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(C) Radius of the hemispherical bowl $R = 180 	ext{ cm}$.Rate of flow of water $\frac{dV}{dt} = 108 	ext{ dm}^3/	ext{min} = 108 	imes 1000 	ext{

Vedclass · Accepted Answer

$\frac{1}{16 \pi} 	ext{ cm/s}$

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(C) Radius of the hemispherical bowl $R = 180 	ext{ cm}$.Rate of flow of water $\frac{dV}{dt} = 108 	ext{ dm}^3/	ext{min} = 108 	imes 1000 	ext{ cm}^3/	ext{min} = 108000 	ext{ cm}^3/	ext{min}$.Converting to seconds: $\frac{dV}{dt} = \frac{108000}{60} 	ext{ cm}^3/	ext{s} = 1800 	ext{ cm}^3/	ext{s}$.Let the depth of water be $x$. The volume of water in a hemispherical bowl is given by $V = \frac{\pi}{3} x^2(3R - x)$.Substituting $R = 180$: $V = \frac{\pi}{3} x^2(540 - x) = 180 \pi x^2 - \frac{\pi}{3} x^3$.Differentiating with respect to time $t$: $\frac{dV}{dt} = (360 \pi x - \pi x^2) \frac{dx}{dt}$.At $x = 120 	ext{ cm}$,we have $1800 = (360 \pi(120) - \pi(120)^2) \frac{dx}{dt}$.$1800 = (43200 \pi - 14400 \pi) \frac{dx}{dt} = 28800 \pi \frac{dx}{dt}$.$\frac{dx}{dt} = \frac{1800}{28800 \pi} = \frac{18}{288 \pi} = \frac{1}{16 \pi} 	ext{ cm/s}$.

Water is running into a hemispherical bowl of radius $180 \text{ cm}$ at the rate of $108 \text{ dm}^3/\text{min}$. How fast is the water level rising when the depth of the water in the bowl is $120 \text{ cm}$? $(1 \text{ dm} = 10 \text{ cm})$

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