Q: 1atm $ = $

$ 76 \mathrm{cm} $ of $ \mathrm{Hg} $. One atmosphere is defined as the pressure exerted by a mercury column of about 76 cm at standard conditions. So the correct equivalence is 76 cm of Hg.

Question 1

If the air density were uniform, then the height of the atmosphere above the sea level to produce a normal atmospheric pressure of $ 1.0 \times 10^{5} $ Pa is (density of air is $ 1.3 \mathrm{kg} / \mathrm{m}^{3}, \mathrm{g}=10 \mathrm{m} / \mathrm{s}^{2} $ ):

Accepted Answer

7.7 km. For a uniform air density, the pressure at sea level is given by the hydrostatic relation P = ρgh. Substituting the given values gives a height of 7.7 km, which matches the correct option.

Question 2

1atm $ = $

Accepted Answer

$ 76 \mathrm{cm} $ of $ \mathrm{Hg} $. One atmosphere is defined as the pressure exerted by a mercury column of about 76 cm at standard conditions. So the correct equivalence is 76 cm of Hg.

Question 3

The atmospheric pressure at is 100,000 pascal.

Accepted Answer

sea level. A pressure of 100,000 pascals is approximately 1 atm, the standard atmospheric pressure measured at sea level. That is why the correct choice is sea level.

Question 4

According to Newton, the viscous force acting between liquid layers of area A and velocity gradient ΔV/ΔZ is given by $ F = -ηA \frac{ΔV}{ΔZ} $ where η is constant called coefficient of viscosity. The dimensional formula of η is

Accepted Answer

ML⁻¹T⁻¹. The viscous force is given by F = ηA(ΔV/ΔZ). Rearranging for η, we get η = F / (A × ΔV/ΔZ). Substituting the dimensional formulas: [F] = MLT⁻², [A] = L², and [ΔV/ΔZ] = T⁻¹, we find [η] = [MLT⁻²] / [L² × T⁻¹] = ML⁻¹T⁻¹.

Question 5

As we know that, surface tension(s) is:

Accepted Answer

[MLT⁻²]/[L] = [MT⁻²]. Surface tension is defined as force per unit length, and its dimensional formula is derived as [MLT⁻²]/[L] = [MT⁻²].

Question 6

The cylindrical tube of a spray pump has radius, R, one end of which has n fine holes, each of radius r. If the speed of the liquid in the tube is V, the speed of the ejection of the liquid through the holes is:

Accepted Answer

V²R² / nr². Using the principle of conservation of volume flow rate, the flow rate in the tube (πR²V) must equal the total flow rate through the holes (n × πr²v). Solving for v, we get v = V(R² / nr²).

Question 7

A small hole of area of cross-section 2 mm² is present near the bottom of a fully filled open tank of height 2 m. Taking g = 10 m/s², the rate of flow of water through the open hole would be nearly:

Accepted Answer

12.6 × 10⁻⁶ m³/s. The rate of flow can be calculated using Torricelli's theorem: velocity v = √(2gh). Substituting g = 10 m/s² and h = 2 m, v = √(2 × 10 × 2) = √40 ≈ 6.32 m/s. The flow rate Q = A × v, where A = 2 × 10⁻⁶ m². Thus, Q ≈ 2 × 10⁻⁶ × 6.32 ≈ 12.6 × 10⁻⁶ m³/s.

Question 8

A small sphere of radius ‘r’ falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to:

Accepted Answer

r⁵. The rate of heat production due to viscous force is equal to the power dissipated, which is given by the product of the viscous force and terminal velocity. Using Stokes' law, the viscous force is proportional to r, and terminal velocity is proportional to r². Thus, the power (rate of heat production) is proportional to r × r² = r⁵.

Question 9

A U tube with both ends open to the atmosphere, is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level (see diagr

Accepted Answer

650 kg/m³. The pressure at the same horizontal level in both arms of the U-tube must be equal. Using the hydrostatic pressure formula, equate the pressure due to the column of oil to the pressure due to the column of water. Solving gives the density of oil as 650 kg/m³.

Question 10

A wind with speed 40 m/s blows parallel to the roof of a house. The area of the roof is 250 m². Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be (ρ_air = 1.2 kg/m³):

Accepted Answer

2.4 × 10⁵ N, upwards. The pressure difference due to the wind is given by Bernoulli's principle: ΔP = 0.5 * ρ_air * v². Substituting ρ_air = 1.2 kg/m³ and v = 40 m/s, we get ΔP = 0.5 * 1.2 * (40)² = 960 Pa. The force is then F = ΔP * A = 960 * 250 = 2.4 × 10⁵ N. Since the pressure outside is lower than inside, the force acts upwards.

Question 11

A fluid is in streamline flow across a horizontal pipe of variable area of cross section. For this, which of the following statements is correct?

Accepted Answer

The velocity is maximum at the narrowest part of the pipe and pressure is maximum at the widest part of the pipe. According to Bernoulli's principle and the equation of continuity, the velocity of the fluid is maximum at the narrowest part of the pipe due to the reduced cross-sectional area, and the pressure is maximum at the widest part of the pipe where the velocity is lower.

Question 12

Two small spherical metal balls, having equal masses, are made from materials of densities ρ₁ and ρ₂ (ρ₁ = 8ρ₂) and have radii of 1 mm and 2 mm, respectively. They are made to fall vertically (from rest) in a viscous medium whose coefficient of viscosity equals η and whose density is 0.1ρ₂

Accepted Answer

79 / 36. The terminal velocity of a sphere in a viscous medium is proportional to the square of its radius and the difference in density between the sphere and the medium. Using the given densities and radii, the ratio of terminal velocities is calculated as (r₁²(ρ₁ - ρ_medium)) / (r₂²(ρ₂ - ρ_medium)) = 79/36.

Question 13

Water rises to a height 'h' in a capillary tube. If the length of capillary tube above the surface of water is made less than 'h' then:

Accepted Answer

water rises upto the top of capillary tube and stays there without overflowing. When the length of the capillary tube is less than the height 'h', water will rise to the top of the tube and stay there due to capillary action, but it will not overflow as the adhesive and cohesive forces balance out.

Question 14

A certain number of spherical drops of a liquid of radius 'r' coalesce to form a single drop of radius 'R' and volume 'V'. If 'T' is the surface tension of the liquid, then:

Accepted Answer

energy = 3VT (1/r - 1/R) is released. When smaller drops coalesce into a larger drop, the total surface area decreases, leading to a reduction in surface energy. The energy released is proportional to the change in surface area, which is given by 3VT(1/r - 1/R).

Question 15

A soap bubble, having radius of 1 mm, is blown from a detergent solution having a surface tension of 2.5 × 10⁻² N/m. The pressure inside the bubble equals at a point Z₀ below the free surface of water in a container. Taking g = 10 m/s², density of water = 10³ kg/m³, the value of Z₀ is:

Accepted Answer

10 cm. The excess pressure inside a soap bubble is given by ΔP = 4T/r, where T is the surface tension and r is the radius. Substituting the values, ΔP = (4 × 2.5 × 10⁻²) / (1 × 10⁻³) = 100 Pa. The pressure at depth Z₀ in water is given by P = ρgZ₀. Equating ΔP = ρgZ₀, we get Z₀ = ΔP / (ρg) = 100 / (10³ × 10) = 0.01 m = 10 cm.

Question 16

The wettability of a surface by a liquid depends primarily on:

Accepted Answer

angle of contact between the surface and the liquid. The wettability of a surface by a liquid is determined by the angle of contact between the liquid and the surface. A smaller contact angle indicates better wettability.

Question 17

A capillary tube of radius r is immersed in water and water rises in it to a height h. The mass of the water in the capillary is 5 g. Another capillary tube of radius 2r is immersed in water. The mass of water that will rise in this tube is:

Accepted Answer

10.0 g. The height of water in a capillary is inversely proportional to the radius of the tube (h ∝ 1/r). The volume of water in the capillary is proportional to the product of the height and the square of the radius (V ∝ h × r²). Doubling the radius (2r) reduces the height to h/2, but the cross-sectional area increases by a factor of 4. Thus, the mass of water (proportional to volume) becomes 2 × 5 g = 10 g.

Question 18

The angle of contact between pure water and pure glass, is:

Accepted Answer

0°. The angle of contact between pure water and pure glass is 0° because water completely wets the glass surface due to strong adhesive forces between water molecules and glass.

Question 19

Inflow rate of volume of the liquid = Outflow rate of volume of the liquid. πR²V = πr²(v) => πR² = πr² => nr².

Accepted Answer

πR²V = πr²v. The equation πR²V = πr²v represents the continuity equation, which states that the inflow rate of volume equals the outflow rate of volume for an incompressible fluid. This matches the given condition.

Question 20

Power = rate of production of heat = F.V = 6πηr Vₜ. Vₜ = 6πηr Vₜ² (∴ F = 6πηr Vₜ stoke's formula). Vₜ ∝ r².

Accepted Answer

Vₜ ∝ r². The terminal velocity (Vₜ) is proportional to the square of the radius (r²) as derived from Stoke's law. This relationship is correctly represented by option D.

NEET Physics: Mechanical Properties of Fluids questions with solutions

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