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A diametrical tunnel is dug across the earth. A ball dropped into the tunnel from one side. The velocity of the ball when it reaches the centre of the earth is [Given: gravitational potential at the centre of earth `= -3//(2GM//R)`]A. `sqrt(R)`B. `sqrt(gR)`C. `sqrt(2.5gR)`D. `sqrt(7.1gR)` |
Answer» Correct Answer - B Gravitational potential at a point on the surface of earth is `-(GM)/R` If the earth is assumed to be a solid sphere, then the gravitational potential at the centre of the earth is `(3/(2GM)//R)`. Decrease in gravitational potential is `R/2xx(GM)/(R^(2))=(Rg)/2` Loss is potential energy is `R/2xx(GM)/(R^(2))xxm` Now gain in kinetic energy `=` loss is potnetial energy. Therefore. `1/2mv^(2)=1/2mgR` or `v=sqrt(gR)` |
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