A) during ascent decreases
B) during decent decreases
C) during ascent increases
D) remain same
Correct Answer: D
Solution :
Force mg is acting at c.m. Hence torque about c.m. zero Hence \[\frac{dy}{dx}=\frac{2}{3}\,{{x}^{-1/3}}+\frac{1}{3}\,{{x}^{-2/3}}\,=\frac{\left( 2{{x}^{1/3}}+1 \right)}{3{{x}^{2}}3}\] constant Hence \[\frac{dy}{dx}=0;\,\,{{x}^{1/3}}\,=\frac{-1}{2}\] remain same
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