Question

Two bodies are in equilibrium when suspended in water from the arms of a balance. The mass of one body is 36 g and its density is 9 g / cm³. If the mass of the other is 48 g, its density in g / cm³is

• A. $\frac { 4 } { 3 }$
• B. $\frac { 3 } { 2 }$
• C. 3
• D. 5

Answer 1

Answer: (D)

5

Answer 2

Apparent weight $= V ( \rho - \sigma ) g = \frac { m } { \rho } ( \rho - \sigma ) g$

where $\rho =$ density of the body and

$\sigma =$ density of water

If two bodies are in equilibrium then their apparent weight

must be equal.

$\therefore$ $\frac { m _ { 1 } } { \rho _ { 1 } } \left( \rho _ { 1 } - \sigma \right) g = \frac { m _ { 2 } } { \rho _ { 2 } } \left( \rho _ { 2 } - \sigma \right) g$

⇒ $\frac { 36 } { 9 } ( 9 - 1 ) = \frac { 48 } { \rho _ { 2 } } \left( \rho _ { 2 } - 1 \right) g$.

By solving we get $\rho _ { 2 } = 3$