In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf 1.5 V is found to be 60 cm. If this cell is replaced by another cell of emf E, the length-of…

Given: emf of first cell E1 = 1.5 V, null point length l1 = 60 cm. The new null point length is l2 = 60 + 40 = 100 cm. Find: the value of x if E = x10 V. In a potentiometer experiment, E1E2 = l1l2 So, 1.5E2 = 6060 + 40 = 610 = 35 Hence, E2 = 52 V Now compare with E2 = x10 V Therefore, x10 = 52 So, x = 25 Therefore, the value of x is 25.

In an experiment to find emf of a cell using potentiometer, the l… — Solution | JEE Physics

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In an experiment to find emf of a cell using potentiometer, the length of null point for a cell of emf $1.5 \, \text{V}$ is found to be $60 \, \text{cm}$ . If this cell is replaced by another cell of emf $E$ , the length-of null point increases by $40 \, \text{cm}$ . The value of $E$ is $\frac{x}{10}$ V. The value of $x$ is _____.

Answer

Correct answer:25

Step-by-step solution

Standard Method

Given: emf of first cell $E_1 = 1.5 \, \text{V}$ , null point length $l_1 = 60 \, \text{cm}$ . The new null point length is $l_2 = 60 + 40 = 100 \, \text{cm}$ .

Find: the value of $x$ if $E = \frac{x}{10} \, \text{V}$ .

In a potentiometer experiment,

\frac{E_1}{E_2} = \frac{l_1}{l_2}

So,

\frac{1.5}{E_2} = \frac{60}{60 + 40} = \frac{6}{10} = \frac{3}{5}

Hence,

E_2 = \frac{5}{2} \, \text{V}

Now compare with

E_2 = \frac{x}{10} \, \text{V}

Therefore,

\frac{x}{10} = \frac{5}{2}

So,

x = 25

Therefore, the value of $x$ is $25$ .

Common mistakes

Using inverse proportionality between emf and balancing length is incorrect here. In a potentiometer, emf is directly proportional to the null point length. Use $\frac{E_1}{E_2} = \frac{l_1}{l_2}$ .
Taking the second length as only $40 \, \text{cm}$ is wrong. The null point increases by $40 \, \text{cm}$ , so the new length is $60 + 40 = 100 \, \text{cm}$ .

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Step-by-step solution

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