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].

Read More

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

Solutions of H.C. Verma’s Concepts of Physics chapter 13 (Fluid Mechanics) are given below. You can download H.C. Verma Solutions in PDF format by simply clicking on pop-out button at right corner and download PDF or clicking on print command and save as PDF.  

Read More

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.

Read More

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

Solutions of H.C. Verma’s Concepts of Physics chapter 25 (Calorimetry) are given below. You can download H.C. Verma Solutions in PDF format by simply clicking on pop-out button at right corner and download PDF or clicking on print command and save as PDF. 

Read More

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.

Read More

HC Verma Concepts Of Physics Exercise Solutions Of Chapter 26 (Laws of Thermodynamics)

HC Verma Concepts of Physics Solutions - Part 2, Chapter 26 - Laws Of Thermodynamics:

EXERCISE

1. A thermally insulated, closed copper vessel contains water at 15°C. When the vessel is shaken vigorously for 15 minutes, the temperature rises to 17°C. The mass of the vessel is 100 g and that of the water is 200 g. The specific heat capacities of copper and water are 420 J/kg-K and 4200 J/kg-K respectively. Neglect any thermal expansion, (a) how much heat is transferred to the liquid—vessel system? (b) How much work has been done on this system? (c) How much is the increase in internal energy of the system?

Sol:

System: water + copper

Given: insulated; closed system; t_i = 15 ⁰c; t_e = 17 ⁰c; mass of copper, m_cu = 100 g = 0.1 kg; mass of water, m_w = 200 g = 0.2 kg; c_cu = 420 J/kg-K and c_w = 4200 J/kg-K.

(a) Due to insulated boundary, heat transfer to the system is zero.

(b) From the 1^st law,

We know, Q = ΔU + W = U₂ – U₁ + W

Or, W = U₁ – U₂              [Q = 0]

Or, W = (U₁ – U₂)_cu + (U₁ – U₂)_w

Or, W = m_cu c_cu (t_i – t_e) + m_w c_w (t_i – t_e)

Or, W = 0.1 * 420 (15 – 17) + 0.2 * 4200 (15 – 17)

Or, W = – 1764 J.

– Ve sign means work done on the system.

(c) The increase in internal energy of the system is ΔU = W = 1764 J.

2. Figure (26-E1) shows a paddle wheel coupled to a mass of 12 kg through fixed frictionless pulleys. The paddle is immersed in a liquid of heat capacity 4200 J/K kept in an adiabatic container. Consider a time interval in which the 12 kg block falls slowly through 70 cm. (a) how much heat is given to the liquid? (b) How much work is done on the liquid? (c) Calculate the rise in the temperature of the liquid neglecting the heat capacity of the container and the paddle.

Sol:

System: liquid + wheel + container

Given: adiabatic system; heat capacity of liquid, C_L = 4200 J/K, mass of block, m = 12 kg, change of height of block, Δh = 70 cm = 0.7 m.

(a) Heat is given to the liquid is zero, because system is adiabatic.

Q = 0

(b) Work is done on the liquid, W = change of potential energy of block

Or, W = mg (Δ h) = 12 * 10 * 0.7 = 84 J.

(c) 1^st law, Q = ΔU + W

Or, 0 = C_L (Δ t) – 84 [W = – 84 work done on the system]

Or, Δ t = 84/4200 = 0.02 ⁰c.

3. A 100 kg block is started with a speed of 2.0 m/s on a long, rough belt kept fixed in a horizontal position. The coefficient of kinetic friction between the block and the belt is 0.20. (a) Calculate the change in the internal energy of the block-belt system as the block comes to a stop on the belt, (b) Consider the situation from a frame of reference moving at 2.0 m/s along the initial velocity of the block. As seen from this frame, the block is gently put on a moving belt and in due time the block starts moving with the belt at 2.0 m/s. Calculate the increase in the kinetic energy of the block as it stops slipping past the belt, (c) Find the work done in this frame by the external force holding the belt.

Sol:

Given: mass of block = 100 kg; u = 2 m/s; μ = 0.2; v = 0.

1^st law: Q = Δu + W

In this case Q = 0

Or, – Δu = W = change in K.E.

Or, Δu = – (½ mv² – ½ mu²)

Or, Δu = ½ * 100 * 2² = 200 J.

(a) The change in the internal energy of the block-belt system is 200 J.

(b) The increase in the kinetic energy of the block is 200 J.

4. Calculate the change in internal energy of a gas kept in a rigid container when 100 J of heat is supplied to it.

Sol:

Given: Q = 100 J; ΔV = 0.

1^st Law: Q = ΔU + W = ΔU + pΔV

Or, 100 = ΔU + 0 → ΔU = 100 J

So, the change in internal energy of a gas is 100 J.

5. The pressure of a gas changes linearly with volume from 10 kPa, 200 cc to 50 kPa, 50 cc. (a) Calculate the work done by the gas. (b) If no heat is supplied or extracted from the gas, what is the change in the internal energy of the gas?

Sol:

Given: Q = 0; v₁ = 200 cc = 2 * 10^-4 m³; v₂ = 50 cc = 0.5 * 10^-4; p₁ = 10 kPa; p₂ = 50 kPa.

(a) Work done during the process 1-2,

W = p₁ (v₂ – v₁) + ½ * (p₂ – p₁)(v₂ – v₁)

Or, W = 10 * (0.5 – 2)*10^-4 + ½ *40 *(0.5 – 2) * 10^-4

Or, W = – 0.0015 – 0.0030 = – 0.0045 kJ

Or, W = – 4.5 J

So, the work done by the gas is – 4.5 J.

(b) 1^st law: Q = ΔU + W

Or, 0 = ΔU + W

Or, ΔU = – W = 4.5 J.

Read More

HC Verma Concepts Of Physics PDF Exercise Solutions Of Chapter 7 (Circular Motion)

Solutions of H.C. Verma’s Concepts of Physics chapter 7 (Circular Motion) are given below. You can download H.C. Verma Solutions in PDF format by simply clicking on pop-out button at right corner and download PDF or clicking on print command and save as PDF. 

Read More

HC Verma Concepts Of Physics Exercise Solutions Of Chapter 7 (Circular Motion)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 7 - Circular Motion:

EXERCISE

1. Find the acceleration of the moon with respect to the earth from the following data: Distance between the earth and the moon = 3.85 * 10⁵ km and the time taken by the moon to complete one revolution around the earth = 27.3 days.

Sol:

Given: Distance between the earth and the moon, r = 3.85 * 10⁵ km = 3.85 * 10⁸ m; T = 27.3 days = 2.36 * 10⁶ s.

Angular speed, ω = 2π/T = 2.66 * 10^-6 rad/s.

Acceleration, a = ω²r = (2.66 * 10^-6)² * 3.85 * 10⁸ = 2.73 * 10^-3 m/s².

2. Find the acceleration of a particle placed on the surface of the earth at the equator due to earth's rotation. The diameter of earth = 12800 km and it takes 24 hours for the earth to complete one revolution about its axis.

Sol:

Given: Distance between the centre of earth and the particle, r = 12800/2 km = 64 * 10⁵ m; T = 24 hrs. = 86400 s.

Angular speed, ω = 2π/T = 7.27 * 10^-5 rad/s.

Acceleration, a = ω²r = (7.27 * 10^-5)² * 64 * 10⁵ = 0.0336 m/s².

3. A particle moves in a circle of radius 1.0 cm at a speed given by v = 2.0 t where v is in cm/s and t in seconds. (a) Find the radial acceleration of the particle at t = 1 s. (b) Find the tangential acceleration at t = 1 s. (c) Find the magnitude of the acceleration at t = 1 s.

Sol:  

Given: radius, r = 1.0 cm; v = 2.0t.

Speed of particle at 1 s, v_{t = 1 s} = 2 cm/s

(a) The radial acceleration of the particle at t = 1 s

→ (a_r) _{t = 1 s} = (v_{t = 1 s})²/r = 4 cm/s².

(b) The tangential acceleration of the particle at t = 1 s 
→ (a_t) _{t = 1 s} = (dv/dt) _{t = 1 s} = d(2t)/dt = 2 cm/s².

(c) The magnitude of the acceleration at t = 1 s

→ a = √ (a_r² + a_t²) = √ (4² + 2²) = √20 cm/s².

4. A scooter weighing 150 kg together with its rider moving at 36 km/hr is to take a turn of radius 30 m. What horizontal force on the scooter is needed to make the turn possible?

Sol:

Given: speed, v = 36 km/hr = 10 m/s; total mass, M = 150 kg; radius of a turn, r = 30 m.

To make the turn possible, horizontal force (F) must be equal to Centripetal force (F_c). Otherwise there will be skidding.

We know, Centripetal force (F_c) = mv²/r = (150 * 10²)/30 = 500 N.

Therefore, horizontal force (F) = 500 N.

5. If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking?

Sol: 

Given: speed, v = 36 km/hr = 10 m/s; total mass, M = 150 kg; radius of a turn, r = 30 m.

From the diagram

R cos θ = Mg ------------------ (1)

R sin θ = F_C

Or, R sin θ = Mv²/r ---------- (2)

Dividing equation (2) by equation (1)

We get, tan θ = v²/rg = 10²/ (30 * 10) = 1/3

Or, θ = tan^-1 (1/3).

6. A park has a radius of 10 m. If a vehicle goes round it at an average speed of 18 km/hr, what should be the proper angle of banking?

Sol: 

Given: speed, v = 18 km/hr = 5 m/s; radius of a turn, r = 10 m.

Angle of banking is given by,

→ tan θ = v²/rg = 5²/ (10 * 10) = ¼

Or, θ = tan^-1 (1/4).

Read More