Question

When 2 resistances P and Q are kept in the left and right gaps of a meter bridge, the null point is obtained at 60 cm. If P is shunted by a resistance equal to half of its value, the shift in null point is

Answer 1

Answer:

Shift of the balance point is by 26.7 cm towards left

Explanation:

As we know that in meter bridge when null deflection point comes then the ratio of resistance of two wires is same as the ratio of two lengths

So here we have

$\frac{R_p}{R_q} = \frac{60}{100 - 60}$

now we shunted the resistance P with another resistance of half of its value

so we will have

$\frac{1}{R} = \frac{1}{R_p} + \frac{2}{R_p}$

$R = \frac{R_p}{3}$

now again we have

$\frac{R_p/3}{R_q} = \frac{L}{100 - L}$

now from first equation

$\frac{1}{3}(\frac{60}{40}) = \frac{L}{100 - L}$

so we have

$2L = 100 - L$

so we have

$L = 33.33 cm$

so the shift of balance point is by

$\Delta x = 60 - 33.33 = 26.7 cm$

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Topic : Meter Bridge

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