Water is running into a hemispherical bowl of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Radius of the hemispherical bowl $R = 180 	ext{ cm}$.Rate of flow $\frac{dV}{dt} = 108 	ext{ dm}^3/	ext{min}$.Since $1 	ext{ dm} = 10 	ext{ c

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Radius of the hemispherical bowl $R = 180 	ext{ cm}$.Rate of flow $\frac{dV}{dt} = 108 	ext{ dm}^3/	ext{min}$.Since $1 	ext{ dm} = 10 	ext{ cm}$,$1 	ext{ dm}^3 = 1000 	ext{ cm}^3$.$\frac{dV}{dt} = 108 	imes 1000 	ext{ cm}^3 / 60 	ext{ sec} = 1800 	ext{ cm}^3/	ext{sec}$.Let the depth of water be $x$. The volume of water in the 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 $t$,we get $\frac{dV}{dt} = (360 \pi x - \pi x^2) \frac{dx}{dt}$.At $x = 120 	ext{ cm}$,$1800 = (360 \pi(120) - \pi(120)^2) \frac{dx}{dt}$.$1800 = (43200 \pi - 14400 \pi) \frac{dx}{dt} = 28800 \pi \frac{dx}{dt}$.$\frac{dx}{dt} = \frac{1800}{28800 \pi} = \frac{18}{288 \pi} = \frac{1}{16 \pi} 	ext{ cm/sec}$.

Water is running into a hemispherical bowl of radius $180 \text{ cm}$ at the rate of $108 \text{ cubic decimetres per minute}$. How fast is the water level rising when the depth of the water level in the bowl is $120 \text{ cm}$? $(1 \text{ decimetre} = 10 \text{ cm})$

Similar Questions

The displacement of a particle at the time $t$ is given by $s = \sqrt{1+t}$. Then its acceleration $a$ is proportional to:

The motion of a stone thrown vertically upward is given by $s = 13.8t - 4.9t^2$,where $s$ is in meters and $t$ is in seconds. Then its velocity at $t = 1$ second is ........ $m/s$.

The point on the curve $y^2 = 18x$,at which the rate of change of $Y$-coordinate is twice the rate of change of $X$-coordinate is . . . . . . . $\left(\frac{dx}{dt} \neq 0\right)$

$A$ stone is dropped into a quiet lake and waves move in circles at the speed of $ 5 \text{ cm s}^{-1} $. At that instant,when the radius of the circular wave is $ 8 \text{ cm} $,how fast is the enclosed area increasing?

Vedclass Test Series

Exam Paper Generator

Online Exam Module