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(B) The frequency of a vibrating wire is given by $n = \frac{1}{2l}\sqrt{\frac{T}{m}}$.Since the wires are identical,$l$ and $m$ are constant,so $n \p

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${T_2}$ was increased

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(B) The frequency of a vibrating wire is given by $n = \frac{1}{2l}\sqrt{\frac{T}{m}}$.Since the wires are identical,$l$ and $m$ are constant,so $n \propto \sqrt{T}$.Let the initial frequencies be $n_1$ and $n_2$ corresponding to tensions $T_1$ and $T_2$ respectively,where $T_1 > T_2$,which implies $n_1 > n_2$.The beat frequency is $n_1 - n_2 = 6$.If the beat frequency remains $6$ after changing the tension,the new frequencies $n_1'$ and $n_2'$ must satisfy $|n_1' - n_2'| = 6$.Case $(i)$: If $n_1$ remains constant,$n_2$ must increase to $n_2'$ such that $n_1 - n_2' = 6$ is not possible unless $n_2$ increases past $n_1$. However,if $n_2$ increases to a value $n_2'$ such that $n_2' - n_1 = 6$,the beat frequency is still $6$. This requires increasing $T_2$.Case $(ii)$: If $n_2$ remains constant,$n_1$ must decrease to $n_1'$ such that $n_1' - n_2 = 6$ (not possible as $n_1 > n_1'$) or $n_2 - n_1' = 6$ (not possible as $n_2 Thus,the only valid physical change to maintain $6$ beats per second while changing one tension is increasing the lower tension $T_2$ such that the new frequency $n_2'$ exceeds $n_1$ by $6$ Hz.

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