In the given figure of meter of bridge experiment, the balancing length AC corresponding to null deflection of the galvanometer is 40 cm. The balancing length, if the radius of the wire AB is doubled, will be _______ cm.
To determine the new balancing length when the radius of the wire AB is doubled, we must consider how the resistance of the wire changes. The resistance R of a wire is given by the formula:
R = ρL/A, where:
ρ is the resistivity of the material.
L is the length of the wire.
A is the cross-sectional area of the wire.
Doubling the radius of the wire increases the cross-sectional area A by a factor of 4 (since A = πr²), thus reducing the resistance by a factor of 4.
In a meter bridge experiment, the balancing condition when the galvanometer shows null deflection is given by:
R1/R2 = L1/(100 - L1)
where R1 and R2 are the resistances connected in the two arms, and L1 is the balancing length (AC).
Initially, the balancing length is 40 cm. Therefore,
R1/R2 = 40/(100 - 40) = 40/60 = 2/3
If the radius of the wire AB is doubled, the resistance of wire AB becomes R/4. Hence, for the length AC, the new condition is:
(R1/4)/R2 = Lnew/(100 - Lnew)
Maintaining the proportion, we have:
R1/R2 = 4Lnew/(100 - Lnew)
Substituting the initial ratio,
2/3 = 4Lnew/(100 - Lnew)
Cross-multiplying and solving for Lnew gives the same length since both sides maintain the relative ratios:
Lnew = 40 cm.
Thus, the new balancing length remains unchanged at 40 cm, consistent with the given range of 40 to 40 cm.
