AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper

Consider the below diagram.Fe‌‌=‌

kq2

r2

...(1)
r‌‌=2l‌cos‌θ
tan‌θ‌‌=‌

Fe

mg

For the value of θ=45∘ and φ=45∘
r‌‌=‌

2l

v2

‌‌=√2l
Substitute the value of r in equation (1)
Fe=‌

kq2

2l

From the geometry of the triangle,
tan‌45∘‌‌=‌

kq2

2lmg

mg‌‌=‌

kq2

2l

...(2)
For the value of θ=30∘ and φ=60∘
r=2l×‌

1

2

=l
Similarly,
Fe′=‌

kq2

l2

So
tan‌30∘=‌

k′q2

l2−m′g

Assuming that the sphere is suspended in a liquid of density ‌

2

3

d
The observed weight of the body is calculated as
m′‌‌=V(d−‌

2d

3

)
‌‌=‌

m

3

Thus,
tan‌30∘‌‌=‌

3k′q2

l2mg

mg‌‌=‌

3√3k′q2

l2

...(3)
From equation (2) and (3)
3√3k′‌‌=‌

k

2

k′‌‌=‌

k

εr

εr‌‌=6√3