A) 7 sq.cm/min
B) 10 sq.cm/min
C) 17.5 sq.cm/min
D) 28 sq.cm/min
Correct Answer: B
Solution :
Given \[\frac{dV}{dt}=35\,cc/\min ,\,\,r=7\,cm\] Volume of sphere \[(V)=\frac{4}{3}\pi \,{{r}^{3}}\] Differentiating V w.r.t t \[\frac{dV}{dt}=4{{\pi }^{2}}\frac{dr}{dt}\] \[\frac{dr}{dt}=\frac{35}{4\pi {{r}^{2}}}\] Surface area of sphere \[(S)=4\pi {{r}^{2}}\] On differentiating w.r.t t \[\frac{dS}{dt}=2\times 4\pi r\frac{dr}{dt}\] \[=2\times 4\pi r\times \frac{35}{4\pi {{r}^{2}}}=\frac{70}{7}\] \[\therefore \] \[\frac{dS}{dt}=10\] sq.cm/min. Hence, rate of increase of surface area of spherical balloon is 10 sq. cm/min.
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