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A hollow metallic sphere, the outer and ...

A hollow metallic sphere, the outer and inner diameters of which are `d_(1) and d_(2)`. floats on the surface of a liquid. The density of metal is `rho_(1)` and the density of liquid is `rho_(2)`. What weight must be added inside the sphere in order for it to float below the level of liquid?

A

`pi[d_(1)^(3)(rho_(2)-rho_(1))+d_(2)^(3)rho_(1)]g`

B

`(pi)/(6)[d_(1)^(3)(rho_(2)-rho_(1))+d_(2)^(3)rho_(1)]g`

C

`(4pi)/(3)[d_(1)^(3)(rho_(2)-rho_(1))+d_(2)^(2)rho_(1)]g`

D

`(4pi)/(3)d_(1)^(3)(rho_(1)-rho_(2))g`

Text Solution

Verified by Experts

The correct Answer is:
B
Weight of liquid displaced `=(4)/(3)pi((d_(1))/(2))^(3)rho_(2)g`
`"Weight of metal sphere" "=[(4)/(3)pi((d_(1))/(2))^(3)-(4)/(3)pi((d_(2))/(2))^(3)]rho_(1)g`
By law of floatation, if x is the addition weight to be placed in the sphere for floating below the level of liquid, then
`[(4)/(3)pi((d_(1))/(2))^(3)-(4)/(3)pi((d_(2))/(2))^(3)]rho_(1)g+x=(4)/(3)pi((d_(1))/(2))^(3)rho_(2)g`
`"Solving, we get, "x=(pi)/(6)[d_(1)^(3)(rho_(2)-rho_(1))+d_(2)^(3)rho_(1)1g`
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