Two bodies are in an 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/cc. If the mass of the other is 48 g, its density in g/cc is

4/3

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3/2

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Solution

The correct option is C 3
$m_{1} = 36 g, ρ_{1} = 9 g / c m^{3}, m_{2} = 48 g, ρ_{2} = ?$
The volume of body 1 $V_{1} = \frac{m_{1}}{ρ_{1}} = \frac{36}{9} = 4 c m^{3}$
weight of the body 1 in water , $W_{1} = m_{1} g - V_{1} ρ_{w} g$
weight of the body 2 in water , $W_{2} = m_{2} g - V_{2} ρ_{w} g$
As the two weights balances each other,
$∴$ $W_{1} = W_{2}$
$m_{g} - V_{1} ρ_{w} g = m_{2} g - V_{2} ρ_{w} g$
$m_{1} - V_{1} ρ_{w} = m_{2} - V_{2} ρ_{w}$
$36 - 4 = 48 - V_{2}$
$V_{2} = 48 - 32 = 16 c m^{3}$
so,the density of body 2 $ρ_{2} = \frac{m_{2}}{V_{2}} = \frac{48}{16} = 3 g / c m^{3}$

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