JEE Main Units and Measurements Previous year Question

11.(JEE Main 2022 (Online) 27th July Morning Shift )

A torque meter is calibrated to reference standards of mass, length and time each with $5 %$ accuracy. After calibration, the measured torque with this torque meter will have net accuracy of :

(A) 15%

(B) 25%

(C) 75%

(D) 5%

Correct answer is (B)

We know that, torque applied on a rotating body,

$\begin{aligned} τ & = Force \times Perpendicular distance \\ \Rightarrow [τ] & = [{MLT}^{- 2}] [L] \Rightarrow [τ] = [{ML}^{2} T^{- 2}] \end{aligned}$

$\Rightarrow \frac{Δ τ}{τ} = \frac{Δ M}{M} + 2 \frac{Δ L}{L} + 2 \frac{Δ T}{T}$

$= 5 \times 5 % = 25 %$

12.(JEE Main 2022 (Online) 25th July Evening Shift )

The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are $1 %, 2 %$ and $3 %$ respectively. The maximum percentage error in the detection of the dissipated heat will be :

(A) 2

(B) 4

(C) 6

(D) 8

Correct answer is (D)

Given, $\frac{Δ R}{R} \times 100 = 1 %$

$\frac{Δ F}{F} \times 100 = 2 % and \frac{Δ t}{t} \times 100 = 3 %$

We know that, heat produced due to current flowing through a resistor $R$ .

$H = I^{2} R t$

$\Rightarrow \frac{Δ H}{H} = \frac{2 Δ I}{I} + \frac{Δ R}{R} + \frac{Δ t}{t}$

$\Rightarrow \frac{Δ H}{H} \times 100 = 2 (\frac{Δ I}{I} \times 100) + \frac{Δ R}{R} \times 100 + \frac{Δ t}{t} \times 100$

$= 2 \times 2 % + 1 % + 3 % = 8 %$

13.(JEE Main 2022 (Online) 27th June Morning Shift )

A silver wire has a mass (0.6 $\pm$ 0.006) g, radius (0.5 $\pm$ 0.005) mm and length (4 $\pm$ 0.04) cm. The maximum percentage error in the measurement of its density will be :

(A) 4%

(B) 3%

(C) 6%

(D) 7%

Correct answer is (A)

$ρ = \frac{m}{V} = \frac{m}{π r^{2} \times l}$

$∴$ % error in $ρ = (\frac{0.006}{0.6} + 2 \times \frac{0.005}{0.5} + \frac{0.04}{4}) \times 100$

$= 4 %$

14.(JEE Main 2022 (Online) 25th June Morning Shift )

If $Z = \frac{A^{2} B^{3}}{C^{4}}$ , then the relative error in Z will be :

(A) $\frac{Δ A}{A} + \frac{Δ B}{B} + \frac{Δ C}{C}$

(B) $\frac{2 Δ A}{A} + \frac{3 Δ B}{B} - \frac{4 Δ C}{C}$

(C) $\frac{2 Δ A}{A} + \frac{3 Δ B}{B} + \frac{4 Δ C}{C}$

(D) $\frac{Δ A}{A} + \frac{Δ B}{B} - \frac{Δ C}{C}$

Correct answer is (C)

$Z = \frac{A^{2} B^{3}}{C^{4}}$

$∴$ $\frac{Δ Z}{Z} = \frac{2 Δ A}{A} + 3 \times \frac{Δ B}{B} + \frac{4 Δ C}{C}$

15. (JEE Main 2022 (Online) 28th July Evening Shift )

In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of $0.5 s$ is measured from time of 100 oscillation with a watch of $1 s$ resolution. If measured value of length is $10 cm$ known to $1 mm$ accuracy, The accuracy in the determination of $g$ is found to be $x %$ . The value of $x$ is ___________.

Correct answer is (5)

$T = 2 π \sqrt{\frac{l}{g}}$

$\frac{d g}{g} \times 100 = \frac{2 d T}{T} \times 100 + \frac{d l}{l} \times 100$

$= 2 \times \frac{1}{50} \times 100 + \frac{1}{100} \times 100 = 5 %$