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Physics Quiz-10

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 person travels along a straight road for the first half of total time with a velocity v_1 and the second half of total time with a velocity v_2. Then the mean velocity v ̅ is give by
  •  v ̅=(v_1+v_2)/2
  •  2/v ̅ =1/v_1 +1/v_2 
  •  v ̅=√(v_1 v_2 )
  •  v ̅=√(v_2/v_1 )
Solution
V_av=(v_1 (t/2)+v_2 (t/2))/t=(v_1+v_2  )/2

Q2.The acceleration versus time graph of a particle is shown in Fig. The respective v-t graph of the particle is











  •  

  •  

  •  

  •  

Solution
From 0 to t_1, acceleration is increasing linearly with time; hence, v-t graph should be parabolic upwards. From t_1 to t_2, acceleration is decreasing linearly with time; hence, the v-t graph should be parabolic downwards

Q3. A police party is chasing a dacoit in a jeep which is moving at a constant speed v. The dacoit is on a motorcycle. When he is at a distance x from the jeep, he accelerates from rest at a constant rate? Which of the following relations is true if the police is able to catch the dacoit?
  •  v^2≤αx
  •  v^2≤2αx
  •  v^2≥2αx
  •  v^2≥αx
Solution
If police is able to catch the dacoit after time t, then
vt=x+1/2 αt^2. This gives α/2 t^2-vt+x=0
or t=(v±√(v^2-2αx))/α
For t to be real, v^2≥2αx

Q4. The displacement-time graph of two bodies A and B is shown in Fig. The ratio of the velocity of A (v_A) to the velocity of B (v_B) is









  •  1/√3
  • √3
  •  1/3
  •  3
Solution
We know that slope of displacement-time graph is equal to velocity. So v_A=tan⁡〖30°=1/√3〗,
v_B=tan⁡〖60°=√3〗 Hence, v_A/v_B=1/3

Q5.An object is vertically thrown upwards. Then the displacement-time graph for the motion is as shown         in
  •  

  •  

  •  

  •  

Solution
Let the particle be thrown up with initial velocity u
Displacement (s) at any time t is S=ut-1/2 gt^2
The graph should be parabolic downwards as shown in option (b)

Q6.  

        Taxies leave station X for station Y every 10 min. Simultaneously, a taxi also leaves station Y for              station X every 10 min. The taxies move at the same constant speed and go from X and Y or vice-              versa in 2 h. How many taxies coming from the other side will meet each taxi enroute from Y and X?

  •  24
  •  23
  •  12
  •  11
Solution
Because one taxi leaves every 10 min, at any instant there will be 11 taxies on the way towards each station, one will be arriving and another leaving the other station. Figure shows the location of taxies going from X and Y at the instant 2.00 PM. The taxi which leaves the station X at 0.00 PM has just arrived at the station Y. Consider the taxi leaving the station Y at 2.00 PM







It will meet all the 11 taxies marked 1 to 11 as well as 12 other taxies which would leave the station X from 2.00 PM to 3.50 PM. When it arrives at the station X at 4.00 PM, there will be one more taxi leaving that station. However, it will not be counted among the taxies crossed by taxi under consideration. That is, it will cross 23 taxies leaving the station X from 0.10 PM to 3.50 PM

Q7.
The displacement s of a particle is proportional to the first power of time t, i.e.,s∝t; then the acceleration of the particle is
  • Infinite
  • Zero
  •  A small finite value
  •  A large finite value
Solution
s=kt. Differentiating s twice to get acceleration, we see that acceleration comes out to be zero.

Q8. A moving car possesses average velocities of  5 ms^(-1),10 ms^(-1) and 15 ms^(-1) in the first, second, and third seconds, respectively. What is the total distance covered by the car in these 3 s?
  •  15 m
  • 30 m
  • 55 m
  •  None of these
Solution
Distance covered =S=v_av× time
For first second: S_1=5×1=5 m
For second: S_2=10×1=10 m
For third second: S_3=15×=15 m
Total distance travelled
S=S_1+S_2+S_3=5+10+15=30 m

Q9.

    A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. The body has fallen for a time of


  •  3 s
  •  5 s
  •  7 s
  •  9 s
    Solution
    Distance covered in first three seconds = Distance covered in last second, i.e.,
    1/2 g(3)^2=g/2 (2t-1)⇒t=5s

Q10. A person travels along a straight road for half the distance with velocity v_1 and the remaining half distance with velocity v_2. Then average velocity is given by
  •  v_1 v_2
  • (v_2^2)/(v_1^2 )
  •  ((v_1+v_2))/2
  • (2v_1 v_2)/((v_1+v_2))
Solution
t_1=(S/2)/v_1 ,t_2  (S/2)/v_2 ,v_av=S/(t_1+t_2 )=(2v_1 v_2)/(v_1+v_2 )


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