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/cm³ is

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$\frac{4}{3}$

$\frac{3}{2}$

answer is C.

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Detailed Solution

Apparent weight = vg ( dsubstance – dliquid)= $\frac{m g}{d_{s}} (d_{s} - d_{l})$
Where m is mass of the body
ds is density of the body
dL is density of water
If two bodies are in equilibrium, then their apparent weight must be equal.
$\begin{array}{l} ∴ \frac{m_{1}}{d s_{1}} (d s_{1} - d_{L}) = \frac{m_{2}}{d s_{2}} (d s_{2} - d_{L}) \end{array} \Rightarrow \frac{36}{9} (9 - 1) = \frac{48}{d s_{2}} (d s_{2} - 1) 4 (8) = \frac{48}{d s_{2}} (d s_{2} - 1) \frac{d s_{2}}{(d s_{2} - 1)} = \frac{6}{4} 4 d s_{2} = 6 d s_{2} - 6 \Rightarrow 2 d s_{2} = 6 \Rightarrow d s_{2} = 3$