KCET 2017 Physics Solved Paper

Time period of oscillations of a simple pendulum is given as T=2√

l

g

where l is length of pendulum.
For pendulum A: Given, completes 10 oscillations is 20 s.
Therefore, time period of single oscillation =

20

10

=2
⇒TA=2
⇒2π√

IA

g

=2 → (1)
For pendulum B: Given, completes 8 oscillations in 10 s.
Therefore, time period of single oscillation =

10

8

=

5

4

⇒TB=‌

5

4

⇒2π√‌

lB

g

=‌

5

4

→(2)
Divide Eq. (1) by Eq. (2), we get
‌

2π√ lA g

2π√‌ lB g

=‌

‌ 2 1 5

4

⇒√‌

lA

lB

=‌

2×4

5

⇒√‌

lA

lB

=‌

8

5

⇒‌

lA

lB

=(‌

8

5

)2=‌

64

25

Therefore, ratio of lengths of pendulums A and B is

64

25