Q.35. If an air bubble of diameter rises steadily through a liquid of density at a rate of , then

Q: What is the correct answer for this exam Physics Q35?

The correct answer is B. For steady rise, the buoyant force ($F_B$) equals the viscous drag force ($F_V$). $F_B = V_{\text{bubble}} \rho_{\text{liquid}} g = \frac{4}{3}\pi R^3 \rho g$. $F_V = 6\pi \eta R v$. Equating the forces: $\frac{4}{3}\pi R^3 \rho g = 6\pi \eta R v$. $\eta = \frac{2 R^2 \rho g}{9 v}$. Given: Diameter $D = 2 \mathrm{~mm} \Rightarrow R = 1 \mathrm{~mm} = 1 \times 10^{-3} \mathrm{~m}$. Density of liqui

Q: Q.35. If an air bubble of diameter rises steadily through a liquid of density at a rate of , then the coefficient of viscosity of liquid is __________ Poise. (Take )

The correct answer is B. For steady rise, the buoyant force ($F_B$) equals the viscous drag force ($F_V$). $F_B = V_{\text{bubble}} \rho_{\text{liquid}} g = \frac{4}{3}\pi R^3 \rho g$. $F_V = 6\pi \eta R v$. Equating the forces: $\frac{4}{3}\pi R^3 \rho g = 6\pi \eta R v$. $\eta = \frac{2 R^2 \rho g}{9 v}$. Given: Diameter $D = 2 \mathrm{~mm} \Rightarrow R = 1 \mathrm{~mm} = 1 \times 10^{-3} \mathrm{~m}$. Density of liqui

Q: What are the options for this this exam Physics question?

A. 88 · B. 8 · C. 8 · D. 088

Question

Q.35. If an air bubble of diameter  rises steadily through a liquid of density  at a rate of , then the coefficient of viscosity of liquid is __________ Poise. (Take )

FirstInTest · Accepted Answer

For steady rise, the buoyant force ($F_B$) equals the viscous drag force ($F_V$).
$F_B = V_{	ext{bubble}} ho_{	ext{liquid}} g = \frac{4}{3}\pi R^3 ho g$.
$F_V = 6\pi \eta R v$.
Equating the forces: $\frac{4}{3}\pi R^3 ho g = 6\pi \eta R v$.
$\eta = \frac{2 R^2 ho g}{9 v}$.
Given: Diameter $D = 2 \mathrm{~mm} \Rightarrow R = 1 \mathrm{~mm} = 1 	imes 10^{-3} \mathrm{~m}$.
Density of liquid $ho = 2000 \mathrm{~kg} / \mathrm{m}^3$.
Velocity $v = 0.5 \mathrm{~cm} / \mathrm{s} = 0.5 	imes 10^{-2} \mathrm{~m} / \mathrm{s}$.
$g = 10 \mathrm{~m} / \mathrm{s}^2$.
$\eta = \frac{2 	imes (1 	imes 10^{-3})^2 	imes 2000 	imes 10}{9 	imes (0.5 	imes 10^{-2})} = \frac{2 	imes 10^{-6} 	imes 20000}{4.5 	imes 10^{-2}} = \frac{40000 	imes 10^{-6}}{4.5 	imes 10^{-2}} = \frac{0.04}{0.045} \approx 0.888 \mathrm{~Pa} \cdot \mathrm{s}$.
To convert $\mathrm{Pa} \cdot \mathrm{s}$ to Poise, multiply by 10 (since 1 Poise = 0.1 $\mathrm{Pa} \cdot \mathrm{s}$).
$\eta = 0.888 	imes 10 = 8.88 \mathrm{~Poise}$.

Answer

88

Answer

8

Answer

8

Answer

088

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