143301
A tank of height $5 \mathrm{~m}$ is full of water. There is a hole of cross sectional area $1 \mathrm{~cm}^{2}$ in its botton. The volume of water that will come out from this hole per second is $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
1 $10^{-3} \mathrm{~m}^{3} / \mathrm{s}$
2 $10^{-4} \mathrm{~m}^{3} / \mathrm{s}$
3 $10 \mathrm{~m}^{3} / \mathrm{s}$
4 $10^{-2} \mathrm{~m}^{3} / \mathrm{s}$
Explanation:
C Given that, Height $(\mathrm{h})=5 \mathrm{~m}$ Cross - sectional area $(\mathrm{A})=1 \mathrm{~cm}^{2}=10^{-4} \mathrm{~m}^{2}$ Acceleration due to gravity $(\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}$ Flow rate $(\mathrm{Q})=\mathrm{A} \sqrt{2 \mathrm{gh}}$ $=10^{-4} \sqrt{2 \times 10 \times 5}$ $=10^{-3} \mathrm{~m}^{3} / \mathrm{s}$
AP EMCET(Medical)-2010
Mechanical Properties of Fluids
143302
The speeds of air-flow on the upper and lower surfaces of a wing of an aeroplane are $v_{1}$ and $v_{2}$ respectively. If $A$ is the cross-sectional area of the wing and ' $\rho$ ' is the density of air, then the upward lift is:
C The difference of air speeds above and below the wings, in according with Bernoulli's principle, creates a pressure difference, due to which on upward force called dynamic lift acts on the plate. Bernoulli's principle, $\frac{\rho v_{1}^{2}}{2}+\rho g h_{1}+P_{1}=\frac{\rho v_{2}^{2}}{2}+\rho g h_{2}+P_{2}$ ${\left[h_{1}=h_{2}=h\right]}$ $\Delta \mathrm{P}=\mathrm{P}_{2}-\mathrm{P}_{1}$ \(\frac{\rho v_1^2}{2}+\rho g h+P_1=\frac{\rho v_2^2}{2}+\rho g h+P_2\) \(\Delta \mathrm{P}=\frac{1}{2} \rho\left(v_1^2-v_2^2\right)\) \(F=\Delta P A\) \(F=\frac{1}{2} \rho A\left(v_1^2-v_2^2\right)\)
AP EAMCET(Medical)-2006
Mechanical Properties of Fluids
143225
Two soap bubbles of different radii are formed at the two ends of a tube and are in communication with each other through the tube. What will happen to them?
1 Smaller bubble will grow until larger one collapses
2 Larger bubble will grow until smaller bubble collapses
3 Smaller bubble will grow till both have same radius
4 Both bubbles will stay as they were
Explanation:
B Air will flow from the smaller bubble into the large one and the larger bubble grow at the expense of the smaller one because excess pressure is inversely proportional to the radius. Then, $\mathrm{P} \propto \frac{1}{\mathrm{R}}$
CG PET- 2010
Mechanical Properties of Fluids
143257
When two capillary tubes of different diameters are dipped vertically, the rise of the liquid is
1 Less in the tube of smaller diameter
2 More in the tube of smaller diameter
3 More in the tube of larger diameter
4 Same in both the tubes
Explanation:
B We know that, Rise of liquid columns in capillaries tube is - $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{g} R}$ $\mathrm{~h} \propto \frac{1}{\mathrm{R}}$ Hence, rise is more for the tube of smaller diameter.
143301
A tank of height $5 \mathrm{~m}$ is full of water. There is a hole of cross sectional area $1 \mathrm{~cm}^{2}$ in its botton. The volume of water that will come out from this hole per second is $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
1 $10^{-3} \mathrm{~m}^{3} / \mathrm{s}$
2 $10^{-4} \mathrm{~m}^{3} / \mathrm{s}$
3 $10 \mathrm{~m}^{3} / \mathrm{s}$
4 $10^{-2} \mathrm{~m}^{3} / \mathrm{s}$
Explanation:
C Given that, Height $(\mathrm{h})=5 \mathrm{~m}$ Cross - sectional area $(\mathrm{A})=1 \mathrm{~cm}^{2}=10^{-4} \mathrm{~m}^{2}$ Acceleration due to gravity $(\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}$ Flow rate $(\mathrm{Q})=\mathrm{A} \sqrt{2 \mathrm{gh}}$ $=10^{-4} \sqrt{2 \times 10 \times 5}$ $=10^{-3} \mathrm{~m}^{3} / \mathrm{s}$
AP EMCET(Medical)-2010
Mechanical Properties of Fluids
143302
The speeds of air-flow on the upper and lower surfaces of a wing of an aeroplane are $v_{1}$ and $v_{2}$ respectively. If $A$ is the cross-sectional area of the wing and ' $\rho$ ' is the density of air, then the upward lift is:
C The difference of air speeds above and below the wings, in according with Bernoulli's principle, creates a pressure difference, due to which on upward force called dynamic lift acts on the plate. Bernoulli's principle, $\frac{\rho v_{1}^{2}}{2}+\rho g h_{1}+P_{1}=\frac{\rho v_{2}^{2}}{2}+\rho g h_{2}+P_{2}$ ${\left[h_{1}=h_{2}=h\right]}$ $\Delta \mathrm{P}=\mathrm{P}_{2}-\mathrm{P}_{1}$ \(\frac{\rho v_1^2}{2}+\rho g h+P_1=\frac{\rho v_2^2}{2}+\rho g h+P_2\) \(\Delta \mathrm{P}=\frac{1}{2} \rho\left(v_1^2-v_2^2\right)\) \(F=\Delta P A\) \(F=\frac{1}{2} \rho A\left(v_1^2-v_2^2\right)\)
AP EAMCET(Medical)-2006
Mechanical Properties of Fluids
143225
Two soap bubbles of different radii are formed at the two ends of a tube and are in communication with each other through the tube. What will happen to them?
1 Smaller bubble will grow until larger one collapses
2 Larger bubble will grow until smaller bubble collapses
3 Smaller bubble will grow till both have same radius
4 Both bubbles will stay as they were
Explanation:
B Air will flow from the smaller bubble into the large one and the larger bubble grow at the expense of the smaller one because excess pressure is inversely proportional to the radius. Then, $\mathrm{P} \propto \frac{1}{\mathrm{R}}$
CG PET- 2010
Mechanical Properties of Fluids
143257
When two capillary tubes of different diameters are dipped vertically, the rise of the liquid is
1 Less in the tube of smaller diameter
2 More in the tube of smaller diameter
3 More in the tube of larger diameter
4 Same in both the tubes
Explanation:
B We know that, Rise of liquid columns in capillaries tube is - $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{g} R}$ $\mathrm{~h} \propto \frac{1}{\mathrm{R}}$ Hence, rise is more for the tube of smaller diameter.
143301
A tank of height $5 \mathrm{~m}$ is full of water. There is a hole of cross sectional area $1 \mathrm{~cm}^{2}$ in its botton. The volume of water that will come out from this hole per second is $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
1 $10^{-3} \mathrm{~m}^{3} / \mathrm{s}$
2 $10^{-4} \mathrm{~m}^{3} / \mathrm{s}$
3 $10 \mathrm{~m}^{3} / \mathrm{s}$
4 $10^{-2} \mathrm{~m}^{3} / \mathrm{s}$
Explanation:
C Given that, Height $(\mathrm{h})=5 \mathrm{~m}$ Cross - sectional area $(\mathrm{A})=1 \mathrm{~cm}^{2}=10^{-4} \mathrm{~m}^{2}$ Acceleration due to gravity $(\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}$ Flow rate $(\mathrm{Q})=\mathrm{A} \sqrt{2 \mathrm{gh}}$ $=10^{-4} \sqrt{2 \times 10 \times 5}$ $=10^{-3} \mathrm{~m}^{3} / \mathrm{s}$
AP EMCET(Medical)-2010
Mechanical Properties of Fluids
143302
The speeds of air-flow on the upper and lower surfaces of a wing of an aeroplane are $v_{1}$ and $v_{2}$ respectively. If $A$ is the cross-sectional area of the wing and ' $\rho$ ' is the density of air, then the upward lift is:
C The difference of air speeds above and below the wings, in according with Bernoulli's principle, creates a pressure difference, due to which on upward force called dynamic lift acts on the plate. Bernoulli's principle, $\frac{\rho v_{1}^{2}}{2}+\rho g h_{1}+P_{1}=\frac{\rho v_{2}^{2}}{2}+\rho g h_{2}+P_{2}$ ${\left[h_{1}=h_{2}=h\right]}$ $\Delta \mathrm{P}=\mathrm{P}_{2}-\mathrm{P}_{1}$ \(\frac{\rho v_1^2}{2}+\rho g h+P_1=\frac{\rho v_2^2}{2}+\rho g h+P_2\) \(\Delta \mathrm{P}=\frac{1}{2} \rho\left(v_1^2-v_2^2\right)\) \(F=\Delta P A\) \(F=\frac{1}{2} \rho A\left(v_1^2-v_2^2\right)\)
AP EAMCET(Medical)-2006
Mechanical Properties of Fluids
143225
Two soap bubbles of different radii are formed at the two ends of a tube and are in communication with each other through the tube. What will happen to them?
1 Smaller bubble will grow until larger one collapses
2 Larger bubble will grow until smaller bubble collapses
3 Smaller bubble will grow till both have same radius
4 Both bubbles will stay as they were
Explanation:
B Air will flow from the smaller bubble into the large one and the larger bubble grow at the expense of the smaller one because excess pressure is inversely proportional to the radius. Then, $\mathrm{P} \propto \frac{1}{\mathrm{R}}$
CG PET- 2010
Mechanical Properties of Fluids
143257
When two capillary tubes of different diameters are dipped vertically, the rise of the liquid is
1 Less in the tube of smaller diameter
2 More in the tube of smaller diameter
3 More in the tube of larger diameter
4 Same in both the tubes
Explanation:
B We know that, Rise of liquid columns in capillaries tube is - $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{g} R}$ $\mathrm{~h} \propto \frac{1}{\mathrm{R}}$ Hence, rise is more for the tube of smaller diameter.
NEET Test Series from KOTA - 10 Papers In MS WORD
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Mechanical Properties of Fluids
143301
A tank of height $5 \mathrm{~m}$ is full of water. There is a hole of cross sectional area $1 \mathrm{~cm}^{2}$ in its botton. The volume of water that will come out from this hole per second is $\left(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}\right)$
1 $10^{-3} \mathrm{~m}^{3} / \mathrm{s}$
2 $10^{-4} \mathrm{~m}^{3} / \mathrm{s}$
3 $10 \mathrm{~m}^{3} / \mathrm{s}$
4 $10^{-2} \mathrm{~m}^{3} / \mathrm{s}$
Explanation:
C Given that, Height $(\mathrm{h})=5 \mathrm{~m}$ Cross - sectional area $(\mathrm{A})=1 \mathrm{~cm}^{2}=10^{-4} \mathrm{~m}^{2}$ Acceleration due to gravity $(\mathrm{g})=10 \mathrm{~m} / \mathrm{s}^{2}$ Flow rate $(\mathrm{Q})=\mathrm{A} \sqrt{2 \mathrm{gh}}$ $=10^{-4} \sqrt{2 \times 10 \times 5}$ $=10^{-3} \mathrm{~m}^{3} / \mathrm{s}$
AP EMCET(Medical)-2010
Mechanical Properties of Fluids
143302
The speeds of air-flow on the upper and lower surfaces of a wing of an aeroplane are $v_{1}$ and $v_{2}$ respectively. If $A$ is the cross-sectional area of the wing and ' $\rho$ ' is the density of air, then the upward lift is:
C The difference of air speeds above and below the wings, in according with Bernoulli's principle, creates a pressure difference, due to which on upward force called dynamic lift acts on the plate. Bernoulli's principle, $\frac{\rho v_{1}^{2}}{2}+\rho g h_{1}+P_{1}=\frac{\rho v_{2}^{2}}{2}+\rho g h_{2}+P_{2}$ ${\left[h_{1}=h_{2}=h\right]}$ $\Delta \mathrm{P}=\mathrm{P}_{2}-\mathrm{P}_{1}$ \(\frac{\rho v_1^2}{2}+\rho g h+P_1=\frac{\rho v_2^2}{2}+\rho g h+P_2\) \(\Delta \mathrm{P}=\frac{1}{2} \rho\left(v_1^2-v_2^2\right)\) \(F=\Delta P A\) \(F=\frac{1}{2} \rho A\left(v_1^2-v_2^2\right)\)
AP EAMCET(Medical)-2006
Mechanical Properties of Fluids
143225
Two soap bubbles of different radii are formed at the two ends of a tube and are in communication with each other through the tube. What will happen to them?
1 Smaller bubble will grow until larger one collapses
2 Larger bubble will grow until smaller bubble collapses
3 Smaller bubble will grow till both have same radius
4 Both bubbles will stay as they were
Explanation:
B Air will flow from the smaller bubble into the large one and the larger bubble grow at the expense of the smaller one because excess pressure is inversely proportional to the radius. Then, $\mathrm{P} \propto \frac{1}{\mathrm{R}}$
CG PET- 2010
Mechanical Properties of Fluids
143257
When two capillary tubes of different diameters are dipped vertically, the rise of the liquid is
1 Less in the tube of smaller diameter
2 More in the tube of smaller diameter
3 More in the tube of larger diameter
4 Same in both the tubes
Explanation:
B We know that, Rise of liquid columns in capillaries tube is - $\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{g} R}$ $\mathrm{~h} \propto \frac{1}{\mathrm{R}}$ Hence, rise is more for the tube of smaller diameter.
