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ICSE Class 10 Physics 2023 Paper

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Question : 45 of 76
Marks: +1, -0
A coin kept inside water [μ=43]\left[\mu = \frac{4}{3}\right] when viewed from air in a vertical direction appears to be raised by 3.0 mm3.0 \text{ mm}. Find the depth of the coin in water.
Solution:  
Given : aμw=43{}_a\mu_w = \frac{4}{3}, Shift =3.0 mm= 3.0 \text{ mm}.
Let the real depth of the coin in water be
=x′ mm= x' \text{ mm}
∴\therefore Apparent depth of the coin =(x−3) mm= (x-3) \text{ mm}.
We know,
aμw= Real depth  Apparent depth {}_a\mu_w = \frac{\text{ Real depth }}{\text{ Apparent depth }}
∴43=xx−3\therefore \frac{4}{3} = \frac{x}{x-3}
⇒4(x−3)=3x\Rightarrow 4(x-3) = 3x
⇒4x−12=3x\Rightarrow 4x - 12 = 3x
⇒4x−3x=12\Rightarrow 4x - 3x = 12
⇒x=12 mm or 1.2 cm\Rightarrow x = 12 \text{ mm} \text{ or } 1.2 \text{ cm}
∴\therefore Real depth of the coin in water
=12 mm or 1.2 cm.= 12 \text{ mm} \text{ or } 1.2 \text{ cm} \text{.}
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