“Everything that is really great and inspiring is created by the individual who can labour in freedom.”

– Albert Einstein

The fractional change in a physical quantity *Q* is (Δ*Q*)/*Q* where Δ*Q* is the actual change in the quantity. The percentage change in the quantity = Fractional change × 100.

Let us consider a few multiple choice questions to make things clear:

(1) A copper wire having resistance 10 Ω is stretched so that its radius decreases by 3%. What is the resistance of this wire after stretching?

(a) 10 Ω (b) 10.3Ω (c) 10.6Ω (d) 11.2Ω (e) 11.8Ω

The resistance *R *of a wire is given by

*R* = *SL/A* where S is the resistivity (specific resistance), *L* is the length and A is the area of cross section. Since *S* is constant for a given material,* R* depends only on *L *and *A*. When the radius of the wire is decreased by 3 %, its cross section area *A* is decreased by 6 % (since *A* = π*r*^{2}). The resistance increases by 6% on account of this. When *A* decreases by 6 %, the length *L* increases by 6 % since the volume of the wire is unchanged. On account of the 6% increase in length there is an additional increase in resistance by 6%. So the total increase in resistance is 12%. The resistance after stretching = 10 + 1.2 =11.2 Ω [Option (d)]

(2) A nichrome wire has resistance 20 Ω. What will be resistance of another nichrome wire whose length is greater by 5 % and radius is greater by 1%?

(a) 20.2 Ω (b) 20.4 Ω (c) 20.6 Ω (d) 20.8 Ω (e) 21 Ω

There is a 5 % *increase* in resistance due to the increase in length and a 2 % *decrease* in resistance due to the increase in radius.

[Note that 1% increase in radius produces 2 % increase in the area of cross section].

The net increase in resistance = 5 % – 2 % = 3 % = 20 × 3/100 = 0.6 Ω. The correct option therefore is (c).

In the above questions, we have considered the changes in certain quantities (length and radius in the above cases). The changes may be positive (increments) or negative (decrements). The final result may increase or decrease because of the individual changes.

(3) A wire has mass *m = *0.2 ± 0.002 g, radius *r = *0.25 ± 0.005 mm and length *L* = 6 ± 0.06 cm. The maximum percentage error in the measurement of the density* ρ* of the material of this wire is

(a) 1 (b) 2 (c) 4 (d) 6 (e) 8

Here the percentage error in mass = (0.002/0.2)×100 = 1 %. Percentage error in radius = (0.005/0.25) × 100 = 2 %. Percentage error in length = (0.06/6)×100 = 1%. Since the density *ρ* = mass/volume = *m*/π*r*^{2}*L*, the maximum percentage error in density = percentage error in mass + 2×percentage error in radius + percentage error in length = 1+ (2×2) + 1 = 6%.

[The percentage error in *r* is multiplied by 2 since the power of *r* is 2. This follows thus:

We have *ρ *= *m*/π*r*^{2}*L*. On taking logarithms and differentiating, we have

d*ρ/**ρ = dm/m* – 0 – (2 d*r/r*) – d*L/L*

d*ρ/**ρ*, *dm/m*, d*r/r* and d*L/L* are respectively the fractional errors in density, mass, radius and length. Therefore we have

Percentage error in density = Percentage error in mass + 2×percentage error in radius + percentage error in length.

Note that we have taken the signs of all errors as positive since we want to make the errors add up to obtain the maximum percentage error].

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