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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Rotational Mechanics

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Rotational Mechanics

A square plate having a mass of 120 g and an edge of 5.0 cm rotates...

A square plate having a mass of 120 g120 \text{ g} and an edge of 5.0 cm5.0 \text{ cm} rotates about one of its edges. Determine the torque acting on the plate if it undergoes a uniform angular acceleration of 0.2 rad/s20.2 \text{ rad/s}^2.

Options

  1. A5.0×10−6 N m5.0 \times 10^{-6} \text{ N m}
  2. B2.0×10−5 N m2.0 \times 10^{-5} \text{ N m}
  3. C1.0×10−5 N m1.0 \times 10^{-5} \text{ N m}
  4. D6.0×10−5 N m6.0 \times 10^{-5} \text{ N m}

Correct answer

B. 2.0×10−5 N m2.0 \times 10^{-5} \text{ N m}

Step-by-step solution

The moment of inertia of a square plate of mass mm and side length aa about one of its edges is given by:

I=ma23I = \dfrac{m a^2}{3}

Given:

m=120 g=0.12 kgm = 120 \text{ g} = 0.12 \text{ kg}

a=5.0 cm=0.05 ma = 5.0 \text{ cm} = 0.05 \text{ m}

Substituting the values:

I=0.12×(0.05)23I = \dfrac{0.12 \times (0.05)^2}{3}

I=0.04×0.0025=1.0×10−4 kg m2I = 0.04 \times 0.0025 = 1.0 \times 10^{-4} \text{ kg m}^2

The torque τ\tau is given by:

τ=Iα\tau = I \alpha

Given α=0.2 rad/s2\alpha = 0.2 \text{ rad/s}^2:

τ=(1.0×10−4)×0.2=2.0×10−5 N m\tau = (1.0 \times 10^{-4}) \times 0.2 = 2.0 \times 10^{-5} \text{ N m}
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