When a known standard resistor of 2.0 is connected to the 0.0cm end of a metre bridge, the balance point is found to be at 55.0cm
1.10Ω
1.64Ω
2.44Ω
27.50Ω
Explanation
Video Explanation
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Discussions (15)
The resistance of the unknown resistor can be calculated using the principle of a meter bridge, which is based on Wheatstone’s bridge. The balance point divides the wire into two sections, and the resistances of these sections are proportional to their lengths.
If the balance point is at 55.0 cm, then the length of the wire on the other side of the balance point is 100.0 cm - 55.0 cm = 45.0 cm.
The formula for a balanced Wheatstone’s bridge is:
R1/R2=L1/L2
where:
R1 is the resistance of the known resistor (2.0 Ω),
R2 is the resistance of the unknown resistor,
L1 is the length of the wire on the same side as R1 (55.0 cm), and
L2 is the length of the wire on the same side as R2 (45.0 cm).
Rearranging the formula to solve for R2 gives:
R2=R1∗(L2/L1)
Substituting the known values gives:
R2=2.0Ω∗(45.0cm/55.0cm)
So, the resistance of the unknown resistor is approximately 1.64 Ω. Therefore, the correct answer is B. 1.64Ω.
This is the correct solution
X/l=R/100-l
Where l is balancing length,x is the unknown resistance and R is the known resistance
X/55=2/100-55
X/55=2/45
X=2*55/45
X=2.44
The correct option is B= 2.44
using this formula
R1/R2=L1/L2
L1=the balance point
The video explanation and the option that was chosen does not tally... please u guys should make things accurate
