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JEE Main Physics Class 11 Waves Part 1 Questions
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© examsnet.com
Question : 65
Total: 100
The total length of a sonometer wire between fixed ends is 110 cm. Two bridges are placed to divide the length of wire in ratio 6 : 3 : 2. The tension in the wire is 400 N and the mass per unit length is 0.01 kg/m. What is the minimum common frequency with which three parts can vibrate?
[Online April 19, 2014]
1100 Hz
1000 Hz
166 Hz
100 Hz
Validate
Solution:
Total length of sonometer wire,
l
=
110
cm = 1.1 m
Length of wire is in ratio, 6 : 3 : 2 i.e. 60 cm, 30 cm, 20 cm.
Tension in the wire, T = 400 N
Mass per unit length, m = 0.01 kg
Minimum common frequency = ?
As we know,
Frequency,
v
=
1
21
√
T
m
=
1000
11
H
z
Similarly,
v
1
=
1000
6
H
z
v
2
=
1000
3
H
z
v
3
=
1000
2
H
z
Hence common frequency
=
1000
H
z
© examsnet.com
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