Answer
Verified
430.2k+ views
Hint: We know that tension on a string is a scalar quantity and is similar to the force which is acting on the string. Since the tension on the string due to the respective length is given, we can find the young’s modulus of the string and use it to calculate the natural length of the wire.
Formula: $Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$
Complete answer:
We know that any bulk material experiences bulk modulus, which gives the elasticity of the material. It is the measure of how strong or weak is any given substance when subjected to some force or tension. We know that the elastic moduli or the Young’s modulus of the material is defined as the ratio of tensile or compressive stress to the longitudinal strain.
i.e. $Y=\dfrac{stress}{strain}$, where stress is given as the force per unit area i.e.$stress=\dfrac{force}{area}$ and strain is given as the ratio of change in size or shape to the original shape or size i.e. $strain=\dfrac{change\; in\; shape}{original\; in\;shape}$.
Then $Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$.Also, Young’s modulus is a constant for any given material.
Here, it is given that the length $l_{1}$ experiences a tension $T_{1}$, and the length $l_{2}$ experiences a tension $T_{2}$, respectively. Let the natural length or the actual length of the wire be $L$, and let $A$ be the cross-sectional area of the given wire.
Then we can say that the young’s modulus due to $T_{1}$ as $Y=\dfrac{T_{1}\times L}{A(l_{1}-L)}$, similarly, the young’s modulus due to $T_{2}$ as $Y=\dfrac{T_{2}\times L}{A(l_{2}-L)}$.
Since the Young’s modulus is a constant for any given material, we can equate the above two equations.
$\implies \dfrac{T_{1}\times L}{A(l_{1}-L)} =\dfrac{T_{2}\times L}{A(l_{2}-L)}$.
$\implies\dfrac{T_{1}}{A(l_{1}-L)} =\dfrac{T_{2}}{A(l_{2}-L)}$.
$\implies T_{2}\times A(l_{1}-L) =T_{1}\times A(l_{2}-L)$.
$\implies T_{2}l_{1}-T_{1}\times l_{2}=T_{2}L-T_{1}L$.
$\therefore L=\dfrac{T_{2}l_{1}-T_{1}\times l_{2}}{T_{2}-T_{1}}$
Hence the correct option is \[C.\dfrac{{{l}_{1}}{{T}_{2}}-{{l}_{2}}{{T}_{1}}}{{{T}_{2}}-{{T}_{1}}}\]
Note:
Any bulk material experiences bulk modulus, which is related to the elasticity of the material. This is a very easy sum, provided one knows the formula of young’s modulus. Here, the length and the respective tensions area given, using which the natural length of the wire is found.
Formula: $Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$
Complete answer:
We know that any bulk material experiences bulk modulus, which gives the elasticity of the material. It is the measure of how strong or weak is any given substance when subjected to some force or tension. We know that the elastic moduli or the Young’s modulus of the material is defined as the ratio of tensile or compressive stress to the longitudinal strain.
i.e. $Y=\dfrac{stress}{strain}$, where stress is given as the force per unit area i.e.$stress=\dfrac{force}{area}$ and strain is given as the ratio of change in size or shape to the original shape or size i.e. $strain=\dfrac{change\; in\; shape}{original\; in\;shape}$.
Then $Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$.Also, Young’s modulus is a constant for any given material.
Here, it is given that the length $l_{1}$ experiences a tension $T_{1}$, and the length $l_{2}$ experiences a tension $T_{2}$, respectively. Let the natural length or the actual length of the wire be $L$, and let $A$ be the cross-sectional area of the given wire.
Then we can say that the young’s modulus due to $T_{1}$ as $Y=\dfrac{T_{1}\times L}{A(l_{1}-L)}$, similarly, the young’s modulus due to $T_{2}$ as $Y=\dfrac{T_{2}\times L}{A(l_{2}-L)}$.
Since the Young’s modulus is a constant for any given material, we can equate the above two equations.
$\implies \dfrac{T_{1}\times L}{A(l_{1}-L)} =\dfrac{T_{2}\times L}{A(l_{2}-L)}$.
$\implies\dfrac{T_{1}}{A(l_{1}-L)} =\dfrac{T_{2}}{A(l_{2}-L)}$.
$\implies T_{2}\times A(l_{1}-L) =T_{1}\times A(l_{2}-L)$.
$\implies T_{2}l_{1}-T_{1}\times l_{2}=T_{2}L-T_{1}L$.
$\therefore L=\dfrac{T_{2}l_{1}-T_{1}\times l_{2}}{T_{2}-T_{1}}$
Hence the correct option is \[C.\dfrac{{{l}_{1}}{{T}_{2}}-{{l}_{2}}{{T}_{1}}}{{{T}_{2}}-{{T}_{1}}}\]
Note:
Any bulk material experiences bulk modulus, which is related to the elasticity of the material. This is a very easy sum, provided one knows the formula of young’s modulus. Here, the length and the respective tensions area given, using which the natural length of the wire is found.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE