JEE Main 26-Aug-2021 Shift 2 Solved Paper

Given, coefficient of static friction, µ = 0.5
Value of acceleration due to gravity, g=10ms−2
For complete system to move together,
F = ma
Here, m is total mass.
F = (1 + 2) a = 3a ...(i)
From free body diagram of 1 kg block, Balance forces in horizontal direction,
Fs=μR=m′a ...(ii)
Balance forces in vertical direction,
R = m′g
Put value of R in Eq. (ii),
µm′g = m′a
⇒ 0.5 × 1 × 10 = 1 × a
⇒a=5ms−2
Put the value of a in Eq. (i), we get
F = 3 × 5 = 15 N
Thus, the maximum horizontal force required to move block together is 15 N.