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Standard XII

Physics

Surface Energy

A layer of be...

Question

A layer of benzene $(μ_{1} = 1.5) 6 cm$ deep floats on water $(μ_{2} = 1.3) 4 cm$ deep. When viewed vertically through air the apparent distance of the bottom of the vessel below the surface of the benzene will be -

14 cm

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7 cm

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21 cm

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3.5 cm

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Solution

The correct option is B $7 cm$
Actual depth $= d$
Apparent depth $= \frac{d}{n}$
Total length for benzene viewed from viewer
$= \frac{6}{3 / 2} = 4 cm$
For water $= \frac{4}{4 / 3} = 3 cm$
So, apparent depth of bottom of vessel below the surface of benzene will be
$4 cm + 3 cm$
$= 7 cm$

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A layer of benzene (μ1=1.5)6 cm deep floats on water (μ2=1.3)4 cm deep. When viewed vertically through air the apparent distance of the bottom of the vessel below the surface of the benzene will be -

The correct option is B 7 cmActual depth =d Apparent depth =dn Total length for benzene viewed from viewer =63/2=4 cm For water =44/3=3 cm So, apparent depth of bottom of vessel below the surface of benzene will be 4 cm+3 cm =7 cm

A layer of benzene $(μ_{1} = 1.5) 6 cm$ deep floats on water $(μ_{2} = 1.3) 4 cm$ deep. When viewed vertically through air the apparent distance of the bottom of the vessel below the surface of the benzene will be -

The correct option is B $7 cm$
Actual depth $= d$
Apparent depth $= \frac{d}{n}$
Total length for benzene viewed from viewer
$= \frac{6}{3 / 2} = 4 cm$
For water $= \frac{4}{4 / 3} = 3 cm$
So, apparent depth of bottom of vessel below the surface of benzene will be
$4 cm + 3 cm$
$= 7 cm$