361395
The velocity of a ball of mass ' \(m\) ' density ' \(d_{1}\) ' when dropped in a container filled with glycerin of density ' \(d_{2}\) ' becomes constant after sometime. The viscous force acting on the ball will be
1 \(m g\left(\dfrac{d_{1}}{d_{2}}\right)\)
2 \(m g\left(1-\dfrac{d_{2}}{d_{1}}\right)\)
3 \(m g\left(\dfrac{d_{1}+d_{2}}{d_{1}}\right)\)
4 \(m g\left(\dfrac{d_{1}+d_{2}}{d_{2}}\right)\)
Explanation:
Viscous force \(=\) Weight - Force of buoyancy \(\begin{gathered}\Rightarrow f_{v}=m g-F_{B}=m g-d_{2} v g \\=m g\left[1-\dfrac{d_{2} v g}{m g}\right]=m g\left[1-\dfrac{d_{2}}{d_{1}}\right]\end{gathered}\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361396
A sphere of mass \(M\) and radius \(R\) is falling in a viscous fluid. The terminal velocity attained by the object will be proportional to:
1 \(R\)
2 \(R^{2}\)
3 \(1 / R^{2}\)
4 \(1 / R\)
Explanation:
The terminal velocity of a sphere of radius \(R\) moving in a liquid of coefficient of viscosity \(\eta \text { is } \quad \Rightarrow v=\dfrac{2}{9} \dfrac{R^{2}(\rho-\sigma) g}{\eta}\) Where \(\rho\) - density of the sphere \(\sigma\)-density of the liquid \(\Rightarrow v \propto R^{2}\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361397
An air bubbles rises from te bottom of a lake of large depth. The rising speed of air bubble will
1 go on increasing till it reaches surface
2 go on decreasing till it reaches surface
3 increase in the beginning then will become constant
4 be constant all throughout
Explanation:
Velocity becomes constant when terminal velocity is attained.
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361398
Eight equal drops of water are falling through air with a steady velocity of \(10\;cm\;{s^{ - 1}}\). If the drops combine to form a single drop big in size, then the terminal velocity of this big drop is
1 \(40\;cm\;{s^{ - 1}}\)
2 \(10\;cm\;{s^{ - 1}}\)
3 \(30\;cm\;{s^{ - 1}}\)
4 \(80\;cm\;{s^{ - 1}}\)
Explanation:
Let \(r\) be radius of each small water droplet and \(R\) be radius of a big drop. Volume of big drop \(=8 \times\) Volume of each \({\text{ small}}\,{\text{droplets}}\frac{4}{3}\pi {R^3}{\text{ }} = 8 \times \frac{4}{3}\pi {r^3}R{\text{ }} = 2r\) Let terminal velocity of small droplet be \(v_{1}\) and of big drop be \(v_{2}\). Terminal velocity, \(v\) is where the symbols have their usual meaning \(v\mu {r^2}\) \(\frac{{{v_2}}}{{{v_1}}} = \frac{{r_2^2}}{{r_1^2}} = {\left( {\frac{{2r}}{r}} \right)^2} = 4\) \(\Rightarrow \quad v_{2}=4 v_{1}\) or \( = 4 \times 10\;cm\;{s^{ - 1}} = 40\;cm\;{s^{ - 1}}\)
361395
The velocity of a ball of mass ' \(m\) ' density ' \(d_{1}\) ' when dropped in a container filled with glycerin of density ' \(d_{2}\) ' becomes constant after sometime. The viscous force acting on the ball will be
1 \(m g\left(\dfrac{d_{1}}{d_{2}}\right)\)
2 \(m g\left(1-\dfrac{d_{2}}{d_{1}}\right)\)
3 \(m g\left(\dfrac{d_{1}+d_{2}}{d_{1}}\right)\)
4 \(m g\left(\dfrac{d_{1}+d_{2}}{d_{2}}\right)\)
Explanation:
Viscous force \(=\) Weight - Force of buoyancy \(\begin{gathered}\Rightarrow f_{v}=m g-F_{B}=m g-d_{2} v g \\=m g\left[1-\dfrac{d_{2} v g}{m g}\right]=m g\left[1-\dfrac{d_{2}}{d_{1}}\right]\end{gathered}\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361396
A sphere of mass \(M\) and radius \(R\) is falling in a viscous fluid. The terminal velocity attained by the object will be proportional to:
1 \(R\)
2 \(R^{2}\)
3 \(1 / R^{2}\)
4 \(1 / R\)
Explanation:
The terminal velocity of a sphere of radius \(R\) moving in a liquid of coefficient of viscosity \(\eta \text { is } \quad \Rightarrow v=\dfrac{2}{9} \dfrac{R^{2}(\rho-\sigma) g}{\eta}\) Where \(\rho\) - density of the sphere \(\sigma\)-density of the liquid \(\Rightarrow v \propto R^{2}\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361397
An air bubbles rises from te bottom of a lake of large depth. The rising speed of air bubble will
1 go on increasing till it reaches surface
2 go on decreasing till it reaches surface
3 increase in the beginning then will become constant
4 be constant all throughout
Explanation:
Velocity becomes constant when terminal velocity is attained.
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361398
Eight equal drops of water are falling through air with a steady velocity of \(10\;cm\;{s^{ - 1}}\). If the drops combine to form a single drop big in size, then the terminal velocity of this big drop is
1 \(40\;cm\;{s^{ - 1}}\)
2 \(10\;cm\;{s^{ - 1}}\)
3 \(30\;cm\;{s^{ - 1}}\)
4 \(80\;cm\;{s^{ - 1}}\)
Explanation:
Let \(r\) be radius of each small water droplet and \(R\) be radius of a big drop. Volume of big drop \(=8 \times\) Volume of each \({\text{ small}}\,{\text{droplets}}\frac{4}{3}\pi {R^3}{\text{ }} = 8 \times \frac{4}{3}\pi {r^3}R{\text{ }} = 2r\) Let terminal velocity of small droplet be \(v_{1}\) and of big drop be \(v_{2}\). Terminal velocity, \(v\) is where the symbols have their usual meaning \(v\mu {r^2}\) \(\frac{{{v_2}}}{{{v_1}}} = \frac{{r_2^2}}{{r_1^2}} = {\left( {\frac{{2r}}{r}} \right)^2} = 4\) \(\Rightarrow \quad v_{2}=4 v_{1}\) or \( = 4 \times 10\;cm\;{s^{ - 1}} = 40\;cm\;{s^{ - 1}}\)
361395
The velocity of a ball of mass ' \(m\) ' density ' \(d_{1}\) ' when dropped in a container filled with glycerin of density ' \(d_{2}\) ' becomes constant after sometime. The viscous force acting on the ball will be
1 \(m g\left(\dfrac{d_{1}}{d_{2}}\right)\)
2 \(m g\left(1-\dfrac{d_{2}}{d_{1}}\right)\)
3 \(m g\left(\dfrac{d_{1}+d_{2}}{d_{1}}\right)\)
4 \(m g\left(\dfrac{d_{1}+d_{2}}{d_{2}}\right)\)
Explanation:
Viscous force \(=\) Weight - Force of buoyancy \(\begin{gathered}\Rightarrow f_{v}=m g-F_{B}=m g-d_{2} v g \\=m g\left[1-\dfrac{d_{2} v g}{m g}\right]=m g\left[1-\dfrac{d_{2}}{d_{1}}\right]\end{gathered}\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361396
A sphere of mass \(M\) and radius \(R\) is falling in a viscous fluid. The terminal velocity attained by the object will be proportional to:
1 \(R\)
2 \(R^{2}\)
3 \(1 / R^{2}\)
4 \(1 / R\)
Explanation:
The terminal velocity of a sphere of radius \(R\) moving in a liquid of coefficient of viscosity \(\eta \text { is } \quad \Rightarrow v=\dfrac{2}{9} \dfrac{R^{2}(\rho-\sigma) g}{\eta}\) Where \(\rho\) - density of the sphere \(\sigma\)-density of the liquid \(\Rightarrow v \propto R^{2}\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361397
An air bubbles rises from te bottom of a lake of large depth. The rising speed of air bubble will
1 go on increasing till it reaches surface
2 go on decreasing till it reaches surface
3 increase in the beginning then will become constant
4 be constant all throughout
Explanation:
Velocity becomes constant when terminal velocity is attained.
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361398
Eight equal drops of water are falling through air with a steady velocity of \(10\;cm\;{s^{ - 1}}\). If the drops combine to form a single drop big in size, then the terminal velocity of this big drop is
1 \(40\;cm\;{s^{ - 1}}\)
2 \(10\;cm\;{s^{ - 1}}\)
3 \(30\;cm\;{s^{ - 1}}\)
4 \(80\;cm\;{s^{ - 1}}\)
Explanation:
Let \(r\) be radius of each small water droplet and \(R\) be radius of a big drop. Volume of big drop \(=8 \times\) Volume of each \({\text{ small}}\,{\text{droplets}}\frac{4}{3}\pi {R^3}{\text{ }} = 8 \times \frac{4}{3}\pi {r^3}R{\text{ }} = 2r\) Let terminal velocity of small droplet be \(v_{1}\) and of big drop be \(v_{2}\). Terminal velocity, \(v\) is where the symbols have their usual meaning \(v\mu {r^2}\) \(\frac{{{v_2}}}{{{v_1}}} = \frac{{r_2^2}}{{r_1^2}} = {\left( {\frac{{2r}}{r}} \right)^2} = 4\) \(\Rightarrow \quad v_{2}=4 v_{1}\) or \( = 4 \times 10\;cm\;{s^{ - 1}} = 40\;cm\;{s^{ - 1}}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361395
The velocity of a ball of mass ' \(m\) ' density ' \(d_{1}\) ' when dropped in a container filled with glycerin of density ' \(d_{2}\) ' becomes constant after sometime. The viscous force acting on the ball will be
1 \(m g\left(\dfrac{d_{1}}{d_{2}}\right)\)
2 \(m g\left(1-\dfrac{d_{2}}{d_{1}}\right)\)
3 \(m g\left(\dfrac{d_{1}+d_{2}}{d_{1}}\right)\)
4 \(m g\left(\dfrac{d_{1}+d_{2}}{d_{2}}\right)\)
Explanation:
Viscous force \(=\) Weight - Force of buoyancy \(\begin{gathered}\Rightarrow f_{v}=m g-F_{B}=m g-d_{2} v g \\=m g\left[1-\dfrac{d_{2} v g}{m g}\right]=m g\left[1-\dfrac{d_{2}}{d_{1}}\right]\end{gathered}\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361396
A sphere of mass \(M\) and radius \(R\) is falling in a viscous fluid. The terminal velocity attained by the object will be proportional to:
1 \(R\)
2 \(R^{2}\)
3 \(1 / R^{2}\)
4 \(1 / R\)
Explanation:
The terminal velocity of a sphere of radius \(R\) moving in a liquid of coefficient of viscosity \(\eta \text { is } \quad \Rightarrow v=\dfrac{2}{9} \dfrac{R^{2}(\rho-\sigma) g}{\eta}\) Where \(\rho\) - density of the sphere \(\sigma\)-density of the liquid \(\Rightarrow v \propto R^{2}\)
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361397
An air bubbles rises from te bottom of a lake of large depth. The rising speed of air bubble will
1 go on increasing till it reaches surface
2 go on decreasing till it reaches surface
3 increase in the beginning then will become constant
4 be constant all throughout
Explanation:
Velocity becomes constant when terminal velocity is attained.
PHXI10:MECHANICAL PROPERTIES OF FLUIDS
361398
Eight equal drops of water are falling through air with a steady velocity of \(10\;cm\;{s^{ - 1}}\). If the drops combine to form a single drop big in size, then the terminal velocity of this big drop is
1 \(40\;cm\;{s^{ - 1}}\)
2 \(10\;cm\;{s^{ - 1}}\)
3 \(30\;cm\;{s^{ - 1}}\)
4 \(80\;cm\;{s^{ - 1}}\)
Explanation:
Let \(r\) be radius of each small water droplet and \(R\) be radius of a big drop. Volume of big drop \(=8 \times\) Volume of each \({\text{ small}}\,{\text{droplets}}\frac{4}{3}\pi {R^3}{\text{ }} = 8 \times \frac{4}{3}\pi {r^3}R{\text{ }} = 2r\) Let terminal velocity of small droplet be \(v_{1}\) and of big drop be \(v_{2}\). Terminal velocity, \(v\) is where the symbols have their usual meaning \(v\mu {r^2}\) \(\frac{{{v_2}}}{{{v_1}}} = \frac{{r_2^2}}{{r_1^2}} = {\left( {\frac{{2r}}{r}} \right)^2} = 4\) \(\Rightarrow \quad v_{2}=4 v_{1}\) or \( = 4 \times 10\;cm\;{s^{ - 1}} = 40\;cm\;{s^{ - 1}}\)
