AP EAPCET 25-Aug-2021 Shift 1 Paper

Given, the coefficient of friction between A and B,µ1=0.35
The coefficient of friction between B and surface, µ2=0.5
Mass of block A,ma=100‌kg
Mass of block B,mb=300‌kg
Frictional force between blocks A and B,
f1=µ1m2g
=0.35×100×9.8
=0.35×980
= 343 N
Frictional force between block B and surface,
f2=µ2(mα+mb)g
=0.5(100+300)9.8
=0.5×4×980=1960N
Minimum force required to move block B,
F=f1+f2=343+1960
⇒ = 2303 N
Which is nearest to 2350N. Hence, option (c) is correct.