CBSE 2023 · Region 1 · Set 2 · Q36 · 4 marks
A tank, as shown in the figure below, formed using a combination of a cylinder and a cone, offers better drainage as compared to a flat bottomed tank.
A tap is connected to such a tank whose conical part is full of water. Water is dripping out from a tap at the bottom at the uniform rate of $\displaystyle 2 \mathrm{~cm}^{3} / \mathrm{s}$. The semi-vertical angle of the conical tank is $\displaystyle 45^{\circ}$. On the basis of given information, answer the following questions :(i)Find the volume of water in the tank in terms of its radius r .(ii)Find rate of change of radius at an instant when $\displaystyle \mathrm{r}=2 \sqrt{2} \mathrm{~cm}$.(iii)Find the rate at which the wet surface of the conical tank is decreasing at an instant when radius $\displaystyle \mathrm{r}=2 \sqrt{2} \mathrm{~cm}$.Find the rate of change of height ' $\displaystyle \mathrm{h}$ ' at an instant when slant height is $\displaystyle 4$ cm . Case Study - $\displaystyle 2$
A tank, as shown in the figure below, formed using a combination of a cylinder and a cone, offers better drainage as compared to a flat bottomed tank.
A tap is connected to such a tank whose conical part is full of water. Water is dripping out from a tap at the bottom at the uniform rate of $\displaystyle 2 \mathrm{~cm}^{3} / \mathrm{s}$. The semi-vertical angle of the conical tank is $\displaystyle 45^{\circ}$. On the basis of given information, answer the following questions :
(i)
Find the volume of water in the tank in terms of its radius r .
(ii)
Find rate of change of radius at an instant when $\displaystyle \mathrm{r}=2 \sqrt{2} \mathrm{~cm}$.
(iii)
Find the rate at which the wet surface of the conical tank is decreasing at an instant when radius $\displaystyle \mathrm{r}=2 \sqrt{2} \mathrm{~cm}$.
Find the rate of change of height ' $\displaystyle \mathrm{h}$ ' at an instant when slant height is $\displaystyle 4$ cm . Case Study - $\displaystyle 2$
Marking-scheme solution
((i) $\displaystyle \mathrm{v}=\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h}=\frac{1}{3} \pi \mathrm{r}^{3} \quad\left[\right.$ as $\displaystyle \theta=45^{\circ}$ gives $\displaystyle \left.\mathrm{r}=\mathrm{h}\right]$
(ii)
$\displaystyle \frac{d \mathrm{v}}{d t}=\pi \mathbf{r}^{2} \frac{d \mathrm{r}}{d t}$
$\displaystyle \Rightarrow\left(\frac{\mathrm{dr}}{\mathrm{dt}}\right)_{\mathrm{r}=2 \sqrt{2}}=-\frac{1}{4 \pi} \mathrm{~cm} / \mathrm{sec}$
(a)
$\displaystyle \mathrm{C}=\pi \mathrm{rl}=\pi \mathrm{r} \sqrt{2} \mathrm{r}=\sqrt{2} \pi \mathrm{r}^{2}$
$\displaystyle \frac{d \mathrm{C}}{d t}=\sqrt{2} \boldsymbol{\pi} \mathbf{2 r} \frac{d \mathrm{r}}{d t}$
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CBSE Class 12 Mathematics past-paper question from the 2023board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.