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← JEE Advanced 2017 · Paper 2
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JEE Advanced 2017 · Paper 2, Physics question 1

Consider an expanding sphere of instantaneous radius RR whose total mass remains constant. The expansion is such that the instantaneous density ρ\rho remains uniform throughout the volume. The rate of fractional change in density (1ρdρdt)\left(\frac{1}{\rho}\frac{d\rho}{dt}\right) is constant. The velocity vv of any point on the surface of the expanding sphere is proportional to

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Answer and worked solution

Official answer: A

Constant mass. M=43πR3ρM=\tfrac43\pi R^3\rho is fixed, so
1ρdρdt=−3RdRdt.\frac1\rho\frac{d\rho}{dt}=-\frac3R\frac{dR}{dt}.

Rate condition. The left side is a constant, so dRdt=v∝R\dfrac{dR}{dt}=v\propto R.

(A)