Question

Four wires of the same length and same material, whose diameters are in the ratio $4:3:2:1$, are clamped in such a way that each wire produces the note of frequency double that of the preceding wire. If the tension in the first wire is $2kg-wt$, then tension in the third string will be ____

• A. $4.5$
• B. $8$
• C. $9$
• D. $16$

Answer 1

The correct option is B $8$

Given : diameter ratio $4:3:2:1$
so, radius ratio $4:3:2:1$
frequency gets double that of preceding wire
so , $f_4=\frac{1}{2}{f}_3=\frac{1}{4}{f}_2=\frac{1}{8}{f}_1$
And f = $\frac{v}{\lambda }$
so, $\left(\frac{v_4}{\lambda }\right)=\frac{1}{4}\left(\frac{v_2}{\lambda }\right)(\because f=\frac{v}{\lambda })$
$\Rightarrow \frac{\sqrt{\frac{2}{\rho \pi (4r{)}^2}}}{\lambda }=\frac{\sqrt{\frac{T}{\rho \pi (2r{)}^2}}}{4\lambda }$
$\Rightarrow \frac{1}{16}=\frac{T}{16\times 4}$
$\Rightarrow T=8$