Units and Measurements Test

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Units and Measurements Test - 4

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following statements is incorrect regarding significant figures?

A
All the non-zero digits are significant

B
All the zeros between two non-zero digits are significant

C
Greater the number of significant figures in a measurement smaller is the percentage error

D
The power of 10 is counted while counting the number of significant figures

Solution

The power of 10 is irrelevant to the determination of significant figures.
Question 2
4 / -1

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

A
35 V

B
35.0 V

C
35.029 V

D
35.03 V

Solution

The final result should be 3 significant figures.
Question 3
4 / -1

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

A
1.7 x 10^-6 m³

B
1.73 x 10^-6 m³

C
1.78 x 10^-6 m³

D
1.732 x 10^-6 m³

Solution

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
Question 4
4 / -1

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

A
11.73 cm³

B
11.736 cm³

C
11.7 cm³

D
117 cm³

Solution

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.
Question 5
4 / -1

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:

A
2.3 kg

B
2.34 kg

C
2.340 kg

D
2.3403 k

Solution

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
Question 6
4 / -1

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?

A
0.02 g

B
0.021 g

C
0.022 g

D
0.024 g

Solution

Difference in masses
= m₂ - m₁ = 20.17 g - 20.15 g
= 0.02 g (correct upto two places of decimal)
Question 7
4 / -1

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

A
3.85 and 3.84

B
3.84 and 3.83

C
3.85 and 3.83

D
3.84 and 3.84

Solution

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.
Question 8
4 / -1

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

A
3, 4, 8

B
4, 4, 8

C
4, 4, 4

D
4, 3, 4

Solution

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.
Question 9
4 / -1

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

A
5 and 6

B
5 and 7

C
2 and 7

D
2 and 6

Solution

As per rule of significant figures, 4.8000 x 10⁴ has 5 significant figures and 48000.50 has 7 significant figures.
Question 10
4 / -1

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

A
(3.5 ± 0.6) m s⁻¹

B
(3.5 ± 0.3) m s⁻¹

C
(6.1 ± 0.6) m s⁻¹

D
(6.1 ± 0.3) m s⁻¹

Solution

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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Units and Measurements Test - 4

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Units and Measurements Test - 4

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Question 1

All the non-zero digits are significant

All the zeros between two non-zero digits are significant

Greater the number of significant figures in a measurement smaller is the percentage error

The power of 10 is counted while counting the number of significant figures

Solution

Question 2

35 V

35.0 V

35.029 V

35.03 V

Solution

Question 3

1.7 x 10-6 m3

1.73 x 10-6 m3

1.78 x 10-6 m3

1.732 x 10-6 m3

Solution

Question 4

11.73 cm3

11.736 cm3

11.7 cm3

117 cm3

Solution

Question 5

2.3 kg

2.34 kg

2.340 kg

2.3403 k

Solution

Question 6

0.02 g

0.021 g

0.022 g

0.024 g

Solution

Question 7

3.85 and 3.84

3.84 and 3.83

3.85 and 3.83

3.84 and 3.84

Solution

Question 8

3, 4, 8

4, 4, 8

4, 4, 4

4, 3, 4

Solution

Question 9

5 and 6

5 and 7

2 and 7

2 and 6

Solution

Question 10

(3.5 ± 0.6) m s−1

(3.5 ± 0.3) m s−1

(6.1 ± 0.6) m s−1

(6.1 ± 0.3) m s−1

Solution

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1.7 x 10^-6 m³

1.73 x 10^-6 m³

1.78 x 10^-6 m³

1.732 x 10^-6 m³

11.73 cm³

11.736 cm³

11.7 cm³

117 cm³

(3.5 ± 0.6) m s⁻¹

(3.5 ± 0.3) m s⁻¹

(6.1 ± 0.6) m s⁻¹

(6.1 ± 0.3) m s⁻¹