JEE Main Units and Measurements Previous year Question

16. (JEE Main 2022 (Online) 25th June Evening Shift )

For $z = a^{2} x^{3} y^{\frac{1}{2}}$ , where 'a' is a constant. If percentage error in measurement of 'x' and 'y' are 4% and 12% respectively, then the percentage error for 'z' will be _______________%.

Correct answer is (18)

% error in $z = 3 \times 4 + \frac{1}{2} \times 12$

$= 12 + 6 = 18 %$

17.(JEE Main 2021 (Online) 1st September Evening Shift )

A student determined Young's Modulus of elasticity using the formula $Y = \frac{M g L^{3}}{4 b d^{3} δ}$ . The value of g is taken to be 9.8 m/s², without any significant error, his observation are as following.

Physical Quantity	Least count of the Equipment used for measurement	Observed value
Mass (M)	1 g	2 kg
Length of bar (L)	1 mm	1 m
Breadth of bar (b)	0.1 mm	4 cm
Thickness of bar (d)	0.01 mm	0.4 cm
Depression ( $δ$ )	0.01 mm	5 mm

Then the fractional error in the measurement of Y is :

(A) 0.0083

(B) 0.0155

(C) 0.155

(D) 0.083

Correct answer is (B)

$y = \frac{M g L^{3}}{4 b d^{3} δ}$

$\frac{Δ y}{y} = \frac{Δ M}{M} + \frac{3 Δ L}{L} + \frac{Δ b}{b} + \frac{3 Δ d}{d} + \frac{Δ δ}{δ}$

$\frac{Δ y}{y} = \frac{10^{- 3}}{2} + \frac{3 \times 10^{- 3}}{1} + \frac{10^{- 2}}{4} + \frac{3 \times 10^{- 2}}{4} + \frac{10^{- 2}}{5}$

$= 10^{- 3} [0.5 + 3 + 2.5 + 7.5 + 2] = 0.0155$

Option (b)

18.(JEE Main 2021 (Online) 26th August Evening Shift )

If the length of the pendulum in pendulum clock increases by 0.1%, then the error in time per day is :

(A) 86.4 s

(B) 4.32 s

(C) 43.2 s

(D) 8.64 s

Correct answer is (C)

$T = 2 π \sqrt{\frac{l}{g}}$

$\frac{Δ T}{T} = \frac{1}{2} \frac{Δ l}{l}$

$Δ T = \frac{1}{2} \times \frac{0.1}{100} \times 24 \times 3600$

$Δ T = 43.2$

19.(JEE Main 2021 (Online) 27th July Evening Shift )

A physical quantity 'y' is represented by the formula $y = m^{2} r^{- 4} g^{x} l^{- \frac{3}{2}}$

If the percentage errors found in y, m, r, l and g are 18, 1, 0.5, 4 and p respectively, then find the value of x and p.

(A) 5 and $\pm$ 2

(B) 4 and $\pm$ 3

(C) $\frac{16}{3}$ and $\pm \frac{3}{2}$

(D) 8 and $\pm$ 2

Correct answer is (C)

$\frac{Δ y}{y} = \frac{2 Δ m}{m} + \frac{4 Δ r}{r} + \frac{x Δ g}{g} + \frac{3}{2} \frac{Δ l}{l}$

$18 = 2 (1) + 4 (0.5) + x p + \frac{3}{2} (4)$

$\Rightarrow$ 8 = xp

By checking from options.

$x = \frac{16}{3}, p = \pm \frac{3}{2}$

20.(JEE Main 2021 (Online) 16th March Evening Shift )

In order to determine the Young's Modulus of a wire of radius 0.2 cm (measured using a scale of least count = 0.001 cm) and length 1m (measured using a scale of least count = 1 mm), a weight of mass 1 kg (measured using a scale of least count = 1 g) was hanged to get the elongation of 0.5 cm (measured using a scale of least count 0.001 cm). What will be the fractional error in the value of Young's Modulus determined by this experiment?

(A) 0.14%

(B) 9%

(C) 1.4%

(D) 0.9%

Correct answer is (C)

$\frac{Δ Y}{Y} = (\frac{Δ m}{m}) + (\frac{Δ g}{g}) + (\frac{Δ A}{A}) + (\frac{Δ l}{l}) + (\frac{Δ L}{L})$

$= (\frac{1 g}{1 k g}) + 0 + 2 (\frac{Δ r}{r}) + (\frac{Δ l}{l}) + (\frac{Δ L}{L})$

$= (\frac{1 g}{1 k g}) + 2 (\frac{0.001 c m}{0.2 c m}) + (\frac{0.001 c m}{0.5 c m}) + (\frac{0.001 m}{1 m})$

$= (\frac{1}{1000}) + 2 (\frac{1 \times 10}{2 \times 10^{3}}) + (\frac{1}{5} \times \frac{10^{2}}{10^{3}}) + (\frac{1}{10^{3}})$

$= \frac{1}{1000} + \frac{1}{100} + \frac{2}{10^{3}} + \frac{1}{10^{3}}$

$= \frac{1 + 10 + 2 + 1}{1000} = \frac{14}{1000} \times 100 %$

$= 1.4 %$