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 is fired straight up from the ground. Its height in feet after $t$ seconds is given by $s(t) = 128t - 16t^2$. The velocity of the particle when it hits the ground is...

If a particle is moving in a straight line so that after $t$ seconds its distance $S$ (in $cm$) from a fixed point on the line is given by $S = f(t) = t^3 - 5t^2 + 8t$, then the acceleration of the particle at $t = 5 \text{ sec}$ is (in $cm/sec^2$).

If the volume of a spherical ball is increasing at the rate of $4 \pi \, cc/sec$,then find the rate of increase of its radius (in $cm/sec$),when the volume is $288 \pi \, cc$.

At present,a firm is manufacturing $2000$ items. It is estimated that the rate of change of production $P$ with respect to the additional number of workers $x$ is given by $\frac{dP}{dx} = 100 - 12\sqrt{x}$. If the firm employs $25$ more workers,then the new level of production of items is:

The rate of change of the diagonal $R$ of a square with respect to its area $A$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module