A bullet is fired vertically upwards with a velocity `upsilon` from the surface of a spherical planet when it reaches its maximum height, its acceleration due to the planet's gravity is `(1)/(4)th` of its value at the surface of the planet. If the escape velocity from the planet is `V_("escape") = upsilon sqrt(N)`, then the value of `N` is : (ignore energy loss due to atmosphere).

Answer

B

Explanation

`g=g_(0)(R^(2))/((R+h)^(2)) = (g_(0))/(4)`
`rArr 4R^(2)=(R+4)^(2)`
`2R=R+hrArr h=R`
So, `-(GMm)/(R )+(1)/(2)mV^(2)=(-GMm)/(2R)+0`
`(1)/(2)mV^(2)=(GMm)/(2R) rArr V = sqrt((GM)/(R )) = (V_(e ))/(sqrt(2))`
as `Ve = sqrt((2GM)/(R ))`
So, `Ve = sqrt(2) V`
`N =2`