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Physics Quiz-8

Dear Readers,

If an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. Motion in a straight line is nothing but linear motion. As the name suggests, it’s in a particular straight line, thus it can be said that it uses only one dimension.


Q1. A point moves with uniform acceleration and  v_1,v_2, and v_3 denote the average velocities in the three successive intervals of time t_1,t_2, and t_3. Which of the following relations is correct?
  •  (v_1-v_2 ):(v_2-v_3 )=(t_1-t_2 ):(t_2+t_3)
  •  (v_1-v_2 ):(v_2-v_3 )=(t_1+t_2 ):(t_2+t_3)
  •  (v_1-v_2 ):(v_2-v_3 )=(t_1-t_2 ):(t_1-t_3)
  •  (v_1-v_2 ):(v_2-v_3 )=(t_1-t_2 ):(t_2-t_3)

Solution
Suppose u be the initial velocity,
Velocity after time t_1:v_11=u+at_1
Velocity after time t_1+t_2:v_22=u+a(t_1+t_2)
Velocity after time t_1+t_2+t_3:
v_33=u+a(t_1+t_2+t_3)
Now v_1=(u+v_11)/2=(u+u+at_1)/2=u+1/2 at_1
v_2=(v_11+v_22)/2=u+at_1+1/2 at_2
v_3=(v_22+v_33)/2=u+at_1+at_2+1/2 at_3
So v_1-v_2=-1/2 a(t_1+t_2)
v_2-v_3=-1/2 a(t_2+t_3)
(v_1-v_2 ):(v_2-v_3 )=(t_1+t_2 ):(t_2+t_3)


Q2.If the displacement of a particle is zero, the distance covered.
  •  Must be zero
  •  May or may not be zero
  •  Cannot be zero
  •  Depends upon the particle
Solution
If the displacement is zero, the distance moved may or may not be zero. For example, if a particle returns to its initial position after moving through a distance, displacement will be zero but distance covered will not be zero.

Q3. A particle slides from rest from the topmost point of a vertical circle of radius r along a smooth chord making an angle θ with the vertical. The time of descent is
  •  Least for θ=0
  •  Maximum for θ=0
  •  Least for θ=45°
  •  Independent of θ
Solution
a=g sin⁡〖a=g sin⁡〖(90°-θ)〗 〗
=g cos⁡θ
l=2R cos⁡θ











Now using s=ut+1/2 at^2, we get
l=0t+1/2 g cos⁡〖θt^2 〗
⇒2R cos⁡〖θ=1/2 g cos⁡〖θt^2 〗 〗⇒t=2 √(R/g)
This is independent of θ


Q4. Which of the following velocity-time graphs is not possible practically?
  •  

  •  

  •  

  •  

Solution
In option (a), there is some part of the graph which is perpendicular to t axis. It indicates infinite acceleration, which is not possible practically.

Q5.The velocity-time graph of a particle moving in a straight line is shown in Fig. The acceleration of the particle at t=9 s is














  •  Zero
  •  5 ms^(-2)
  •  -5 ms^(-2)
  •  -2 ms^(-2)
Solution
Acceleration between 8 to 10 s (or at t=9 s):
a=(v_2-v_1)/(t_2-t_1 )=(5-15)/(10-8)=-5 ms^(-2)

Q6. 

        A particle is dropped from rest from a large height. Assume g to be constant throughout the motion. The time taken by it to fall through successive distance of 1 m each will be

  •  All equal, being equal to √(2/g) second
  •  In the ratio of the square roots of the integers 1, 2, 3,…,
  •  In the ratio of the difference in the square root of the integers, i.e., √1,(√2-√1),(√3-√2),(√4-            √3),……
  •  In the ratio of the reciprocals of the square roots of the integers, i.e., 1/√1,1/√2,1/√3,…
Solution
Time taken to cover first n metre is given by
n=1/2 gt_n^2 or t_n=√(2n/g)
Time taken to cover first (n+1) m is give by
t_(n+1)=√((2(n+1))/g)
So time taken to cover (n+1)^th m is given by
t_(n+1)-t_n=√((2(n+1))/g)-√(2n/g)=√(2/g) [√(n+1)-√n]
This given the required ratio as
√1,(√2-√1),(√3-√2),…etc (starting from n=0)

Q7.
The velocity-time relation of an electron starting from rest is given by v=kt where k=2 ms^(-2). The distance traversed in first 3 s is
  •  9 m
  •  16 m
  •  27 m
  •  36 m
Solution
Given that u=0 (the electron starts from rest), at any time t:v=kt=2t
a=dv/dt=2 ms^(-2)  (constant acceleration)
Now s=ut+1/2 at^2=0×3+1/2×2×(3)^2=9 m

Q8. The acceleration versus the graph of a particle moving in a straight line is shown in Fig. The velocity-time graph of the particle would be













  •  A straight line
  •  A parabola
  •  A circle
  •  An ellipse
Solution
a=-4/2 t+4 ⇒a=-2t+4
v=∫▒〖a dt+C〗=∫▒〖(-2t+4)dt+C〗=-t^2+4t+C
Hence, graph will be parabolic

Q9.

    A person is throwing two balls in the air one after the other. He throws the second ball when the first ball is at the highest point. If he is throwing the balls every second, how high do they rise?


  •  5 m
  •  3.75 m
  •  2.50 m
  •  1.25 m

  • Solution
    Time of ascent =1 s ⇒u/g=1⇒u=10 ms^(-1)
    Maximum height attained =u^2/2g=10^2/(2×10)=5 m
Q10. B_1,B_2 and B_3, are three balloons ascending with velocities v,2v, and 3v, respectively. If a bomb is dropped from each when they are at the same height, then
  •  Bomb from B_1 reaches ground first
  •  Bomb from B_2 reaches ground first
  •  Bomb from B_3 reaches ground first
  •  They reach the ground simultaneously 
Solution
Bomb B_1 will have less velocity upward on dropping, so it will reach ground first


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