A) \[432\,c{{m}^{3}}/\sec \]
B) \[2160\,c{{m}^{3}}/\sec \]
C) \[180\,c{{m}^{3}}/\sec \]
D) None of these
Correct Answer: B
Solution :
Let velocity \[v=5\,cm/\sec \] (Increasing the rate/sec is called the velocity) \[\frac{da}{dt}=5\] .....(i) Where a is distance and t is time. But if a is edge of a cube, then \[V={{a}^{3}}\] Differentiating w.r.t. time t, so \[\frac{dV}{dt}=3{{a}^{2}}\frac{da}{dt}=3{{a}^{2}}.5\,=\,15{{a}^{2}}=15\times {{(12)}^{2}}\] \[=2160\,\,c{{m}^{3}}/\sec \] (\[\because \]edge \[a=12\,cm)\].
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