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In a metre bridge experiment the balance point is obtained if the gaps are closed by 2Ω and 3Ω. A shunt of XΩ is added to 3Ω resistor to shift the balancing point by 22.5 cm. The value of X is -

Option: 1

2


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & \frac{2}{\ell_1}=\frac{3}{100-\ell_1} \\ & 200-2 \ell_1=3 \ell_1 \end{aligned}

\begin{aligned} & 200=5 \ell_1 \\ & \ell_1=40 \mathrm{~cm} \end{aligned}
Now

 \ell_2=\ell_1+22.5\\ \ell_2=40+22.5=62.5 \mathrm{~cm}

\text{So, }\frac{2}{62.5}=\frac{\left(\frac{3 \cdot x}{3+x}\right)}{37.5} \Rightarrow(37.5) \times 2=\frac{(62.5)(3 x)}{3+x} \\ \begin{aligned} & 3+x=\frac{(62.5)}{25} x \\ & 3+x=2.5 x \\ & 3=1.5 x \Rightarrow x=2 \end{aligned}

 

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Rishabh

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