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Mathematics · 2026 · 4 marks
CBSE 2026 · Region 4 · Set 1 · Q36
At a birthday party, children are being served orange juice in conical cups, as shown in the figure. Each cup is 15 cm deep and has a radius 5 cm.…
At a birthday party, children are being served orange juice in conical cups, as shown in the figure.
Each cup is $\displaystyle 15$ cm deep and has a radius $\displaystyle 5$ cm. The juice is being poured into this cup at a rate of $\displaystyle 0 \cdot 1 \mathrm{~cm}^{3} / \mathrm{s}$. On the basis of the above information, answer the following questions :
(i)
Establish a relation between the height $\displaystyle h$ of the juice in the cup and radius $\displaystyle r$ of the surface of the juice in the cup, if the semi-vertical angle of the cone is $\displaystyle \alpha$.
(ii)
At what rate is the juice level in the cup rising when the juice is $\displaystyle 6$ cm deep?
(iii)
(a)
When the juice is $\displaystyle 6$ cm deep, then find at what rate is the upper surface area of juice increasing?
OR
(b)
When the juice is $\displaystyle 6$ cm deep, then find the rate at which the wetted surface area of the cup is increasing. Case Study - $\displaystyle 2$
Marking-scheme solution
(i)
$\displaystyle \tan \alpha=\dfrac{r}{h}=\dfrac{5}{15} \Rightarrow r=\dfrac{h}{3}$
(ii)
Volume, $\displaystyle V=\dfrac{1}{27} \pi h^{3} \Rightarrow \dfrac{dV}{dt}=\dfrac{1}{9} \pi h^{2} \dfrac{dh}{dt} \Rightarrow \dfrac{dh}{dt}=\dfrac{1}{40 \pi}$ cm/sec
(iii)
(a)
Upper surface area of juice, $\displaystyle A=\pi r^{2}=\dfrac{\pi h^{2}}{9}$
$\displaystyle \dfrac{dA}{dt}=\dfrac{2 \pi h}{9} \dfrac{dh}{dt}=\dfrac{1}{30}$ cm$\displaystyle ^{2}$/sec
OR
(b)
Wetted surface area, $\displaystyle S=\pi r l=\dfrac{\sqrt{10} \pi h^{2}}{9}$
$\displaystyle \dfrac{dS}{dt}=\dfrac{2 \sqrt{10} \pi h}{9} \dfrac{dh}{dt}=\dfrac{\sqrt{10}}{30}$ cm$\displaystyle ^{2}$/sec
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