A sphere of diameter 7 , cm and mass 266.5 , | vedclass

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Question

(A) The volume of the sphere is $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (3.5)^3 \approx 179.59 \, cm^3$.By the law of floatation,at $35^{\circ} C$,

Vedclass · Accepted Answer

$8.486 	imes 10^{-4} \, ^{\circ}C^{-1}$

Vedclass · Answer

(A) The volume of the sphere is $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (3.5)^3 \approx 179.59 \, cm^3$.By the law of floatation,at $35^{\circ} C$,the weight of the sphere equals the weight of the displaced liquid:$m = V 	imes ho_{35} \implies 266.5 = 179.59 	imes ho_{35}$.$ho_{35} = \frac{266.5}{179.59} \approx 1.4839 \, g/cm^3$.The density of a liquid at temperature $T$ is given by $ho_T = \frac{ho_0}{1 + \gamma T}$.$1.4839 = \frac{1.527}{1 + \gamma(35)}$.$1 + 35\gamma = \frac{1.527}{1.4839} \approx 1.02904$.$35\gamma = 0.02904$.$\gamma = \frac{0.02904}{35} \approx 8.3 	imes 10^{-4} \, ^{\circ}C^{-1}$.Rounding to the provided options,the closest value is $8.486 	imes 10^{-4} \, ^{\circ}C^{-1}$.

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