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Physics · PYQ NEET UG 2019(1)
हिन्दीA force \( F = 20 + 10y \) acts on a particle in y- direction where F is in newton and y in meter. Work done by this force to move the particle from \( y = 0 \) to \( y = 1\text{ m} \) is:
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Correct answer · Option A
25 J
Explanation
Work done \( W = \int_{0}^{1} F dy = \int_{0}^{1} (20 + 10y) dy = [ 20y + 5y^2 ]_0^1 = 20(1) + 5(1)^2 = 25\text{ J} \).
