A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration, if the kinetic energy of the particle becomes equal to 8 x 10-4 J by the end of the second revolution after the beginning of the motion?

  • 0.15 m/s2

  • 0.18 m/s2

  • 0.2 m/s2

  • 0.2 m/s2


D.

0.2 m/s2

Given, mass of particle. m = 0.01 kg

Radius of circle along which particle is moving , r = 6.4 cm

Kinetic energy of particle, K.E. = 8 x 10-4 J



Given that, KE of particle is equal to 8 x 10-4 J by the end of second revolution after the beginning of the motion of particle.

It means, initial velocity (u) is 0 m/s at this moment.

Now, using the Newton's 3rd equation of motion,

v2 = u2 + 2as

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If the velocity of a particle is v = At + Bt2, where A and B are constants, then the distance travelled by it between 1 s and 2 s is,

  • 3A + 7B


B.

Velocity of the particles is given as,
v = At + Bt2, where a and B are constants.


Integrating both sides, we get


Therefore, distance travelled between 1 s and 2 s is,

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A particle  moves so that its position vector is given by r = cos   where   is a constant. Which of the following is true?

  • velocity and acceleration both are parallel to r

  • velocity is perpendicular to r and acceleration is directed towards the origin

  • velocity is perpendicular to r and acceleration is directed away from the origin

  • velocity and acceleration both are perpendicular to r


B.

velocity is perpendicular to r and acceleration is directed towards the origin

Position vector of the article is given by,

where  is a constant.
Velocity of the particle is,



The particle is at a point P. That is, its position vector is directed as shown below:

Therefore, acceleration is directed towards -r, that is towards "O"

We have

 

        


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A particle of unit mass undergoes one-dimensional motion such that its velocity according to
V(x) =  βx-2n where β and n are constants and x is the position of the particle. The acceleration of the particle as a function of x is given by 

  • -2nβ2 x-2n-1

  • -2nβx-4n-1

  • -2β x-2n+1

  • -2β x-2n+1


B.

-2nβx-4n-1

Given, v = βx-2n

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What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

  • square root of 2 gR end root
  • square root of 3 gR end root
  • square root of 5 gR end root
  • square root of 5 gR end root


C.

square root of 5 gR end root

The question is illustrated in the figure below,

Let, the tension at point A be TA.

Using Newton's second law, we have


Energy at point A = 
Energy at point C is,


At point C, using Newton's second law,


In order to complete a loop, Tc 
So,
 

From equation (i) and (ii)

Using the principle of conservation of energy,

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