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 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`.