Question

What should be the velocity of rotation of earth due to rotation about its own axis so that the weight of a person becomes (3)/(5) of the present weight at the equator. Equatorial radius of earth is 6400km.

• A. \(8.7 \times 10^{-4}\,\text{rad/s}\)
• B. \(7.8 \times 10^{-4}\,\text{rad/s}\)
• C. \(6.7 \times 10^{-4}\,\text{rad/s}\)
• D. 7.4 × 10⁻3rad/s

Hint:

At the equator, effective gravity is reduced due to centrifugal force: gₑff = g - ω² R Always use this relation when Earth’s rotation affects weight.

Answer 1

The Correct Option is B

Step 1: At the equator, effective weight of a person is W = m(g - ω² R)
Step 2: Given that the new weight becomes (3)/(5) of the original weight: m(g - ω² R) = (3)/(5)mg
Step 3: Simplifying: g - ω² R = (3)/(5)g ω² R = (2)/(5)g
Step 4: Substituting g = 9.8m/s² and R = 6400 × 10³m: ω² = (2 × 9.8)/(5 × 6.4 × 10⁶) ω² = 6.125 × 10⁻7
Step 5: Taking square root: ω = 7.8 × 10⁻4rad/s