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Quiz Description :
Name: Angular Motion objective question answer quiz
Subject: Physics
Topic: Motion
Questions: 10 Objective Type
Time Allowed: 15 Minutes
Important for: 11th & 12th School students, Engineering and Medical Entrance exams, B Sc / M Sc Entrance and University exam.
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 Question 1 of 10
1. Question
1 pointsA couple produces
CorrectWe know that two equal and unlike parallel forces acting on a body form a couple. And moment of the couple is equal to the product of the magnitude of either force and perpendicular distance between the lines of action of the forces.
This couple tends the body to have purely rotational motion.IncorrectWe know that two equal and unlike parallel forces acting on a body form a couple. And moment of the couple is equal to the product of the magnitude of either force and perpendicular distance between the lines of action of the forces.
This couple tends the body to have purely rotational motion.UnattemptedWe know that two equal and unlike parallel forces acting on a body form a couple. And moment of the couple is equal to the product of the magnitude of either force and perpendicular distance between the lines of action of the forces.
This couple tends the body to have purely rotational motion.  Question 2 of 10
2. Question
1 pointsThe angular velocity of a wheel is 70 rads^{1}. If radius of the wheel is 0.5m, then linear velocity of the wheel is
CorrectGiven: Angular velocity of wheel (ω) = 70 rads^{1} and radius of wheel (r) = 0.5m.
We know that linear velocity of the wheel (v) = ω r = 70 × 0.5 = 35 ms^{1}IncorrectGiven: Angular velocity of wheel (ω) = 70 rads^{1} and radius of wheel (r) = 0.5m.
We know that linear velocity of the wheel (v) = ω r = 70 × 0.5 = 35 ms^{1}UnattemptedGiven: Angular velocity of wheel (ω) = 70 rads^{1} and radius of wheel (r) = 0.5m.
We know that linear velocity of the wheel (v) = ω r = 70 × 0.5 = 35 ms^{1}  Question 3 of 10
3. Question
1 pointsA dics of radius 1m is rotating with an angular velocity of 10 rads^{1}. The linear velocity of the dics on a point on a point 0.5m from its centre is
CorrectGiven: Radius of dics (r) = 1m; Angular velocity of dics (ω) = 10 rads^{1} and distance of the point from its centre (d) = 0.5m.
We know that linear velocity of dice on its extreme point (v) = d ω = 0.5 × 10 = 5 ms^{1}.IncorrectGiven: Radius of dics (r) = 1m; Angular velocity of dics (ω) = 10 rads^{1} and distance of the point from its centre (d) = 0.5m.
We know that linear velocity of dice on its extreme point (v) = d ω = 0.5 × 10 = 5 ms^{1}.UnattemptedGiven: Radius of dics (r) = 1m; Angular velocity of dics (ω) = 10 rads^{1} and distance of the point from its centre (d) = 0.5m.
We know that linear velocity of dice on its extreme point (v) = d ω = 0.5 × 10 = 5 ms^{1}.  Question 4 of 10
4. Question
1 pointsA point on the rim of a wheel of diameter 400 cm has a velocity of 16 ms^{1}. The angular velocity of the wheel is
CorrectGiven that diameter of wheel (d) = 400cm = 4m or Radius (r) = 2m and velocity of the point on the rim (v) = 16 ms^{1}.
We know that angular velocity of wheel (ω) = v/r = 16/2 = 8 rads^{1}.IncorrectGiven that diameter of wheel (d) = 400cm = 4m or Radius (r) = 2m and velocity of the point on the rim (v) = 16 ms^{1}.
We know that angular velocity of wheel (ω) = v/r = 16/2 = 8 rads^{1}.UnattemptedGiven that diameter of wheel (d) = 400cm = 4m or Radius (r) = 2m and velocity of the point on the rim (v) = 16 ms^{1}.
We know that angular velocity of wheel (ω) = v/r = 16/2 = 8 rads^{1}.  Question 5 of 10
5. Question
1 pointsIf a flywheel makes 120 revmin^{1}, then its angular speed will be
CorrectGiven that angular frequency of flywheel (N) = 120 revmin^{1} = 2 revs^{1}.
We know that angular speed of the flywheel (ω) = 2 π N = 2 π × 2 = 4 π rads^{1}.IncorrectGiven that angular frequency of flywheel (N) = 120 revmin^{1} = 2 revs^{1}.
We know that angular speed of the flywheel (ω) = 2 π N = 2 π × 2 = 4 π rads^{1}.UnattemptedGiven that angular frequency of flywheel (N) = 120 revmin^{1} = 2 revs^{1}.
We know that angular speed of the flywheel (ω) = 2 π N = 2 π × 2 = 4 π rads^{1}.  Question 6 of 10
6. Question
1 pointsAn engine flywheel of diameter 1m rotates with an angular velocity of 600 revmin^{1}. The linear velocity of a particle on the periphery of the wheel will be
CorrectGiven that diameter of flywheel (d) = 1m or Radius (r) = 0.5m and angular frequency of flywheel (N) = 600 revmin^{1} = 10 revs^{1}.
We know that angular speed of flywheel (ω) = 2 π N = 2 π × 10 = 20 π ms^{1}.
Therefore linear velocity of a particle on the periphery of the wheel (v) = ωr
= 0.5 × 20 π = 10 π ms^{1}.IncorrectGiven that diameter of flywheel (d) = 1m or Radius (r) = 0.5m and angular frequency of flywheel (N) = 600 revmin^{1} = 10 revs^{1}.
We know that angular speed of flywheel (ω) = 2 π N = 2 π × 10 = 20 π ms^{1}.
Therefore linear velocity of a particle on the periphery of the wheel (v) = ωr
= 0.5 × 20 π = 10 π ms^{1}.UnattemptedGiven that diameter of flywheel (d) = 1m or Radius (r) = 0.5m and angular frequency of flywheel (N) = 600 revmin^{1} = 10 revs^{1}.
We know that angular speed of flywheel (ω) = 2 π N = 2 π × 10 = 20 π ms^{1}.
Therefore linear velocity of a particle on the periphery of the wheel (v) = ωr
= 0.5 × 20 π = 10 π ms^{1}.  Question 7 of 10
7. Question
1 pointsA fan is making 600 revmin^{1}. If it makes 1200 revm^{1}, then increase in its angular velocity is
CorrectGiven that initially angular frequency (N_{1}) = 600 revm^{1} = 10 revs^{1} and final angular frequency (N_{2}) = 1200 revmin^{1} = 20 revs^{1}.
We know that increase in angular velocity (∆ω) = ω_{2} – ω_{1} = 2 π N_{2} – 2 π N_{1} = (2π × 20) – (2π × 10)
= (40 π × 20 π) = 20 π rads^{1}.IncorrectGiven that initially angular frequency (N_{1}) = 600 revm^{1} = 10 revs^{1} and final angular frequency (N_{2}) = 1200 revmin^{1} = 20 revs^{1}.
We know that increase in angular velocity (∆ω) = ω_{2} – ω_{1} = 2 π N_{2} – 2 π N_{1} = (2π × 20) – (2π × 10)
= (40 π × 20 π) = 20 π rads^{1}.UnattemptedGiven that initially angular frequency (N_{1}) = 600 revm^{1} = 10 revs^{1} and final angular frequency (N_{2}) = 1200 revmin^{1} = 20 revs^{1}.
We know that increase in angular velocity (∆ω) = ω_{2} – ω_{1} = 2 π N_{2} – 2 π N_{1} = (2π × 20) – (2π × 10)
= (40 π × 20 π) = 20 π rads^{1}.  Question 8 of 10
8. Question
1 pointsA flywheel starts form rest and gains a speed of 540 revmin^{1} in 6s. Angular acceleration of the flywheel will be
CorrectGiven that initially angular frequency (N_{1}) = 0 (because its starts from rest); Final angular frequency (N_{2}) = 540 revmin^{1} = 9 revs^{1} and time taken (t) = 6s.
We know that angular acceleration (α) = [(ω_{2} – ω_{1}) / t] = [(2 π N_{2} – 2 π N_{1}) / t] = [2π (N_{2} – N_{1})/ 6]
= (2 π ( 9 – 0) / 6 = 3 π rads^{2}.IncorrectGiven that initially angular frequency (N_{1}) = 0 (because its starts from rest); Final angular frequency (N_{2}) = 540 revmin^{1} = 9 revs^{1} and time taken (t) = 6s.
We know that angular acceleration (α) = [(ω_{2} – ω_{1}) / t] = [(2 π N_{2} – 2 π N_{1}) / t] = [2π (N_{2} – N_{1})/ 6]
= (2 π ( 9 – 0) / 6 = 3 π rads^{2}.UnattemptedGiven that initially angular frequency (N_{1}) = 0 (because its starts from rest); Final angular frequency (N_{2}) = 540 revmin^{1} = 9 revs^{1} and time taken (t) = 6s.
We know that angular acceleration (α) = [(ω_{2} – ω_{1}) / t] = [(2 π N_{2} – 2 π N_{1}) / t] = [2π (N_{2} – N_{1})/ 6]
= (2 π ( 9 – 0) / 6 = 3 π rads^{2}.  Question 9 of 10
9. Question
1 pointsA wheel is rotated at 900 revmin1 about its axis. When power is cut off, it comes to rest in 1 min. The angular retardation is
CorrectGiven that initial angular frequency (N_{1}) = 900 revmin^{1} = 15 revs^{1}; Final angular frequency (N_{2}) = 0 (because it comes to rest) and time taken (t) = 1 min = 60s.
We know that angular acceleration (α) = [(ω_{1} – ω_{2}) / t] = [(2 π N_{1} – 2 π N_{2}) / t] = [2π (N_{1} – N_{2})/ 60]
= (2 π (15 – 0) / 60 = 30 π / 60 = π / 2 rads^{2}.IncorrectGiven that initial angular frequency (N_{1}) = 900 revmin^{1} = 15 revs^{1}; Final angular frequency (N_{2}) = 0 (because it comes to rest) and time taken (t) = 1 min = 60s.
We know that angular acceleration (α) = [(ω_{1} – ω_{2}) / t] = [(2 π N_{1} – 2 π N_{2}) / t] = [2π (N_{1} – N_{2})/ 60]
= (2 π (15 – 0) / 60 = 30 π / 60 = π / 2 rads^{2}.UnattemptedGiven that initial angular frequency (N_{1}) = 900 revmin^{1} = 15 revs^{1}; Final angular frequency (N_{2}) = 0 (because it comes to rest) and time taken (t) = 1 min = 60s.
We know that angular acceleration (α) = [(ω_{1} – ω_{2}) / t] = [(2 π N_{1} – 2 π N_{2}) / t] = [2π (N_{1} – N_{2})/ 60]
= (2 π (15 – 0) / 60 = 30 π / 60 = π / 2 rads^{2}.  Question 10 of 10
10. Question
1 pointsA body of mass m is fixed to one end of a spring of force constant k and length l. The other end of the spring is fixed to a rod and the system is rotated with an angular velocity of ω, in a gravity free space as shown in the figure. The increase in length of the spring will be
CorrectGiven that Mass of the body = m; Spring constant = k; length of the spring = l and the angular velocity of the system = ω.
We know that when the system is rotated, it will cause some centrifugal force, which will increase the length of the spring.
We also know that centrifugal force (F) = mω^{2} (l + x). [where x = Extension in the spring]
Therefore increase in the length of spring (x) = Centrifugal force / Spring constant = F / k
or, F = kx.
Equating two values of F, we get kx = mω^{2} (l + x) = mω^{2}l + m^{2}x
or, mω^{2}l = kx – mω^{2}x = x (k – mω^{2})
or, x = (mω^{2}l / k) – mω^{2}.IncorrectGiven that Mass of the body = m; Spring constant = k; length of the spring = l and the angular velocity of the system = ω.
We know that when the system is rotated, it will cause some centrifugal force, which will increase the length of the spring.
We also know that centrifugal force (F) = mω^{2} (l + x). [where x = Extension in the spring]
Therefore increase in the length of spring (x) = Centrifugal force / Spring constant = F / k
or, F = kx.
Equating two values of F, we get kx = mω^{2} (l + x) = mω^{2}l + m^{2}x
or, mω^{2}l = kx – mω^{2}x = x (k – mω^{2})
or, x = (mω^{2}l / k) – mω^{2}.UnattemptedGiven that Mass of the body = m; Spring constant = k; length of the spring = l and the angular velocity of the system = ω.
We know that when the system is rotated, it will cause some centrifugal force, which will increase the length of the spring.
We also know that centrifugal force (F) = mω^{2} (l + x). [where x = Extension in the spring]
Therefore increase in the length of spring (x) = Centrifugal force / Spring constant = F / k
or, F = kx.
Equating two values of F, we get kx = mω^{2} (l + x) = mω^{2}l + m^{2}x
or, mω^{2}l = kx – mω^{2}x = x (k – mω^{2})
or, x = (mω^{2}l / k) – mω^{2}.