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A small bulb is placed at the bottom of a tank containing water to a depth of 7 m. The refractive index of water is 4/3 . The area of the surface of water through which light from the bulb can emerge out is xπ m2. The value of x is_______.

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detailed solution

Correct option is I

C : Criticle angle

tanC=rhr=htanCsinC=1μ=34tanC=37r=7×37=3
Area of surface = πr2 = 9πm2

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detailed solution

Correct answer is 9

C : Criticle angle

tanC=rhr=htanCsinC=1μ=34tanC=37r=7×37=3
Area of surface = πr2 = 9πm2

Talk to our academic expert!

+91

Are you a Sri Chaitanya student?

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