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An aluminum sphere is dipped into water. Which of the following is true?
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The correct answer is:
Buoyancy will be less in water at $0^{\circ} \mathrm{C}$ than that in water at $4^{\circ} \mathrm{C}$
Buoyancy will be less in water at $0^{\circ} \mathrm{C}$ than that in water at $4^{\circ} \mathrm{C}$
As we know that, the Buoyant force $(F)$ on a body volume $(V)$ and density of $(\rho)$, when immersed in liquid of density $\left(\rho_l\right)$ is $=V^{\prime} \rho_l g$ where $V^{\prime}=$ volume of displaced liquid by dipped body $(V)$.
Let volume of the sphere is $\mathrm{V}$ and $\rho$ is its density, then we can write buoyant force
$$
\begin{array}{ll}
F=V \rho_s G & \text { (for liquid) } \\
F \propto \rho_l & \left(\because \rho_{4^{\circ} \mathrm{C}}>\rho_{0^{\circ} \mathrm{C}}\right) \\
\frac{F_{4^{\circ} \mathrm{C}}}{F_{0^{\circ} \mathrm{C}}}=\frac{\rho_{4^{\circ} \mathrm{C}}}{\rho_{0^{\circ} \mathrm{C}}}>1 & \\
F_{4^{\circ} \mathrm{C}}>F_{0^{\circ} \mathrm{C}} &
\end{array}
$$
Hence, buoyancy will be less in water at $0^{\circ} \mathrm{C}$ than that in water at $4^{\circ} \mathrm{C}$.
Let volume of the sphere is $\mathrm{V}$ and $\rho$ is its density, then we can write buoyant force
$$
\begin{array}{ll}
F=V \rho_s G & \text { (for liquid) } \\
F \propto \rho_l & \left(\because \rho_{4^{\circ} \mathrm{C}}>\rho_{0^{\circ} \mathrm{C}}\right) \\
\frac{F_{4^{\circ} \mathrm{C}}}{F_{0^{\circ} \mathrm{C}}}=\frac{\rho_{4^{\circ} \mathrm{C}}}{\rho_{0^{\circ} \mathrm{C}}}>1 & \\
F_{4^{\circ} \mathrm{C}}>F_{0^{\circ} \mathrm{C}} &
\end{array}
$$
Hence, buoyancy will be less in water at $0^{\circ} \mathrm{C}$ than that in water at $4^{\circ} \mathrm{C}$.
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