The fundamental frequency of a sonometer wire increases by $6$ $Hz$ if its tension is increased by $44\%$ keeping the length constant. The change in the fundamental frequency of the sonometer wire in $Hz$ when the length of the wire is increased by $20\%$, keeping the original tension in the wire will be :-
Diffcult
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Fundamental frequency of sonometer
${\rm{n}} = \frac{1}{{2l}}\sqrt {\frac{{\rm{T}}}{{\rm{m}}}} $
when tension is increased by $44 \%$ then $n^{\prime}=n+6$
$ \Rightarrow {\rm{n}} + 6 = \frac{1}{{2l}}\sqrt {\frac{{1.44{\rm{T}}}}{{\rm{m}}}} = 1.2{\rm{n}} \Rightarrow {\rm{n}} = 30\,{\rm{Hz}}$
When length of the wire is increased by $20 \%$ then
${{\rm{n}}^\prime } = \frac{1}{{2{l^\prime }}}\sqrt {\frac{{\rm{T}}}{{\rm{m}}}} = \frac{1}{{2(1.2l)}}\sqrt {\frac{{\rm{T}}}{{\rm{m}}}} = \left( {\frac{1}{{1.2}}} \right)(30) = 25\,{\rm{Hz}}$
Change in frequency $=30-25=5 \mathrm{\,Hz}$
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