When a resistor of `5 Omega` is connected across a cell, its terminal potential differnce is balanced by 140 cm of potentiometer wire and when a resistance of `8 Omega` is connected across the cell, the terminal potential difference is balanced by 160 cm of the potentiometer wire. Find the internal resistance of the cell.

When a resister of 10Omega is connected across a cell, its terminal potential difference is balanced by 150cm of potentiometer wire and when 20Omega resistance is connected across the cell, terminal potential difference is balanced by 175cm of the same potentiometer wire. Find the balancing length when the cell is in open circuit and also find the internal resistance of the cell.

When a resistor of 5 Omega resistance is connected across a cell,its terminal potential difference is balanced by 150 cm of the wire and when a resistor of 10 Omega resistance is connected across cell, the terminal p.d. is balanced by 175 cm of same potentiometer wire. Find the balancing length when the cell is in open circuit and the internal resistance of acell.

When a resistance of 12 ohm is connected across a cell, its terminal potential difference is balanced by 120 cm length of potentiometer wire. When the resistance of 18 ohm is connected across the same cell, the balancing length is 150 cm. Find the balancing length when the cell is in open circuit. Also calculate the internal resistance of the cell.

In fig. battery E is balanced on 55cm length of potentiometer wire but when a resistance of 10Omega is connected in parallel with the battery then it balances on 50cm length of the potentiometer wire then internal resistance r of the battery is:-

In fig. battery E is balanced on 55cm length of potentiometer wire but when a resistance of 10Omega is connected in parallel with the battery then it balances on 50cm length of the potentiometer wire then internal resistance r of the battery is:-

In the given figure, battery E is balanced on 55 cm length of potentiometer wire but when a resistance of 10Omega is connected in parallel with the battery then it balances on 50 cm length of the potentiometer wire then internal resistance r of the battery is

In fig. battery E is balanced on 55cm length of potentiometer wire but when a resistance of 10Omega is connected in parallel with the battery then it balances on 50cm length of the potentiometer wire then internal resistance r of the battery is:-

The e.m.f. of a cell is balanced by fall of potential along 281 cm of potentiometer wire. If a resistance of 8 Omega is connected across the cell , the balancing length is found to be 151 cm. Calculate internal resistance of a cell.

The e.m.f. of a standard cell balances across 150 cm length of a wire of potentiometer. When a resistance of 2Omega is connected as a shunt with the cell, the balance point is obtained at 100cm . The internal resistance of the cell is