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(D) The frequency of a vibrating wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$. Since the wires are identical,$L$ and $\mu$ are constant,so

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$T_1$ is decreased or $T_2$ is increased

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(D) The frequency of a vibrating wire is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$. Since the wires are identical,$L$ and $\mu$ are constant,so $f \propto \sqrt{T}$.Given $T_1 > T_2$,the frequencies are $f_1$ and $f_2$ such that $f_1 > f_2$. The beat frequency is $f_1 - f_2 = 6$.If we change the tension slightly,the beat frequency remains $6$ if the difference between the frequencies remains the same.Let $f_1 = k\sqrt{T_1}$ and $f_2 = k\sqrt{T_2}$.If $T_1$ is decreased,$f_1$ decreases,moving closer to $f_2$. If $T_2$ is increased,$f_2$ increases,moving closer to $f_1$. In both cases,the difference $f_1 - f_2$ decreases.However,if we decrease $T_1$ such that $f_1$ becomes less than $f_2$,the beat frequency $|f_2 - f_1|$ can remain $6$ if the new difference is equal to the original difference.Specifically,if $T_1$ is decreased or $T_2$ is increased,the frequencies approach each other and then cross. The condition for the beat frequency to remain unchanged after a slight change is that the new difference $|f_1' - f_2'|$ equals the original difference $6$.

Two identical straight wires are stretched so as to produce $6$ beats per second when vibrating simultaneously with tensions $T_1$ and $T_2$ respectively. On changing the tension slightly in one of them,the beat frequency remains unchanged. This will happen when (Given $\rightarrow T_1 > T_2$)

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