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Q. No.A submarine is designed to withstand an absolute pressure of \(100\) atm. How deep can it go below the water surface?
(consider the density of water\(=1000 ~\text{kg m}^{-3},\) \(1 ~\text{atm}=1\times 10^{5} ~\text{Pa}\) and gravitational acceleration \(g=10~\text{m/s}^2\))
1. \(9900~\text{m}\)
2. \(99~\text{m}\)
3. \(9000~\text{m}\)
4. \(990~\text{m}\)
(consider the density of water\(=1000 ~\text{kg m}^{-3},\) \(1 ~\text{atm}=1\times 10^{5} ~\text{Pa}\) and gravitational acceleration \(g=10~\text{m/s}^2\))
1. \(9900~\text{m}\)
2. \(99~\text{m}\)
3. \(9000~\text{m}\)
4. \(990~\text{m}\)
Subtopic: Pressure |
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Level 2: 60%+
NEET - 2026
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Water flows in a streamline motion through a horizontal pipe of circular cross-section, as shown in the figure. The pressure difference of water between \(P\) and \(Q\) is \(15~\text{Nm}^{-2}.\) The areas of cross-section at \(P\) and \(Q\) are \(40~\text{cm}^2\) and \(20~\text{cm}^2,\) respectively. The rate of flow of water through the pipe, in \(\text{cm}^{3} \text{s}^{-1},\) is:
(take the density of water\(=1000~\text{kg}~\text{m}^{-3}\))
(take the density of water\(=1000~\text{kg}~\text{m}^{-3}\))
| 1. | \(400\) | 2. | \(100\) |
| 3. | \(200\) | 4. | \(300\) |
Subtopic: Bernoulli's Theorem |
Level 3: 35%-60%
NEET - 2026
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In the measurement of the viscosity of liquids using the terminal velocity experiment, spherical balls of the same radius but having different densities are used. The variation of the terminal velocity \((v)\) with the ratio of the density of the spherical ball \((\sigma)\) to the density of the liquid \((\rho),\) is best represented by:
| 1. |  |
2. |  |
| 3. |  |
4. |  |
Subtopic: Viscosity |
Level 4: Below 35%
NEET - 2026
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Consider a water tank shown in the figure. It has one wall at \(x=L\) and can be taken to be very wide in the \(z\) direction. When filled with a liquid of surface tension \(S\) and density \(\rho,\) the liquid, surface makes angle \(\theta_{0}\left(\theta_0 \ll 1\right)\) with the \(x\text-\)axis at \(x=L.\) If \(y(x)\) is the height of the surface then the equation for \(y(x)\) is:
(take \(\theta(x)=\sin \theta(x)=\tan \theta(x)=\dfrac{d y}{d x}, g\) is the acceleration due to gravity)
(take \(\theta(x)=\sin \theta(x)=\tan \theta(x)=\dfrac{d y}{d x}, g\) is the acceleration due to gravity)
| 1. | \(\dfrac{d^2 y}{d x^2}=\sqrt{\dfrac{\rho g}{S}}\) | 2. | \(\dfrac{d y}{d x}=\sqrt{\dfrac{\rho g}{S}} x\) |
| 3. | \(\dfrac{d^2 y}{d x^2}=\dfrac{\rho g}{S} x\) | 4. | \(\dfrac{d^2 y}{d x^2}=\dfrac{\rho g}{S} y\) |
Subtopic: Surface Tension |
Level 4: Below 35%
NEET - 2025
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A thin flat circular disc of radius \(4.5~\text {cm}\) is placed gently over the surface of water. If the surface tension of water is \(0.07~\text{Nm}^{-1},\) then the excess force required to take it away from the surface is:
1. \(198~\text{N}\)
2. \(1.98~\text{mN}\)
3. \(99~\text{N}\)
4. \(19.8~\text{mN}\)
1. \(198~\text{N}\)
2. \(1.98~\text{mN}\)
3. \(99~\text{N}\)
4. \(19.8~\text{mN}\)
Subtopic: Surface Tension |
53%
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NEET - 2024
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A wire of length \(L\) and radius \(r(r\ll L)\) is kept floating on the surface of a liquid of density \(\rho.\) The maximum radius of the wire for which it may not sink is:
(the surface tension of liquid is \(T\))
(the surface tension of liquid is \(T\))
| 1. | \(\sqrt { \dfrac{T}{\rho g}}\) | 2. | \(\sqrt { \dfrac{2T}{\rho g}}\) |
| 3. | \(\sqrt{\dfrac{2T\rho}{\pi g}}\) | 4. | \(\sqrt{\dfrac{2T} {\pi \rho g}}\) |
Subtopic: Surface Tension |
59%
Level 3: 35%-60%
NEET - 2024
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The pressure experienced by a swimmer \(20~\text{m}\) below the water surface in a lake is appropriately:
(Given density of water = \(10^3 ~\text{kgm}^{-3},~ g=10 ~\text{ms}^{-2} \) and \(1~\text{atm} = 10^5~\text{Pa}\))
(Given density of water = \(10^3 ~\text{kgm}^{-3},~ g=10 ~\text{ms}^{-2} \) and \(1~\text{atm} = 10^5~\text{Pa}\))
| 1. | \(1~\text{atm}\) | 2. | \(2~\text{atm}\) |
| 3. | \(3~\text{atm}\) | 4. | \(4~\text{atm}\) |
Subtopic: Pressure |
68%
Level 2: 60%+
NEET - 2024
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An ideal fluid is flowing in a non-uniform cross-sectional tube \(XY\) (as shown in the figure) from end \(X\) to end \(Y.\) If \(K_1\) and \(K_2\) are the kinetic energies per unit volume of the fluid at \(X\) and \(Y\) respectively, the correct relationship between \(K_1\) and \(K_2\) is:
| 1. | \(K_1=K_2\) | 2. | \({2K}_1={K}_2\) |
| 3. | \({K}_1>{K}_2\) | 4. | \({K}_1<{K}_2\) |
Subtopic: Bernoulli's Theorem |
69%
Level 2: 60%+
NEET - 2024
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The venturi-meter works on:
| 1. | The principle of perpendicular axes |
| 2. | Huygen's principle |
| 3. | Bernoulli's principle |
| 4. | The principle of parallel axes |
Subtopic: Bernoulli's Theorem |
86%
Level 1: 80%+
NEET - 2023
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The amount of energy required to form a soap bubble of radius \(2~\text{cm}\) from a soap solution is nearly:
\((\)the surface tension of soap solution \(0.03~\text{Nm}^{-1})\)
1. \(50.1 \times 10^{-4}~\text{J}\)
2. \(30.16 \times 10^{-4}~ \text{J}\)
3. \(5.06 \times 10^{-4} ~\text{J}\)
4. \(3.01\times 10^{-4} ~\text{J}\)
\((\)the surface tension of soap solution \(0.03~\text{Nm}^{-1})\)
1. \(50.1 \times 10^{-4}~\text{J}\)
2. \(30.16 \times 10^{-4}~ \text{J}\)
3. \(5.06 \times 10^{-4} ~\text{J}\)
4. \(3.01\times 10^{-4} ~\text{J}\)
Subtopic: Surface Tension |
63%
Level 2: 60%+
NEET - 2023
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