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  • A star suddenly shrinks and its density becomes times its orginal value. The value of acceleration due to gra

A star suddenly shrinks and its density becomes 10^{9} times its orginal value. The value of acceleration due to gravity on its surface will increase by a factor of

Option: 1

10^{-9}



 


Option: 2

10^{-6}
 


Option: 3

10^6


Option: 4

10^9


Answers (1)

best_answer

\mathrm{g}=\frac{\mathrm{GM}}{\mathrm{R}^2}=\frac{\mathrm{GV} \rho}{\mathrm{R}^2}=\mathrm{G} \frac{4}{3} \pi \frac{\mathrm{R}^3 \rho}{\mathrm{R}^2}=\mathrm{G} \frac{4}{3} \pi R \rho

\mathrm{ \mathrm{M}=\frac{4}{3} \pi R^3 \rho=\frac{4}{3} \pi R^{\prime 3} \rho^{\prime}=\frac{4}{3} \pi R^{\prime 3} \times 10^9 \rho }

\mathrm{ \left(\mathrm{R}^{\prime} \times 10^3\right)^3=\mathrm{R}^3 ; \mathrm{R}^{\prime}=\frac{\mathrm{R}}{1000}=\mathrm{R} \times 10^{-3} }

\mathrm{ \therefore \frac{\mathrm{g}^{\prime}}{\mathrm{g}}=\frac{\mathrm{G} \frac{4}{3} \pi \mathrm{R}^{\prime} \rho^{\prime}}{\mathrm{G} \frac{4}{3} \pi R \rho}=\frac{\mathrm{R} \times 10^{-3}}{\mathrm{R}} \times \frac{10^9 \rho}{1(\rho)}=10^6 }

Hence option 3 is correct.


 

Posted by

vishal kumar

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