Resonance DPP

TARGET : JEE (Main + Advanced) 2017 O

Course : VIKAAS (JA)

Date : 25-05-2015

DPP No. : 13 (JEE-ADVANCED) Total Marks : 38

Max. Time : 40 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.3 Multiple choice objective ('–1' negative marking) Q.4 to Q.6 Comprehension ('–1' negative marking) Q.7 to Q.9 Match the Following (no negative marking) (2 × 4) Q.10

(3 marks 3 min.) (4 marks 4 min.) (3 marks 3 min.) (8 marks 10 min.)

[09, 09] [12, 12] [09, 09] [08, 10]

ANSWER KEY OF DPP No. : 13 1. 8.

(C) (D)

2. 9.

(C) (B)

3. 10.

1.

Find area under the curve y = 5ex and the x-axis, between x = 0 and x = 2 : x = 0  x = 2 y = 5ex x-  : (A) 5e2

(C) 4. (A,C) 5. (A,C) 6. (A) – p,r,s ; (B) – q,r ; (C) – q,s ; (D) – p

(B) e2 – 1

(C*) 5(e2 – 1)

(D)

(C,D)

7.

(C)

e2 1 5

2

5e x dx

Sol.

5( e 2 1)

0

2.

A car starts from rest and moves with constant acceleration. The ratio of the distance covered in the nth second to distance covered in n seconds is :    n   n  

 (A)

2 n

–

2

1 n

1 a ( 2n 1) 2 1 2 an 2

(B) 2n

1

n2

n2

2 n

2

+ 2 n

1 n

(C*)

Sol.

SN S

3.

If A and B are two non–zero vectors such that | A

2 1 – 2 n n

(D)

1 1 + 2 n n

1 n2 B| =

|A

B| 2

and | A | = 2 | B | then the angle

between A and B is :

 A  B    | A B | =

|A

B| 2

 | A | = 2 | B |   A  B 

 (A) 37º

(B) 53º

(C*) cos–1(–3/4)

(D) cos–1(–4/3)

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PAGE NO.-1

Sol.

1 (A2 + B2 – 2AB cos 4 3A2 + 3B2 + 10 AB cos = 0 or

(A2 + B2 + 2AB cos ) =

15B2 + 20B2 cos 4.

=0

cos

) 12B2 + 3B2 + 10(2B) (B) cos = 0 =–

3 4

A particle is moving in a straight line. Its displacement at time t is given by s =

4t2 + 2t, then its

velocity and acceleration at time t = 1 second are 2

  t s =

4t2 + 2t,   t = 1 

2

 (A*) velocity of the particle is 2 m/s (B) velocity of the particle is m/s 2 (C*) acceleration of the particle is 8 m/s 2 (D) acceleration of the particle is m/s (A*)   2 m/s  (B)   m/s  2

(C*)  8 m/s  (D)  m/s2  5.

From the top of a tower of height 200 m, a ball A is projected up with 10 m s 1 and two seconds later another ball B is projected vertically down with the same speed. Then : (Take g = 10 m/s2) (200 m      10 m/s        B ) (g = 10 m/s2 ) (A*) both A and B will reach the ground simultaneously (A B   ) (B) the ball A will hit the ground 2 seconds later than B hitting the ground ( B    A) (C*) both the balls will hit the ground with the same velocity (   ) (D) none of these ()

Sol.

s = ut +

at 2 2

0 = 10t –

10 t 2 2

t=2s Hence A & C are correct. 6.

The magnitude of the displacement is equal to the distance covered in a given interval of time if the particle. (A) moves with constant acceleration along any path (B) moves with constant speed (C*) moves in same direction with constant velocity or with variable velocity (D*) have acceleration and velocity in same direction.


PAGE NO.-2

 (A)   (B)   (C*)      (D*)   Sol.

To have distance equal to magnitude of displacement the particle has to move in same direction in straight line path.

    Comprehension : If a man has a velocity varying with time given as v = 3t2 , v is in m/s and t in sec then : v = 3t2 (v , m/s t )

7.

Sol. 8.

Find out the velocity of the man after 3 sec. 3   (A) 18 m/s (B) 9 m/s v = 3t2 v = 3(3)2 = 27 m/s

Find out his displacement after 2 seconds of his start : 2   (A) 10 m (B) 6 m (C) 12 m s

Sol.

10.



(D*) 8 m

3 t 2 dt

dS =

S=

Sol.

(D) 36 m/s

2

0

9.

(C*) 27 m/s

0 2 t3 0

=8

Find out his acceleration after 3 seconds : 3   (A) 9 m/s2 (B*) 18 m/s2 dv f= = 6t dt f = 6 × 3 = 18 m/s2

(C) 12 m/s2

(D) 6 m/s2

A particle is moving along x-axis. Its position (x) varies with time (t) is shown in the graph. Points A,B,C,D and E are also given in the graph. Here positive values of velocity and acceleration means they are towards +x direction and their negative values means that they are towards –x direction. Match the column–I, according the correct results given in column–II. (Centre of curvature of BD is below the graph but centre of curvature of AB and DE is above the graph)  x   (x) (t)    A,B, C, DE +x        –x  I     II   BD  AB DE


PAGE NO.-3

Sol.

Coulmn–I Column–II (A) From point A to B (p) acceleration of the particle is positive (B) From point B to C (q) acceleration of the particle is negative (C) From point C to D (r) velocity of the particle is positive or zero (D) From point D to E (s) speed is increasing (A)  A  B  (p)  (B)  B  C  (q)  (C) C  D  (r)     (D)  D  E  (s)  Ans. (A) – p,r,s ; (B) – q,r ; (C) – q,s ; (D) – p v = slope of xt graph (xt ) a = curvature of xt graph. (xt )


PAGE NO.-4

TARGET : JEE (Main + Advanced) 2017 O

Course : VIKAAS (JA)

Date : 25-05-2015

DPP No. : 14 (JEE-MAIN) Total Marks : 60

Max. Time : 60 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.20

(3 marks 3 min.)

[60, 60]

ANSWER KEY OF DPP No. : 14 1. 8. 15.

(A) (D) (C)

2 9. 16.

(B) (A) (B)

3. 10. 17.

(A) (B) (B)

4. 11. 18.

(A) (C) (C)

5. 12. 19.

(B) (A) (B)

6. 13. 20.

(A) (D) (A)

7. 14.

(D) (B)

/3

1.

sin( 3 x )dx = /6

(A*) Sol.

2 .

1 3

cos 3 x 3

(B) /3

= /6

1 3

(C) 1

(D) 3

1 1 (–1 – 0) = 3 3

If y = sinx n(3x) dy will be : dx  y = sinx n(3x)

then

dy : dx 3 cos x (A) x (C) sin (x2) + cosx n(3x)



Sol.

dy dx

sin x + (cosx) n(3x) x (D) 3xsinx + 3cosx n(x)

(B*)

sin x + n(3x) cosx x


PAGE NO.-5

3.

Sol.

4.

The acceleration–time graph of a particle moving on a straight line is as shown in figure. The velocity of the particle at time t = 0 is 2m/s. The velocity after 2 seconds will be       t = 0  2 m/s   2  

(A*) 6 m/s (B) 4 m/s (C) 2 m/s v(2) = v(0) + area under a–t graph from t = 0 to t = 2 1 = 2+ (2) (4) = 6 m/s. 2

The v elocity of particle (whose displacement time graph is shown) at t = 3s, is: t = 3  

(A*) 4m/s 5.

(D) 8 m/s

(B) 6 m/s

(C) 8m/s

(D) 2m/s

A person starts from origin and for his linear motion velocity is given as shown in figure. Find displacement of the person from t = 15 sec to t = 35 sec.

     t = 15 sec  t = 35 sec.

(A) 75 m Sol.

(B*) 0

(C) –75 m

(D) 150 m

Area under velocity time graph gives displacement. t2

S=

35

vdt t1

vdt 15

1 1 × (25 – 15) × 15 – (35 – 25) × 15 2 2 S = 75 – 75 = 0

=

6.

The velocity of a car moving on a straight road increases linearly according to equation, v = a + b x, where a & b are positive constants. The acceleration in the course of such motion: (x is the displacement)   , v = a + b x,  a  b   (x ) (A*) increases (B) decreases (C) stay constant (D) becomes zero (A*)   (B)  (C)  (D)  


PAGE NO.-6

Sol.

V = a + bx (V increases as x increases) dV dx = b; =V dx dt dV so, acceleration = V = V.b dx hence acceleration increases as V increases with x.

7.

Figure shows the velocity time graph of a particle moving along straight line (v is m/s and t is in seconds). Its average velocity in 20 seconds will be:  v m/s  t,  20  

Sol.

8.

(A) 10 m/s (B) zero   Displaceme nt  Vav = time  Displacement = area of curve =  1 (10 20 ) 10 30 Vav = 2 = m/s. 20 4

1 sec. 2

10t =



(x

(C) Let

4

2

10 t

10 3 2

x3 7)8

(C)

3/ 2 sec.

(B)

1 1 7 ( x 4 7 )7

(D*)

3 / 2 sec.

u2 4

t=

3 . 2

dx =

1 1 28 ( x 4 7 ) 7

(A*)

Sol.

(B) 1 sec.

u cos60º = 5, Vy = u sin60º – 10t V2 = (u sin60º 10t)2 + (u cos60º) u2 3 = u 4 2

9.

(D*) 7.5 m/s

A particle is projected with speed 10 m/s at angle 60° with the horizontal. Then the time after which its speed becomes half of initial (g = 10m/s2)   60º  10 m/s      (g = 10m/s2)  (A)

Sol.

(C) 3.75 m/s

+C

1 1 +C 4 ( x 4 7) 7 u = x4 + 7 du = 4x3dx

1 du 4 u8

1 u 7 4 7

(D)

1 (x

4

7 )7

+C

+C

1 1 4 28 ( x 7 )7 Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.) - 324005 Website: www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555 | CIN: U80302RJ2007PTC024029

PAGE NO.-7

10.



 Sol.

A particle has initial velocity, v = 3î 4ˆj and a constant force F = 4î 3ˆj acts on the particle. The path of the particle is :  v = 3î 4ˆj  F = 4î 3ˆj  : (A) straight line (B*) parabolic (C) circular (D) elliptical  (A)  (B*)  (C)  (D)  (B) For constant acceleration if initial velocity makes an oblique angle with acceleration then path will be parabolic. 

 

11.

A particle travels according to the equation x = at3, y = bt3. The equation of the trajectory is x = at3  y = bt3  (A) y

12.

ax 2 b

(B) y

bx 2 a

(C*) y

bx a

(D) y

bx 3 a

A particle moves in a plane from A to E along the shown path. It is given that AB = BC = CD = DE = 10 metre. Then the magnitude of net displacement of particle is :   A  E    AB = BC = CD = DE = 10    D 108° E

108°

C

108°







Sol.

Given figure represents a regular pentagan so magnitude of AE = 10 metre.  AE 10 m 

13.

The magnitude of the displacement is equal to the distance covered in a given interval of time if the particle. (A) moves with constant acceleration along any path (B) moves with constant speed (C) moves with variable velocity (D*) have acceleration and velocity in same direction.

(A*) 10 m





A

(B) 15 m

B

(C) 5 m

(D) 20 m

 (A)   (B)   (C)   (D*)   Sol.

To have distance equal to magnitude of displacement the particle has to move in same direction in straight line path.

   


PAGE NO.-8

14.

A particle travels from A to B path shown in figure, then the displacement of particle is :   A  B :

(A) 2 2 m (C) 52 m 15.

 Sol.

16.

Sol.

17.



18.



(B*) 4 2 m (D) None of these 

A stone is projected with speed 20 m/s at angle 37° with the horizontal and it hits the ground with speed 12 m/s due to air resistance. Assume the effect of air resistance is to reduce only horizontal component of velocity. Then the time of flight will be (g = 10 m/s2)   37°   20 m/s      12 m/s    -      (g = 10 m/s2)(A) greater than 2.4 sec (B) less than 2.4 sec. (C*) 2.4 sec. (D) depends on other data  (A) 2.4  (B) 2.4  (C*) 2.4  (D)  Time of flight only depends on the vertical component of the velocity, 2u sin Hence, T = = 2.4 s. g A particle is projected under gravity at an angle of projection 45° with horizontal. Its horizontal range is 36 m. Find maximum Height attained by particle.   45°     36 m.  (A) 8 meter (B*) 9 meter (C) 10 meter (D) 20 meter H tan R 4 = 45° & R = 36 m H= 9m A ball is projected horizontally with a velocity of 5 m s–1 from the top of a building 19.6 m high. How long will the ball take to hit the ground ? (g = 9.8 m/s2)  5 m s–1 19.6 m   ? (g = 9.8 m/s2) (A) 1s (B*) 2 s (C) 4s (D) 3 s A bomb is dropped from an aeroplane when it is at a height h directly above the target. If the aeroplane is moving horizontally with a speed v, the distance by which the bomb will miss the target is given by   v      h   

  (A) 2v

h g

(B) v

h g

(C*) v

2h g

(D) v

h 2g


PAGE NO.-9

19.

A train starts from station A at 1 pm and reaches station B at 3 pm, after tarvelling 108 km. It halts at station B for one hour and then starts for station C which is 42 km from station B. It reaches station C at 7pm on the same day. The average speed of the train for the whole journey from A to C is 1 pm  A   108 km     3 pm  B   B  C   B  42 km       C    7 PM    A  C     

 Sol.

20.

(A) 68 km/hr (B*) 25 km/hr S = 108 + 42 = 150 km t = 2 + 1 + 3 = 6 hr 150 = = 25 km/hr 6

(D) 35 km/hr

The projection of A 2iˆ 2ˆj 5kˆ on to B î ˆj is : A 2iˆ 2ˆj 5kˆ  B î ˆj   : (A*) Zero

Sol.

(C) 30 km/hr

(B)

Projection of A onto B =

A

4

(C)

2 . B B

4 2

(D) 4

0


PAGE NO.-10

Resonance DPP

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