An ice layer of uniform thickness is deposite | vedclass

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

Question

(D) Let $r$ be the thickness of the ice layer at any time $t$ and $V$ be the volume of the ice layer.Given that the rate of change of volume is $\frac

Vedclass · Accepted Answer

$1/(18\pi)$

Vedclass · Answer

(D) Let $r$ be the thickness of the ice layer at any time $t$ and $V$ be the volume of the ice layer.Given that the rate of change of volume is $\frac{dV}{dt} = -50 	ext{ cm}^3/	ext{min}$.We need to find $\frac{dr}{dt}$ when $r = 5 	ext{ cm}$.The radius of the sphere including the ice layer is $(r + 10) 	ext{ cm}$.The volume of the ice layer $V$ is the volume of the sphere with ice minus the volume of the iron sphere:$V = \frac{4}{3}\pi(r + 10)^3 - \frac{4}{3}\pi(10)^3$.Differentiating both sides with respect to $t$:$\frac{dV}{dt} = 4\pi(r + 10)^2 \frac{dr}{dt}$.Substituting the given values $\frac{dV}{dt} = -50$ and $r = 5$:$-50 = 4\pi(5 + 10)^2 \frac{dr}{dt}$.$-50 = 4\pi(15)^2 \frac{dr}{dt}$.$-50 = 4\pi(225) \frac{dr}{dt}$.$-50 = 900\pi \frac{dr}{dt}$.$\frac{dr}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} 	ext{ cm/min}$.Since the question asks for the rate at which the thickness is decreasing,the rate is $1/(18\pi) 	ext{ cm/min}$.

An ice layer of uniform thickness is deposited on an iron sphere of radius $10 \text{ cm}$. The ice melts at a rate of $50 \text{ cm}^3/\text{min}$. When the thickness of the ice layer is $5 \text{ cm}$,the rate at which the thickness of the ice layer is decreasing is ...... $\text{cm/min}$.

Similar Questions

The equation of motion of a particle moving along a straight line is $s = 2t^3 - 9t^2 + 12t$,where the units of $s$ and $t$ are $cm$ and $sec$. The acceleration of the particle will be zero after:

Moving along the $x$-axis are two points with $x = 10 + 6t$ and $x = 3 + t^2$. The speed with which they are receding from each other at the time of encounter is ........... $cm/sec$. ($x$ is in $cm$ and $t$ is in seconds)

The radius of a cylinder is increasing at the rate of $2 \text{ cm/sec}$ and its height is decreasing at the rate of $3 \text{ cm/sec}$. Find the rate of change of volume when the radius is $3 \text{ cm}$ and the height is $5 \text{ cm}$.

$A$ particle is moving in a straight line according to the formula $s = t^2 + 8t + 12$. If $s$ is measured in meters and $t$ is measured in seconds,then the average velocity of the particle in the third second is .......... $m/s$.

The radius of the base of a cone is increasing at the rate $3 \text{ cm/min}$ and the altitude is decreasing at the rate $4 \text{ cm/min}$. The rate at which the lateral surface area is changing,when the radius is $7 \text{ cm}$ and altitude is $24 \text{ cm}$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module