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  • The circuit diagram given in the figure shows the experimental setup for the measurement of unknown resistance by using a meter bridge. The wire connected between the points P&Q has non-uniform

The circuit diagram given in the figure shows the experimental setup for the measurement of unknown resistance by using a meter bridge. The wire connected between the points P&Q has non-uniform resistance such that resistance per unit length varies directly as the distance from the point P. Null point is obtained with the jockey J with R_{1} and R_{2} in the given position. On interchanging the positions R_{1} and R_{2} in the gaps the jockey has to be displaced through a distance \Delta from the previous  position along the wire to establish the null point. If the ratio of \frac{R_{1}}{R_{2}}=3, and \Delta =5x then the value of x (in cm) is. Ignore any end corrections.[Take\sqrt{3}=1.7 ]

 

Option: 1

2 cm


Option: 2

3 cm


Option: 3

5 cm


Option: 4

7 cm


Answers (1)

best_answer

Total resistance=\int_0^L \lambda x d x=\frac{\lambda}{2} L^2

For balance condition \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{(\lambda / 2) l_1^2}{(\lambda / 2)\left(\mathrm{L}^2-l_1^2\right)}

Similarly, \quad \frac{R_2}{R_1}=\frac{I_2^2}{L^2-I_2^2}

\therefore \frac{R_1}{R_2}=\frac{l_1^2}{L^2-l_1^2}=\frac{L^2-I_2^2}{L_2^2}=n

Solving, L_1=L \sqrt{\frac{n}{n+1}}\; and \; L_2=L \sqrt{\frac{1}{n+1}} ; \quad

\therefore \quad \Delta=L_1-I_2=\frac{L}{\sqrt{n+1}}(\sqrt{n}-1)

Putting L = 100 cm and n = 3

\begin{aligned} & \left.\Delta=\frac{100}{2}(\sqrt{3}-1)=35 \mathrm{~cm}\right] \\ & x=7 \mathrm{~cm} \end{aligned}

 

Posted by

Rakesh

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