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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

 

d2rdt2=-43πGDr

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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