A) \[1\]
B) \[\frac{2}{3}\]
C) \[2\]
D) \[\frac{3}{2}\]
Correct Answer: D
Solution :
[d] Let at time t, radius of top surface be x and depth of the water be h. \[{{x}^{2}}={{10}^{2}}-{{(10-h)}^{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}=20h-{{h}^{2}}\] \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,2x\frac{dx}{dt}=(20-2h)\frac{dh}{dt}\] At \[h=4;\] \[x=\sqrt{80-16}=8\] \[\therefore \,\,\,\,\,16\frac{dx}{dt}=(20-2\times 4)\times (-2)\] \[\Rightarrow \,\,\,\,\,16\frac{dx}{dt}=-24\] \[\Rightarrow \,\,\,\,\,\frac{dx}{dt}=-\frac{3}{2}\]
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