Class 11th Moment of Inertia Physics System of Particles and Rotational Motion Maths NCERT Exemplar Solutions Class 11th Chapter Eleven Physics

Moment of inertia (M.I) of four bodies, having same mass and radius, and reported as:

I₁ = M.I. of thin circular ring about its diameter

I₂ = M.I. of circular disc about an axis perpendicular to disc and going through the centre,

I₃ = M.I. of solid cylinder about its axis and

I₄ = M.I. of solid sphere about its diameter.

Option 1 - <p>I<sub>1</sub> + I<sub>3</sub> < I<sub>2</sub> + I<sub>4</sub></p>

Option 2 - <p>I<sub>1</sub> = I<sub>2</sub> = I<sub>3</sub> < I<sub>4</sub></p>

Option 3 - <p>I<sub>1</sub> = I<sub>2</sub> = I<sub>3</sub> > I<sub>4</sub></p>

Option 4 - <p><span class="mathml" contenteditable="false"> <math> <mrow> <msub> <mrow> <mi>I</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>I</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> <mo>=</mo> <msub> <mrow> <mi>I</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msub> <mo>+</mo> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <msub> <mrow> <mi>I</mi> </mrow> <mrow> <mn>4</mn> </mrow> </msub> </mrow> </math> </span></p>

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Alok Kumar Singh | Contributor-Level 10

5 months ago

Correct Option - 1

Detailed Solution:

$l_{1} = \frac{1}{2} m r^{2}$

$\begin{array}{l} l_{2} = \frac{1}{2} m r^{2} \\ l_{3} = \frac{1}{2} m r^{2} \\ l_{4} = \frac{2}{5} m r^{2} \end{array}$

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Two solid spheres each of mass 2 kg and radius 75 cm are arranged as shown. Find MOI of the system about the given axis.

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$I = (\frac{2}{5} M R^{2} + M R^{2}) \times 2$

$= \frac{14}{5} \times 2 \times \frac{9}{16}$

$= \frac{63}{20}$

= 3.15 kg m²

Two particles of masses 5 kg and 10 kg are placed along x-axis at positions x = 1 m and x = 3 m. The moment of inertia of the system about an axis respectively passing through centre of mass and parallel to y-axis is

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Distance of CM from 5 kg = 10*2/15 = 4/3m

Two circular rings (of same material) are of same thickness. The diameter of first ring is twice that of second. The moment of inertia of first ring as compared to that of second is

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Moment of inertia of a system made from three uniform rods each of mass m and length l joined to each other, as shown in the figure about one end O and perpendicular to its plane will be

The momentum of inertia of a uniform thin rod about a perpendicular axis passing through on end is I1. The same rod is into a ring and its moment of inertia about a diameter is I2. If $\frac{I_{1}}{I_{2}} i s \frac{X π^{2}}{3},$ $\frac{_{}}{_{}} \frac{^{}}{}$ then the value of x will be__________.

According to question, we can write

$l = 2 π r \Rightarrow r = \frac{l}{2 π}$

$I_{1} = \frac{m l^{2}}{3}, a n d I_{2} = \frac{m r^{2}}{2} = \frac{m l^{2}}{8 π^{2}}$

$\Rightarrow \frac{I_{1}}{I_{2}} = \frac{m l^{2}}{3} \times \frac{8 π^{2}}{m l^{2}} = \frac{8 π^{2}}{3}$

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Physics System of Particles and Rotational Motion 2025

Physics System of Particles and Rotational Motion 2025

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