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JEE Advanced 2023 Paper 2 Solutions
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© examsnet.com
Question : 10
Total: 52
A string of length
1
m
and mass
2
×
10
−
5
kg
is under tension
T
. When the string vibrates, two successive harmonics are found to occur at frequencies
750
Hz
and
1000
Hz
. The value of tension
T
is _____________ newton.
Your Answer:
Validate
Solution:
I
=
1
m
,
m
=
2
×
10
−
5
kg
,
T
: Tension in the string.
because
quad
Successive frequencies are being given
therefore
quad
It is the case of both ends fixed.Now,
f
n
+
1
−
f
n
=
1000
−
750
⇒
(
n
+
1
)
2
l
√
T
µ
−
n
2
l
√
T
µ
=
250
⇒
1
2
l
√
T
µ
=
250
⇒
√
T
2
×
10
−
5
=
250
×
2
×
1
⇒
T
2
×
10
−
5
=
25
×
10
−
4
⇒
T
=
50
×
10
−
1
T
=
5
N
© examsnet.com
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