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# Test: Significant Figures (NCERT)

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## 10 Questions MCQ Test Physics Class 11 | Test: Significant Figures (NCERT)

Test: Significant Figures (NCERT) for NEET 2022 is part of Physics Class 11 preparation. The Test: Significant Figures (NCERT) questions and answers have been prepared according to the NEET exam syllabus.The Test: Significant Figures (NCERT) MCQs are made for NEET 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Significant Figures (NCERT) below.
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Test: Significant Figures (NCERT) - Question 1

### Which of the following statements is incorrect regarding significant figures?

Detailed Solution for Test: Significant Figures (NCERT) - Question 1

The power of 10 is irrelevant to the determination of significant figures.

Test: Significant Figures (NCERT) - Question 2

### The value of resistance is 10.845 Ω and the current is 3.23 A. On multiplying, we get the potential difference is 35.02935 V. The value of potential difference in terms of significant figures would be

Detailed Solution for Test: Significant Figures (NCERT) - Question 2

The final result should be 3 significant figures.

Test: Significant Figures (NCERT) - Question 3

### A cube has a side of length 1.2 x 10-2m. Its volume upto correct significant figures is

Detailed Solution for Test: Significant Figures (NCERT) - Question 3

Here
Length of the cube, L = 1.2 x 10-2 m
Volume of the cube, V = (1.2 x 10-2m)3 = 1.728 x 10-6 m3
As the result can have only two significant figures, therefore, on rounding off, we get, V = 1.7 x 10-6 m3

Test: Significant Figures (NCERT) - Question 4

The radius of a sphere is 1.41cm. Its volume to an appropriate number of significant figures is

Detailed Solution for Test: Significant Figures (NCERT) - Question 4

Radius of the sphere, r = 1.41cm
(3 significant figures)
Volume of the sphere, = 11.736 cm3
Rounded off upto 3 significant figures = 11.7 cm3.

Test: Significant Figures (NCERT) - Question 5

The mass of a box measured by a grocer's balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. The total mass of the box is:

Detailed Solution for Test: Significant Figures (NCERT) - Question 5

Here, mass of the box, m = 2.3 kg
Mass of one gold piece, m1 = 20.15 g = 0.02015 kg
Mass of other gold piece, m2 = 20.17 g = 0.02017 kg
∴ Total mass = m + m1 + m2
= 2.3 kg + 0.02015 kg + 0.02017 kg = 2.34032 kg
As the result is correct only upto one place of decimal, therefore, on rounding off, we get
Total mass = 2.3 kg

Test: Significant Figures (NCERT) - Question 6

The mass of a box measured by a grocer's balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box.

What is the difference in the masses of the pieces?

Detailed Solution for Test: Significant Figures (NCERT) - Question 6

Difference in masses
= m2 - m1 = 20.17 g - 20.15 g
= 0.02 g (correct upto two places of decimal)

Test: Significant Figures (NCERT) - Question 7

The numbers 3.845 and 3.835 on rounding off to 3 significant figures will give

Detailed Solution for Test: Significant Figures (NCERT) - Question 7

The number 3.845 rounded off to three significant figures becomes 3.84 since the preceding digit is even. On the other hand, the number 3.835 rounded off to three significant figures becomes 3.84 since the preceding digit is odd.

Test: Significant Figures (NCERT) - Question 8

The respective number of significant figures for the numbers 6.320, 6.032, 0.0006032 are

Detailed Solution for Test: Significant Figures (NCERT) - Question 8

According to the rules of significant figures 6.320 has four significant figures.
6.032 has four significant figures. 0.0006032 has four significant figures.

Test: Significant Figures (NCERT) - Question 9

The number of significant figures in the numbers 4.8000 x 104 and 48000.50 are respectively

Detailed Solution for Test: Significant Figures (NCERT) - Question 9

As per rule of significant figures, 4.8000 x 104 has 5 significant figures and 48000.50 has 7 significant figures.

Test: Significant Figures (NCERT) - Question 10

A body travels uniformly a distance of (13.8 ± 0.2) m in a time (4.0 ± 0.3) s. Its velocity with error limits is

Detailed Solution for Test: Significant Figures (NCERT) - Question 10

Here, s = (13.8 ± 0.2) m, t = (4.0 ± 0.3)s
∴ veliocity, v = s/t = 13.8/4.0 = 3.45 m s−1
(Rounded off to first place of decimal)  = 0.0865
or Δv = v × 0.0865 = 3.45 × 0.0865
= 0.3087.
∴ velocity =(3.5 ± 0.3) m s−1.

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## Physics Class 11

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