JEE Mains: Mechanical Properties of Fluids Test
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1. In an experiment to verify Stokes law, a small spherical ball of radius \(r\) and density (denoted by \(\rho\)) falls under gravity through a distance \(h\) in air before entering a tank of water. If the terminal velocity of the ball inside water is the same as its velocity just before entering the water, then \(h\) is proportional to:
Detailed Solution:
When falling in air (ignoring air resistance), the speed just before entering the water is given by:
\(v_{\text{air}}=\sqrt{2gh}\).
In water, under Stokes law, the terminal velocity is:
\(v_{\text{water}}=\frac{2}{9}\frac{(\rho_{\text{ball}}-\rho_{\text{water}})g\,r^2}{\mu}\),
where \(\mu\) is the dynamic viscosity.
Equating \(v_{\text{air}}=v_{\text{water}}\) gives:
\(\sqrt{2gh}\propto r^2\) ⟹ \(h\propto r^4\).
Thus, the correct answer is (a) \(r^4\).
2. A solid sphere of radius \(R\) attains a terminal velocity \(v_1\) when falling through a viscous fluid (viscosity \(\eta\)). The sphere is then broken into 27 identical smaller spheres. If each smaller sphere attains a terminal velocity \(v_2\) in the same fluid, the ratio \(\frac{v_1}{v_2}\) equals:
Detailed Solution:
The terminal velocity under Stokes law is proportional to \(r^2\). For the original sphere, \(v_1\propto R^2\).
When divided into 27 equal spheres, the volume of each is \(\frac{1}{27}\) of the original. Since volume \(\propto r^3\), the new radius is:
\(r=\frac{R}{3}\).
Thus, \(v_2\propto \left(\frac{R}{3}\right)^2=\frac{R^2}{9}\), and the ratio is:
\(\frac{v_1}{v_2}=\frac{R^2}{R^2/9}=9\).
The correct answer is (a) 9.
3. The velocity of water in a river near the surface is \(18\, \text{km/hr}\). If the river is \(5\, \text{m}\) deep, find the shearing stress between the horizontal layers of water. (Coefficient of viscosity of water = \(10^{-2}\) poise) [Online April 19, 2014]
Detailed Solution:
First, convert \(18\, \text{km/hr}\) to m/s: \(18\times\frac{1000}{3600}=5\, \text{m/s}\).
The shear rate between the surface and the bottom is approximately \(\frac{5\, \text{m/s}}{5\, \text{m}}=1\, \text{s}^{-1}\).
Note: \(1\, \text{poise}=0.1\, \text{Pa·s}\), so \(10^{-2}\) poise \(=10^{-3}\, \text{Pa·s}\).
Therefore, the shearing stress is:
\(\tau=\mu\frac{du}{dy}=10^{-3}\times1=10^{-3}\, \text{N/m}^2\).
The correct answer is (c) \(10^{-3}\, \text{N/m}^2\).
4. The average mass of raindrops is \(3.0\times10^{-5}\, \text{kg}\) and their average terminal velocity is \(9\, \text{m/s}\). Calculate the energy transferred by rain to each square metre of the surface at a place receiving \(100\, \text{cm}\) of rain in a year. [Online April 11, 2014]
Detailed Solution:
\(100\, \text{cm}\) of rain is equivalent to a \(1\, \text{m}\) water column, meaning each square metre receives \(1\, \text{m}^3\) of water, which has a mass of approximately \(1000\, \text{kg}\).
However, using the given raindrop mass:
Number of drops per m\(^2\)= \(\frac{1000}{3.0\times10^{-5}}\approx 3.33\times10^{7}\).
Energy per drop \(=\frac{1}{2}\,m\,v^2=\frac{1}{2}\times3.0\times10^{-5}\times9^2\approx1.215\times10^{-3}\, \text{J}\).
Total energy per m\(^2\)= \(3.33\times10^{7}\times1.215\times10^{-3}\approx4.05\times10^{4}\, \text{J}\).
The correct answer is (b) \(4.05\times10^{4}\, \text{J}\).
5. A tank is filled with water and kerosene (specific gravity 0.8). The height of water is \(3\, \text{m}\) and that of kerosene is \(2\, \text{m}\). When the hole at the bottom is opened, the velocity of the fluid coming out is nearly (take \(g=10\, \text{m/s}^2\) and density of water \(=10^3\, \text{kg/m}^3\)):
Detailed Solution:
The pressure head is due to both liquids. The water contributes \(3\, \text{m}\) and the kerosene contributes an equivalent water column of \(2\times0.8=1.6\, \text{m}\) (since specific gravity \(=0.8\)).
Total effective height \(=3+1.6=4.6\, \text{m}\).
Using Torricelli’s law, \(v=\sqrt{2gH}=\sqrt{2\times10\times4.6}\approx\sqrt{92}\approx9.6\, \text{m/s}\).
The correct answer is (b) \(9.6\, \text{m/s}\).
6. In an experiment, a small steel ball falls through a liquid at a constant speed of \(10\, \text{cm/s}\). If the ball is then pulled upward with a force equal to twice its effective weight, how fast will it move upward? [Online April 25, 2013]
Step-by-Step Solution
1. Define the Effective Weight and Drag Force
Let \( W_{\text{eff}} \) denote the effective weight of the ball (i.e., the gravitational force minus the buoyant force). When the ball falls at terminal speed \( v_d = 10\,\text{cm/s} \), the drag force, \( F_D \), exactly balances \( W_{\text{eff}} \). Thus,
\(\displaystyle W_{\text{eff}} = F_D(10) \)
For small spheres moving slowly in a viscous fluid, the drag force is given by Stokes' law and is proportional to the velocity. Therefore, we can write:
\(\displaystyle F_D(v) = k\, v \)
where \( k \) is a constant that depends on the properties of the fluid and the size of the sphere.
2. Determine the Constant \( k \)
At terminal velocity while falling:
\(\displaystyle W_{\text{eff}} = k \times 10 \)
This equation tells us that the constant \( k \) multiplied by 10 cm/s gives the effective weight.
3. Analyze the Upward Motion
Now, the ball is pulled upward with a force:
\(\displaystyle F_{\text{pull}} = 2\,W_{\text{eff}} \)
When moving upward, two forces oppose the pulling force:
- The effective weight \( W_{\text{eff}} \) (which still acts downward), and
- The drag force \( F_D(v_u) \) which, by the nature of drag, always opposes the direction of motion.
Thus, the net force \( F_{\text{net}} \) when the ball moves upward at speed \( v_u \) is:
\(\displaystyle F_{\text{net}} = F_{\text{pull}} - W_{\text{eff}} - F_D(v_u) = 2W_{\text{eff}} - W_{\text{eff}} - F_D(v_u) = W_{\text{eff}} - F_D(v_u) \)
4. Find the Upward Terminal Speed
At the new terminal speed \( v_u \), the net force is zero. Therefore:
\(\displaystyle W_{\text{eff}} - F_D(v_u) = 0 \quad \Longrightarrow \quad F_D(v_u) = W_{\text{eff}} \)
Substitute the drag force expression \( F_D(v) = k\, v \):
\(\displaystyle k\, v_u = W_{\text{eff}} \)
But we already know that \( W_{\text{eff}} = k \times 10 \) from the falling condition. Thus:
\(\displaystyle k\, v_u = k \times 10 \)
Cancel \( k \) (assuming \( k \neq 0 \)):
\(\displaystyle v_u = 10\,\text{cm/s} \)
Conclusion
The upward terminal speed of the ball is 10 cm/s. Therefore, the correct answer is (c) 10 cm/s.
7. The terminal velocity of a small sphere of radius \(a\) in a viscous liquid is proportional to:
Detailed Solution:
According to Stokes law, the terminal velocity is given by:
\(v=\frac{2}{9}\frac{(\rho_{\text{ball}}-\rho_{\text{fluid}})g\,a^2}{\mu}\),
so it is directly proportional to \(a^2\). Hence, the correct answer is (a) \(a^2\).
8. A ball of steel (density \(7.8\, \text{g/cm}^3\)) attains a terminal velocity of \(10\, \text{cm/s}\) when falling in water (viscosity \(8.5\times10^{-4}\, \text{Pa·s}\)). What is its approximate terminal velocity in glycerine (density \(1.2\, \text{g/cm}^3\), viscosity \(13.2\, \text{Pa·s}\))? [2011 RS]
Step-by-Step Detailed Solution
1. Terminal Velocity by Stokes' Law
For a small sphere moving at low speeds in a viscous fluid, the terminal velocity \(v\) is given by Stokes' law:
\(\displaystyle v = \frac{2}{9} \frac{(\rho_{\text{sphere}} - \rho_{\text{fluid}}) g a^2}{\eta}\)
Here, \(a\) is the radius of the sphere, \(g\) is the acceleration due to gravity, and \(\eta\) is the viscosity of the fluid.
2. Expressing Terminal Velocities in Two Media
For water, the terminal velocity is:
\(\displaystyle v_{\text{water}} = \frac{2}{9} \frac{(\rho_{\text{steel}} - \rho_{\text{water}}) g a^2}{\eta_{\text{water}}}\)
And for glycerine, it is:
\(\displaystyle v_{\text{gly}} = \frac{2}{9} \frac{(\rho_{\text{steel}} - \rho_{\text{gly}}) g a^2}{\eta_{\text{gly}}}\)
Note that the factor \(\frac{2}{9}g\,a^2\) is common in both cases.
3. Ratio of Terminal Velocities
Taking the ratio of \(v_{\text{gly}}\) to \(v_{\text{water}}\) gives:
\(\displaystyle \frac{v_{\text{gly}}}{v_{\text{water}}} = \frac{(\rho_{\text{steel}} - \rho_{\text{gly}})}{(\rho_{\text{steel}} - \rho_{\text{water}})} \times \frac{\eta_{\text{water}}}{\eta_{\text{gly}}}\)
Rearranging, the terminal velocity in glycerine is:
\(\displaystyle v_{\text{gly}} = v_{\text{water}} \times \frac{(\rho_{\text{steel}} - \rho_{\text{gly}})}{(\rho_{\text{steel}} - \rho_{\text{water}})} \times \frac{\eta_{\text{water}}}{\eta_{\text{gly}}}\)
4. Converting Densities to SI Units
Since \(1\,\text{g/cm}^3 = 1000\,\text{kg/m}^3\):
- \(\rho_{\text{steel}} = 7.8\,\text{g/cm}^3 = 7800\,\text{kg/m}^3\)
- \(\rho_{\text{water}} = 1.0\,\text{g/cm}^3 = 1000\,\text{kg/m}^3\)
- \(\rho_{\text{gly}} = 1.2\,\text{g/cm}^3 = 1200\,\text{kg/m}^3\)
5. Substituting the Given Values
The given terminal velocity in water is \(v_{\text{water}} = 10\,\text{cm/s}\). In SI units, that is:
\(10\,\text{cm/s} = 0.1\,\text{m/s}\)
Now calculate the density differences:
- \(\rho_{\text{steel}} - \rho_{\text{water}} = 7800 - 1000 = 6800\,\text{kg/m}^3\)
- \(\rho_{\text{steel}} - \rho_{\text{gly}} = 7800 - 1200 = 6600\,\text{kg/m}^3\)
And the ratio of viscosities:
\(\displaystyle \frac{\eta_{\text{water}}}{\eta_{\text{gly}}} = \frac{8.5 \times 10^{-4}\,\text{Pa·s}}{13.2\,\text{Pa·s}}\)
Calculating this:
\(\displaystyle \frac{8.5 \times 10^{-4}}{13.2} \approx 6.44 \times 10^{-5}\)
6. Computing the Terminal Velocity in Glycerine
Substitute the values into the ratio:
\(\displaystyle v_{\text{gly}} = 0.1\,\text{m/s} \times \frac{6600}{6800} \times 6.44 \times 10^{-5}\)
First, compute the density ratio:
\(\displaystyle \frac{6600}{6800} \approx 0.9706\)
Then multiply:
\(0.1 \times 0.9706 \approx 0.09706\,\text{m/s}\)
Now multiply by the viscosity ratio:
\(0.09706 \times 6.44 \times 10^{-5} \approx 6.25 \times 10^{-6}\,\text{m/s}\)
Finally, converting \(6.25 \times 10^{-6}\,\text{m/s}\) to cm/s (recall \(1\,\text{m/s} = 100\,\text{cm/s}\)):
\(6.25 \times 10^{-6}\,\text{m/s} = 6.25 \times 10^{-4}\,\text{cm/s}\)
Conclusion
The terminal velocity of the steel ball in glycerine is approximately \(6.25 \times 10^{-4}\,\text{cm/s}\). Therefore, the correct answer is (a) \(6.25 \times 10^{-4}\,\text{cm/s}\).
9. If the terminal speed of a sphere of gold (density \(19.5\, \text{kg/m}^3\)) is \(0.2\, \text{m/s}\) in a viscous liquid (density \(1.5\, \text{kg/m}^3\)), what is the terminal speed of a sphere of silver (density \(10.5\, \text{kg/m}^3\)) of the same size in the same liquid? [2006]
Detailed Solution:
The terminal velocity in Stokes flow is proportional to the effective density difference, i.e. \(v\propto (\rho_{\text{sphere}}-\rho_{\text{fluid}})\).
For gold: \(19.5-1.5=18\). For silver: \(10.5-1.5=9\).
Thus, \(\frac{v_{\text{silver}}}{0.2}=\frac{9}{18}=0.5\) ⟹ \(v_{\text{silver}}=0.1\, \text{m/s}\).
The correct answer is (c) \(0.1\, \text{m/s}\).
10. Spherical balls of radius \(R\) are falling in a viscous fluid (viscosity \(\eta\)) with velocity \(v\). The retarding viscous force acting on a sphere is:
Detailed Solution:
According to Stokes law, the drag force on a sphere is given by:
\(F=6\pi\eta Rv\),
which shows that the force is directly proportional to both \(R\) and \(v\).
The correct answer is (b) Directly proportional to both \(R\) and \(v\).
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