A body falls freely for 10 s. Its average velocity during this journey is (Take, g = 10 ms-2)

  • 100 ms-1

  • 10 ms-1

  • 50 ms-1

  • 5 ms-1


C.

50 ms-1

Average velocity = Total DisplacementTime interval

In 10 s, the distance covered by the body

          s = 12 gt2Here,  t = 10 s          g = 10 ms-2      s = 12 10 × 102 = 500 m       uav = 50010       uav = 50 ms-1


From the top of a tower a stone is thrown up which reaches the ground in a time t1. A second stone thrown down with the same speed reaches the ground in a time t2. A third stone released from rest from the same between reaches the ground in a time t3. Then

  • 1t3 = 1t2 - 1t1

  • t32 = t12 - t22

  • t3 = t1 + t22

  • t3 = t1 t2


D.

t3 = t1 t2

When stone is thrown up

      h = ut1 + 12gt12                                ....... (i)

When thrown down,

     h = ut2 + 12 gt22                            ....... (ii)

When released

               h = 12 gt32            ht2 = - ut1t2 + 12 gt1 2t2            ht1 = + ut1t2 + 12 gt22 t1h t2 + t1 = + 12 gt1t2 t1 + t2              h = 12 gt1t2                                        ...... (iv)Comparing (iii) and (iv)             t3 = t1t2


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Two identical charged spheres suspended from a  common point by two massless strings of lengths l, are initially at a distance d (d < l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then, v varies as a function of the distance x between the sphere, as

  • vx

  • vx-1


B.

Two identical charged spheres suspended from a common point by two massless strings of length L.



In △ABC,





The charge begins to leak from both the sphere at a constant rate. As a result, the spheres approach each other with velocity v.

Therefore, equation (i) can be rewritten as,



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If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is,

  • 90o

  • 45o

  • 180o

  • 0o


A.

90o

There are two vectors P and Q.
It is given that,


Let, angle between P and Q is  ,


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The velocity-time graph for two bodies A and B are shown in figure. Then, the acceleration of Aand B are in the ratio

                

  • sin 25° to sin 50°

  • tan 25° to tan 40°

  • cos 25° to cos 50°

  • tan 25° to tan 50°


D.

tan 25° to tan 50°

Acceleration deals with a change in speed magnitude, a change in direction or change in both.

Acceleration represents the slope of velocity-time graph

  aAaB = tan θ1tan θ2   aAaB = tan 25°tan 50°


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