Answer: [tex]0.00680\ m^3[/tex]
Explanation:
Given
The weight of the object, when submerged in the water is [tex]980\ N[/tex]
When it is submerged in the bromine liquid, it weighs [tex]840\ N[/tex]
Suppose,
[tex]\rho=\text{Density of object}\\\rho_w=\text{Density of water}\\\rho_b=\text{Density of bromine}\\V=\text{Volume of the object}[/tex]
for water,
[tex]\Rightarrow V(\rho -\rho_w)g=980\quad \ldots(i)[/tex]
For bromine
[tex]\Rightarrow V(\rho-\rho_b)g=840\quad \ldots(ii)[/tex]
Divide (i) and (ii)
[tex]\Rightarrow \dfrac{\rho-1000}{\rho-3100}=\dfrac{980}{840}\\\\\Rightarrow 840\rho -840\times 1000=980\rho-980\times 3100\\\\\Rightarrow 140\rho=(3038-840)\cdot 1000\\\\\Rightarrow \rho=15,700\ kg/m^3[/tex]
Put the density value in equation (i)
[tex]\Rightarrow V(15,700-1000)\cdot 9.8=980\\\\\Rightarrow V=\dfrac{100}{14,700}\\\\\Rightarrow V=0.00680\ m^3[/tex]
