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Two bodies are in equilibrium when suspended in water from the arms of balance. The mass of one body is 36 g and its density is `9 g// cm^3` If the mass of the other is 46 g, its density in `g//cm^3` isA. `(4)/(3)`B. `(3)/(2)`C. `3`D. `5` |
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Answer» Correct Answer - C Apparent weight `=V(rho-sigma)g = (m)/(rho) (rho-sigma)g` where m = mass of the body, rho = density of the body and `sigma` = density of water If two bodies are in equilibrium then their apparent weight must be equal. `therefore (m_(1))/(rho_(1))(rho_(1)-sigma)g = (m_(2))/(rho_(2))(rho_(2)-sigma)g implies (36)/(9)(9-1)=(48)/(rho_(2))(rho_(2)-1)g`. By solving we get `rho_(2) =3`. |
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