Layer of benzene (μ1 = 1.5) 6 cm deep floats on water (μ2 = 1.33) 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 -
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4 (1 - μ2/μ1)
= 4(1 - 8/9)
= 4 (1/9)
= 4/9 cm shift from surface of water
6 (1 - 2/3)
= 2cm shift from benzene
total shift = 2 + 4/9
= 22/9 cm
= 4(1 - 8/9)
= 4 (1/9)
= 4/9 cm shift from surface of water
6 (1 - 2/3)
= 2cm shift from benzene
total shift = 2 + 4/9
= 22/9 cm
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