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JEE Main 2021 (27 Aug, Shift 1) · Work by a Constant Force

Work by a Constant Force question from JEE Main 2021 (27 Aug, Shift 1)

Q14JEE Main 2021 · 27 AUG · SHIFT 1Easy

Work by a Constant Force

Two persons A and B perform same amount of work in moving a body through a certain distance d with application of forces acting at angle 45∘{}^\circ and 60∘{}^\circ with the direction of displacement respectively. The ratio of force applied by person A to the force applied by person B is 1x\frac{1}{\sqrt{x}}. The value of x is .....................

Numerical answer type. Work it out, then open the answer.

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Answer: 2

  1. Same work over the same dd: FAdcos⁡45∘=FBdcos⁡60∘F_A d\cos 45^\circ = F_B d\cos 60^\circ
  2. FAFB=cos⁡60∘cos⁡45∘=1/21/2=12\frac{F_A}{F_B} = \frac{\cos 60^\circ}{\cos 45^\circ} = \frac{1/2}{1/\sqrt{2}} = \frac{1}{\sqrt{2}}
  3. Comparing with 1x\frac{1}{\sqrt{x}}, x=2x = 2.
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