A ball of mass $m$ and radius $ r $ is gently released in a viscous liquid. The mass of the liquid displaced by it is $m' $ such that $m > m'$. The terminal velocity is proportional to
A ball of mass $m$ and radius $ r $ is gently released in a viscous liquid. The mass of the liquid displaced by it is $m' $ such that $m > m'$. The terminal velocity is proportional to
- A
$\frac{{m - m'}}{r}$
- B
$\frac{{m + m'}}{r}$
- C
$\frac{{(m + m')}}{{{r^2}}}$
- D
$(m - m') r^2$
Similar Questions
How coefficient of liquid and gas depend on temperature ?
How coefficient of liquid and gas depend on temperature ?
If a ball of steel (density $\rho=7.8 \;gcm ^{-3}$) attains a terminal velocity of $10 \;cms ^{-1}$ when falling in a tank of water (coefficient of viscosity $\eta_{\text {water }}=8.5 \times 10^{-4} \;Pa - s$ ) then its terminal velocity in glycerine $\left(\rho=12 gcm ^{-3}, \eta=13.2\right)$ would be nearly
If a ball of steel (density $\rho=7.8 \;gcm ^{-3}$) attains a terminal velocity of $10 \;cms ^{-1}$ when falling in a tank of water (coefficient of viscosity $\eta_{\text {water }}=8.5 \times 10^{-4} \;Pa - s$ ) then its terminal velocity in glycerine $\left(\rho=12 gcm ^{-3}, \eta=13.2\right)$ would be nearly
- [AIEEE 2011]
A table tennis ball has radius $(3 / 2) \times 10^{-2} m$ and mass $(22 / 7) \times 10^{-3} kg$. It is slowly pushed down into a swimming pool to a depth of $d=0.7 m$ below the water surface and then released from rest. It emerges from the water surface at speed $v$, without getting wet, and rises up to a height $H$. Which of the following option(s) is (are) correct?
[Given: $\pi=22 / 7, g=10 ms ^{-2}$, density of water $=1 \times 10^3 kg m ^{-3}$, viscosity of water $=1 \times 10^{-3} Pa$-s.]
$(A)$ The work done in pushing the ball to the depth $d$ is $0.077 J$.
$(B)$ If we neglect the viscous force in water, then the speed $v=7 m / s$.
$(C)$ If we neglect the viscous force in water, then the height $H=1.4 m$.
$(D)$ The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500 / 9$.
A table tennis ball has radius $(3 / 2) \times 10^{-2} m$ and mass $(22 / 7) \times 10^{-3} kg$. It is slowly pushed down into a swimming pool to a depth of $d=0.7 m$ below the water surface and then released from rest. It emerges from the water surface at speed $v$, without getting wet, and rises up to a height $H$. Which of the following option(s) is (are) correct?
[Given: $\pi=22 / 7, g=10 ms ^{-2}$, density of water $=1 \times 10^3 kg m ^{-3}$, viscosity of water $=1 \times 10^{-3} Pa$-s.]
$(A)$ The work done in pushing the ball to the depth $d$ is $0.077 J$.
$(B)$ If we neglect the viscous force in water, then the speed $v=7 m / s$.
$(C)$ If we neglect the viscous force in water, then the height $H=1.4 m$.
$(D)$ The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500 / 9$.
- [IIT 2024]
Which of the following graphs best represents the motion of a raindrop?
Which of the following graphs best represents the motion of a raindrop?
The terminal velocity of a copper ball of radius $2.0 \;mm$ falling through a tank of oll at $20\,^{\circ} C$ is $6.5 \;cm s ^{-1} .$ Compute the viscosity of the oil at $20\,^{\circ} C .$ Density of oil is $1.5 \times 10^{3} \;kg m ^{-3},$ density of copper is $8.9 \times 10^{3} \;kg m ^{-3}$
The terminal velocity of a copper ball of radius $2.0 \;mm$ falling through a tank of oll at $20\,^{\circ} C$ is $6.5 \;cm s ^{-1} .$ Compute the viscosity of the oil at $20\,^{\circ} C .$ Density of oil is $1.5 \times 10^{3} \;kg m ^{-3},$ density of copper is $8.9 \times 10^{3} \;kg m ^{-3}$