Question

When a current of $\left( 2.50\pm 0.5 \right)A$ flows through a wire, it develops a potential difference of $\left( 20\pm 1 \right)V$. The resistance of the wire is:
$\begin{aligned} & \left( A \right)\left( 8\pm 2 \right)\Omega \\\ & \left( B \right)\left( 10\pm 3 \right)\Omega \\\ & \left( C \right)\left( 18\pm 4 \right)\Omega \\\ & \left( D \right)\left( 20\pm 6 \right)\Omega \\\ \end{aligned}$

Answer 1

Apply Ohm’s law. Ohm’s law states that the potential difference or voltage across a conductor is directly proportional to the current flowing through it and the constant of proportionality is called resistance. Then by rearranging the equation and substituting the values we will get the resistance. Then calculate the error in resistance using the error in voltage and current. Thus the resistance of the wire will be the sum or difference between the calculated resistance and the error value.

Formula used:
According to Ohm’s law,
$V=IR$
where, V is the potential difference
I is the current
R is the resistance.

Complete step by step answer:
Given that the value of current is $\left( 2.50\pm 0.5 \right)A$ flows through a wire and a potential difference of $\left( 20\pm 1 \right)V$.
According to Ohm’s law,
$V=IR$
where, V is the potential difference
I is the current
R is the resistance.
Rearranging the equation we get,
$R=\dfrac{V}{I}$
Substituting the values we get,
$\begin{aligned} & R=\dfrac{20}{2.5} \\\ & \therefore R=8\Omega \\\ \end{aligned}$
$\begin{aligned} & \Rightarrow \dfrac{\vartriangle R}{R}=\dfrac{\vartriangle V}{V}+\dfrac{\vartriangle I}{I} \\\ & \Rightarrow \dfrac{\vartriangle R}{R}=\dfrac{1}{20}+\dfrac{0.5}{2.5} \\\ \end{aligned}$
$\begin{aligned} & \Rightarrow \dfrac{\vartriangle R}{R}=\dfrac{1}{20}+\dfrac{1}{5} \\\ & \Rightarrow \dfrac{\vartriangle R}{R}=\dfrac{1+4}{20} \\\ & \Rightarrow \dfrac{\vartriangle R}{R}=\dfrac{5}{20} \\\ & \Rightarrow \dfrac{\vartriangle R}{R}=\dfrac{1}{4} \\\ & \therefore \dfrac{\vartriangle R}{R}=0.25 \\\ \end{aligned}$
Then,
$\begin{aligned} & \vartriangle R=0.25\times R \\\ & \Rightarrow \vartriangle R=0.25\times 8 \\\ & \therefore \vartriangle R=2 \\\ \end{aligned}$
Thus the resistance of the wire is,
$\left( R\pm \vartriangle R \right)=(8\pm 2)\Omega$

Therefore, option (A) is correct.

Additional information:
Measurement is the foundation of all experimental science and technology. Thus uncertainty in a measurement is said to be the error. The accuracy of a measurement is a measure of how close the measured value is to the true value of quantity. Precision is the resolution or limit the quantity is measured.

Note:
The systematic errors are those errors that tend to be in one direction, either positive or negative. Instrumental errors arise from the errors due to imperfect design or calibration of measuring instruments. Another error arises due to the imperfection in experimental technique or procedure. And personal error arises due to lack of proper setting of the apparatus or carelessness.