JEE Main 2014PhysicsLaws of MotionMediumMCQ

JEE Main 2014Laws of Motion Question with Solution

JEE Main 2014 (19 Apr Online)

Question

A heavy box is to be dragged along a rough horizontal floor. To do so, the person A pushes it at an angle 30° from the horizontal and requires a minimum force FA, while the person B pulls the box at an angle 60° from the horizontal and needs minimum force FB. If the coefficient of friction between the box and the floor is 35, the ratio FAFB is

Choose an option

Show full solutionCorrect option: B
Correct answer
B23

Step-by-step explanation

FAcos30°=μmg+FAsin30°

FBcos60°=μmg-FBsin60°

FAcos30°-μsin30°=μmg

FBcos60°+μsin60°=μmg

Divide the equation,

we get,

FAFB=μmg/cos30°-μsin30°μmg/cos60°+μsin60°=23

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About this question

This is a previous-year question from JEE Main 2014, covering the Laws of Motion chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.