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SOUND WAVES
EXERCISES- Q-51 to Q-60
51. A 30.0 cm long wire having a mass of 10.0 g is fixed at two ends and is vibrated in its fundamental mode. A 50.0 cm long closed organ pipe, placed with its open end near the wire, is set up into resonance in its fundamental mode by vibrating wire. Find the tension in the wire. The speed of sound in air = 340 m/s.
ANSWER: Let the tension in the wire = F. The length of the wire, l = 30 cm = 0.30 m. Mass of the wire, m =10.0 g = 0.010 kg.
Mass per unit length, µ = m/l =0.010/0.30 kg/m =1/30 kg/m.
The fundamental frequency of the wire,
ν = (1/2l)√(F/µ) =(1/2*0.30)√(30 F)
→ν = (5/3)√(30 F)
Diagram for Q-51
This same frequency sets the air column in the closed organ pipe (L = 0.50 m) into the fundamental mode which is V/4L. Equating,
(5/3)√(30 F) = V/4L = 340/4*0.50 =170
→√(30 F) =3*170/5 =3*34 =102
→30 F = 102²
→F ≈ 347 N
52. Show that if the room temperature changes by a small amount from T to T+ΔT, the fundamental frequency of an organ pipe changes from ν to ν+Δν, where
Δν/ν = ½(ΔT/T).
ANSWER: If the length of an open or closed organ pipe is kept constant, the fundamental frequency varies proportionally to the speed of the sound in air. But the speed of sound varies proportionally to the square root of the absolute temperature.
ν ∝ V and V ∝ √T
Hence ν ∝ √T
→ν = K√T, where K is a constant of proportionality.
Let there be a small change in temperature by ΔT, due to it the change in the frequency is Δν (Given).
So now ν+Δν = K√(T+ΔT). Hence,
ν+Δν/ν = √{(T+ΔT)/T}
→1+Δν/ν = √(1+ΔT/T)
→1+Δν/ν = 1+½(ΔT/T)
{Expanding as a binomial theorem in the form of √(1+x) and neglecting (ΔT/T)² and the higher powers}
→Δν/ν = ½ (ΔT/T)
Hence proved.
53. The fundamental frequency of a closed pipe is 293 Hz when the air in it is at a temperature of 20°C. What will be the fundamental frequency when the temperature changes to 22°C?
ANSWER: The fundamental frequency of the closed organ pipe at 20°C = ν₀ and at 22°C = ν₁. Since the frequency is proportional to the speed of sound V and the speed of sound is proportional to the square root of the temperature, hence
ν₀/ν₁ =√(T₀/T₁)
Given ν₀ = 293 Hz, T₀ = 273+20 =293 K, T₁ = 273+22 =295 K.
→293/ν₁ = √(293/295)
→ν₁ = 293*√(295/293) =√(295*293) =294 Hz
54. A Kundt's tube apparatus has a copper rod of length 1.0 m clamped at 25 cm from one of the ends. The tube contains air in which the speed of sound is 340 m/s. The powder collects in heaps separated by a distance of 5.0 cm. Find the speed of sound waves in copper.
ANSWER: Given, V = 340 m/s and Δl =5.0 cm =0.05 m. The rod length clamped, L =25 cm =0.25 m. The clamped point of the rod becomes a pressure antinode and the end a pressure node. Hence the wavelength of the wave in the rod 𝜆' = 4*0.25 m = 1.0 m. The wavelength of the sound wave in air 𝜆 =2*0.05 m = 0.10 m. The frequency of the waves ν is the same. If the speed of the sound waves in the copper = V', then
V' = ν𝜆' and V = ν𝜆
→V'/V = 𝜆'/𝜆
→V' = V𝜆'/𝜆 = 340*1.0/0.10 m/s =3400 m/s
55. A Kundt's tube apparatus has a steel rod of length 1.0 m clamped at the center. It is vibrated in its fundamental mode at a frequency of 2600 Hz. The lycopodium powder dispersed in the tube collects into heaps separated by 6.5 cm. Calculate the speed of sound in steel and in the air.
ANSWER: Given that the frequency, ν = 2600 Hz and length of the rod, L= 1.0 m. The steel rod acts as an open organ pipe in which frequency of the fundamental mode of vibration =V'/2L, where V' is the speed of the sound in the steel rod.
→V'/2L =ν
→V' =2νL = 2*2600*1.0 m/s =5200 m/s
The separation of heaps in the air, l = 6.5 cm =0.065 m. Hence the wavelength of sound in air, 𝜆 = 2l =2*0.065 m =0.13 m.
The speed of sound in air, V = ν𝜆 =2600*0.13 m/s
=338 m/s
56. A source of sound with adjustable frequency produces 2 beats per second with a tuning fork when its frequency is either 476 Hz or 480 Hz. What is the frequency of the tuning fork?
ANSWER: If the frequency of the tuning fork = ν, then either ν - 476 = 2 or 480 - ν = 2. Equating them,
ν - 476 = 480 - ν
→2ν = 956
→ν = 478 Hz
57. A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first one is now loaded with a little wax and the beat frequency is found to increase to 6 per second. What was the original frequency of the tuning fork?
ANSWER: With the loading of wax the frequency of a tuning fork decreases. Since the beat frequency is the difference of the two frequencies, the frequency of the first tuning fork is less than the second because after wax loading the first the beat frequency increases (i.e. the difference of the two frequencies increases). Thus the original frequency of the tuning fork = 256 - 4 Hz = 252 Hz.
58. Calculate the frequency of beats produced in air when two sources of sound are activated, one emitting a wavelength, of 32 cm and the other 32.2 cm. The speed of sound in air is 350 m/s.
ANSWER: 𝜆 = 32 cm = 0.32 m, ν = V/𝜆 =350/0.32 Hz = 1094 Hz.
𝜆' = 32.2 cm = 0.322 m, ν' = V/𝜆' =350/0.322 =1087 Hz.
The frequency of beats =|ν - ν'| =|1094 - 1087|
= 7 per second.
59. A tuning fork of unknown frequency makes 5 beats per second with another tuning fork which can cause a closed organ pipe of length 40 cm to vibrate in its fundamental mode. The beat frequency decreases when the first tuning fork is slightly loaded with wax. Find its original frequency. The speed of sound in air is 320 m/s.
ANSWER: Since with the wax loading of the first tuning-fork decreases the beat frequency, the difference of two frequencies decreases. It means when the frequency of the first tuning fork is reduced the difference is reduced. It can happen if the frequency of the first tuning fork is greater than the second.
Let the frequency of the second tuning fork = ν. This frequency causes a closed organ pipe of 40 cm ( = 0.40 m) to vibrate in its fundamental mode, hence
ν = V/4L =320/(4*0.40) =80/0.40 Hz = 200 Hz.
Beat frequency = 5 per second, hence the frequency of the first tuning-fork = 200 + 5 Hz = 205 Hz.
60. A piano wire A vibrates at a fundamental frequency of 600 Hz. A second identical wire B produces 6 beats per second with it when the tension in A is slightly increased. Find the ratio of the tension in A to the tension in B.
ANSWER: Let the tension in A = F and in B = F'. The fundamental frequency of the vibration in the wire B
(1/2L)√(F'/µ) = 600
and in A, (1/2L)√(F/µ) = ν
Hence √(F/F') = ν/600
→F/F' = ν²/600²
Since the tension in A is more than the tension in B, thus ν > 600. The beat produced = 6 beats per second. Hence ν = 600 + 6 = 606 Hz.
→F/F' = 606²/600² =(606/600)² =(1.01)² =1.02
51. A 30.0 cm long wire having a mass of 10.0 g is fixed at two ends and is vibrated in its fundamental mode. A 50.0 cm long closed organ pipe, placed with its open end near the wire, is set up into resonance in its fundamental mode by vibrating wire. Find the tension in the wire. The speed of sound in air = 340 m/s.
ANSWER: Let the tension in the wire = F. The length of the wire, l = 30 cm = 0.30 m. Mass of the wire, m =10.0 g = 0.010 kg.
Mass per unit length, µ = m/l =0.010/0.30 kg/m =1/30 kg/m.
The fundamental frequency of the wire,
ν = (1/2l)√(F/µ) =(1/2*0.30)√(30 F)
→ν = (5/3)√(30 F)
This same frequency sets the air column in the closed organ pipe (L = 0.50 m) into the fundamental mode which is V/4L. Equating,
(5/3)√(30 F) = V/4L = 340/4*0.50 =170
→√(30 F) =3*170/5 =3*34 =102
→30 F = 102²
→F ≈ 347 N
52. Show that if the room temperature changes by a small amount from T to T+ΔT, the fundamental frequency of an organ pipe changes from ν to ν+Δν, where
Δν/ν = ½(ΔT/T).
ANSWER: If the length of an open or closed organ pipe is kept constant, the fundamental frequency varies proportionally to the speed of the sound in air. But the speed of sound varies proportionally to the square root of the absolute temperature.
ν ∝ V and V ∝ √T
Hence ν ∝ √T
→ν = K√T, where K is a constant of proportionality.
Let there be a small change in temperature by ΔT, due to it the change in the frequency is Δν (Given).
So now ν+Δν = K√(T+ΔT). Hence,
ν+Δν/ν = √{(T+ΔT)/T}
→1+Δν/ν = √(1+ΔT/T)
→1+Δν/ν = 1+½(ΔT/T)
{Expanding as a binomial theorem in the form of √(1+x) and neglecting (ΔT/T)² and the higher powers}
→Δν/ν = ½ (ΔT/T)
Hence proved.
53. The fundamental frequency of a closed pipe is 293 Hz when the air in it is at a temperature of 20°C. What will be the fundamental frequency when the temperature changes to 22°C?
ANSWER: The fundamental frequency of the closed organ pipe at 20°C = ν₀ and at 22°C = ν₁. Since the frequency is proportional to the speed of sound V and the speed of sound is proportional to the square root of the temperature, hence
ν₀/ν₁ =√(T₀/T₁)
Given ν₀ = 293 Hz, T₀ = 273+20 =293 K, T₁ = 273+22 =295 K.
→293/ν₁ = √(293/295)
→ν₁ = 293*√(295/293) =√(295*293) =294 Hz
54. A Kundt's tube apparatus has a copper rod of length 1.0 m clamped at 25 cm from one of the ends. The tube contains air in which the speed of sound is 340 m/s. The powder collects in heaps separated by a distance of 5.0 cm. Find the speed of sound waves in copper.
ANSWER: Given, V = 340 m/s and Δl =5.0 cm =0.05 m. The rod length clamped, L =25 cm =0.25 m. The clamped point of the rod becomes a pressure antinode and the end a pressure node. Hence the wavelength of the wave in the rod 𝜆' = 4*0.25 m = 1.0 m. The wavelength of the sound wave in air 𝜆 =2*0.05 m = 0.10 m. The frequency of the waves ν is the same. If the speed of the sound waves in the copper = V', then
V' = ν𝜆' and V = ν𝜆
→V'/V = 𝜆'/𝜆
→V' = V𝜆'/𝜆 = 340*1.0/0.10 m/s =3400 m/s
55. A Kundt's tube apparatus has a steel rod of length 1.0 m clamped at the center. It is vibrated in its fundamental mode at a frequency of 2600 Hz. The lycopodium powder dispersed in the tube collects into heaps separated by 6.5 cm. Calculate the speed of sound in steel and in the air.
ANSWER: Given that the frequency, ν = 2600 Hz and length of the rod, L= 1.0 m. The steel rod acts as an open organ pipe in which frequency of the fundamental mode of vibration =V'/2L, where V' is the speed of the sound in the steel rod.
→V'/2L =ν
→V' =2νL = 2*2600*1.0 m/s =5200 m/s
The separation of heaps in the air, l = 6.5 cm =0.065 m. Hence the wavelength of sound in air, 𝜆 = 2l =2*0.065 m =0.13 m.
The speed of sound in air, V = ν𝜆 =2600*0.13 m/s
=338 m/s
56. A source of sound with adjustable frequency produces 2 beats per second with a tuning fork when its frequency is either 476 Hz or 480 Hz. What is the frequency of the tuning fork?
ANSWER: If the frequency of the tuning fork = ν, then either ν - 476 = 2 or 480 - ν = 2. Equating them,
ν - 476 = 480 - ν
→2ν = 956
→ν = 478 Hz
57. A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first one is now loaded with a little wax and the beat frequency is found to increase to 6 per second. What was the original frequency of the tuning fork?
ANSWER: With the loading of wax the frequency of a tuning fork decreases. Since the beat frequency is the difference of the two frequencies, the frequency of the first tuning fork is less than the second because after wax loading the first the beat frequency increases (i.e. the difference of the two frequencies increases). Thus the original frequency of the tuning fork = 256 - 4 Hz = 252 Hz.
58. Calculate the frequency of beats produced in air when two sources of sound are activated, one emitting a wavelength, of 32 cm and the other 32.2 cm. The speed of sound in air is 350 m/s.
ANSWER: 𝜆 = 32 cm = 0.32 m, ν = V/𝜆 =350/0.32 Hz = 1094 Hz.
𝜆' = 32.2 cm = 0.322 m, ν' = V/𝜆' =350/0.322 =1087 Hz.
The frequency of beats =|ν - ν'| =|1094 - 1087|
= 7 per second.
59. A tuning fork of unknown frequency makes 5 beats per second with another tuning fork which can cause a closed organ pipe of length 40 cm to vibrate in its fundamental mode. The beat frequency decreases when the first tuning fork is slightly loaded with wax. Find its original frequency. The speed of sound in air is 320 m/s.
ANSWER: Since with the wax loading of the first tuning-fork decreases the beat frequency, the difference of two frequencies decreases. It means when the frequency of the first tuning fork is reduced the difference is reduced. It can happen if the frequency of the first tuning fork is greater than the second.
Let the frequency of the second tuning fork = ν. This frequency causes a closed organ pipe of 40 cm ( = 0.40 m) to vibrate in its fundamental mode, hence
ν = V/4L =320/(4*0.40) =80/0.40 Hz = 200 Hz.
Beat frequency = 5 per second, hence the frequency of the first tuning-fork = 200 + 5 Hz = 205 Hz.
60. A piano wire A vibrates at a fundamental frequency of 600 Hz. A second identical wire B produces 6 beats per second with it when the tension in A is slightly increased. Find the ratio of the tension in A to the tension in B.
ANSWER: Let the tension in A = F and in B = F'. The fundamental frequency of the vibration in the wire B
(1/2L)√(F'/µ) = 600
and in A, (1/2L)√(F/µ) = ν
Hence √(F/F') = ν/600
→F/F' = ν²/600²
Since the tension in A is more than the tension in B, thus ν > 600. The beat produced = 6 beats per second. Hence ν = 600 + 6 = 606 Hz.
→F/F' = 606²/600² =(606/600)² =(1.01)² =1.02
ANSWER: Let the tension in the wire = F. The length of the wire, l = 30 cm = 0.30 m. Mass of the wire, m =10.0 g = 0.010 kg.
Mass per unit length, µ = m/l =0.010/0.30 kg/m =1/30 kg/m.
The fundamental frequency of the wire,
ν = (1/2l)√(F/µ) =(1/2*0.30)√(30 F)
→ν = (5/3)√(30 F)
 |
| Diagram for Q-51 |
This same frequency sets the air column in the closed organ pipe (L = 0.50 m) into the fundamental mode which is V/4L. Equating,
(5/3)√(30 F) = V/4L = 340/4*0.50 =170
→√(30 F) =3*170/5 =3*34 =102
→30 F = 102²
→F ≈ 347 N
52. Show that if the room temperature changes by a small amount from T to T+ΔT, the fundamental frequency of an organ pipe changes from ν to ν+Δν, where
Δν/ν = ½(ΔT/T).
ANSWER: If the length of an open or closed organ pipe is kept constant, the fundamental frequency varies proportionally to the speed of the sound in air. But the speed of sound varies proportionally to the square root of the absolute temperature.
ν ∝ V and V ∝ √T
Hence ν ∝ √T
→ν = K√T, where K is a constant of proportionality.
Let there be a small change in temperature by ΔT, due to it the change in the frequency is Δν (Given).
So now ν+Δν = K√(T+ΔT). Hence,
ν+Δν/ν = √{(T+ΔT)/T}
→1+Δν/ν = √(1+ΔT/T)
→1+Δν/ν = 1+½(ΔT/T)
{Expanding as a binomial theorem in the form of √(1+x) and neglecting (ΔT/T)² and the higher powers}
→Δν/ν = ½ (ΔT/T)
Hence proved.
53. The fundamental frequency of a closed pipe is 293 Hz when the air in it is at a temperature of 20°C. What will be the fundamental frequency when the temperature changes to 22°C?
ANSWER: The fundamental frequency of the closed organ pipe at 20°C = ν₀ and at 22°C = ν₁. Since the frequency is proportional to the speed of sound V and the speed of sound is proportional to the square root of the temperature, hence
ν₀/ν₁ =√(T₀/T₁)
Given ν₀ = 293 Hz, T₀ = 273+20 =293 K, T₁ = 273+22 =295 K.
→293/ν₁ = √(293/295)
→ν₁ = 293*√(295/293) =√(295*293) =294 Hz
54. A Kundt's tube apparatus has a copper rod of length 1.0 m clamped at 25 cm from one of the ends. The tube contains air in which the speed of sound is 340 m/s. The powder collects in heaps separated by a distance of 5.0 cm. Find the speed of sound waves in copper.
ANSWER: Given, V = 340 m/s and Δl =5.0 cm =0.05 m. The rod length clamped, L =25 cm =0.25 m. The clamped point of the rod becomes a pressure antinode and the end a pressure node. Hence the wavelength of the wave in the rod 𝜆' = 4*0.25 m = 1.0 m. The wavelength of the sound wave in air 𝜆 =2*0.05 m = 0.10 m. The frequency of the waves ν is the same. If the speed of the sound waves in the copper = V', then
V' = ν𝜆' and V = ν𝜆
→V'/V = 𝜆'/𝜆
→V' = V𝜆'/𝜆 = 340*1.0/0.10 m/s =3400 m/s
55. A Kundt's tube apparatus has a steel rod of length 1.0 m clamped at the center. It is vibrated in its fundamental mode at a frequency of 2600 Hz. The lycopodium powder dispersed in the tube collects into heaps separated by 6.5 cm. Calculate the speed of sound in steel and in the air.
ANSWER: Given that the frequency, ν = 2600 Hz and length of the rod, L= 1.0 m. The steel rod acts as an open organ pipe in which frequency of the fundamental mode of vibration =V'/2L, where V' is the speed of the sound in the steel rod.
→V'/2L =ν
→V' =2νL = 2*2600*1.0 m/s =5200 m/s
The separation of heaps in the air, l = 6.5 cm =0.065 m. Hence the wavelength of sound in air, 𝜆 = 2l =2*0.065 m =0.13 m.
The speed of sound in air, V = ν𝜆 =2600*0.13 m/s
=338 m/s
56. A source of sound with adjustable frequency produces 2 beats per second with a tuning fork when its frequency is either 476 Hz or 480 Hz. What is the frequency of the tuning fork?
ANSWER: If the frequency of the tuning fork = ν, then either ν - 476 = 2 or 480 - ν = 2. Equating them,
ν - 476 = 480 - ν
→2ν = 956
→ν = 478 Hz
57. A tuning fork produces 4 beats per second with another tuning fork of frequency 256 Hz. The first one is now loaded with a little wax and the beat frequency is found to increase to 6 per second. What was the original frequency of the tuning fork?
ANSWER: With the loading of wax the frequency of a tuning fork decreases. Since the beat frequency is the difference of the two frequencies, the frequency of the first tuning fork is less than the second because after wax loading the first the beat frequency increases (i.e. the difference of the two frequencies increases). Thus the original frequency of the tuning fork = 256 - 4 Hz = 252 Hz.
58. Calculate the frequency of beats produced in air when two sources of sound are activated, one emitting a wavelength, of 32 cm and the other 32.2 cm. The speed of sound in air is 350 m/s.
ANSWER: 𝜆 = 32 cm = 0.32 m, ν = V/𝜆 =350/0.32 Hz = 1094 Hz.
𝜆' = 32.2 cm = 0.322 m, ν' = V/𝜆' =350/0.322 =1087 Hz.
The frequency of beats =|ν - ν'| =|1094 - 1087|
= 7 per second.
59. A tuning fork of unknown frequency makes 5 beats per second with another tuning fork which can cause a closed organ pipe of length 40 cm to vibrate in its fundamental mode. The beat frequency decreases when the first tuning fork is slightly loaded with wax. Find its original frequency. The speed of sound in air is 320 m/s.
ANSWER: Since with the wax loading of the first tuning-fork decreases the beat frequency, the difference of two frequencies decreases. It means when the frequency of the first tuning fork is reduced the difference is reduced. It can happen if the frequency of the first tuning fork is greater than the second.
Let the frequency of the second tuning fork = ν. This frequency causes a closed organ pipe of 40 cm ( = 0.40 m) to vibrate in its fundamental mode, hence
ν = V/4L =320/(4*0.40) =80/0.40 Hz = 200 Hz.
Beat frequency = 5 per second, hence the frequency of the first tuning-fork = 200 + 5 Hz = 205 Hz.
60. A piano wire A vibrates at a fundamental frequency of 600 Hz. A second identical wire B produces 6 beats per second with it when the tension in A is slightly increased. Find the ratio of the tension in A to the tension in B.
ANSWER: Let the tension in A = F and in B = F'. The fundamental frequency of the vibration in the wire B
(1/2L)√(F'/µ) = 600
and in A, (1/2L)√(F/µ) = ν
Hence √(F/F') = ν/600
→F/F' = ν²/600²
Since the tension in A is more than the tension in B, thus ν > 600. The beat produced = 6 beats per second. Hence ν = 600 + 6 = 606 Hz.
→F/F' = 606²/600² =(606/600)² =(1.01)² =1.02
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
CHAPTER- 15 - Wave Motion and Waves on a String
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q1 TO Q10
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
EXERCISES -Q 31 TO 39
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 16 - Sound Waves
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES -Q-21 TO Q-30
EXERCISES -Q-31 TO Q-40
EXERCISES -Q-41 TO Q-50
CHAPTER- 15 - Wave Motion and Waves on a String
EXERCISES - Q-1 TO Q-10
EXERCISES - Q-11 TO Q-20
EXERCISES - Q-21 TO Q-30
EXERCISES - Q-31 TO Q-40
EXERCISES - Q-41 TO Q-50
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
EXERCISES- Q11 TO Q20
EXERCISES- Q21 TO Q30
EXERCISES- Q31 TO Q40
EXERCISES- Q41 TO Q50
EXERCISES- Q51 TO Q58 (2-Extra Questions)
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
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CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 6 - Friction
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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - The Forces
The Forces-
"Questions for short Answers"
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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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