Q118 / 150 · MHT CET 2024 (09 May Shift 1) · Physics
Three immiscible transparent liquids with uniform refractive indices $\frac{3}{2}, \frac{4}{3}$ and $\frac{6}{5}$ are arranged one on top of another. The depths of the liquids are 3 cm 4 cm and 6 cm respectively. The apparent depth of the vessel is
- Correct answerA10 cm
- B9 cm
- C8 cm
- D7 cm
Solution
$\mu=\frac{\text { Real depth }(R)}{\text { Apparent depth }} \Rightarrow \text { Apparent depth }=\frac{R}{\mu}$
For vessel, $\begin{aligned} \text { Apparent depth } & =\frac{R_1}{\mu_1}+\frac{R_2}{\mu_2}+\frac{R_3}{\mu_3} \\ & =\frac{3}{3 / 2}+\frac{4}{4 / 3}+\frac{6}{6 / 5} \\ & =2+3+5 \\ \text { Apparent depth } & =10 \mathrm{~cm} \end{aligned}$
For vessel, $\begin{aligned} \text { Apparent depth } & =\frac{R_1}{\mu_1}+\frac{R_2}{\mu_2}+\frac{R_3}{\mu_3} \\ & =\frac{3}{3 / 2}+\frac{4}{4 / 3}+\frac{6}{6 / 5} \\ & =2+3+5 \\ \text { Apparent depth } & =10 \mathrm{~cm} \end{aligned}$
Hence, the correct answer is option: 'A'.