A layer of benzene (μ=1.5) 6 cm deep floats on water (μ=(4/3)), 4 cm deep. When viewed vertically through air, the apparent distance of the bottom of the vessel below the free surface of benzene is

Question

A layer of benzene (μ=1.5) 6 cm deep floats on water (μ=(4/3)), 4 cm deep. When viewed vertically through air, the apparent distance of the bottom of the vessel below the free surface of benzene is (A) 14 cm (B) 7 cm (C) 21 cm (D) 3.5 cm

Tardigrade Admin · Accepted Answer

Apparent depth =(tb/μb)+(tw/μw) =(6 cm /1.5)+(4 cm /4 / 3)=(4+3) cm =7 cm

Q. A layer of benzene $(μ = 1.5) 6 c m$ deep floats on water $(μ = \frac{4}{3}), 4 c m$ deep. When viewed vertically through air, the apparent distance of the bottom of the vessel below the free surface of benzene is

14 cm

7 cm

21 cm

3.5 cm

Apparent depth $= \frac{t _{b}}{μ _{b}} + \frac{t _{w}}{μ _{w}}$
$= \frac{6 c m}{1.5} + \frac{4 c m}{4/3} = (4 + 3) c m = 7 c m$

Q. A layer of benzene (μ=1.5)6cm deep floats on water (μ=34​),4cm deep. When viewed vertically through air, the apparent distance of the bottom of the vessel below the free surface of benzene is