Mechanical Properties of Solids Practice Questions - DPP for JEE

 Page 1


1. (a) , where W = load, l = length of beam and I is
geometrical moment of inertia for rectangular beam,
  where b = breadth and d = depth
For square beam b = d
? 
For a beam of circular cross-section, 
? (for sq. cross section)
and  
(for circular cross-section)
Now   
( i.e., they have same cross-sectional area)
2. (a) Energy / volume =  = 3600 J m
–3
[Strain = 0.06 × 10
–2
]
3. (b) Using Hooke’s law, F = kx we can write
4=k(a – l
0
) … (i)
and 5=k(b – l
0
) … (ii)
If l be the length under tension 9N, then
9=k( l – l
0
) … (iii)
Page 2


1. (a) , where W = load, l = length of beam and I is
geometrical moment of inertia for rectangular beam,
  where b = breadth and d = depth
For square beam b = d
? 
For a beam of circular cross-section, 
? (for sq. cross section)
and  
(for circular cross-section)
Now   
( i.e., they have same cross-sectional area)
2. (a) Energy / volume =  = 3600 J m
–3
[Strain = 0.06 × 10
–2
]
3. (b) Using Hooke’s law, F = kx we can write
4=k(a – l
0
) … (i)
and 5=k(b – l
0
) … (ii)
If l be the length under tension 9N, then
9=k( l – l
0
) … (iii)
After solving above equations, we get
l=(5b – 4a).
4. (c)
For a beam, the depression at the centre is given by, 
[f, L, b, d are constants for a particular beam]
i.e. 
5. (a) Given: F = 100 kN = 10
5
 N
Y = 2 × 10
11
 Nm
–2
l
0
 = 1.0 m
radius r = 10 mm = 10
– 2 
m
From formula, Y = 
?
= 
= 
Therefore % strain = = 0.16%
6. (b)
7. (c) The given graph does not obey Hooke’s law. and there is no well
defined plastic region. So the graph represents elastomers.
8. (a)
Initial volume = Final volume
Page 3


1. (a) , where W = load, l = length of beam and I is
geometrical moment of inertia for rectangular beam,
  where b = breadth and d = depth
For square beam b = d
? 
For a beam of circular cross-section, 
? (for sq. cross section)
and  
(for circular cross-section)
Now   
( i.e., they have same cross-sectional area)
2. (a) Energy / volume =  = 3600 J m
–3
[Strain = 0.06 × 10
–2
]
3. (b) Using Hooke’s law, F = kx we can write
4=k(a – l
0
) … (i)
and 5=k(b – l
0
) … (ii)
If l be the length under tension 9N, then
9=k( l – l
0
) … (iii)
After solving above equations, we get
l=(5b – 4a).
4. (c)
For a beam, the depression at the centre is given by, 
[f, L, b, d are constants for a particular beam]
i.e. 
5. (a) Given: F = 100 kN = 10
5
 N
Y = 2 × 10
11
 Nm
–2
l
0
 = 1.0 m
radius r = 10 mm = 10
– 2 
m
From formula, Y = 
?
= 
= 
Therefore % strain = = 0.16%
6. (b)
7. (c) The given graph does not obey Hooke’s law. and there is no well
defined plastic region. So the graph represents elastomers.
8. (a)
Initial volume = Final volume
? 
9. (c) According to questions,
As, 
 
 
? 
10. (b) As Y = 
But V = Al so A = 
Therefore ?l = 
Hence graph of ?l versus l
2
 will give a straight line.
Page 4


1. (a) , where W = load, l = length of beam and I is
geometrical moment of inertia for rectangular beam,
  where b = breadth and d = depth
For square beam b = d
? 
For a beam of circular cross-section, 
? (for sq. cross section)
and  
(for circular cross-section)
Now   
( i.e., they have same cross-sectional area)
2. (a) Energy / volume =  = 3600 J m
–3
[Strain = 0.06 × 10
–2
]
3. (b) Using Hooke’s law, F = kx we can write
4=k(a – l
0
) … (i)
and 5=k(b – l
0
) … (ii)
If l be the length under tension 9N, then
9=k( l – l
0
) … (iii)
After solving above equations, we get
l=(5b – 4a).
4. (c)
For a beam, the depression at the centre is given by, 
[f, L, b, d are constants for a particular beam]
i.e. 
5. (a) Given: F = 100 kN = 10
5
 N
Y = 2 × 10
11
 Nm
–2
l
0
 = 1.0 m
radius r = 10 mm = 10
– 2 
m
From formula, Y = 
?
= 
= 
Therefore % strain = = 0.16%
6. (b)
7. (c) The given graph does not obey Hooke’s law. and there is no well
defined plastic region. So the graph represents elastomers.
8. (a)
Initial volume = Final volume
? 
9. (c) According to questions,
As, 
 
 
? 
10. (b) As Y = 
But V = Al so A = 
Therefore ?l = 
Hence graph of ?l versus l
2
 will give a straight line.
11. (d) W
1
=
and W
2
=
? W=
=
12. (c) Poisson’s ratio, 
For material like copper, s = 0.33
And, Y = 3k (1 – 2 s)
Also, 
Y = 2n (1+ s)
Hence, n < Y < k
13. (b) Bulk modulus ....(1)
and or    ...(2)
From eqs. (1) and (2),  or 
14. (a)
15. (c) Using the usual expression for the Young’s modulus, the force
constant for the wire can be written as k =  where the
symbols have their usual meanings. Now the two wires together
will have an effective force constant . Substituting the
corresponding lengths and the Young’s moduli we get the answer.
Page 5


1. (a) , where W = load, l = length of beam and I is
geometrical moment of inertia for rectangular beam,
  where b = breadth and d = depth
For square beam b = d
? 
For a beam of circular cross-section, 
? (for sq. cross section)
and  
(for circular cross-section)
Now   
( i.e., they have same cross-sectional area)
2. (a) Energy / volume =  = 3600 J m
–3
[Strain = 0.06 × 10
–2
]
3. (b) Using Hooke’s law, F = kx we can write
4=k(a – l
0
) … (i)
and 5=k(b – l
0
) … (ii)
If l be the length under tension 9N, then
9=k( l – l
0
) … (iii)
After solving above equations, we get
l=(5b – 4a).
4. (c)
For a beam, the depression at the centre is given by, 
[f, L, b, d are constants for a particular beam]
i.e. 
5. (a) Given: F = 100 kN = 10
5
 N
Y = 2 × 10
11
 Nm
–2
l
0
 = 1.0 m
radius r = 10 mm = 10
– 2 
m
From formula, Y = 
?
= 
= 
Therefore % strain = = 0.16%
6. (b)
7. (c) The given graph does not obey Hooke’s law. and there is no well
defined plastic region. So the graph represents elastomers.
8. (a)
Initial volume = Final volume
? 
9. (c) According to questions,
As, 
 
 
? 
10. (b) As Y = 
But V = Al so A = 
Therefore ?l = 
Hence graph of ?l versus l
2
 will give a straight line.
11. (d) W
1
=
and W
2
=
? W=
=
12. (c) Poisson’s ratio, 
For material like copper, s = 0.33
And, Y = 3k (1 – 2 s)
Also, 
Y = 2n (1+ s)
Hence, n < Y < k
13. (b) Bulk modulus ....(1)
and or    ...(2)
From eqs. (1) and (2),  or 
14. (a)
15. (c) Using the usual expression for the Young’s modulus, the force
constant for the wire can be written as k =  where the
symbols have their usual meanings. Now the two wires together
will have an effective force constant . Substituting the
corresponding lengths and the Young’s moduli we get the answer.
16. (b) [?p = constant]
17. (a) From the graph, it is clear that for the same value of load,
elongation is maximum for wire OA. Hence OA is the thinnest wire
among the four wires.
18. (c) We know that Young’s modulus
Since Y, F are same for both the wires, we have,
or, = 
or,
So,   l
1
 : l
2
 = 1 : 8
19. (c)
20. (d) At extension l
1
, the stored energy = 
At extension l
2
, the stored energy = 
Work done in increasing its extension from l
1
 to l
2
21. (4) Young’s modulus of elasticity is