Q: According to Stoke's law, the viscous drag force, on an oil drop is proportional to:

$ V $. Stoke's law describes the drag force on a sphere moving through a viscous fluid as directly proportional to its velocity.

Q: Atmospheric pressure at sea level is

1.013 105 $ \mathrm{Pa} $. Standard atmospheric pressure at sea level is defined as 1 atmosphere, which equals about 1.013 × 10^5 pascals. That matches the option with 10^5 Pa.

Q: A tank, which is open at the top, contains a liquid up to a height $ \boldsymbol{H} . \mathbf{A} $ small hole is made in the side of a tank at a distance y below the liquid surface. The liquid emerging from the hole lands at a distance $ x $ from the tank. his question has multiple cor

x is maximum for $ y=H / 2 $.. The jet leaves the hole with speed $\sqrt{2gy}$ and falls a vertical distance $H-y$, so the time of flight is $\sqrt{2(H-y)/g}$. Thus $x = \sqrt{2gy}\,\sqrt{2(H-y)/g} = 2\sqrt{y(H-y)}$, which is maximized when $y=H/2$.

Q: A thin tube of uniform cross-section is sealed at both ends. It lies horizontally. The middle $ 5 c m $ contains $ \mathrm{Hg} $ and two equal ends contain air at the same pressure $ P_{0} . $ When the tube is held at an angle of $ 60^{\circ} $ with the vertical, the length of the

$ 70.4 \mathrm{cm} $ of $ \mathrm{Hg} $. Each air pocket is compressed/expanded isothermally, so its pressure changes inversely with its length. The two pressures differ by the hydrostatic head of the mercury column along the vertical direction, which lets you solve for the original pressure.

Q: Water is filled upto same height in two identical closed containers $ A $ and $ B $. Container $ A $ has vacuum over the water while container $ B $ has air over the water. At the same depth of both the containers there is an opening on which identical balloons $ A $ and $ B $

Balloon - $ B $ will bulge more than balloon - $ A $. At the same depth, the pressure at the opening equals the pressure on the water surface plus the hydrostatic term $\rho gh$. Container A has vacuum above the water, so its opening pressure is lower. Container B has air above the water, so its opening pressure is higher, making balloon B bulge more.

Q: Under same depth, atmospheric pressure is water pressure.

less than. Pressure in a fluid at depth is given by $P = P_0 + \rho gh$. Since water has a much larger density than air, the pressure increase with the same depth is much greater in water. Therefore atmospheric pressure is less than water pressure at the same depth.

Q: The area of cross section of the wider tube shown in figure is $ 900 \mathrm{cm}^{2} $. If the boy standing on the piston weighs 45 kg, the difference in the levels of water in the two tubes.

$ 75 \mathrm{cm} $. The boy adds pressure on the wider piston equal to his weight divided by the piston area. In a connected fluid, that extra pressure is balanced by the pressure from a vertical water column, so the level difference comes from equating these two pressures.

Question 1

According to Stoke's law, the viscous drag force, on an oil drop is proportional to:

Accepted Answer

$ V $. Stoke's law describes the drag force on a sphere moving through a viscous fluid as directly proportional to its velocity.

Question 2

Which of the following factors affect pressure? This question has multiple correct options

Accepted Answer

Force. Pressure is defined as force per unit area, so changing the force changes the pressure directly. Area also matters in the formula, but among the given options the correct factor is force.

Question 3

Atmospheric pressure at sea level is

Accepted Answer

1.013 105 $ \mathrm{Pa} $. Standard atmospheric pressure at sea level is defined as 1 atmosphere, which equals about 1.013 × 10^5 pascals. That matches the option with 10^5 Pa.

Question 4

A tank, which is open at the top, contains a liquid up to a height $ \boldsymbol{H} . \mathbf{A} $ small hole is made in the side of a tank at a distance y below the liquid surface. The liquid emerging from the hole lands at a distance $ x $ from the tank. his question has multiple cor

Accepted Answer

x is maximum for $ y=H / 2 $.. The jet leaves the hole with speed $\sqrt{2gy}$ and falls a vertical distance $H-y$, so the time of flight is $\sqrt{2(H-y)/g}$. Thus $x = \sqrt{2gy}\,\sqrt{2(H-y)/g} = 2\sqrt{y(H-y)}$, which is maximized when $y=H/2$.

Question 5

A thin tube of uniform cross-section is sealed at both ends. It lies horizontally. The middle $ 5 c m $ contains $ \mathrm{Hg} $ and two equal ends contain air at the same pressure $ P_{0} . $ When the tube is held at an angle of $ 60^{\circ} $ with the vertical, the length of the

Accepted Answer

$ 70.4 \mathrm{cm} $ of $ \mathrm{Hg} $. Each air pocket is compressed/expanded isothermally, so its pressure changes inversely with its length. The two pressures differ by the hydrostatic head of the mercury column along the vertical direction, which lets you solve for the original pressure.

Question 6

Water is filled upto same height in two identical closed containers $ A $ and $ B $. Container $ A $ has vacuum over the water while container $ B $ has air over the water. At the same depth of both the containers there is an opening on which identical balloons $ A $ and $ B $

Accepted Answer

Balloon - $ B $ will bulge more than balloon - $ A $. At the same depth, the pressure at the opening equals the pressure on the water surface plus the hydrostatic term $\rho gh$. Container A has vacuum above the water, so its opening pressure is lower. Container B has air above the water, so its opening pressure is higher, making balloon B bulge more.

Question 7

Under same depth, atmospheric pressure is water pressure.

Accepted Answer

less than. Pressure in a fluid at depth is given by $P = P_0 + \rho gh$. Since water has a much larger density than air, the pressure increase with the same depth is much greater in water. Therefore atmospheric pressure is less than water pressure at the same depth.

Question 8

The area of cross section of the wider tube shown in figure is $ 900 \mathrm{cm}^{2} $. If the boy standing on the piston weighs 45 kg, the difference in the levels of water in the two tubes.

Accepted Answer

$ 75 \mathrm{cm} $. The boy adds pressure on the wider piston equal to his weight divided by the piston area. In a connected fluid, that extra pressure is balanced by the pressure from a vertical water column, so the level difference comes from equating these two pressures.

Question 9

The tangential forces per unit area of the liquid layer required to maintain unit velocity gradient is known as:

Accepted Answer

coefficient of viscosity of the liquid. By Newton’s law of viscosity, the tangential force per unit area needed to maintain a unit velocity gradient is the fluid’s viscosity. This is exactly the coefficient of viscosity, which quantifies internal friction in a liquid.

Question 10

To measure the radius of the drop Millikan used $ _{-1-}- $ law of freely falling drops.

Accepted Answer

stoke's. Stokes’ law gives the viscous drag on a small spherical drop moving through a fluid, which lets Millikan relate the drop’s terminal speed to its radius. The other laws are unrelated to this falling-drop measurement.

Question 11

Fluid friction is exerted by

Accepted Answer

liquids and gases. Fluid friction is the resistance an object experiences while moving through a fluid. Since both liquids and gases are fluids, they exert fluid friction; solids do not in this context.

Question 12

A large cylindrical tank has a hole of area $ A $ at its bottom and water is poured into the tank through a tube of cross-sectional area $ \boldsymbol{A} $ ejecting water at the speed $ v $. Which of the following is true?

Accepted Answer

water level will rise to a height $ \frac{v^{2}}{2 g} $ and then stop.. The water level rises until the hydrostatic pressure at the hole makes the outflow speed equal the inflow speed. Setting Torricelli’s law $\sqrt{2gh}=v$ gives $h=\frac{v^2}{2g}$, so after that the inflow and outflow balance. Therefore the level stops rising at that height.

Question 13

At certain temperature radius of an air bubble is doubled when it comes to the top from bottom of a mercury column of height H if the pressure is:

Accepted Answer

10.64. At constant temperature, the gas obeys Boyle’s law, so pressure is inversely proportional to volume. Since the radius doubles, the bubble’s volume becomes 8 times larger, and combining this with the pressure drop across the mercury column gives the required pressure ratio of 10.64.

NEET Physics: Fluid Mechanics questions with solutions

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