Resistance in two gaps of a meter bridge are $10$ and $30$ respectively if the resistance are interchanged the balance point shifts by:

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Solution

$S = (\frac{100 - l}{l})$
Intially $30 = (\frac{100 - l}{l}) \times 10 ⟹ l = 25$
Finally $10 = (\frac{100 - l}{l}) \times 30 ⟹ l = 75$
So, shift $= 50 c m$

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Resistance in two gaps of a meter bridge are 10 and 30 respectively if the resistance are interchanged the balance point shifts by:

S=(100−ll)Intially 30=(100−ll)×10⟹l=25Finally 10=(100−ll)×30⟹l=75 So, shift =50cm

Resistance in two gaps of a meter bridge are $10$ and $30$ respectively if the resistance are interchanged the balance point shifts by:

$S = (\frac{100 - l}{l})$
Intially $30 = (\frac{100 - l}{l}) \times 10 ⟹ l = 25$
Finally $10 = (\frac{100 - l}{l}) \times 30 ⟹ l = 75$
So, shift $= 50 c m$