When a load of $80 \mathrm{~N}$ is suspended from a string, its length is $101 \mathrm{~mm}$. If a load of $100 \mathrm{~N}$ is suspended, its length is $102 \mathrm{~mm}$. If a load of $160 \mathrm{~N}$ is suspended from it, then length of the string is (Assume the area of crosssection unchanged)

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Natural length }(l)=\frac{l_1 T_2-l_2 T_1}{T_2-T_1} \ & =\frac{101 	imes 100-102 	imes 80}{(100-80)}\end{aligned}$ $\begin{aligned} & =\frac{101 	imes 100-102 	imes 80}{20} \ & =\frac{101 	imes 100}{20}-\frac{102 	imes 80}{20} \ & =505-408=97 \mathrm{~cm} \ & 	ext { Again } \frac{T_1}{T_2}=\frac{4}{l_3-97} \ & \frac{80}{160}=\frac{4}{l_3-97} 	ext { or, } l_3-97=8 \ & 	herefore \quad l_3=105 \mathrm{~mm} 	ext { or } 10.5 \mathrm{~cm}\end{aligned}$

When a load of $80 \mathrm{~N}$ is suspended from a string, its length is $101 \mathrm{~mm}$. If a load of…

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