When a load of 80 ,N is suspended from a stri | vedclass

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(D) According to Hooke's Law,the extension $\Delta l$ is proportional to the applied force $T$. Let $l_0$ be the natural length and $k$ be the force c

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$10.5$

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(D) According to Hooke's Law,the extension $\Delta l$ is proportional to the applied force $T$. Let $l_0$ be the natural length and $k$ be the force constant of the string.The length $l$ under tension $T$ is given by $l = l_0 + \frac{T}{k}$.For $T_1 = 80 \,N$,$l_1 = 101 \,mm = l_0 + \frac{80}{k}$  --- $(1)$For $T_2 = 100 \,N$,$l_2 = 102 \,mm = l_0 + \frac{100}{k}$ --- $(2)$Subtracting $(1)$ from $(2)$: $102 - 101 = \frac{100-80}{k} \implies 1 = \frac{20}{k} \implies k = 20 \,N/mm$.Substituting $k$ into $(1)$: $101 = l_0 + \frac{80}{20} \implies 101 = l_0 + 4 \implies l_0 = 97 \,mm$.Now,for $T_3 = 160 \,N$,the length $l_3 = l_0 + \frac{T_3}{k} = 97 + \frac{160}{20} = 97 + 8 = 105 \,mm$.Converting to cm: $105 \,mm = 10.5 \,cm$.

When a load of $80 \,N$ is suspended from a string,its length is $101 \,mm$. If a load of $100 \,N$ is suspended,its length is $102 \,mm$. If a load of $160 \,N$ is suspended from it,then the length of the string is (Assume the area of cross-section is unchanged). (in $\,cm$)

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