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A block of iron contains a hollow cavity as shown below. The block weighs $6000 \mathrm{~N}$ in air and $4000 \mathrm{~N}$ in water. If the density of iron and water are $6 \mathrm{~g} / \mathrm{cm}^3$ and $1 \mathrm{~g} / \mathrm{cm}^3$, then the volume of the cavity is (assume, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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The correct answer is:
$0.1 \mathrm{~m}^3$
Weight of block in air, $w_{\text {air }}=6000 \mathrm{~N}$
Weight of block in water, $w_{\text {water }}=4000 \mathrm{~N}$
According to figure given,
volume of cavity,
$V_1=$ volume of water displaced - volume of iron
$=\frac{\text { weight of water displaced }}{\rho_{\text {water }} \times g}-\frac{\text { weight of iron }}{\rho_{\text {iron }} \times g}$
$=\frac{6000-4000}{10^3 \times 10}-\frac{6000}{6 \times 10^3 \times 10}$
$\left[\begin{array}{c}\because \rho_{\text {iron }}=6 \mathrm{~g} / \mathrm{cm}^3=6 \times 10^3 \mathrm{~kg} / \mathrm{m}^3 \\ \rho_{\text {water }}=1 \mathrm{~g} / \mathrm{cm}^3=1 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\end{array}\right]$
$=0.2-0.1=0.1 \mathrm{~m}^3$
Weight of block in water, $w_{\text {water }}=4000 \mathrm{~N}$
According to figure given,
volume of cavity,
$V_1=$ volume of water displaced - volume of iron
$=\frac{\text { weight of water displaced }}{\rho_{\text {water }} \times g}-\frac{\text { weight of iron }}{\rho_{\text {iron }} \times g}$
$=\frac{6000-4000}{10^3 \times 10}-\frac{6000}{6 \times 10^3 \times 10}$
$\left[\begin{array}{c}\because \rho_{\text {iron }}=6 \mathrm{~g} / \mathrm{cm}^3=6 \times 10^3 \mathrm{~kg} / \mathrm{m}^3 \\ \rho_{\text {water }}=1 \mathrm{~g} / \mathrm{cm}^3=1 \times 10^3 \mathrm{~kg} / \mathrm{m}^3\end{array}\right]$
$=0.2-0.1=0.1 \mathrm{~m}^3$
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