AP EAPCET 18 May 2024 Shift 1 Solved Paper

Time period of oscillation of a pendulum in air is T.
Density of the bob, ρ=‌

5000

3

kg∕m3.
Density of water, σ=1000kg∕m3.
We need to find the ratio ‌

T

t

, where t is the time period of the pendulum's oscillation in water.
The time period of oscillation of a pendulum in water can be expressed as:
t=‌

T

√1−‌ σ ρ

Substitute the given values:
t=‌

T

√1−‌ 1000×3 5000

This simplifies to:
t=‌

T

√1−‌ 3000 5000

Simplifying further:
‌t=‌

T

√1−0.6

‌t=‌

T

√0.4

The ratio ‌

T

t

is:
‌

T

t

‌=√‌

1

0.4

‌

T

t

‌=√‌

5

2

Thus, the value of ‌

T

t

is √‌

5

2

.