TS EAMCET 19-July-2022 Shift 2 Solved Paper

Initial length of wire, l1=0.4m
Frequency of stretched wire,
f1=250‌vib∕ s=250‌Hz
Final length of wire after stretching,
l2‌=l1+0.1m
‌=0.4+0.1=0.5m
Initial force (tension), T1=T
Final force (tension), T2=‌

T

4

We know that, frequency of wave produced due to stretched string.
f=‌

l

2l

√‌

T

m

⇒‌‌‌

f2

f1

=‌

2

5

⇒‌‌f2=‌

2

5

f1=‌

2

5

×250=100‌Hz.