If a tunnel is cut at any orientation through | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$g_{d}=g\left(1-\frac{d}{R}\right)$

$g_{d}=\frac{g}{R} x$

$F_{R}=-m g_{d} \Rightarrow F_{R}=-\frac{m g}{R} x$

$T=2 \pi \sqrt{\frac{m R}{m g}}

Vedclass · Accepted Answer

$42.3$

Vedclass · Answer

$g_{d}=g\left(1-\frac{d}{R}\right)$

$g_{d}=\frac{g}{R} x$

$F_{R}=-m g_{d} \Rightarrow F_{R}=-\frac{m g}{R} x$

$T=2 \pi \sqrt{\frac{m R}{m g}} \Rightarrow T=2 \pi \sqrt{\frac{R}{g}}$

$T=84.6 \mathrm{min}$

$t=\frac{T}{2} \Rightarrow t=42.3 \mathrm{min}$

If a tunnel is cut at any orientation through earth, then a ball released from one end will reach the other end in time ........ $\min$ (neglect earth rotation)

Similar Questions

What is weightlessness ? Explain with example.

The acceleration of a body due to the attraction of the earth (radius $R$) at a distance $2 \,R$ from the surface of the earth is ($g =$ acceleration due to gravity at the surface of the earth)

Let $g$ be the acceleration due to gravity at earth's surface and $K$ be the rotational kinetic energy of the earth. Suppose the earth's radius decreases by $2\%$ keeping all other quantities same, then

If the radius of the earth be increased by a factor of $5,$ by what factor its density be changed to keep the value of $g$ the same ?

Average density of the earth