Two sitar strings, $A$ and $B,$ playing the note $'Dha'$ are slightly out of tune and produce beats and frequency $5\,Hz.$ The tension of the string $B$ is slightly increased and the beat frequency is found to decrease by $3\,Hz$ . If the frequency of $A$ is $425\,Hz,$ the original frequency of $B$ is ... $Hz$
JEE MAIN 2018, Diffcult
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$\mathrm{n}_{\mathrm{A}}=425 \mathrm{Hz}, \mathrm{n}_{\mathrm{B}}=?$
Beat frequency $x=5$ $Hz$ which is decreasing $(5 \rightarrow 3)$ after increasing the tension of the string $B$.
Also tension of string $\mathrm{B}$ increasing so
$\mathrm{n}_{\mathrm{B}} \uparrow(\because \mathrm{n} \propto \sqrt{\mathrm{T}})$
Hence $\quad \mathrm{n}_{\mathrm{A}}-\mathrm{n}_{\mathrm{B}} \uparrow=x \downarrow \longrightarrow$ correct
$\mathrm{n}_{\mathrm{B}} \uparrow-\mathrm{n}_{\mathrm{A}}=\mathrm{x} \downarrow \longrightarrow \text { incorrect }$
$\therefore \mathrm{n}_{\mathrm{B}}=\mathrm{n}_{\mathrm{A}}-\mathrm{x}=425-5=420 \mathrm{Hz}$
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