A) \[mg\,\cos \theta \]
B) \[\frac{mg\,}{\cos \theta }\]
C) \[mg\left[ \frac{1}{\cos \theta }-1 \right]\]
D) \[mg\left[ \frac{1}{\cos \theta }+1 \right]\]
Correct Answer: C
Solution :
[c] Let l be the length of the cylinder, when vertical, in water. Let A be the cross-sectional area of the cylinder. Equating weight of the cylinder with the up thrust, we get \[Mg=Al\rho g\,\,or\text{ }m=Al\rho \] When the cylinder is tilted through an angle \[\theta \], length of Cylinder in water \[=\frac{1}{\cos \theta }\] Weight of water displaced \[=\frac{l}{\cos \theta }A\rho g\] Restoring force \[=\frac{lA\rho g}{\cos \theta }-lA\rho g\] \[=lA\rho g\left[ \frac{1}{\cos \theta }-1 \right]=mg\left[ \frac{1}{\cos \theta }-1 \right]\]
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