Previous Year Questions (2014-21): Motion in a Straight Line Notes | Study Physics Class 11 - NEET

Q.1. A small block slides down on a smooth inclined plane starting from rest at time t=0. Let Sn be the distance travelled by the block in the interval t= n-1 to t=n. Then the ratio  is:       [2021]
A:
B:
C:
D:
Ans: D
Solution:
Sn = Distance in n^th sec. 
i.e. t = n – 1 to t = n
S_{n + 1} = Distance in (n + 1)^th sec.
i.e. t = n to t = n + 1
So as we know,


Q.2. A ball is thrown vertically downward with a velocity of 20m/s from the top of a tower. It hits the ground after some time with a velocity of 80m/s. The height of the tower is : (g = 10m/s²)       [2020]
A:  320 m 
B: 300 m 
C: 360 m 
D: 340 m
Ans: B

v² – u² = 2as
(80)²  – (–20)² = 2(–10) × s
6400 – 400 = 2 × (–10) × s

s = – 300 m
Height of tower h = 300 m

Q.3. Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time t₁. On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time t₂. The time taken by her to walk up on the moving escalator will be     [2017]
A:
B:
C: t₁ – t₂ 
D:
Ans: B
Solution:
As we learnt in
Case of Relative velocity -
When A & B are moving along with straight line in opposite direction.

Relative velocity of A with respect to B is.


Q.4. A particle moves so that its position vector is given by r = cosωtx + sinωty. Where ω is a constant. Which of the following is true?
A: Velocity is perpendicular to r and acceleration is directed away from the origin.
B: Velocity and acceleration both are perpendicular to r
C: Velocity is acceleration both are parallel to r
D: Velocity is perpendicular to r and acceleration is directed towards the origin.    [2016]
Ans: D
Solution:


Q.5. If the velocity of a particle is  , where A and B are constants, then the distance travelled by it between 1s and 2s is    [2016]
A: A/2 + B/3
B: 3A/2 + B
C: 3A + 7B
D: 3A/2 + 7B/3
Ans: D
Solution:


Q.6. A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to v(x) = βx^-2n, where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x, is given by :    [2015]
A: -2nβ² e^{-4n + 1} 
B: -2nβ² x^{-2n - 1} 
C: -2nβ² x^{-4n - 1} 
D: -2nβ² x^{-2n + 1} 
Ans: C
Solution: