Q4. Assertion: a metre bridge gives its most reliable reading when the balance point is near the middle of the wire. Reason: the relative error from a fixed uncertainty Δl is smallest there, since Δl/l + Δl/(100 − l) is least at 50 cm. Are both true, and does the reason explain the assertion?
Both are true within this comparison of length-reading errors. Unequal end corrections can still leave systematic bias, even near the midpoint.
Q5. A 2.0 Ω coil is in the right gap. What resistance R in the left gap brings the balance point to (a) 50.0 cm, (b) 40.0 cm from the end next to R?
From R/S = l/(100 − l): (a) R = 2.0 × 50.0/50.0 = 2.0 Ω; (b) R = 2.0 × 40.0/60.0 ≈ 1.3 Ω.
Q6. A wire 0.80 m long and 0.50 mm in diameter has a resistance of 4.0 Ω, found with a metre bridge. Find the resistivity of its material.
A = πd/4 = 3.14 × (0.50 × 10)/4 ≈ 1.96 × 10 m, so ρ = SA/L = 4.0 × 1.96 × 10/0.80 ≈ 9.8 × 10 Ω m (NCERT Eq. (3.9), page 84).
Q7. The bridge wire is usually made of an alloy such as manganin or constantan (standard laboratory practice). Why is such an alloy a good choice?
The resistivity of these alloys depends only very weakly on temperature, so the resistance of each length of the wire stays almost the same even if the wire warms a little while the current flows (NCERT page 90). Their resistivity is also much higher than copper’s, so the wire’s resistance is large compared with that of the copper strips and joins (standard laboratory practice).
Q8. In a metre bridge the balance point is 40.0 cm from the end next to R. When a 12 Ω resistor is joined in parallel with S, the balance point moves to 50.0 cm. Find R and S.
Before: R/S = 40.0/60.0, so R = 2S/3. After: S in parallel with 12 Ω is 12S/(S + 12), and with the balance at the middle this equals R. So 12S/(S + 12) = 2S/3, which gives S + 12 = 18. Hence S = 6.0 Ω and R = 4.0 Ω.
Q9. A balance point is read as 20.0 cm ± 0.1 cm. Find the percentage error in S, and compare it with a reading of 50.0 cm ± 0.1 cm.
To first order, the maximum fractional uncertainty is ΔS/S ≈ Δl/l + Δl/(100 − l) = 0.1/20.0 + 0.1/80.0 = 0.625 %, about 0.6 %. At 50.0 cm it is 0.1/50.0 + 0.1/50.0 = 0.40 %, the smallest possible.