
If voltage \[V=(100\pm 5)V\] and current $I=(10\pm 0.2)A$ , the percentage error in resistance $R$ is
$A.5.2%$
$B.25%$
$C.7%$
$D.10%$
Answer
484.8k+ views
Hint: We will apply the basic formula of voltage in terms of electric current and resistance to get the resistance first and then use the relationship of error and then in the mode of percentage. Percentage error is the difference between the estimated value and true value with the use of percentage.
Formula Used:
We will use the following formulae to get the correct answer:-
$R=\dfrac{V}{I}$ and $\dfrac{\Delta R}{R}=\pm \left( \dfrac{\Delta V}{V}+\dfrac{\Delta I}{I} \right)$
Complete answer:
From the question above we have the following parameters we have:-
Voltage, $V=100V$
Error in voltage, $\Delta V=\pm 5V$
Electric current, $I=10A$
Error in electric current, $\Delta I=\pm 0.2A$
To find the resistance we will use the following:-
$R=\dfrac{V}{I}$ …………. (i)
Putting values in equation $(i)$ we get
$R=\dfrac{100}{10}$
$\Rightarrow R=10\Omega $ …………… $(ii)$
Now, error of resistance we will use the following formula:-
$\dfrac{\Delta R}{R}=\pm \left( \dfrac{\Delta V}{V}+\dfrac{\Delta I}{I} \right)$ ………….. $(iii)$
Putting the values in $(iii)$
$\dfrac{\Delta R}{R}=\pm \left( \dfrac{5}{100}+\dfrac{0.2}{10} \right)$
$\Rightarrow \dfrac{\Delta R}{R}=\pm \left( \dfrac{5+2}{100} \right)$
$\Rightarrow \dfrac{\Delta R}{R}=\pm \dfrac{7}{100}$ ………….. $(iv)$
Now, error in percentage in $R$ will be as follows:-
$\dfrac{\Delta R}{R}\times 100$ ……………… $(v)$
Using the equation $(iv)$ in $(v)$ we get
$\dfrac{\pm 7}{100}\times 100$
$\pm 7%$ .
Therefore, the percentage error in resistance $R$ is $\pm 7%$.
Hence, option $(C)$ is correct.
Additional Information:
We should know the basics of percentage error. Percentage error tells us about how big our errors are when we measure a data during the process of analysis. Small percentage error indicates that our measurement is correct to the actual value while large percentage error indicates that our measurement is not very accurate.
Note:
We should remember that percentage error is always represented in terms of per cent. It should be noted that errors in measurements are unavoidable because of many reasons and factors. It should also be noted that when relative error is expressed as percentage then we get percentage error.
Formula Used:
We will use the following formulae to get the correct answer:-
$R=\dfrac{V}{I}$ and $\dfrac{\Delta R}{R}=\pm \left( \dfrac{\Delta V}{V}+\dfrac{\Delta I}{I} \right)$
Complete answer:
From the question above we have the following parameters we have:-
Voltage, $V=100V$
Error in voltage, $\Delta V=\pm 5V$
Electric current, $I=10A$
Error in electric current, $\Delta I=\pm 0.2A$
To find the resistance we will use the following:-
$R=\dfrac{V}{I}$ …………. (i)
Putting values in equation $(i)$ we get
$R=\dfrac{100}{10}$
$\Rightarrow R=10\Omega $ …………… $(ii)$
Now, error of resistance we will use the following formula:-
$\dfrac{\Delta R}{R}=\pm \left( \dfrac{\Delta V}{V}+\dfrac{\Delta I}{I} \right)$ ………….. $(iii)$
Putting the values in $(iii)$
$\dfrac{\Delta R}{R}=\pm \left( \dfrac{5}{100}+\dfrac{0.2}{10} \right)$
$\Rightarrow \dfrac{\Delta R}{R}=\pm \left( \dfrac{5+2}{100} \right)$
$\Rightarrow \dfrac{\Delta R}{R}=\pm \dfrac{7}{100}$ ………….. $(iv)$
Now, error in percentage in $R$ will be as follows:-
$\dfrac{\Delta R}{R}\times 100$ ……………… $(v)$
Using the equation $(iv)$ in $(v)$ we get
$\dfrac{\pm 7}{100}\times 100$
$\pm 7%$ .
Therefore, the percentage error in resistance $R$ is $\pm 7%$.
Hence, option $(C)$ is correct.
Additional Information:
We should know the basics of percentage error. Percentage error tells us about how big our errors are when we measure a data during the process of analysis. Small percentage error indicates that our measurement is correct to the actual value while large percentage error indicates that our measurement is not very accurate.
Note:
We should remember that percentage error is always represented in terms of per cent. It should be noted that errors in measurements are unavoidable because of many reasons and factors. It should also be noted that when relative error is expressed as percentage then we get percentage error.
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