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Rotational Motion question

2008 · Shift 0 · Q83
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  5. /2008 · Shift 0 · Q83

Rotational Motion question

2008 · Shift 0 · Q83

JEE MainPhysicsRotational MotionMCQ+4 / −1
Consider a uniform square plate of side ′a′' a '′a′ and mass ′m′'m'′m′. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is
  1. A
    56ma2{5 \over 6}m{a^2}65​ma2
  2. B
    112ma2{1 \over 12}m{a^2}121​ma2
  3. C
    712ma2{7 \over 12}m{a^2}127​ma2
  4. D
    23ma2{2 \over 3}m{a^2}32​ma2
View written solutionFree

Correct answer: D

  1. Moment of inertia of a square plate about its center

For a rectangular lamina of sides aaa and aaa (i.e. a square), the moment of inertia about an axis perpendicular to the plane and passing through its center is

Icenter=112m(a2+a2)=16ma2.I_{\text{center}}=\frac{1}{12}m(a^2+a^2)=\frac{1}{6}ma^2.Icenter​=121​m(a2+a2)=61​ma2.

  1. Distance between center and corner

The center of the square is at equal distance from all corners.

If side of square is aaa, then coordinates of a corner relative to the center may be taken as

(a2,a2).(\tfrac{a}{2},\tfrac{a}{2}).(2a​,2a​).

So the distance from center to corner is

=\sqrt{\frac{a^2}{4}+\frac{a^2}{4}} =\sqrt{\frac{a^2}{2}}= rac{a}{\sqrt{2}}.$$ Hence, $$d^2=\frac{a^2}{2}.$$ 3. **Apply parallel axis theorem** The required axis is perpendicular to the plane and passes through a corner. By parallel axis theorem, $$I_{\text{corner}}=I_{\text{center}}+md^2.$$ Substitute the values: $$I_{\text{corner}}=\frac{1}{6}ma^2+m\left(\frac{a^2}{2}\right).$$ $$I_{\text{corner}}=\frac{1}{6}ma^2+\frac{1}{2}ma^2 =\left(\frac{1}{6}+\frac{3}{6}\right)ma^2 =\frac{4}{6}ma^2 =\frac{2}{3}ma^2.$$ 4. **Check options** - A: $\frac{5}{6}ma^2$ ❌ - B: $\frac{1}{12}ma^2$ ❌ - C: $\frac{7}{12}ma^2$ ❌ - D: $\frac{2}{3}ma^2$ ✅ Therefore, the correct answer is **Option D**.
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