JEE MAIN - Physics (2016 - 10th April Morning Slot - No. 17)
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 :
M n2 R2 t
M n R2 t
M n R2 t2
$${1 \over 2}$$ M n2 R2 t2
설명
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
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