Q.47. The terminal velocity of a metallic ball of radius 6 mm in a viscous fluid is 20 cm/s. The ter

Q: What is the correct answer for JEE Main Physics Q47?

The correct answer is 5. The terminal velocity $V_T$ of a spherical ball in a viscous fluid is given by the formula:$$V_T = \frac{2r^2g(\rho_s - \rho_f)}{9\eta}$$where $r$ is the radius of the ball, $g$ is the acceleration due to gravity, $\rho_s$ is the density of the ball, $\rho_f$ is the density of the fluid, and $\eta$ is the viscosity of the fluid.For two balls of the same material in the same fluid, $g$, $\rho_s$, $

Q: Q.47. The terminal velocity of a metallic ball of radius 6 mm in a viscous fluid is 20 cm/s. The terminal velocity of another ball of same material and having radius 3 mm in the same fluid will be…

The correct answer is 5. The terminal velocity $V_T$ of a spherical ball in a viscous fluid is given by the formula:$$V_T = \frac{2r^2g(\rho_s - \rho_f)}{9\eta}$$where $r$ is the radius of the ball, $g$ is the acceleration due to gravity, $\rho_s$ is the density of the ball, $\rho_f$ is the density of the fluid, and $\eta$ is the viscosity of the fluid.For two balls of the same material in the same fluid, $g$, $\rho_s$, $

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Q.47. The terminal velocity of a metallic ball of radius 6 mm in a viscous fluid is 20 cm/s. The terminal velocity of another ball of same material and having radius 3 mm in the same fluid will be .......... cm/s.

FirstInTest · Accepted Answer

The terminal velocity $V_T$ of a spherical ball in a viscous fluid is given by the formula:$$V_T = \frac{2r^2g(ho_s - ho_f)}{9\eta}$$where $r$ is the radius of the ball, $g$ is the acceleration due to gravity, $ho_s$ is the density of the ball, $ho_f$ is the density of the fluid, and $\eta$ is the viscosity of the fluid.For two balls of the same material in the same fluid, $g$, $ho_s$, $ho_f$, and $\eta$ are all constant.Therefore, $V_T \propto r^2$.So, we can write the ratio of terminal velocities as:$$\frac{V_{T2}}{V_{T1}} = \frac{r_2^2}{r_1^2} = (\frac{r_2}{r_1})^2$$Given:Radius of first ball $r_1 = 6 	ext{ mm}$.Terminal velocity of first ball $V_{T1} = 20 	ext{ cm/s}$.Radius of second ball $r_2 = 3 	ext{ mm}$.We need to find $V_{T2}$.$$\frac{V_{T2}}{20 	ext{ cm/s}} = (\frac{3 	ext{ mm}}{6 	ext{ mm}})^2 = (\frac{1}{2})^2 = \frac{1}{4}$$$$V_{T2} = 20 	ext{ cm/s} 	imes \frac{1}{4} = 5 	ext{ cm/s}$$

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