Today we will discuss a few questions on errors in measurement. Students are generally found to commit mistake in working out questions involving the calculation of error. The following question appeared in IIT – JEE 2008 question paper:

Students I, II and III perform an experiment for measuring the acceleration due to gravity (*g*) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table.

Least count for length = 0.1 cm.

Least count for time = 0.1 s.

__Student .Length of pend(cm)__

__.__

__No.of osc (n).Total time (s)..__

__Time period (s)__

....__I.............64.0............ .........8...................128.0 ...............16.0.... __

__....II ...........64.0......................4.....................64.0................16.0....__

__....III..........20.0......................4.....................36.0..................9.0.....__

If E_{I}, E_{II} and E_{III} are the percentage errors in *g*, i.e., (∆*g/g*)×100 for students I, II and III respectively,

(a) E_{I} = 0

(b) E_{I} is minimum

(c) E_{I} = E_{II}

(d) E_{II} is maximum

The period of oscillation (*T*) of a simple pendulum of length *ℓ* is given by

*T = *2π√(*ℓ/g*)

Therefore, *g* = 4π^{2} *ℓ/T ^{2}* so that the

**in**

*fractional error**g*is given by

∆*g/g = *(∆*ℓ/ℓ*) + 2(∆*T/T*)

[The above expression is obtained by taking logarithm of both sides and then differentiating. Note that the sign of the second term on the RHS is changed from negative to positive since we have to consider the *maximum *possible error].

Here ∆*ℓ = *0.1 cm and ∆*T = *0.1 s

The ** percentage error **is 100 times the

*fractional error*so that

E_{I} = ∆*g/g =* [(0.1/64) + 2(0.1/128)]×100 = 5/16 %,

E_{II} = ∆*g/g =* [(0.1/64) + 2(0.1/64)]×100 = 15/32 % and

E_{III} = ∆*g/g =* [(0.1/20) + 2(0.1/36)]×100 = 19/18 %

Thus E_{I} is_{ }minimum so that the correct option is (b).

Now, consider the following MCQ:

A firm manufacturing electronic watches claims that the maximum error in the time indicated by their watches is 5 seconds in six months. What will be maximum possible difference between the times indicated by two watches in one year?

(a) 5 s

(b) 10 s

(c) 20 s

(d) 40 s

(e) zero** **

This is a very simple question. But you must remember that both *positive* and *negative* errors are possible. The maximum possible difference between the times indicated by the watches will be obtained when one watch loses time (slow) and the other watch gains time (fast).

The maximum time lost in one year is 10 s. The maximum time gained in one year also is 10 s. Therefore, the maximum possible difference between the times indicated by the clocks in one year will be 20 s.

Here is another question:

A 200 Ω carbon film resistor with 1% tolerance is connected in series with a 100 Ω carbon film resistor with 5% tolerance. The effective resistance of the combination is (in Ω)

(a) 300 ± 7 Ω

(b) 300 ± 6 Ω

(c) 300 ± 3 Ω

(d) 300 + 18 Ω

(e)** **300 +7 Ω** **

The effective resistance (*R*) is given by

*R = R*_{1}* + R*_{2} where *R*_{1}* =* (200 ± 2) Ω and * R*_{2} = (100 ± 5) Ω

Thus *R =*(200 ± 2) Ω + (100 ± 5) Ω = 300 ± 7 Ω

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