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Question Bank [ 2022 ] — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsQuestion Bank [ 2022 ]3 Marks
Question
Two wires of the same material and same cross-section are stretched on a sonometer. One wire is loaded with $1\ kg$ and another is loaded with $9\ kg.$ The vibrating length of the first wire is $60\ cm$ and its fundamental frequency of vibration is the same as that of the second wire. Calculate the vibrating length of the other wire.
✓

Answer

Given:
$M_1=1 kg , M_2=9 kg$,
$A_1=A_2$
$m _1= m _2$ (where $m$ is linear density of wire)
$L_1=60 cm =0.6 m$
$\left( n _0\right)_1=\left( n _0\right)_2$
To find: Vibrating length of wire $\left(L_2\right)$
Formula:
$n =\frac{1}{2 L } \sqrt{\frac{ T }{ m }}$
Calculation:
From formula,
$( n )_1=\frac{1}{2 L _1} \sqrt{\frac{ T _1}{ m _1}}$
$( n )_2=\frac{1}{2 L _2} \sqrt{\frac{ T _2}{ m _2}}$
$\because(n)_1=(n)_2$
$ \therefore \frac{1}{2 L _1} \sqrt{\frac{ T _1}{ m _1}}=\frac{1}{2 L _2} \sqrt{\frac{ T _2}{ m _2}}$
$\therefore \frac{1}{4 L _1^2} \frac{ T _1}{ m _1}=\frac{1}{4 L _2^2} \frac{ T _2}{ m _2}$
$\therefore L _2=\sqrt{ L _1^2 \times \frac{ T _2}{ m _2} \times \frac{ m _1}{ T _1}}$
$=\sqrt{(0.6)^2 \times \frac{9 g }{1 g }}$
$=0.6 \times 3$
$= 1 . 8\ m $
The vibrating length of the wire is $1.8\ m$.

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