## Physics Quiz-1

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 train 100 m long travelling at 40 ms^(-1) starts overtaking another train 200 m long travelling at 30 ms^(-1). The time taken by the first train to pass the second train completely is
•  30s
•  40s
•  50s
•  60s
Solution
Relative velocity of overtaking =40-30=10 ms^(-1). Total relative distance covered with this relative velocity during overtaking =100+200=300 m
So time taken =300/10=30 s

Q2.Water drops fall from a tap on the floor 5 m below at regular intervals of time, the first drop striking the floor when the fifth drop begins to fall. The height at which the third drop will be from ground (at the instant when the first drop strikes the ground), will be (g =10 ms^-2)
•  1.25m
•  2.15m
•  2.75m
•  3.75m
Solution
By the time 5th water drop starts falling, the first water drop reaches the ground
u=0,h=1/2 gt^2  ⇒5=1/2×10×t^2⇒t=1 s
Hence, the interval of falling of each water drop is "1s/4 = 0.25 s"
When the 5th drop starts its journey towards ground, the third drop travels in air for
0.25+0.25=0.5
Therefore, height (distance) covered by 3rd drop in air is
h_1=1/2 gt^2=1/2×10×(0.5)^2=5×0.25=1.25 m
The third water drop will be at a height of
5-1.25=3.75 m

Q3.  The position  of a particle varies with time(t) as x=at^2-bt^3. The acceleration at time "t" of the particle will be equal to zero, where "t" is equal to
•  2a/3b
•  a/b
•  a/3b
•  Zero
Solution
x= at^2-bt^3
Velocity = dx/dt=2at-3bt^2
and acceleration = (d^2 x) / (dt^2 ) = 2a-6bt
Acceleration will be zero if
2a-6bt = 0 ⇒ t = 2a/6b = a/3b

Q4.
An object accelerates from rest to a velocity 27.5 ms^-1 in 10 s. Find the distance covered by the object during the next 10 s
• 412.5 m
•  137.5 m
•  550 m
•  275 m
Solution

Q5.Two balls are dropped from the top of a high tower with a time interval of  "t" second, where "t" is smaller than the time taken by the first ball to reach the floor, which is perfectly inelastic. The distance "s"   between the two balls, plotted against the time lapse "t1"   from the instant of dropping the second ball is best represented by
•

•

•

Solution
Before the second ball is dropped, the first ball would have travelled some distance say So=1/2g(to)^2
. After dropping the second ball, the relative acceleration of both balls becomes zero. So distance between them increases linearly. After some time, the first ball will collide with the ground and the distance between them will start decreasing and the magnitude of relative velocity will be increasing for this time. Option (d) represents all these clearly

Q6.
 A body sliding on a smooth inclined plane requires 4 s to reach the bottom, starting from rest at the top. How much time does it take to cover one-fourth the distance starting from rest at the top?
•  1s
•  2s
• 4s
•  16s
Solution
When a body slides on an inclined plane, component of weight along the plane produces an acceleration
a=(mg sin⁡Î¸)/m=g sin⁡Î¸=constant
If s is the length of the inclined plane, then
s= 0+1/2 at^2 = 1/2 g sin⁡Î¸×t^2
s'^/s= t'^2/t^2  or s/(s')^' = t^2/(t^')^2
Given  t=4 s and s^'=s/4
t^'=t√(s^'/s)=4√(s/4s)=4/2=2 s

Q7. A ball is thrown from the top of a tower in vertically upward direction. Velocity at a point "h"   metre below the point of projection is twice of the velocity at a point "h" metre about the point of projection. Find the maximum height reached by the ball above the top of tower
•  2h
•  3h
•  (5/3)h
•  (4/3)h
Solution

H=u^2/2g; given v_2=2v_1   …(i)
A to B: v_1^2=u^2-2gh      …(ii)
A to C: v_2^2=u^2-2g(-h)      …(iii)
Solving (i), (ii) and (iii), we get the value of u^2 as 10gh/3 and then we get the value of H by using H=u^2/2g

Q8. Drops of water fall at regular intervals from roof of a building of height "H=16m", the first drop striking the ground at the same moment as the fifth drop detaches itself from the roof. The distance between separate drops in air as the first drop reaches the ground are
•  1m,5m,7m,3m
•  1m,3m,5m,7m
•  1m,3m,7m,5m
•  None of the above
Solution
S1+S2+S3+S4 = 16 m,S1:S2:S3:S4=1:3:5:7
Solve to get S1=1 m,S2=3 m,S3=5 m,S4=7 m

Q9.
 Which of the following velocity-time graphs shows a realistic situation for a body in motion?
•

•

•

•

Solution
In options (a), (c) and (d), we can find from the graphs that more than one velocity can be possible at a single time, which is not possible practically

Q10. A drunkard is walking along a straight road. He takes 5 steps forward and 3 steps backward and so on. Each step is 1 m long and takes 1s. There is a pit on the road 11 m away from the starting point. The drunkard will fall into the pit after
•  29s
•  21s
•  37s
•  31s
Solution
Since the last five steps covering 5 m land the drunkard fell into the pit, the displacement prior to this as (11-5) m=6 m
Time taken for first eight steps (displacement in first eight steps =5-3=2 m) =8 s. Then time taken to cover first 6 m of journey =6/2×8=24 s
Time taken to cover last 5 m = 5 s
Total time =24+5=29 s

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