HC Verma Concepts Of Physics Exercise Solutions Of Chapter 14 (Some Mechanical Properties Of Matter)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 14 - Some Mechanical Properties Of Matter:

EXERCISE

1. A load of 10 kg is suspended by a metal wire 3 m long and having a cross-sectional area 4 mm². Find (a) the stress (b) the strain and (c) the elongation. Young's modulus of the metal is 2.0 * 10¹¹ N/m².

Sol:

Given: mass, m = 10 kg; length of wire, L = 3 m; cross-sectional area of wire, A = 4 mm² = 4 * 10^–6 m²; Y = 2.0 * 10¹¹ N/m².

Force, F = mg = 10 * 10 = 100 N.

(a) We know,

→ Stress, σ = Force/area = 100/ (4 * 10^–6)

Or, Stress, σ = 2.5 * 10⁷ N/m².

(b) We know,

→ Stress = Young's modulus * strain

Or, σ = Y * ε

Or, ε = σ/Y = (2.5 * 10⁷)/ (2.0 * 10¹¹)

Or, strain, ε = 1.25 * 10^–4.

(c) We know,

→ Strain, ε = (elongation of wire, ΔL)/ (length of wire, L)

Or, elongation of wire, ΔL = ε * L

Or, elongation of wire, ΔL = 1.25 * 10^–4 * 3

Or, elongation of wire, ΔL = 3.75 * 10^–4 m.

2. A vertical metal cylinder of radius 2 cm and length 2 m is fixed at the lower end and a load of 100 kg is put on it. Find (a) the stress (b) the strain and (c) the compression of the cylinder. Young's modulus of the metal = 2 * 10¹¹ N/m².

Sol:

Given: radius of cylinder, r = 2 cm = 0.02 m; length, L = 2 m; mass of load, m = 100 kg; Y = 2 * 10¹¹ N/m²; Area, A = πr² = 0.001257 m².

Force applied on cylinder, F = mg

Or, F = 100 * 10 = 1000 N

(a) We know,

→ Stress, σ = Force/area
Or, σ = 1000/ (0.001257)

Or, Stress, σ = 7.96 * 10⁵ N/m².

(b) We know,

→ Stress = Young's modulus * strain

Or, σ = Y * ε

Or, ε = σ/Y = (7.96 * 10⁵)/ (2 * 10¹¹)

Or, strain, ε = 4 * 10^-6.

(c) We know,

→ Strain, ε = (compression of cylinder, ΔL)/ (length, L)

Or, compression of cylinder, ΔL = ε * L

Or, compression of cylinder, ΔL = 4 * 10^–6 * 2

Or, compression of cylinder, ΔL = 8 * 10^-6 m. 

3. The elastic limit of steel is 8 * 10⁸ N/m² and its Young's modulus 2 * 10¹¹ N/m². Find the maximum elongation of a half-meter steel wire that can be given without exceeding the elastic limit.

Sol:

Given: stress of steel upto elastic limit, σ = 8 * 10⁸ N/m²; Y = 2 * 10¹¹ N/m²; length of steel wire, L = 0.5 m.

We know,

→ Stress = Young's modulus * strain

Or, σ = Y * ε

Or, ε = σ/Y = (8 * 10⁸)/ (2 * 10¹¹)

Or, strain, ε = 4 * 10^-3.

And, strain, ε = (ΔL)/L

Or, ΔL = ε * L = 4 * 10^-3 * 0.5

Or, ΔL = 2 * 10^-3 m = 2 mm.

4. A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of (a) the stresses developed in the two wires and (b) the strains developed. Y of steel = 2 * 10¹¹ N/m². Y of copper = 1.3 * 10¹¹ N/m².

Sol: 

Given: Length of copper, L_c = Length of steel, L_s; sectional area of copper, A_c = sectional area of steel, A_s; Y_s = 2 * 10¹¹ N/m²; Y_c = 1.3 * 10¹¹ N/m².

(a) Stress developed in copper wire is = σ_c = P/ A_c

And, Stress developed in steel wire is = σ_s = P/ A_s

So, the ratio of the stresses developed in the two wires is = σ_c/ σ_s = 1 [since A_c = A_s]

(b) Strain developed in copper wire is = ε_c = σ_c / Y_c

And, strain developed in steel wire is = ε_s = σ_s/ Y_s 

So, the ratio of the strains developed in the two wires is = ε_c / ε_s = (σ_c / Y_c)/ (σ_s / Y_s) = Y_s/ Y_c = (2 * 10¹¹)/ (1.3 * 10¹¹) = 20/13.

5. In figure (14-E1) the upper wire is made of steel and the lower of copper. The wires have equal cross-section. Find the ratio of the longitudinal strains developed in the two wires.

Sol:

Given: Length of copper, L_c = Length of steel, L_s; sectional area of copper, A_c = sectional area of steel, A_s; Y_s = 2 * 10¹¹ N/m²; Y_c = 1.3 * 10¹¹ N/m².

Stress developed in copper wire is = σ_c = P/ A_c

And, Stress developed in steel wire is = σ_s = P/ A_s

So, the ratio of the stresses developed in the two wires is = σ_c/ σ_s = 1 [since A_c = A_s]

Strain developed in copper wire is = ε_c = σ_c / Y_c

And, Strain developed in steel wire is = ε_s = σ_s/ Y_s 

So, the ratio of the strains developed in the two wires is = ε_c / ε_s = (σ_c / Y_c)/ (σ_s / Y_s) = Y_s/ Y_c = (2 * 10¹¹)/ (1.3 * 10¹¹) = 1.54. 

6. The two wires shown in figure (14-E2) are made of the same material which has a breaking stress of 8 * 10⁸ N/m². The area of cross-section of the upper wire is 0.006 cm² and that of the lower wire is 0.003 cm². The mass m₁ = 10 kg, m₂ = 20 kg and the hanger is light, (a) Find the maximum load that can be put on the hanger without breaking a wire. Which wire will beak first if the load is increased? (b) Repeat the above part if m₁ = 10 kg and m₂ = 36 kg.

Sol:

Given: breaking stress of Material, σ_b = 8 * 10⁸ N/m²; area of cross-section of the upper wire, A_u = 0.006 cm²; area of cross-section of the lower wire, A_l = 0.003 cm²; m₁ = 10 kg; m₂ = 20 kg.

(a) Let, the maximum load be m kg.

Tension in upper wire, T₁ = (m₁ + m₂ + m) g

And, stress in upper wire, σ₁ = (m₁ + m₂ + m) g/A_u

Tension in lower wire, T₂ = (m₁ + m) g

And, stress in upper wire, σ₂ = (m₁ + m) g/A_l

Both wire has same breaking stress

So, (m₁ + m₂ + m) g/A_u = 8 * 10⁸

Or, (10 + 20 + m) * 10/ (0.006 * 10^-4) = 8 * 10⁸

Or, m = 18 kg

And, (m₁ + m) g/A_l = 8 * 10⁸

Or, (10 + m) * 10/ (0.003 * 10^-4) = 8 * 10⁸

Or, m = 14 kg

For same breaking stress, the maximum load that can be put is 14 kg. If the load is increased the lower wire will break first.

(b) Given: m₁ = 10 kg and m₂ = 36 kg.

 Let, the maximum load be m kg.

Tension in upper wire, T₁ = (m₁ + m₂ + m) g

And, stress in upper wire, σ₁ = (m₁ + m₂ + m) g/A_u

Tension in lower wire, T₂ = (m₁ + m) g

And, stress in upper wire, σ₂ = (m₂ + m) g/A_l

Both wire has same breaking stress

So, (m₁ + m₂ + m) g/A_u = 8 * 10⁸

Or, (10 + 36 + m) * 10/ (0.006 * 10^-4) = 8 * 10⁸

Or, m = 2 kg

And, (m₁ + m) g/A_l = 8 * 10⁸

Or, (10 + m) * 10/ (0.003 * 10^-4) = 8 * 10⁸

Or, m = 14 kg

For same breaking stress, the maximum load that can be put is 2 kg. If the load is increased the upper wire will break first.

7. Two persons pull a rope towards themselves. Each person exerts a force of 100 N on the rope. Find Young's modulus of the material of the rope if it extends in length by 1 cm. Original length of the rope = 2 m and the area of cross-section = 2 cm².

Sol:

Given: Force, F = 100 N; area of cross-section, A = 2 cm² = 2 * 10^-4 m²; length of the rope, L = 2 m; extends in length, ΔL = 1 cm = 0.01 m.

We know,

Stress, σ = F/A = 100/ (2 * 10^-4) 
Or, σ = 50 * 10⁴ N/m²

And, strain, ε = ΔL/L = 0.01/2 = 5 * 10^-3

From Hooke’s law

→ σ = Y ε

Or, Y = σ/ε = (50 * 10⁴)/ (5 * 10^-3)

Or, Y = 10⁸ N/m² 

So, Young's modulus of the material of the rope is 10⁸ N/m².

8. A steel rod of cross-sectional area 4 cm² and length 2 m shrinks by 0.1 cm as the temperature decreases in night. If the rod is clamped at both ends during the day hours, find the tension developed in it during night hours. Young's modulus of steel = 1.9 * 10¹¹ N/m².

Sol:

Given: cross-sectional area of rod, A = 4 cm² = 4 * 10^-4 m²; length of the rod, L = 2 m; change in length, ΔL = 0.1 cm = 0.001 m; Y = 1.9 * 10¹¹ N/m².

From Hooke’s law

→ σ = Y ε

Or, F/A = Y (ΔL/L)

Or, F = YA (ΔL/L)

Or, F = 1.9 * 10¹¹ * 4 * 10^-4 * (0.001/2)

Or, F = 3.8 * 10⁴ N.

So, the tension developed in it during night hours is 3.8 * 10⁴ N.

9. Consider the situation shown in figure (14-E3). The few F is equal to the m₂g/2. If the area of cross-section of the string is A and its Young's modulus Y, find the strain developed in it. The string is light and there is no friction anywhere.

Sol:

Given: F = m₂g/2; area of cross-section of the string = A; Young's modulus = Y.

Let, tension in the string be T.

Equation of motion of block m₁:

→ m₁a = T – F

Or, m₁a = T – m₂g/2 ----------- (1)

Equation of motion of block m₂:

→ m₂a = m₂g – T ------------- (2)

Solving equation (1) and (2) we get,

Tension in the string, T = [m₂g (2m₁ + m₂)]/ [2(m₁ + m₂)]

Therefore, strain developed in the string is ε = T/AY = [m₂g (2m₁ + m₂)]/ [2AY (m₁ + m₂)].

10. A sphere of mass 20 kg is suspended by a metal of unstretched length 4 m and diameter 1 mm. When in equilibrium, there is a clear gap of 2 mm between sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle θ with the vertical is released. Find the maximum value of θ so that sphere does not rub the floor. Young's modulus of metal of the wire is 2.0 * 10¹¹ N/m². Make appropriate approximations.

Sol:

Given: mass of sphere, m = 20 kg; length, L = 4 m; diameter of metal, d = 1 mm = 0.001 m; ΔL = 0.002 m; Y = 2.0 * 10¹¹ N/m².

Cross section area, A = πd²/4 = 0.78 * 10^-6 m².

At equilibrium → T = mg

When it moves to an angle θ and released, the tension the T’ at lowest point is

→ T’ = mg + mv²/L

The change in tension is due to centrifugal force ΔT = mv²/L … (1)

→ Again, by work energy principle,

→ ½ mv² – 0 = mgL (1 – cos θ)

Or, v² = 2gL (1 – cos θ) … (2)

So, ΔT = m [2gL (1 – cos θ)]/L

Or, ΔT = 2mg (1 – cos θ)

→ F = ΔT

From Hooke’s law

→ σ = Y ε

Or, F/A = Y (ΔL/L)

Or, F = YA (ΔL/L)

So, YA (ΔL/L) = 2mg (1 – cos θ)

Or, cos θ = 1 – (YA ΔL/2mgL)

Or, cos θ = 1 – [(2 * 10¹¹ * 0.78 * 10^-6 * 0.002)/ (2 * 20 * 10 * 4)]

Or, cos θ = 0.805

Or, θ = 36.4⁰.

12. A copper wire of cross-sectional area 0.01 cm² is under a tension of 20 N. Find the decrease in the cross-sectional area. Young's modulus of copper = 1.1 * 10¹¹ N/m² and Poisson's ratio = 0.32. [Hint: ΔA/A = 2Δr/r]

Sol:

Given: cross-sectional area of copper wire, A = 0.01 cm²; Force, F = 20 N, Y = 1.1 * 10¹¹ N/m²; Poisson's ratio, ν = 0.32.

We know, Y = FL/AΔL

Or, ΔL/L = F/AY

And, ν = (ΔD/D)/ (ΔL/L)

Or, ν = (Δr/r)/ (ΔL/L)

Or, (Δr/r) = ν * (ΔL/L) = ν * (F/AY)

Again, ΔA/A = 2Δr/r

Or, ΔA = 2AΔr/r = 2 * A * ν * (F/AY)

Or, ΔA = 2νF/Y = (2 * 0.32 * 20)/ (1.1 * 10¹¹)

Or, ΔA = 1.164 * 10^-10 m² = 1.164 * 10^-6 cm².

So, the decrease in the cross-sectional area is 1.164 * 10^-6 cm².

13. Find the increase in pressure required to decrease the volume of a water sample by 0.01%. Bulk modulus of water = 2.1 * 10³ N/m².

Sol:

Given: ΔV/V = – 0.01 % = – 0.01 * 100 = – 1; Bulk modulus of water, K = 2.1 * 10³ N/m².

We know,

Bulk modulus of water, K = – ΔP/ (ΔV/V)

Or, ΔP = – K * (ΔV/V)

Or, ΔP = – 2.1 * 10³ * (– 1)

Or, ΔP = 2.1 * 10³ N/m².

Therefore, the increase in pressure required is 2.1 * 10³ N/m².

14. Estimate the change in the density of water in ocean at a depth of 400 m below the surface. The density of water at the surface = 1030 kg/m³ and the bulk modulus of water = 2 * 10⁹ N/m².

Sol:

Given: depth of ocean, h = 400 m; density of water at the surface, ρ = 1030 kg/m³; bulk modulus of water, K = 2 * 10⁹ N/m².

Change in pressure at the depth, ΔP = ρgh = 1030 * 10 * 400 = 412 * 10⁴ N/m².

We know, K = – ΔP/ (ΔV/V) = ΔP/ (Δρ/ρ)

Or, Δρ = (ΔP * ρ) / (K)

Or, Δρ = (412 * 10⁴ * 1030)/ (2 * 10⁹)

Or, Δρ = 2.1 kg/m³.

Therefore, the change in the density of water 2.1 kg/m³.

15. A steel plate of face-area 4 cm² and thickness 0.5 cm is fixed rigidly at the lower surface. A tangential force of 10 N is applied on the upper surface. Find the lateral displacement of the upper surface with respect to the lower surface. Rigidity modulus of steel = 8.4 * 10¹⁰ N/m².

Sol:

Given: Rigidity modulus of steel, G = 8.4 * 10¹⁰ N/m²; tangential force, F = 10 N; face-area of plate, A = 4 cm² = 4 * 10^-4 m²; thickness, t = 0.5 cm = 0.005 m.

Let we assume the lateral displacement of the upper surface be X.

Shear stress, τ = F/A

Or, τ = 10/ (4 * 10^-4)

Or, τ = 2.5 * 10⁴ N/m².

We know, τ = Gγ

Or, γ = τ/G = (2.5 * 10⁴)/ (8.4 * 10¹⁰)

Or, γ = 0.298 * 10^-6 radian.

And, tan γ = X/t

Or, γ = X/t       [tan γ = γ, since γ is very small and γ in radian]

Or, (0.298 * 10^-6) = X/ (0.005)

Or, X = 1.5 * 10^-9 m.

So, the lateral displacement of the upper surface is 1.5 * 10^-9 m.

16. A 5.0 cm long straight piece of thread is kept on the surface of water. Find the force with which the surface on one side of the thread pulls it. Surface tension of water = 0.076 N/m.

Sol:

Given: length of the thread, L = 5 cm = 0.05 m; Surface tension of water, S = 0.076 N/m.

We know, Surface tension, S = F/L

Or, pull force, F = SL

Or, F = 0.076 * 0.05

Or, F = 3.8 * 10^-3 N.

So, the force with which the surface on one side of the thread pulls it is 3.8 * 10^-3 N.

17. Find the excess pressure inside (a) a drop of mercury of radius 2 mm (b) a soap bubble of radius 4 mm and (c) an air bubble of radius 4 mm formed inside a tank of water. Surface tension of mercury, soap solution and water are 0.465 N/m, 0.03 N/m and 0.076 N/m respectively.

Sol:

Given: Surface tension of mercury, S_Hg = 0.465 N/m; Surface tension of soap solution, S_ss = 0.03 N/m and Surface tension of water, S_w = 0.076 N/m.

(a) Surface tension of mercury, S = 0.465 N/m, Radius of mercury drop, r = 2 mm = 0.002 m.

The excess pressure inside the mercury drop, ΔP = 2S/r

Or, ΔP = (2 * 0.465)/ 0.002 = 465 N/m².

(b) Surface tension of mercury, S = 0.03 N/m, Radius of mercury drop, r = 4 mm = 0.004 m.

The excess pressure inside a soap bubble, ΔP = 4S/r

Or, ΔP = (4 * 0.03)/ 0.004 = 30 N/m².

(c) Surface tension of water, S = 0.076 N/m, Radius of mercury drop, r = 4 mm = 0.004 m. 

The excess pressure inside an air bubble, ΔP = 2S/r

Or, ΔP = (2 * 0.076)/ 0.004 = 38 N/m².

18. Consider a small surface area of 1 mm² at the top of a mercury drop of radius 4.0 mm. Find the force exerted on this area (a) by the air above it (b) by the mercury below it and (c) by the mercury surface in contact with it. Atmospheric pressure = 1.0 * 10⁵ Pa and surface tension of mercury = 0.465 N/m. Neglect the effect of gravity. Assume all numbers to be exact.

Sol:

Given: surface tension of mercury, S = 0.465 N/m; Atmospheric pressure, P₀ = 1.0 * 10⁵ Pa; small area, a = 1 mm² = 10^-6 m²; radius of the drop, r = 4 mm = 0.004 m.

(a) The force exerted on small area by the air, F = P₀a

Or, F = 1.0 * 10⁵ * 10^-6 = 0.1 N.

(b) Pressure of the mercury on the small area, P’ = P₀ + 2S/r

Therefore, force exerted on small area by the mercury is = P’a = (P₀ + 2S/r) * a = (10⁵ + 2*0.465/0.004) * 10^-6 = 0.10023 N.

(c) P = 2S/r = (2*0.465/0.004) = 232.5 N/m².

So, force exerted on small area by the mercury surface in contact with it, F = Pa = 232.5 * 10^-6 = 0.00023 N.

19. The capillaries shown in figure (14-E4) have inner radii 0.5 mm, 1.0 mm and 1.5 mm respectively. The liquid in the beaker is water. Find the heights of water level in the capillaries. The surface tension of water is 7.5 * 10^-2 N/m.

Sol:

Given: surface tension of water, S = 7.5 * 10^-2 N/m; Contact angle (water with glass), θ = 0⁰.

(i) Inner radius of tube A, r = 0.5 mm = 0.0005 m.

We know,

Height of water rise, h = (2S cos θ)/ (rρg)

Or, h = (2 * 7.5 * 10^-2 * cos 0)/ (0.0005 * 1000 * 10)

Or, h = 3 * 10^-2 m = 3 cm.

(ii) Inner radius of tube B, r = 1 mm = 0.001 m.

We know,

Height of water rise, h = (2S cos θ)/ (rρg)

Or, h = (2 * 7.5 * 10^-2 * cos 0)/ (0.001 * 1000 * 10)

Or, h = 1.5 * 10^-2 m = 1.5 cm.

(iii) Inner radius of tube C, r = 1.5 mm = 0.0015 m.

We know,

Height of water rise, h = (2S cos θ)/ (rρg)

Or, h = (2 * 7.5 * 10^-2 * cos 0)/ (0.0015 * 1000 * 10)

Or, h = 1 * 10^-2 m = 1 cm.

20. The lower end of a capillary tube is immersed in mercury. The level of mercury in the tube is found to be 2 cm below the outer level. If the same tube is immersed in water, upto what height will the water rise in the capillary?

Sol:

Given: capillary depression, h_d = – 2 cm = – 0.02 m; contact angle (Mercury with glass), θ_Hg,g = 140⁰; contact angle (Water with glass), θ_w,g = 0⁰; density of mercury, ρ_Hg = 13600 kg/m³; surface tension of water, S_w = 0.075 N/m; surface tension of mercury, S_Hg = 0.465 N/m.

We know, h_d = (2S_Hg cos θ_Hg,g)/ (rρ_Hgg)

Or, – 0.02 = (2 * 0.465 * cos 140⁰)/ (r * 13600 * 10)

Or, r = 0.0002 m = 0.2 mm.

Again, h_r = (2S_w cos θ_w,g)/ (rρ_wg)

Or, h_r = (2 * 0.075 * cos 0⁰)/ (0.0002 * 1000 * 10)

Or, h_r = 0.0573 m = 5.73 cm.

Therefore, the height of the water raised in the capillary is 5.73 cm.

21. A barometer is constructed with its tube having radius 0.1 mm. Assume that the surface of mercury in the tube is spherical in shape. If the atmospheric pressure is equal to 76 cm of mercury, what will be the height raised in the barometer tube. The contact angle of mercury with glass = 135° and surface tension of mercury = 0.465 N/m. Density of mercury = 13600 kg/m³.

Sol:

Given: radius of tube, r = 1.0 mm = 0.001 m; pressure, p = 76 cm of Hg; contact angle of mercury with glass, θ = 135°; surface tension of mercury, S = 0.465 N/m; Density of mercury, ρ = 13600 kg/m³.

We have, h = (2S cos θ)/ (rρg)

Or, h = (2 * 0.465 * cos 135⁰)/ (0.001 * 13600 * 10)

Or, h = – 0.005 m = – 0.5 cm.

So, the height raised in the barometer tube is (76 – 0.5) = 75.5 cm.

23. Find the surface energy of water kept in a cylindrical vessel of radius 6.0 cm. Surface tension of water = 0.075 J/m².

Sol:

Given: radius of cylindrical vessel, r = 6 cm = 0.06 m; Surface tension of water, S = 0.075 J/m².

Area of surface, A = πr² = π * (0.06)² = 0.0113 m².

We know,

→ Surface energy = Surface tension * Area

Or, Surface energy = 0.075 * 0.0113

Or, Surface energy = 0.00085 = 8.5 * 10^-4 J.

24. A drop of mercury of radius 2 mm is split into 8 identical droplets. Find the increase in surface energy. Surface tension of mercury = 0.465 J/m².

Sol:

Given: radius of mercury drop, r = 2 mm = 0.002 m; Surface tension of mercury, S = 0.465 J/m².

Drop of mercury split into 8 identical droplets.

Let, the radius of new droplet be r’.

Volume of water remain constant.

So, volume of old droplet = 8 * volume of new droplet

Or, (4/3) πr³ = 8 * (4/3) π (r’)³

Or, r’ = r/2 = 0.002/2 = 0.001 m.

The increase in surface energy = [8 * 4π (r’)² – 4πr²] * S = 4π * (8 * 0.001² – 0.002²) * 0.465 = 23.4 * 10^-6 J = 23.4 µJ.

25. A capillary tube of radius 1 mm is kept vertical with the lower end in water, (a) Find the height of water raised in the-capillary, (b) If the length of the capillary tube is half the answer of part (a), find the angle θ made by the water surface in the capillary with the wall.

Sol:

Given: radius of tube, r = 1 mm = 0.001 m; contact angle (water with glass), θ = 0⁰; density of water, ρ = 1000 kg/m³; Surface tension of water, S = 0.075 N/m.

(a) We know,

→ Height of water rise, h = (2S cos θ)/ (rρg)

Or, h = (2 * 0.075 * cos 0⁰)/ (0.001 * 1000 * 10)

Or, h = 0.015 m = 1.5 cm.

(b) New height, H = h/2 = 0.015/2 = 0.0075 m.

We know,

→ Height of water rise, H = (2S cos θ)/ (rρg)

Or, 0.0075 = (2 * 0.075 * cos θ)/ (0.001 * 1000 * 10)

Or, cos θ = ½

Or, θ = 60⁰.

26. The lower end of a capillary tube of radius 1 mm is dipped vertically into mercury, (a) Find the depression of mercury column in the capillary, (b) If the length dipped inside is half the answer of part (a), find the angle made by the mercury surface at the end of the capillary with the vertical. Surface tension of mercury = 0.465 N/m and the contact angle of mercury with glass = 135°.

Sol:                                       

Given: radius of tube, r = 1 mm = 0.001 m; contact angle (mercury with glass), θ = 135⁰; density of mercury, ρ = 13600 kg/m³; Surface tension of mercury, S = 0.465 N/m.

(a) We know, 

→ Height of mercury, h = (2S cos θ)/ (rρg)

Or, h = (2 * 0.465 * cos 135⁰)/ (0.001 * 13600 * 10)

Or, h = – 0.00534 m = – 5.34 mm.

So, the depression of mercury column in the capillary is 5.34 mm.

(b) New height, H = h/2 = 0.00534 /2 = 0.00267 m.

We know,

→ Height of mercury dip, H = (2S cos θ)/ (rρg)

Or, – 0.00267 = (2 * 0.075 * cos θ)/ (0.001 * 1000 * 10)

Or, cos θ = – 0.3746

Or, θ = 112⁰.

27. Two large glass plates are placed vertically and parallel to each other inside a tank of water with separation between the plates equal to 1 mm. Find the rise of water in the space between the plates. Surface tension of water = 0.075 N/m.

Sol:

Given: separation between the plates, L = 0.001 m; Surface tension of water, S = 0.075 N/m, contact angle, θ = 0⁰.

Let the rise of water in the space between the plates be h and b be the width of the plate.

Now,

→ Weight of water rise = force on liquid surface by plate surface

Or, (0.001 * b) hρg = 2 * b * S cos θ

Or, h = (2 * 0.075 * 1)/ (1000 * 10 * 0.001)

Or, h = 0.015 m = 1.5 cm.

28. Consider an ice cube of edge 1.0 cm kept in a gravity free hall. Find the surface area of the water when the ice melts. Neglect the difference in densities of ice and water.

Sol:

Given: edge of cube, L = 1 cm.

In the gravity free hall, the ice melts takes the spherical shape due to surface tension.

Since, the difference in densities of ice and water is neglected.

So, volume of ice cube = volume of water

Or, L³ = (4/3) πr³

Or, r = (3/4π)^1/3 cm.

Therefore, the surface area of the water = 4 πr² = 4 * π * [(3/4π)^1/3]² = (36π)^1/3 cm².

29. A wire forming a loop is dipped into soap solution and taken out so that a film of soap solution is formed. A loop of 6.28 cm long thread is gently put on the film and the film is pricked with a needle inside the loop. The thread loop takes the shape of a circle. Find the tension in the thread. Surface tension of soap solution = 0.030 N/m.

Sol:

Given: length of the thread, L = 6.28 cm = 0.0628 m; Surface tension of soap solution, S = 0.03 N/m.

→ Perimeter of circle = 2πr = 0.0628

Or, r = 0.01 m.

The length of small element = r dθ

→ dF = S * rdθ

Considering symmetric elements,

→ dF_y = 2 S rdθ * Sin θ [dF_x = 0]

So, F = 2Sr ∫ (sin θ) dθ

Limit → 0 to π/2

Or, F = – 2Sr [cos π/2 – cos 0]

Or, F = 2Sr

Tension → 2T = 2Sr

Or, T = 0.03 * 0.01 = 0.0003 N

Or, T = 3 * 10^-4 N.

30. A metal sphere of radius 1 mm and mass 50 mg falls vertically in glycerine. Find (a) the viscous force exerted by the glycerine on the sphere when the speed of the sphere is 1 cm/s, (b) the hydrostatic force exerted by the glycerine on the sphere and (c) the terminal velocity with which the sphere will move down without acceleration. Density of glycerine = 1260 kg/m³ and its coefficient of viscosity at room temperature = 8.0 poise.

Sol:

Given: radius of metal sphere, r = 1 mm = 0.001 m; mass, m = 50 mg = 5 * 10^-5 kg; Density of glycerine, ρ = 1260 kg/m³; coefficient of viscosity, μ = 8 poise = 0.8 Pa-s.

(a) The speed of the sphere, V = 1 cm/s = 0.01 m/s.

We know, the viscous force = 6πμrv

Or, the viscous force = 6 * π * 0.8 * 0.001 * 0.01

Or, the viscous force = 1.5 * 10^-4 N.

(b) Hydrostatic force = Buoyancy force

Or, Hydrostatic force = (4/3) πr³ρg

Or, Hydrostatic force = (4/3) * π * 0.001³ * 1260 * 10

Or, Hydrostatic force = 5.28 * 10^-5 N.

(c) Since the acceleration of sphere is zero, so the sphere is at equilibrium (summation of all the force acting on the sphere is zero).

→ 6πμrv + (4/3) πr³ρg = mg

Or, 6 * π * 0.8 * 0.001 * v + 5.28 * 10^-5 = 5 * 10^-5 * 10

Or, v = 0.029 m/s = 2.9 cm/s.

31. Estimate the speed of vertically falling raindrops from the following data. Radius of the drops = 0.02 cm. viscosity of air = 1.8 * 10^-4 poise, g = 9.9 m/s² and density of water = 1000 kg/m³.

Sol:

Given: Radius of the drops, r = 0.02 cm = 0.0002 m; viscosity of air, μ = 1.8 * 10^-4 poise = 1.8 * 10^-5 Pa-s.

To find the terminal velocity of rain drops, the forces acting on the drop are,

i) The weight (4/3) πr³ρ_wg downward.

ii) Force of buoyancy (4/3) πr³ρ_ag upward.

iii) Force of viscosity 6πμrv upward.

Because, ρ_a of air is very small, the force of buoyancy may be neglected.

Thus, 6πμrv = (4/3) πr³ρ_wg

Or, v = 2r²ρ_wg/9μ

Or, v = (2 * 0.0002² * 1000 * 9.9)/ (9 * 1.8 * 10^-5)

Or, v = 4.88 m/s ≈ 5 m/s.

32. Water flows at a speed of 6 cm/s through a tube of radius 1 cm. Coefficient of viscosity of water at room temperature is 0.01 poise. Calculate the Reynolds number. Is it a steady flow?

Sol:

Given: speed of flowing water, V = 6 cm/s = 0.06 m/s; radius of tube, r = 1 cm = 0.01 m; Coefficient of viscosity, μ = 0.01 poise = 0.001 Pa-s.

We know, Reynolds number = VρD/μ

Or, Reynolds number = (0.06 * 1000 * 0.02)/ 0.001

Or, Reynolds number = 1200

If the Reynolds number is less than 2000, then the flow is steady.

Therefore, it is a steady flow.

Discussion - If you have any Query or Feedback comment below.

Tagged With: Previous year Chapter-wise JEE Questions Solutions, Chapter-wise previous year solutions of IIT JEE, AIEEE, Other JEE Exams, HC Verma Solutions, Concepts ofPhysics Volume 1, Concepts of Physics Volume 2, HC Verma Solutions Part 1 andPart 2, Chapter wise solutions of hc verma’s Concepts of Physics, HC Verma, HCVerma Part 1 PDF Solution Download, HCVERMA SOLUTION (CHAPTERWISE), HC Verma Part 2 PDF Solution Download, hcverma part 1 solutions, hc verma part 2 solutions, hc verma objectivesolutions, Concepts of Physics solutions download, Solutions of HC VermaConcepts of Physics Volume 1 & 2.