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Chapter-wise Previous Year's Questions With Solutions

Step by step detail solutions of previous years questions of various JE Exams, such as JEE Main, JEE Advance, IIT JEE, AIEEE, WBJEE, EAMCET, Karnataka CET, CPMT, Kerala CET, MP PMT and Other Exams.

Friday, 20 April 2018

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

HC-Verma-Concepts-Of-Physics-Some-Mechanical-Properties-Of-Matter-Chapter-14-Solution


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 * 106 m2; Y = 2.0 * 1011 N/m2.
Force, F = mg = 10 * 10 = 100 N.

(a) We know,
→ Stress, σ = Force/area = 100/ (4 * 106)
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 * 104.

(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 * 106 * 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, Lc = Length of steel, Ls; sectional area of copper, Ac = sectional area of steel, As; Ys = 2 * 1011 N/m2; Yc = 1.3 * 1011 N/m2.

(a) Stress developed in copper wire is = σc = P/ Ac
And, Stress developed in steel wire is = σs = P/ As
So, the ratio of the stresses developed in the two wires is = σc/ σs = 1 [since Ac = As]

(b) Strain developed in copper wire is = εc = σc / Yc
And, strain developed in steel wire is = εs = σs/ Ys 
So, the ratio of the strains developed in the two wires is = εc / εs = (σc / Yc)/ (σs / Ys) = Ys/ Yc = (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, Lc = Length of steel, Ls; sectional area of copper, Ac = sectional area of steel, As; Ys = 2 * 1011 N/m2; Yc = 1.3 * 1011 N/m2.
Stress developed in copper wire is = σc = P/ Ac
And, Stress developed in steel wire is = σs = P/ As
So, the ratio of the stresses developed in the two wires is = σc/ σs = 1 [since Ac = As]
Strain developed in copper wire is = εc = σc / Yc
And, Strain developed in steel wire is = εs = σs/ Ys 
So, the ratio of the strains developed in the two wires is = εc / εs = (σc / Yc)/ (σs / Ys) = Ys/ Yc = (2 * 1011)/ (1.3 * 1011) = 1.54

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Tuesday, 17 April 2018

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

HC-Verma-Concepts-Of-Physics-Work-and-Energy-Chapter-8-Solution


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 = r1r2
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]do 
Or, w = [ad + bd2/2]
Or, w = d [a + bd/2].

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Saturday, 7 April 2018

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

HC-Verma-Concepts-Of-Physics-Fluid-Mechanics-Chapter-13-PDF-Solution


Solutions of H.C. Verma’s Concepts of Physics chapter 13 (Fluid Mechanicsare 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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