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← JEE Main 2022 · 25 Jun · Shift 1
PhysicsInteger+4 / -1Question 60 of 90

JEE Main 2022 · 25 Jun · Shift 1, Physics question 60

A force on an object of mass 100 g is (10i^+5j^)\left(10\hat i+5\hat j\right) N. The position of that object at t = 2 s is (ai^+bj^)\left(a\hat i+b\hat j\right) m after starting from rest. The value of ab\dfrac ab will be ______.

Answer and worked solution

Official answer: 2

From rest r⃗=12a⃗t2\vec r=\frac12\vec at^2, so ab=axay=FxFy=105=2\dfrac ab=\dfrac{a_x}{a_y}=\dfrac{F_x}{F_y}=\dfrac{10}5=2.

Answer: 2