CBSE 2022 · Region 4 · Set 3 · Q11 · 3 marks
A diver looking up through water $\displaystyle \left(\mu=\frac{4}{3}\right)$ sees the outside world contained in a circular area on the surface of water. If the diver's eyes are $\displaystyle \sqrt{7} \mathrm{~m}$ below the surface of water, then calculate the area of the circle.
Marking-scheme solution
$\displaystyle \operatorname{Sin} \mathrm{C}=\frac{1}{\mu}$
Sin $\displaystyle \mathrm{C}=\frac{3}{4}$
$\displaystyle \operatorname{Sin} \mathrm{C}=\frac{R}{\sqrt{R^{2}+7}}$\begin{aligned}
\frac{3}{4} & =\frac{R}{\sqrt{R^{2}+7}}
R & =3 \mathrm{~m}
\text { Area } & =\pi R^{2}
& =3.14 \times(3)^{2}
& =28.26 \mathrm{~m}^{2}
\end{aligned}
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CBSE Class 12 Physics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.