Question 725: Current Electricity

Question

During an experiment with a metre bridge, the galvanometer shall a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90$$\Omega$$, as shown in the figure. The least count of the scale used in the meter bridge is 1 mm. The unknown resistance is

Accepted Answer

when a metre bridge is balanced, then
$${{{R_1}} \over x} = {{{R_2}} \over {(100 - x)}}$$ ..... (i)
From figure,
R
1
= R, R
2
= 90 $$\Omega$$, x = 40 cm
Then, $${R \over {40}} = {{90} \over {(100 - 40)}} = {{90} \over {60}}$$
$$\Rightarrow$$ R = 60 $$\Omega$$
Now, for $$\Delta$$R taking natural log on both sides of eqn. (i),
$$\ln {R_1} = \ln {R_2} + \ln x - \ln (100 - x)$$
or, $$\ln R = \ln x - \ln (100 - x) + \ln (90)$$
On differentiating,
$${{\Delta R} \over R} = {{\Delta x} \over x} - {{\Delta (100 - x)} \over {(100 - x)}}$$
$$ \Rightarrow {{\Delta R} \over R} = {{\Delta x} \over x} + {{\Delta x} \over {(100 - x)}}$$
$$ \Rightarrow \Delta R = \left( {{{0.1} \over {40}} + {{0.1} \over {60}}} ight) 	imes 60 = {{0.5} \over {120}} 	imes 60 = 0.25\Omega $$
$$	herefore$$ Required value of R = (60 $$\pm$$ 0.25) $$\Omega$$

Answer

60 $$\pm$$ 0.15 $$\Omega$$

Answer

135 $$\pm$$ 0.56 $$\Omega$$

Answer

60 $$\pm$$ 0.25 $$\Omega$$

Answer

135 $$\pm$$ 0.23 $$\Omega$$

Current Electricity

During an experiment with a metre bridge, the galvanometer shall a null point when the jockey is pressed at 40.0 cm using a standard resistance of 90$\Omega$, as shown in the figure. The least count of the scale used in the meter bridge is 1 mm. The unknown resistance is

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