A potentiometer wire of length 10m and resistance 20 $\Omega$ is connected in series with a 25 V battery and an external resistance $30\Omega$ . A cell of emf E in secondary circuit is balanced by 250cm long potentiometer wire. The value of E (in volt) is $\frac{x}{10}.$ The value of x is ______.

AB = 10 m

$R_{AB}=20\Omega$

$L_{AC}=250cm$

= 2.5 m

$l=\frac{25}{20+30}$

$E=\frac{x}{10}=\frac{25}{10}$

<p>AB = 10 m</p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1753239946phpq6ndMQ.jpeg" alt="" width="116" height="105" loading="lazy"></picture></div></div><p><math><mrow><msub><mrow><mi>R</mi></mrow><mrow><mi>A</mi><mi>B</mi></mrow></msub><mo>=</mo><mn>2</mn><mn>0</mn><mi>Ω</mi></mrow></math></p><p><math><mrow><msub><mrow><mi>L</mi></mrow><mrow><mi>A</mi><mi>C</mi></mrow></msub><mo>=</mo><mn>2</mn><mn>5</mn><mn>0</mn><mtext> </mtext><mtext> </mtext><mi>c</mi><mi>m</mi></mrow></math></p><p>= 2.5 m</p><p><math><mrow><mi>l</mi><mo>=</mo><mfrac><mrow><mn>2</mn><mn>5</mn></mrow><mrow><mn>2</mn><mn>0</mn><mo>+</mo><mn>3</mn><mn>0</mn></mrow></mfrac></mrow></math></p><p><math><mrow><mi>E</mi><mo>=</mo><mfrac><mrow><mi>x</mi></mrow><mrow><mn>1</mn><mn>0</mn></mrow></mfrac><mo>=</mo><mfrac><mrow><mn>2</mn><mn>5</mn></mrow><mrow><mn>1</mn><mn>0</mn></mrow></mfrac></mrow></math></p>

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