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Standard XII

Physics

Meter Bridge

In a potentio...

Question

In a potentiometer of one metre length, an unknown emf voltage source is balanced at $60 c m$ length of potentiometer wire, while a $3 V$ battery is balanced at $45 c m$ length. Then, the emf of the unknown voltage source is:

$3 V$

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$2.25 V$

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$4 V$

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$4.5 V$

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Solution

The correct option is C
$4 V$

We know,
emf $\propto$ balanced length:

$E_{1} = k l_{1} E_{2} = k l_{2} E_{2} = 3 V l_{1} = 60 c m, l_{2} = 45 c m \frac{E}{3} = \frac{k \times 60}{k \times 45} E = 4 V$

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Q. In a potention meter of one metre length an unknown emf voltage source is balanced at 60 cm length of potention meter wire, while a 3v battery is balanced at 45 cm length. Then, the emf of the unknown voltage source is :

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In a potentiometer of one metre length, an unknown emf voltage source is balanced at 60 cm length of potentiometer wire, while a 3 V battery is balanced at 45 cm length. Then, the emf of the unknown voltage source is:

The correct option is C 4 V We know, emf ∝ balanced length: E1=kl1E2=kl2E2=3 Vl1=60 cm,l2=45 cmE3=k×60k×45E=4 V

In a potentiometer of one metre length, an unknown emf voltage source is balanced at $60 c m$ length of potentiometer wire, while a $3 V$ battery is balanced at $45 c m$ length. Then, the emf of the unknown voltage source is:

The correct option is C
$4 V$

We know,
emf $\propto$ balanced length:

$E_{1} = k l_{1} E_{2} = k l_{2} E_{2} = 3 V l_{1} = 60 c m, l_{2} = 45 c m \frac{E}{3} = \frac{k \times 60}{k \times 45} E = 4 V$