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SOME MECHANICAL PROPERTIES OF MATTER:--
EXERCISES, Q-11 To Q-20
11. A steel wire of original length 1 m and cross-sectional area 4.00 mm² is clamped at the two ends so that it lies horizontally and without tension. If a load of 2.16 kg is suspended from the middle point of the wire, what would be its vertical depression?
Y of steel = 2.0 x 10¹¹ N/m². Take g = 10 m/s.
ANSWER: The length of wire = L. Let the tension developed in the wire be T and the length increased = L'.
Strain = L'/L
Stress = T/A = T/4x10⁻⁶ N/m² =2.5x10⁵T N/m²
The relationship between stress and strain
Y*L'/L = 2.5x10⁵T
L' = 2.5x10⁵TL/Y .................... (i)
Diagram for Q-11
Let the depression of middle point = CD =d cm =0.01d m
The load hung W = 2.16g =21.6 N
This load is balanced by the component of tension.
2T.sinα =21.6
T = 21.6/(2.sinα) =10.8*CB/CD =10.8{(L+L')/2}/0.01d
→T = 540(1+L')/d
Putting in (i)
L' = 2.5x10⁵*540(1+L')L/Yd = 2.5x10⁵*540(1+L')/Yd
→L'Yd = 1.35x10⁸ + 1.35x10⁸L'
→L'(Yd-1.35x10⁸) =1.35x10⁸
→L'(2.0x10¹¹d-1.35x10⁸) = 1.35x10⁸ ------------- (ii)
Now in the right angled triangle BCD
BC² = CD² + BD²
→(L+L')²/4 = 1x10⁻⁴d² + L²/4
→L²+2LL'+L'² = 4x10⁻⁴d² +L²
→2LL' = 4x10⁻⁴d² {neglecting L'²}
L' = 2x10⁻⁴d²
Now putting it in (ii)
4x10⁷d³ - 2.7x10⁴d² =1.35x10⁸
→d³ - 6.75x10⁻⁴d² - 3.375=0
{the middle term is much smaller than other two. Neglecting it we get}
→d³ = 3.375
→d = 1.5 cm
12. A copper wire of cross-sectional area 0.01 cm² is under a tension of 20 N. Find the decrease in the cross-sectional area. Young's modulus of copper = 1.1x 10¹¹ N/m² and Poisson's ratio = 0.32
ANSWER: Cross-sectional area A =0.01 cm²
→A = 0.01/10000 =1x10⁻⁶ m²
Stress = 20/1x10⁻⁶ N/m² =2x10⁷ N/m²
Hence the longitudinal strain =Stress/Y =2x10⁷/1.1x10¹¹ =1.82x10⁻⁴
Poisson's ratio = 0.32
Lateral strain = 0.32*1.82x10⁻⁴ =5.82x10⁻⁵
Taking the cross-section as circular with diameter D.
(ΔD/D)/(ΔL/L) =σ
so, ΔD/D = Lateral strain = 5.82x10⁻⁵
We have area A = πD²/4
dA =2πD.dD/4
Hence dA/A =2.dD/D
→dA =2A(dD/D)
→dA =2*0.01*(5.82x10⁻⁵) cm² =1.164x10⁻⁶ cm²
13. Find the increase in pressure required to decrease the volume of a water sample by 0.01%. The bulk modulus of water = 2.1x10⁹ N/m².
ANSWER: Let the original volume =V when the pressure = 0. At a certain pressure P, the volume is V'.
Change in volume = -(V-V'),
Bulk strain = -(V-V')/V
Bulk modulus = B =2.1x10⁹ N/m²
Henve P = -B(V-V')/V = -B(1-V'/V) ......... (i)
→dP = B.dV'/V
But given dV'/V' = 0.01/100
Increase in Bulk stress = pressure increase
=dP = B.(dV'/V')*(V'/V)
=2.1x10⁹(0.01/100)*1 N/m² =2.1x10⁵ N/m²
(Taking V'/V ≈ 1)
14. Estimate the change in density of water in the ocean at a depth of 400 m below the surface. The density of water at the surface =1030 kg/m³ and the bulk modulus of water = 2x10⁹ N/m².
ANSWER: At a depth x m below the surface the pressure P =ρgx
Bulk strain = P/B =ρgx/B
→ΔV/V =ρgx/B
→ΔV = ρgxV/B
Density at x m below the surface, ρ'= ρV/V'
{V' is the compressed volume V, V' =V-ΔV}
=ρV/(V-ρgxV/B) =ρB/(B-ρgx)
=1030*2x10⁹/(2x10⁹ - 1030*10*400) =2.06x10¹²/1.9958x10⁹
=1032 kg/m³ {Putting x=400 m}
The change in density, dρ =ρ'-ρ =1032-1030 = 2 kg/m³
15. A steel plate of face area 4 cm² and thickness 0.5 cm is fixed rigidly at the lower surface. A tangential force of 10 N is applied on the upper surface. Find the lateral displacement of the upper surface with respect to the lower surface. Rigidity modulus of steel = 8.4x10¹⁰ N/m².
ANSWER: Let the lateral displacement of the upper surface with respect to lower surface =Δx
The thickness of the plate, d = 0.5 cm = 0.005 m
Shearing Strain = Δx/d = Δx/0.005 =200*Δx
Shearing stress = Force /area =10/(4/10000) N/m² =25000 N/m²
Shearing stress/shearing strain = Rigidity modulus
→25000/(200*Δx) =8.4x10¹⁰ N/m²
→Δx =25000/(200*8.4x10¹⁰) m
→Δx ≈ 1.5x10⁻⁹ m
16. A 5.0 cm long straight piece of thread is kept on the surface of the water. Find the force with which the surface on one side of the thread pulls it. Surface tension of water = 0.076 N/m.
ANSWER: Length of the thread, L = 5.0 cm =0.05 m
The surface tension of the water, S = 0.076 N/m
Hence the force of surface tension on one side of the thread =S*L
=0.076*0.05 N =0.0038 N = 3.8x10⁻³ N
11. A steel wire of original length 1 m and cross-sectional area 4.00 mm² is clamped at the two ends so that it lies horizontally and without tension. If a load of 2.16 kg is suspended from the middle point of the wire, what would be its vertical depression?
Y of steel = 2.0 x 10¹¹ N/m². Take g = 10 m/s.
ANSWER: The length of wire = L. Let the tension developed in the wire be T and the length increased = L'.
Strain = L'/L
Stress = T/A = T/4x10⁻⁶ N/m² =2.5x10⁵T N/m²
The relationship between stress and strain
Y*L'/L = 2.5x10⁵T
L' = 2.5x10⁵TL/Y .................... (i)
 |
| Diagram for Q-11 |
Let the depression of middle point = CD =d cm =0.01d m
The load hung W = 2.16g =21.6 N
This load is balanced by the component of tension.
2T.sinα =21.6
T = 21.6/(2.sinα) =10.8*CB/CD =10.8{(L+L')/2}/0.01d
→T = 540(1+L')/d
Putting in (i)
L' = 2.5x10⁵*540(1+L')L/Yd = 2.5x10⁵*540(1+L')/Yd
→L'Yd = 1.35x10⁸ + 1.35x10⁸L'
→L'(Yd-1.35x10⁸) =1.35x10⁸
→L'(2.0x10¹¹d-1.35x10⁸) = 1.35x10⁸ ------------- (ii)
Now in the right angled triangle BCD
BC² = CD² + BD²
→(L+L')²/4 = 1x10⁻⁴d² + L²/4
→L²+2LL'+L'² = 4x10⁻⁴d² +L²
→2LL' = 4x10⁻⁴d² {neglecting L'²}
L' = 2x10⁻⁴d²
Now putting it in (ii)
4x10⁷d³ - 2.7x10⁴d² =1.35x10⁸
→d³ - 6.75x10⁻⁴d² - 3.375=0
{the middle term is much smaller than other two. Neglecting it we get}
→d³ = 3.375
→d = 1.5 cm
12. A copper wire of cross-sectional area 0.01 cm² is under a tension of 20 N. Find the decrease in the cross-sectional area. Young's modulus of copper = 1.1x 10¹¹ N/m² and Poisson's ratio = 0.32
ANSWER: Cross-sectional area A =0.01 cm²
→A = 0.01/10000 =1x10⁻⁶ m²
Stress = 20/1x10⁻⁶ N/m² =2x10⁷ N/m²
Hence the longitudinal strain =Stress/Y =2x10⁷/1.1x10¹¹ =1.82x10⁻⁴
Poisson's ratio = 0.32
Lateral strain = 0.32*1.82x10⁻⁴ =5.82x10⁻⁵
Taking the cross-section as circular with diameter D.
(ΔD/D)/(ΔL/L) =σ
so, ΔD/D = Lateral strain = 5.82x10⁻⁵
We have area A = πD²/4
dA =2πD.dD/4
Hence dA/A =2.dD/D
→dA =2A(dD/D)
→dA =2*0.01*(5.82x10⁻⁵) cm² =1.164x10⁻⁶ cm²
13. Find the increase in pressure required to decrease the volume of a water sample by 0.01%. The bulk modulus of water = 2.1x10⁹ N/m².
ANSWER: Let the original volume =V when the pressure = 0. At a certain pressure P, the volume is V'.
Change in volume = -(V-V'),
Bulk strain = -(V-V')/V
Bulk modulus = B =2.1x10⁹ N/m²
Henve P = -B(V-V')/V = -B(1-V'/V) ......... (i)
→dP = B.dV'/V
But given dV'/V' = 0.01/100
Increase in Bulk stress = pressure increase
=dP = B.(dV'/V')*(V'/V)
=2.1x10⁹(0.01/100)*1 N/m² =2.1x10⁵ N/m²
(Taking V'/V ≈ 1)
14. Estimate the change in density of water in the ocean at a depth of 400 m below the surface. The density of water at the surface =1030 kg/m³ and the bulk modulus of water = 2x10⁹ N/m².
ANSWER: At a depth x m below the surface the pressure P =ρgx
Bulk strain = P/B =ρgx/B
→ΔV/V =ρgx/B
→ΔV = ρgxV/B
Density at x m below the surface, ρ'= ρV/V'
{V' is the compressed volume V, V' =V-ΔV}
=ρV/(V-ρgxV/B) =ρB/(B-ρgx)
=1030*2x10⁹/(2x10⁹ - 1030*10*400) =2.06x10¹²/1.9958x10⁹
=1032 kg/m³ {Putting x=400 m}
15. A steel plate of face area 4 cm² and thickness 0.5 cm is fixed rigidly at the lower surface. A tangential force of 10 N is applied on the upper surface. Find the lateral displacement of the upper surface with respect to the lower surface. Rigidity modulus of steel = 8.4x10¹⁰ N/m².
ANSWER: Let the lateral displacement of the upper surface with respect to lower surface =Δx
The thickness of the plate, d = 0.5 cm = 0.005 m
Shearing Strain = Δx/d = Δx/0.005 =200*Δx
Shearing stress = Force /area =10/(4/10000) N/m² =25000 N/m²
Shearing stress/shearing strain = Rigidity modulus
→25000/(200*Δx) =8.4x10¹⁰ N/m²
→Δx =25000/(200*8.4x10¹⁰) m
→Δx ≈ 1.5x10⁻⁹ m
16. A 5.0 cm long straight piece of thread is kept on the surface of the water. Find the force with which the surface on one side of the thread pulls it. Surface tension of water = 0.076 N/m.
ANSWER: Length of the thread, L = 5.0 cm =0.05 m
The surface tension of the water, S = 0.076 N/m
Hence the force of surface tension on one side of the thread =S*L
=0.076*0.05 N =0.0038 N = 3.8x10⁻³ N
17. Find the excess pressure inside (a) a drop of mercury of radius 2 mm(b) a soap bubble of radius 4 mm and (c) an air bubble of radius 4 mm formed inside a tank of water. The surface tension of mercury, soap solution and water are 0.465 N/m, 0.03 N/m and 0.076 N/m respectively.
ANSWER: Excess pressure inside a drop is given as p = 2S/r.
(a) The radius of mercury drop r = 2 mm =0.002 m
The surface tension of mercury Sₘ =0.465 N/m
Excess pressure inside = 2Sₘ/r =2*0.465/0.002 N/m²
= 465 N/m²
(b) The radius of the soap bubble = r'=4 mm =0.004 m
The surface tension of the soap surface Sₓ =0.03 N/m
Excess pressure inside a soap bubble = 4Sₓ/r'
{Since it has two liquid surfaces}
=4*0.03/0.004 N/m² = 30 N/m²
(c) The surface tension of water, Sᵥ =0.076 N/m
The radius of the water bubble r = 4 mm =0.004 m
Excess pressure inside the water bubble in the tank = 2Sᵥ/r
=2*0.076/0.004 N/m² = 38 N/m²
18. Consider a small surface area of 1 mm² at the top of a mercury drop of radius 4.0 mm. Find the force exerted on this area (a) by the air above it (b) by the mercury below it and (c) by the mercury surface in contact with it. Atmospheric pressure = 1.0 x 10⁵ Pa and surface tension of mercury = 0.465 N/m. Neglect the effect of gravity. Assume all numbers to be exact.
ANSWER: The area of the given surface, A = 1 mm² =1x10⁻⁶ m².
(a) The force exerted by the air above it =A*atmospheric pressure
=1x10⁻⁶ m²*1.0x10⁵ N/m²
=0.10 N
(b) The surface tension of the mercury, S =0.465 N/m
The excess pressure inside the drop, Δp = 2S/r
→Δp =2*0.465/(4/1000) =465/2 = 232.5 N/m²
Total pressure inside the drop =Δp + Atmospheric pressure
=232.5 +1.0x10⁵ N/m²
Force on the area by the mercury =A*mercury pressure
=1x10⁻⁶(232.5 +1.0x10⁵) N/m²
=2.325x10⁻⁴+0.10 N/m²
=0.00023 + 0.10 N/m²
=0.10023 N
(c) The mercury surface in contact with the area has a net pressure equal to the pressure in mercury minus the atmospheric pressure outside, i.e. excess pressure inside the drop =2S/r
=2*0.465/(4/1000) N/m²
=465/2 N/m²
=232.5 N/m²
Hence the force = Area*Pressure
=1x10⁻⁶*232.5 N
=0.00023 N
19. The capillaries shown in figure (14-E4) have inner radii 0.5 mm, 1.0 mm and 1.5 mm respectively. The liquid in the beaker is water. Find the heights of water level in the capillaries. The surface tension of water is 7.5 x 10⁻² N/m.
Figure for Q-19
ANSWER: The rise in a capillary = 2S.cosθ/rρg
Here S = 7.5x10⁻² N/m, ρ = 1000 kg/m², Let g = 10 m/s²,
For water θ = 0°.
Hence for A, r = 0.5 mm =0.0005 m
Rise = 2* 7.5x10⁻²*cos0°/(0.0005*1000*10)
=15/500 m
=3/100 m
=3 cm
For B, r = 1.0 mm =0.001 m
Rise = 2* 7.5x10⁻²*cos0°/(0.001*1000*10)
=15/1000 m
=15*100/1000 cm
=1.5 cm
For C, r = 1.5 mm =0.0015 m
Rise = 2* 7.5x10⁻²*cos0°/(0.0015*1000*10)
=15/1500 m
=1/100 m
=1.0 cm
20. The lower end of a capillary tube is immersed in mercury. The level of mercury in the tube is found to be 2 cm below the outer level. If the same tube is immersed in water, up to what height will the water rise in the capillary?
ANSWER: Rise/depression = 2S.cosθ/rρg
For mercury -0.02 = 2S'.cosθ/rρ'g ------- (i)
For water rise, h = 2S.cos0°/rρg ------- (ii)
Dividing (ii) by (i)
h/0.02 = -Sρ'/(S'*cosθ*ρ)
→h = -0.02*Sρ'/(1000S'.cosθ)
S' =7.5 x 10⁻² N/m. S = 0.465 N/m
ρ' = 13600 kg/m³
θ = 140°
So, h = -0.02*7.5x10⁻²*13600/(1000*0.465*cos140°)
→h = 0.0573 m = 5.73 cm
17. Find the excess pressure inside (a) a drop of mercury of radius 2 mm(b) a soap bubble of radius 4 mm and (c) an air bubble of radius 4 mm formed inside a tank of water. The surface tension of mercury, soap solution and water are 0.465 N/m, 0.03 N/m and 0.076 N/m respectively.
ANSWER: Excess pressure inside a drop is given as p = 2S/r.
(a) The radius of mercury drop r = 2 mm =0.002 m
The surface tension of mercury Sₘ =0.465 N/m
Excess pressure inside = 2Sₘ/r =2*0.465/0.002 N/m²
= 465 N/m²
(b) The radius of the soap bubble = r'=4 mm =0.004 m
The surface tension of the soap surface Sₓ =0.03 N/m
Excess pressure inside a soap bubble = 4Sₓ/r'
{Since it has two liquid surfaces}
=4*0.03/0.004 N/m² = 30 N/m²
(c) The surface tension of water, Sᵥ =0.076 N/m
The radius of the water bubble r = 4 mm =0.004 m
Excess pressure inside the water bubble in the tank = 2Sᵥ/r
=2*0.076/0.004 N/m² = 38 N/m²
18. Consider a small surface area of 1 mm² at the top of a mercury drop of radius 4.0 mm. Find the force exerted on this area (a) by the air above it (b) by the mercury below it and (c) by the mercury surface in contact with it. Atmospheric pressure = 1.0 x 10⁵ Pa and surface tension of mercury = 0.465 N/m. Neglect the effect of gravity. Assume all numbers to be exact.
ANSWER: The area of the given surface, A = 1 mm² =1x10⁻⁶ m².
(a) The force exerted by the air above it =A*atmospheric pressure
=1x10⁻⁶ m²*1.0x10⁵ N/m²
=0.10 N
(b) The surface tension of the mercury, S =0.465 N/m
The excess pressure inside the drop, Δp = 2S/r
→Δp =2*0.465/(4/1000) =465/2 = 232.5 N/m²
Total pressure inside the drop =Δp + Atmospheric pressure
=232.5 +1.0x10⁵ N/m²
Force on the area by the mercury =A*mercury pressure
=1x10⁻⁶(232.5 +1.0x10⁵) N/m²
=2.325x10⁻⁴+0.10 N/m²
=0.00023 + 0.10 N/m²
=0.10023 N
(c) The mercury surface in contact with the area has a net pressure equal to the pressure in mercury minus the atmospheric pressure outside, i.e. excess pressure inside the drop =2S/r
=2*0.465/(4/1000) N/m²
=465/2 N/m²
=232.5 N/m²
Hence the force = Area*Pressure
=1x10⁻⁶*232.5 N
=0.00023 N19. The capillaries shown in figure (14-E4) have inner radii 0.5 mm, 1.0 mm and 1.5 mm respectively. The liquid in the beaker is water. Find the heights of water level in the capillaries. The surface tension of water is 7.5 x 10⁻² N/m.
 |
| Figure for Q-19 |
ANSWER: The rise in a capillary = 2S.cosθ/rρg
Here S = 7.5x10⁻² N/m, ρ = 1000 kg/m², Let g = 10 m/s²,
For water θ = 0°.
Hence for A, r = 0.5 mm =0.0005 m
Rise = 2* 7.5x10⁻²*cos0°/(0.0005*1000*10)
=15/500 m
=3/100 m
=3 cm
For B, r = 1.0 mm =0.001 m
Rise = 2* 7.5x10⁻²*cos0°/(0.001*1000*10)
=15/1000 m
=15*100/1000 cm
=1.5 cm
For C, r = 1.5 mm =0.0015 m
Rise = 2* 7.5x10⁻²*cos0°/(0.0015*1000*10)
=15/1500 m
=1/100 m
=1.0 cm
20. The lower end of a capillary tube is immersed in mercury. The level of mercury in the tube is found to be 2 cm below the outer level. If the same tube is immersed in water, up to what height will the water rise in the capillary?
ANSWER: Rise/depression = 2S.cosθ/rρg
For mercury -0.02 = 2S'.cosθ/rρ'g ------- (i)
For water rise, h = 2S.cos0°/rρg ------- (ii)
Dividing (ii) by (i)
h/0.02 = -Sρ'/(S'*cosθ*ρ)
→h = -0.02*Sρ'/(1000S'.cosθ)
S' =7.5 x 10⁻² N/m. S = 0.465 N/m
ρ' = 13600 kg/m³
θ = 140°
So, h = -0.02*7.5x10⁻²*13600/(1000*0.465*cos140°)
→h = 0.0573 m = 5.73 cm
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CHAPTER- 14 - Fluid Mechanics
Questions for Short Answers
OBJECTIVE-I
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EXERCISES- Q-1 TO Q-10
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CHAPTER- 11 - Gravitation
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CHAPTER- 5 - Newton's Laws of Motion
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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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