A length of $150 c m$ of a potentiometer wire is required to balance the emf of a cell. When a $5 Ω$ resistor is connected across the terminals of the cell, the length required for balance is $125 c m$ . The internal resistance of the cell is

1 o h m

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1.1 o h m

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1.2 o h m

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1.3 o h m

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Solution

The correct option is D $1 o h m$
Given,
Length of the potentiometer wire $l_{1} = 150 c m$
Resistance $R = 5 Ω$
balancing length is obtained at $l_{2} = 125 c m$
Internal resistance $r = ?$
we have,

$r = R [\frac{l_{1}}{l_{2}} - 1]$

$= 5 [\frac{150}{125} - 1]$

$= 5 [\frac{25}{125}]$

$= 1 Ω$

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A length of 150 cm of a potentiometer wire is required to balance the emf of a cell. When a 5Ω resistor is connected across the terminals of the cell, the length required for balance is 125 cm. The internal resistance of the cell is

The correct option is D 1 ohmGiven,Length of the potentiometer wirel1=150cmResistance R=5Ωbalancing length is obtained at l2=125cmInternal resistance r=?we have,r=R[l1l2−1]=5[150125−1]=5[25125]=1Ω

A length of $150 c m$ of a potentiometer wire is required to balance the emf of a cell. When a $5 Ω$ resistor is connected across the terminals of the cell, the length required for balance is $125 c m$ . The internal resistance of the cell is

The correct option is D $1 o h m$
Given,
Length of the potentiometer wire $l_{1} = 150 c m$
Resistance $R = 5 Ω$
balancing length is obtained at $l_{2} = 125 c m$
Internal resistance $r = ?$
we have,

$r = R [\frac{l_{1}}{l_{2}} - 1]$

$= 5 [\frac{150}{125} - 1]$

$= 5 [\frac{25}{125}]$

$= 1 Ω$