Question

A particle moves along a circle of a fixed radius with a variable centripetal acceleration $a_n=k{t}^n$, where $k$ is a constant and $t$ is time. If slope of $P-t$ graph on log-log scale is $2$, then value of $n$ is

(Here, $P$ is instantaneous power)

• A. $1$
• B. $2$
• C. $-1$
• D. $3$

Answer 1

The correct option is D $3$

$a_n=\frac{v^2}{r}=k{t}^n\Rightarrow v=\sqrt{kr{t}^n}$

Power, $P=\left(m\frac{dv}{dt}\right)v=\left(m\sqrt{kr{t}^n}\right)\left(\sqrt{kr}\right)\left(\frac{n}{2}\right)t^{\frac{n}{2}-1}=\left(\frac{mkm}{2}\right)t^{n-1}$

$\Rightarrow \log P=\log \left(\frac{mkm}{2}\right)+(n-1)\log t$

Slope $=n-1=2$ $\Rightarrow$ $n=3$