Work Power & Energy

JEE Physics · 82 questions · Page 9 of 9 · Click an option or "Show Solution" to reveal answer

Q81

A particle of mass M is moving in a circle of fixed radius R in such a way that its centripetal acceleration at time t is given by n2 R t2 where n is a constant. The power delivered to the particle by the force acting on it, is :

A M n2 R2 t

B M n R2 t

C M n R2 t2

{1 \over 2}

M n2 R2 t2

Correct Answer

Option A

Solution

We know, centripetal acceleration =

{{{V^2}} \over R}

$\therefore$ According to question,

{{{V^2}} \over R}

{n^2}R{t^2}

$\Rightarrow$ V2 = n2 R2 t2 $\Rightarrow$ V = nRt $\Rightarrow$

{{dV} \over {dt}}

= nR Power (P) = Force (F) $\times$ Velocity (V) = M

{{dV} \over {dt}}

(V) = M (nR) (nRt) = Mn2R2t

Q82

A wire suspended vertically from one of its ends is stretched by attaching a weight of

200N

to the lower end. The weight stretches the wire by

1

mm.

Then the elastic energy stored in the wire is

0.2

J

10

J

20

J

0.1

J

Correct Answer

Option D

Solution

The elastic potential energy

= {1 \over 2} \times

Force $\times$ extension

= {1 \over 2} \times 200 \times 0.001 = 0.1\,J

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