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