Question

With an unknown resistance X in the left gap and a resistance of 30  in the right gap of meter-bridge the null point is obtained at 40 cm from the left end of the wire. Find (i) the unknown resistance and (ii) the shift in the position of the null point (a)when the resistances in both the gaps are increased by 15  and (b)when the resistance in each gap is shunted by a resistance of 8  (Ans : (i) 20 , (ii) (a) 3.75 cm towards right, (b) 7.5 cm towards right) ​

Answer 1

Given :

Resistance in the left  gap of the meter bridge = X ohm

Resistance in the right gap of the meter bridge = 30 ohm

Position of null point = 40 cm from left end of the wire

To Find :

(i) The unknown resistance X

(ii) The shift in the position of the null point for the given two cases

(a)when the resistances in both the gaps are increased by 15 ohm  

(b)when the resistance in each gap is shunted by a resistance of 8 ohm

Solution :

The circuit diagram is shown in the attached fig .

(i) Now , we know that in a ,meter bridge :

$\frac{R_1}{R_2} = \frac{l_1}{l_2}$

So putting the given values here we get :

$\frac{X}{30}= \frac{40}{60}$    ( since the meter bridge wire is of 100 cm )

So, $X = \frac{40}{60} \times 30$

          = 20 ohm

Hence , the unknown resistance is 20 ohm .

(ii) . (a)

When each resistance is increased by 15 ohm , we have :

$\frac{R_1}{R_2} =\frac{l_1}{l_2}$

Or, $\frac{X + 15 }{30 + 15 }= \frac{l_1}{100-l_1}$

Or, $\frac{35}{45 } = \frac{l_1}{100-l_1}$

Or, $3500 - 35l_1 = 45l_1$

Or , $l_1 = \frac{3500}{80}$

So, $l_1 = 43.75$

So , shift = (43.75 - 40  ) cm = 3.75 cm

Hence , after increasing both the resistances by 15 ohm the shift in position of null point is 3 75 cm .

(ii) . (b)

Now , when both resistances with 8 ohm , we have :

$R_1 = \frac{X \times 8 }{X + 8} =\frac{160}{28}$

And , $R_2= \frac{30 \times 8 }{30 + 8} = \frac{240}{38}$

So now again  , $\frac{R_1}{R_2} =\frac{l_1}{l_2}$

Or, $\frac{(\frac{160}{28})}{(\frac{240}{38})} = \frac{l_1}{100-l_1}$

Or, $\frac{160 \times 38}{28 \times 240 }= \frac{l_1}{100- l_1}$

Or , $\frac{6080}{6720} = \frac{l_1}{100-l_1}$

Or, $0.9 (100 - l_1) = l_1$

Or , $l_1 = \frac{90}{1.9}$

           = 47.36 cm

So , shift = (47.36- 40 ) cm

              = 7.36 cm

Hence , after shunting both the resistors by 8 ohm the shift in the position of the null point is 7.36 cm .