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SOME MECHANICAL PROPERTIES OF MATTER:--
EXERCISES, Q-1 To Q-10
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 2.0x10¹¹ N/m².
ANSWER: Length of wire L= 3 m,
Load F = 10*10 N =100 N, {taking g = 10 m/s²}
Area of cross-section A = 4 mm²
→A = 4x10⁻⁶ m²
(a) The stress = Load (force) on unit area of cross-section =F/A
= 100/4x10⁻⁶ N/m²
=25 x 10⁶ N/m²
=2.5 x10⁷ N/m²
(b) Let the elongation of the wire under this stress be l,
The strain =l/L, Young's modulus of the metal Y =2.0 x10¹¹ N/m²
We have Stress/strain =Y (constant)
→(F/A)/Strain =Y
→Strain =(F/A)/Y =2.5x10⁷/2.0 x10¹¹
→Strain = 1.25 x10⁻⁴ m
(c) Strain = l/L = 1.25 x10⁻⁴
→l = 3.0 * 1.25 x10⁻⁴ m =3.75 x10⁻⁴ m
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 2.0x10¹¹ N/m².
ANSWER: Length of wire L= 3 m,
Load F = 10*10 N =100 N, {taking g = 10 m/s²}
Area of cross-section A = 4 mm²
→A = 4x10⁻⁶ m²
(a) The stress = Load (force) on unit area of cross-section =F/A
= 100/4x10⁻⁶ N/m²
=25 x 10⁶ N/m²
=2.5 x10⁷ N/m²
(b) Let the elongation of the wire under this stress be l,
The strain =l/L, Young's modulus of the metal Y =2.0 x10¹¹ N/m²
We have Stress/strain =Y (constant)
→(F/A)/Strain =Y
→Strain =(F/A)/Y =2.5x10⁷/2.0 x10¹¹
→Strain = 1.25 x10⁻⁴ m
(c) Strain = l/L = 1.25 x10⁻⁴
→l = 3.0 * 1.25 x10⁻⁴ m =3.75 x10⁻⁴ 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) strain and (c) the compression of the cylinder. Young's modulus of the metal is 2.0x10¹¹ N/m².
ANSWER: (a) Length of the cylinder L = 2 m
Area of cross-section A =πr² =π(2/100)² m² =4π/10000 m²
Force on the cylinder F = 100*10 N =1000 N
Hence the stress =F/A =1000/(4π/10000) N/m²
=(1/4π)*10⁷ N/m²
=0.0796 x10⁷ N/m²
=7.96 x10⁵ N/m²
(b) Strain = Stress/Y =(F/A)/Y
={7.96 x10⁵}/2.0 x10¹¹
=3.98 x10⁻⁶ ≈4 x10⁻⁶
(c) Let the compression of the cylinder = l m
The strain =l/L = l/2
→4 x10⁻⁶ =l/2
→l =8 x10⁻⁶ 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) strain and (c) the compression of the cylinder. Young's modulus of the metal is 2.0x10¹¹ N/m².
ANSWER: (a) Length of the cylinder L = 2 m
Area of cross-section A =πr² =π(2/100)² m² =4π/10000 m²
Force on the cylinder F = 100*10 N =1000 N
Hence the stress =F/A =1000/(4π/10000) N/m²
=(1/4π)*10⁷ N/m²
=0.0796 x10⁷ N/m²
=7.96 x10⁵ N/m²
(b) Strain = Stress/Y =(F/A)/Y
={7.96 x10⁵}/2.0 x10¹¹
=3.98 x10⁻⁶ ≈4 x10⁻⁶
(c) Let the compression of the cylinder = l m
The strain =l/L = l/2
→4 x10⁻⁶ =l/2
→l =8 x10⁻⁶ m
3. The elastic limit of steel is 8x10⁸ N/m² and its Young's modulus 2.0x10¹¹ N/m². Find the maximum elongation of a half-meter steel wire that can be given without exceeding the elastic limit.
ANSWER: It means the maximum stress in the wire σ= 8 x10⁸ N/m²
Y =2.0 x10¹¹ N/m²
Length of the wire L =0.50 m
Let corresponding maximum elongation = l m
Strain ε= l/L =l/0.50 =2l
we have σ/ε =Y
→ε =σ/Y
→2l = 8 x10⁸/2.0 x10¹¹
→l =2/1000 m = 2 mm
3. The elastic limit of steel is 8x10⁸ N/m² and its Young's modulus 2.0x10¹¹ N/m². Find the maximum elongation of a half-meter steel wire that can be given without exceeding the elastic limit.
ANSWER: It means the maximum stress in the wire σ= 8 x10⁸ N/m²
Y =2.0 x10¹¹ N/m²
Length of the wire L =0.50 m
Let corresponding maximum elongation = l m
Strain ε= l/L =l/0.50 =2l
we have σ/ε =Y
→ε =σ/Y
→2l = 8 x10⁸/2.0 x10¹¹
→l =2/1000 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 x 10¹¹ N/m². Y of copper = 1.3 x 10¹¹ N/m².
ANSWER: (a) Since both types of wires are joined end to end, the same load will be on the cross-sections of them. If the load = F and the area of cross-section =A, then stress developed in steel wire σ =F/A and the stress developed in copper wire σ¹ =F/A.
The ratio σ/σ¹ =(F/A)/(F/A) = 1
(b) Strain in copper wire = Stress in copper wire/Y =F/AY
Similarly the strain in steel wire =F/AY'
Where Y = Modulus of copper
Y' = Modulus of steel
strain in copper wire/strain in steel wire =(F/AY)/(F/AY')
=Y'/Y
=2.0 x10¹¹/1.3 x10¹¹
=2.0/1.3
= 20/13
5. In figure (14-E1) the upper wire is made of steel and the lower of copper. The wires have an equal cross-section. Find the ratio of the longitudinal strains developed in the two wires.
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 x 10¹¹ N/m². Y of copper = 1.3 x 10¹¹ N/m².
ANSWER: (a) Since both types of wires are joined end to end, the same load will be on the cross-sections of them. If the load = F and the area of cross-section =A, then stress developed in steel wire σ =F/A and the stress developed in copper wire σ¹ =F/A.
The ratio σ/σ¹ =(F/A)/(F/A) = 1
(b) Strain in copper wire = Stress in copper wire/Y =F/AY
Similarly the strain in steel wire =F/AY'
Where Y = Modulus of copper
Y' = Modulus of steel
strain in copper wire/strain in steel wire =(F/AY)/(F/AY')
=Y'/Y
=2.0 x10¹¹/1.3 x10¹¹
=2.0/1.3
= 20/13
5. In figure (14-E1) the upper wire is made of steel and the lower of copper. The wires have an equal cross-section. Find the ratio of the longitudinal strains developed in the two wires.
Figure for Q-5
ANSWER: Since both wires have equal cross-sections, the stress in both wires will be the same. Let the elongation in steel wire be l and in copper wire =l'. Since both the wires have the same length , say L, the strain in steel wire =l/L and the strain in copper wire =l'/L. If the modulus of elasticity of steel wire is Y and the modulus of elasticity of copper = Y', then
strain in steel wire/strain in copper wire =(stress in steel wire/Y)/(stress in copper wire/Y')
=Y'/Y
=2.0 x10¹¹/1.3 x10¹¹ {taking the values from the previous problem}
= 2.0/1.3
= 20/13
= 1.54
|
| Figure for Q-5 |
ANSWER: Since both wires have equal cross-sections, the stress in both wires will be the same. Let the elongation in steel wire be l and in copper wire =l'. Since both the wires have the same length , say L, the strain in steel wire =l/L and the strain in copper wire =l'/L. If the modulus of elasticity of steel wire is Y and the modulus of elasticity of copper = Y', then
strain in steel wire/strain in copper wire =(stress in steel wire/Y)/(stress in copper wire/Y')
=Y'/Y
=2.0 x10¹¹/1.3 x10¹¹ {taking the values from the previous problem}
= 2.0/1.3
= 20/13
= 1.54
6. The two wires shown in figure (14-E2) are made of the same material which has a breaking stress of 8 x 10⁸ N/m². The area of the cross-section of the upper wire is 0.006 cm² and that of the lower wire is 0.003 cm². The mass m₁ = 10 kg, m₂ = 20 kg and the hanger is light. (a) Find the maximum load that can be put on the hanger without breaking a wire. Which wire will break first if the load is increased? (b) Repeat the above part if m₁ =10 kg and m₂ =36 kg.
Figure for Q-6
ANSWER: Breaking stress = 8 x 10⁸ N/m²
C/S Area of the upper wire A= 0.006 cm² =0.006/10000 m²
C/S area of the lower wire A' = 0.003 cm² =0.003/10000 m²
(a) Let the length of each wire =L and the maximum load on the hanger W
Total load on the lower wire =W+m₁
The stress on the lower wire =(W+m₁)g/A'
= (W+10)g/(0.003/10000) N/m²
= (W+10)g*10⁷/3 N/m²
If we equate it to breaking stress,
(W+10)g*10⁷/3 = 8 x10⁸
→W =(3*80/g)-10 kg =240/10 -10 =24-10 =14 kg
The load on the upper wire =W+m₁+m₂ =W+10+20 =W+30 kg
The stress on the upper wire =(W+30)g/A
=(W+30)g/(0.006/10000) N/m²
=(W+30)g*10⁷/6 N/m²
Equating it to breaking stress,
(W+30)g*10⁷/6 = 8 x10⁸
→W =6*80/10-30 =48-30 =18 kg
Since the lower wire reaches the breaking stress with a lower weight, the maximum weight W =14 kg. The lower wire will break first.
(b) If m₂ = 36 kg
The stress on upper wire =(W+10+36)g/A
=(W+46)g/(0.006/10000)
=(W+46)g*10⁷/6
Equating with breaking stress,
(W+46)g*10⁷/6 =8 x10⁸
→W =6*80/10-46 =48-46 =2 kg
So now the maximum load on the hanger = 2 kg and the upper wire breaks first.
7. Two persons pull a rope towards themselves. Each person exerts a force of 100 N on the rope. Find Young's modulus of the material of the rope if it extends in length by 1 cm. The original length of the rope = 2 m and the area of cross-section = 2 cm².
6. The two wires shown in figure (14-E2) are made of the same material which has a breaking stress of 8 x 10⁸ N/m². The area of the cross-section of the upper wire is 0.006 cm² and that of the lower wire is 0.003 cm². The mass m₁ = 10 kg, m₂ = 20 kg and the hanger is light. (a) Find the maximum load that can be put on the hanger without breaking a wire. Which wire will break first if the load is increased? (b) Repeat the above part if m₁ =10 kg and m₂ =36 kg.
 |
| Figure for Q-6 |
ANSWER: Breaking stress = 8 x 10⁸ N/m²
C/S Area of the upper wire A= 0.006 cm² =0.006/10000 m²
C/S area of the lower wire A' = 0.003 cm² =0.003/10000 m²
(a) Let the length of each wire =L and the maximum load on the hanger W
Total load on the lower wire =W+m₁
The stress on the lower wire =(W+m₁)g/A'
= (W+10)g/(0.003/10000) N/m²
= (W+10)g*10⁷/3 N/m²
If we equate it to breaking stress,
(W+10)g*10⁷/3 = 8 x10⁸
→W =(3*80/g)-10 kg =240/10 -10 =24-10 =14 kg
The load on the upper wire =W+m₁+m₂ =W+10+20 =W+30 kg
The stress on the upper wire =(W+30)g/A
=(W+30)g/(0.006/10000) N/m²
=(W+30)g*10⁷/6 N/m²
Equating it to breaking stress,
(W+30)g*10⁷/6 = 8 x10⁸
→W =6*80/10-30 =48-30 =18 kg
Since the lower wire reaches the breaking stress with a lower weight, the maximum weight W =14 kg. The lower wire will break first.
(b) If m₂ = 36 kg
The stress on upper wire =(W+10+36)g/A
=(W+46)g/(0.006/10000)
=(W+46)g*10⁷/6
Equating with breaking stress,
(W+46)g*10⁷/6 =8 x10⁸
→W =6*80/10-46 =48-46 =2 kg
So now the maximum load on the hanger = 2 kg and the upper wire breaks first.
7. Two persons pull a rope towards themselves. Each person exerts a force of 100 N on the rope. Find Young's modulus of the material of the rope if it extends in length by 1 cm. The original length of the rope = 2 m and the area of cross-section = 2 cm².
ANSWER: Area of cross-section A = 2 cm² =2/10000 m²
Force F =100 N
Stress σ =F/A =100/(2/10000) =1000000/2 =500000 =5*10⁵ N/m²
l=1 cm =1/100 m =0.01 m
L =2 m
Strain ε =l/L =0.01/2 =0.005
Young's modulus of the material Y =σ/ε
=5*10⁵/0.005 N/m²
=1 x10⁸ N/m²
ANSWER: Area of cross-section A = 2 cm² =2/10000 m²
Force F =100 N
Stress σ =F/A =100/(2/10000) =1000000/2 =500000 =5*10⁵ N/m²
l=1 cm =1/100 m =0.01 m
L =2 m
Strain ε =l/L =0.01/2 =0.005
Young's modulus of the material Y =σ/ε
=5*10⁵/0.005 N/m²
=1 x10⁸ N/m²
8. A Steel rod of cross-sectional area 4 cm² and length 2 m shrinks by 0.1 cm as the temperature decreases in the night. If the rod is clamped at both ends during the day hours, find the tension developed in it during night hours. Young's modulus of steel = 1.9 x 10¹¹ N/m².
ANSWER: Length of the rod =L =2 m
The length increased during the night = l =0.1 cm= 0.001 m
The strain developed during the night =l/L =0.001/2 =0.0005
If the tension developed = T N
The stress developed = T/A N/m²
= T/0.0004 N/m²
But stress = strain*Y
T/0.0004 = 0.0005*1.9 ×10¹¹ N
→T = 0.0004*0.0005*1.9×10¹¹ N
→T= 2×10⁻⁷*1.9×10¹¹ N
→T= 3.8x10⁴ N
8. A Steel rod of cross-sectional area 4 cm² and length 2 m shrinks by 0.1 cm as the temperature decreases in the night. If the rod is clamped at both ends during the day hours, find the tension developed in it during night hours. Young's modulus of steel = 1.9 x 10¹¹ N/m².
ANSWER: Length of the rod =L =2 m
The length increased during the night = l =0.1 cm= 0.001 m
The strain developed during the night =l/L =0.001/2 =0.0005
If the tension developed = T N
The stress developed = T/A N/m²
= T/0.0004 N/m²
But stress = strain*Y
T/0.0004 = 0.0005*1.9 ×10¹¹ N
→T = 0.0004*0.0005*1.9×10¹¹ N
→T= 2×10⁻⁷*1.9×10¹¹ N
→T= 3.8x10⁴ N
9. Consider the situation shown in figure (14-E3). The force F is equal to the m₂g/2. If the area of the cross-section of the string is A and its Young's modulus of Y, find the strain developed in it. The string is light and there is no friction anywhere.
Figure for Q-9
ANSWER: Let the tension in the string = T and the acceleration of the blocks =a
Diagram for Q-9
For the hanging block
m₂g-T = m₂a
→a = g-T/m₂
For the block on the table
T -F = m₁a
→T =m₁a+F
→T= m₁a+m₂g/2
→T= m₁g-m₁T/m₂+m₂g/2 {Putting the value of a}
→T(m₁ +m₂)/m₂ = (2m₁+m₂)g/2
→T = m₂ (2m₁+m₂)g/2(m₁+m₂)
The strain = stress/Y
= T/AY
= m₂g(2m₁+m₂)/2AY(m₁+m₂)
9. Consider the situation shown in figure (14-E3). The force F is equal to the m₂g/2. If the area of the cross-section of the string is A and its Young's modulus of Y, find the strain developed in it. The string is light and there is no friction anywhere.
 |
| Figure for Q-9 |
ANSWER: Let the tension in the string = T and the acceleration of the blocks =a
|
| Diagram for Q-9 |
For the hanging block
m₂g-T = m₂a
→a = g-T/m₂
For the block on the table
T -F = m₁a
→T =m₁a+F
→T= m₁a+m₂g/2
→T= m₁g-m₁T/m₂+m₂g/2 {Putting the value of a}
→T(m₁ +m₂)/m₂ = (2m₁+m₂)g/2
→T = m₂ (2m₁+m₂)g/2(m₁+m₂)
The strain = stress/Y
= T/AY
= m₂g(2m₁+m₂)/2AY(m₁+m₂)
10. A Sphere of mass 20 kg is suspended by a metal wire of unstretched length 4 m and diameter 1 mm. When in equilibrium there is a clear gap of 2 mm between the sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle θ with the vertical and is released. Find the maximum value of θ so that the sphere does not rub the floor. Young's modulus of the metal of the wire is 2.0x10¹¹ N/m². Make appropriate approximations.
ANSWER: Let us assume that when oscillating the sphere traverses a circular path. If the elongation at equilibrium position be l, then
l/L = mg/AY
l = mgL/AY
The length in the equilibrium position L' = L+l
Diagram for Q-10
When oscillating the sphere will have a horizontal speed v at vertical position of the wire. This v will depend on the angle θ. P.E. of the sphere at this displacement angle =mg(L'-L'.cosθ)
{Taking the lowest position as zero potential energy level}
K.E. of the sphere at the vertical position of the wire =mv²/2
Equating both we get
mv²/2 = mgL'(1-cosθ)
mv² = 2mgL'(1-cosθ)
The total tension in the wire at vertical position = mg + mv²/r
= mg + 2mgL'(1-cosθ)/r
= mg{r + 2L'(1-cosθ)}/r
The stress in the wire at this time
= Tension in the wire/cross-sectional area
= (mg/rA){r+2L'(1-cosθ)}
Allowed elongation= l + 0.002 m (so that the sphere does not touch the floor)
Strain at this time = (l + 0.002)/L
Hence (l + 0.002)/L = (mg/rAY){r+2L'(1-cosθ)}
→mgL/AY + 0.002 = (mgL/rAY){r+2L'(1-cosθ)}
[substituting l with its expression]
→{r+2L'(1-cosθ)} = r + 0.002*rAY/mgL
→1-cosθ = 0.001*rAY/mgLL'
→cosθ = 1 -0.001*rAY/mgLL'
→cosθ=1-0.001*(L+l+0.002){π(0.001)²/4}*2x10¹¹/{20*10*4*(L+l)}
{Let us consider the 0.002 m length negligible in comparision to L', so L+l+0.002≈L+l, i.e r ≈ L'}
→cosθ = 1 - 0.196
→cosθ = 0.804
→θ = 36.4°
10. A Sphere of mass 20 kg is suspended by a metal wire of unstretched length 4 m and diameter 1 mm. When in equilibrium there is a clear gap of 2 mm between the sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle θ with the vertical and is released. Find the maximum value of θ so that the sphere does not rub the floor. Young's modulus of the metal of the wire is 2.0x10¹¹ N/m². Make appropriate approximations.
ANSWER: Let us assume that when oscillating the sphere traverses a circular path. If the elongation at equilibrium position be l, then
l/L = mg/AYl = mgL/AY
The length in the equilibrium position L' = L+l
 |
| Diagram for Q-10 |
When oscillating the sphere will have a horizontal speed v at vertical position of the wire. This v will depend on the angle θ. P.E. of the sphere at this displacement angle =mg(L'-L'.cosθ)
{Taking the lowest position as zero potential energy level}
K.E. of the sphere at the vertical position of the wire =mv²/2
Equating both we get
mv²/2 = mgL'(1-cosθ)
mv² = 2mgL'(1-cosθ)
The total tension in the wire at vertical position = mg + mv²/r
= mg + 2mgL'(1-cosθ)/r
= mg{r + 2L'(1-cosθ)}/r
The stress in the wire at this time
= Tension in the wire/cross-sectional area
= (mg/rA){r+2L'(1-cosθ)}
Allowed elongation= l + 0.002 m (so that the sphere does not touch the floor)
Strain at this time = (l + 0.002)/L
Hence (l + 0.002)/L = (mg/rAY){r+2L'(1-cosθ)}
→mgL/AY + 0.002 = (mgL/rAY){r+2L'(1-cosθ)}
[substituting l with its expression]
[substituting l with its expression]
→{r+2L'(1-cosθ)} = r + 0.002*rAY/mgL
→1-cosθ = 0.001*rAY/mgLL'
→cosθ = 1 -0.001*rAY/mgLL'
→cosθ=1-0.001*(L+l+0.002){π(0.001)²/4}*2x10¹¹/{20*10*4*(L+l)}
{Let us consider the 0.002 m length negligible in comparision to L', so L+l+0.002≈L+l, i.e r ≈ L'}
→cosθ = 1 - 0.196
→cosθ = 0.804
→θ = 36.4°
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Click here for → EXERCISES (21-30)
Click here for → OBJECTIVE-I
Click here for → OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → EXERCISES (11-20)
Click here for → EXERCISES (21-30)
CHAPTER- 6 - Friction
Click here for → Questions for Short Answer
Click here for → Questions for Short Answer
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
For more practice on problems on friction solve these- "New Questions on Friction".
Click here for → OBJECTIVE-I
Click here for → Friction - OBJECTIVE-II
Click here for → EXERCISES (1-10)
Click here for → Exercises (11-20)
Click here for → EXERCISES (21-31)
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CHAPTER- 5 - Newton's Laws of Motion
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for → QUESTIONS FOR SHORT ANSWER
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→ Newton's laws of motion - Objective - I
Click here for → Newton's Laws of Motion - Objective -II
Click here for → Newton's Laws of Motion-Exercises(Q. No. 1 to 12)
Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
Click here for "The Forces" - OBJECTIVE-I
Click here for "The Forces" - OBJECTIVE-II
Click here for "The Forces" - Exercises
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CHAPTER- 3 - Kinematics - Rest and Motion
Click here for "Questions for short Answers"
Click here for "OBJECTIVE-I"
Click here for EXERCISES (Question number 1 to 10)
Click here for EXERCISES (Question number 11 to 20)
Click here for EXERCISES (Question number 21 to 30)
Click here for EXERCISES (Question number 31 to 40)
Click here for EXERCISES (Question number 41 to 52)
CHAPTER- 2 - "Vector related Problems"
CHAPTER- 2 - "Vector related Problems"
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