Question

A $2$ resistor is connected in series with $R$ resistor. This combination is connected across a cell. When the potential difference across $2$ resistor is balanced on potentiometer wire, null point is obtained at length of $300cm$. When the same procedure is repeated for $R$ resistor, null point is obtained at length $350cm$, value of $R$ is

• A. $5$
  
• B. $3.33$
  
• C. $4.6$
  
• D. $2.33$
  

Answer 1

The correct option is D $2.33$

Let the length of the potentiometer wire be $L$.

When potential difference across $2$ is balanced, null point is obtained at $l_1=300cm$

Using balanced Wheat-stone bridge equation, $\frac{2}{R}=\frac{l_1}{L-l_1}$

Thus we get $\frac{2}{R}=\frac{300}{L-300}$

$⟹$ $L=150R+300$

When potential difference across $R$ is balanced, null point is obtained at $l_2=350cm$

Using $\frac{R}{2}=\frac{l_2}{L-l_2}$

Or $\frac{R}{2}=\frac{350}{(150R+300)-350}$

We get $150R^2-50r-700=0$

Or $3R^2-R-14=0$

Thus we get $R=\frac{1\pm \sqrt{1+4\left(14\right)\left(3\right)}}{2\times 3}$

Or $R=\frac{1\pm 13}{2\times 3}$

$⟹$ $R=2.33$