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(D) The upthrust $U$ on a body immersed in a liquid is given by $U = V_b \rho_l g$,where $V_b$ is the volume of the ball and $\rho_l$ is the density o

Vedclass · Accepted Answer

$0.05$

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(D) The upthrust $U$ on a body immersed in a liquid is given by $U = V_b ho_l g$,where $V_b$ is the volume of the ball and $ho_l$ is the density of the liquid.Since the mass of the liquid displaced is $m = V_b ho_l$,and the mass of the liquid remains constant,we have $ho_l = m / V_b$.Thus,$U = V_b (m / V_b) g = m g$. However,the volume of the ball $V_b$ changes with temperature as $V_b' = V_b(1 + \gamma_b \Delta T)$.The density of the liquid changes as $ho_l' = ho_l / (1 + \gamma_l \Delta T)$.The new upthrust is $U' = V_b' ho_l' g = V_b(1 + \gamma_b \Delta T) \cdot \frac{ho_l}{1 + \gamma_l \Delta T} \cdot g = U \frac{1 + \gamma_b \Delta T}{1 + \gamma_l \Delta T}$.Using the binomial approximation $(1+x)^{-1} \approx 1-x$ for small $x$,$U' \approx U(1 + \gamma_b \Delta T)(1 - \gamma_l \Delta T) \approx U(1 + (\gamma_b - \gamma_l) \Delta T)$.The change in upthrust is $\Delta U = U' - U = U(\gamma_b - \gamma_l) \Delta T$.The percentage change is $\frac{\Delta U}{U} 	imes 100 = (\gamma_b - \gamma_l) \Delta T 	imes 100$.Given $\gamma_b = 3 	imes 10^{-6} / ^\circ C$,$\gamma_l = 8 	imes 10^{-6} / ^\circ C$,and $\Delta T = 100 ^\circ C$.Percentage change $= (3 	imes 10^{-6} - 8 	imes 10^{-6}) 	imes 100 	imes 100 = (-5 	imes 10^{-6}) 	imes 10^4 = -0.05 \%$.The magnitude of the percentage change is $0.05 \%$.

$A$ solid ball is completely immersed in a liquid. The coefficients of volume expansion of the ball and liquid are $3 \times 10^{-6} / ^\circ C$ and $8 \times 10^{-6} / ^\circ C$ respectively. The percentage change in upthrust when the temperature is increased by $100 ^\circ C$ is ....... $\%$

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