Assume that a frictionless tunnel is made in the earth along its diameter, a particle in projected in the tunnel from the surface of the earth with an initial speed `v_(0)=sqrt(gR)`, where g is the acceleration due to gravity on the earth's surface & R is the radius of the earth, if time taken by the particle to reach the centre of the earth is `(pi)/(n)sqrt((4R)/(g))` the value of `n` is?

Answer

8

Explanation

From conservation of energy
`(1)/(2)m(gR)-(GMm)/(R)=(1)/(2)mv^(2)-(3GMm)/(2R)`
`implies(mgR)/(2)+(GMm)/(2R)=(1)/(2)mv^(2)`
`impliesV=sqrt(2gR)`
`(g=(GM)/(R^(2)))`
Let at `t=0` the particle is at centre of the earth `v=v_(0)cosomegat`
`impliessqrt(gR)=sqrt(2gR)cosomegaimpliesomegat=(pi)/(4)`
`impliest=(pi)/(4omega)=(pi)/(4)sqrt((R)/(g))=(pi)/(8)sqrt((4R)/(g))`
`becausen=8`