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(C) Let $h$ be the height of the water level when the coin is first visible. The light ray from the coin travels to the water surface and refracts tow

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(C) Let $h$ be the height of the water level when the coin is first visible. The light ray from the coin travels to the water surface and refracts towards the observer $O$.From the geometry,the angle of refraction $r$ is given by $	an r = \frac{L-R}{H-d} = \frac{1.5-1}{5.75-4} = \frac{0.5}{1.75} = \frac{2}{7}$.Let $x$ be the horizontal distance from the coin to the point where the ray hits the water surface. Then $	an r = \frac{x}{d-h} \Rightarrow x = (d-h) 	an r$.Also,$	an i = \frac{R-x}{h}$.Using Snell's law,$n \sin i = \sin r$. Since $n = 4/3$,we have $\sin r = \frac{4}{3} \sin i$.Using $	an r = 2/7$,$\sin r = \frac{2}{\sqrt{2^2+7^2}} = \frac{2}{\sqrt{53}}$.Thus,$\sin i = \frac{3}{4} \sin r = \frac{3}{4} 	imes \frac{2}{\sqrt{53}} = \frac{3}{2\sqrt{53}}$.Then $	an i = \frac{\sin i}{\sqrt{1-\sin^2 i}} = \frac{3/2\sqrt{53}}{\sqrt{1-9/(4 	imes 53)}} = \frac{3/2\sqrt{53}}{\sqrt{203}/(2\sqrt{53})} = \frac{3}{\sqrt{203}}$.Equating $x = R - h 	an i = (d-h) 	an r$,we get $h = \frac{d 	an r - R}{	an r - 	an i} = \frac{4(2/7) - 1}{2/7 - 3/\sqrt{203}} = \frac{1/7}{0.2857 - 0.2105} \approx 1.92 \,m$.The volume of water $V = \pi R^2 h = \pi (1)^2 (1.92) = 1.92 \pi \approx 6.03 \,m^3$.Since $V = Qt$,$t = V/Q = 6.03 / 0.1 = 60.3 \,s$. The closest option is $63 \,s$.

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