Kerala Engineering Entrance [KEAM (Engineering)] 2007 Question on Percentage Error

In continuation of the previous post (dated 23^rd March 2011) related to errors in measurements, I give you today the following question which appeared in KEAM (Engineering) 2007 question paper:

The value of two resistors are R₁ = (6 ± 0.3) kΩ and R₂ =(10 ± 0.2) kΩ. The percentage error in the equivalent resistance when they are connected in parallel is

(a) 5.125 % (b) 2 % (c) 3.125 %

(d) 7 % (e) 10.125 %

When resistances R₁ and R₂ are connected in parallel the equivalent resistance R is given by

R = R₁R₂/(R₁ + R₂)

The fractional error in R is ∆R/R, which you can find by taking logarithm and differentiating:

∆R/R = (∆R₁/R₁) + (∆R₂/R₂) + ∆(R₁ + R₂)/(R₁ + R₂)

[Here ∆R₁ is the error in R₁, ∆R₂ is the error in R₂ and ∆(R₁ + R₂) is the error in (R₁ + R₂). On differentiation the third term is negative but we have taken the sign of the third term as positive since we have to consider the worst case in which the maximum possible error is obtained].

Therefore, ∆R/R = (0.3/6) + (0.2/10) + (0.5/16)

Percentage error in R = percentage error in R₁ + percentage error in R₂ + percentage error in (R₁ + R₂) = (0.3/6)×100 + (0.2/10)×100 + (0.5/16)×100 = 5 % + 2 % + 3.125 % = 10.125 %.

As an extension of the above problem consider the following question:

In the above question what will be the percentage error in the heat produced per second when a current of (4 ± 0.1) ampere is passed through the parallel combination of R₁ and R₂?

(a) 3.5 %

(b) 7 %

(c) 10.125 %

(d) 15.125 %

(e) 20.25 %

The heat H produced is given by H = I²R, where I is the current.

The fractional error in the heat produced per second is

∆H/H = 2×(∆I/I) + ∆R/R

Percentage error in H is 2×(∆I/I) ×100 + (∆R/R) ×100

Therefore, percentage error in H = 2×(0.1/4)×100 + 10.125 = 15.125%

Percentage Change and Percentage Error in a Physical Quantity

“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 (ΔQ)/Q where ΔQ 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 = πr²). 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/πr²L, 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/πr²L. On taking logarithms and differentiating, we have

dρ/ρ = dm/m – 0 – (2 dr/r) – dL/L

dρ/ρ, dm/m, dr/r and dL/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].

Multiple Choice Practice Questions on Electrostatics

“Live as if you were to die tomorrow. Learn as if you were to live for ever.”

– Mahatma Gandhi

If you grasp the fundamental principles well, you will be able to answer multiple choice questions within the stipulated time. If you are confused about the fundamental principles, you will be tempted to waste your time while dealing with even simple questions just because of some simple distractions. See the following questions:

(1) A battery of emf 12 V is connected in series with two initially uncharged capacitors, each of value 100 μF (Fig.). The capacitors are fully charged. If the total energy spent by the battery for charging the capacitors is E joule, the energy of each capacitor is

(a) E

(b) E/2

(c) 2 E

(d) E/4

(e) E/8

If the total charge supplied by the battery is Q coulomb, the total energy E supplied by the capacitor is given by

E = VQ joule where V is the emf of the battery (which is 12 volt).

The two identical 100 μF capacitors in series is equivalent to a single capacitor of capacitance C = 50 μF. The energy of this charged capacitor combination is ½ CV² joule = ½ VQ joule since Q = CV.

Therefore, the energy of the charged capacitor combination is ½ E.

Since there are two identical capacitors in the combination, the energy of each capacitor is E/4.

[So the capacitors have gained a total energy of E/2, even though the battery has delivered a total energy E. What happens to the other half of the energy? Well, it is irrecoverably lost as heat generated in the resistance of the circuit].

(2) Charges of equal magnitude are fixed at the diagonally opposite corners A and C of a non conducting square frame ABCD (Fig.). In case (i) the charges at A and C are positive and a third free negative charge q is moved from B to D. In case (ii) the charges at A and C are negative and a third free negative charge q is moved from B to D. In case (iii) the charges at A and C are of opposite sign and a third free negative charge q is moved from B to D. Pick out the correct statement regarding the work done for moving the charge q:

(a) In case (i) the work done is positive

(b) In case (ii) the work done is positive

(c) In case (i) the work done is negative

(d) In case (iii) the work done is positive

(e) In all cases the work done is zero

In case (i) the potentials at B and D are positive, but of equal value. In case (ii) the potentials at B and D are negative, but of equal value. In case (iii) the potentials at B and D are zero. Therefore, in all cases the potential difference between points B and D is zero so that the work done in moving the charge q from B to D is zero.

You will find some useful questions (MCQ) with solution here.