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IIT JEE Advanced 2009 Paper 2
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© examsnet.com
Question : 16
Total: 57
A 20 cm long string, having a mass of 1.0 g, is fixed at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string.
Your Answer:
Validate
Solution:
The distance between the successive nodes is λ / 2. Therefore,
λ
2
=
V
2
v
=
1
2
×
100
√
T
l
m
=
1
200
√
0.5
×
0.2
10
−
3
=
1
20
m
= 5 cm
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