The figure shows experimental set up of a meter bridge. When the two unknown resistances X and Y are inserted, the null point D is obtained 40 cm from the end A. When a resistance of 10 Ω is connected in series with X, the null point shifts by 10 cm. Find the position of the null point when the 10 Ω resistance is instead connected in series with resistance ‘Y’. Determine the values of the resistances X and Y.CBSE DELHI - SET 1 2009
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$\frac{X}{Y} =\frac{40}{60}\Rightarrow\text{X} = \frac{2}{3}\text{Y} $......................................(i}$\frac{\text{X} + 10}{\text{Y}} = \frac{50}{50}$....................................................(ii)
$\Rightarrow\text{X} +10 = \text{Y}$
From (i) and (ii)
$\frac{2}{3}\text{Y}+ 10 =\text{Y} $ $\Rightarrow \text{Y} = 30\Omega$
As $\text{X} =\frac{2}{3}\text{Y}$ $\Rightarrow 20 \Omega$
When 10Ω resistor is connected in series with resistor Y, we have
$\frac{\text{X}}{\text{Y} +10} = \frac{l}{100 - l}$ $\Rightarrow \frac{20}{40} = \frac{l}{10 - l}$
$\Rightarrow2l = 100 - l$
$3l = 100$
$l = \frac{100}{3} = 33.33\text{cm }$.
$\Rightarrow\text{X} +10 = \text{Y}$
From (i) and (ii)
$\frac{2}{3}\text{Y}+ 10 =\text{Y} $ $\Rightarrow \text{Y} = 30\Omega$
As $\text{X} =\frac{2}{3}\text{Y}$ $\Rightarrow 20 \Omega$
When 10Ω resistor is connected in series with resistor Y, we have
$\frac{\text{X}}{\text{Y} +10} = \frac{l}{100 - l}$ $\Rightarrow \frac{20}{40} = \frac{l}{10 - l}$
$\Rightarrow2l = 100 - l$
$3l = 100$
$l = \frac{100}{3} = 33.33\text{cm }$.
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