# Class 11 Physics MCQ – Significant Figures

This set of Class 11 Physics Chapter 2 Multiple Choice Questions & Answers (MCQs) focuses on “Significant Figures”.

1. The length and breadth of a rectangle are 4.5 mm and 5.9 mm. Keeping the number of significant figures in mind, its area in mm2 is ____
a) 22.55
b) 26.55
c) 26.6
d) 27

Explanation: Area of rectangle is length multiplied by breadth. The area of this rectangle is 4.5×5.9 mm2 = 26.55 mm2. However, when multiplying/dividing, the answer should have the same number of significant figures as the limiting term. The limiting term is the number with the least number of significant figures. In our example, both the numbers have 2 significant figures. So, the result should also have 2 significant figures. Hence, 26.55 will be rounded to 2 significant figures and it will become 27.

2. The number of significant digits in 1559.00 is ____
a) 6
b) 5
c) 3
d) 4

Explanation: The number of significant digits is 6. The zeros after the decimal are also significant.

3. The number 0.005900, in standard scientific form can be expressed as ______
a) 5.9×103
b) 59×104
c) 5.9×102
d) 5.9×104

Explanation: Here the number of significant digits is 4. In standard scientific form, we need not consider the zeros after 9 and need to bring the value between 1 and 9 in the first place.

4. Number of significant digits in 0.0028900 is ______
a) 5
b) 6
c) 7
d) 8

Explanation: The correct answer is 5. While calculating the number of significant digits, we need not consider the zero preceding 2.

5. What is the number 75.66852 rounded off to 5 significant digits?
a) 75.67
b) 75.669
c) 75.668
d) 75.667

Explanation: According to the rounding off rules, if the digit to be removed is 5 and the following digit is not zero, then the last remaining digit is increased by one.
Given number = 75.66852
After applying rounding rules, 852 rounds off to 9
Hence, the final answer will be “75.669”.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

6. The length of a cube is 2.3 cm. What is its volume rounded off to 4 significant digits in cm3?
a) 12.67
b) 12.167
c) 12.17
d) 13

Explanation: Volume of a cube is equaled to its length cubed, therefore, required volume = 12.167 cm3. The 67 in the end will be rounded off to 7, hence the final number becomes 12.17.

7. How many significant digits are there in 25.33600?
a) 7
b) 8
c) 5
d) 6

Explanation: The correct answer is 7. The zeros after 6 in the decimal part are also significant.

8. The number of significant digits in 5.002 is ____
a) 5
b) 4
c) 3
d) 6

Explanation: The correct answer is 4. All the digits in the given number are significant.

9. The volume of a box, 10m wide, 4.5m long and 2.3m high up to 3 significant digits in m3 is ____
a) 104
b) 103.5
c) 103
d) 100

Explanation: As per the rule of multiplication and division, the result should have the same significant figure as the number with least significant figure. Among the 3 numbers: 10, 4.5 and 2.3, the number 10 has only 1 significant figure. The volume of the box is length x width x height = 103.5.

As per the actual rule, 103.5 would be rounded up to 100 and the answer would be 100 (as the number 100 has only 1 significant figure). However, we are asking for 3 significant figures in the question, 103.5 will be rounded off to 3 significant figures and the final answer will become 104.

10. How many significant digits are there in 002.5001?
a) 1
b) 5
c) 7
d) 6

Explanation: The correct answer is 5. The zeros preceding 2 are not significant.

Sanfoundry Global Education & Learning Series – Physics – Class 11.

To practice all chapters and topics of class 11 Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.