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A $20 \mathrm{~g}$ copper block is suspended by a vertical spring causing $1 \mathrm{~cm}$ elongation over the natural length of spring. If a beaker of water is placed below the block so that the copper block is completely immersed in the liquid, the elongation of the spring is
(Density of copper $9000 \mathrm{~kg} \mathrm{~m}^{-3}$, Density of water 1000 $\mathrm{kg} \mathrm{m}^{-3}, \mathrm{~g}=10 \mathrm{~ms}^{-2}$ )
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(Density of copper $9000 \mathrm{~kg} \mathrm{~m}^{-3}$, Density of water 1000 $\mathrm{kg} \mathrm{m}^{-3}, \mathrm{~g}=10 \mathrm{~ms}^{-2}$ )
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Verified Answer
The correct answer is:
$0.89 \mathrm{~cm}$
Mass of Block, $\mathrm{m}=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}$ Elongation, $x=1 \mathrm{~cm}=0.01 \mathrm{~m}$
Spring elongation after immersed in water, $\mathrm{x}^{\prime}=$ ?
Density, $\rho=9000 \mathrm{~kg} / \mathrm{m}^3$
Density of water, $\rho_\omega=1000 \mathrm{~kg} / \mathrm{m}^3$
At equilibrium, $\mathrm{kx}=\mathrm{mg}$
$$
\mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}}=\frac{0.002 \times 10}{0.01}=2 \mathrm{~N} / \mathrm{m}
$$
when mass is immersed in water
$$
\begin{aligned}
& \mathrm{mg}=\mathrm{kx}^{\prime}+\left(\rho_\omega \mathrm{V}\right) \mathrm{g} \\
& \mathrm{mg}=\mathrm{kx}^{\prime}+\rho_\omega\left[\frac{\mathrm{m}}{\rho_{\mathrm{c}}}\right] \mathrm{g} \\
& 20 \times 10^{-3} \times 10=2 \times \mathrm{x}^{\prime}+\frac{1000}{9000} \times 20 \times 10^{-3} \times 10 \\
& 0.2=2 \mathrm{x}^{\prime}+\frac{1}{9} \times 0.2 \\
& \frac{0.1}{10}=\mathrm{x}^{\prime}+\frac{1}{90} \\
& \mathrm{x}^{\prime}=\frac{1}{10}-\frac{1}{90}=0.89 \\
& \mathrm{x}^{\prime}=0.89
\end{aligned}
$$
Spring elongation after immersed in water, $\mathrm{x}^{\prime}=$ ?
Density, $\rho=9000 \mathrm{~kg} / \mathrm{m}^3$
Density of water, $\rho_\omega=1000 \mathrm{~kg} / \mathrm{m}^3$
At equilibrium, $\mathrm{kx}=\mathrm{mg}$
$$
\mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}}=\frac{0.002 \times 10}{0.01}=2 \mathrm{~N} / \mathrm{m}
$$
when mass is immersed in water
$$
\begin{aligned}
& \mathrm{mg}=\mathrm{kx}^{\prime}+\left(\rho_\omega \mathrm{V}\right) \mathrm{g} \\
& \mathrm{mg}=\mathrm{kx}^{\prime}+\rho_\omega\left[\frac{\mathrm{m}}{\rho_{\mathrm{c}}}\right] \mathrm{g} \\
& 20 \times 10^{-3} \times 10=2 \times \mathrm{x}^{\prime}+\frac{1000}{9000} \times 20 \times 10^{-3} \times 10 \\
& 0.2=2 \mathrm{x}^{\prime}+\frac{1}{9} \times 0.2 \\
& \frac{0.1}{10}=\mathrm{x}^{\prime}+\frac{1}{90} \\
& \mathrm{x}^{\prime}=\frac{1}{10}-\frac{1}{90}=0.89 \\
& \mathrm{x}^{\prime}=0.89
\end{aligned}
$$
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