Ncert Exercises & Solutions

In an experiment with a potentiometer, $V_{B}$ = 10V. R is adjusted to be 50 $Ω$ (figure). A student wanting to measure the voltage $E_{1}$ of a battery (approx. 8V) 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.

Hint: The potential drop in the potentiometer wire is directly proportional to the length of the wire.

Step 1: Let R' be the resistance of the potentiometer wire.

Effective resistance of potentiometer and a variable resistor (R = 50

Ω

) is given by = 50

Ω

+ R'

The effective voltage applied across potentiometer = 10V

The current through the main Circuit,

I = \frac{V}{50 Ω + R'} = \frac{10}{50 Ω + R'}

The potential difference across the wire of potentiometer =

IR' = \frac{10 R'}{50 Ω + R'}

Step 2: Since with a 50

Ω

resistor, the null point is not obtained. It is possible only when;

\frac{10 \times R'}{50 + R'} < 8

10 R' < 400 + 8 R'

\Rightarrow 2 R' < 400 or R' < 200 Ω

Similarly, with a 10

Ω

resistor, a null point is obtained. It's possible only when;

\frac{10 \times R^{'}}{10 + R^{'}} > 8

\Rightarrow 2 R^{'} > 80

\Rightarrow R^{'} > 40

And, \frac{10 \times \frac{3}{4} R'}{10 + R'} < 8

\Rightarrow 7 .5 R^{'} < 80 + 8 R^{'}

R' > 160

\Rightarrow 160 < R^{'} < 200

Any R' between 160

Ω

and 200

Ω

will work.

Step 3: Since the null point is on the last (4th) segment of the potentiometer, therefore potential drop across 400 cm of wire > 8V.

This implies that the potential gradient

kx 400 cm > 8V
kx 4m > 8V
or k>2V/m

Similarly, potential drop across 300 cm wire < 8V
or kx 300cm < 8V
kx 3m < 8V

k < 2 \frac{2}{3} V / m

Thus 2 \frac{2}{3} V / m > k > 2 V / m

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions