DC Pandey Solutions (JEE Advance): Motion in One Dimension- 5 | DC Pandey Solutions for JEE Physics PDF Download

Objective Questions (Level 1) 
Single Correct Option
Ques 1: A packet is released from a rising balloon accelerating upward with acceleration a. The acceleration of the stone just after the release is 
(a) a upward 
(b) g downward 
(c) (g - a) downward 
(d) (g + a) downward
Ans:  g downward
Sol: As the packet is detached from rising balloon its acceleration will be g in the downward direction.
Option (b) is correct.

Ques 2: A ball is thrown vertically upwards from the ground. If T₁ and T₂ are the respective time taken in going up and coming down, and the air resistance is not ignored, then 
(a) T₁ > T₂ 
(b) T₁ = T2 
(c) T₁ < T₂ 
(d) nothing can be said
Ans: T₁ < T₂ 
Sol: While going up:


Using v² = u² + 2as

i.e.,
While coming down


⇒
Using Eq. (i) and Eq. (ii)

⇒ T₂ > T₁
Option (c) is correct.

Ques 3: The length of a seconds hand in watch is 1 cm. The change in velocity of its tip in 15 s is
(a) zero 



Ans:
Sol: Angular speed (ω) of seconds hand

Speed of the tip of seconds hand
 cm/s   [∵ v = rω and r = 1 cm]
As in 15s the seconds hand rotates through 90°, the change in velocity of its tip in 15 s will be

Option (d) is correct.

Ques 4: A particle moving along a straight line travels half of the distance with uniform speed 30 ms^-1 and the rem aining half of the distance with speed 60 ms^-1. The average speed of the particle is 
(a) 45 ms^-1 
(b) 42 ms^-1 
(c) 40 ms^-1 
(d) 50 ms^-1
Ans: 40 ms^-1
Sol: 

Option (c) is correct.

Ques 5: A boat is moving with a velocity  with respect to ground. The water in the river is moving with a velocity  with respect to ground. The relative velocity of the boat with respect to water is



(d) zero
Sol: Relative velocity of boat w.r.t. water

Option (b) is correct.

Ques 6: During the first 18 min of a 60 min trip, a car has an average speed of 11 ms 1. What should be the average speed for remaining 42 min so that car is having an average speed of 21 ms -1 for the entire trip ? 
(a) 25.3 ms^-1 
(b) 29.2 ms^-1 
(c) 31 ms^-1 
(d) 35.6 ms^-1
Ans: 25.3 ms^-1 
Sol:

Ques 7: A particle moves along a straight line. Its position at any instant is given by  where x is in metre and t in second. Find the acceleration of the particle at the instant when particle is at rest, 
(a) -16 ms^-2 
(b) -32 ms^-2 
(c) 32 ms^-2 
(d) 16 ms^-2
Ans: -32 ms^-2 
Sol: 

  …(i)
∴ Particle is at rest when
32 - 8 t² = 0
i.e., t = 2 s
Differentiating Eq. (i) w.r.t. time t

∴ a at time t = 2 s (when particle is at rest)
= - (16) x (2) = -32 m/s² 
Option (b) is correct.

Ques 8: A car starts from rest and accelerates at constant rate in a straight line. In the first second the car covers a distance of 2 m. The velocity of the car at the end of second (sec) will be 
(a) 4.0 ms^-1 
(b) 8.0 ms^-1 
(c) 16 ms^-1 
(d) None of these
Ans: 8.0 ms^-1 
Sol: For first one second

⇒ a = 4 m/s² 
Velocity at the end of next second
v = (4) x (2) = 8 m/s
Option (b) is correct.

Ques 9: A particle is moving along x-axis whose position is varying with time according to the relation x = - 3t + t³ where x is in metre and t is in second. The displacement of particle for t = 1 s to t = 3 s is 
(a) +16 m 
(b) -16 m 
(c) +20 m 
(d) -20 m
Ans: +20 m
Sol: x = - 3t + t³
∴ displacement at time t (= 1 s)
= - 3(1) + (1)³ = - 2 m
and displacement at time t (= 3 s)
= - 3(3) + (3)³ = 18 m
And as such displacement in the time interval (t = 1 s to t = 3 s)
= (18 m) - (- 2 m)
= + 20 m
Option (c) is correct.

Ques 10: The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v₀. The distance travelled by the particle in time t will be




Ans:
Sol: Acceleration α = bt
∴
or  
or
Now, at t = 0, v = v₀
∴ C = v₀
i.e.,
or
or
or
At t = 0, s = 0,
∴ k = 0
⇒
Option (a) is correct.

Ques 11: Water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap, the instant the first drop touches the ground. How far above the ground is the second drop at that instant, (g = 1 0 ms^-2) 
(a) 1.25 m 
(b) 2.50 m 
(c) 3.75 m 
(d) 4.00 m
Ans: 3.75 m
Sol:
 
∴
∴  Height of 2nd drop  from ground

Option (c) is correct.

Ques 12: A stone is dropped from the top of a tower and one second later, a second stone is thrown vertically downward with a velocity 20 ms^-1. The second stone will overtake the first after travelling a distance of (g = 10 ms^-2) 
(a) 13 m 
(b) 15 m 
(c) 11.25 m 
(d) 19.5 m
Ans: 11.25 m
Sol: 

  [At time = t - 1]

or    t² = 4 (t - 1) + (t² - 2t + 1)
or    2t - 3 = 0
or
∴
= 11.25 m
Option (c) is correct.

Ques 13: When a ball is thrown up vertically with velocity v₀, it reaches a maximum height of h. If one wishes to triple the maximum height then the ball should be thrown with velocity

(b) 3 v₀ 
(c) 9 v₀ 
(d) 3/2 v₀
Ans:
Sol:  and u² = 2g (3h)
∴
or  u = v₀√3
Option (a) is correct.

Ques 14: A particle moves in the x-y plane with velocity v_x = 8t - 2 and v_y = 2. If it passes through the point x = 14 and y = 4 at t = 2 s, the equation of the path is 
(a) x = y² - y + 2
(b) x = y² - 2
(c) x = y² + y - 6
(d) None of these
Ans: x = y² - y + 2
Sol:  v_x = 8t - 2
∴
or       ∫ dx = ∫ (8 t - 2) dt
or        x = 4t² - 2t + k₁
Now at t = 2,  x = 14
∴ 14 = 4 × 2² - (2) 2 + k₁ 
i.e.,        k₁ = 2
Thus, x = 4 t² - 2t + 2 …(i)
Further, v_y = 2
i.e.,
or ∫ dy = ∫ 2 dt
or  y = 2t + k₂
Now, at t = 2, y = 4
∴ k₂ = 0
Thus, y = 2t   …(ii)
Substituting t = y/2 in Eq. (i),

or   x = y² - y + 2
Option (a) is correct.

Ques 15: The horizontal and vertical displacements of a particle moving along a curved line are given by x = 5t and y = 2t² + 1. Time after which its velocity vector makes an angle o f 45° with the horizontal is 
(a) 0.5 s 
(b) 1 s 
(c) 2 s 
(d) 1.5 s
Ans: 1 s
Sol: x = 5t and y = 2t² + t
∴
Now,  


⇒ t = 1 s
Option (b) is correct.

Ques 16: The height y and the distance x along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by y = (8t - 5 t²) metre and x = 6t metre where t is in second. The velocity of projection is 
(a) 8 ms^-1 
(b) 6 ms^-1 
(c) 10 ms^-1 
(d) data insufficient
Ans: 10 ms^-1 
Sol: y = 8t - 5t² and x = 6t
∴
At t = 0

∴ Velocity of projection

Option (c) is correct.

Ques 17: A ball is released from the top of a tower of height h metre. It takes T second to reach the ground. What is the position of the ball in T/3 second?
(a) h/9 metre from the ground
(b) (7h/9) metre from the ground
(c) (8h/9) metre from the ground
(d) (17h/18)metre form the ground
Ans: (8h/9) metre from the ground
Sol: 

∴

Option (c) is correct.

Ques 18: An ant is at a comer of a cubical room of side a. The ant can move with a constant speed u. The minimum time taken to reach the farthest comer of the cube is
(a) 3a/u 



Ans:
Sol: Distance of farthest corner from one corner
= a + a + a = 3 a
∴ Time taken = 3a/u
Option (a) is correct.

Ques 19: A lift starts from rest. Its acceleration is plotted against time. When it comes to rest its height above its starting point is

(a) 20 m
(b) 64 m
(c) 32 m
(d) 36 m
Ans: 64 m
Sol: Time-velocity graph of the given time-acceleration graph will be

∴ Height of lift above the starting point when it comes to rest
= Area under t-v graph

Option (b) is correct.

Ques 20: A lift performs the first part of its ascent with uniform acceleration a and the remaining with uniform retardation 2a. If t is the time of ascent, find the depth of the shaft.




Ans:
Sol: 

In time interval t₁

In time interval t₂

∴
i.e.,   t₁ = 2t₂
Now   t₁ + t₂ = t
∴  2t₂ + t₂ = t
or  t₂ = t/3
Depth of shaft = Displacement of lift

Option (b) is correct.