In a potentiometer circuit a cell of EMF <math> <mn>1.5</mn> <mtext> </mtext> <mi mathvariant="normal">V</mi> </math> gives balance point at <math> <mn>36</mn> <mtext> </mtext> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">m</mi> </math> length of wire. If another cel) of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs?

E 1 E 2 = ? l 1 ? l 2 1.5 2.5 = 36 l 2 ⇒ l 2 = 36 × 5 3 = 60 c m

In a potentiometer circuit a cell of EMF <math> <mn>1.5</mn> <mtext> </mtext> <mi mathvariant="normal">V</mi> </math> gives balance point at <math> <mn>36</mn> <mtext> </mtext> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">m</mi> </math> &nbsp;length of wire. If another cel) of EMF 2.5 V replaces the first cell, then at what length of the wire, the balance point occurs? &lt;!-- [if !supportLineBreakNewLine]--&gt; &lt;!--[endif]--&gt;

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Option 1 -

$60 c m$

Option 2 -

$21.6 c m$

Option 3 -

$64 c m$

Option 4 -

$62 c m$

2 Views | Posted 4 weeks ago

Answered by

alok kumar singh | Contributor-Level 10

4 weeks ago

Correct Option - 1

Detailed Solution:

$\frac{E_{1}}{E_{2}} = \frac{? l_{1}}{? l_{2}}$

$\frac{1.5}{2.5} = \frac{36}{l_{2}} \Rightarrow l_{2} = 36 \times \frac{5}{3} = 60 c m$