A) \[\frac{\rho \operatorname{g}}{3}\]
B) \[\frac{\rho \operatorname{g}}{2}\]
C) \[\frac{\rho \operatorname{g}}{6}\]
D) None
Correct Answer: C
Solution :
Net torque above hinge is zero \[\int\limits_{\operatorname{o}}^{1/2}{\rho \operatorname{g}}\left[ \operatorname{x}+\frac{1}{2} \right].\operatorname{xdx}-\int\limits_{{}}^{1/2}{\rho \operatorname{g}}\left[ \frac{1}{2}-\operatorname{x} \right]\operatorname{xdx}=F.\frac{1}{2}\] \[\rho \operatorname{g}{{\left[ \frac{{{\operatorname{x}}^{3}}}{3}+\frac{1}{2}\frac{{{\operatorname{x}}^{2}}}{2} \right]}_{o}}^{1/2}-\rho \operatorname{g}{{\left[ \frac{{{\operatorname{x}}^{2}}}{2}-\frac{{{\operatorname{x}}^{3}}}{3} \right]}_{o}}^{1/2}=\frac{F}{2}\] On solving we get \[\operatorname{F}=\frac{\rho g}{6}\]
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