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If the volume of an expanding cube is increasing at the rate of $4cm^{3}/sec $ then the rate of change of surface area when the volume of the cube is $8$ cubic cm is

\[\begin{array}{1 1}(1)8cm^{2}/sec&(2)16cm^{2}/sec\\(3)2cm^{2}/sec&(4)4cm^{2}/sec\end{array}\]

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Let V be the volume ad a be the side of the cube for time 't' Volume of the cube. $V= a^2$
Given $\large\frac{dV}{dt}$$=4 \;cm^2 sec $
$V= 8\;cm^2$
When $V=g => a^3 =8=2^3$
$S= 6a^2$
$V= a^3$
$\large\frac{dV}{dt} $$=3a^2. \large\frac{da}{dt}$
$4=3a^2 \large\frac{da}{dt}$
$\large\frac{4}{3a^2} =\frac{da}{dt}$
$S= 6a^2$
$\large\frac{dS}{dt} $$=12 a \large\frac{da}{dt}$
$\quad= 12 a \times \large\frac{4}{3a^2}$
$\qquad= \large\frac{16}{2}$
$\qquad= 8 \;cm^2 sec$
Hence 1 is the correct answer.
answered May 19, 2014 by meena.p

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