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Standard XII

Physics

Galvanometer

In a potentio...

Question

In a potentiometer experiment when a battery of emf $2 V$ is included in the secondary circuit the balancing point is at $500 cm$ . The balancing length(in $cm$ ) from the same end when a cadmium cell of emf $1.018 V$ is connected to the second circuit is

254.500

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254.5

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

254.50

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Solution

$E_{1} = 2 V, E_{2} = 1.018 V$

$l_{1} = 500 cm, l_{2} = ?$

$E \propto l$

$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow l_{2} = \frac{E_{2}}{E_{1}} l_{1}$

$l_{2} = \frac{1.018}{2} \times 500 = 0.509 \times 500 = 254.5 cm$

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In a potentiometer experiment when a battery of emf 2 V is included in the secondary circuit the balancing point is at 500 cm. The balancing length(in cm) from the same end when a cadmium cell of emf 1.018 V is connected to the second circuit is

E1=2 V,E2=1.018 V l1=500 cm,l2=? E∝l E1E2=l1l2⇒l2=E2E1l1 l2=1.0182×500=0.509×500=254.5 cm

In a potentiometer experiment when a battery of emf $2 V$ is included in the secondary circuit the balancing point is at $500 cm$ . The balancing length(in $cm$ ) from the same end when a cadmium cell of emf $1.018 V$ is connected to the second circuit is

$E_{1} = 2 V, E_{2} = 1.018 V$

$l_{1} = 500 cm, l_{2} = ?$

$E \propto l$

$\frac{E_{1}}{E_{2}} = \frac{l_{1}}{l_{2}} \Rightarrow l_{2} = \frac{E_{2}}{E_{1}} l_{1}$

$l_{2} = \frac{1.018}{2} \times 500 = 0.509 \times 500 = 254.5 cm$