1)

On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10cm. The resistance of their series combination is 1kΩ. How much was the resistance on the left slot before interchanging the resistances?


A) 990Ω

B) 505Ω

C) 550Ω

D) 910Ω

Answer:

Option C

Explanation:

We have X+Y=1000Ω

48201952_POINT.JPG

Initially,   $\frac{X}{l}=\frac{1000-X}{100-l}$         ..............(i)

 When X and Y  are interchanged, then

482019334_poit.JPG

 $\frac{1000-X}{l-10}=\frac{X}{100-(l-10)}$

OR    $\frac{1000-X}{l-10}=\frac{X}{110-l}$        .........(ii)

 From Eqs. (i) and (ii), we get

$\frac{100-l}{l}=\frac{l-10}{110-l}$

$\left(100-l\right)\left(110-l\right)$ = $\left(l-10\right)l$

$11000-100l -110l+l^{2}=l^{2}-10l$

$\Rightarrow$        $11000=200l$

$\therefore$             $l=55 cm$

Substituting the value of l in Eq. (i), we get

$\frac{X}{55}=\frac{1000-55}{100-55}$

$\Rightarrow$          20X=11000

                                 X= 550Ω