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Rotational Motion question

2020 · 4 Sep · Shift 2 · Q47
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Rotational Motion question

2020 · 4 Sep · Shift 2 · Q47

JEE MainPhysicsRotational MotionMCQ+4 / −1
For a uniform rectangular sheet shown in the figure, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through O (the centre of mass) and O' (corner point) is : JEE Main 2020 (Online) 4th September Evening Slot Physics - Rotational Motion Question 138 English
  1. A
    12{1 \over 2}21​
  2. B
    14{1 \over 4}41​
  3. C
    18{1 \over 8}81​
  4. D
    23{2 \over 3}32​
View written solutionFree

Correct answer: B

  1. Moment of inertia about the axis through the centre OOO

For a uniform rectangular sheet of sides aaa and bbb, the moment of inertia about an axis perpendicular to the sheet through its centre of mass OOO is

IO=M12(a2+b2).I_O = \frac{M}{12}(a^2+b^2).IO​=12M​(a2+b2).

  1. Moment of inertia about the axis through the corner O′O'O′

The axis through corner O′O'O′ is also perpendicular to the sheet. Use the parallel axis theorem:

IO′=IO+Md2,I_{O'} = I_O + Md^2,IO′​=IO​+Md2,

where ddd is the distance between the centre and the corner.

For a rectangle,

d=(a2)2+(b2)2d = \sqrt{\left(\frac a2\right)^2+\left(\frac b2\right)^2}d=(2a​)2+(2b​)2​

so

d2=a2+b24.d^2 = \frac{a^2+b^2}{4}.d2=4a2+b2​.

Hence,

IO′=M12(a2+b2)+M⋅a2+b24.I_{O'} = \frac{M}{12}(a^2+b^2) + M\cdot \frac{a^2+b^2}{4}.IO′​=12M​(a2+b2)+M⋅4a2+b2​.

Take common terms:

= M(a^2+b^2)\left(\frac{1+3}{12}\right) = \frac{M}{3}(a^2+b^2).$$ 3. **Required ratio** The question asks for the ratio of moments of inertia about axes through $O$ and $O'$: $$\frac{I_O}{I_{O'}} = \frac{\frac{M}{12}(a^2+b^2)}{\frac{M}{3}(a^2+b^2)} = \frac{1}{4}. $$ 4. **Option matching** $$\boxed{\frac{1}{4}}$$ So the correct option is **B**.
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