Question

If a $0.1\% $ increase in length due to stretching, the percentage increase in its resistance will be
A. $0.2\% $
B. $2\% $
C. $1\% $
D. $0.1\% $

Answer 1

Hint: Here we know the percentage increase in the length. When a substance is stretched its length increases but the volume remains the same. Using this concept we will find the percentage increase in resistance.

Formula Used:
\[{A_f}{l_f} = {A_i}{l_i}\]
The formula of resistance is,
$\Rightarrow R = \dfrac{{\rho l}}{A}$
The expression of change in resistance is,
$\Rightarrow \vartriangle R = \dfrac{{{R_f} - {R_i}}}{{{R_i}}} \times 100$
Where \[A\] is the area of cross section, $l$ is the length, $R$ is the resistance and $\rho $ is the resistivity of the substance.

Complete step by step solution:
Given: ${l_f} = \left( {l + 0.1\% \,l} \right)$ where ${l_i} = l$ is the initial length and ${l_f}$ is the final length.
Solving this we get, ${l_f} = \left( {l + \dfrac{{0.1\,l}}{{100}}} \right)$
That is, \[{l_f} = \left( {\dfrac{{100.1\,l}}{{100}}} \right)\] ...(1)
Now, we know that when a substance is stretched, its length increases but the volume is conserved.
This implies that, \[{A_f}{l_f} = {A_i}{l_i}\]
Let ${A_i} = A$ be the initial area of the cross section.
We get, ${A_f} = \left( {\dfrac{l}{{{l_f}}}A} \right)$...(2)
Substituting equation (1) in equation (2), we get
${A_f} = \dfrac{{100\,l}}{{100.1\,l}}A$
Solving this we get,
${A_f} = \dfrac{{100}}{{100.1}}A$...(3)

Now we know the relationship between resistance, resistivity, area of cross section and length of the substance.
It is $R = \dfrac{{\rho l}}{A}$ ...(4)
This is the initial resistance. Also, the value of resistivity does not change on stretching. It remains the same for a given substance. Final resistance is,
${R_f} = \dfrac{{\rho {l_f}}}{{{A_f}}}$...(5)
Substituting equations (1) and (3) in equation (5) we get, ${R_f} = \dfrac{{\rho \left( {\dfrac{{100.1\,l}}{{100}}} \right)}}{{\left( {\dfrac{{100}}{{100.1}}A} \right)}}$ ...(6)
Solving this we get, ${R_f} = {\left( {\dfrac{{100.1}}{{100}}} \right)^2}\dfrac{{\rho l}}{A}$
Substitute equation (4) in equation (6), we get
${R_f} = {\left( {\dfrac{{100.1}}{{100}}} \right)^2}R$

We know that to find percentage we use the following formula:
$\vartriangle R = \dfrac{{{R_f} - {R_i}}}{{{R_i}}} \times 100 \\ $
Where $\vartriangle R$ is the percentage change in the resistance.
We get,
$\vartriangle R = \dfrac{{\left[ {{{\left( {\dfrac{{100.1}}{{100}}} \right)}^2}R} \right] - R}}{R} \times 100 \\ $
Solving further,
$\vartriangle R = \left[ {{{\left( {\dfrac{{100.1}}{{100}}} \right)}^2} - 1} \right] \times 100 \\ $
We can write $100.1 = 100 + 0.1$ , we get;
$\vartriangle R = \left[ {{{\left( {\dfrac{{100 + 0.1}}{{100}}} \right)}^2} - 1} \right] \times 100 \\ $
Use the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ , we get
\[\vartriangle R = \left[ {\dfrac{{{{\left( {100} \right)}^2} + {{\left( {0.1} \right)}^2} + \left( {2 \times 100 \times 0.1} \right) - {{\left( {100} \right)}^2}}}{{{{\left( {100} \right)}^2}}}} \right] \times 100 \\ \]
Now ${\left( {0.1} \right)^2}$ can be ignored in the numerator because it is very small as compared to the other values in the addition. We get,
\[\vartriangle R = \left[ {\dfrac{{\left( {2 \times 100 \times 0.1} \right)}}{{{{\left( {100} \right)}^2}}}} \right] \times 100 =0.2\%\]

Hence, option A is the answer.

Note: While solving this question, to find the percentage change in the area of cross section of the substance because when a substance is stretched its area of cross section is not conserved.