A spherical iron ball of radius 10 , cm is co | vedclass

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

Question

(D) Let $x$ be the thickness of the ice layer. The radius of the sphere including the ice is $r = (10 + x) \, cm$.The volume of the ice layer is $V =

Vedclass · Accepted Answer

$1 / (18\pi) \, cm/min$

Vedclass · Answer

(D) Let $x$ be the thickness of the ice layer. The radius of the sphere including the ice is $r = (10 + x) \, cm$.The volume of the ice layer is $V = \frac{4}{3}\pi (10 + x)^3 - \frac{4}{3}\pi (10)^3$.Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4\pi (10 + x)^2 \frac{dx}{dt}$.Given that the ice melts at a rate of $50 \, cm^3/min$,we have $\frac{dV}{dt} = -50 \, cm^3/min$ (since volume is decreasing).Substituting $x = 5$ and $\frac{dV}{dt} = -50$ into the equation:$-50 = 4\pi (10 + 5)^2 \frac{dx}{dt}$$-50 = 4\pi (15)^2 \frac{dx}{dt}$$-50 = 4\pi (225) \frac{dx}{dt}$$-50 = 900\pi \frac{dx}{dt}$$\frac{dx}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} \, cm/min$.The rate at which the thickness decreases is $\frac{1}{18\pi} \, cm/min$.

$A$ spherical iron ball of radius $10 \, cm$ is coated with a layer of ice of uniform thickness that melts at a rate of $50 \, cm^3/min$. When the thickness of the ice is $5 \, cm$,find the rate at which the thickness of the ice decreases.

Similar Questions

$A$ particle moves in a straight line such that $s = \sqrt{t}$,then its acceleration is proportional to

Let $x$ and $y$ be the sides of two squares such that $y = x - x^2$. The rate of change of the area of the second square with respect to the area of the first square is

$A$ balloon,which always remains spherical,is being inflated by pumping in $900 \ cm^3/sec$ of gas. Find the rate at which the radius of the balloon is increasing when the radius is $15 \ cm$.

$A$ man of height $2 \ m$ walks at a uniform speed of $7 \ m/min$ away from a lamp post of height $9 \ m$. The rate (in $m/min$) at which the length of his shadow increases is

The volume of a spherical balloon is increasing at the rate of $2 \ cm^3/sec$. When its radius is $4 \ cm$,the rate of change of its surface area (in $cm^2/sec$) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module