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Q.

When a hole is drilled along the diameter of the earth and if a body is dropped into it, it executes SHM with a period T equal to (R&D are radius and density of the earth)

a) $\large 2\pi \sqrt {\frac{R}{g}}$ $\large b)\,\,\sqrt {\frac{{3\pi }}{{GD}}}$

$\large c)\,\,2\pi \sqrt {\frac{\pi }{{GD}}}$ $\large d)\,\,2\pi \sqrt {\frac{R}{{2g}}}$

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a

a and b are true

b

c and d correct

c

a and c are true

d

b and c are true

answer is A.

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Detailed Solution

At a radial distance r,

\large g\, = \,\frac{4}{3}\pi DG.r\,

\large \therefore \,m\frac{{{d^2}r}}{{d{t^2}}}\, = \, - mg\, = \, - m \times \frac{4}{3}\pi DG.r

$\frac{d^{2} r}{d t^{2}} = - \{\frac{4}{3} π G D\} r$

This equation represents SHM. whose time period is

\large T\, = \,2\pi \sqrt {\frac{3}{{4\pi DG}}} \, = \,\sqrt {\frac{{3\pi }}{{GD}}}

Also, at a radial distance,

\large m\frac{{{d^2}r}}{{d{t^2}}}\, = \, - mg.\frac{r}{R}\, \Rightarrow \,\frac{{{d^2}r}}{{d{t^2}}}\, = \, - \frac{g}{R}.r

\large \therefore

The above equation represents SHM whose time period is

\large T\, = \,\frac{{2\pi }}{{\sqrt {\frac{g}{R}} \,}}\, = \,2\pi \sqrt {\frac{R}{g}}

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