(A) The equation of motion for a sphere falling in a viscous medium is given by $m \frac{dv}{dt} = mg - 6\pi \eta rv$. At terminal velocity $v_t$,the

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(A) The equation of motion for a sphere falling in a viscous medium is given by $m \frac{dv}{dt} = mg - 6\pi \eta rv$. At terminal velocity $v_t$,the acceleration is zero,so $mg = 6\pi \eta rv_t$. Substituting this,we get $m \frac{dv}{dt} = 6\pi \eta r (v_t - v)$. Rearranging,$\frac{dv}{v_t - v} = \frac{6\pi \eta r}{m} dt$. Integrating from $v=0$ to $v=0.63v_t$,we find the time constant $\tau = \frac{m}{6\pi \eta r}$. Checking dimensions: $[\tau] = [T]$. Dimension of $\frac{m}{6\pi \eta r} = \frac{[M]}{[ML^{-1}T^{-1}][L]} = \frac{[M]}{[MT^{-1}]} = [T]$. Thus,option $(a)$ is dimensionally correct.

Class 11 Physics Fluid Mechanics and Surface Tension Viscosity and Stoke's Law and Terminal Velocity

Medium

$A$ spherical body of mass $m$ and radius $r$ is allowed to fall in a medium of viscosity $\eta$. The time in which the velocity of the body increases from zero to $0.63$ times the terminal velocity $(v_t)$ is called the time constant $(\tau)$. Dimensionally,$\tau$ can be represented by:

AIIMS 1987 All AIIMS

A
$\frac{m}{6\pi \eta r}$
B
$\sqrt{\frac{6\pi mr\eta}{g^2}}$
C
$\frac{m}{6\pi \eta rv_t}$
D
None of the above

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Fluid Mechanics and Surface Tension

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Viscosity and Stoke's Law and Terminal Velocity

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An object falling through a fluid is observed to have an acceleration given by $a = g - bv$,where $g$ is the gravitational acceleration and $b$ is a constant. After a long time of release,it is observed to fall with a constant speed. The value of this constant speed is:

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Find the viscosity of glycerine $($having density $1.3 \ g \ cm^{-3})$ if a steel ball of $2 \ mm$ radius $($density $8 \ g \ cm^{-3})$ acquires a terminal velocity of $4 \ cm \ s^{-1}$ in falling freely in the tank of glycerine. $(g = 980 \ cm \ s^{-2})$ (in $\text{poise}$)

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$A$ spherical ball of radius $1 \ mm$ and density $10.5 \ g/cc$ is dropped in glycerine of coefficient of viscosity $9.8 \ \text{poise}$ and density $1.5 \ g/cc$. The viscous force on the ball when it attains constant velocity is $3696 \times 10^{-x} \ N$. The value of $x$ is $\text{(Given, } g = 9.8 \ m/s^2 \text{ and } \pi = \frac{22}{7}\text{)}$.

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The terminal velocity of a sphere of radius $3 \ mm$ falling in a viscous liquid is $10 \ cm/s$. The terminal velocity of a ball of radius $6 \ mm$ of the same material falling in the same liquid will be $...... \ cm/s$.

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$A$ spherical body of mass $m$ and radius $r$ is allowed to fall in a medium of viscosity $\eta$. The time in which the velocity of the body increases from zero to $0.63$ times the terminal velocity $(v_t)$ is called the time constant $(\tau)$. Dimensionally,$\tau$ can be represented by:

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An object falling through a fluid is observed to have an acceleration given by $a = g - bv$,where $g$ is the gravitational acceleration and $b$ is a constant. After a long time of release,it is observed to fall with a constant speed. The value of this constant speed is:

Find the viscosity of glycerine $($having density $1.3 \ g \ cm^{-3})$ if a steel ball of $2 \ mm$ radius $($density $8 \ g \ cm^{-3})$ acquires a terminal velocity of $4 \ cm \ s^{-1}$ in falling freely in the tank of glycerine. $(g = 980 \ cm \ s^{-2})$ (in $\text{poise}$)

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