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 Multiple Choice QuestionsMultiple Choice Questions

1.

All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.


C.


2.

A body of mass m is accelerated uniformly from rest to a speed v in a time T. The instantaneous power delivered to the body as a function time is given by

  • fraction numerator mv squared over denominator straight T squared. end fraction straight t
  • mv squared over straight T squared. straight t squared
  • 1 half mv squared over straight T squared. straight t
  • 1 half mv squared over straight T squared. straight t

A.

fraction numerator mv squared over denominator straight T squared. end fraction straight t
equals space ma squared straight t
equals space straight m straight V squared over straight T squared straight t
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3.

A ball is thrown from a point with a speed ν0 at an angle of projection θ. From the same point and at the same instant person starts running with a constant speed ν0/2 to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?

  • yes, 60°

  • yes, 30°

  • no

  • no


A.

yes, 60°

Man will catch the ball if the horizontal component of velocity becomes equal to the constant speed of man i.e.

straight v subscript 0 space cos space straight theta space equals space straight v subscript 0 over 2
cos space straight theta space equals space 1 half
cos space straight theta space equals space cos space 60 to the power of straight o
straight theta space equals space 60 to the power of straight o

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4.

A particle moves in a straight line with retardation proportional to its displacement. Its loss of kinetic energy for any displacement x is proportional to

  • x2

  • ex

  • x

  • x


A.

x2

In this problem acceleration (a) is given in terms of displacement (x)  to determine the velocity with respect to position or displacement we have to apply integration method.
From given information a =-kx, where a is acceleration, x is displacement and k is proportionality constant. 

vdx over dx space equals space minus space kx space open square brackets therefore space dv over dt space equals space fraction numerator begin display style dv end style over denominator dx end fraction open parentheses dx over dt close parentheses space equals space straight v dv over dx close square brackets
straight v space dv space equals space minus kx space dx
Let for any displacement from 0 to x , the velocity changes from vo to v

integral subscript straight v subscript straight o end subscript superscript straight v space vdv space equals space minus integral subscript 0 superscript straight x space kx space dx
rightwards double arrow space integral subscript 0 superscript straight x space kx space dx
rightwards double arrow space fraction numerator straight v squared minus straight v subscript 0 superscript 2 over denominator 2 end fraction space equals space minus space kx squared over 2
rightwards double arrow space straight m space open parentheses fraction numerator straight v squared minus space straight v subscript 0 superscript 2 over denominator 2 end fraction close parentheses space equals space fraction numerator negative mkx squared over denominator 2 end fraction
increment straight K space proportional to space straight x squared

493 Views