JEE Main 17 March 2021 Shift 2 Solved Paper

Given, mass of the body, m=1kg
Coefficient of static friction, µ=1/√3
Let's draw the free body diagram of the block Using the condition of the equilibrium
In x-direction,
F‌cos‌θ=f=φ=µN...(i)
In y-direction,
F‌sin‌θ+N=mg
⇒‌‌N=mg−F‌sin‌θ....(ii)
Substituting the value of N in Eq. (i), we get
F=‌

µmg

cos‌θ+µ‌sin‌θ

⇒F=‌

µmg

√1+µ2

Substituting the values in the above equation, we get
F=‌

‌ 1 √3 ×10

√1+(‌ 1 √3 )2

⇒F=5N
Hence, the body move by applying minimum possible force of 5N.
So, the value of F will be 5 .