A wire of variable mass per unit length mu = | vedclass

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Question

${\rm{m}} = \int_0^x {{\mu _0}} {\rm{xdx}} = \frac{{{\mu _0}{{\rm{x}}^2}}}{2}$

$\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{\mu_{0} \mathr

Vedclass · Accepted Answer

$\sqrt {\frac{{8\,{l_0}}}{g}} $

Vedclass · Answer

${m{m}} = \int_0^x {{\mu _0}} {m{xdx}} = \frac{{{\mu _0}{{m{x}}^2}}}{2}$

$\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mu}}=\sqrt{\frac{\mu_{0} \mathrm{x}^{2} \mathrm{g}}{2. \mu_{0} \mathrm{x}}}$

$v=\sqrt{\frac{x g}{2}}$

$a=\frac{v d v}{d x}=\sqrt{\frac{x g}{2}} \sqrt{\frac{g}{2}} \cdot \frac{1}{2 \sqrt{x}}=\frac{g}{4} \,m / s^{2}$

${l_0} = \frac{1}{2} 	imes \frac{{m{g}}}{4}{{m{t}}^2} \Rightarrow {m{t}} = \sqrt {\frac{{8{l_0}}}{{m{g}}}} $

A wire of variable mass per unit length $\mu = \mu _0x$ , is hanging from the ceiling as shown in figure. The length of wire is $l_0$ . A small transverse disturbance is produced at its lower end. Find the time after which the disturbance will reach to the other ends

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