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#thin ring

15 public questions tagged with this topic.

A thin ring of mass 1kg and radius 0.4m has a kinetic energy of 8J. What is its angular speed?

I = MR2 = 1×(0.4)2 = 0.16kg m2. K = 12Iω2⇒8 = 12×0.16×ω2⇒16 = 0.16ω2⇒ω2 = 100⇒ω = 10rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10 rad/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A thin ring of mass 3kg and radius 0.4m rotates about its center at 5rad/s. What is its rotational kinetic energy?

I = MR2 = 3×(0.4)2 = 0.48kg m2. K = 12Iω2 = 12×0.48×(5)2 = 0.24×25 = 6J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.0 J. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A thin ring of mass 1kg and radius 0.2m rotates about its center. If its kinetic energy is 4J, what is its angular speed

KE: K = 12Iω2. For a ring: I = MR2 = 1×(0.2)2 = 0.04kg m2. 4 = 12×0.04×ω2⇒8 = 0.04ω2⇒ω2 = 200⇒ω = 200≈14.14rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 14 rad/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A thin ring of mass 2kg and radius 0.4m has a kinetic energy of 8J. What is its angular speed?

I = MR2 = 2×(0.4)2 = 0.32kg m2. K = 12Iω2⇒8 = 12×0.32×ω2⇒16 = 0.32ω2⇒ω2 = 50⇒ω = 50≈7.07rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 7 rad/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A thin ring of mass 5kg and radius 0.5m has an angular momentum of 12.5kg m2/s. What is its angular velocity?

I = MR2 = 5×(0.5)2 = 1.25kg m2. ω = LI = 12.51.25 = 10rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10 rad/s. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A thin ring of mass 4kg and radius 0.3m rotates about its center at 8rad/s. What is its rotational kinetic energy?

I = MR2 = 4×(0.3)2 = 0.36kg m2. K = 12Iω2 = 12×0.36×(8)2 = 0.18×64 = 11.52J. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 11.52 J. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A thin ring of mass 2kg and radius 0.5m has a kinetic energy of 10J. What is its angular speed?

I = MR2 = 2×(0.5)2 = 0.5kg m2. K = 12Iω2⇒10 = 12×0.5×ω2⇒20 = 0.5ω2⇒ω2 = 40⇒ω = 40≈6.32rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6.32 rad/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.

A thin ring of mass 6kg and radius 0.4m has an angular momentum of 9.6kg m2/s. What is its angular velocity?

I = MR2 = 6×(0.4)2 = 0.96kg m2. ω = LI = 9.60.96 = 10rad/s. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 10 rad/s. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Thermal Properties and Gravitation.