A) \[\frac{mg}{3}\]
B) \[\frac{2}{3}mg\]
C) \[\frac{mg}{2}\]
D) \[\frac{mg}{4}\]
Correct Answer: C
Solution :
 |
| Force on an elemental length dy of rod due to earth is |
| \[dF=dm\cdot g\] |
| \[=\frac{m}{R}\cdot dy\cdot gy=\frac{m}{R}\cdot dy\cdot g\left( \frac{{{R}^{2}}}{{{(R+y)}^{2}}} \right)\] |
| \[\Rightarrow \]\[dF=\frac{mg}{R}\left( \frac{1}{\left( 1+\frac{y}{R} \right)} \right)dy\] |
| Total force on complete length of rod is |
| \[F=\int_{0}^{R}{dF=\frac{mg}{R}\left( \int_{0}^{R}{\frac{dy}{\left( 1+\frac{y}{R} \right)}} \right)}=\frac{mg}{2}\] |
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