Question

A particle is moving with a uniform speed in a circular orbit of radius R and a centripetal force which is directly proportional to the nth power of R acts in the particle. If the time period is T, and T² proportional to R^5/2, find n.​

Answer 1

Hope this helps you.....

Refer to the ⬆️ above attachment.....

Answer 2

Given that,

$\sf F_c \propto R^n$

$\longrightarrow \sf F_c = k \times R^n$

$\longrightarrow \sf \dfrac{mv^2}{R} = k \times R^n$

$\longrightarrow \sf v^2 = \dfrac{k}{m} \times R^{(n + 1)}$

$\longrightarrow \sf v = \bigg(\dfrac{k}{m}\bigg)^{ \frac{1}{2}} \times (R^{n+1})^{ \frac{1}{2} }$

Let $\sf \bigg(\dfrac{k}{m}\bigg)^{\frac{1}{2}}$ be $\sf a$ (as this value is of no use in further steps). And $\sf a$ is also a constant.

$\longrightarrow \sf v = a \times R^{\big( \scriptsize\dfrac{n+1}{2}\big)}$

$\sf Time\:\:period = T = \dfrac{2\pi R}{V}$

$\longrightarrow \sf T = \Bigg[\dfrac{2\pi R}{\bigg(a \times R^{\big( \scriptsize\dfrac{n+1}{2}\big)}\bigg) }\Bigg]$

Let $\sf \dfrac{2\pi}{a}$ be $\sf b.$

$\longrightarrow \sf T = b \times \dfrac{R}{ R^{\big( \scriptsize\dfrac{n+1}{2}\big)}}$

$\longrightarrow \sf T = b \times R^{\bigg[ 1 - \big(\dfrac{n+1}{2}\big)\bigg]}$

$\longrightarrow \sf T = b \times R^{\bigg[ 1 - \dfrac{1}{2} - \dfrac{n}{2}\bigg]}$

$\longrightarrow \sf T = b \times R^{\bigg( \dfrac{1-n}{2}\bigg)}$

$\longrightarrow \sf T \propto R^{\bigg( \dfrac{1-n}{2}\bigg)}$

$\longrightarrow \sf T^2 \propto R^{(1-n)}\quad\quad \dots (1)$

But, given that,

$\sf T^2 \propto R^{\scriptsize \dfrac{5}{2}}\quad\quad \dots (2)$

From (1) and (2),

$\sf 1 - n = \dfrac{5}{2}$

$\longrightarrow \sf n = 1 - \dfrac{5}{2}$

$\longrightarrow \underline{\underline{\sf{\green{n = -\dfrac{3}{2}}}}}$