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FLUID MECHANICS:--
EXERCISES Q-21 TO Q-30
21. Find the ratio of the weights, as measured by a spring balance, of a 1 kg block of iron and 1 kg block of wood. Density of iron =7800 kg/m³, density of wood =800 kg/m³ and density of air =1.293 kg/m³.
ANSWER: The spring balance measures the weight of an object. It will show the same weight for the given iron and the wood block in the vacuum but in the air, a force of buoyancy will act on both the blocks which will be equal to the weight of the air displaced.
The volume of iron block = 1/7800 m³
The volume of wooden block = 1/800 m³
The weight of air displaced by iron block =1.293*1/7800 kg
=1.293/7800 kg
The weight of air displaced by wooden block =1.293*1/800 kg
=1.293/800 kg
The apparent weight of the iron block shown in the spring balance =1-1.293/7800 kg
=(7800-1.293)/7800 kg
= 7798.707/7800 kg
The apparent weight of the wooden block shown in the spring balance =1-1.293/800 kg
=(800-1.293)/800 kg
= 798.707/800 kg
Hence the ratio of the apparent weights =(7798.707/7800)/(798.707/800)
=(7798.707*800)/(7800*798.707)
=1.0015
22. A cylindrical object of outer diameter 20 cm and mass 2 kg floats in water with its axis vertical. If it is slightly depressed and then released find the time period of the resulting simple harmonic motion of the object.
ANSWER: Cross-sectional area of the cylinder =π(0.10)² m²
If the cylinder is depressed to a depth of X meter, the net force on the cylinder upward =πX(0.10)² * 1000 kg-weight.
=10πX kg-weight
=10πgX Newton
Mass of the cylinder =2 kg
Acceleration =Force/mass =10πgX/2 =5πgX m/s²
But acceleration is also =⍵²X, equating both we get,
⍵²X =5πgX
→⍵ =√(5πg)
Hence the time period of the simple harmonic motion =2π/⍵
=2π/√(5πg) =2*3.14/√(5π*9.8) =0.50 s
23. A cylindrical object of outer diameter 10 cm, height 20 cm and density 8000 kg/m³ is supported by a vertical spring and is half dipped in water as shown in the figure (13-E6). (a) Find the elongation of the spring in the equilibrium condition. (b) If the object is slightly depressed and released, find the time period of resulting oscillations of the object. The spring constant =500 N/m.
Figure for Q-23
ANSWER: The volume of the object =π*5²*20 cm³ =500π cm³
The mass of the object =500π*8000/10⁶ kg
=4π kg
Since the object is half dipped, the volume of the water displaced =500π/2 =250π cm³
The mass of the water displaced =250π*1000/10⁶ kg
=0.25π kg
The force of buoyancy =0.25πg N
Hence the apparent weight of the object =(4π-0.25π)g N
=3.75πg N
The spring constant k =500 N/m
(a) Hence the elongation of the spring =3.75πg/500 m
=3.75*3.14*10/500 m {Taking g = 10 m/s²}
=0.235 m
=23.5 cm
(b) Let the object be dipped to a distance X m. Assuming the axis of the cylindrical object vertical, the extra volume displaced =π*5²*X/10000 m³
The buoyancy force =(π*5²*X/10000 m³)*1000g N
=2.5πgX N
The spring force =kX N
Hence the total upward force =2.5πgX+500X N
Hence the vertical acceleration at the moment =Force/mass
=(2.5πgX+500X)/4π m/s²
=(2.5πg+500)X/4π m/s²
But also this acceleration =⍵²X m/s²
Equating, ⍵²X = (2.5πg+500)X/4π =46.04X
→⍵ =√(46.04) =6.785 s⁻¹
So, the time period =2π/⍵
=2π/6.785 s =0.93 s
24. A wooden block of mass 0.5 kg and density 800 kg/m³ is fastened to the free end of a vertical spring of spring constant 50 N/m fixed at the bottom. If the entire system is completely immersed in water, find (a) the elongation (or compression) of the spring in equilibrium and (b) the time period of vertical oscillations of the block when it is slightly depressed and released.
ANSWER: (a) The volume of wooden block = 0.5/800 m³
=0.000625 m³
The force of buoyancy =0.000625*1000 kg =0.625 kg
The Net force on the block in the water =0.625-0.500 kg-weight
=0.125*10 N {Taking g = 10 m/s²}
=1.25 N upward.
Spring constant k=50 N/m
Hence the elongation of the spring =(1.25/50)*100 cm
=2.5 cm
(b) Let the block be depressed by X cm.
The force on the block by the spring =50X N
Hence the acceleration =50X/0.5 =100X m/s²
So, ⍵²X =100X
→⍵ = 10 s⁻¹
Hence time period T = 2π/⍵ =2π/10 s
= π/5 s
25. A cube of ice of edge 4 cm is placed in an empty cylindrical glass of inner diameter 6 cm. Assume that the ice melts uniformly from each side so that it always retains its cubical shape. Remembering that ice is lighter than water, find the length of the edge of the ice cube at the instant it just leaves contact with the bottom of the glass.
ANSWER:
Figure for Q-25
Let the edge of the cube at the instant it just leaves the bottom =X cm and the height of water in the glass =Y cm. If the density of water and the ice be ρ and ρᵢ g/cm³ respectively, the weight of ice cube =X³ρᵢ gram and the weight of the water displaced =X²Yρ gram. These two will be equal.
→X³ρᵢ =X²Yρ
→Xρᵢ=Yρ
→Y =ρᵢX/ρ
The volume of the water in the glass =(π*3²-X²)*Y =(9π-X²)Y ..(i)
The volume of ice that melted =4³-X³, this volume of melted ice becomes the corresponding volume of water =(4³-X³)*ρᵢ/ρ .....(ii)
Equating (i) and (ii)
(9π-X²)Y = (4³-X³)*ρᵢ/ρ
→(9π-X²)*ρᵢX/ρ =(4³-X³)*ρᵢ/ρ
→9πX-X³ = 4³-X³
→9πX =4³
→X =64/9π
→X =2.26 cm
26. A U-tube containing a liquid is accelerated horizontally with a constant acceleration a₀. If the separation between the vertical limbs is l, find the difference in the heights of the liquid in the two arms.
ANSWER: Let the pressure at the base of the back limb at A =Pₐ and the pressure at the base of the front limb at B = Pᵦ. The separation between the vertical limbs AB = l (given). When the U-tube is accelerated at an acceleration a₀, the difference of pressure between horizontal points A and B is given as,
Pₐ-Pᵦ = lρa₀
Figure for Q-26
where ρ is the density of the liquid. Due to this pressure difference, the liquid in limb A will rise to a height h more than in limb B. The pressure difference due to this height difference balances the difference of pressure between points A and B. The pressure due to this height h =ρgh. Equating we get,
ρgh = Pₐ-Pᵦ = lρa₀
→gh = la₀
→h = a₀l/g
27. At Deoprayag river Alaknanda mixes with the river Bhagirathi and becomes river Ganga. Suppose Alaknanda has a width of 12 m, Bhagirathi has a width of 8 m and Ganga has a width of 16 m. Assume that the depth of water is the same in the three rivers. Let the average speed of water in Alaknanda be 20 km/h and in Bhagirathi be 16 km/h. Find the average speed of the water in the river Ganga.
ANSWER: If the average depth of rivers is h meter, then the flow volume of water in the river Alaknanda =20*h*12000/3600 m³/s
=66.67h m³/s
The flow volume of water in the river Bhagirathi =8*h*16000/3600 m³/s
=35.56h m³/s
Total flow volume in the rivers before mixing
=66.67h + 35.56h =102.23h m³/s
If the average speed in the river Ganga be V km/h, then flow volume =16*h*V*1000/3600 m³/s
=4.44hV m³/s
This would be equal to the flow volume before mixing, so
4.44hV =102.23h
→V = 102.23/4.44 =23 km/h
28. Water flows through a horizontal tube of a variable cross-section (figure 13-E7). The area of the cross-section at A and B are 4 mm² and 2 mm² respectively. If 1 cc of water enters per second through A, find (a) the speed of the water at A, (b) the speed of the water at B and (c) the pressure difference Pₐ-Pᵦ.
Figure for Q-28
ANSWER: (a) The area of the cross-section at A =4 mm² =4/100 cm²
=0.04 cm²
The volume of water flowing per second Q=1 cm³
Hence the speed at A, v =1/0.04 cm/s =100/4 cm/s =25 cm/s
(b) The area of the cross-section at B =2 mm² =2/100 cm²
=0.02 cm²
The volume of water flowing per second Q=1 cm³
Hence the speed at B, v' =1/0.02 cm/s =100/2 cm/s =50 cm/s
(c) From the Bernoulli's theorem
Pₐ+½ρv² = Pᵦ+½ρv'²
→Pₐ-Pᵦ =½ρ(v'²-v²)
=½*1000*(0.50²-0.25²) N/m²
=½*1000*0.75*0.25 N/m²
=93.75 N/m² ≈94 N/m²
29. Suppose the tube in the previous problem is kept vertical with A upward but the other conditions remain the same. The separation between the cross sections at A and B is 15/16 cm. Repeat parts (a), (b) and (c) of the previous problem. Take g = 10 m/s².
ANSWER: Since the discharge through the tube remains the same i.e. 1 cc/s, the speeds at A and B will be the same.
(a) The speed at A, v =1/0.04 cm/s =100/4 cm/s =25 cm/s
(b) The speed at B, v' =1/0.02 cm/s =100/2 cm/s =50 cm/s
(c) From the Bernoulli's theorem
Pₐ+ρgh+½ρv² = Pᵦ+½ρv'²
Taking the point B as a reference level for the height. Given that h=15/16 cm
→Pₐ-Pᵦ =½ρ(v'²-v²) -ρgh
=½*1000*(0.50²-0.25²) -1000*10*15/1600 N/m²
=½*1000*0.75*0.25 - 1500/16 N/m²
=93.75 - 93.75 N/m² = zero
30. Suppose the tube in the previous problem is kept vertical with B upward. Water enters through B at the rate of 1 cm³/s. Repeat parts (a), (b) and (c). Note that the speed decreases as the water falls down.
ANSWER: Since the discharge through the tube remains the same i.e. 1 cm³/s, the speeds at A and B will be the same.
(a) The speed at A, v =1/0.04 cm/s =100/4 cm/s =25 cm/s
(b) The speed at B, v' =1/0.02 cm/s =100/2 cm/s =50 cm/s
(c) Since the end B is upward, we take the end A as a reference level for the height. So the height of end A =0 and the height of end B = h =15/16 cm =15/1600 m.
From the Bernoulli's theorem
Pₐ+½ρv² = Pᵦ+½ρv'²+ρgh
→Pₐ-Pᵦ =½ρ(v'²-v²) +ρgh
=½*1000*(0.50²-0.25²) +1000*10*15/1600 N/m²
=½*1000*0.75*0.25 + 1500/16 N/m²
=93.75 + 93.75 N/m²
=187.5 N/m² ≈ 188 N/m²
21. Find the ratio of the weights, as measured by a spring balance, of a 1 kg block of iron and 1 kg block of wood. Density of iron =7800 kg/m³, density of wood =800 kg/m³ and density of air =1.293 kg/m³.
ANSWER: The spring balance measures the weight of an object. It will show the same weight for the given iron and the wood block in the vacuum but in the air, a force of buoyancy will act on both the blocks which will be equal to the weight of the air displaced.
The volume of iron block = 1/7800 m³
The volume of wooden block = 1/800 m³
The weight of air displaced by iron block =1.293*1/7800 kg
=1.293/7800 kg
The weight of air displaced by wooden block =1.293*1/800 kg
=1.293/800 kg
The apparent weight of the iron block shown in the spring balance =1-1.293/7800 kg
=(7800-1.293)/7800 kg
= 7798.707/7800 kg
The apparent weight of the wooden block shown in the spring balance =1-1.293/800 kg
=(800-1.293)/800 kg
= 798.707/800 kg
Hence the ratio of the apparent weights =(7798.707/7800)/(798.707/800)
=(7798.707*800)/(7800*798.707)
=1.0015
22. A cylindrical object of outer diameter 20 cm and mass 2 kg floats in water with its axis vertical. If it is slightly depressed and then released find the time period of the resulting simple harmonic motion of the object.
ANSWER: Cross-sectional area of the cylinder =π(0.10)² m²
If the cylinder is depressed to a depth of X meter, the net force on the cylinder upward =πX(0.10)² * 1000 kg-weight.
=10πX kg-weight
=10πgX Newton
Mass of the cylinder =2 kg
Acceleration =Force/mass =10πgX/2 =5πgX m/s²
But acceleration is also =⍵²X, equating both we get,
⍵²X =5πgX
→⍵ =√(5πg)
Hence the time period of the simple harmonic motion =2π/⍵
=2π/√(5πg) =2*3.14/√(5π*9.8) =0.50 s
23. A cylindrical object of outer diameter 10 cm, height 20 cm and density 8000 kg/m³ is supported by a vertical spring and is half dipped in water as shown in the figure (13-E6). (a) Find the elongation of the spring in the equilibrium condition. (b) If the object is slightly depressed and released, find the time period of resulting oscillations of the object. The spring constant =500 N/m.
 |
| Figure for Q-23 |
ANSWER: The volume of the object =π*5²*20 cm³ =500π cm³
The mass of the object =500π*8000/10⁶ kg
=4π kg
Since the object is half dipped, the volume of the water displaced =500π/2 =250π cm³
The mass of the water displaced =250π*1000/10⁶ kg
=0.25π kg
The force of buoyancy =0.25πg N
Hence the apparent weight of the object =(4π-0.25π)g N
=3.75πg N
The spring constant k =500 N/m
=3.75*3.14*10/500 m {Taking g = 10 m/s²}
=0.235 m
=23.5 cmThe buoyancy force =(π*5²*X/10000 m³)*1000g N
=2.5πgX N
The spring force =kX N
Hence the total upward force =2.5πgX+500X N
Hence the vertical acceleration at the moment =Force/mass
=(2.5πgX+500X)/4π m/s²
=(2.5πg+500)X/4π m/s²
But also this acceleration =⍵²X m/s²
Equating, ⍵²X = (2.5πg+500)X/4π =46.04X
→⍵ =√(46.04) =6.785 s⁻¹
So, the time period =2π/⍵
=2π/6.785 s =0.93 s
24. A wooden block of mass 0.5 kg and density 800 kg/m³ is fastened to the free end of a vertical spring of spring constant 50 N/m fixed at the bottom. If the entire system is completely immersed in water, find (a) the elongation (or compression) of the spring in equilibrium and (b) the time period of vertical oscillations of the block when it is slightly depressed and released.
ANSWER: (a) The volume of wooden block = 0.5/800 m³
=0.000625 m³
The force of buoyancy =0.000625*1000 kg =0.625 kg
The Net force on the block in the water =0.625-0.500 kg-weight
=0.125*10 N {Taking g = 10 m/s²}
=1.25 N upward.
Spring constant k=50 N/m
Hence the elongation of the spring =(1.25/50)*100 cm
=2.5 cm
(b) Let the block be depressed by X cm.
The force on the block by the spring =50X N
Hence the acceleration =50X/0.5 =100X m/s²
So, ⍵²X =100X
→⍵ = 10 s⁻¹
Hence time period T = 2π/⍵ =2π/10 s
= π/5 s
25. A cube of ice of edge 4 cm is placed in an empty cylindrical glass of inner diameter 6 cm. Assume that the ice melts uniformly from each side so that it always retains its cubical shape. Remembering that ice is lighter than water, find the length of the edge of the ice cube at the instant it just leaves contact with the bottom of the glass.
ANSWER:
|
| Figure for Q-25 |
Let the edge of the cube at the instant it just leaves the bottom =X cm and the height of water in the glass =Y cm. If the density of water and the ice be ρ and ρᵢ g/cm³ respectively, the weight of ice cube =X³ρᵢ gram and the weight of the water displaced =X²Yρ gram. These two will be equal.
→X³ρᵢ =X²Yρ
→Xρᵢ=Yρ
→Y =ρᵢX/ρ
The volume of the water in the glass =(π*3²-X²)*Y =(9π-X²)Y ..(i)
The volume of ice that melted =4³-X³, this volume of melted ice becomes the corresponding volume of water =(4³-X³)*ρᵢ/ρ .....(ii)
Equating (i) and (ii)
(9π-X²)Y = (4³-X³)*ρᵢ/ρ
→(9π-X²)*ρᵢX/ρ =(4³-X³)*ρᵢ/ρ
→9πX-X³ = 4³-X³
→9πX =4³
→X =64/9π
→X =2.26 cm
26. A U-tube containing a liquid is accelerated horizontally with a constant acceleration a₀. If the separation between the vertical limbs is l, find the difference in the heights of the liquid in the two arms.
ANSWER: Let the pressure at the base of the back limb at A =Pₐ and the pressure at the base of the front limb at B = Pᵦ. The separation between the vertical limbs AB = l (given). When the U-tube is accelerated at an acceleration a₀, the difference of pressure between horizontal points A and B is given as,
Pₐ-Pᵦ = lρa₀
 |
| Figure for Q-26 |
where ρ is the density of the liquid. Due to this pressure difference, the liquid in limb A will rise to a height h more than in limb B. The pressure difference due to this height difference balances the difference of pressure between points A and B. The pressure due to this height h =ρgh. Equating we get,
ρgh = Pₐ-Pᵦ = lρa₀
→gh = la₀
→h = a₀l/g27. At Deoprayag river Alaknanda mixes with the river Bhagirathi and becomes river Ganga. Suppose Alaknanda has a width of 12 m, Bhagirathi has a width of 8 m and Ganga has a width of 16 m. Assume that the depth of water is the same in the three rivers. Let the average speed of water in Alaknanda be 20 km/h and in Bhagirathi be 16 km/h. Find the average speed of the water in the river Ganga.
ANSWER: If the average depth of rivers is h meter, then the flow volume of water in the river Alaknanda =20*h*12000/3600 m³/s
=66.67h m³/s
The flow volume of water in the river Bhagirathi =8*h*16000/3600 m³/s
=35.56h m³/s
Total flow volume in the rivers before mixing
=66.67h + 35.56h =102.23h m³/s
If the average speed in the river Ganga be V km/h, then flow volume =16*h*V*1000/3600 m³/s
=4.44hV m³/s
This would be equal to the flow volume before mixing, so
4.44hV =102.23h
→V = 102.23/4.44 =23 km/h
28. Water flows through a horizontal tube of a variable cross-section (figure 13-E7). The area of the cross-section at A and B are 4 mm² and 2 mm² respectively. If 1 cc of water enters per second through A, find (a) the speed of the water at A, (b) the speed of the water at B and (c) the pressure difference Pₐ-Pᵦ.
 |
| Figure for Q-28 |
ANSWER: (a) The area of the cross-section at A =4 mm² =4/100 cm²
=0.04 cm²
The volume of water flowing per second Q=1 cm³
Hence the speed at A, v =1/0.04 cm/s =100/4 cm/s =25 cm/s
=0.02 cm²
The volume of water flowing per second Q=1 cm³
Pₐ+½ρv² = Pᵦ+½ρv'²
→Pₐ-Pᵦ =½ρ(v'²-v²)
=½*1000*(0.50²-0.25²) N/m²
=½*1000*0.75*0.25 N/m²
=93.75 N/m² ≈94 N/m²
29. Suppose the tube in the previous problem is kept vertical with A upward but the other conditions remain the same. The separation between the cross sections at A and B is 15/16 cm. Repeat parts (a), (b) and (c) of the previous problem. Take g = 10 m/s².
ANSWER: Since the discharge through the tube remains the same i.e. 1 cc/s, the speeds at A and B will be the same.
(a) The speed at A, v =1/0.04 cm/s =100/4 cm/s =25 cm/s
(b) The speed at B, v' =1/0.02 cm/s =100/2 cm/s =50 cm/s
(c) From the Bernoulli's theorem
Pₐ+ρgh+½ρv² = Pᵦ+½ρv'²
Taking the point B as a reference level for the height. Given that h=15/16 cm
→Pₐ-Pᵦ =½ρ(v'²-v²) -ρgh
=½*1000*(0.50²-0.25²) -1000*10*15/1600 N/m²
=½*1000*0.75*0.25 - 1500/16 N/m²
=93.75 - 93.75 N/m² = zero
30. Suppose the tube in the previous problem is kept vertical with B upward. Water enters through B at the rate of 1 cm³/s. Repeat parts (a), (b) and (c). Note that the speed decreases as the water falls down.
ANSWER: Since the discharge through the tube remains the same i.e. 1 cm³/s, the speeds at A and B will be the same.
(a) The speed at A, v =1/0.04 cm/s =100/4 cm/s =25 cm/s
(b) The speed at B, v' =1/0.02 cm/s =100/2 cm/s =50 cm/s
(c) Since the end B is upward, we take the end A as a reference level for the height. So the height of end A =0 and the height of end B = h =15/16 cm =15/1600 m.
From the Bernoulli's theorem
From the Bernoulli's theorem
Pₐ+½ρv² = Pᵦ+½ρv'²+ρgh
→Pₐ-Pᵦ =½ρ(v'²-v²) +ρgh
=½*1000*(0.50²-0.25²) +1000*10*15/1600 N/m²
=½*1000*0.75*0.25 + 1500/16 N/m²
=93.75 + 93.75 N/m²
=187.5 N/m² ≈ 188 N/m²
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CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
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CHAPTER- 11 - Gravitation
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HC Verma's Concepts of Physics, Chapter-8, WORK AND ENERGY
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These solutions are the best with no errors and perfect explaination.
ReplyDeleteThanks, Abhishek
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