Laws of Motion

JEE Main level, single correct, medium

100 of 210

Question 100 of 210

A particle is moving with uniform speed along the circumference of a circle of radius RR under the action of a central fictitious force FF which is inversely proportional to R3{R}^{3}. Its time period of revolution will be given by :

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Answer: (D) T∝R2T\propto {R}^{2}

F∝1R3F\propto \dfrac{1}{{R}^{3}}
KR3=mω2R\dfrac{K}{{R}^{3}}=m{\omega }^{2}R
ω2=Km×1R4{\omega }^{2}=\dfrac{K}{m}\times \dfrac{1}{{R}^{4}}
(2πT)2=Km×1R4{\left(\dfrac{2\pi }{T}\right)}^{2}=\dfrac{K}{m}\times \dfrac{1}{{R}^{4}}
T2∝R4{T}^{2}\propto {R}^{4}
T∝R2T\propto {R}^{2}

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