Question

Suppose that water is emptied from a spherical tank of radius 10 cm. If the depth of the water in the tank is 4 cm and is decreasing at the rate of 2 cm/sec, them the radius of the top surface of water is decreasing at the rate of

• A. 1
• B. $\frac{2}{3}$
• C. $\frac{3}{2}$
• D. 2

Answer 1

The correct option is C

$\frac{3}{2}$

$\frac{dh}{dt}=-2;r=10cm$

We have to find $\frac{dx}{dt}$ when h = 4
where x is the radius of the top surface.
From the figure $r^2=x^2+(10-h{)}^2$
$\therefore 2x\frac{dx}{dt}=-2(10-h)\frac{dh}{dt}$
$\Rightarrow \frac{dx}{dt}=-\frac{(10-h)}{x}(-2)\Rightarrow \frac{dx}{dt}=\frac{2(10-4)}{x}=\frac{12}{x}\cdots \left(i\right)whenh=4,then{x}^2={10}^2-6^2=64orx=8\therefore \frac{dx}{dt}=\frac{12}{8}=\frac{3}{2}$
Hence the correct option is c