
The length of a rubber cord is ${{l}_{1}}$ meters when the tension in it is 4N and ${{l}_{2}}$ meters when the tension is 5N. Then the length in meters when the tension is 9N is
A. $3{{l}_{2}}+4{{l}_{1}}$
B. $3{{l}_{2}}+2{{l}_{1}}$
C. $5{{l}_{2}}-4{{l}_{1}}$
D. $3{{l}_{2}}-2{{l}_{1}}$
Answer
555k+ views
Hint: As a first step, you could apply the expression for Young’s modulus for all the three cases. Thus we get three equations in terms of${{l}_{1}},{{l}_{2}},{{l}_{3}}$. Now, various combinations of this could result in two equations. We could solve these equations to get the value of ${{l}_{3}}$ in terms of ${{l}_{1}}$ and ${{l}_{2}}$.
Formula used:
Expression for Young’s modulus, $Y=\dfrac{FL}{A\Delta L}$
Complete answer:
In the question, we are said that the length of a rubber cord becomes ${{l}_{1}}$ when the tension in the cord is 4N and it becomes ${{l}_{2}}$ when the tension becomes 5N. We are asked to find the length of the same cord when the tension becomes 9N.
For longitudinal extension of wire, we have the Young’s modulus which is given by,
$Y=\dfrac{FL}{A\Delta L}$
Where, L is the original length of the rubber cord and $\Delta L$ is the change in length due to the tension in it. Also, we know that the young’s modulus is constant for the same material.
For a tension of 4N we have,
$Y=\dfrac{4\times L}{A\left( {{l}_{1}}-L \right)}$ …………………………………… (1)
For a tension of 5N we have,
$Y=\dfrac{5\times L}{A\left( {{l}_{2}}-L \right)}$……………………………………. (2)
For a tension of 9N let the resultant length be${{l}_{3}}$, then,
$Y=\dfrac{9\times L}{A\left( {{l}_{3}}-L \right)}$……………………………………… (3)
Combining (1) and (2), we have,
$\dfrac{9\times L}{A\left( {{l}_{3}}-L \right)}=\dfrac{4\times L}{A\left( {{l}_{1}}-L \right)}$
$\Rightarrow 9\left( {{l}_{1}}-L \right)=4\left( {{l}_{3}}-L \right)$
$\Rightarrow 4{{l}_{3}}+5L=9{{l}_{1}}$
Multiplying both sides by 4 we get,
$16{{l}_{3}}+20L=36{{l}_{1}}$ …………………………………………. (4)
Similarly, combining (2) and (3), we have,
$\dfrac{9\times L}{A\left( {{l}_{3}}-L \right)}=\dfrac{5\times L}{A\left( {{l}_{2}}-L \right)}$
$\Rightarrow 9\left( {{l}_{2}}-L \right)=5\left( {{l}_{3}}-L \right)$
$\Rightarrow 5{{l}_{3}}+4L=9{{l}_{2}}$
Multiplying both sides by 5 we get,
$25{{l}_{3}}+20L=45{{l}_{2}}$ …………………………………………. (5)
Subtracting (4) from (5),
$9{{l}_{3}}=45{{l}_{2}}-36{{l}_{1}}$
$\Rightarrow {{l}_{3}}=\dfrac{45}{9}{{l}_{2}}-\dfrac{36}{9}{{l}_{1}}$
$\therefore {{l}_{3}}=5{{l}_{2}}-4{{l}_{1}}$
Therefore, we found the length in meters when the tension is 9N to be $5{{l}_{2}}-4{{l}_{1}}$.
Hence, option (C) is found to be the correct answer.
Note:
Young’s modulus by definition is the ratio of longitudinal stress to that of longitudinal strain. It actually defines the tensile stiffness of a solid material. The tensile stiffness of a solid material is very different from strength, geometric stiffness, hardness and toughness of a solid material and hence, we shouldn’t get confused with these properties.
Formula used:
Expression for Young’s modulus, $Y=\dfrac{FL}{A\Delta L}$
Complete answer:
In the question, we are said that the length of a rubber cord becomes ${{l}_{1}}$ when the tension in the cord is 4N and it becomes ${{l}_{2}}$ when the tension becomes 5N. We are asked to find the length of the same cord when the tension becomes 9N.
For longitudinal extension of wire, we have the Young’s modulus which is given by,
$Y=\dfrac{FL}{A\Delta L}$
Where, L is the original length of the rubber cord and $\Delta L$ is the change in length due to the tension in it. Also, we know that the young’s modulus is constant for the same material.
For a tension of 4N we have,
$Y=\dfrac{4\times L}{A\left( {{l}_{1}}-L \right)}$ …………………………………… (1)
For a tension of 5N we have,
$Y=\dfrac{5\times L}{A\left( {{l}_{2}}-L \right)}$……………………………………. (2)
For a tension of 9N let the resultant length be${{l}_{3}}$, then,
$Y=\dfrac{9\times L}{A\left( {{l}_{3}}-L \right)}$……………………………………… (3)
Combining (1) and (2), we have,
$\dfrac{9\times L}{A\left( {{l}_{3}}-L \right)}=\dfrac{4\times L}{A\left( {{l}_{1}}-L \right)}$
$\Rightarrow 9\left( {{l}_{1}}-L \right)=4\left( {{l}_{3}}-L \right)$
$\Rightarrow 4{{l}_{3}}+5L=9{{l}_{1}}$
Multiplying both sides by 4 we get,
$16{{l}_{3}}+20L=36{{l}_{1}}$ …………………………………………. (4)
Similarly, combining (2) and (3), we have,
$\dfrac{9\times L}{A\left( {{l}_{3}}-L \right)}=\dfrac{5\times L}{A\left( {{l}_{2}}-L \right)}$
$\Rightarrow 9\left( {{l}_{2}}-L \right)=5\left( {{l}_{3}}-L \right)$
$\Rightarrow 5{{l}_{3}}+4L=9{{l}_{2}}$
Multiplying both sides by 5 we get,
$25{{l}_{3}}+20L=45{{l}_{2}}$ …………………………………………. (5)
Subtracting (4) from (5),
$9{{l}_{3}}=45{{l}_{2}}-36{{l}_{1}}$
$\Rightarrow {{l}_{3}}=\dfrac{45}{9}{{l}_{2}}-\dfrac{36}{9}{{l}_{1}}$
$\therefore {{l}_{3}}=5{{l}_{2}}-4{{l}_{1}}$
Therefore, we found the length in meters when the tension is 9N to be $5{{l}_{2}}-4{{l}_{1}}$.
Hence, option (C) is found to be the correct answer.
Note:
Young’s modulus by definition is the ratio of longitudinal stress to that of longitudinal strain. It actually defines the tensile stiffness of a solid material. The tensile stiffness of a solid material is very different from strength, geometric stiffness, hardness and toughness of a solid material and hence, we shouldn’t get confused with these properties.
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