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SOUND WAVES
EXERCISES- Q-41 to Q-50
41. In a resonance column experiment, a tuning fork of frequency 400 Hz is used. The first resonance is observed when the air column has a length of 20 cm and the second resonance is observed when the air column has a length of 62.0 cm. (a) Find the speed of sound in air. (b) How much distance above the open end does the pressure node form?
ANSWER: (a) Given that frequency ν = 400 Hz.
The length of air column for the fundamental frequency of resonance =L₁ = 20.0 cm.
The length of air column for the first overtone frequency of resonance =L₂ = 62.0 cm.
If the pressure node forms at d above the open end,
Hence (L₂+d) - (L₁+d) = 𝜆/2 {See the diagram below}
Diagram for Q - 41
→62.0 - 20.0 cm =𝜆/2
→𝜆 = 2*42.0 cm =84.0 cm =0.84 m
Hence the speed of sound in air V = ν𝜆 =400*0.84 m/s =336 m/s
(b) Since L₁+d =𝜆/4 {From figure for fundamental vibration}
Hence, d = 𝜆/4-L₁ =84/4 -20 =21 -20 cm = 1.0 cm.
42. The first overtone frequency of a closed organ pipe P₁ is equal to the fundamental frequency of an open organ pipe P₂. If the length of pipe P₁ is 30 cm, what will be the length of P₂?
ANSWER: The first overtone frequency of the closed organ pipe ν =3V/4L₁
The fundamental frequency of an open organ pipe
ν =V/2L₂
Equating,
3V/4L₁ = V/2L₂
→L₂ = 2L₁/3 {Given L₁ = 30 cm}
→L₂ = 2*30/3 cm =20 cm.
43. A copper rod of length 1.0 m is clamped at its middle point. Find the frequencies between 20 Hz - 20000 Hz at which standing longitudinal waves can be set up in the rod. The speed of sound in copper is 3.8 km/s.
ANSWER: For the longitudinal sound waves, the rod will act as an open organ pipe. The frequencies for the standing longitudinal waves will be
ν = nV/2L.
→ν = n*3800/(2*1.0) Hz, where n = 1, 2, 3, ....
{Given the speed of sound in copper, V = 3.8 km/s =3800 m/s.}
→ν = 1900n Hz
Thus the minimum frequency for is for n = 1, i.e. =1900 Hz =1.9 kHz.
Maximum frequency is for n = 10, i.e. =19000 Hz =19 kHz. The value for n=11 will give frequency =20.9 kHz which is more than 20000 Hz. Hence the standing longitudinal wave frequencies are
νₙ = 1.9n kHz where n = 1, 2, 3, 4, ......, 10.
44. Find the greatest length of an organ pipe open at both ends that will have its fundamental frequency in the normal hearing range (20-20000 Hz). The speed of sound in air = 340 m/s.
ANSWER: The fundamental frequency in an open organ pipe, ν₀ = V/2L.
For the greatest length of the pipe to have its fundamental frequency, ν₀ = 20 Hz
→V/2L = 20
→V = 40L
→L = V/40
{Given V = 340 m/s.}
→L = 340/40 m
→L = 8.5 m.
45. An open organ pipe has a length of 5 cm. (a) Find the fundamental frequency of vibration of this pipe. (b) What is the highest harmonic of such a tube that is in the audible range? The speed of sound in air is 340 m/s and the audible range is 20-20000 Hz.
ANSWER: (a) The fundamental frequency of an open organ pipe ν₀ = V/2L
Here L= 5 cm =0.05 m and V = 340 m/s.
ν₀ = 340/(2*0.05) Hz = 3400 Hz =3.4 kHz.
(b) The frequencies for higher harmonics are given as,
νₙ = nV/2L, where n = 1, 2, 3, ...
For the highest harmonic to be in the audible range
νₙ ≤ 20000 Hz
→nV/2L ≤ 20000
→n ≤ 40000L/V
→n ≤ 40000*0.05/340
→n ≤ 5.88
Since n is a whole number, n = 5.
So the highest harmonic for this tube is the 5th overtone.
46. An electrically driven loudspeaker is placed near the open end of a resonance column apparatus. The length of the air column in the tube is 80 cm. The frequency of the loudspeaker can be varied between 20 Hz -2 kHz. Find the frequencies at which the column will resonate. The speed of sound in air =320 m/s.
ANSWER: The resonance column acts as a closed organ pipe for which the resonant frequencies are given as
νₙ = (2n+1)V/4L, where n = 0, 1, 2, 3, ....
Here V = 320 m/s, L = 80 cm = 0.80 m
→νₙ = (2n+1)*320/(4*0.80) Hz
→νₙ = 100(2n+1) Hz
Putting n = 0, ν = 100 Hz > 20 Hz, Acceptable.
for n = 10, ν = 2100 Hz > 2kHz, Not acceptable.
For n = 9, ν = 1900 Hz < 2kHz, acceptable.
Hence the frequencies at which the column will resonate
= 100(2n+1) Hz where n = 0, 1, 2, 3,....., 9
47. Two successive resonant frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speed of sound in air =324 m/s.
ANSWER: The resonant frequencies in an open organ pipe is given as
νₙ = nV/2L
Let the given frequencies are for n and n+1 harmonics, thus
nV/2L = 1944 and
(n+1)V/2L = 2592
→nV/2L + V/2L =2592
→1944+V/2L = 2592
→V/2L = 2592 - 1944 = 648
→L = V/(2*648) = 324/1296 m = 0.25 m
→L = 25 cm
48. A piston is fitted in a cylindrical tube of small cross-section with the other end of the tube open. The tube resonates with a tuning fork of frequency 512 Hz. The piston is gradually pulled out of the tube and it is found that a second resonance occurs when the piston is pulled out through a distance of 32.0 cm. Calculate the speed of sound in the air of the tube.
ANSWER: The piston acts as the closed end of the organ pipe. At the piston, there will be pressure antinode while at the open end there will be a pressure node. Since in both the cases the frequency of the vibration (or the wavelength of the sound) is same = 512 Hz, the next resonance will occur when the length of the air column is increased to 𝜆/2 because one more node will be induced in the air column. From the given problem,
𝜆/2 = 32 cm = 0.32 m
→𝜆 = 0.64 m
ν = 512 Hz
Hence the speed of the sound in air V = ν𝜆
→V = 512*0.64 m/s ≈ 328 m/s.
49. A U-tube having unequal arm lengths has water in it. A tuning fork of frequency 440 Hz can set up the air in the shorter arm in its fundamental mode of vibration and the same tuning fork can set up the air in the longer arm in its first overtone vibrations. Find the length of the air columns. Neglect any end effect and assume that the speed of sound in air = 330 m/s.
ANSWER: Given that V = 330 m/s, the air columns in the U-tube are the closed organ pipes. The fundamental frequency
V/4L = 440 Hz
→L = V/(4*440) m = 330/(4*440) m
→L = 3/16 m = 0.188 m = 18.8 cm
The first overtone frequency is
3V/4L = 440 Hz
→L = 3V/(4*440) m =3*330/(4*440) m
→L = 9/16 m = 0.563 m = 56.3 cm
50. Consider the situation in figure (16-E9). The wire which has a mass of 4.00 g oscillates in its second harmonic and sets the air column in the tube into vibrations in its fundamental mode. Assuming that the speed of sound in air is 340 m/s, find the tension in the wire.
The figure for Q-50
ANSWER: The frequency of the waves in the string will be the same as the frequency of the sound wave in the air column. The air column vibrates as in a closed organ pipe and its frequency in the fundamental mode is given as ν = V/4L. Given V= 340 m/s and L = 1.0 m, hence the frequency
ν = 340/(4*1.0) = 85 Hz.
Let the tension in the string be F.
The length of the wire l = 40 cm =0.40 m.
The mass of the wire m = 4 g = 0.004 kg.
Hence the linear mass density of the wire µ = m/l
=0.004/0.40 kg/m =0.01 kg/m.
Since the string oscillates in the second harmonic, its frequency ν = (2/2l)√(F/µ)
=(1/0.40)√(F/0.01) Hz
=2.5*10√F = 25√F
But it is equal to 85 Hz, hence
25√F =85
→√F = 85/25 =3.4
→F = 11.56 N
41. In a resonance column experiment, a tuning fork of frequency 400 Hz is used. The first resonance is observed when the air column has a length of 20 cm and the second resonance is observed when the air column has a length of 62.0 cm. (a) Find the speed of sound in air. (b) How much distance above the open end does the pressure node form?
ANSWER: (a) Given that frequency ν = 400 Hz.
The length of air column for the fundamental frequency of resonance =L₁ = 20.0 cm.
The length of air column for the first overtone frequency of resonance =L₂ = 62.0 cm.
If the pressure node forms at d above the open end,
Hence (L₂+d) - (L₁+d) = 𝜆/2 {See the diagram below}
→62.0 - 20.0 cm =𝜆/2
→𝜆 = 2*42.0 cm =84.0 cm =0.84 m
Hence the speed of sound in air V = ν𝜆 =400*0.84 m/s =336 m/s
42. The first overtone frequency of a closed organ pipe P₁ is equal to the fundamental frequency of an open organ pipe P₂. If the length of pipe P₁ is 30 cm, what will be the length of P₂?
ANSWER: The first overtone frequency of the closed organ pipe ν =3V/4L₁
The fundamental frequency of an open organ pipe
ν =V/2L₂
Equating,
3V/4L₁ = V/2L₂
→L₂ = 2L₁/3 {Given L₁ = 30 cm}
→L₂ = 2*30/3 cm =20 cm.
43. A copper rod of length 1.0 m is clamped at its middle point. Find the frequencies between 20 Hz - 20000 Hz at which standing longitudinal waves can be set up in the rod. The speed of sound in copper is 3.8 km/s.
ANSWER: For the longitudinal sound waves, the rod will act as an open organ pipe. The frequencies for the standing longitudinal waves will be
ν = nV/2L.
→ν = n*3800/(2*1.0) Hz, where n = 1, 2, 3, ....
{Given the speed of sound in copper, V = 3.8 km/s =3800 m/s.}
→ν = 1900n Hz
Thus the minimum frequency for is for n = 1, i.e. =1900 Hz =1.9 kHz.
Maximum frequency is for n = 10, i.e. =19000 Hz =19 kHz. The value for n=11 will give frequency =20.9 kHz which is more than 20000 Hz. Hence the standing longitudinal wave frequencies are
νₙ = 1.9n kHz where n = 1, 2, 3, 4, ......, 10.
44. Find the greatest length of an organ pipe open at both ends that will have its fundamental frequency in the normal hearing range (20-20000 Hz). The speed of sound in air = 340 m/s.
ANSWER: The fundamental frequency in an open organ pipe, ν₀ = V/2L.
For the greatest length of the pipe to have its fundamental frequency, ν₀ = 20 Hz
→V/2L = 20
→V = 40L
→L = V/40
{Given V = 340 m/s.}
→L = 340/40 m
→L = 8.5 m.
45. An open organ pipe has a length of 5 cm. (a) Find the fundamental frequency of vibration of this pipe. (b) What is the highest harmonic of such a tube that is in the audible range? The speed of sound in air is 340 m/s and the audible range is 20-20000 Hz.
ANSWER: (a) The fundamental frequency of an open organ pipe ν₀ = V/2L
Here L= 5 cm =0.05 m and V = 340 m/s.
ν₀ = 340/(2*0.05) Hz = 3400 Hz =3.4 kHz.
(b) The frequencies for higher harmonics are given as,
νₙ = nV/2L, where n = 1, 2, 3, ...
For the highest harmonic to be in the audible range
νₙ ≤ 20000 Hz
→nV/2L ≤ 20000
→n ≤ 40000L/V
→n ≤ 40000*0.05/340
→n ≤ 5.88
Since n is a whole number, n = 5.
So the highest harmonic for this tube is the 5th overtone.
46. An electrically driven loudspeaker is placed near the open end of a resonance column apparatus. The length of the air column in the tube is 80 cm. The frequency of the loudspeaker can be varied between 20 Hz -2 kHz. Find the frequencies at which the column will resonate. The speed of sound in air =320 m/s.
ANSWER: The resonance column acts as a closed organ pipe for which the resonant frequencies are given as
νₙ = (2n+1)V/4L, where n = 0, 1, 2, 3, ....
Here V = 320 m/s, L = 80 cm = 0.80 m
→νₙ = (2n+1)*320/(4*0.80) Hz
→νₙ = 100(2n+1) Hz
Putting n = 0, ν = 100 Hz > 20 Hz, Acceptable.
for n = 10, ν = 2100 Hz > 2kHz, Not acceptable.
For n = 9, ν = 1900 Hz < 2kHz, acceptable.
Hence the frequencies at which the column will resonate
= 100(2n+1) Hz where n = 0, 1, 2, 3,....., 9
47. Two successive resonant frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speed of sound in air =324 m/s.
ANSWER: The resonant frequencies in an open organ pipe is given as
νₙ = nV/2L
Let the given frequencies are for n and n+1 harmonics, thus
nV/2L = 1944 and
(n+1)V/2L = 2592
→nV/2L + V/2L =2592
→1944+V/2L = 2592
→V/2L = 2592 - 1944 = 648
→L = V/(2*648) = 324/1296 m = 0.25 m
→L = 25 cm
48. A piston is fitted in a cylindrical tube of small cross-section with the other end of the tube open. The tube resonates with a tuning fork of frequency 512 Hz. The piston is gradually pulled out of the tube and it is found that a second resonance occurs when the piston is pulled out through a distance of 32.0 cm. Calculate the speed of sound in the air of the tube.
ANSWER: The piston acts as the closed end of the organ pipe. At the piston, there will be pressure antinode while at the open end there will be a pressure node. Since in both the cases the frequency of the vibration (or the wavelength of the sound) is same = 512 Hz, the next resonance will occur when the length of the air column is increased to 𝜆/2 because one more node will be induced in the air column. From the given problem,
𝜆/2 = 32 cm = 0.32 m
→𝜆 = 0.64 m
ν = 512 Hz
Hence the speed of the sound in air V = ν𝜆
→V = 512*0.64 m/s ≈ 328 m/s.
49. A U-tube having unequal arm lengths has water in it. A tuning fork of frequency 440 Hz can set up the air in the shorter arm in its fundamental mode of vibration and the same tuning fork can set up the air in the longer arm in its first overtone vibrations. Find the length of the air columns. Neglect any end effect and assume that the speed of sound in air = 330 m/s.
ANSWER: Given that V = 330 m/s, the air columns in the U-tube are the closed organ pipes. The fundamental frequency
V/4L = 440 Hz
→L = V/(4*440) m = 330/(4*440) m
→L = 3/16 m = 0.188 m = 18.8 cm
The first overtone frequency is
3V/4L = 440 Hz
→L = 3V/(4*440) m =3*330/(4*440) m
→L = 9/16 m = 0.563 m = 56.3 cm
50. Consider the situation in figure (16-E9). The wire which has a mass of 4.00 g oscillates in its second harmonic and sets the air column in the tube into vibrations in its fundamental mode. Assuming that the speed of sound in air is 340 m/s, find the tension in the wire.
ANSWER: The frequency of the waves in the string will be the same as the frequency of the sound wave in the air column. The air column vibrates as in a closed organ pipe and its frequency in the fundamental mode is given as ν = V/4L. Given V= 340 m/s and L = 1.0 m, hence the frequency
ν = 340/(4*1.0) = 85 Hz.
Let the tension in the string be F.
The length of the wire l = 40 cm =0.40 m.
The mass of the wire m = 4 g = 0.004 kg.
Hence the linear mass density of the wire µ = m/l
=0.004/0.40 kg/m =0.01 kg/m.
Since the string oscillates in the second harmonic, its frequency ν = (2/2l)√(F/µ)
=(1/0.40)√(F/0.01) Hz
=2.5*10√F = 25√F
But it is equal to 85 Hz, hence
25√F =85
→√F = 85/25 =3.4
→F = 11.56 N
ANSWER: (a) Given that frequency ν = 400 Hz.
The length of air column for the fundamental frequency of resonance =L₁ = 20.0 cm.
The length of air column for the first overtone frequency of resonance =L₂ = 62.0 cm.
If the pressure node forms at d above the open end,
Hence (L₂+d) - (L₁+d) = 𝜆/2 {See the diagram below}
 |
| Diagram for Q - 41 |
→62.0 - 20.0 cm =𝜆/2
→𝜆 = 2*42.0 cm =84.0 cm =0.84 m
Hence the speed of sound in air V = ν𝜆 =400*0.84 m/s =336 m/s
(b) Since L₁+d =𝜆/4 {From figure for fundamental vibration}
Hence, d = 𝜆/4-L₁ =84/4 -20 =21 -20 cm = 1.0 cm.
42. The first overtone frequency of a closed organ pipe P₁ is equal to the fundamental frequency of an open organ pipe P₂. If the length of pipe P₁ is 30 cm, what will be the length of P₂?
ANSWER: The first overtone frequency of the closed organ pipe ν =3V/4L₁
The fundamental frequency of an open organ pipe
ν =V/2L₂
Equating,
3V/4L₁ = V/2L₂
→L₂ = 2L₁/3 {Given L₁ = 30 cm}
→L₂ = 2*30/3 cm =20 cm.
43. A copper rod of length 1.0 m is clamped at its middle point. Find the frequencies between 20 Hz - 20000 Hz at which standing longitudinal waves can be set up in the rod. The speed of sound in copper is 3.8 km/s.
ANSWER: For the longitudinal sound waves, the rod will act as an open organ pipe. The frequencies for the standing longitudinal waves will be
ν = nV/2L.
→ν = n*3800/(2*1.0) Hz, where n = 1, 2, 3, ....
{Given the speed of sound in copper, V = 3.8 km/s =3800 m/s.}
→ν = 1900n Hz
Thus the minimum frequency for is for n = 1, i.e. =1900 Hz =1.9 kHz.
Maximum frequency is for n = 10, i.e. =19000 Hz =19 kHz. The value for n=11 will give frequency =20.9 kHz which is more than 20000 Hz. Hence the standing longitudinal wave frequencies are
νₙ = 1.9n kHz where n = 1, 2, 3, 4, ......, 10.
44. Find the greatest length of an organ pipe open at both ends that will have its fundamental frequency in the normal hearing range (20-20000 Hz). The speed of sound in air = 340 m/s.
ANSWER: The fundamental frequency in an open organ pipe, ν₀ = V/2L.
For the greatest length of the pipe to have its fundamental frequency, ν₀ = 20 Hz
→V/2L = 20
→V = 40L
→L = V/40
{Given V = 340 m/s.}
→L = 340/40 m
→L = 8.5 m.
45. An open organ pipe has a length of 5 cm. (a) Find the fundamental frequency of vibration of this pipe. (b) What is the highest harmonic of such a tube that is in the audible range? The speed of sound in air is 340 m/s and the audible range is 20-20000 Hz.
ANSWER: (a) The fundamental frequency of an open organ pipe ν₀ = V/2L
Here L= 5 cm =0.05 m and V = 340 m/s.
ν₀ = 340/(2*0.05) Hz = 3400 Hz =3.4 kHz.
(b) The frequencies for higher harmonics are given as,
νₙ = nV/2L, where n = 1, 2, 3, ...
For the highest harmonic to be in the audible range
νₙ ≤ 20000 Hz
→nV/2L ≤ 20000
→n ≤ 40000L/V
→n ≤ 40000*0.05/340
→n ≤ 5.88
Since n is a whole number, n = 5.
So the highest harmonic for this tube is the 5th overtone.
46. An electrically driven loudspeaker is placed near the open end of a resonance column apparatus. The length of the air column in the tube is 80 cm. The frequency of the loudspeaker can be varied between 20 Hz -2 kHz. Find the frequencies at which the column will resonate. The speed of sound in air =320 m/s.
ANSWER: The resonance column acts as a closed organ pipe for which the resonant frequencies are given as
νₙ = (2n+1)V/4L, where n = 0, 1, 2, 3, ....
Here V = 320 m/s, L = 80 cm = 0.80 m
→νₙ = (2n+1)*320/(4*0.80) Hz
→νₙ = 100(2n+1) Hz
Putting n = 0, ν = 100 Hz > 20 Hz, Acceptable.
for n = 10, ν = 2100 Hz > 2kHz, Not acceptable.
For n = 9, ν = 1900 Hz < 2kHz, acceptable.
Hence the frequencies at which the column will resonate
= 100(2n+1) Hz where n = 0, 1, 2, 3,....., 9
47. Two successive resonant frequencies in an open organ pipe are 1944 Hz and 2592 Hz. Find the length of the tube. The speed of sound in air =324 m/s.
ANSWER: The resonant frequencies in an open organ pipe is given as
νₙ = nV/2L
Let the given frequencies are for n and n+1 harmonics, thus
nV/2L = 1944 and
(n+1)V/2L = 2592
→nV/2L + V/2L =2592
→1944+V/2L = 2592
→V/2L = 2592 - 1944 = 648
→L = V/(2*648) = 324/1296 m = 0.25 m
→L = 25 cm
48. A piston is fitted in a cylindrical tube of small cross-section with the other end of the tube open. The tube resonates with a tuning fork of frequency 512 Hz. The piston is gradually pulled out of the tube and it is found that a second resonance occurs when the piston is pulled out through a distance of 32.0 cm. Calculate the speed of sound in the air of the tube.
ANSWER: The piston acts as the closed end of the organ pipe. At the piston, there will be pressure antinode while at the open end there will be a pressure node. Since in both the cases the frequency of the vibration (or the wavelength of the sound) is same = 512 Hz, the next resonance will occur when the length of the air column is increased to 𝜆/2 because one more node will be induced in the air column. From the given problem,
𝜆/2 = 32 cm = 0.32 m
→𝜆 = 0.64 m
ν = 512 Hz
Hence the speed of the sound in air V = ν𝜆
→V = 512*0.64 m/s ≈ 328 m/s.
49. A U-tube having unequal arm lengths has water in it. A tuning fork of frequency 440 Hz can set up the air in the shorter arm in its fundamental mode of vibration and the same tuning fork can set up the air in the longer arm in its first overtone vibrations. Find the length of the air columns. Neglect any end effect and assume that the speed of sound in air = 330 m/s.
ANSWER: Given that V = 330 m/s, the air columns in the U-tube are the closed organ pipes. The fundamental frequency
V/4L = 440 Hz
→L = V/(4*440) m = 330/(4*440) m
→L = 3/16 m = 0.188 m = 18.8 cm
The first overtone frequency is
3V/4L = 440 Hz
→L = 3V/(4*440) m =3*330/(4*440) m
→L = 9/16 m = 0.563 m = 56.3 cm
50. Consider the situation in figure (16-E9). The wire which has a mass of 4.00 g oscillates in its second harmonic and sets the air column in the tube into vibrations in its fundamental mode. Assuming that the speed of sound in air is 340 m/s, find the tension in the wire.
 |
| The figure for Q-50 |
ANSWER: The frequency of the waves in the string will be the same as the frequency of the sound wave in the air column. The air column vibrates as in a closed organ pipe and its frequency in the fundamental mode is given as ν = V/4L. Given V= 340 m/s and L = 1.0 m, hence the frequency
ν = 340/(4*1.0) = 85 Hz.
Let the tension in the string be F.
The length of the wire l = 40 cm =0.40 m.
The mass of the wire m = 4 g = 0.004 kg.
Hence the linear mass density of the wire µ = m/l
=0.004/0.40 kg/m =0.01 kg/m.
Since the string oscillates in the second harmonic, its frequency ν = (2/2l)√(F/µ)
=(1/0.40)√(F/0.01) Hz
=2.5*10√F = 25√F
But it is equal to 85 Hz, hence
25√F =85
→√F = 85/25 =3.4
→F = 11.56 N
===<<<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
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
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)
CHAPTER- 7 - Circular Motion
Click here for → Questions for Short Answer
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 → 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)
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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