In a practical wheat stone bridge circuit as shown, when one more resistance of `100 Omega` is connected in parallel with unknown resistance 'x', then ratio `l_1//l_2` become '2'`l_1` is balance length , AB is a uniform wire. Then value of 'x' must be :

The resistance in two arms of the meter bridge is 10 Omega and x respectively. The balance length is l_1 from left. Now the resistance of 10 Omega is shunted with another resistor of resistance equal to unknown resistance x then new balance point is 0.75l_1 The value of resistance x is

A potentiometer wire of length L and a resistance r are connected in series with a battery of emf E_(0) and resistance 'r_(1)' . An unknown emf E is balanced at a length 'l' of the potentiometer wire. The emf E will be given by

In a meter bridge, the value of resistance in the resistance box is 10Omega . The balancing length is l_(1)=55 cm . Find the value of dunknown resistance.

In the circuit a meter bridge a shown in the balanced state .The meter bridge wire has a resistance of 1 Omega //cm . Calculate the unknown resistance Y and the current drawn from the battery of negiligible internal resistance