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Question: The length of elastic string, obeying Hooke's law is \({l_1}\) meters when the tension \[4\,N\] and ...

The length of elastic string, obeying Hooke's law is l1{l_1} meters when the tension 4 N4\,N and l2{l_2} metres when the tension is 5 N5\,N. The length in meters when the tension is 9 N9\,N is :
A. 5l1−4l25{l_1} - 4{l_2}
B. 9l1−8l29{l_1} - 8{l_2}
C. 5l2−4l15{l_2} - 4{l_1}
D. 9l2−8l19{l_2} - 8{l_1}

Explanation

Solution

Stress and strain have a direct proportional relationship up to an elastic limit. The association between stress and strain is explained by Hooke's rule. Hooke's law states that the pressure in a solid is proportional to the applied stress, which must be within the solid's elastic limit.

Complete step by step answer:
The elastic modulus is the proportionality constant in this relationship. The general relationship between stress and strain is linear in the linear limit of low stress values.Hence, by Hooke's law we can state the formula as:
Elastic modulus =stressstrain = \dfrac{{stress}}{{strain}}

Stress is characterised as the force applied to an object that causes it to change shape, while strain is defined as the change in shape of an object caused by stress.
Hence, stress can be calculated as:
stress=FAstress = \dfrac{F}{A} where FF is force and AA is area.
On the other hand strain can be calculated using the following formula:
strain=l1−llstrain = \dfrac{{{l_1} - l}}{l}

Substituting the above values to the main equation we get,
Elastic modulus (γ)=FA×ll1−l(\gamma ) = \dfrac{F}{A} \times \dfrac{l}{{{l_1} - l}}
When 4N4N tension is applied,
γ=4A×ll1−l−(i)\gamma = \dfrac{4}{A} \times \dfrac{l}{{{l_1} - l}} - (i)
When 5 N5\,N tension is applied,
γ=5A×ll2−l−(ii)\gamma = \dfrac{5}{A} \times \dfrac{l}{{{l_2} - l}} - (ii)

Dividing (i)(i) and (ii)(ii) equation, we get,
\Rightarrow \dfrac{{{\gamma } = \dfrac{4}{{{A}}} \times \dfrac{{{l}}}{{{l_1} - l}}}}{{{\gamma } = \dfrac{5}{{{A}}} \times \dfrac{{{l}}}{{{l_2} - l}}}} \\\
⇒l1−ll2−l=45 ⇒5(l1−l)=4(l2−l) ⇒5l1−5l=4l2−4l ⇒5l1−4l2=l \Rightarrow \dfrac{{{l_1} - l}}{{{l_2} - l}} = \dfrac{4}{5} \\\ \Rightarrow 5({l_1} - l) = 4({l_2} - l) \\\ \Rightarrow 5{l_1} - 5l = 4{l_2} - 4l \\\ \Rightarrow 5{l_1} - 4{l_2} = l \\\
Now, for 9 N9\,N tension , we can write,
γ=9A×ll′−l−(iii)\gamma = \dfrac{9}{A} \times \dfrac{l}{{l' - l}} - (iii)
We will now compare (i)(i) and (iii)(iii)
\Rightarrow \dfrac{{{\gamma } = \dfrac{4}{{{A}}} \times \dfrac{{{l}}}{{{l_1} - l}}}}{{{\gamma } = \dfrac{9}{{{A}}} \times \dfrac{{{l}}}{{l' - l}}}} \\\
⇒l1−ll′−l=49 ⇒9(l1−l)=4(l′−l) ⇒9l1−9l=4l′−4l ⇒9l1−5l=4l′ \Rightarrow \dfrac{{{l_1} - l}}{{l' - l}} = \dfrac{4}{9} \\\ \Rightarrow 9({l_1} - l) = 4(l' - l) \\\ \Rightarrow 9{l_1} - 9l = 4l' - 4l \\\ \Rightarrow 9{l_1} - 5l = 4l' \\\
Now , substituting ll value we get,
9l1−5(5l1−4l2)=4l′ ⇒9l1−25l1+20l2=4l′ ⇒−16l1+20l2=4l′ ∴l′=5l2−4l1 9{l_1} - 5(5{l_1} - 4{l_2}) = 4l' \\\ \Rightarrow 9{l_1} - 25{l_1} + 20{l_2} = 4l' \\\ \Rightarrow - 16{l_1} + 20{l_2} = 4l' \\\ \therefore l' = 5{l_2} - 4{l_1} \\\
Hence, the correct option is C.

Note: A strain is a dimensionless quantity with no variable, while stress is a measurable quantity with a unit. The terms "stress" and "strain" come from the Latin words "strictus" and "stringere," respectively, which mean "to draw close" and "to tie firmly." Stress may exist without strain, but strain does not exist in the absence of stress.