
A wire is loaded by 6kg at its one end, the increase in length is 12mm. If the radius of the wire is doubled and all other magnitudes are unchanged, then increase in length will be
A. 6mm
B. 3mm
C. 24mm
D. 48mm
Answer
163.2k+ views
Hint: To solve this question, we need to use the formula for the young’s modulus of a string in terms of its geometrical parameters. Then, putting the values given in the question, we can get the required value of the increase in length of the wire.
Formula used:
The formula which has been used to solve this question is given by
$Y = \dfrac{{FL}}{{Al}}$
Here Y is the young’s modulus of a string of length L and area of cross section A , F is the force applied on it due to which its length gets changed by l .
Complete answer:
We know that the Young’s modulus for a wire can be written as
$Y = \dfrac{{FL}}{{Al}}$
$ \Rightarrow l = \dfrac{{FL}}{{\pi {R^2}Y}}$
Since all other magnitudes are unchanged,
$ \Rightarrow l \propto \dfrac{1}{{{R^2}}}$
Then we can say that
$ \Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{R_2}^2}}{{{R_1}^2}}$
Given that ${R_2} = 2{R_1}$and ${l_1} = 12mm$
Then
$ \Rightarrow \dfrac{{12}}{{{l_2}}} = \dfrac{{4{R_1}^2}}{{{R_1}^2}}$
Now we can say that ${l_2} = 3mm$. So, the correct option is B.
Note: We should not forget to convert the values of the quantities given in the question into their respective SI units. In this question the length of the wire was given in millimetres, which is not an SI unit. So, it was supposed to be converted into meters.
Formula used:
The formula which has been used to solve this question is given by
$Y = \dfrac{{FL}}{{Al}}$
Here Y is the young’s modulus of a string of length L and area of cross section A , F is the force applied on it due to which its length gets changed by l .
Complete answer:
We know that the Young’s modulus for a wire can be written as
$Y = \dfrac{{FL}}{{Al}}$
$ \Rightarrow l = \dfrac{{FL}}{{\pi {R^2}Y}}$
Since all other magnitudes are unchanged,
$ \Rightarrow l \propto \dfrac{1}{{{R^2}}}$
Then we can say that
$ \Rightarrow \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{R_2}^2}}{{{R_1}^2}}$
Given that ${R_2} = 2{R_1}$and ${l_1} = 12mm$
Then
$ \Rightarrow \dfrac{{12}}{{{l_2}}} = \dfrac{{4{R_1}^2}}{{{R_1}^2}}$
Now we can say that ${l_2} = 3mm$. So, the correct option is B.
Note: We should not forget to convert the values of the quantities given in the question into their respective SI units. In this question the length of the wire was given in millimetres, which is not an SI unit. So, it was supposed to be converted into meters.
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