HC Verma Concepts Of Physics Exercise Solutions Of Chapter 44 (X-Rays)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 44 - X-Rays:

EXERCISE

Planck constant, h = 4.14 * 10^–15 eV-s or, 6.63 * 10^–34 J-s; speed of light, c = 3 * 10⁸ m/s.

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1. Find the energy, the frequency and the momentum of an X-ray photon of wavelength 0.10 nm.

Sol:

Given: wavelength, λ = 0.10 nm = 1 * 10^–10 m.

We know,

→ λ = c/ν

Or, ν = c/λ

Or, ν = (3 * 10⁸)/ (1 * 10^–10)

Or, ν = 3 * 10¹⁸ s^–1 or Hz.

So, the frequency of wavelength is 3 * 10¹⁸ s^–1 or Hz.

→ Energy, E = hν

Or, E = 4.14 * 10^–15 * 3 * 10¹⁸

Or, E = 12.4 * 10³ eV = 12.4 keV.

→ Momentum, p = E/c

Or, p = (12.4 * 10³ * 1.6 * 10^–19)/ (3 * 10⁸)

Or, p = 6.62 * 10^–24 kg m/s.

2. Iron emits K_α X-ray of energy 6.4 keV and calcium emits K_α X-ray of energy 3.69 keV. Calculate the times taken by an iron K_α photon and a calcium K_α photon to cross through a distance of 3 km.

Sol:

Given: distance, S = 3 km = 30000 m.

Speed of both the photon are same and equal to speed of light, c = 3 * 10⁸ m/s.

So, the times taken by both = S/c = 3000/ (3 * 10⁸) = 10 µs.

3. Find the cut-off wavelength for the continuous X-rays coming from an X-ray tube operating at 30 kV.

Sol:

Given: potential, V = 30 kV = 30000 V.

We know,

→ λ_min = hc/eV

Or, λ_min = (4.14 * 10^–15 eV-s* 3 * 10⁸ m/s)/ (30000 eV)

Or, λ_min = 41.4 * 10^–12 m = 41.4 pm.

4. What potential difference should be applied across an X-ray tube to get X-ray of wavelength not less than 0.10 nm? What is the maximum energy of a photon of this X-ray in joule?

Sol:

Given: λ_min = 0.10 nm = 0.1 * 10^–9 m.

We know,

→ λ_min = hc/eV

Or, V = hc/ (eλ_min)

Or, V = (4.14 * 10^–15 eV-s * 3 * 10⁸ m/s)/ (e * 0.1 * 10^–9)

Or, V = 12.4 * 10³ V = 12.4 kV.

→ Maximum energy, E = eV

Or, E = 1.6 * 10^–19 * 12.4 * 10³

Or, E = 1.98 * 10^–15 J.

5. The X-ray coming from a Coolidge tube has a cut-off wavelength of 80 pm. Find the kinetic energy of the electrons hitting the target.

Sol:

Given: λ = 80 pm = 0.8 * 10^–10 m.

We know,

→ λ = hc/E

Or, E = hc/λ

Or, E = (4.14 * 10^–15 * 3 * 10⁸)/ (0.8 * 10^–10)

Or, E = 15.5 * 10³ eV = 15.5 keV.

So, the kinetic energy of the electrons is 15.5 keV.

6. If the operating potential in an X-ray tube is increased by 1%, by what percentage does the cutoff wavelength decrease?

Sol:

We know,

Wave length, λ = hc/eV

Now, potential, V is increased by 1%.

New wave length, λ’ = hc/1.01eV = λ/1.01

→ Δλ = λ – λ’ = (0.01/1.01) λ

So, % change of wave length = (Δλ/λ)*100

Or, % change of wave length = 1/1.01

Or, % change of wave length = 0.99 = 1 %.

7. The distance between the cathode (filament) and the target in an X-ray tube is 1.5 m. If the cutoff wavelength is 30 pm, find the electric field between the cathode and the target.

Sol:

Given: distance, d = 1.5 m; wavelength, λ = 30 pm = 0.3 * 10^–10 m. 

We know, E = hc/λ

Or, E = (4.14 * 10^–15 * 3 * 10⁸)/ (0.3 * 10^–10)

Or, E = 41.4 * 10³ eV

Or, potential, V = E/e = 41.4 * 10³ V

Now, Electric field = V/d

Or, Electric field = 41.4 * 10³/1.5

Or, Electric field = 27.6 * 10³ V/m

Or, Electric field = 27.6 kV/m.

8. The short-wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage?

Sol:
Given: λ’ = λ – 26 * 10^–12; V’ = 1.5 V.

Original wavelength, λ = hc/eV

New wavelength, λ’ = hc/eV’

→ λV = λ’V’

Or, λV = (λ – 26 * 10^–12) * 1.5 V

Or, λ = (1.5 * 26 * 10^–12)/0.5

Or, λ = 78 * 10^–12 m.

So, original voltage, V = hc/eλ

Or, V = (4.14 * 10^–15 eV-s * 3 * 10⁸ m/s)/ (e * 78 * 10^–12)

Or, V = 15.93 * 10³ V = 15.93 kV.

9. The electron beam in a colour TV is accelerated through 32 kV and then strikes the screen. What is the wavelength of the most energetic X-ray photon?

Sol:

Given: voltage, V = 32 kV = 32000 V.

We know, λ = hc/eV

Or, λ = (4.14 * 10^–15 eV-s* 3 * 10⁸ m/s)/ (320000 eV)

Or, λ = 38.8 * 10^–12 m = 38.8 pm.

10. When 40 kV is applied across an X-ray tube, X-ray is obtained with a maximum frequency of 9.7 * 10¹⁸ Hz. Calculate the value of Planck constant from these data.

Sol:

Given: voltage, V = 40 kV = 40 * 10³ V; frequency, ν = 9.7 * 10¹⁸ Hz.

We know, E = hν

Or, eV = hν

Or, h = eV/ν

Or, h = 40 * 10³/9.7 * 10¹⁸ eV-s

Or, h = 4.12 * 10^–15 eV-s.

11. An X-ray tube operates at 40 kV. Suppose the electron converts 70% of its energy into a photon at each collision. Find the lowest three wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.

Sol:

Given: voltage, V = 40 kV = 40 * 10³ V.

Energy, E = eV = 40 * 10³ eV.

→ Energy utilized = (70/100) * 40 * 10³ eV.

Or, Energy utilized = 28 * 10³ eV.

Now, λ = hc/E

Or, λ = (4.14 * 10^–15 * 3 * 10⁸)/ (28 * 10³)

Or, λ = 44.35 * 10^–12 = 44.35 pm.

For 2^nd wavelength:

→ Energy, E = 70% of left energy

Or, E = (70/100) * (40 – 28) * 10³ eV.

Or, E = 84 * 10² eV.

Now, λ = hc/E

Or, λ = (4.14 * 10^–15 * 3 * 10⁸)/ (84 * 10²)

Or, λ = 148 * 10^–12 = 148 pm.

For 3^rd wavelength:

→ Energy, E = 70% of left energy

Or, E = (70/100) * (12 – 8.4) * 10³ eV.

Or, E = 25.2 * 10² eV.

Now, λ = hc/E

Or, λ = (4.14 * 10^–15 * 3 * 10⁸)/ (25.2 * 10²)

Or, λ = 493 * 10^–12 = 493 pm.

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

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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. 

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HC Verma Concepts Of Physics Exercise Solutions Of Chapter 8 (Work and Energy)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 8 - Work and Energy:

EXERCISE

1. The mass of cyclist together with the bike is 90 kg. Calculate the increase in kinetic energy if the speed increases from 6.0 km/h to 12 km/h.

Sol:

Given: total mass, M = 90 kg; initial speed, u = 6 km/h = 5/3 m/s; final velocity, v = 12 km/h = 10/3 m/s.

Increase in kinetic energy,

→ ΔK.E. = ½ M (v² – u²)

Or, ΔK.E. = ½ * 90 * [(10/3)² – (5/3)²]

Or, ΔK.E. = 375 J.

2. A block of mass 2.00 kg moving at a speed of 10.0 m/s accelerates at 3.00 m/s² for 5.00 s. Compute its final kinetic energy.

Sol:

Given: mass, m = 2 kg; initial speed, u = 10 m/s; acceleration, a = 3 m/s²; time, t = 5 s.

→ Final velocity, v = u + at

Or, v = 10 + 3 * 5

Or, v = 25 m/s

Therefore, Final kinetic energy = ½ mv²

Or, Final kinetic energy = ½ * 2 * 25

Or, Final kinetic energy = 25 J.

3. A box is pushed through 4.0 m across a floor offering 100 N resistance. How much work is done by the resisting force?

Sol:

Given: resisting force, F = 100 N; displacement, s = 4 m.

→ Work is done, w = F * s

Or, w = 100 * 4 = 400 J.

4. A block of mass 5.0 kg slides down an incline of inclination 30° and length 10 m. Find the work done by the force of gravity.

Sol:

Given: mass, m = 5 kg; length of inclination, l = 10 m; angle of inclination, θ = 30⁰; g = 9.8 m/s².

Force along inclination, F = mg sin θ

Or, F = 5 * 9.8 * sin 30⁰

Or, F = 24.5 N

So, Work done by the force of gravity = F * l = 24.5 * 10 = 245 J.

5. A constant force of 2.50 N accelerates a stationary particle of mass 15 g through a displacement of 2.50 m. Find the work done and the average power delivered.

Sol:

Given: force, F = 2.5 N; displacement, s = 2.5 m; mass, m = 15 g = 0.015 kg; initial velocity, u = 0; a = 2.5/0.015 = 500/3 m/s².

The work done, w = F * s

Or, w = 2.5 * 2.5 = 6.25 J.

Applying work-energy principle,

→ ½ mv² – ½ mu² = w

Or, ½ * 0.015 * v² – 0 = 6.25

Or, v = 28.86 m/s

We know, t = (v – u)/a

Or, t = (28.86 – 0) * 3/500

Or, t = 0.173 s.

So, time taken to travel this distance is 0.173 s.

Therefore, average power delivered is = w/t = 36.1 watt.

6. A particle moves from a point r₁ = (2 m) i + (3 m) j to another point r₂ = (3 m) i + (2 m) j during which a certain force F = (5 N) i + (5 N) j acts on it. Find the work done by the force on the particle during the displacement.

Sol:

Displacement, r = r₁ – r₂

Or, r = [(2 m) i + (3 m) j] – [(3 m) i + (2 m) j]

Or, r = (– 1 m) i + (1 m) j

The work done, w = F. r

Or, w = [(5 N) i + (5 N) j]. [(– 1 m) i + (1 m) j]

Or, w = – 5 + 5 = 0

So, the work done by the force on the particle during the displacement is zero.

7. A man moves on a straight horizontal road with a block of mass 2 kg in his hand, If he covers a distance of 40 m with an acceleration of 0.5 m/s², find the work done by the man on the block during the motion.

Sol:

Given: mass of block, m = 2 kg; acceleration, a = 0.5 m/s²; distance covered, s = 40 m.

Work done, w = F * s

Or, w = m * a * s

Or, w = 2 * 0.5 * 40 = 40 J.

8. A force F = a + bx acts on a particle in the x-direction, where a and b are constants. Find the work done by this force during a displacement from x = 0 to x = d.

Sol:                                     

Work done, dw = F.dx

Or, w = ∫dw = ∫F.dx

Or, w = ∫ (a + bx) dx

Limit of integration: from 0 to d.

Or, w = [ax + bx²/2]^d_o 

Or, w = [ad + bd²/2]

Or, w = d [a + bd/2].

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HC Verma Concepts Of Physics Exercise Solutions Of Chapter 13 (Fluid Mechanics)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 13 - Fluid Mechanics:

EXERCISE

1. The surface of water in a water tank on the top of a house is 4 m above the tap level. Find the pressure of water at the tap when the tap is closed. Is it necessary to specify that the tap is closed? Take g = 10 m/s².

Sol:

Given: height, h = 4 m; g = 10 m/s²; density, ρ = 1000 kg/m³.

We know, pressure due to hydrostatic is given by

→ Pressure, p = hρg = 4 * 10 * 1000 = 40000 N/m² or Pa.

It is necessary to specify that the tap is closed. Otherwise pressure will gradually decrease as h decrease. Because, of the tap is open, the pressure at the tap is atmospheric.

2. The heights of mercury surfaces in the two arms of the manometer shown in figure (13-E1) are 2 cm and 8 cm.  Atmospheric pressure = 1.01 * 10⁵ N/m². Find (a) the pressure of the gas in the cylinder and (b) the pressure of mercury at the bottom of the U tube.

Sol:

Given: Atmospheric pressure, p_atm = 1.01 * 10⁵ N/m²; density of mercury, ρ_Hg = 13600 kg/m³; height difference between mercury surface, Δh = (8 – 2) = 6 cm = 0.06 m.

(a) We know, pressure at the same horizontal level in a continuous fluid are same.

So, p_A = p_B

Or, p_g = ρ_Hg *g*Δh + p_atm

Or, p_g = 13600 * 10 * 0.06 + 1.01 * 10⁵

Or, p_g = (0.0816 + 1.01) * 10⁵ = 1.0916 * 10⁵ N/m².

(b) The pressure of mercury at the bottom of the U tube is

= p_atm + ρ_Hg *g*h

= 1.01 * 10⁵ + 13600 * 10 * 0.08

= (1.01 + 0.11) * 10⁵ = 1.12 * 10⁵ N/m².

3. The area of cross-section of the wider tube shown in figure (13-E2) is 900 cm². If the boy standing on the piston weighs 45 kg, find the difference in the levels of water in the two tubes.

Sol:

Given: mass of man, m = 45 kg; area of wider tube, A = 900 cm² = 0.09 m²; density of water, ρ = 1000 kg/m³.

Let the difference in the levels of water in the two tubes be Δh.

We know, pressure at the same horizontal level in a continuous fluid are same.

So, p_A = p_B

Or, ρg (Δh) = mg/A

Or, Δh = m/Aρ

Or, Δh = 45/ (0.09 * 1000) = 0.5 m = 50 cm.

So, the difference in the levels of water in the two tubes is 50 cm.

4. A glass full of water has a bottom of area 20 cm², top of area 20 cm², height 20 cm and volume half a litre.

(a) Find the force exerted by the water on the bottom.

(b) Considering the equilibrium of the water, find the resultant force exerted by the sides of the glass on the water. Atmospheric pressure = 1.0 * 10⁵ N/m². Density of water = 1000 kg/m³ and g = 10 m/s². Take all numbers to be exact.

Sol:

Given: area of top of glass, A_t = area of bottom of glass, A_b = 20 cm² = 0.002 m²; height, h = 20 cm = 0.2 m; Atmospheric pressure, p_atm = 1.0 * 10⁵ N/m²; Density of water, ρ = 1000 kg/m³; g = 10 m/s²; mass of 0.5 litre water = 0.5 * 10^-3 * 1000 = 0.5 kg.

(a) Force exerted at the bottom

= Force due to cylindrical water column + atm. Force

= A_b * h * ρ * g + p_atm * A

= A_b * (h * ρ * g + p_atm)

= 0.002 * (0.2 * 1000 * 10 + 10⁵)

= 204 N.

(b) Let the resultant force exerted by the sides of the glass be F_side.

From the free body diagram of water inside the glass

→ F_side + F_bottom – p_a * A – mg = 0

Or, F_side + 204 – 10⁵ * 0.002 – 0.5 * 10 = 0

Or, F_side = 205 – 204 = 1 N upward.

5. Suppose the glass of the previous problem is covered by a jar and the air inside the jar is completely pumped out. (a) What will be the answers to the problem? (b) Show that the answers do not change if a glass of different shape is used provided the height, the bottom area and the volume are unchanged.

Sol:

Given: area of top of glass, A_t = area of bottom of glass, A_b = 20 cm² = 0.002 m²; height, h = 20 cm = 0.2 m; Atmospheric pressure, p_atm = 1.0 * 10⁵ N/m²; Density of water, ρ = 1000 kg/m³; and g = 10 m/s².

(a) Force exerted at the bottom.

= Force due to cylindrical water column + atm. Force

= A_b * h * ρ * g

= 0.002 * 0.2 * 1000 * 10

= 4 N.

(b) Let the resultant force exerted by the sides of the glass be F_side.

From the free body diagram of water inside the glass

→ F_side + F_bottom – mg = 0

Or, F_side + 4 – 0.5 * 10 = 0

Or, F_side = 5 – 4 = 1 N upward.

6. If water be used to construct a barometer, what would be the height of water column at standard atmospheric pressure (76 cm of mercury)?

Sol:

Given: Density of water, ρ_w = 1000 kg/m³; Density of mercury, ρ_Hg = 13600 kg/m³; height of mercury column, h_Hg = 76 cm.

Let the height of water column be h_w.

As the atmospheric pressure is same for both the cases.

Therefore, h_w * ρ_w * g = h_Hg * ρ_Hg * g

Or, h_w * 1000 = 76 * 13600

Or, h_w = 1033.6 cm.

7. Find the force exerted by the water on a 2 m² plane surface of a large stone placed at the bottom of a sea 500 m deep. Does the force depend on the orientation of the surface? Neglect the size of the stone in comparison to the depth of the sea.

Sol:

Given: area of the plane surface, A = 2 m²; height of water, h = 500 m; density of the water, ρ = 1000 kg/m³.

(a) The force exerted by the water, F = P * A

Or, F = (h * ρ * g) * A

Or, F = 500 * 1000 * 10 * 2

Or, F = 10⁷ N.

(b) The force does not depend on the orientation of the rock as long as the surface area remains same. [No]

8. Water is filled in a rectangular tank of size 3 m * 2 m * 1 m. (a) Find the total force exerted by the water on the bottom surface of the tank, (b) Consider a vertical side of area 2 m * 1 m. Take a horizontal strip of width δx metre in this side, situated at a depth of x metre from the surface of water. Find the force by the water on this strip, (c) Find the torque of the force calculated in part (b) about the bottom edge of this side. (c) Find the torque of the force calculated in part (b) about the bottom edge of this side. (d) Find the total force by the water on this side. (e) Find the total torque by the water on the side about the bottom edge. Neglect the atmospheric pressure and take g = 10 m/s².

Sol:

Given: volume of the tank, V = 3 * 2 * 1 = 6 m; density of water, ρ = 1000 kg/m³.

(a) The total force exerted by the water on the bottom surface of the tank, F = ρ * V * g = 1000 * 6 * 10 = 6000 N.

(b) The force exerted by water on the strip of width δx is

→ dF = p_{at x} * δA

Or, dF = (xρg) * 2 * δx

Or, dF = x * 1000 * 10 * 2 * δx = 20000 x δx.

(c) The torque of the force about the bottom edge is

= dF * (1 – x) = 20000 x (1 – x) δx.

(d) The total force by the water on this side (from 0 to 1) is

= ∫20000 x δx = 10000 N. 

(e) The total torque by the water on the side about the bottom edge (from 0 to 1) is

= ∫20000 x(1 – x) δx = 10000/3 N-m.

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HC Verma Concepts Of Physics Exercise Solutions Of Chapter 25 (Calorimetry)

HC Verma Concepts of Physics Solutions - Part 2, Chapter 25 (Calorimetry):

EXERCISE

1. An aluminium vessel of mass 0.5 kg contains 0.2 kg of water at 20°C. A block of iron of mass 0.2 kg at 100°C is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium, iron and water are 910 J/kg-K, 470 J/kg-K and 4200 J/kg-K respectively.

Sol:

Given: Mass of aluminium m_a = 0.5kg; Mass of water m_w = 0.2 kg; Mass of Iron m_i = 0.2 kg; Temperature of aluminium and water = 20°C = 297 K; Temperature of Iron = 100°C = 373 K; Specific heat of aluminium = 910 J/kg-k; Specific heat of Iron = 470 J/kg-k; Specific heat of water = 4200J/kg-k.

Assumption: heat interaction outside of the boundary is zero.

∴ Heat loss by iron block = heat gain by aluminium vessel and water

Or, 0.2 * 470 * (373 – T) = (T – 293) (0.5 * 910 + 0.2 * 4200)

Or, 94 * (373 – T) = (T – 293) (455 + 840)

Or, 373 – T = (T – 293) * 13.8

Or, T = 298 K = 25 ⁰C.

2. A piece of iron of mass 100 g is kept inside a furnace for a long time and then put in a calorimeter of water equivalent 10 g containing 240 g of water at 20°C. The mixture attains an equilibrium temperature of 60°C. Find the temperature of the furnace. Specific heat capacity of iron = 470 J/kg-°C.

Sol:

Given: mass of iron, m_iron = 100 g = 0.1 kg; mass of water, m_w = 240 g = 0.24 kg; water equivalent of calorimeter, m_eq = 10 g = 0.01 kg; final temp of mixture, T_f = 60⁰ C; temp of water, T_w = 20⁰ C; Specific heat capacity of iron, C_iron = 470 J/kg-°C; C_w = 4184 J/kg-°C.

Let the furnace temperature be T.

Now, heat loss by the iron = heat gain by the water and calorimeter

Or, m_iron * C_iron * (T – T_f) = (m_w + m_eq) * C_w * (T_f – T_w)

Or, 0.1 * 470 * (T – 60) = (0.24 +0.01) * 4184 * (60 – 20)

Or, T = 950⁰ C.

So, the temperature of the furnace is 950⁰ C. 

3. The temperatures of equal masses of three different liquids A, B and C are 12°C, 19°C and 28°C respectively. The temperature when A and B are mixed is 16°C, and when B and C are mixed, it is 23°C. What will be the temperature when A and C are mixed?

Sol:

Given: T_A = 12⁰C; T_B = 19⁰C; T_C = 28⁰C; m_A = m_B = m_C = m; temperature of mixture A and B is 16⁰C; temperature of mixture B and C is 23⁰C.

For mixture A and B: temperature of mixture, T = 16⁰C.

→Heat gain by A = heat loss by B

Or, m_A C_A (T – T_A) = m_B C_B (T_B – T)

Or, m C_A (16 – 12) = m C_B (19 – 16)

Or, 4 C_A = 3 C_B

For mixture B and C: temperature of mixture, T = 23⁰C.

→Heat gain by B = heat loss by C

Or, m_B C_B (T – T_B) = m_C C_C (T_C – T)

Or, m C_B (23 – 19) = m C_C (28 – 23)

Or, 4 C_B = 5 C_C

For mixture A and C: temperature of mixture, T.

→Heat gain by A = heat loss by C

Or, m_A C_A (T – T_A) = m_C C_C (T_C – T)

Or, m (3/4) C_B (T – 12) = m (4/5) C_B (28 – T)

Or, (3/4) (T – 12) = (4/5) (28 – T)

Or, 15 T – 180 = 448 – 16 T

Or, T = 628/31 = 20.3⁰C.

4. Four 2 cm * 2 cm * 2 cm cubes of ice are taken out from a refrigerator and are put in 200 ml of a drink at 10°C. (a) Find the temperature of the drink when thermal equilibrium is attained in it. (b) If the ice cubes do not melt completely, find the amount melted. Assume that no heat is lost to the outside of the drink and that the container has negligible heat capacity. Density of ice = 900 kg/m³, density of the drink = 1000 kg/m³, specific heat capacity of the drink = 4200 J/kg-K, latent heat of fusion of ice = 3.4 * 10⁵ J/kg.

Sol:

Given: Density of ice, ρ_i = 900 kg/m³; specific heat capacity of ice, C_i = 2108 J/kg-K; density of the drink, ρ_d = 1000 kg/m³; specific heat capacity of the drink, C_d = 4200 J/kg-K; latent heat of fusion of ice, L = 3.4 * 10⁵ J/kg; volume of ice, V_i = 4 * 2³ = 32 cm³ = 32 * 10^-6 m³; volume of drink, V_d = 200 ml = 2 * 10^-4 m³.

Mass of the ice, m_i = ρ_i V_i = 900 * 32 * 10^-6 = 288 * 10^-4 kg

Mass of the drink, m_d = ρ_d V_d = 1000 * 2 * 10^-4 = 0.2 kg

Heat required to melt ice completely is

→ Q_m = m_iL = 288 * 10^-4 * 3.4 * 10⁵ = 9792 J.

Heat release when drink comes from 10⁰ C to 0⁰ C is

→ Q_d = m_dC_dΔT = 0.2 * 4200 * (10 – 0) = 8400 J.

Since, Q_d is greater than Q_m. So complete ice will not melt.

(a) The temperature of the drink when thermal equilibrium is attained is 0⁰ C.

(b) Let the amount ice melt be m.

→ Latent heat of melted ice = Heat loss by the drink

Or, mL = 8400

Or, 3.4 * 10⁵ * m = 8400

Or, m = 8400/ (3.4 * 10⁵) = 0.025 kg = 25 g.

5. Indian style of cooling drinking water is to keep it in a pitcher having porous walls. Water comes to the outer surface very slowly and evaporates. Most of the energy needed for evaporation is taken from the water itself and the water is cooled down. Assume that a pitcher contains 10 kg of water and 0.2 g of water comes out per second. Assuming no backward heat transfer from the atmosphere to the water, calculate the time in which the temperature decreases by 5°C. Specific heat capacity of water = 4200 J/kg-°C and latent heat of vaporization of water = 2.27 * 10⁶ J/kg.

Sol:

Given: mass of water, m_w = 10 kg; C_w = 4200 J/kg-°C; latent heat of vaporization, L = 2.27 * 10⁶ J/kg; rate at which water comes, m’ = 0.2 g/s = 0.2 * 10^-3 kg/s; ΔT = 5⁰ C. let time = t.

According to the question,

→ Heat loss by the water = latent heat of vaporization of water

Or, m_w * C_w * ΔT = m’ * t * L

Or, 10 * 4200 * 5 = 0.2 * 10^-3 * 2.27 * 10⁶ * t

Or, t = 462.55 s = 7.7 min.

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