î . P l w a x = 20 × 4 = – 80 m p w o w h Vy = 10 – 10 (t 2) = 10 –10(4) = − 30 jˆ B , : ) After second collision r e i t i S . s Vx = – 20 î K b . e R ˆ Vy = + 15 j . w S ( m a 2 × 20 ×15 y i o Range = r r 10 a f K e . = 60 m g R Net:→ 60 m + 80 m a g a k = 140 m ] h c u a S : P Question No.56 to 57 (2 questions) s y h t d A projectile of mass "m" is projected from ground with a speed of 50 m/s at an angle of 53° with the a u t horizontal. It breaks up into two equal parts at the highest point of the trajectory. One particle coming to M , s S e rest immediately after the explosion. s d s Q.56 The ratio of the radii of curvatures of the moving particle just before and just after the explosion are: a a mom l o C l (A*) 1 : 4 (B) 1 : 3 (C) 2 : 3 (D) 4 : 9 o n k w e T o D E[Sol. E R F 2 2


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50 × = 30 m 5 o c . m g m(30) = m (0) + (v’) a 2 h u v’ = 60 S y v12 v 22 B g= g= s r r2 1 h t a 2 2 v2 v1 M . r2 = r1 = w g g w 2 w r2 v 2 4 = = ] & r1 v12 1 m o c . sQ.57mom The distance between the pieces of the projectile when they reach the ground are: e (A*) 240 (B) 360 (C) 120 (D) none s s u sin 53° 50 4 a l × = 4 sec. [Sol. z = = g C 10 5 o xrel = vrel × t = 4 × 60 = 240 m] k e T . Question No. 58 & 59 (2 Questions) w w Two blocks (from very far apart) are approaching towards each w other with velocities as shown in figure. The coefficient : of friction for both the blocks is µ = 0.2 e t i Q.58 Linear momentum of the system is s 13mom b (A) conserved all the time (B) never conserved e (C*) is conserved upto 5 seconds (D) none of these w till = 5 second m[Sol. Fnet = 0 o r f e g a 20 − 10 k Vcm = = 5 m/s ] c 2 a P y dQ.59 How much distance will centre of mass travel before coming permanently to rest 13mom u t (A) 25 m (B*) 37.5 m (C) 42.5 m (D) 50 m S d 1 a[Sol. (m1 + m2) ∆X cm = m1 X1 + m2 X 2 X1 = –10(5) + × 2(5)2 = 25 2 o l n 1 w 2 ( ) = –25 + 100 = –75 = 20 (10) – × 2 × (10) 2 = 100 ∆ X X cm o 2 2 D | ∆X cm | = 37.5 m ] E E R F Successful People Replace the words like; "wish", "try" & "should" with "I Will". Ineffective People don't.

3 2

e g a p


Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com Q.6013mom From a thin circular disc of radius R, a circular hole of radius 4R/5 is cut as shown. The distance of the centre of mass of remaining disc, from the centre of the original disc m o is c . (A) 15R/40 (B) R/3 g a (C) R/4 (D*) 16R/45

h u S[Sol. y B s h t a M . w w w & m o c . s e s s a l C o k e T . w w w : e t i s b e w m o r f e g a k c a P y d u t S d a o l n w o D E E R F

σ =

4 2

m

e g a p

πR 2

 4R  × π   m’= πR 2  5  m

m’ =

16 25

2

m

 m 16  R     25  5 

0×0 −  Xcm =

m−

16 m 25

 16    125 

− mR 

Xcm =

⇒ Xcm =

9m 25

− 16 R 45

R–

=–R

4R 5

=

R 5

16 × 25 125

]



Get Solution of These Packages & Learn by Video Tutorials on www.MathsBySuhag.com m o c . gQ.1 a h uQ.8 S y BQ.15 s h t Q.22 a M . wQ.29 w wQ.36 & mQ.43 o c . sQ.50 e s sQ.57 a l C o k e T . w w w : e t i s b e w m o r f e g a k c a P y d u t S d a o l n w o D E E R F

ANSWER KEY DIWALI ASSIGNMENT

A

Q.2

C

Q.3

A

Q.4

C

Q.5

A

Q.6

D

Q.7

C

A

Q.9

D

Q.10

C

Q.11

D

Q.12

B

Q.13

C

Q.14

A,D

C

Q.16

C

Q.17

A

Q.18

D

Q.19

D

Q.20

C

Q.21

B

B,C

Q.23

D

Q.24

A

Q.25

C

Q.26

D

Q.27

C

Q.28

A

A,B

Q.30

D

Q.31

B

Q.32

B

Q.33

C

Q.34

B

Q.35

D

C

Q.37

A

Q.38

A

Q.39

D

Q.40

D

Q.41

C

Q.42

D

A

Q.44

A

Q.45

D

Q.46

A

Q.47

B

Q.48

A

Q.49

C

B

Q.51

B

Q.52

B

Q.53

C

Q.54

B

Q.55

C

Q.56

A

A

Q.58

C

Q.59

B

Q.60

D


5 2

e g a p


Class 11 Diwali Assignment Physics Solution

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