Answer
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Hint: In this question, we need to determine the error in the measurement of H following the relation $H = {I^2}Rt$ such that error in the measurement of I, R, and t are 3%, 4%, and 6%, respectively. For this, we will sum up the individual error in H due to each variable.
Complete step by step answer:
The product of the resistance of the coil and square of the current flowing through the coil and the time for which the current is flowing in the coil results in the heat produced by the coil due to the current and the resistance present in the coil. Mathematically, $H = {I^2}Rt$ where ‘I’ is the current flowing through the coil, ‘R’ is the resistance of the coil, and ‘t’ is the time for which the current is flowing.
The error raised by a variable in a measuring instrument is given by $\dfrac{{\vartriangle x}}{x}$.
Here $H = {I^2}Rt$ is the relation where we need to determine the total error occurs in the measuring instrument, so we can write
$
\Rightarrow H = {I^2}Rt \\
\Rightarrow\dfrac{{\vartriangle H}}{H} = 2\left( {\dfrac{{\vartriangle I}}{I}} \right) + \dfrac{{\vartriangle R}}{R} + \dfrac{{\vartriangle t}}{t} - - - - (i) \\
$
Now, according to the question, errors in the measurement of I, R, and t are 3%, 4%, and 6%, respectively. So, substituting the values in the equation (i), we get
$
\Rightarrow\dfrac{{\vartriangle H}}{H} = 2\left( {\dfrac{{\vartriangle I}}{I}} \right) + \dfrac{{\vartriangle R}}{R} + \dfrac{{\vartriangle t}}{t} \\
\Rightarrow\dfrac{{\vartriangle H}}{H}= \left( {2 \times 3} \right) + 4 + 6 \\
\therefore\dfrac{{\vartriangle H}}{H}= 16\% \\
$
Hence, the error in the measurement of H following the relation $H = {I^2}Rt$ such that error in the measurement of I, R, and t is 3%, 4%, and 6%, respectively, is 16%.
Note: Error in the measuring instrument is classified into two parts, i.e., random error and systematic error. It is interesting to note here that the variable which is raised with the positive powers will enhance the error in the resultant measuring constant.
Complete step by step answer:
The product of the resistance of the coil and square of the current flowing through the coil and the time for which the current is flowing in the coil results in the heat produced by the coil due to the current and the resistance present in the coil. Mathematically, $H = {I^2}Rt$ where ‘I’ is the current flowing through the coil, ‘R’ is the resistance of the coil, and ‘t’ is the time for which the current is flowing.
The error raised by a variable in a measuring instrument is given by $\dfrac{{\vartriangle x}}{x}$.
Here $H = {I^2}Rt$ is the relation where we need to determine the total error occurs in the measuring instrument, so we can write
$
\Rightarrow H = {I^2}Rt \\
\Rightarrow\dfrac{{\vartriangle H}}{H} = 2\left( {\dfrac{{\vartriangle I}}{I}} \right) + \dfrac{{\vartriangle R}}{R} + \dfrac{{\vartriangle t}}{t} - - - - (i) \\
$
Now, according to the question, errors in the measurement of I, R, and t are 3%, 4%, and 6%, respectively. So, substituting the values in the equation (i), we get
$
\Rightarrow\dfrac{{\vartriangle H}}{H} = 2\left( {\dfrac{{\vartriangle I}}{I}} \right) + \dfrac{{\vartriangle R}}{R} + \dfrac{{\vartriangle t}}{t} \\
\Rightarrow\dfrac{{\vartriangle H}}{H}= \left( {2 \times 3} \right) + 4 + 6 \\
\therefore\dfrac{{\vartriangle H}}{H}= 16\% \\
$
Hence, the error in the measurement of H following the relation $H = {I^2}Rt$ such that error in the measurement of I, R, and t is 3%, 4%, and 6%, respectively, is 16%.
Note: Error in the measuring instrument is classified into two parts, i.e., random error and systematic error. It is interesting to note here that the variable which is raised with the positive powers will enhance the error in the resultant measuring constant.
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