JEE Main Units and Measurements Previous year Question

1. (JEE Main 2024 (Online) 8th April Morning Shift)

In an expression $a \times 10^{b}$ :

A. $a$ is order of magnitude for $b \leq 5$

B. $b$ is order of magnitude for $a \leq 5$

C. $b$ is order of magnitude for $a \geq 5$

D. $b$ is order of magnitude for $a \geq 5$

Correct option is (B)

In expression $a \times 10^{b}$ , If $a \leq 5; a \approx 1$ by round off

$\Rightarrow$ Order $B$

2. (JEE Main 2024 (Online) 5th April Morning Shift)

Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as $4.62 s, 4.632 s, 4.6 s$ and $4.64 s$ . The arithmetic mean of these readings in correct significant figure is :

A.5s

B.4.6s

C.4.62s

D.4.623s

Correct option is (B)

To find the arithmetic mean of the time periods recorded, we need to sum up the values and then divide by the number of readings. Let's calculate the sum first:

$4.62 s + 4.632 s + 4.6 s + 4.64 s$

Adding these values together:

$4.62 + 4.632 + 4.6 + 4.64 = 18.492 s$

Now, we divide this sum by the number of readings, which is 4:

$\frac{18.492 s}{4} = 4.623 s$

So, the arithmetic mean of these readings is $4.623 s$ . However, we need to consider the significant figures. The least number of significant figures among the readings is 2 (from 4.6 s). Hence, the mean should also be represented with 2 significant figures.

In this case, the correct answer with proper significant figures is:

Option B

4.6 s

3. (JEE Main 2024 (Online) 1st February Evening Shift)

Match List - I with List - II.

List I (Number)	List II (Significant figure)
(A) 1001	(I) 3
(B) 010.1	(II) 4
(C) 100.100	(III) 5
(D) 0.0010010	(IV) 6

Choose the correct answer from the options given below :

A. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

B. (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

C. (A)-(IV), (B)-(III), (C)-(I), (D)-(II)

D. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

Correct option is (A)

Significant figures in a number represent the digits that carry meaning contributing to its precision. This includes all digits except:

All leading zeros.

Trailing zeros only when they are merely placeholders to indicate the scale of the number (they must not be significant if there is no decimal point).

Here is the explanation for the significant figures of each number in List I:

(A) 1001 - All the digits in this number are significant because there are no leading or trailing zeros acting as placeholders. Hence, this number has 4 significant figures.

(B) 010.1 - The leading zero is not significant, but the zero between 1 and the decimal point, the 1 itself, and the 1 after the decimal point are all significant. So, this number has 3 significant figures.

(C) 100.100 - Here, the number has zeros which are significant since they are between significant digits or after the decimal point. This makes all the zeros and the 1s significant. Therefore, the number has 6 significant figures.

(D) 0.0010010 - The leading zeros are not significant, but the three zeros within and at the end of the number are significant because they are between significant digits or at the end of the number after the decimal. Therefore, this number has 5 significant figures.

Given List I and List II, the correct matching based on the explanation of significant figures is as follows:

Option A

(A) - II (1001 has 4 significant figures)

(B) - I (010.1 has 3 significant figures)

(C) - IV (100.100 has 6 significant figures)

(D) - III (0.0010010 has 5 significant figures)

Thus, the correct answer is Option A.

4.(JEE Main 2020 (Online) 9th January Evening Slot )

For the four sets of three measured physical quantities as given below. Which of the following options is correct ?
(i) A₁ = 24.36, B₁ = 0.0724, C₁ = 256.2
(ii) A₂ = 24.44, B₂ = 16.082, C₂ = 240.2
(iii) A₃ = 25.2, B₃ = 19.2812, C₃ = 236.183
(iv) A₄ = 25, B₄ = 236.191, C₄ = 19.5

(A) A₁ + B₁ + C₁ = A₂ + B₂ + C₂ = A₃ + B₃ + C₃ = A₄ + B₄ + C₄

(B) A₄ + B₄ + C₄ $<$ A₁ + B₁ + C₁ $<$ A₃ + B₃ + C₃ $<$ A₂ + B₂ + C₂

(C) A₄ + B₄ + C₄ $<$ A₁ + B₁ + C₁ = A₂ + B₂ + C₂ = A₃ + B₃ + C₃

(D) A₄ + B₄ + C₄ $>$ A₃ + B₃ + C₃ = A₂ + B₂ + C₂ $>$ A₁ + B₁ + C₁

Correct answer is (D)

A₁ + B₁ + C₁ = 24.36 + 0.0724 + 256.2
= 280.6324 = 280.6 (After rounding off)

A₂ + B₂ + C₂ = 24.44 + 16.082 + 240.2
= 280.722 = 280.7 (After rounding off)

A₃ + B₃ + C₃ = 25.2 + 19.2812 + 236.183
= 280.6642 = 280.7 (After rounding off)

A₄ + B₄ + C₄ = 25 + 236.191 + 19.5
= 280.691 = 281 (After rounding off)

5.(JEE Main 2019 (Online) 9th April Evening Slot )

The area of a square is 5.29 cm². The area of 7 such squares taking into account the significant figures is :-

(A) 37.0 cm²

(B) 37 cm²

(C) 37.030 cm²

(D) 37.03 cm²

Correct answer is (A)

The area of one square is 5.29 cm².

To find the area of 7 such squares, we can simply multiply the area of one square by 7:

7 x 5.29 cm² = 37.03 cm²

Since the given area has three significant figures, we need to round our answer to three significant figures as well.

Therefore, the area of 7 such squares, taking into account the significant figures, is 37.0 cm².