JEE Main Units and Measurements Previous year Question

36.(JEE Main 2017 (Online) 9th April Morning Slot )

A physical quantity P is described by the relation

P = a $^{1 / 2}$ b² c³ d^{$-$ 4}

If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and 5%, then the relative error in P will be :

(A) 8%

(B) 12%

(C) 32%

(D) 25%

Correct answer is (C)

Given,

P = a $^{\frac{1}{2}}$ b² c³ d^{$-$ 4}

Relative error =

$\frac{Δ P}{P}$ $\times$ 100 = ( $\frac{1}{2}$ $\times$ $\frac{Δ a}{a}$ + 2 $\frac{Δ b}{b}$ + 3 $\frac{Δ c}{c}$ + 4 $\frac{Δ d}{d}$ ) $\times$ 100

= $\frac{1}{2}$ $\times$ 2 + 2 $\times$ 1 + 3 $\times$ 3 + 4 $\times$ 5

= 32%

37.(JEE Main 2017 (Offline) )

The following observations were taken for determining surface tension T of water by capillary method:
diameter of capillary, D = 1.25 $\times$ 10^-2 m
rise of water, h = 1.45 $\times$ 10^-2m
Using g = 9.80 m/s² and the simplified relation T = $\frac{r h g}{2} \times 10^{3} N / m$ , the possible error in surface tension is closest to :

(A) 10 %

(B) 0.15 %

(C) 1.5 %

(D) 2.4 %

Correct answer is (C)

Surface tension,

T = $\frac{r h g}{2} \times 10^{3} N / m$

Relative error,

$\frac{Δ T}{T} = \frac{Δ r}{r} + \frac{Δ h}{h}$

Percentage error,

$\frac{Δ T}{T} \times 100 = \frac{Δ r}{r} \times 100 + \frac{Δ h}{h} \times 100$

$\frac{Δ T}{T} \times 100 = (\frac{10^{- 2} \times 0.01}{1.25 \times 10^{- 2}} + \frac{10^{- 2} \times 0.01}{1.45 \times 10^{- 2}}) \times 100$

= (0.8 + 0.689) = 1.489 % = 1.5 %

38.(JEE Main 2016 (Offline) )

A student measures the time period of 100 oscillations of a simple pendulum four times. The data set is 90 s, 91 s, 95 s and 92 s. If the minimum division in the measuring clock is 1 s, then the reported mean time should be :

(A) 92 $\pm$ 1.8 s

(B) 92 $\pm$ 3 s

(C) 92 $\pm$ 2 s

(D) 92 $\pm$ 5.0 s

Correct answer is (C)

Here t₁ = 90 s, t₂ = 91 s, t₃ = 95 s, t₄ = 92 s

Mean(t) = $\frac{t_{1} + t_{2} + t_{3} + t_{4}}{4}$

= $\frac{90 + 91 + 95 + 92}{4}$ = 92 s

Now mean deviation

= $\frac{2 + 1 + 3 + 0}{4}$ = 1.5 s

Since least count of clock is one second, so reported mean time

= (92 $\pm$ 2) s

39.(AIEEE 2012 )

Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then error in the value of resistance of the wire is

(A) 6 %

(B) zero

(C) 1 %

(D) 3 %

Correct answer is (A)

We know R = $\frac{V}{I}$

$∴$ $\frac{Δ R}{R} = \frac{Δ V}{V} + \frac{Δ I}{I}$

Percentage error in R =

$\frac{Δ R}{R} \times 100 = \frac{Δ V}{V} \times 100 + \frac{Δ I}{I} \times 100$

$∴$ $\frac{Δ R}{R} \times 100 =$ 3% + 3 % = 6%