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The acceleration due to gravity on the s...

The acceleration due to gravity on the surface of a pulsar of mass `M = 1 . 98 xx 10^(30)` kg and radius R = 12 km rotating with time period T =0 . 0 41 seconds is ` x xx 10^(11) m s^(-2) ` where the value of x is _____
`(G = 6 . 67 xx 10^(-11) MKS)`

A

`9.2xx10^(11)m//s^2`

B

`8.15xx10^(11)m//s^2`

C

`7.32xx10^(11)m//s^2`

D

`6.98xx10^(11)m//s^2`

Text Solution

Verified by Experts

The correct Answer is:
A
`g=(GM)/(R^2)=(6.67xx10^(-11)xx1.98xx10^(30))/((12xx10^3)^2)`
`~~9.2xx10^(11)m//s^2`
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