Question 539: Current Electricity

Question

Resistances in the left gap and right gap of a meter bridge are $10 \Omega$ and $30 \Omega$ respectively. If the resistances in the two gaps are interchanged, the balance point will shift to right by

Accepted Answer

In a meter bridge, the balance point is determined based on the principle of a Wheatstone bridge. With a resistance $R_1 = 10 \, \Omega$ in the left gap and $R_2 = 30 \, \Omega$ in the right gap, the initial position of the balance point divides the bridge wire into lengths $l_1$ and $l_2$, where:
$$ \frac{R_1}{R_2} = \frac{l_1}{l_2} $$
Given that the total length of the wire is 100 cm:
Initial Balance Point:
$    \frac{10}{30} = \frac{l_1}{100 - l_1} \implies \frac{1}{3} = \frac{l_1}{100 - l_1}    $
Solving for $ l_1 $:
$    3l_1 = 100 - l_1 \    4l_1 = 100 \    l_1 = 25 \, 	ext{cm}    $
Thus, the initial balance point is at 25 cm from the left side.
After Interchanging the Resistances:
When the resistances are interchanged, the new principle becomes:
$    \frac{30}{10} = \frac{l_1'}{100 - l_1'}    $
Solving for $ l_1' $:
$    3 = \frac{l_1'}{100 - l_1'} \    3(100 - l_1') = l_1' \    300 - 3l_1' = l_1' \    300 = 4l_1' \    l_1' = 75 \, 	ext{cm}    $
Thus, the new balance point is at 75 cm from the left side.
Shift in the Balance Point:
The shift in the balance point is $ l_1' - l_1 $:
$    	ext{Shift} = 75 \, 	ext{cm} - 25 \, 	ext{cm} = 50 \, 	ext{cm}    $
Thus, the balance point shifts to the right by
50 cm
.
Answer:
Option C: 50 cm.

Answer

30 cm

Answer

40 cm

Answer

50 cm

Answer

60 cm

Current Electricity

Resistances in the left gap and right gap of a meter bridge are 10 \Omega and 30 \Omega respectively. If the resistances in the two gaps are interchanged, the balance point will shift to right by

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