In an experiment with a potentiometer, VB = 10 V. R is adjusted to be  50 Ω (figure). A student wanting to measure voltage E1of a battery (approx. 8 V) finds no null point possible. He then diminishes R to 10 Ω  and is able to locate the null point on the-last (4th) segment of the potentiometer. Find the resistance of the potentiometer wire and potential drop per unit length across the wire in the second case.

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    Answered by

    alok kumar singh | Contributor-Level 10

    4 months ago

    This is a Long Answer Type Questions as classified in NCERT Exemplar

    Explanation – let R' be the resistance of potentiometer wire.

    Effective resistance of potentiomter and variable resistor r=50ohm is 500+R'

    Effective voltage across potentiometer = 10V

    The current through main circuit I= V 50 + R = 10 50 + R

    Potential difference across wire of potentiometer

     IR'= 10 R ' 50 + R

    Since with 50 ohm resistor, null point is not obtained it is possible when

    10 * R ' 50 + R < 8

    10R' < 400 + 8 R '

    2R'<400 or R'<200 ohm

    Similarly with 10 ohm resistor, null point is obtained its is only possible when

    10 * R ' 50 + R < 8

    2R'>40

    R'>40

    10 * 3 4 R ' 10 + R ' < 8

    7.5R'<80+8R'

    R'>160

    160

    Any R' between 160 ohm and 200 ohm will achieve.

    Since the null point on the last 4th segment of

    ...more

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