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Q. No.Water from a tap emerges vertically downwards with an initial speed of 1.0 ms–1. The cross-sectional area of the tap is 10–4 m2. Assume that the pressure is constant throughout the stream of water and that the flow is streamlined. The cross-sectional area of the stream, 0.15 m below the tap would be: (Take g = 10 ms–2)
1. 5 × 10–4m2
2. 2 × 10–5m2
3. 5 × 10–5m2
4. 1 × 10–5 m2
Subtopic: Bernoulli's Theorem |
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A fluid is flowing through a horizontal pipe of varying cross-sections, with speed \(v\) ms-1 at a point where the pressure is \(P\) pascal. At another point where pressure is \(\frac{P}{2}\) pascal, its speed is \(v\) ms-1. If the density of the fluid is \(\rho\) kg-m-3 and the flow is streamlined, then \(v\) is equal to:
| 1. | \(\sqrt{\frac{P}{2\rho }+v^{2}}\) | 2. | \(\sqrt{\frac{P}{\rho }+v^{2}}\) |
| 3. | \(\sqrt{\frac{2P}{\rho }+v^{2}}\) | 4. | \(\sqrt{\frac{P}{\rho }+v^{}}\) |
Subtopic: Bernoulli's Theorem |
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An ideal fluid of density 800 kgm-3, flows smoothly through a bent pipe (as shown in the figure) that tapers in cross-sectional area from a to \(a \over 2\). The pressure difference between the wide and narrow sections of the pipe is 4100 Pa. At the wider section, the velocity of the fluid is \({ \sqrt x \over 6}ms^{-1}\). The value of x is: (Given: g = 10 m-2)
1. 124
2. 236
3. 363
4. 432
1. 124
2. 236
3. 363
4. 432
Subtopic: Bernoulli's Theorem |
From NCERT
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The area of the cross-section of a large tank is \(0.5~\text{m}^2\). It has a narrow opening near the bottom having an area of cross-section \(1~\text{cm}^2\). A load of \(25~\text{kg}\) is applied on the water at the top of the tank. Neglecting the speed of water in the tank, the velocity of the water, coming out of the opening at the time when the height of the water level in the tank is \(40~\text{cm}\) above the bottom, will be: [Take \(g = 10~\text{ms}^{-2}\)]
1. \(1~\text{ms}^{-1}\)
2. \(2~\text{ms}^{-1}\)
3. \(3~\text{ms}^{-1}\)
4. \(4~\text{ms}^{-1}\)
1. \(1~\text{ms}^{-1}\)
2. \(2~\text{ms}^{-1}\)
3. \(3~\text{ms}^{-1}\)
4. \(4~\text{ms}^{-1}\)
Subtopic: Bernoulli's Theorem |
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A liquid of density \(750~\text{kgm}^{-3}\) flows smoothly through a horizontal pipe that tapers in cross-sectional area from \(A_1 = 1.2 \times 10^{-2}~\text{m}^2\) to \(A_2 = \frac{A_1}{2}\). The pressure difference between the wide and narrow sections of the pipe is \(4500~\text{Pa}\). The rate of flow of liquid is:
1. \(20\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
2. \(30\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
3. \(28\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
4. \(24\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
1. \(20\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
2. \(30\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
3. \(28\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
4. \(24\times 10^{-3}~\text{m}^{-3}\text{s}^{-1}\)
Subtopic: Bernoulli's Theorem |
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Water is flowing inside the conical type tube having a ratio of area of cross-section \(6:1\). If the speed of the water outlet through a smaller area is \(60~\text{m/s}\), then the pressure difference across these two cross-sections is:
(assume incompressible fluid, density of water = \(1000~\text{kg/m}^3\) )
1. \(175\times 10^4~ \text{Pa}\)
2. \(175\times 10^3~ \text{Pa}\)
3. \(250\times 10^4~ \text{Pa}\)
4. \(250\times 10^3~ \text{Pa}\)
(assume incompressible fluid, density of water = \(1000~\text{kg/m}^3\) )
1. \(175\times 10^4~ \text{Pa}\)
2. \(175\times 10^3~ \text{Pa}\)
3. \(250\times 10^4~ \text{Pa}\)
4. \(250\times 10^3~ \text{Pa}\)
Subtopic: Bernoulli's Theorem |
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The pressures at the ends of a horizontal pipe are given for water. The speed \({v}\) at end \(2\) if the speed at end \(1\) is \(10~\text{m/s}.\) (density of water \(=1000~\text{kg/m}^3\)). Then the speed \({v}\) (in \(\text{m/s}\)) is:
1. \(22\)
2. \(33\)
3. \(11\)
4. \(55\)
1. \(22\)
2. \(33\)
3. \(11\)
4. \(55\)
Subtopic: Bernoulli's Theorem |
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Correct Bernoulli's equation is (symbols have their usual meaning) :
1. \(P+\rho g h+\frac{1}{2} \rho v^2=\text { constant }\)
2. \(\mathrm{P}+\rho \mathrm{gh}+\rho \mathrm{v}^2=\mathrm{constant}\)
3. \(P+\mathrm{mgh}+\frac{1}{2} \mathrm{mv}^2=\text { constant }\)
4. \(\mathrm{P}+\frac{1}{2} \rho g h+\frac{1}{2} \rho \mathrm{v}^2=\text { constant }\)
1. \(P+\rho g h+\frac{1}{2} \rho v^2=\text { constant }\)
2. \(\mathrm{P}+\rho \mathrm{gh}+\rho \mathrm{v}^2=\mathrm{constant}\)
3. \(P+\mathrm{mgh}+\frac{1}{2} \mathrm{mv}^2=\text { constant }\)
4. \(\mathrm{P}+\frac{1}{2} \rho g h+\frac{1}{2} \rho \mathrm{v}^2=\text { constant }\)
Subtopic: Bernoulli's Theorem |
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