The value of tension in a long thin metal wir | vedclass

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(A) According to Hooke's Law,the tension $T$ in a wire is related to its extension by $T = k(l - l_{0})$,where $k$ is the force constant,$l$ is the st

Vedclass · Accepted Answer

$\frac{T_{1} l_{2}-T_{2} l_{1}}{T_{1}-T_{2}}$

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(A) According to Hooke's Law,the tension $T$ in a wire is related to its extension by $T = k(l - l_{0})$,where $k$ is the force constant,$l$ is the stretched length,and $l_{0}$ is the actual (natural) length.For tension $T_{1}$,the length is $l_{1}$: $T_{1} = k(l_{1} - l_{0})$  --- $(i)$For tension $T_{2}$,the length is $l_{2}$: $T_{2} = k(l_{2} - l_{0})$  --- $(ii)$Dividing equation $(i)$ by equation $(ii)$:$\frac{T_{1}}{T_{2}} = \frac{l_{1} - l_{0}}{l_{2} - l_{0}}$Cross-multiplying:$T_{1}(l_{2} - l_{0}) = T_{2}(l_{1} - l_{0})$$T_{1}l_{2} - T_{1}l_{0} = T_{2}l_{1} - T_{2}l_{0}$Rearranging to solve for $l_{0}$:$T_{1}l_{2} - T_{2}l_{1} = T_{1}l_{0} - T_{2}l_{0}$$T_{1}l_{2} - T_{2}l_{1} = l_{0}(T_{1} - T_{2})$$l_{0} = \frac{T_{1}l_{2} - T_{2}l_{1}}{T_{1} - T_{2}}$Thus,the actual length of the wire is $\frac{T_{1}l_{2} - T_{2}l_{1}}{T_{1} - T_{2}}$.

Column-$I$	Column-$II$
$(a)$ Stress is proportional to strain.	$(i)$ Elastic limit
$(b)$ When the load on the wire is removed, the body regains its original dimension.	$(ii)$ Limit of proportionality
	$(iii)$ Plastic deformation

The value of tension in a long thin metal wire has been changed from $T_{1}$ to $T_{2}$. The lengths of the metal wire at two different values of tension $T_{1}$ and $T_{2}$ are $l_{1}$ and $l_{2}$ respectively. The actual length of the metal wire is:

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