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**Quiz Description** :

**Name: **Significant figures and Rounding off Test

**Subject: **Physics

**Topic: **Significant figures and Rounding off

**Questions: **11 Objective Type

**Time Allowed: **10 minutes

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- Question 1 of 11
##### 1. Question

1 pointsThe significant figures of the number 6.0023 are

CorrectGiven then number is 6.0023

We know that all zeroes occurring between two non-zero digit are significant.

Therefore significant figures of the given number are 5.IncorrectGiven then number is 6.0023

We know that all zeroes occurring between two non-zero digit are significant.

Therefore significant figures of the given number are 5. - Question 2 of 11
##### 2. Question

1 pointsThe significant figures in 3400 are

CorrectGiven the number is 3400.

We know that all zeros to the right of the last non-zero digit are non-significant.

Therefore significant figures of the given number are 2.IncorrectGiven the number is 3400.

We know that all zeros to the right of the last non-zero digit are non-significant.

Therefore significant figures of the given number are 2. - Question 3 of 11
##### 3. Question

1 pointsThe no. of significant figures in 5418000 is

CorrectThe given number is 5418000.

We know that all zeroes to the right of the last non-zero digit are non-significant. Therefore no. of significant figures of the given number is 4.IncorrectThe given number is 5418000.

We know that all zeroes to the right of the last non-zero digit are non-significant. Therefore no. of significant figures of the given number is 4. - Question 4 of 11
##### 4. Question

1 pointsThe no. of significant figures in 40.00 is

CorrectThe given number is 40.00.

We know that all zeroes to the right of a decimal point are significant, if they are not followed by a non zero digits. Therefore no. of significant figures of the given number is 4.IncorrectThe given number is 40.00.

We know that all zeroes to the right of a decimal point are significant, if they are not followed by a non zero digits. Therefore no. of significant figures of the given number is 4. - Question 5 of 11
##### 5. Question

1 pointsThe no. significant figures in 0.0003125 is

CorrectThe given number is 0.0003125.

We know that the number of significant figure is counted from the first non-zero digit on the left to the last digit on the right.

Therefore no. of significant figures in the given number is 4.IncorrectThe given number is 0.0003125.

We know that the number of significant figure is counted from the first non-zero digit on the left to the last digit on the right.

Therefore no. of significant figures in the given number is 4. - Question 6 of 11
##### 6. Question

1 pointsGiven: P=0.0030 m, Q=2.40 m, and R=3,000 m. The significant figures in P, Q and R are respectively.

CorrectGiven that P=0.0030 m, Q=2.40 m and R=3,000 m.

We know that in P (0.0030) initial zeroes after the decimal point are not significant. Therefore significant figure in P (0.0030) are 2.

Similarly, in Q(2.40) the final zero is significant. Therefore significant figures in Q are 3. And in R(3,000) all the digits are significant. Therefore significant figures in P are 4.IncorrectGiven that P=0.0030 m, Q=2.40 m and R=3,000 m.

We know that in P (0.0030) initial zeroes after the decimal point are not significant. Therefore significant figure in P (0.0030) are 2.

Similarly, in Q(2.40) the final zero is significant. Therefore significant figures in Q are 3. And in R(3,000) all the digits are significant. Therefore significant figures in P are 4. - Question 7 of 11
##### 7. Question

1 pointsThe length and breadth of a metal sheet are 3.124 m and 3.002 m respectively. The area of this sheet up to four significant figures is

CorrectGiven that the length (

*l*)=3.124 m and breadth (b) = 3.002 m.

We know that area of the sheet (*A*) =*l**× b*= 3.124 × 3.002 = 9.378248 m^{2}.

Therefore area of the sheet after rounding off to four significant figures is 9.378 m^{2}.IncorrectGiven that the length (

*l*)=3.124 m and breadth (b) = 3.002 m.

We know that area of the sheet (*A*) =*l**× b*= 3.124 × 3.002 = 9.378248 m^{2}.

Therefore area of the sheet after rounding off to four significant figures is 9.378 m^{2}. - Question 8 of 11
##### 8. Question

1 pointsA cube has a side of length 1.2×10

^{-2}m. Its volume is significant figures isCorrectGiven that the side of cube (a) = 1.2 × 10

^{-2}m.

We know that volume of cube (*V*) = a^{3}= (1.2 × 10^{-2})^{3}= 1.728 × 10^{-6}m^{2}.

Since side o the cube has two significant figures, therefore volume of cube will also have two significant figures. Thus volume of cube is 1.7 × 10^{-6}m^{2}.IncorrectGiven that the side of cube (a) = 1.2 × 10

^{-2}m.

We know that volume of cube (*V*) = a^{3}= (1.2 × 10^{-2})^{3}= 1.728 × 10^{-6}m^{2}.

Since side o the cube has two significant figures, therefore volume of cube will also have two significant figures. Thus volume of cube is 1.7 × 10^{-6}m^{2}. - Question 9 of 11
##### 9. Question

1 pointsIn a circuit, the resistance is 10.845 Ω and the current is 3.23 A. Its potential difference in four significant figures after rounding off, will be

CorrectGiven that the Resistance (

*R*) = 10.845 and current (*I*) = 3.23 A.

We know that potential difference (*V*) =*I. R*= 3.23 × 10.845 = 35.02935 V.

We also know that when digit 2 (in the potential difference of 35.02935) is omitted which is less than 5, then zero in left remains unchanged.

Thus value of the potential difference in four significant figures will be 35.00 V.IncorrectGiven that the Resistance (

*R*) = 10.845 and current (*I*) = 3.23 A.

We know that potential difference (*V*) =*I. R*= 3.23 × 10.845 = 35.02935 V.

We also know that when digit 2 (in the potential difference of 35.02935) is omitted which is less than 5, then zero in left remains unchanged.

Thus value of the potential difference in four significant figures will be 35.00 V. - Question 10 of 11
##### 10. Question

1 pointsThe volume of a sphere is 1.76 cm

^{3}. The volume of 25 such spheres, taking into a account the significant figures, will beCorrectGiven that the volume of sphere = 1.76 cm

^{3}and number of spheres = 25.

We know that volume of 25 spheres = 25 × 1.76 = 44 cm^{3}.

Since volume of the sphere has three significant figures, therefore volume of the 25 spheres will also have three significant figures.

Therefore in this case, zero is knowingly added after the decimal point on the right taking into account the significant figures. Thus total volume is 44.0 cm^{3}.IncorrectGiven that the volume of sphere = 1.76 cm

^{3}and number of spheres = 25.

We know that volume of 25 spheres = 25 × 1.76 = 44 cm^{3}.

Since volume of the sphere has three significant figures, therefore volume of the 25 spheres will also have three significant figures.

Therefore in this case, zero is knowingly added after the decimal point on the right taking into account the significant figures. Thus total volume is 44.0 cm^{3}. - Question 11 of 11
##### 11. Question

1 pointsIf radius of a circle is 2.14 m, then area of the circle, with due regards for significant figures, will be

CorrectGiven that the radius of the circle (

*r*) = 2.14 m.

We know that the area of the circle (*A*) = πr^{2}= 3.142 × (2.14)^{2}=14.38910 m^{2}.

Since radius of the circle has three significant figures, therefore area of the circle will also have three significant figures.

We also know that the fourth digit 8 (in the area of 14.38910) is greater than 5.

Therefore the earlier digit 3 in the area is increased by 1 with regards for significant figures. Thus the area of the circle is 14.4 m^{2}.IncorrectGiven that the radius of the circle (

*r*) = 2.14 m.

We know that the area of the circle (*A*) = πr^{2}= 3.142 × (2.14)^{2}=14.38910 m^{2}.

Since radius of the circle has three significant figures, therefore area of the circle will also have three significant figures.

We also know that the fourth digit 8 (in the area of 14.38910) is greater than 5.

Therefore the earlier digit 3 in the area is increased by 1 with regards for significant figures. Thus the area of the circle is 14.4 m^{2}.