Moment of Inertia Conservation of Angular Momentum Numerical Class-11 Nootan ISC Physics Ch-11. Step by step Solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

## Moment of Inertia Conservation of Angular Momentum Numerical Class-11 Nootan ISC Physics

Board | ISC |

Class | 11 |

Subject | Physics |

Writer | Kumar and Mittal |

Publication | Nageen Prakashan |

Chapter-11 | Rotational Motion of Rigid Body: Moment of Inertia |

Topics | Numericals on Moment of Inertia, K.E. of Rotation and Law of Conservation of Angular Momentum |

Academic Session | 2024-2025 |

### Numericals on Moment of Inertia, K.E. of Rotation and Law of Conservation of Angular Momentum

Ch-11 Rotational Motion of Rigid Body: Moment of Inertia Class-11 ISC Nootan Solutions of Kumar and Mittal Physics , Nageen Prakashan

**Q-8: A body of mass 50 g is revolving about an axis in a circular path. The distance of the centre of mass of the body from the axis of rotation is 50 cm. Find the moment of inertia of the body.**

**Ans- **I = m r²

=> 50 x 10^-3 x (50 x 10^-2)²

=> **1.25 x 10^-2 kg m²**

**Q-9: Find the moment of inertia of the hydrogen molecule about an axis passing through its centre of mass and perpendicular to the inter-nuclear axis. Given : mass of the hydrogen atom = 1.7 x 10^-27 kg, inter-atomic distance = 4 × 10^-10 m.**

**Ans- **I = m1 r1² + m2 r2²

=> 1.7 x 10^-27 x (2 x 10^-10)² + 1.7 x 10^-27 x (2 x 10^-10)²

=> 2 x 1.7 x 10^-27 x 4 x 10^-20

=> **13.6 x 10^-47 kg m²**

**Q-10: A nitrogen molecule has two nitrogen atoms separated by a distance of 1.3 × 10^-10 m. Find its moment of inertia about the centre of mass. If its rotational kinetic energy is 4 x 10^-21 J, then compute its frequency of rotation about the centre of mass. The mass of a nitrogen atom is 14 amu (1 amu = 1.66 × 10^-27 kg).**

**Ans- **I = m1 r1² + m2 r2²

=> 14 x 1.66 x 10^-27 x (0.65 x 10^-10)² x 2

=> **19.6 x 10^-46 kg m²**

again k = 1/2 I ω²

=> ω² = 2k / I

=> ω = √2k / I

=> 2 π n = √2k / I

=> n = (1/2π) √2k / I = 1 / (2 x 3.14) √(2 x 4 x 10^-21) / (1.96 x 10^-46)

=> **1.0 x 10^12 Hz**

**Q-11: Three particles (each of mass 10 g) are situated at the three corners of an equilateral triangle of side 5 cm. Determine the moment of inertia of this system about an axis passing through one corner of the triangle and perpendicular to the plane of the triangle.**

**Ans- **Moment of Inertia due to mass passing through the axis is zero.

Moment of inertia due to other two mass = 2 m r²

=> 2 x 10^-2 x 25 x 10^-4 = **5 x 10^-5 kg m²**

**Q-12: The moment of inertia of a flywheel is 4 kg m². What angular acceleration will be produced in it by applying a torque of 10 N-m on it?**

**Ans-12 **τ = I α

=> α = 10 / 4 = **2.5 rad s^-1**

**Q-13: A torque of 2.0 x10^-4 N m is to be applied to produce an angular acceleration of 4 rad s^-2 in a rotating body. What is the moment of inertia of the body?**

**Ans- **τ = I α

=> 2 x 10^-4 = I x 4

=> I = 2 x 10^-4 / 4 = **0.5 x 10^-4 kg m²**

**Q-14: The moment of inertia of a body is 2.5 kg m². Calculate the torque required to produce an angular acceleration of 18 rad s^-2 in the body. **

**Ans- **τ = I α

=> 2.5 x 18 = 45 N m

**Q-15: The angular momentum of a body is 31.4 J s and its rate of revolution is 10 cycles per second. Calculate the moment of inertia of the body about the axis of rotation.**

**Ans- **ω = 2 π n = 2π x 10

now L = I ω

=> I = L / ω = 31.4 / 2 π x 10

=> 31.4 / 2 x 3.14 x 10 = **0.5 kg m²**

**Q-16: A body of mass 1 kg is revolved in a horizontal circle by attaching to one end of 1 metre long string. If the frequency of revolution is 20 rev/s, find about the axis of rotation (i) moment of inertia, (ii) angular momentum, (iii) rotational kinetic energy of the body and (iv) centripetal force acting on the body.**

**Ans- (i) **I = m r²

=> 1 x 1² = **1 kg m²**

**(ii) **L = I ω

=> 1 x 2 π n

=> 1 x 2 π x 20 = **40 π kg m² s^-1**

**(iii) **k = I ω²

=> 1/2 x 1 x (2 π x 20)²

=> 1/2 x 4π² x 400 = **800 π² J**

**(iv) **a = r ω²

=> 1 x (2π x 20)² = **1600 π² N**

**Q-17: A disc of diameter 0.4 m and of mass 10 kg is rotating about its axis at the rate of 1200 rev/minute. Find (i) angular momentum and (ii) rotational kinetic energy of the disc. **

**Ans- **I of disc = M R² / 2

=> 10 x (0.2)² / 2 = 0.4 / 2 = 0.2 kg m²

n= 1200 / 60 = 20

ω = 2 π n = 40 π

**(i) **L = I ω

=> 0.2 x 40 π

=> 8 x 3.14 = **25.14 kg m² s^-1**

**(ii) **k = 1/2 I ω²

=> k = 1/2 x 0.2 x (40 π)² = **1580 J**

**Q-18: A wheel is rotating at a rate of 1000 rotations/minute and its kinetic energy of rotation is 10^6 J. Determine the moment of inertia of the wheel about the axis of rotation.**

**Ans- **n = 1000 / 60

ω= 2 π n = 2000 π / 60

k = 10^6 J

now k = 1/2 I ω²

=> I = 2k / ω²

=> (2 x 10^6 x 60 x 60) / (2000 x 2000 x π²) = **182 kg m²**

**Q-19: A person taking in his hands spheres each of mass 2 kg is standing on a table revolving with an angular velocity of 10 rad/s. His arms are extended and each sphere is at a distance of 1 m from the axis of rotation. If the moment of inertia of the person and the table about the axis of rotation be 1 kg m² then what will be the kinetic energy of rotation of the whole system? If the person pulls in his arms so that each sphere is now at a distance of 0.3 m from the axis of rotation, then what would be the new kinetic energy? Explain this difference.**

**Ans- **I1 = 1 + 2 x 2(1)² = 5 kg m²

I2 = 1 + 2 x 2(0.3)² = 1.36 kg m²

now I1 ω1 = I2 ω2

=> 5 x 10 = 1.36 x ω2

=> ω2 = 50 / 1.36

k1 = 1/2 x 5 x 10² = **250 J**

I2 = 1/2 π x 1.36 x (50 / 1.36)² = **919 J**

**Q-20: A ball tied to a string takes 4 seconds in one complete revolution in a horizontal circle. If, by pulling the cord, the radius of the circle is reduced to half of the previous value, then how much time the ball will now take in one revolution?**

**Ans- ** I1 ω1 = I2 ω2

=> m r² x 2 π / 4 = m (r / 2)² x 2 π / T2

=> T2 = **1 s**

**Q-21: Moment of Inertia of a ring is 3 kg m². It is rotated for 20 s from its rest position by a torque of 6 N-m. Calculate the work done.**

**Ans- **τ = 6 N m I = 3 kg m²

=> α = τ / I = 2 rad /s²

now ω = α t

=> 2 x 20 = 40 rad/s

again work done = K.E. gained

W = 1/2 x 3 x (40)²

=> 1/2 x 3 x 1600 = **2400 J**

**Q-22: A disc of mass 1.0 kg and radius 0.1 m is rotating at the rate of 2 rpm about an axis passing through the centre and normal to its plane : (i) what is moment of inertia of the disc about axis of rotation? (ii) calculate its angular momentum. (iii) find its K.E. of rotation.**

**Ans- (i) **I of disc = M R² / 2

=> 1 x (0.1)² / 2 = **0.005 kg m²**

**(ii) **L = I ω

=> I x 2 π n

=> 0.005 x 2 x 3.14 x 2 / 60 = **1.05 x 10^-3 kg m² s^-1**

**(iii) **k = 1/2 I ω²

=> 1/2 x 0.005 x (2 x 3.14 x 2 / 60)² = **1.1 x 10^-4 J**

**Q-23: When 100 J of work is done on a flywheel, its angular speed increases from 60 rpm to 180 rpm. Calculate its moment of inertia.**

**Ans- **W = Δk

=> 100 = 1/2 I (ω2² – ω1²)

=> 200 = I x 4 π² (n2² – n1²)

=> I = 200 / 4 π² ÷ {(180 / 60)² – (60 / 60)²}

=> 50 / π² x 8 = **0.633 kg m²**

**Q-24: Calculate the ratio of radii of gyration of a circular ring and a disc of the same radius about the axis passing through their centres and perpendicular to their planes.**

**Ans- **Radius of gyration k = √I / M

now (k)ring : (k)disc = √M R² / M : √M R² / 2 M

=> 1 : 1 / √2 = **√2 : 1**

**Q-25: Calculate the moment of inertia of the earth about its diameter, taking it to be a sphere of 10^25 kg and diameter 12800 km.**

**Ans- **I = 2/5 M R² (solid sphere)

=> 2/5 x 10^25 x (12800 x 10³ / 2)²

=> **1.64 x 10^38 kg m²**

—: end of Moment of Inertia Conservation of Angular Momentum Numerical Class-11 Nootan ISC Ch-11 Solutions :—

Return to : **– Nootan Solutions for ISC Physics Class-11 Nageen Prakashan**

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