A resistance $\text{R}=4\Omega$ is connected to one of the gaps in a meter bridge, which uses a wire of length 1m. An unknown resistance $\text{x}>4\Omega$ is connected in the other gap as shown in the figure. The balance point is noticed at ‘l’ cm from the positive end of the battery. On interchanging R and X, it is found that the balance point further shifts by 20cm (away from the end A). Neglecting the end correction calculate the value of unknown resistance ‘X’ used.
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From 'meter bridge' formula,
$\frac{\text{R}}{\text{X}}=\frac{\text{l}}{100-\text{l}}$
$\Rightarrow\text{X}=\frac{100-\text{l}}{\text{l}}\text{R}$
Given, $\text{R}=4\Omega$
$\therefore\text{X}=\frac{(100-\text{l})}{\text{l}}\times4\Omega\ \dots\text{(i)}$
On interchanging R and X, the balance point is obtained at a distance (l + 20)cm from end A, so
$\frac{\text{X}}{\text{R}}=\frac{\text{l}+20}{100-(\text{l}+20)}$
$\Rightarrow\text{X}=\frac{\text{l}+20}{80-\text{l}}\times4\Omega\ \dots\text{(ii)}$
Equating (i) and (ii)
$\frac{(100-\text{l})}{\text{l}}\times4=\frac{\text{l}+20}{80-\text{l}}\times4$
Solving we get l = 40cm
$\therefore$ Unknown resistance, $\text{X}=\frac{100-\text{l}}{\text{l}}\times\text{4}\Omega=\frac{100-40}{40}\times4\Omega$
$\Rightarrow\text{X}=6\Omega$
$\frac{\text{R}}{\text{X}}=\frac{\text{l}}{100-\text{l}}$
$\Rightarrow\text{X}=\frac{100-\text{l}}{\text{l}}\text{R}$
Given, $\text{R}=4\Omega$
$\therefore\text{X}=\frac{(100-\text{l})}{\text{l}}\times4\Omega\ \dots\text{(i)}$
On interchanging R and X, the balance point is obtained at a distance (l + 20)cm from end A, so
$\frac{\text{X}}{\text{R}}=\frac{\text{l}+20}{100-(\text{l}+20)}$
$\Rightarrow\text{X}=\frac{\text{l}+20}{80-\text{l}}\times4\Omega\ \dots\text{(ii)}$
Equating (i) and (ii)
$\frac{(100-\text{l})}{\text{l}}\times4=\frac{\text{l}+20}{80-\text{l}}\times4$
Solving we get l = 40cm
$\therefore$ Unknown resistance, $\text{X}=\frac{100-\text{l}}{\text{l}}\times\text{4}\Omega=\frac{100-40}{40}\times4\Omega$
$\Rightarrow\text{X}=6\Omega$
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