Question

Find the moment of inertia of a plane rectangular lamina of length 20 cm, breadth 10 cm, and mass 500 g about an axis passing through its centre and perpendicular to its plane. (A) 0.002083 kgm^2 (B) 0.001083 kgm^2 (C) 0.003083 kgm^2 (D) 0.004083 kgm^2

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Answer


The correct option is A.

Given data:

  • Mass (m) of the lamina: [ 500 , \text{g} = 0.5 , \text{kg} ]

  • Length (l) of the lamina: [ 20 , \text{cm} = 0.2 , \text{m} ]

  • Breadth (b) of the lamina: [ 10 , \text{cm} = 0.1 , \text{m} ]

To find the moment of inertia (I) of a plane rectangular lamina about an axis passing through its center and perpendicular to its plane, we use the formula: [ I = m \left( \frac{l^2 + b^2}{12} \right) ]

Substituting the given values: [ l = 0.2 , \text{m}, , b = 0.1 , \text{m}, , \text{and} , m = 0.5 , \text{kg} ]

We get: [ I = 0.5 \left( \frac{(0.2)^2 + (0.1)^2}{12} \right) ]

First, calculate ( l^2 + b^2 ): [ (0.2)^2 + (0.1)^2 = 0.04 + 0.01 = 0.05 ]

Now, find: [ \frac{0.05}{12} = 0.0041667 , \text{m}^2 ]

Multiplying by the mass: [ I = 0.5 \times 0.0041667 , \text{kgm}^2 = 0.00208335 , \text{kgm}^2 ]

Rounding to significant figures: [ I \approx 0.002083 , \text{kgm}^2 ]

Therefore, the moment of inertia is: [ \boxed{0.002083 , \text{kgm}^2} ]


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