Question

Two resistances are connected in two gaps of a meter bridge. The null point is obtained at 20 cm from zero end. A resistance of 15 is connected in series with smaller of the two. The null point shifts to 40 cm. The smaller resistance used in is ______.

Options

• 3

• 6

• 9

• 12

Answer 1

Two resistances are connected in two gaps of a meter bridge. The null point is obtained at 20 cm from zero end. A resistance of 15 is connected in series with smaller of the two. The null point shifts to 40 cm. The smaller resistance used in is 9.

Explanation:

Let X be the smaller resistance in the metre bridge \[l_{\mathrm{X}}=20\mathrm{cm}\]

\[\therefore\quad l_{\mathrm{R}}=100-20=80\mathrm{cm}\]

As the bridge is balanced,

\[\frac{l_\mathrm{x}}{l_\mathrm{R}}=\frac{\mathrm{X}}{\mathrm{R}}\]

\[\therefore\quad\frac{20}{80}=\frac{\mathrm{X}}{\mathrm{R}}\]

\[\therefore\quad\frac{\mathrm{X}}{\mathrm{R}}=\frac{1}{4}\]

\[\therefore\] R = 4X                        ...(i)

From second condition,

\[\frac{\mathrm{X}+15}{\mathrm{R}}=\frac{40}{100-40}\]

\[\therefore\quad\frac{\mathrm{X}+15}{\mathrm{R}}=\frac{40}{60}\]

\[\therefore\quad\frac{\mathrm{X}+15}{\mathrm{R}}=\frac{2}{3}\]

\[\therefore\] 2R = 3X + 45

\[\therefore\quad\mathrm{R}=\frac{3\mathrm{X}+45}{2}\quad....(\mathrm{ii})\]

Equating (i) and (ii) we get,

\[\frac{3\mathrm{X}+45}{2}=4\mathrm{X}\]

\[\therefore\] 8X = 3X + 45

\[5X=45\Rightarrow X=9\Omega\]