A leak-proof cylinder of length 1 ; m, made o | vedclass

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(A) Let $A$ be the cross-sectional area of the cylinder and $\rho_0$ and $\rho_4$ be the densities of water at $0^{\circ} C$ and $4^{\circ} C$ respect

Vedclass · Accepted Answer

$1.01$

Vedclass · Answer

(A) Let $A$ be the cross-sectional area of the cylinder and $ho_0$ and $ho_4$ be the densities of water at $0^{\circ} C$ and $4^{\circ} C$ respectively.At $0^{\circ} C,$ the submerged length of the cylinder is $h_0 = 100 \; cm - 20 \; cm = 80 \; cm.$By the principle of flotation,the weight of the cylinder equals the weight of the displaced water:$mg = A 	imes h_0 	imes ho_0 	imes g \implies m = A 	imes 80 	imes ho_0.$At $4^{\circ} C,$ the submerged length of the cylinder is $h_4 = 100 \; cm - 21 \; cm = 79 \; cm.$Since the mass of the cylinder remains constant:$m = A 	imes 79 	imes ho_4.$Equating the two expressions for $m$:$A 	imes 80 	imes ho_0 = A 	imes 79 	imes ho_4.$Therefore,the ratio of the density at $4^{\circ} C$ to the density at $0^{\circ} C$ is:$\frac{ho_4}{ho_0} = \frac{80}{79} \approx 1.0126.$Rounding to the nearest given option,the value is close to $1.01$.

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