Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

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NEET : Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

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Objective Questions (Level 2)
Single Correct Option

Ques 1: When a man moves down the inclined plane with a constant speed 5 ms-1 which makes an angle of 37° with the horizontal, he finds that the rain is falling vertically downward. When he moves up the same inclined plane with the same speed, he finds that the rain makes an angleMotion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev with the horizontal. The speed of the rain is
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
(c) 5 ms-1
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans:Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: Applying sine for mula in Δ AOB
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Applying sine formula in
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Δ AOB'
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Comparing Eqs. (i) and (ii)
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
sin α - cos α tan 37°
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or 4 sin α - 3 cos α = 32 cos α - 28 sin α + 24 sin α + 21 cos α
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or 32 cos α + 21 cos α + 3 cos α = 28 sin α - 24 sin α + 4 sin α
or 56 cos α = 8 sin α
tan α = 7/1
From Eq. (i)
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Option (b) is correct.

Ques 2: Equation of motion of a body is dv/dt = -4v + 8  where v is the velocity in ms-1 and t is the time in second.
Initial velocity of the particle was zero. Then 
(a) the initial rate of change of acceleration of the particle is 8 ms-2 
(b) the terminal speed is 2 ms-1 
(c) Both (a) and (b) are correct 
(d) Both (a) and (b) are wrong
Ans: the terminal speed is 2 ms-1 
Sol: 
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
At the time the body acquires terminal speed its acceleration (a) must be zero.
Thus
- 4v + 8 = 0
v = 2 m/s
Terminal speed = 2 m/s.
Option (b) is correct.

Ques 3: Two particles A and B are placed in gravity free space at (0, 0, 0)m and (30, 0, 0)m respectively. Particle A is projected with a velocity Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev while particle B is projected with a velocity Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

simultaneously. Then
(a) they will collide at (10, 20, 10)m
(b) they will collide at (10, 10, 10) m
(c) they will never collide
(d) None of the above

Ans: they will never collide
Sol: Particles will collide if

Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
∴ Option (c) is correct.

Ques 4: Velocity of the river with respect to ground is given by v0. Width of the river is d. A swimmer swims (with respect to water) perpendicular to the current with acceleration a = 2t (where t is time) starting from rest from the origin O at t = 0. The equation of trajectory of the path followed by the swimmer is
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: Displacement along y-axis at time t
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or  ∫ dvy = ∫ 2t dt
or vy = t2 + C1 
At t = 0, vy = 0 (given)
∴ C1 = 0
∴ vy = t2
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or ∫ dy = ∫ t2 + C2
or Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
At t = 0, Y = 0  (given)
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev  …(i)
Displacement along x-axis at time t:
x = v0t
or Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Substituting above value of t in Eq. (i)
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Option (a) is correct.

Ques 5: The relation between time t and displacement x is t = αx2 + βx where α and β are constants. The retardation is 
(a) 2αv3 
(b) 2βv3 
(c) 2αβv3 
(d) 2β2v3
Ans: 2αv3
Sol: t = ax2 + bx
Differentiating w.r.t. time t
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or v (β + 2α × x ) = 1
or v = (β + 2α × x )-1
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
∴ Retardation = 2αv3
Option (a) is correct.

Ques 6: A street car moves rectilinearly from station A to the next station B with an acceleration varying according to the law f = a — bx, where a and b are constants and x is the distance from station A. The distance between the two stations and the maximum velocity are
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: f = a - bx
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
As at x = 0 car is at rest C = 0
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
For v to be maximum.
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
i.e., a - bx = 0
or  x = a/b
Substituting x = a/b in Eq. (i),
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Car will come to rest when
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
x = 2a/b
∴ Distance between two stations = 2a/b
Option (a) is correct.

Ques 7: A particle of mass m moves on positive x-axis under the influence of force acting towards the origin given byMotion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRevIf the particle starts from rest at x = a, the speed it will attain when it crosses the origin is
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ans:Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: Force = - kx2
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Now at x = a, v = 0
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Thus Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
∴ Velocity at x = 0
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

Option (d) is correct.

Ques 8: A particle is moving mX-Y plane such that vx = 4 + 4t and vy = 4t. If the initial position of the particle is (1, 2). Then the equation of trajectory will be 
(a) y2 = 4x
(b) y = 2x
(c) x2 = y/2
(d) None of these

Sol: 
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or ∫ dx = ∫(4 + 4t) dt
and  ∫ dy = ∫ 4t dt
or x = 4t + 2t2 + C1 
∴ C1 = 1
Thus x = 4t + 2t2 + 1 …(i)
and y = 2t2 + C2 
At t = 0 x = 1
and at t = 0, y = 2
∴ C2 = 0
Thus y = 2t2 + 2 …(ii)
Substituting value of t in Eq. (i) in terms of y using Eq. (ii) we won’t get relationship as mentioned in option (a), (b) or (c).
∴ Option (d) is correct.

Ques 9: A particle is moving along a straight line whose velocity-displacement graph is as shown in the figure. What is the magnitude of acceleration when displacement is 3 m?
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: 

Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
∴ Magnitude of acceleration Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Option (a) is correct.

Ques 10: A particle is falling freely under gravity. In first t second it covers distance x1 and in the next t second it covers distance x2, then t is given by
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: 
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Option (a) is correct.

Ques 11: A rod AB is shown in figure. End A of the rod is fixed on the ground. Block is moving with velocity 2 ms-1 towards right. The vertical com ponent of velocity of end B of rod at the instant shown in figure is
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
(b) 2 ms-1
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
(d) 4 ms-1
Ans:Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: y2 + x2 = l2
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Option (c) is correct.

Ques 12: A thief in a stolen car passes through a police check post at his top speed of 90 kmh-1. A motorcycle cop, reacting after 2 s, accelerates from rest at 5 ms-2. His top speed beingl 08 kmh-1. Find the maximum separation between policemen and thief. 
(a) 112.5 m 
(b) 115 m 
(c) 116.5 m 
(d) None of these
Ans: 112.5 m
Sol: Maximum separation (xmax) between the police and the thief will be at time t (shown in figure)
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
i.e.,      t = 7 s
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Option (a) is correct.

Ques 13: Sachin (S) hits a ball along the ground with a speed u in a direction which makes an angle 30° with the line joining him and the fielder Prem (P). Prem runs to intercept the ball with a speed 2u/3. At what angle θ should he run to intercept the ball ?
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: If meeting time is t
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
i.e.,  sin θ = 3/4
or Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Option (c) is correct.

Ques 14: A car is travelling on a straight road. The maximum velocity the car can attain is 24 ms-1. The maximum acceleration and deceleration it can attain are 1 ms-2 and 4 ms-2 respectively. The shortest time the car takes from rest to rest in a distance of 200 m is, 
(a) 22.4 s 
(b) 30 s 
(c) 11.2s 
(d) 5.6 s
Ans: 22.4 s
Sol:  Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
= 22.36 s
Option (a) is correct.

Ques 15: A car is travelling on a road. The maximum velocity the car can attain is 24 ms 1 and the maximum deceleration is 4 ms-2. If car starts from rest and comes to rest after travelling 1032 m in the shortest time of 56 s, the maximum acceleration that the car can attain is 
(a) 6 ms-2 
(b) 1.2 ms-2 
(c) 12 ms-2 
(d) 3.6 ms-2
Ans: 1.2 ms-2 
Sol: Let BC = t'
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
⇒ t' = 6 s
∴ OB = 50 s
Let OA = t
∴ AB = 56 - t
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
⇒ t = 20 s
∴ maximum acceleration Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
= 1.2 m/s2 
Option (b) is correct.

Ques 16: Two particles are moving along two long straight lines, in the same plane with same speed equal to 20 cms-1. The angle between the two lines is 60° and their intersection point is O. At a certain moment, the two particles are located at distances 3m and 4m from O and are moving towards O. Subsequently, the shortest distance between them will be
(a) 50 cm
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans:Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: From Δ OAB
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
∴ OA2 + OB2 - AB2 = OA × OB
i.e., AB2 = OA2 + OB2 - OA × OB
= (4 - 0.2t)2 + (3 - 0.2t)2 + (4 - 0.2t) (3 - 0.2t)
or AB2 = 16 + 0.04t2 - 1.6t + 9 + 0.04t2 - 1.2t + 12 - 1.4t + 0.04t2
or   AB2 = 0.12t2 - 4.2t + 37
For AB to be minimum
0.24t - 4.2 = 0
or t = 17.5 s
∴ (AB)2min = 0.12(17.5)2 - 4.2 × (17.5) + 37
= 36.75 - 73.5 + 37
(AB)2min = 0.75 m2 
= 7500 cm2
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Option (d) is correct.

Ques 17: An elevator without a ceiling is ascending up with an acceleration of 5 ms-2. A boy on the elevator shoots a ball in vertical upward direction from a height of 2 m above the floor of elevator. At this instant the elevator is moving up with a velocity of 10 ms-1, and floor of the elevator is at a height of 50 m from the ground. The initial speed of the ball is 15 ms-1 with respect to the elevator. Consider the duration for which the ball strikes the floor of elevator in answering following questions, (g = 10 ms-2)
The time in which the ball strikes the floor of elevator is given by
(a) 2.13 s 
(b) 2.0 s 
(c) 1.0 s 
(d) 3.12 s
Ans: 2.13 s
Sol: Velocity of ball w.r.t. elevator = 15 m/s
Acceleration of ball w.r.t. elevator
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Final displacement of ball w.r.t. elevator = - 2m
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
= 2.13 s
Option (a) is correct.

Ques 18: An elevator without a ceiling is ascending up with an acceleration of 5 ms-2. A boy on the elevator shoots a ball in vertical upward direction from a height of 2 m above the floor of elevator. At this instant the elevator is moving up with a velocity of 10 ms-1, and floor of the elevator is at a height of 50 m from the ground. The initial speed of the ball is 15 ms-1 with respect to the elevator. Consider the duration for which the ball strikes the floor of elevator in answering following questions, (g = 10 ms-2)
The maximum height reached by ball, as measured from the ground would be 
(a) 73.65 m 
(b) 116.25 m 
(c) 82.56 m 
(d) 63.25 m
Ans: 82.56 m
Sol: Velocity of ball w.r.t. ground (vBE)
= (15 + 10) m/s
= 25 m/s
Now   v2 = u2 + 2as
∴ 02 = (25)2 + 2(- 10) H
or    H = 31.25 m
Maximum height by ball as measured from ground = 31.25 + 2 + 50
= 83.25 m
Option (c) is correct.

Ques 19: An elevator without a ceiling is ascending up with an acceleration of 5 ms-2. A boy on the elevator shoots a ball in vertical upward direction from a height of 2 m above the floor of elevator. At this instant the elevator is moving up with a velocity of 10 ms-1, and floor of the elevator is at a height of 50 m from the ground. The initial speed of the ball is 15 ms-1 with respect to the elevator. Consider the duration for which the ball strikes the floor of elevator in answering following questions, (g = 10 ms-2)
Displacement of ball with respect to ground during its flight would be 
(a) 16.25 m 
(b) 8.76 m 
(c) 20.24 m 
(d) 30.56 m
Ans: 30.56 m
Sol: Displacement of ball w.r.t. ground during its flight = H = 31.25 m
Option (d) is correct.

Ques 20: An elevator without a ceiling is ascending up with an acceleration of 5 ms-2. A boy on the elevator shoots a ball in vertical upward direction from a height of 2 m above the floor of elevator. At this instant the elevator is moving up with a velocity of 10 ms-1, and floor of the elevator is at a height of 50 m from the ground. The initial speed of the ball is 15 ms-1 with respect to the elevator. Consider the duration for which the ball strikes the floor of elevator in answering following questions, (g = 10 ms-2)
The maximum separation between the floor of elevator and the ball during its flight would be 
(a) 12 m 
(b) 15 m 
(c) 9.5 m 
(d) 7.5 m
Ans: 12 m
Sol: Displacement of ball w.r.t. floor of elevator at time t
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
will be maximum, when
ds/dt = 0
i.e., 15 - 15t = 0
i.e., t = 1s
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
= 12 m
Option (a) is correct.

Ques 21: A situation is shown in which two objects A and B start their motion from same point in same direction. The graph of their velocities against time is drawn. uA and uB are the initial velocities of A and B respectively. T is the time at which their velocities become equal after start of motion. You can not use the data of one question while solving another question of the same set. So all the questions are independent of each other.
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
If the value of T is 4 s, then the times after which A will meet B is 
(a) 12 s 
(b) 6 s 
(c) 8 s 
(d) data insufficient
Ans: 8 s 
Sol: Let the particles meet at time t
i.e., displacement of the particles are equal and which is possible when
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Area of Δ ACB = Area of Δ A' B' C
[Area OBCA' θO being common]
or  Area of Δ APC = Area A' P' C
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or     AP × PC = A' P' × P' C
or   (PC tan θ) PC = (P' C tan θ) P' C
or     PC = P' C'
∴OR = PC + P' C' = 2 PC = 8 s.
Option (c) is correct.

Ques 22: A situation is shown in which two objects A and B start their motion from same point in same direction. The graph of their velocities against time is drawn. uA and uB are the initial velocities of A and B respectively. T is the time at which their velocities become equal after start of motion. You can not use the data of one question while solving another question of the same set. So all the questions are independent of each other.
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Let vA and vB be the velocities of the particles A and B respectively at the moment A and B meet after start of the motion. If uA = 5 ms-1 and uB = 15 ms-1 , then the magnitude of the difference of velocities vA and vB is
(a) 5 ms-1
(b) 10 ms-1 
(c) 15 ms-1 
(d) data insufficient

Ans: 10 ms-1
Sol: Area of Δ A' B' O = Area of Δ ABO
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
i.e., vB - vA = uA - uB 
= 5 - 15 = - 10
⇒ vA - vB = 10 ms-1
Option (b) is correct.

Ques 23: A situation is shown in which two objects A and B start their motion from same point in same direction. The graph of their velocities against time is drawn. uA and uB are the initial velocities of A and B respectively. T is the time at which their velocities become equal after start of motion. You can not use the data of one question while solving another question of the same set. So all the questions are independent of each other.
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
After 10 s of the start of motion of both objects A and B, find the value of velocity of A if uA = 6 ms-1, uB = 12 ms-1 and at T velocity of A is 8 ms-1 and T = 4 s 
(a) 12 ms-1 
(b) 10 ms-1 
(c) 15 ms-1 
(d) None of these
Sol:  uA = 6 ms-1, uB = 12 ms-1 
and at t = 4 s common velocity = 8 ms-1 
To find velocity of A at t = 10 s
Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or  Motion in One Dimension: JEE Advance(Part- 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
∴ vA = 5 ms-1
Option (d) is correct.

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