AP EAPCET 04-Jul-2022 Shift 2 Paper

Given, volume of block A=0.25m3
Let the density of iron be ρ‌kg∕m3.
Thus, mass of iron block, A=0.25ρ‌kg
When block A is attached to the spring elongation produced (x)=0.2m.
Thus, force on spring of spring constant k is
F=−kx(-ve sign denotes force is in opposite direction of elongation)
mg‌=−k(0.2)
0.25ρg‌=−0.2k⇒k=1.25.ρgN∕m
Now, the same spring is attached to another block B.
Volume of block B=0.75m3
Mass of block B=0.75ρ‌kg
FBD of the spring system on inclined surface is given as
Force on block B,F=−(kx)
Mgsin‌θ‌=−kx
0.75ρgsin‌30∘‌=−(−1.25ρg)x
‌

0.75

1.25

sin‌30∘‌=x⇒‌

0.75

1.25

×‌

1

2

=x
⇒‌‌x‌=0.3m
Thus, the distance of block from the top = original length of spring + elongation
=1m+0.3m=1.3m