A) 24
B) 12
C) 6\[\sqrt{2}\]
D) 3
Correct Answer: C
Solution :
If d is the density of the material of earth, then \[g=\frac{GM}{{{R}^{2}}}=\frac{G.\frac{4}{3}\pi {{R}^{3}}d}{{{R}^{2}}}\] \[=\frac{4}{3}G\pi Rd\] or \[g\propto d\] Now, \[T=2\pi \sqrt{\frac{l}{g}}\] \[\Rightarrow \] \[T=2\pi \sqrt{\frac{l}{d}}\] or \[T\propto \frac{1}{\sqrt{d}}\] \[\therefore \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{{{d}_{2}}}{{{d}_{1}}}}\] Given, \[{{d}_{1}}=d,\,{{d}_{2}}=8d\] \[\Rightarrow \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=\sqrt{\frac{8d}{d}}\] \[\Rightarrow \] \[\frac{{{T}_{1}}}{{{T}_{2}}}=2\sqrt{2}\] \[\Rightarrow \] \[{{T}_{2}}=\frac{{{T}_{1}}}{2\sqrt{2}}=\frac{24}{2\sqrt{2}}\] \[=6\sqrt{2}\,h\]
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