The total length of the wire of a potentiometer is 10 m. A potential gradient of 0.0015 V/cm is obtained when a steady current is passed through this wire . Calculate: a) The distance of the null point on connecting standard cell of 1.018 V. b) The unknown p.d. if the null point is obtained at a distance of 940 cm c) The maximum p.d. which can be measured by this instrument.

Given,

The length of the potentiometer $(L) = 10\;m$

Potential gradient $(K) = 0.0015\; V/cm = 0.15 \; V/m$

a) Distance of null point $(l_1) = ?$

Voltage of the cell $(V) = 1.018 \; V$

From the principle of the potentiometer, We have,

$V \propto l$

or, $V = k\;l_1$

or, $1.018 = 0.15 * l_1$

or, $l_1 = 6.8 \; m$

b) Potential Difference $(V) = ?$

Length of null point $(l_2) = 940 \; cm = 9.4 \; m$

Again from the principle of the potentiometer,

$V \propto l_2$

or, $V = k * l_2 = 0.15 * 9.4 = 1.41\;V$

c) Maximum P.d. $(V_{max}) = ?$

$V_{max} = k * L = 0.15 * 10 = 1.5 \; V$

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