Question

Sand is poured at a rate of $1000\text{c.c/sec}$ on the ground(from a certain height), which is always in the form of cone whose height $\left(h\right)$ is equal to the base radius $r$. If $l$ is its slant height and $S$ is the curved surface area and $h=40\text{cm}$(at an instant), then which of the following(s) is/are true

• A. $\frac{dS}{dt}=50\sqrt{2}\text{sq.cm/sec}$
• B. $\frac{dh}{dt}=\frac{5}{8\pi }\text{cm/sec}$
• C. $\frac{dl}{dt}=\frac{5\sqrt{2}}{8\pi }\text{cm/sec}$
• D. $\frac{dh}{dt}=\frac{5}{16\pi }\text{cm/sec}$

Answer 1

Answer: (A), (B), (C)

$\frac{dl}{dt}=\frac{5\sqrt{2}}{8\pi }\text{cm/sec}$

Let $V$ be the volume of sand(cone) poured.
Given, $\frac{dV}{dt}=1000\text{c.c/sec},h=40\text{cm}$
$h=r$ at any instance.
We know, $V=\frac{\pi r^{2}h}{3}=\frac{\pi h^3}{3}$
$\therefore \frac{dV}{dt}=\pi h^2\cdot \frac{dh}{dt}$
$\Rightarrow 1000=\pi \times 1600\cdot \frac{dh}{dt}$
$\Rightarrow \frac{dh}{dt}=\frac{5}{8\pi }\text{cm/sec}$
Now, $l=\sqrt{h^2+r^2}=\sqrt{2}h$
$\Rightarrow \frac{dl}{dt}=\sqrt{2}\cdot \frac{dh}{dt}=\frac{5\sqrt{2}}{8\pi }\text{cm/sec}$
We know, $S=\pi rl=\pi hl$
$\Rightarrow \frac{dS}{dt}=\pi (h\frac{dl}{dt}+l\frac{dh}{dt})$
$\therefore \frac{dS}{dt}=50\sqrt{2}\text{sq.cm/sec}$