A sphere of diameter 7,, cm and mass 266.5 ,,gm floats in a bath of a liquid. As temperature is rais

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10-1.Thermometry, Thermal Expansion and Calorimetry

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A sphere of diameter $7\,\, cm$ and mass $266.5 \,\,gm$ floats in a bath of a liquid. As the temperature is raised, the sphere just begins to sink at a temperature $35^o C$. If the density of a liquid at $0^o C$ is $1.527 \,\,gm/cc$, then neglecting the expansion of the sphere, the coefficient of cubical expansion of the liquid is$f$ :

A

$8.486 × 10^{-4}\,\, per ^0C$

B

$8.486 × 10^{-5}\,\, per ^0C$

C

$8.486 × 10^{-6}\,\, per^ 0C$

D

$8.486 × 10^{-3}\,\, per^ 0C$

Solution

By the law of floatation

$266.5=\frac{4 \pi}{3}\left(\frac{7}{2}\right)^{3} \times \rho_{35}$

where $\rho_{35}=$ density of the liquid at $35^{\circ} \mathrm{C}$

$\rho_{35}=1.4839 \frac{g}{c m^{3}}$

But $\rho_{0}=\rho_{35}(1+\gamma \times 35)$

$1.527=1.4839(1+35 \gamma)=8.3 \times 10^{-4\circ} C^{-1}$

Standard 11

Physics

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