Gravitation

JEE Main level, single correct, medium

28 of 223

Question 28 of 223

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the nth{n}^{th} power of R. If the period of rotation of the particle is T, then:

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Answer: (D) T∝Rn+12T\propto {R}^{\dfrac{n+1}{2}}

F=KRnF=\dfrac{K}{{R}^{n}}⇒mV2r=KRn\Rightarrow \dfrac{m{V}^{2}}{r}=\dfrac{K}{{R}^{n}}⇒V2∝KRn−1\Rightarrow {V}^{2}\propto \dfrac{K}{{R}^{n-1}}
⇒V∝1Rn−12\Rightarrow V\propto \dfrac{1}{{R}^{\dfrac{n-1}{2}}}⇒T=2πRV∝RV\Rightarrow T=\dfrac{2\pi R}{V}\propto \dfrac{R}{V}
⇒T∝R1Rn−12\Rightarrow T\propto \dfrac{R}{\dfrac{1}{{R}^{\dfrac{n-1}{2}}}}⇒T∝Rn+12\Rightarrow T\propto {R}^{\dfrac{n+1}{2}}

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