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Suppose that water is emptied from a spherical tank of radius $10 c m$ . If the depth of the water in the tank is $4 c m$ and is decreasing at the rate of $2 c m / s e c$ , then the radius of the top surface of water is decreasing at the rate of

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$1$

$2 / 3$

$3 / 2$

$2$

answer is C.

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Detailed Solution

$\frac{d h}{d t} = - 2; r = 10 c m$

$\frac{d x}{d t} = ?$ where $h = 4$

where $x$ is the radius of the top surface.

now $r^{2} = x^{2} + {(10 - h)}^{2}$

$2 x \frac{d x}{d t} = - 2 (10 - h) \frac{d h}{d t} \frac{d x}{d t} = - \frac{(10 - h)}{x} (- 2) \frac{d x}{d t} = \frac{2 (10 - 4)}{x} = \frac{12}{x} . . . . (1)$

where $h = 4$ then $x^{2} = 10^{2} - 6^{2} = 64$

$x = 8 ∴ \frac{d x}{d t} = \frac{12}{8} = \frac{3}{2}$

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