A uniform string resonates with a tuning fork, at a maximum tension of $32 \,N$. If it is divided into two segments by placing a wedge at a distance one-fourth of length from one end, then to resonance with same frequency the maximum value of tension for string will be ........... $N$
Medium
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(a)
Let length be $l$ .
$f=\sqrt{\frac{T}{\mu}} \times \frac{1}{2 l} \quad \dots (i)$
$f=\sqrt{\frac{T}{\mu}} \times \frac{4}{2 l} \quad \dots (ii)$
or $f=\sqrt{\frac{T}{\mu}} \times \frac{4}{6 l} \quad \dots (iii)$.
Equating $(i)$ $(ii)$ and $(i)$ and $(iii)$
$\sqrt{\frac{T}{T_1}}=4$ and $\sqrt{\frac{T}{T_2}}=\frac{4}{3}$
Put $T=32 \,N$
$\frac{32}{16}=T_1 \frac{9}{16} \times 32=T_2$
$T_1=2 N T_2=18 \,N$
of the options on $T_1$ is right.
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