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NCERT Exemplar: Mechanical Properties of Solids
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Page 1
Mechanical Properties of Solids
Multiple Choice Questions – I
1. Modulus of rigidity of ideal liquids is
(a) infinity.
(b) zero.
(c) unity.
(d) some finite small non-zero constant value.
Ans: Correct option (b)
Because since we know from given condition that liquid is ideal, thus it does not
have frictional force among its layers. Hence, tangential forces are zero since no
stress is developed there.
2. The maximum load a wire can withstand without breaking when its length
is reduced to half of its original length, will
(a) be double.
(b) be half.
(c) be four times.
(d) remain same.
Ans: Correct option (d)
Because Breaking stress
Breaking force
=
Area of crosss ection
Page 2
Mechanical Properties of Solids
Multiple Choice Questions – I
1. Modulus of rigidity of ideal liquids is
(a) infinity.
(b) zero.
(c) unity.
(d) some finite small non-zero constant value.
Ans: Correct option (b)
Because since we know from given condition that liquid is ideal, thus it does not
have frictional force among its layers. Hence, tangential forces are zero since no
stress is developed there.
2. The maximum load a wire can withstand without breaking when its length
is reduced to half of its original length, will
(a) be double.
(b) be half.
(c) be four times.
(d) remain same.
Ans: Correct option (d)
Because Breaking stress
Breaking force
=
Area of crosss ection
As breaking force does not depend on length. Thus, changing the cross section has
no effect which implies breaking force remain same.
3. The temperature of a wire is doubled. The Young's modulus of elasticity
(a) will also double.
(b) will become four times.
(c) will remain same.
(d) will decrease.
Ans: Correct option (d)
Because as we know that
t0
L = L (1 + a ) t ?
t 0 0
L - L = a L Lt ? ? ?
00
stress FL FL
Y = = =
strain A A(1 + a ) Lt ??
1
Y
t
?
?
Thus, when temperature increases then Young’s modules decreases.
4. A spring is stretched by applying a load to its free end. The strain produced
in the spring is
(a) volumetric.
(b) shear.
(c) longitudinal and shear.
(d) longitudinal.
Page 3
Mechanical Properties of Solids
Multiple Choice Questions – I
1. Modulus of rigidity of ideal liquids is
(a) infinity.
(b) zero.
(c) unity.
(d) some finite small non-zero constant value.
Ans: Correct option (b)
Because since we know from given condition that liquid is ideal, thus it does not
have frictional force among its layers. Hence, tangential forces are zero since no
stress is developed there.
2. The maximum load a wire can withstand without breaking when its length
is reduced to half of its original length, will
(a) be double.
(b) be half.
(c) be four times.
(d) remain same.
Ans: Correct option (d)
Because Breaking stress
Breaking force
=
Area of crosss ection
As breaking force does not depend on length. Thus, changing the cross section has
no effect which implies breaking force remain same.
3. The temperature of a wire is doubled. The Young's modulus of elasticity
(a) will also double.
(b) will become four times.
(c) will remain same.
(d) will decrease.
Ans: Correct option (d)
Because as we know that
t0
L = L (1 + a ) t ?
t 0 0
L - L = a L Lt ? ? ?
00
stress FL FL
Y = = =
strain A A(1 + a ) Lt ??
1
Y
t
?
?
Thus, when temperature increases then Young’s modules decreases.
4. A spring is stretched by applying a load to its free end. The strain produced
in the spring is
(a) volumetric.
(b) shear.
(c) longitudinal and shear.
(d) longitudinal.
Ans: Correct option is (c) because when spring is stretched by load then its shape
and length changes which produces shearing and longitudinal strain.
5. A rigid bar of mass M is supported symmetrically by three wires each of
length 1. Those at each end are of copper and the middle one is of iron. The
ratio of their diameters, if each is to have the same tension, is equal to
(a)
copper
iron
Y
Y
(b)
iron
copper
Y
Y
(c)
2
iron
2
copper
Y
Y
(d)
iron
copper
Y
Y
Ans: Correct option is (b)
Because
2
Stress FL FL
Y = = =
Strain A
D
p
2
L
L
?
??
?
??
??
2
4FL
Y =
p D L ?
L = l
Copper iron( ) L L g ? ? ?
F = weight of rod same in both cases
Page 4
Mechanical Properties of Solids
Multiple Choice Questions – I
1. Modulus of rigidity of ideal liquids is
(a) infinity.
(b) zero.
(c) unity.
(d) some finite small non-zero constant value.
Ans: Correct option (b)
Because since we know from given condition that liquid is ideal, thus it does not
have frictional force among its layers. Hence, tangential forces are zero since no
stress is developed there.
2. The maximum load a wire can withstand without breaking when its length
is reduced to half of its original length, will
(a) be double.
(b) be half.
(c) be four times.
(d) remain same.
Ans: Correct option (d)
Because Breaking stress
Breaking force
=
Area of crosss ection
As breaking force does not depend on length. Thus, changing the cross section has
no effect which implies breaking force remain same.
3. The temperature of a wire is doubled. The Young's modulus of elasticity
(a) will also double.
(b) will become four times.
(c) will remain same.
(d) will decrease.
Ans: Correct option (d)
Because as we know that
t0
L = L (1 + a ) t ?
t 0 0
L - L = a L Lt ? ? ?
00
stress FL FL
Y = = =
strain A A(1 + a ) Lt ??
1
Y
t
?
?
Thus, when temperature increases then Young’s modules decreases.
4. A spring is stretched by applying a load to its free end. The strain produced
in the spring is
(a) volumetric.
(b) shear.
(c) longitudinal and shear.
(d) longitudinal.
Ans: Correct option is (c) because when spring is stretched by load then its shape
and length changes which produces shearing and longitudinal strain.
5. A rigid bar of mass M is supported symmetrically by three wires each of
length 1. Those at each end are of copper and the middle one is of iron. The
ratio of their diameters, if each is to have the same tension, is equal to
(a)
copper
iron
Y
Y
(b)
iron
copper
Y
Y
(c)
2
iron
2
copper
Y
Y
(d)
iron
copper
Y
Y
Ans: Correct option is (b)
Because
2
Stress FL FL
Y = = =
Strain A
D
p
2
L
L
?
??
?
??
??
2
4FL
Y =
p D L ?
L = l
Copper iron( ) L L g ? ? ?
F = weight of rod same in both cases
2
1
Y
D
?
2
11
D , D
YY
??
copper
iron
iron copper
D
Y
=
DY
6. A mild steel wire of length 2L and cross-sectional area A is stretched, well
within elastic limit, horizontally between two pillars (Figure). A mass m is
suspended from the mid-point of the wire. Strain in the wire is
(a)
2
2
x
2l
(b)
l
x
Page 5
Mechanical Properties of Solids
Multiple Choice Questions – I
1. Modulus of rigidity of ideal liquids is
(a) infinity.
(b) zero.
(c) unity.
(d) some finite small non-zero constant value.
Ans: Correct option (b)
Because since we know from given condition that liquid is ideal, thus it does not
have frictional force among its layers. Hence, tangential forces are zero since no
stress is developed there.
2. The maximum load a wire can withstand without breaking when its length
is reduced to half of its original length, will
(a) be double.
(b) be half.
(c) be four times.
(d) remain same.
Ans: Correct option (d)
Because Breaking stress
Breaking force
=
Area of crosss ection
As breaking force does not depend on length. Thus, changing the cross section has
no effect which implies breaking force remain same.
3. The temperature of a wire is doubled. The Young's modulus of elasticity
(a) will also double.
(b) will become four times.
(c) will remain same.
(d) will decrease.
Ans: Correct option (d)
Because as we know that
t0
L = L (1 + a ) t ?
t 0 0
L - L = a L Lt ? ? ?
00
stress FL FL
Y = = =
strain A A(1 + a ) Lt ??
1
Y
t
?
?
Thus, when temperature increases then Young’s modules decreases.
4. A spring is stretched by applying a load to its free end. The strain produced
in the spring is
(a) volumetric.
(b) shear.
(c) longitudinal and shear.
(d) longitudinal.
Ans: Correct option is (c) because when spring is stretched by load then its shape
and length changes which produces shearing and longitudinal strain.
5. A rigid bar of mass M is supported symmetrically by three wires each of
length 1. Those at each end are of copper and the middle one is of iron. The
ratio of their diameters, if each is to have the same tension, is equal to
(a)
copper
iron
Y
Y
(b)
iron
copper
Y
Y
(c)
2
iron
2
copper
Y
Y
(d)
iron
copper
Y
Y
Ans: Correct option is (b)
Because
2
Stress FL FL
Y = = =
Strain A
D
p
2
L
L
?
??
?
??
??
2
4FL
Y =
p D L ?
L = l
Copper iron( ) L L g ? ? ?
F = weight of rod same in both cases
2
1
Y
D
?
2
11
D , D
YY
??
copper
iron
iron copper
D
Y
=
DY
6. A mild steel wire of length 2L and cross-sectional area A is stretched, well
within elastic limit, horizontally between two pillars (Figure). A mass m is
suspended from the mid-point of the wire. Strain in the wire is
(a)
2
2
x
2l
(b)
l
x
(c)
2
x
l
(d)
2
x
2l
Ans: Correct option is (a) because
l (AO BO) AB ? ? ? ?
l 2AO 2I 2[AO l] ? ? ? ? ?
? ?
1
22
2
2 l x 1
??
? ? ?
??
??
1
2
2
2
x
21 1 1
l
??
??
??
? ? ?
??
??
??
??
2 2 2
22
2 1 1 2
22
x x x
ll
l l l
??
? ? ? ? ? ?
??
??
Read MoreFAQs on NCERT Exemplar: Mechanical Properties of Solids
| 1. What are the mechanical properties of solids? |  |
Ans. The mechanical properties of solids refer to the behavior of solid materials when subjected to mechanical forces. These properties include elasticity, plasticity, tensile strength, compressive strength, hardness, and toughness. They determine how a material deforms and fails under different load conditions.
| 2. How is Young's modulus defined, and why is it important? | |
Ans. Young's modulus is defined as the ratio of tensile stress to tensile strain in a material within the elastic limit. It is important because it quantifies a material's stiffness, helping engineers and designers select appropriate materials for structural applications based on their ability to withstand deformation.
| 3. What is the difference between elastic and plastic deformation? | |
Ans. Elastic deformation is the temporary change in shape or size of a material when a force is applied, which is fully recovered once the force is removed. Plastic deformation, on the other hand, is a permanent change in shape that occurs when the applied stress exceeds the yield strength of the material, and it does not return to its original form after the load is removed.
| 4. What factors affect the tensile strength of a solid material? | |
Ans. The tensile strength of a solid material is influenced by several factors, including the material's composition, temperature, rate of loading, and the presence of defects or impurities. Additionally, the microstructure and grain size of the material also play a crucial role in determining its tensile strength.
| 5. How do temperature and strain rate affect the mechanical properties of solids? | |
Ans. Temperature and strain rate significantly affect the mechanical properties of solids. As temperature increases, materials often become more ductile and less brittle, which can lead to lower yield strength. Conversely, an increase in strain rate typically leads to higher strength and hardness, as materials may exhibit a more brittle behavior under rapid loading conditions.
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Apr 18, 2026 Last updated
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