Solved Example: Mechanical Properties of Solid | Physics for JEE Main & Advanced PDF Download

 Page 1


JEE Solved Example on Mechanical Properties of 
Solid 
JEE Mains 
Q1: A compressive force, ?? is applied at the two ends of a long thin steel rod. It is heated, 
simultaneously, such that its temperature increases by ?? ?? . The net change in its length is zero. Let 
?? be the length of the rod, ?? is its area of crosssection. ?? is Young's modulus, and ?? is its 
coefficient of linear expansion. Then, ?? is equal to 
a. ?? ?? ???? ?? ?? 
b. ???? /?? ?? ?? 
d. LAYa ?? ?? 
Ans:  (c) AYa?T 
Thermal expansion, ?L = L?? ?T - - - (1) 
Let ??? '
 be the compression produced by applied force 
?? = ???? /?? ??? '
? ?? = ???? ??? '
/?? - - - - - (2) 
Net change in length = 0 ? ??? '
= ??? - -(3) 
From (1),(2) and (3) 
?? = ?????? (???? ??? )/?? = ?????? ??? 
 
Q2: A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to 
the lower end. The weight stretches the wire by 1mm. Then the elastic energy stored in the wire is 
a. 0.2 J 
b. 10 J 
c. 20 J 
d. 0.1 J 
Ans: (d) 0.1 J 
 
Elastic energy per unit volume = ½ x stress x strain 
Elastic Energy = ½ x stress x strain x volume 
= ½ x F/A x (?L /L) x (AL) 
= ½ x F?L 
= ½ x 200 x 10
-3
 
Elastic Energy = 0.1 J 
Page 2


JEE Solved Example on Mechanical Properties of 
Solid 
JEE Mains 
Q1: A compressive force, ?? is applied at the two ends of a long thin steel rod. It is heated, 
simultaneously, such that its temperature increases by ?? ?? . The net change in its length is zero. Let 
?? be the length of the rod, ?? is its area of crosssection. ?? is Young's modulus, and ?? is its 
coefficient of linear expansion. Then, ?? is equal to 
a. ?? ?? ???? ?? ?? 
b. ???? /?? ?? ?? 
d. LAYa ?? ?? 
Ans:  (c) AYa?T 
Thermal expansion, ?L = L?? ?T - - - (1) 
Let ??? '
 be the compression produced by applied force 
?? = ???? /?? ??? '
? ?? = ???? ??? '
/?? - - - - - (2) 
Net change in length = 0 ? ??? '
= ??? - -(3) 
From (1),(2) and (3) 
?? = ?????? (???? ??? )/?? = ?????? ??? 
 
Q2: A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to 
the lower end. The weight stretches the wire by 1mm. Then the elastic energy stored in the wire is 
a. 0.2 J 
b. 10 J 
c. 20 J 
d. 0.1 J 
Ans: (d) 0.1 J 
 
Elastic energy per unit volume = ½ x stress x strain 
Elastic Energy = ½ x stress x strain x volume 
= ½ x F/A x (?L /L) x (AL) 
= ½ x F?L 
= ½ x 200 x 10
-3
 
Elastic Energy = 0.1 J 
Q3: A rod of length L at room temperature and uniform area of cross-section A, Is made of a metal 
having a coefficient of linear expansion a. It is observed that an external compressive force F is 
applied to each of its ends, prevents any change in the length of the rod when its temperature 
rises by ?T K. Young’s modulus, Y for this metal is 
a.F/A a?T 
b.F/Aa(?T – 273) 
c. F/2Aa? 
d. 2F/Aa?T 
Ans: (a) F/Aa?T 
Young’s Modulus Y = stress/strain = (F/A)/(? l/ l) 
Substituting the coefficient of linear expansion 
a =? l /( l?T) 
? l / l= a?T 
Y= (F/Aa?T) 
Q4: Young’s moduli of two wires A and B are in the ratio 7:4. Wire A is 2m long and has radius R. 
Wire B is 1.5 m long and has a radius of 2mm. If the two wires stretch by the same length for a 
given load, then the value of R is close t 
a. 1.5 mm 
b. 1.9 mm 
c. 1.7 mm 
d. 1.3 mm 
Ans: (c) 1.7 mm 
? 1= ? 2 
(Fl 1/pr 1
2
y 1) = (Fl 2/pr 2
2
y 2) 
2/(R
2
 x 7)= 1.5/(2
2
x 4) 
R= 1.75 mm 
 
 
Page 3


JEE Solved Example on Mechanical Properties of 
Solid 
JEE Mains 
Q1: A compressive force, ?? is applied at the two ends of a long thin steel rod. It is heated, 
simultaneously, such that its temperature increases by ?? ?? . The net change in its length is zero. Let 
?? be the length of the rod, ?? is its area of crosssection. ?? is Young's modulus, and ?? is its 
coefficient of linear expansion. Then, ?? is equal to 
a. ?? ?? ???? ?? ?? 
b. ???? /?? ?? ?? 
d. LAYa ?? ?? 
Ans:  (c) AYa?T 
Thermal expansion, ?L = L?? ?T - - - (1) 
Let ??? '
 be the compression produced by applied force 
?? = ???? /?? ??? '
? ?? = ???? ??? '
/?? - - - - - (2) 
Net change in length = 0 ? ??? '
= ??? - -(3) 
From (1),(2) and (3) 
?? = ?????? (???? ??? )/?? = ?????? ??? 
 
Q2: A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to 
the lower end. The weight stretches the wire by 1mm. Then the elastic energy stored in the wire is 
a. 0.2 J 
b. 10 J 
c. 20 J 
d. 0.1 J 
Ans: (d) 0.1 J 
 
Elastic energy per unit volume = ½ x stress x strain 
Elastic Energy = ½ x stress x strain x volume 
= ½ x F/A x (?L /L) x (AL) 
= ½ x F?L 
= ½ x 200 x 10
-3
 
Elastic Energy = 0.1 J 
Q3: A rod of length L at room temperature and uniform area of cross-section A, Is made of a metal 
having a coefficient of linear expansion a. It is observed that an external compressive force F is 
applied to each of its ends, prevents any change in the length of the rod when its temperature 
rises by ?T K. Young’s modulus, Y for this metal is 
a.F/A a?T 
b.F/Aa(?T – 273) 
c. F/2Aa? 
d. 2F/Aa?T 
Ans: (a) F/Aa?T 
Young’s Modulus Y = stress/strain = (F/A)/(? l/ l) 
Substituting the coefficient of linear expansion 
a =? l /( l?T) 
? l / l= a?T 
Y= (F/Aa?T) 
Q4: Young’s moduli of two wires A and B are in the ratio 7:4. Wire A is 2m long and has radius R. 
Wire B is 1.5 m long and has a radius of 2mm. If the two wires stretch by the same length for a 
given load, then the value of R is close t 
a. 1.5 mm 
b. 1.9 mm 
c. 1.7 mm 
d. 1.3 mm 
Ans: (c) 1.7 mm 
? 1= ? 2 
(Fl 1/pr 1
2
y 1) = (Fl 2/pr 2
2
y 2) 
2/(R
2
 x 7)= 1.5/(2
2
x 4) 
R= 1.75 mm 
 
 
Q5: The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it 
is to support a 400 N load without exceeding its elastic limit? 
a. 1 mm 
b. 1.15 mm 
c. 0.90 mm 
d. 1.36 mm 
Ans: (b) 1.15 mm 
Stress = F/A 
Stress = 400 x 4/pd
2
 
= 379 x 10
6
 N/m
2
 
d
2 
= (400 x 4)/(379 x 10
6
p) 
d = 1.15 mm 
Q6: A uniform cylindrical rod of length L and radius r, is made from a material whose Young’s 
modulus of Elasticity equals Y . When this rod is heated by temperature T and simultaneously 
subjected to a net longitudinal compressional force F, its length remains unchanged. The 
coefficient of volume expansion, of the material of the rod is nearly equal to 
a. 9F/(pr
2
YT) 
b. 6F/(pr
2
YT) 
c. 3F/(pr
2
YT) 
d. F/(3pr
2
YT) 
Ans: (c) 3F/pr
2
YT 
Y = (F/pr
2
) x L/?L 
?L = F l/pr
2
Y——–(1) 
Change in length due to temperature change 
?L=La?T————(2) 
From equa (1) and (2) 
L a?T = FL/AY 
a= F/AY?T 
Page 4


JEE Solved Example on Mechanical Properties of 
Solid 
JEE Mains 
Q1: A compressive force, ?? is applied at the two ends of a long thin steel rod. It is heated, 
simultaneously, such that its temperature increases by ?? ?? . The net change in its length is zero. Let 
?? be the length of the rod, ?? is its area of crosssection. ?? is Young's modulus, and ?? is its 
coefficient of linear expansion. Then, ?? is equal to 
a. ?? ?? ???? ?? ?? 
b. ???? /?? ?? ?? 
d. LAYa ?? ?? 
Ans:  (c) AYa?T 
Thermal expansion, ?L = L?? ?T - - - (1) 
Let ??? '
 be the compression produced by applied force 
?? = ???? /?? ??? '
? ?? = ???? ??? '
/?? - - - - - (2) 
Net change in length = 0 ? ??? '
= ??? - -(3) 
From (1),(2) and (3) 
?? = ?????? (???? ??? )/?? = ?????? ??? 
 
Q2: A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to 
the lower end. The weight stretches the wire by 1mm. Then the elastic energy stored in the wire is 
a. 0.2 J 
b. 10 J 
c. 20 J 
d. 0.1 J 
Ans: (d) 0.1 J 
 
Elastic energy per unit volume = ½ x stress x strain 
Elastic Energy = ½ x stress x strain x volume 
= ½ x F/A x (?L /L) x (AL) 
= ½ x F?L 
= ½ x 200 x 10
-3
 
Elastic Energy = 0.1 J 
Q3: A rod of length L at room temperature and uniform area of cross-section A, Is made of a metal 
having a coefficient of linear expansion a. It is observed that an external compressive force F is 
applied to each of its ends, prevents any change in the length of the rod when its temperature 
rises by ?T K. Young’s modulus, Y for this metal is 
a.F/A a?T 
b.F/Aa(?T – 273) 
c. F/2Aa? 
d. 2F/Aa?T 
Ans: (a) F/Aa?T 
Young’s Modulus Y = stress/strain = (F/A)/(? l/ l) 
Substituting the coefficient of linear expansion 
a =? l /( l?T) 
? l / l= a?T 
Y= (F/Aa?T) 
Q4: Young’s moduli of two wires A and B are in the ratio 7:4. Wire A is 2m long and has radius R. 
Wire B is 1.5 m long and has a radius of 2mm. If the two wires stretch by the same length for a 
given load, then the value of R is close t 
a. 1.5 mm 
b. 1.9 mm 
c. 1.7 mm 
d. 1.3 mm 
Ans: (c) 1.7 mm 
? 1= ? 2 
(Fl 1/pr 1
2
y 1) = (Fl 2/pr 2
2
y 2) 
2/(R
2
 x 7)= 1.5/(2
2
x 4) 
R= 1.75 mm 
 
 
Q5: The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it 
is to support a 400 N load without exceeding its elastic limit? 
a. 1 mm 
b. 1.15 mm 
c. 0.90 mm 
d. 1.36 mm 
Ans: (b) 1.15 mm 
Stress = F/A 
Stress = 400 x 4/pd
2
 
= 379 x 10
6
 N/m
2
 
d
2 
= (400 x 4)/(379 x 10
6
p) 
d = 1.15 mm 
Q6: A uniform cylindrical rod of length L and radius r, is made from a material whose Young’s 
modulus of Elasticity equals Y . When this rod is heated by temperature T and simultaneously 
subjected to a net longitudinal compressional force F, its length remains unchanged. The 
coefficient of volume expansion, of the material of the rod is nearly equal to 
a. 9F/(pr
2
YT) 
b. 6F/(pr
2
YT) 
c. 3F/(pr
2
YT) 
d. F/(3pr
2
YT) 
Ans: (c) 3F/pr
2
YT 
Y = (F/pr
2
) x L/?L 
?L = F l/pr
2
Y——–(1) 
Change in length due to temperature change 
?L=La?T————(2) 
From equa (1) and (2) 
L a?T = FL/AY 
a= F/AY?T 
a = F/pr
2
YT 
Coefficient of volume expansion 
3? = 3F/pr
2
YT 
Q7: The following four wires are made of the same material. Which of these will have the largest 
extension when the same tension is applied? 
(a) length = 200 cm, diameter = 2 mm 
(b) length = 300 cm, diameter = 3 mm 
(c) length = 50 cm, diameter = 0.5 mm 
(b) length = 100 cm, diameter = 1 mm 
Answer: (c) length = 50 cm, diameter = 0.5 mm 
Since all four wires are made from the same material Young’s modulus will be the same. 
?L ? L/D
2
 
In (a) L/D
2
 = 200/(0.2)
2
 = 5 x 10
3
 cm
-1
 
In (b) L/D
2
 = 300/(0.3)
2
 = 3.3 x 10
3
 cm
-1
 
In (c) L/D
2
 = 50/(0.5)
2
 = 20 x 10
3 
cm
-1
 
In (d) L/D
2
 = 100/(0.1)
2 
= 10 x 10
3 
cm
-1
 
Q8: A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming 
that his density remains the same, the stress in the leg will change by a factor of 
(a) 1/9 
(b) 81 
(c) 1/81 
(d) 9 
Answer: (d) 9 
Stress = Force/Area 
Stress = Force/L
2
 
Now, dimensions increases by a factor of 9 
Now, S = (volume x density) x g /L
2
 
Page 5


JEE Solved Example on Mechanical Properties of 
Solid 
JEE Mains 
Q1: A compressive force, ?? is applied at the two ends of a long thin steel rod. It is heated, 
simultaneously, such that its temperature increases by ?? ?? . The net change in its length is zero. Let 
?? be the length of the rod, ?? is its area of crosssection. ?? is Young's modulus, and ?? is its 
coefficient of linear expansion. Then, ?? is equal to 
a. ?? ?? ???? ?? ?? 
b. ???? /?? ?? ?? 
d. LAYa ?? ?? 
Ans:  (c) AYa?T 
Thermal expansion, ?L = L?? ?T - - - (1) 
Let ??? '
 be the compression produced by applied force 
?? = ???? /?? ??? '
? ?? = ???? ??? '
/?? - - - - - (2) 
Net change in length = 0 ? ??? '
= ??? - -(3) 
From (1),(2) and (3) 
?? = ?????? (???? ??? )/?? = ?????? ??? 
 
Q2: A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to 
the lower end. The weight stretches the wire by 1mm. Then the elastic energy stored in the wire is 
a. 0.2 J 
b. 10 J 
c. 20 J 
d. 0.1 J 
Ans: (d) 0.1 J 
 
Elastic energy per unit volume = ½ x stress x strain 
Elastic Energy = ½ x stress x strain x volume 
= ½ x F/A x (?L /L) x (AL) 
= ½ x F?L 
= ½ x 200 x 10
-3
 
Elastic Energy = 0.1 J 
Q3: A rod of length L at room temperature and uniform area of cross-section A, Is made of a metal 
having a coefficient of linear expansion a. It is observed that an external compressive force F is 
applied to each of its ends, prevents any change in the length of the rod when its temperature 
rises by ?T K. Young’s modulus, Y for this metal is 
a.F/A a?T 
b.F/Aa(?T – 273) 
c. F/2Aa? 
d. 2F/Aa?T 
Ans: (a) F/Aa?T 
Young’s Modulus Y = stress/strain = (F/A)/(? l/ l) 
Substituting the coefficient of linear expansion 
a =? l /( l?T) 
? l / l= a?T 
Y= (F/Aa?T) 
Q4: Young’s moduli of two wires A and B are in the ratio 7:4. Wire A is 2m long and has radius R. 
Wire B is 1.5 m long and has a radius of 2mm. If the two wires stretch by the same length for a 
given load, then the value of R is close t 
a. 1.5 mm 
b. 1.9 mm 
c. 1.7 mm 
d. 1.3 mm 
Ans: (c) 1.7 mm 
? 1= ? 2 
(Fl 1/pr 1
2
y 1) = (Fl 2/pr 2
2
y 2) 
2/(R
2
 x 7)= 1.5/(2
2
x 4) 
R= 1.75 mm 
 
 
Q5: The elastic limit of brass is 379 MPa. What should be the minimum diameter of a brass rod if it 
is to support a 400 N load without exceeding its elastic limit? 
a. 1 mm 
b. 1.15 mm 
c. 0.90 mm 
d. 1.36 mm 
Ans: (b) 1.15 mm 
Stress = F/A 
Stress = 400 x 4/pd
2
 
= 379 x 10
6
 N/m
2
 
d
2 
= (400 x 4)/(379 x 10
6
p) 
d = 1.15 mm 
Q6: A uniform cylindrical rod of length L and radius r, is made from a material whose Young’s 
modulus of Elasticity equals Y . When this rod is heated by temperature T and simultaneously 
subjected to a net longitudinal compressional force F, its length remains unchanged. The 
coefficient of volume expansion, of the material of the rod is nearly equal to 
a. 9F/(pr
2
YT) 
b. 6F/(pr
2
YT) 
c. 3F/(pr
2
YT) 
d. F/(3pr
2
YT) 
Ans: (c) 3F/pr
2
YT 
Y = (F/pr
2
) x L/?L 
?L = F l/pr
2
Y——–(1) 
Change in length due to temperature change 
?L=La?T————(2) 
From equa (1) and (2) 
L a?T = FL/AY 
a= F/AY?T 
a = F/pr
2
YT 
Coefficient of volume expansion 
3? = 3F/pr
2
YT 
Q7: The following four wires are made of the same material. Which of these will have the largest 
extension when the same tension is applied? 
(a) length = 200 cm, diameter = 2 mm 
(b) length = 300 cm, diameter = 3 mm 
(c) length = 50 cm, diameter = 0.5 mm 
(b) length = 100 cm, diameter = 1 mm 
Answer: (c) length = 50 cm, diameter = 0.5 mm 
Since all four wires are made from the same material Young’s modulus will be the same. 
?L ? L/D
2
 
In (a) L/D
2
 = 200/(0.2)
2
 = 5 x 10
3
 cm
-1
 
In (b) L/D
2
 = 300/(0.3)
2
 = 3.3 x 10
3
 cm
-1
 
In (c) L/D
2
 = 50/(0.5)
2
 = 20 x 10
3 
cm
-1
 
In (d) L/D
2
 = 100/(0.1)
2 
= 10 x 10
3 
cm
-1
 
Q8: A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming 
that his density remains the same, the stress in the leg will change by a factor of 
(a) 1/9 
(b) 81 
(c) 1/81 
(d) 9 
Answer: (d) 9 
Stress = Force/Area 
Stress = Force/L
2
 
Now, dimensions increases by a factor of 9 
Now, S = (volume x density) x g /L
2
 
S = L
3
 x ? g /L
2 
= L ? g 
Stress S ? L 
S 2/S 1 = L 2/L 1 = 9L 1/L 1 = 9 
Q9. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in 
a cylindrical container. A massless piston of the area a floats on the surface of the liquid, covering 
an entire cross -section of the cylindrical containe r. When a mass m is placed on the surface of the 
piston to compress the liquid, the fractional decrement in the radius of the sphere is Mg/ aAB. Find 
the value of a. 
a. 4 
b. 5 
c. 3 
d. 2 
Ans: (c) 3 
Increase in pressure is ?p = Mg/A 
Bulk modulus is B = ?p /(?V/V) 
?V/V = ?p / B = Mg/AB——(1) 
The volume of the sphere is V = (4/3)pR
3
 
?V/V = 3(?R/R) 
From equation (1) we get 
Mg/AB = 3(?R/R) 
?R/R = Mg/3AB 
Therefore a = 3 
 
Q10: A steel wire having a radius of 2.0 mm, carrying a load of 4 kg, is hanging from a ceiling. Given 
that g = 3.1pm/s
2
, what will be the tensile stress that would be developed in the wire? 
a. 4.8 x 10
6
 N/m
2
 
b. 3.1 x 10
6
 N/m
2
 
c. 6.2 x 10
6
 N/m
2
 
d. 5.2 x 10
6
 N/m
2