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Two spheres of equal masses but radii r1 and r2 are allowed to fall in a liquid of infinite column. The ratio of their terminal velocities are
- a)1
- b)r1 : r2
- c)r2 : r1
- d)√r1 : √r2
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
Two spheres of equal masses but radii r1 and r2 are allowed to fall in...
$\sqrt{r2} : \sqrt{r1}$
The terminal velocity of a sphere falling in a liquid is given by the formula:
$v_t = \frac{2}{9} \frac{(\rho_s - \rho_l)gr^2}{\eta}$
Where $\rho_s$ is the density of the sphere, $\rho_l$ is the density of the liquid, $g$ is the acceleration due to gravity, $r$ is the radius of the sphere, and $\eta$ is the viscosity of the liquid.
Since both spheres have the same mass, we can assume they have the same density. Therefore, $\rho_s$ is constant for both spheres. We can also assume that the liquid has a constant density and viscosity throughout the column.
Using the above formula, we can write:
$\frac{v_{t1}}{v_{t2}} = \frac{r_1^2}{r_2^2}$
Substituting $r_2 = k^2 r_1$, where $k$ is a constant:
$\frac{v_{t1}}{v_{t2}} = \frac{r_1^2}{k^4 r_1^2} = \frac{1}{k^4}$
We want to find the ratio of the terminal velocities, not the inverse ratio. So we take the reciprocal of both sides:
$\frac{v_{t2}}{v_{t1}} = k^4 = (\sqrt{k^2})^4 = (\sqrt{r_2/r_1})^4 = r_2/r_1$
Therefore, the answer is (c) $r_2 : r_1$.
The terminal velocity of a sphere falling in a liquid is given by the formula:
$v_t = \frac{2}{9} \frac{(\rho_s - \rho_l)gr^2}{\eta}$
Where $\rho_s$ is the density of the sphere, $\rho_l$ is the density of the liquid, $g$ is the acceleration due to gravity, $r$ is the radius of the sphere, and $\eta$ is the viscosity of the liquid.
Since both spheres have the same mass, we can assume they have the same density. Therefore, $\rho_s$ is constant for both spheres. We can also assume that the liquid has a constant density and viscosity throughout the column.
Using the above formula, we can write:
$\frac{v_{t1}}{v_{t2}} = \frac{r_1^2}{r_2^2}$
Substituting $r_2 = k^2 r_1$, where $k$ is a constant:
$\frac{v_{t1}}{v_{t2}} = \frac{r_1^2}{k^4 r_1^2} = \frac{1}{k^4}$
We want to find the ratio of the terminal velocities, not the inverse ratio. So we take the reciprocal of both sides:
$\frac{v_{t2}}{v_{t1}} = k^4 = (\sqrt{k^2})^4 = (\sqrt{r_2/r_1})^4 = r_2/r_1$
Therefore, the answer is (c) $r_2 : r_1$.
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Community Answer
Two spheres of equal masses but radii r1 and r2 are allowed to fall in...
$\sqrt{\frac{r_2}{r_1}} : \sqrt{\frac{r_1}{r_2}}$
The terminal velocity of a sphere falling in a liquid is given by:
$v_t = \frac{2}{9} \frac{\left(\rho_s - \rho_l\right)gr^2}{\eta}$
where $r$ is the radius of the sphere, $\rho_s$ is the density of the sphere, $\rho_l$ is the density of the liquid, $g$ is the acceleration due to gravity, and $\eta$ is the viscosity of the liquid.
The masses of the two spheres are equal, so we can write:
$\rho_s V_1 = \rho_s V_2$
where $V_1$ and $V_2$ are the volumes of the two spheres.
Since the densities are equal, we can write:
$V_1 = \frac{4}{3} \pi r_1^3$
and
$V_2 = \frac{4}{3} \pi r_2^3$
Substituting these expressions into the equation above, we get:
$r_1^3 = \left(\frac{r_2}{r_1}\right)^{3/2} r_2^3$
Simplifying, we get:
$r_1^2 = r_2^{5/2}$
or
$\frac{r_2}{r_1} = \left(\frac{r_1}{r_2}\right)^{2/5}$
Now, we can write the ratio of the terminal velocities as:
$\frac{v_{t1}}{v_{t2}} = \frac{\frac{2}{9} \frac{\left(\rho_s - \rho_l\right)gr_1^2}{\eta}}{\frac{2}{9} \frac{\left(\rho_s - \rho_l\right)gr_2^2}{\eta}} = \frac{r_1^2}{r_2^2} = \left(\frac{r_1}{r_2}\right)^{4/5}$
Substituting $\frac{r_2}{r_1} = \left(\frac{r_1}{r_2}\right)^{2/5}$, we get:
$\frac{v_{t1}}{v_{t2}} = \frac{\left(\frac{r_1}{r_2}\right)^{4/5}}{\left(\frac{r_2}{r_1}\right)^{4/5}} = \sqrt{\frac{r_2}{r_1}} : \sqrt{\frac{r_1}{r_2}}$
Therefore, the answer is (d) $\sqrt{\frac{r_2}{r_1}} : \sqrt{\frac{r_1}{r_2}}$.
The terminal velocity of a sphere falling in a liquid is given by:
$v_t = \frac{2}{9} \frac{\left(\rho_s - \rho_l\right)gr^2}{\eta}$
where $r$ is the radius of the sphere, $\rho_s$ is the density of the sphere, $\rho_l$ is the density of the liquid, $g$ is the acceleration due to gravity, and $\eta$ is the viscosity of the liquid.
The masses of the two spheres are equal, so we can write:
$\rho_s V_1 = \rho_s V_2$
where $V_1$ and $V_2$ are the volumes of the two spheres.
Since the densities are equal, we can write:
$V_1 = \frac{4}{3} \pi r_1^3$
and
$V_2 = \frac{4}{3} \pi r_2^3$
Substituting these expressions into the equation above, we get:
$r_1^3 = \left(\frac{r_2}{r_1}\right)^{3/2} r_2^3$
Simplifying, we get:
$r_1^2 = r_2^{5/2}$
or
$\frac{r_2}{r_1} = \left(\frac{r_1}{r_2}\right)^{2/5}$
Now, we can write the ratio of the terminal velocities as:
$\frac{v_{t1}}{v_{t2}} = \frac{\frac{2}{9} \frac{\left(\rho_s - \rho_l\right)gr_1^2}{\eta}}{\frac{2}{9} \frac{\left(\rho_s - \rho_l\right)gr_2^2}{\eta}} = \frac{r_1^2}{r_2^2} = \left(\frac{r_1}{r_2}\right)^{4/5}$
Substituting $\frac{r_2}{r_1} = \left(\frac{r_1}{r_2}\right)^{2/5}$, we get:
$\frac{v_{t1}}{v_{t2}} = \frac{\left(\frac{r_1}{r_2}\right)^{4/5}}{\left(\frac{r_2}{r_1}\right)^{4/5}} = \sqrt{\frac{r_2}{r_1}} : \sqrt{\frac{r_1}{r_2}}$
Therefore, the answer is (d) $\sqrt{\frac{r_2}{r_1}} : \sqrt{\frac{r_1}{r_2}}$.
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Two spheres of equal masses but radii r1 and r2 are allowed to fall in a liquid of infinite column. The ratio of their terminal velocities area)1b)r1: r2c)r2 : r1d)√r1: √r2Correct answer is option 'C'. Can you explain this answer?
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