Mechanical Properties of Solids Practice Questions - DPP for NEET

 Page 1


1. (a). Y = 
2
MgL
r pDl
but Mg/pr
2
 = 20 × 10
8
  & Dl = L then
 Y = 20 × 10
8
 N/m
2
2. (b). F = Y a DtA
A = 2 × 10
–6
 m
2
, Y = 2 × 11 N/m
2
a = 1.1 × 10
–5
, t = 50 – 30 = 20°C
F = 2 × 10
11
 × 1.1 × 10
–5
 × 20 × 2 × 10
–6
 = 88N.
3. (d). Work done on the wire
W =
1
2
 F × l = 
1
2
 × stress x volume x strain
W =
1
2
 × Y × strain
2
 × volume
W = 
1
2
 × Y × 
2
2
L
D l
 × AL = 
2
YAL
2L
D
W = 
11 66
2 10 10 10
21
--
´ ´´
´
 = 0.1 J
4. (d). W = 
1
2
 ×  load × elongation
W = 
1
2
 × 5.4 × 10
6
 × 3
W = 8.1 × 10
6
 ergs
5. (d). By Hook's law
Y = 
F / A FL
/LA
=
ll
Y = 
82
161
(4 10 ) (0.2 10 )
--
´
´´
 = 2 × 10
11
 N/m
2
6. (a).  B = – 
PV
V
D
D
Given, DP = hdg = 200 x 10
3
 x 10
DP = 2 × 10
5
 N/m
2
V 0.1
V 100
D
= = 10
–3
\ B = 
6
3
2 10
10
-
´
 = 2 × 10
9
 N/m
2
7. (b). Y = 
stress F/ A FL
strain / L A
==
ll
\ l = 
3 2 11
FL 209.84
AY
(10 ) 1.96 10
-
´´
=
p´ ´´
  = 1.27 × 10
–3
 m = 1.27 mm
8. (a). Limiting stress = 4.0 × 10
8
 N/m
2
F 400
AA
= = 4.0 × 10
8
or   A = 10
–6
 m
2
\  D = 
1/2
1/2 6
4 A 4 10
-
æö
´ æö
=
ç÷ ç÷
èø
pp
èø
       = 1.13 × 10
–3
 m = 1.13 mm
9. (c). Stress = 
3
42
F 4.8 10N
A
1.2 10 m
-
´
=
´
 = 4.0 × 10
7
 N/m
2
10. (c). Strain = 
3
1 10
2
-
D´
=
l
l
 = 5 × 10
–4
, longitudinal
11. (c). F = Y A a Dt
       = 2 × 10
11
 × 3 × 10
–6
 × 10
-5
 × 20
    F = 120 N.
12. (c).   Compressibility
c = 
1V
K Vp
D
=-
D
 = 5 × 10
–10
\ Fractional decrease in volume
= –
V
V
D
= c D p = 5 × 10
–10
 × 15 × 10
6
           = 7.5 × 10
–3
13. (c). Increase in length on heating Dl = a L DT
To annul this increase if pressure applied is p then
  p = Y 
L
D l
= Ya DT
= 2 × 10
11
 × 1.1 × 10
–5
 × 100 = 2.2 × 10
8
 N/m
2
14. (c).  y = 2h (1 + s)
y = 2.4 × h
2.4 h = 2h (1 + s)
(1 + s) = 1.2
s = 0.2
15. (c).  Stress = F/A = 10/(2 × 10
-6
)  = 5 × 10
6
 N/m
2
Strain = 
6
11
Stress 5 10
Y
2 10
´
=
´
 = 2.5 × 10
–5
l = L × strain = 1 × 2.5 × 10
–5
l = 2.5 × 10
–5
 m
16. (b).   
12
Mg 1000 980 100
L AY
10 0.01
D ´´
==
´
l
Dl = 0.0098 cm.
17. (a). V olume = Mass/density
Area of cross-section = volume/length
Page 2


1. (a). Y = 
2
MgL
r pDl
but Mg/pr
2
 = 20 × 10
8
  & Dl = L then
 Y = 20 × 10
8
 N/m
2
2. (b). F = Y a DtA
A = 2 × 10
–6
 m
2
, Y = 2 × 11 N/m
2
a = 1.1 × 10
–5
, t = 50 – 30 = 20°C
F = 2 × 10
11
 × 1.1 × 10
–5
 × 20 × 2 × 10
–6
 = 88N.
3. (d). Work done on the wire
W =
1
2
 F × l = 
1
2
 × stress x volume x strain
W =
1
2
 × Y × strain
2
 × volume
W = 
1
2
 × Y × 
2
2
L
D l
 × AL = 
2
YAL
2L
D
W = 
11 66
2 10 10 10
21
--
´ ´´
´
 = 0.1 J
4. (d). W = 
1
2
 ×  load × elongation
W = 
1
2
 × 5.4 × 10
6
 × 3
W = 8.1 × 10
6
 ergs
5. (d). By Hook's law
Y = 
F / A FL
/LA
=
ll
Y = 
82
161
(4 10 ) (0.2 10 )
--
´
´´
 = 2 × 10
11
 N/m
2
6. (a).  B = – 
PV
V
D
D
Given, DP = hdg = 200 x 10
3
 x 10
DP = 2 × 10
5
 N/m
2
V 0.1
V 100
D
= = 10
–3
\ B = 
6
3
2 10
10
-
´
 = 2 × 10
9
 N/m
2
7. (b). Y = 
stress F/ A FL
strain / L A
==
ll
\ l = 
3 2 11
FL 209.84
AY
(10 ) 1.96 10
-
´´
=
p´ ´´
  = 1.27 × 10
–3
 m = 1.27 mm
8. (a). Limiting stress = 4.0 × 10
8
 N/m
2
F 400
AA
= = 4.0 × 10
8
or   A = 10
–6
 m
2
\  D = 
1/2
1/2 6
4 A 4 10
-
æö
´ æö
=
ç÷ ç÷
èø
pp
èø
       = 1.13 × 10
–3
 m = 1.13 mm
9. (c). Stress = 
3
42
F 4.8 10N
A
1.2 10 m
-
´
=
´
 = 4.0 × 10
7
 N/m
2
10. (c). Strain = 
3
1 10
2
-
D´
=
l
l
 = 5 × 10
–4
, longitudinal
11. (c). F = Y A a Dt
       = 2 × 10
11
 × 3 × 10
–6
 × 10
-5
 × 20
    F = 120 N.
12. (c).   Compressibility
c = 
1V
K Vp
D
=-
D
 = 5 × 10
–10
\ Fractional decrease in volume
= –
V
V
D
= c D p = 5 × 10
–10
 × 15 × 10
6
           = 7.5 × 10
–3
13. (c). Increase in length on heating Dl = a L DT
To annul this increase if pressure applied is p then
  p = Y 
L
D l
= Ya DT
= 2 × 10
11
 × 1.1 × 10
–5
 × 100 = 2.2 × 10
8
 N/m
2
14. (c).  y = 2h (1 + s)
y = 2.4 × h
2.4 h = 2h (1 + s)
(1 + s) = 1.2
s = 0.2
15. (c).  Stress = F/A = 10/(2 × 10
-6
)  = 5 × 10
6
 N/m
2
Strain = 
6
11
Stress 5 10
Y
2 10
´
=
´
 = 2.5 × 10
–5
l = L × strain = 1 × 2.5 × 10
–5
l = 2.5 × 10
–5
 m
16. (b).   
12
Mg 1000 980 100
L AY
10 0.01
D ´´
==
´
l
Dl = 0.0098 cm.
17. (a). V olume = Mass/density
Area of cross-section = volume/length
60
DPP/ P 20
=
mass
density length ´
 =
3
15.6 10
7800 2.5
-
´
´
 = 8 × 10
–7
 m
2
Y = 
73
F 8 9.8 2.5
AL
(8 10 ) 1.25 10
--
´´
=
D
´ ´´
l
Y = 1.96 × 10
11
 N/m
2
18. (c). Potential energy per unit volume = u
                 = 
1
2
× stress × strain
But    Y = 
stress
strain
\ stress = Y x strain = Y x S
\  Potential energy per unit volume = u
 = 
1
2
 × (YS)S = 
1
2
 YS
2
19. (d). 
2
1 12
2
2
21
Lr
Lr
=
l
l
L
1
 = L, L
2
 = 2L, r
1
 = 2R., r
2
 = R
\  
2
1
2
2
LR1
.
2L8
4R
==
l
l
20. (c). stress = 
2
ForceF
Area
r
=
p
\ stress S µ
2
1
r
\ 
2
12
21
Sr
Sr
æ ö æö
=
ç ÷ ç÷
è ø èø
Given 
1
2
r 2
r1
=
  \
1
2
S 1
S4
=
21. (c) F YA FA = aDq\µ
22. (a) For twisting, angle of shear 
1
L
fµ
i.e. if L is more then f will be small.
23. (a) ( ) Y 21 = h +s
24. (a) 
( )
0.5Y
Y 21
-h
= h +s Þ s=
h
25. (a) Tensile stress 
2
Fcos Fcos
a/cosa
qq
==
q
26. (a) Tensile stress is maximum when 
2
cos q
 is maximum,
i.e., 
0 q=°
27. (b) Shearing stress
Fsin Fsin cos
a/cosa
q qq
==
q
Fsin2
2a
q
=
28. (a) Elasticity is a measure of tendency of the body to regain
its original configuration. As steel is deformed less than
rubber therefore steel is more elastic than rubber.
29. (a) Bulk modulus of elasticity measures how good the body
is to regain its original volume on being compressed.
Therefore, it represents incompressibility of the material.
PV
K
V
-
=
D
 where P is increase in pressure, DV is change
in volume.
30. (a) A bridge during its use undergoes alternating strains
for a large number of times each day , depending upon the
movement of vehicles on it when a bridge is used for long
time, it losses its elastic strength. Due to which the amount
of strain in the bridge for a given stress will become large
and ultimately, the bridge may collapse. This may not
happen, if the bridges are declared unsafe after long use.