Question

Four wires of 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 note of frequency double that of the preceding wire. If the tension in the first wire is $2kg-wt$, then tension in the second wire will be (in $kg-wt$)

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

Answer 1

The correct option is A $4.5$

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, $\frac{v_4}{\lambda }=\frac{1}{2}\frac{v_3}{\lambda }$
$\frac{\sqrt{\frac{2}{\rho \pi (4r{)}^2}}}{\lambda }=\frac{1}{2}\frac{\sqrt{\frac{T}{\rho \pi (3r{)}^2}}}{\lambda }$
$\frac{2}{16}=\frac{T}{4\times 9}$
$\Rightarrow \frac{9}{2}=T$
$\Rightarrow T=4.5$