A potentiometer wire 10 m long and having -20- - resistance is connected in series with a battery and a resistance R - 2160 - A Daniel cell of e-m-f 1-08 V is balanced across the P-D of the resistance -R- The e-m-f of thermocouple is balanced across 300 cm of the wire- The e-m-f of thermocouple is

P=R/L=20/10=2Ω/m i=V/R_1=1.08/2160=0.05× 10^ 2A E=iPl =0.05× 10^ 2× 2 × 3 =3× 10^ 3V =3mV ...

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Byju's Answer

Standard XII

Physics

Potentiometer

A potentiomet...

Question

A potentiometer wire 10 m long and having $‘ 20 Ω^{'}$ resistance is connected in series with a battery and a resistance $R = 2160 Ω$ . A Daniel cell of e.m.f 1.08 V is balanced across the P.D of the resistance ‘R’. The e.m.f of thermocouple is balanced across 300 cm of the wire. The e.m.f of thermocouple is

2 mV

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4 mV

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3 m/V

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1 mV

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Solution

The correct option is C 3 m/V
$P = \frac{R}{L} = \frac{20}{10} = 2 Ω / m i = \frac{V}{R_{1}} = \frac{1.08}{2160} = 0.05 \times 10^{- 2} A E = i P l = 0.05 \times 10^{- 2} \times 2 \times 3 = 3 \times 10^{- 3} V = 3 m V$

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The correct option is C 3 m/VP=RL=2010=2Ω/mi=VR1=1.082160=0.05×10−2AE=iPl=0.05×10−2×2×3=3×10−3V=3mV

The correct option is C 3 m/V
$P = \frac{R}{L} = \frac{20}{10} = 2 Ω / m i = \frac{V}{R_{1}} = \frac{1.08}{2160} = 0.05 \times 10^{- 2} A E = i P l = 0.05 \times 10^{- 2} \times 2 \times 3 = 3 \times 10^{- 3} V = 3 m V$