HC Verma Concepts Of Physics Objective Solutions Of Chapter 3 (Rest and Motion:Kinematics)

HC Verma Concepts of Physics Solutions - Part 1, Chapter 3 (Rest and Motion: Kinematics):

OBJECTIVE –I

1. A motor car is going due north at a speed of 50 km/h. It makes a 90° left turn without changing the speed. The change in the velocity of the car is about

(a) 50 km/h towards west

(b) 70 km/h towards south-west

(c) 70 km/h towards north-west

(d) Zero.

Sol: (b)

Change in velocity, Δ V

→ Δ V = V₂ – V₁

Or, Δ V = √ [50² + (– 50)²]

Or, Δ V = √ (5000)

Or, Δ V = 70.7 ≈ 70 km/h towards south-west.

2. Figure (3-Q2) shows the displacement-time graph of a particle moving on the X-axis.

(a) The particle is continuously going in positive x direction

(b) The particle is at rest

(c) The velocity increases up to a time t₀, and then becomes constant

(d) The particle moves at a constant velocity up to a time t₀, and then stops.

Sol: (d)

Slope of the tangent of a curve at a point in displacement-time graph is equal to the velocity. Here up to a time t₀, slope of the curve remain constant. After that slope is zero. So, the particle moves at a constant velocity up to a time t₀, and then stops.

3. A particle has a velocity u towards east at t = 0. Its acceleration is towards west and is constant. Let X_A and X_B be the magnitude of displacements in the first 10 seconds and the next 10 seconds

(a) X_A < X_B

(b) X_A = X_B

(c) X_A > X_B

(d) The information is insufficient to decide the relation of x_A with x_B.

Sol: (d)

Data is insufficient to decide the relation of x_A with x_B.

4. A person travelling on a straight line moves with a uniform velocity v₁ for some time and with uniform velocity v₂ for the next equal time. The average velocity v is given by

(a) v = (v₁ + v₂)/2

(b) v = √( v₁ * v₂)

(c) 2/v = (1/v₁) + (1/v₂)

(d) 1/v = (1/v₁) + (1/v₂)

Sol: (a)

Given: t₁ = t₂ = t.

Total distance travelled, S = v₁t + v₂t

Total time taken, T = t + t = 2t

So, the average velocity, v = S/T

Or, v = (v₁ + v₂)/2.

5. A person travelling on a straight line moves with a uniform velocity v₁ for a distance x and with a uniform velocity v₂ for the next equal distance. The average velocity v is given by

(a) v = (v₁ + v₂)/2

(b) v = √( v₁ * v₂)

(c) 2/v = (1/v₁) + (1/v₂)

(d) 1/v = (1/v₁) + (1/v₂)

Sol: (c)

Time taken by person for travelling 1^st x distance, t₁ = x/v₁.

Time taken by person for travelling next x distance, t₂ = x/v₂.

Total time taken, t = t₁ + t₂ = x {(1/v₁) + (1/v₂)}

Total distance travelled, S = x + x = 2x

So, the average velocity, v = S/t

Or, v = 2/ {(1/v₁) + (1/v₂)}

Or, 2/v = (1/v₁) + (1/v₂).

6. A stone is released from an elevator going up with an acceleration a. The acceleration of the stone after the release is

(a) a upward
(b) (g – a) upward

(c) (g – a) downward
(d) g downward.

Sol: (d)

After the release of the stone, only acceleration due to gravity acts downward.

7. A person standing near the edge of the top of a building throws two balls A and B. The ball A is thrown vertically upward and B is thrown vertically downward with the same speed. The ball A hits the ground with a speed v_A and the ball B hits the ground with a speed v_B. We have

(a) v_A > v_B 

(b) v_A < v_B 

(c) v_A = v_B

(d) the relation between v_A and v_B depends on height of the building above the ground.
Sol: (c) 

Let height of the building be h.

Ball A projected upwards with velocity u falls back to building top with velocity u downwards. It completes its journey to ground under gravity.

Therefore, (v_A)² = u² + 2gh ------- (1)

Ball B starts with downwards velocity u and reaches ground after travelling a vertical distance h.

Therefore, (v_B)² = u² + 2gh ------- (2)

From equation 1 and 2

We get, v_A = v_B.

8. In a projectile motion the velocity

(a) is always perpendicular to the acceleration

(b) is never perpendicular to the acceleration

(c) is perpendicular to the acceleration for one instant only

(d) is perpendicular to the acceleration for two instants.

Sol: (c) 

Only one instant velocity is perpendicular to the acceleration.

9. Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit the ground first?

(a) the faster one

(b) the slower one

(c) both will reach simultaneously

(d) depends on the masses.

Sol: (c) 

Both will reach simultaneously, because both have same vertical motion.

Initial velocity and acceleration in vertical direction are same for both the cases.

10. The range of a projectile fired at an angle of 15° is 50 m. If it is fired with the same speed at an angle of 45°, its range will be

(a) 25 m 
(b) 37 m
(c) 50 m 
(d) 100 m.

Sol: (d) 

Given: for both the cases initial velocity (u) is same.

We know,

Range, R = u² sin 2θ/g

For 1^st case:

→ 50 = u² sin 2*15 /g

Or, u²/g = 100

For 2^nd case:

→ R = u² sin (2*45)/g

Or, R = u²/g = 100 m.

11. Two projectiles A and B are projected with angle of projection 15° for the projectile A and 45° for the projectile B. If R_A and R_B be the horizontal range for the two projectiles, then
(a) R_A < R_B

(b) R_A = R_B

(c) R_A > R_B

(d) the information is insufficient to decide the relation of R_A with R_B.

Sol: (d) 

We have required the value of initial velocity (u) to decide the relation of R_A with R_B.

12. A river is flowing from west to east at a speed of 5 metres per minute. A man on the south bank of the river, capable of swimming at 10 metres per minute in still water, wants to swim across the river in the shortest time. He should swim in a direction

(a) due north

(b) 30° east of north

(c) 30° north of west

(d) 60° east of north.

Sol: (a) 

He should swim in a direction due north to across the river in the shortest time. 

13. In the arrangement shown in figure (3-Q3), the ends P and Q of an inextensible string move downwards with uniform speed u. Pulleys A and B are fixed. The mass M moves upwards with a speed

(a) 2u cos θ

(b) u/cos θ

(c) 2u/cos θ

(d) u cos θ.

Sol: (b) 

OBJECTIVE –II

1. Consider the motion of the tip of the minute hand of a clock. In one hour

(a) the displacement is zero

(b) the distance covered is zero

(c) the average speed is zero

(d) the average velocity is zero.

Sol: (a), (d)

Initial and final position of the tip of the minute hand of a clock is same in one hour. So, the displacement of the tip of the minute hand is zero.

And, velocity = displacement/time = zero.

2. A particle moves along the X-axis as

x = u (t – 2s) + a(t – 2 s)².

(a) the initial velocity of the particle is u

(b) the acceleration of the particle is a

(c) the acceleration of the particle is 2a

(d) at t = 2 s particle is at the origin.

Sol: (c), (d)

Given: x = u (t – 2s) + a (t – 2 s)².

We have, dx/dt = u + 2a (t – 2 s)

And, d²x/dt² = 2a

So, initial velocity = dx/dt|_{t = 0} = u – 4a.

Therefore, option (a) is wrong.

Acceleration, a = d²x/dt² = 2a

Therefore, option (b) is wrong and option (c) is correct.

Position of particle at t = 2 s is given by

→ x = u (t – 2s) + a(t – 2 s)²

Or, x = u (2 s – 2s) + a(2 s – 2 s)²

Or, x = 0

Therefore, option (d) is correct.

3. Pick the correct statements:

(a) Average speed of a particle in a given time is never less than the magnitude of the average velocity.

(b) It is possible to have a situation in which |dv/dt| ≠ 0 but d|v|/dt = 0.

(c) The average velocity of a particle is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval.

(d) The average velocity of a particle moving on a straight line is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval. (Infinite accelerations are not allowed.)

Sol: (a), (b), (c)

4. An object may have

(a) varying speed without having varying velocity

(b) varying velocity without having varying speed

(c) nonzero acceleration without having varying velocity

(d) nonzero acceleration without having varying speed.

Sol: (b), (d)

If speed of an object is changes then velocity also changes. So option (a) is wrong. In case of uniform circular motion speed remain constant but direction changes. So it velocity changes. Therefore, option (b) is correct.

Rate of change of velocity is acceleration. So, option (c) is wrong.

In case of uniform circular motion speed remain constant but it has an acceleration. So, option (d) is correct.

5. Mark the correct statements for a particle going on a straight line:

(a) If the velocity and acceleration have opposite sign, the object is slowing down.

(b) If the position and velocity have opposite sign, the particle is moving towards the origin.

(c) If the velocity is zero at an instant, the acceleration should also be zero at that instant.

(d) If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

Sol: (a), (b), (d)

6. The velocity of a particle is zero at t = 0.

(a) The acceleration at t = 0 must be zero.

(b) The acceleration at t = 0 may be zero.

(c) If the acceleration is zero from t = 0 to t = 10s speed is also zero in this interval.

(d) If the speed is zero from t = 0 to t = 10s the acceleration is also zero in this interval.

Sol: (b), (c), (d)

7. Mark the correct statements:

(a) The magnitude of the velocity of a particle is equal to its speed.

(b) The magnitude of average velocity in an interval is equal to its average speed in that interval.

(c) It is possible to have a situation in which the speed of a particle is always zero but the average speed is not zero.

(d) It is possible to have a situation in which the speed of the particle is never zero but the average speed in an interval is zero.

Sol: (a)

8. The velocity-time plot for a particle moving on a straight line is shown in the figure (3-Q4).

(a) The particle has a constant acceleration.

(b) The particle has never turned around.

(c) The particle has zero displacement.

(d) The average speed in the interval 0 to 10 s is the same as the average speed in the interval 10 s to 20 s.

Sol: (a), (d)

9. Figure (3-Q5) shows the position of a particle moving on the X-axis as a function of time.

(a) The particle has come to rest 6 times.

(b) The maximum speed is at t = 6 s.

(c) The velocity remains positive for t = 0 to t = 6 s.

(d) The average velocity for the total period shown is negative.

Sol: (a)

10. The accelerations of a particle as seen from two frames S₁ and S₂ have equal magnitude 4 m/s².

(a) The frames must be at rest with respect to each other.

(b) The frames may be moving with respect to each other but neither should be accelerated with respect to the other.

(c) The acceleration of S₂, with respect to S₁ may either be zero or 8 m/s².

(d) The acceleration of S₂ with respect to S₁ may be anything between zero and 8 m/s².

Sol: (d)

Previous year’s chapter-wise questions and solutions of Kinematics: Click Here

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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 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.  

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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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