A) 10 J
B) 20 J
C) 30 J
D) 40 J
Correct Answer: C
Solution :
\[\frac{{{m}_{2}}g}{(4{{m}_{1}}+{{m}_{2}})}\] So angular retardation \[\frac{2{{m}_{2}}g}{(4{{m}_{1}}+{{m}_{2}})}\] Now, angular speed after 2s \[\frac{2{{m}_{1}}g}{({{m}_{1}}+4{{m}_{2}})}\] Work done by torque in 2 s = loss in kinetic energy \[\frac{2{{m}_{1}}g}{({{m}_{1}}+{{m}_{2}})}\] \[F(x)=-kx+a{{x}^{3}}\] \[x\ge 0,\]
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