A wire having a linear density of 0.05 g/cm is stretched between two rigid supports with a tension of 4.5×102N it is observed that the wire resonance at a frequency of 420 Hz. The next higher frequency at which the wire resonates is 490 Hz. Determine the length of the wire.
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
2.14 m
b
3.2 m
c
4 m
d
8 m
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Expression for Frequency when both ends are tied f = n2LTμ Now, 420=n2LTμ→(i) 490=n+12LTμ→(ii) Divide equation i÷iin=6Putting in equation (i)420=62L4.5×1020.05×10−310−2⇒ L=2.14 m