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` is

Options

A. `4/3`
B. `3/2`
C. 3
D. 5

Answer

C

Explanation

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 appraent weight must be equal .
`:. (m_1)/(rho_1) (rho_(1)-sigma) g = (m_2)/(rho_2) (rho_(2)-sigma) g`
implies 36/9 (9-1) = 48/(rho_(2)-1)g`
By solving we get `rho_(2) = 3` .