1.

The resistance of a 10 m long potentiometer wire is 20 Omega . It is connected in series with a 3 V battery and 10 Omegaresistor. The potential difference between two points separated by distance 30 cm is equal to .......

Answer»

0.02 V
0.06 V
0.1 V
1.2 V

Solution :0.06 V
`L rho = 20 OMEGA , L = 10 ` m
`therefore rho - 2 (Omega)/(m)`
Now, `V_(l) = I rho `l
`therefore I =(E rho l )/(I rho + R + r)= (3 XX 2 xx 0.3)/(20 + 10 + 0) = (1.8)/(30)`
`therefore V_(l) = 0.06 ` V


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