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SOUND WAVES
EXERCISES- Q-1 to Q-10
1. A steel tube of length 1.00 m is struck at one end. A person with his ear close to the other end hears the sound of the blow twice, one traveling through the body of the tube and other through the air in the tube. Find the time gap between the two hearings. Use the table in the text for the speed of sound in various substances.
ANSWER: The speed of sound in steel, V = 5200 m/s
The speed of sound in air, V' = 332 m/s
The distance from the source to the observer, D = 1.00 m
The time by sound wave in the steel t = D/V =1.00/5200 s =1x1000/5200 ms =5/26 ms =0.192 ms.
The time by sound wave in the air in the tube t' = D/V' =1.00/332 s =1x1000/332 ms =3.012 ms
Hence the time gap between the two hearings =t' - t
=3.012 -0.192 ms =2.82 ms
2. At a prayer meeting, the disciples sing JAI-RAM JAI-RAM. The sound amplified by a loudspeaker comes back after reflection from a building at a distance of 80 m from the meeting. What maximum time interval can be kept between JAI-RAM and the next JAI-RAM so that the echo does not disturb a listener sitting in the meeting. The speed of sound in air is 320 m/s.
ANSWER: The maximum time interval will be the time of speaking JAI-RAM and the echo coming back after reflection. The sound waves will travel a distance 2 x 80 m =160 m. The speed of the sound V =320 m/s.
Hence the required time = 160/320 s =0.5 s
3. A man stands before a large wall at a distance of 50.0 m and claps his hands at regular intervals. Initially, the interval is large. He gradually reduces the interval and fixes it at a value when the echo of a clap merges with the next clap. If he has to clap 10 times during every 3 seconds, find the velocity of sound in air.
ANSWER: When the echo comes back traveling a distance 2 x 50 m = 100 m, just then the next clap is executed. The time between the consecutive claps
= 3/10 s =0.30 s. If the velocity of the sound be V, then
V = 100/0.3 m/s = 333 m/s
4. A person can hear sound waves in the frequency range 20 Hz to 20 kHz. Find the minimum and maximum wavelengths of sound that is audible to the person. The speed of sound is 360 m/s.
ANSWER: The speed of sound V = 360 m/s.
The wavelength 𝛌 = V/ν, where ν - frequency.
The wavelength corresponding to the frequency 20 kHz =360/(20*1000) m =0.018 m =18 mm
The wavelength corresponding to the frequency 20 Hz =360/20 m =18 m.
Hence the minimum audible wavelength =18 mm
and maximum audible wavelength = 18 m.
5. Find the minimum and maximum wavelengths of sound in water that is in the audible range (20 - 20000 Hz) for an average human ear. The speed of sound in water = 1450 m/s.
ANSWER: The minimum wavelength will be corresponding to the maximum audible frequency and vice versa because the wavelength is inversely proportional to the frequency the velocity being constant in a medium. 𝛌 = V/ν.
Here V = 1450 m/s, so the minimum audible wavelength of the sound in the water is corresponding to the frequency 20000 Hz
= 1450/20000 m =0.0725 m = 7.25 cm
The maximum audible wavelength of the sound in the water is corresponding to the frequency 20 Hz
= 1450/20 m =72.5 m
6. Sound waves from a loudspeaker spread nearly uniformly in all directions if the wavelength of the sound is much larger than the diameter of the loudspeaker.
(a) Calculate the frequency for which the wavelength of sound in air is ten times the diameter of the speaker if the diameter is 20 cm.
(b) The sound is essentially transmitted in the forward direction if the wavelength is much shorter than the diameter of the speaker. Calculate the frequency at which the wavelength of the sound is one-tenth of the diameter of the speaker described above. Take the speed of the sound to be 340 m/s.
ANSWER: (a) The speed of sound V = 340 m/s
The diameter of the speaker d = 20 cm
The wavelength of the sound wave =10*20 cm =200 cm =2.0 m
Hence the frequency of the sound wave ν =V/𝛌
=340/2 Hz =170 Hz
(b) In the second case the wavelength of the sound =20/10 cm =2 cm =0.02 m
Hence the frequency of the sound waves ν' =V/𝛌 = 340/0.02 Hz =17000 Hz =17 kHz
7. Ultrasonic waves of frequency 4.5 MHz are used to detect a tumour in soft tissues. The speed of sound in tissue is 1.5 km/s and that in the air 340 m/s. Find the wavelength of this ultrasonic wave in the air and in tissue.
ANSWER: The wavelength 𝛌 = V/ν
where V - the speed of sound in a medium and ν - frequency of the sound waves.
Here the frequency ν = 4.5 MHz =4.5 x 10⁶ Hz
In air, V =340 m/s
Hence the wavelength in the air =340/(4.5 x 10⁶) m
≈7.6 x 10⁻⁵ m,
In the tissue, V = 1.5 km/s =1500 m/s
Hence the wavelength in the tissue
=1500/(4.5 x 10⁶) m =3.3 x 10⁻⁴ m
8. The equation of a traveling sound wave is
y = 6.0 sin(600 t - 1.8 x)
where y is measured in 10⁻⁵ m, t in second and x in meter.
(a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave.
(b) Find the ratio of the velocity amplitude of the particle to the wave speed.
ANSWER: (a) Comparing the given equation with the general equation y =A sin(⍵t-kx), A =6.0x10⁻⁵ m and,
k =1.8 m⁻¹
→2π/𝛌 = 1.8 m⁻¹
→𝛌 = 2π/1.8 m =3.49 m
Hence A/𝛌 =6.0x10⁻⁵/3.49 =1.7x10⁻⁵
(b) The equation for the velocity of the particle can be found out by differentiating y with respect to the time t.
V =dy/dt = A⍵ cos(⍵t-kx)
The given equation will be
V =(6.0*600) cos(600 t - 1.8 x)
Hence the velocity amplitude V¹ =6.0*600x10⁻⁵ m/s
=0.036 m/s
if V' is the speed of the wave then k =⍵/V'
→V' =⍵/k =600/1.8 m/s =333 m/s
Hence the ratio V¹/V' = 0.036/333 =1.1 x 10⁻⁴
9. A sound wave of frequency 100 Hz is traveling in air. The speed of sound in air is 350 m/s.
(a) By how much is the phase changed at a given point in 2.5 ms?
(b) What is the phase difference at a given instant between two points separated by a distance 10.0 cm along the direction of propagation?
ANSWER: (a) The wavelength of the sound wave
𝛌 =speed/frequency =350/100 m =3.5 m
Time t = 2.5 ms =2.5/1000 s =0.0025 s
The small distance traveled in this time Δx =Vt
=350*0.0025 m =0.875 m
Hence the phase difference
ⲫ=2π*Δx/𝛌 =2π*0.875/3.5
=0.5*π =π/2
(b) Here Δx = 10.0 cm =0.10 m
Hence the phase difference ⲫ=2π*Δx/𝛌
=2π*0.10/3.5 =2π/35
10. Two point sources of sound are kept at a separation of 10 cm. They vibrate in phase to produce waves of wavelength 5.0 cm. What would be the phase difference between the two waves arriving at a point 20 cm from one source (a) on the line joining the sources and (b) on the perpendicular bisector of the line joining the sources?
ANSWER: (a) This point will be 20 cm from one source and 20+10 =30 cm from the other. Hence the path difference the two waves, Δx = 30 - 20 =10 cm.
Hence the phase difference ⲫ =2π*Δx/𝛌
=2π*10/5 =2*2π
Since the phase difference is a multiple of 2π, hence they are in the same phase and the phase difference is zero.
Diagram for Q-10
(b) Since each point on the perpendicular bisector of a line-segment is equidistant from both ends of the line segment, the sound from each source in the given problem travel the same distance to reach a point on the bisector. Hence the path difference Δx=0.
Hence the phase difference ⲫ =2π*Δx/𝛌
=2π*0/𝛌 = 0 (zero)
1. A steel tube of length 1.00 m is struck at one end. A person with his ear close to the other end hears the sound of the blow twice, one traveling through the body of the tube and other through the air in the tube. Find the time gap between the two hearings. Use the table in the text for the speed of sound in various substances.
ANSWER: The speed of sound in steel, V = 5200 m/s
The speed of sound in air, V' = 332 m/s
The distance from the source to the observer, D = 1.00 m
The time by sound wave in the steel t = D/V =1.00/5200 s =1x1000/5200 ms =5/26 ms =0.192 ms.
The time by sound wave in the air in the tube t' = D/V' =1.00/332 s =1x1000/332 ms =3.012 ms
Hence the time gap between the two hearings =t' - t
=3.012 -0.192 ms =2.82 ms
2. At a prayer meeting, the disciples sing JAI-RAM JAI-RAM. The sound amplified by a loudspeaker comes back after reflection from a building at a distance of 80 m from the meeting. What maximum time interval can be kept between JAI-RAM and the next JAI-RAM so that the echo does not disturb a listener sitting in the meeting. The speed of sound in air is 320 m/s.
ANSWER: The maximum time interval will be the time of speaking JAI-RAM and the echo coming back after reflection. The sound waves will travel a distance 2 x 80 m =160 m. The speed of the sound V =320 m/s.
Hence the required time = 160/320 s =0.5 s
3. A man stands before a large wall at a distance of 50.0 m and claps his hands at regular intervals. Initially, the interval is large. He gradually reduces the interval and fixes it at a value when the echo of a clap merges with the next clap. If he has to clap 10 times during every 3 seconds, find the velocity of sound in air.
ANSWER: When the echo comes back traveling a distance 2 x 50 m = 100 m, just then the next clap is executed. The time between the consecutive claps
= 3/10 s =0.30 s. If the velocity of the sound be V, then
V = 100/0.3 m/s = 333 m/s
4. A person can hear sound waves in the frequency range 20 Hz to 20 kHz. Find the minimum and maximum wavelengths of sound that is audible to the person. The speed of sound is 360 m/s.
ANSWER: The speed of sound V = 360 m/s.
The wavelength 𝛌 = V/ν, where ν - frequency.
The wavelength corresponding to the frequency 20 kHz =360/(20*1000) m =0.018 m =18 mm
The wavelength corresponding to the frequency 20 Hz =360/20 m =18 m.
Hence the minimum audible wavelength =18 mm
and maximum audible wavelength = 18 m.
5. Find the minimum and maximum wavelengths of sound in water that is in the audible range (20 - 20000 Hz) for an average human ear. The speed of sound in water = 1450 m/s.
ANSWER: The minimum wavelength will be corresponding to the maximum audible frequency and vice versa because the wavelength is inversely proportional to the frequency the velocity being constant in a medium. 𝛌 = V/ν.
Here V = 1450 m/s, so the minimum audible wavelength of the sound in the water is corresponding to the frequency 20000 Hz
= 1450/20000 m =0.0725 m = 7.25 cm
The maximum audible wavelength of the sound in the water is corresponding to the frequency 20 Hz
= 1450/20 m =72.5 m
6. Sound waves from a loudspeaker spread nearly uniformly in all directions if the wavelength of the sound is much larger than the diameter of the loudspeaker.
(a) Calculate the frequency for which the wavelength of sound in air is ten times the diameter of the speaker if the diameter is 20 cm.
(b) The sound is essentially transmitted in the forward direction if the wavelength is much shorter than the diameter of the speaker. Calculate the frequency at which the wavelength of the sound is one-tenth of the diameter of the speaker described above. Take the speed of the sound to be 340 m/s.
ANSWER: (a) The speed of sound V = 340 m/s
The diameter of the speaker d = 20 cm
The wavelength of the sound wave =10*20 cm =200 cm =2.0 m
Hence the frequency of the sound wave ν =V/𝛌
=340/2 Hz =170 Hz
(b) In the second case the wavelength of the sound =20/10 cm =2 cm =0.02 m
Hence the frequency of the sound waves ν' =V/𝛌 = 340/0.02 Hz =17000 Hz =17 kHz
7. Ultrasonic waves of frequency 4.5 MHz are used to detect a tumour in soft tissues. The speed of sound in tissue is 1.5 km/s and that in the air 340 m/s. Find the wavelength of this ultrasonic wave in the air and in tissue.
ANSWER: The wavelength 𝛌 = V/ν
where V - the speed of sound in a medium and ν - frequency of the sound waves.
Here the frequency ν = 4.5 MHz =4.5 x 10⁶ Hz
In air, V =340 m/s
Hence the wavelength in the air =340/(4.5 x 10⁶) m
≈7.6 x 10⁻⁵ m,
In the tissue, V = 1.5 km/s =1500 m/s
Hence the wavelength in the tissue
=1500/(4.5 x 10⁶) m =3.3 x 10⁻⁴ m
8. The equation of a traveling sound wave is
y = 6.0 sin(600 t - 1.8 x)
where y is measured in 10⁻⁵ m, t in second and x in meter.
(a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave.
(b) Find the ratio of the velocity amplitude of the particle to the wave speed.
ANSWER: (a) Comparing the given equation with the general equation y =A sin(⍵t-kx), A =6.0x10⁻⁵ m and,
k =1.8 m⁻¹
→2π/𝛌 = 1.8 m⁻¹
→𝛌 = 2π/1.8 m =3.49 m
Hence A/𝛌 =6.0x10⁻⁵/3.49 =1.7x10⁻⁵
(b) The equation for the velocity of the particle can be found out by differentiating y with respect to the time t.
V =dy/dt = A⍵ cos(⍵t-kx)
The given equation will be
V =(6.0*600) cos(600 t - 1.8 x)
Hence the velocity amplitude V¹ =6.0*600x10⁻⁵ m/s
=0.036 m/s
if V' is the speed of the wave then k =⍵/V'
→V' =⍵/k =600/1.8 m/s =333 m/s
Hence the ratio V¹/V' = 0.036/333 =1.1 x 10⁻⁴
9. A sound wave of frequency 100 Hz is traveling in air. The speed of sound in air is 350 m/s.
(a) By how much is the phase changed at a given point in 2.5 ms?
(b) What is the phase difference at a given instant between two points separated by a distance 10.0 cm along the direction of propagation?
ANSWER: (a) The wavelength of the sound wave
𝛌 =speed/frequency =350/100 m =3.5 m
Time t = 2.5 ms =2.5/1000 s =0.0025 s
The small distance traveled in this time Δx =Vt
=350*0.0025 m =0.875 m
Hence the phase difference
ⲫ=2π*Δx/𝛌 =2π*0.875/3.5
=0.5*π =π/2
(b) Here Δx = 10.0 cm =0.10 m
Hence the phase difference ⲫ=2π*Δx/𝛌
=2π*0.10/3.5 =2π/35
10. Two point sources of sound are kept at a separation of 10 cm. They vibrate in phase to produce waves of wavelength 5.0 cm. What would be the phase difference between the two waves arriving at a point 20 cm from one source (a) on the line joining the sources and (b) on the perpendicular bisector of the line joining the sources?
ANSWER: (a) This point will be 20 cm from one source and 20+10 =30 cm from the other. Hence the path difference the two waves, Δx = 30 - 20 =10 cm.
Hence the phase difference ⲫ =2π*Δx/𝛌
=2π*10/5 =2*2π
Since the phase difference is a multiple of 2π, hence they are in the same phase and the phase difference is zero.
(b) Since each point on the perpendicular bisector of a line-segment is equidistant from both ends of the line segment, the sound from each source in the given problem travel the same distance to reach a point on the bisector. Hence the path difference Δx=0.
Hence the phase difference ⲫ =2π*Δx/𝛌
=2π*0/𝛌 = 0 (zero)
ANSWER: The speed of sound in steel, V = 5200 m/s
The speed of sound in air, V' = 332 m/s
The distance from the source to the observer, D = 1.00 m
The time by sound wave in the steel t = D/V =1.00/5200 s =1x1000/5200 ms =5/26 ms =0.192 ms.
The time by sound wave in the air in the tube t' = D/V' =1.00/332 s =1x1000/332 ms =3.012 ms
Hence the time gap between the two hearings =t' - t
=3.012 -0.192 ms =2.82 ms
2. At a prayer meeting, the disciples sing JAI-RAM JAI-RAM. The sound amplified by a loudspeaker comes back after reflection from a building at a distance of 80 m from the meeting. What maximum time interval can be kept between JAI-RAM and the next JAI-RAM so that the echo does not disturb a listener sitting in the meeting. The speed of sound in air is 320 m/s.
ANSWER: The maximum time interval will be the time of speaking JAI-RAM and the echo coming back after reflection. The sound waves will travel a distance 2 x 80 m =160 m. The speed of the sound V =320 m/s.
Hence the required time = 160/320 s =0.5 s
3. A man stands before a large wall at a distance of 50.0 m and claps his hands at regular intervals. Initially, the interval is large. He gradually reduces the interval and fixes it at a value when the echo of a clap merges with the next clap. If he has to clap 10 times during every 3 seconds, find the velocity of sound in air.
ANSWER: When the echo comes back traveling a distance 2 x 50 m = 100 m, just then the next clap is executed. The time between the consecutive claps
= 3/10 s =0.30 s. If the velocity of the sound be V, then
V = 100/0.3 m/s = 333 m/s
4. A person can hear sound waves in the frequency range 20 Hz to 20 kHz. Find the minimum and maximum wavelengths of sound that is audible to the person. The speed of sound is 360 m/s.
ANSWER: The speed of sound V = 360 m/s.
The wavelength 𝛌 = V/ν, where ν - frequency.
The wavelength corresponding to the frequency 20 kHz =360/(20*1000) m =0.018 m =18 mm
The wavelength corresponding to the frequency 20 Hz =360/20 m =18 m.
Hence the minimum audible wavelength =18 mm
and maximum audible wavelength = 18 m.
5. Find the minimum and maximum wavelengths of sound in water that is in the audible range (20 - 20000 Hz) for an average human ear. The speed of sound in water = 1450 m/s.
ANSWER: The minimum wavelength will be corresponding to the maximum audible frequency and vice versa because the wavelength is inversely proportional to the frequency the velocity being constant in a medium. 𝛌 = V/ν.
Here V = 1450 m/s, so the minimum audible wavelength of the sound in the water is corresponding to the frequency 20000 Hz
= 1450/20000 m =0.0725 m = 7.25 cm
The maximum audible wavelength of the sound in the water is corresponding to the frequency 20 Hz
= 1450/20 m =72.5 m
6. Sound waves from a loudspeaker spread nearly uniformly in all directions if the wavelength of the sound is much larger than the diameter of the loudspeaker.
(a) Calculate the frequency for which the wavelength of sound in air is ten times the diameter of the speaker if the diameter is 20 cm.
(b) The sound is essentially transmitted in the forward direction if the wavelength is much shorter than the diameter of the speaker. Calculate the frequency at which the wavelength of the sound is one-tenth of the diameter of the speaker described above. Take the speed of the sound to be 340 m/s.
ANSWER: (a) The speed of sound V = 340 m/s
The diameter of the speaker d = 20 cm
The wavelength of the sound wave =10*20 cm =200 cm =2.0 m
Hence the frequency of the sound wave ν =V/𝛌
=340/2 Hz =170 Hz
(b) In the second case the wavelength of the sound =20/10 cm =2 cm =0.02 m
Hence the frequency of the sound waves ν' =V/𝛌 = 340/0.02 Hz =17000 Hz =17 kHz
7. Ultrasonic waves of frequency 4.5 MHz are used to detect a tumour in soft tissues. The speed of sound in tissue is 1.5 km/s and that in the air 340 m/s. Find the wavelength of this ultrasonic wave in the air and in tissue.
ANSWER: The wavelength 𝛌 = V/ν
where V - the speed of sound in a medium and ν - frequency of the sound waves.
Here the frequency ν = 4.5 MHz =4.5 x 10⁶ Hz
In air, V =340 m/s
Hence the wavelength in the air =340/(4.5 x 10⁶) m
≈7.6 x 10⁻⁵ m,
In the tissue, V = 1.5 km/s =1500 m/s
Hence the wavelength in the tissue
=1500/(4.5 x 10⁶) m =3.3 x 10⁻⁴ m
8. The equation of a traveling sound wave is
y = 6.0 sin(600 t - 1.8 x)
where y is measured in 10⁻⁵ m, t in second and x in meter.
(a) Find the ratio of the displacement amplitude of the particles to the wavelength of the wave.
(b) Find the ratio of the velocity amplitude of the particle to the wave speed.
ANSWER: (a) Comparing the given equation with the general equation y =A sin(⍵t-kx), A =6.0x10⁻⁵ m and,
k =1.8 m⁻¹
→2π/𝛌 = 1.8 m⁻¹
→𝛌 = 2π/1.8 m =3.49 m
Hence A/𝛌 =6.0x10⁻⁵/3.49 =1.7x10⁻⁵
(b) The equation for the velocity of the particle can be found out by differentiating y with respect to the time t.
V =dy/dt = A⍵ cos(⍵t-kx)
The given equation will be
V =(6.0*600) cos(600 t - 1.8 x)
Hence the velocity amplitude V¹ =6.0*600x10⁻⁵ m/s
=0.036 m/s
if V' is the speed of the wave then k =⍵/V'
→V' =⍵/k =600/1.8 m/s =333 m/s
Hence the ratio V¹/V' = 0.036/333 =1.1 x 10⁻⁴
9. A sound wave of frequency 100 Hz is traveling in air. The speed of sound in air is 350 m/s.
(a) By how much is the phase changed at a given point in 2.5 ms?
(b) What is the phase difference at a given instant between two points separated by a distance 10.0 cm along the direction of propagation?
ANSWER: (a) The wavelength of the sound wave
𝛌 =speed/frequency =350/100 m =3.5 m
Time t = 2.5 ms =2.5/1000 s =0.0025 s
The small distance traveled in this time Δx =Vt
=350*0.0025 m =0.875 m
Hence the phase difference
ⲫ=2π*Δx/𝛌 =2π*0.875/3.5
=0.5*π =π/2
(b) Here Δx = 10.0 cm =0.10 m
Hence the phase difference ⲫ=2π*Δx/𝛌
=2π*0.10/3.5 =2π/35
10. Two point sources of sound are kept at a separation of 10 cm. They vibrate in phase to produce waves of wavelength 5.0 cm. What would be the phase difference between the two waves arriving at a point 20 cm from one source (a) on the line joining the sources and (b) on the perpendicular bisector of the line joining the sources?
ANSWER: (a) This point will be 20 cm from one source and 20+10 =30 cm from the other. Hence the path difference the two waves, Δx = 30 - 20 =10 cm.
Hence the phase difference ⲫ =2π*Δx/𝛌
=2π*10/5 =2*2π
Since the phase difference is a multiple of 2π, hence they are in the same phase and the phase difference is zero.
 |
| Diagram for Q-10 |
(b) Since each point on the perpendicular bisector of a line-segment is equidistant from both ends of the line segment, the sound from each source in the given problem travel the same distance to reach a point on the bisector. Hence the path difference Δx=0.
Hence the phase difference ⲫ =2π*Δx/𝛌
=2π*0/𝛌 = 0 (zero)
===<<<O>>>===
Links to the Chapters
===<<<O>>>===
Links to the Chapters
CHAPTER- 16 - Sound Waves
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
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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Click here for → Exercise(43-54)
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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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For more practice on problems on friction solve these- "New Questions on Friction".
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CHAPTER- 5 - Newton's Laws of Motion
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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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