Answer: B. 20 ohms
An unknown resistance R1 is connected in series with a resistance of 10Ω. This combination is connected to one gap of a meter bridge while a resistance R2 is connected to the other gap. The balance point is at 50 cm. Now when the 10Ω resistance is removed, the balance point shifts to 40 cm. The value of R1 is (in ohms):20 ohms.
Solution:
Given an unknown resistance, R1 is connected in series with a resistance of 10Ω .
The balance point l1=50 cm
it is a meter bridge,
so l1+l2=100 cm
l2=100-l1=50 cm
(R1+10)/R2=l1/l2=l1/(100-l1)
(R1+10)/R2=50/(100-50)
R1+10=R2________(1)
When the 10-ohm resistor is removed, The balance point l1=40 cm.
l2=100-l1=100-40=60
R1/R2=l1/l2=40/60
R1=2/3R2 _________(2)
By solving equations (1) &(2)
R1=20 Ohms
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