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All questions of Significant Figures for EmSAT Achieve Exam
Which of the following physical quantity has different dimension than others?
- a)Energy
- b)Watt hour
- c)Momentum
- d)Work
Correct answer is option 'C'. Can you explain this answer?
Which of the following physical quantity has different dimension than others?
a)
Energy
b)
Watt hour
c)
Momentum
d)
Work
|
Riya Banerjee answered |
Dimension of work and energy are the same i.e. [ML²T⁻²].
Power = Energy/Time,
Energy = Power × Time (in terms of hour instead of second),
Energy = Watt × Hour,
= [ML²T⁻³] [T],
= [ML²T⁻²].
But, Momentum(p) = Mass × Velocity,
p = [M] [LT⁻¹],
p = [MLT⁻¹].
Hence the Dimension of work, energy and watt-hour is the same i.e. [ML²T⁻²] but the dimension of momentum is different ([MLT⁻¹]).
Power = Energy/Time,
Energy = Power × Time (in terms of hour instead of second),
Energy = Watt × Hour,
= [ML²T⁻³] [T],
= [ML²T⁻²].
But, Momentum(p) = Mass × Velocity,
p = [M] [LT⁻¹],
p = [MLT⁻¹].
Hence the Dimension of work, energy and watt-hour is the same i.e. [ML²T⁻²] but the dimension of momentum is different ([MLT⁻¹]).
Three physical quantity having dimensions [ M L-1 T-2] are.- a)Stress, pressure, density
- b)Stress, pressure, work
- c)Stress, pressure, coefficient of elasticity
- d)Stress, pressure, torque
Correct answer is option 'C'. Can you explain this answer?
Three physical quantity having dimensions [ M L-1 T-2] are.
a)
Stress, pressure, density
b)
Stress, pressure, work
c)
Stress, pressure, coefficient of elasticity
d)
Stress, pressure, torque
|
Preeti Iyer answered |
Pressure = Force/Area = ML1T-2/ L2 = [M L-1 T-2]
Stress = Force/Area = ML1T-2/ L2 =[M L-1 T-2]
Coefficient of elasticity = Stress/strain = [M L-1 T-2]
The velocity (in cm s-1) of a particle is given in terms of time t (in s) by the equation 
The dimensions of a, b and c are- a)L T 2, L T, L
- b)L T -2, L, T
- c)L2, T, L T -2
- d)L, L T, T 2
Correct answer is 'B'. Can you explain this answer?
The velocity (in cm s-1) of a particle is given in terms of time t (in s) by the equation
The dimensions of a, b and c area)
L T 2, L T, L
b)
L T -2, L, T
c)
L2, T, L T -2
d)
L, L T, T 2
|
|
Raghav Bansal answered |
Volume of water = Volume of the tank = Length * breadth * height = 5.6 * 8.2 * 12.8 = 587.776
Volume, correct up to 4 significant figure = 587.8
The numbers of significant figures in 1.84 x10-27 kg are:- a)five
- b)thirty
- c)three
- d)four
Correct answer is option 'C'. Can you explain this answer?
The numbers of significant figures in 1.84 x10-27 kg are:
a)
five
b)
thirty
c)
three
d)
four
|
Suresh Iyer answered |
As the given number is itself written in the format of powers of ten, it is very obvious that all the non significant zeros are considered by the power of the and hence all the rest digits that are 3 in number are significant.
A dimensionless quantity- a)has no unit
- b)must have a unit
- c)cannot exist
- d)might have a unit
Correct answer is option 'D'. Can you explain this answer?
A dimensionless quantity
a)
has no unit
b)
must have a unit
c)
cannot exist
d)
might have a unit
|
|
Rajeev Saxena answered |
A unit need not always mean a physical unit. So a dimensionless quantity will always be independent of the base units — meter, second, kilograms, kelvin, candela, moles, ampere.
But there are other quantities that are dimensionless, but have a unit. For example, radian is the unit for an angle. But it is dimensionless because it is defined to be the ratio of two lengths.
Dimensions cannot be used to- a)check dimensional correctness of a formula
- b)convert units
- c)find value of the constant of proportionality in an equation
- d)deduce a relation among physical quantities
Correct answer is option 'C'. Can you explain this answer?
Dimensions cannot be used to
a)
check dimensional correctness of a formula
b)
convert units
c)
find value of the constant of proportionality in an equation
d)
deduce a relation among physical quantities
|
|
Preeti Iyer answered |
- Dimensional Equations are used to derive the Relations between the Physical quantities . Also, they are used to check the correctness of the Equation by the Principle of the homogeneity of the dimensions. Hence, Option (a) and option (b) are true for the Dimensional equation. Thus, It is not correct for the given questions.
- Dimensional Equations are not valid for the Pure numbers since they are non-dimensional constants.and also they do not provide the Value of the Constant of the Proportionality.
- Thus, we can also say that all equations which are correct must be dimensionally correct but all those equations which are dimensionally correct may or may not be accurately correct. Thus, Option Option (c). is correct answer for the Given question.
- Dimensional Equations are used for the conversions of the units thus this option can not be correct for the given question.
Find the option with 5 significant figures- a)26.08mm
- b)0.02608m
- c)2.608cm
- d)2.6008 cm
Correct answer is option 'D'. Can you explain this answer?
Find the option with 5 significant figures
a)
26.08mm
b)
0.02608m
c)
2.608cm
d)
2.6008 cm
|
|
Preeti Iyer answered |
In option a, b and c, we only have 4 significant figures in the number
In option d, the number has 5 significant figures.
Find the option with 3 significant figures.- a)0.0268
- b)2.068
- c)2.608
- d)26.08
Correct answer is option 'A'. Can you explain this answer?
Find the option with 3 significant figures.
a)
0.0268
b)
2.068
c)
2.608
d)
26.08
|
|
Raghav Bansal answered |
0.0268 have three significant figures where as 2.608 cm, 26.08 mm and 2.068 cm have four significant figures.
Can you explain the answer of this question below:The numbers of significant figures in 9.1 x10-31 kg are:
- A:
three
- B:
two
- C:
ten
- D:
twenty one
The answer is b.
The numbers of significant figures in 9.1 x10-31 kg are:
three
two
ten
twenty one
|
Om Desai answered |
As the power of 10 is irrelevant to the determination of significant figures hence, the numbers of significant figures are 2 (9 and 1).
Numbers of significant figures in the volume of a cube of side 6.103 m are:- a)2
- b)4
- c)3
- d)1
Correct answer is option 'B'. Can you explain this answer?
Numbers of significant figures in the volume of a cube of side 6.103 m are:
a)
2
b)
4
c)
3
d)
1
|
|
Raghav Bansal answered |
- As per rule, zero between two non zero digits are significant.
- The number of a significant figure in the length of the side is 4.
- So the calculated volume should be rounded off to 4 significant figures. Multiplying numbers do not increase significant figures.
The number of significant figures in the distance of one light year , 9.4605 × 1015 m is- a)thirty
- b)five
- c)four
- d)three
Correct answer is option 'B'. Can you explain this answer?
The number of significant figures in the distance of one light year , 9.4605 × 1015 m is
a)
thirty
b)
five
c)
four
d)
three
|
|
Preeti Iyer answered |
As the power of 10 is irrelevant in the determination of significant figures hence, the numbers of significant figures are 5.
A 2.16 g of a substance occupies 1.5 cm3. Its density to appropriate significant figure is:- a)1.444
- b)1.44
- c)1.4
- d)1.5
Correct answer is option 'D'. Can you explain this answer?
A 2.16 g of a substance occupies 1.5 cm3. Its density to appropriate significant figure is:
a)
1.444
b)
1.44
c)
1.4
d)
1.5
|
Hansa Sharma answered |
The multiplication of the numbers:
10.610 × 0.210 = 2.2281
10.610 × 0.210 = 2.2281
Since, the numbers 10.610 and 0.210 have 5 and 3 significant digits, respectively. So, the final result must also have 3 significant digits.
The final answer is 2.23.
The order of magnitude of a number expressed in scientific notation is- a)number of decimal places
- b)number of digit before decimal
- c)all digits in the number
- d)power of ten
Correct answer is 'D'. Can you explain this answer?
The order of magnitude of a number expressed in scientific notation is
a)
number of decimal places
b)
number of digit before decimal
c)
all digits in the number
d)
power of ten
|
Lavanya Menon answered |
- To convert a regular number to scientific notation, we first rewrite it as a decimal, then multiply it by a power of 10.
- There is an infinite number of ways to do this for any given number, but we always prefer the one that has only a single digit in front of the decimal point.
Can you explain the answer of this question below:After decimal point, number is less than 5 which is to be neglected,then retained number
- A:
gets increased
- B:
get decreased
- C:
remains unchanged
- D:
may decreased
The answer is c.
After decimal point, number is less than 5 which is to be neglected,then retained number
gets increased
get decreased
remains unchanged
may decreased
|
|
Preeti Iyer answered |
In rounding off numbers, the last figure kept should be unchanged if the first figure dropped is less than 5.
Example: If only one decimal is to be kept, then 6.422 becomes 6.4.
Example: If only one decimal is to be kept, then 6.422 becomes 6.4.
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
Correct answer is option 'D'. Can you explain this answer?
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
|
Mira Joshi answered |
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.
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
Correct answer is option 'B'. Can you explain this answer?
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
|
Dev Patel answered |
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.
∴ 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.
The order of magnitude of a number expressed in scientific notation is- a)number of decimal places
- b)number of digit before decimal
- c)all digits in the number
- d)power of ten
Correct answer is option 'D'. Can you explain this answer?
The order of magnitude of a number expressed in scientific notation is
a)
number of decimal places
b)
number of digit before decimal
c)
all digits in the number
d)
power of ten
|
Naina Bansal answered |
To convert a regular number to scientific notation, we first rewrite it as a decimal, then multiply it by a power of 10. There is an infinite number of ways to do this for any given number, but we always prefer the one that has only a single digit in front of the decimal point.
Which of the following physical quantity is dimensionless?- a)stress
- b)strain
- c)area
- d)pressure
Correct answer is option 'B'. Can you explain this answer?
Which of the following physical quantity is dimensionless?
a)
stress
b)
strain
c)
area
d)
pressure
|
Shivani answered |
Strain has no dimensions...as it is the ratio of same quantities..
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
Correct answer is option 'B'. Can you explain this answer?
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
|
Jyoti Sengupta answered |
The final result should be 3 significant figures.
Three students measured length of a needle with meter rod and recorded as :
(i) 0.2145m
(ii) 0.21m
(iii) 0.214m
Which one is correct record?- a)Both (i) and (ii)
- b)Only (iii)
- c)Only (ii)
- d)Only (i)
Correct answer is option 'C'. Can you explain this answer?
Three students measured length of a needle with meter rod and recorded as :
(i) 0.2145m
(ii) 0.21m
(iii) 0.214m
Which one is correct record?
(i) 0.2145m
(ii) 0.21m
(iii) 0.214m
Which one is correct record?
a)
Both (i) and (ii)
b)
Only (iii)
c)
Only (ii)
d)
Only (i)
|
EduRev NEET answered |
The correct answer is C.
Only the measurement (ii) is accurate because:
- A meter rod typically measures up to two decimal points.
- Measurements (i) and (iii) exceed this precision.
Thus, the only valid record is 0.21m.
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
Correct answer is option 'C'. Can you explain this answer?
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
|
Anjali Sharma answered |
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.
6.032 has four significant figures. 0.0006032 has four significant figures.
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
Correct answer is option 'A'. Can you explain this answer?
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
|
|
Dev Patel answered |
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
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
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
Correct answer is option 'A'. Can you explain this answer?
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
|
Vivek Patel answered |
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
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
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
Correct answer is option 'D'. Can you explain this answer?
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
|
|
Ajay Yadav answered |
The power of 10 is irrelevant to the determination of significant figures.
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
Correct answer is option 'C'. Can you explain this answer?
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
|
|
Aurora Institute answered |
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.
The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, respectively. The area of the sheet in appropriate significant figures and error would be, respectively,- a)163.6 ± 2.6 cm2
- b)163.62 ± 2.6 cm2
- c)164 ± 3 cm2
- d)163.62 ± 3 cm2
Correct answer is option 'C'. Can you explain this answer?
The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, respectively. The area of the sheet in appropriate significant figures and error would be, respectively,
a)
163.6 ± 2.6 cm2
b)
163.62 ± 2.6 cm2
c)
164 ± 3 cm2
d)
163.62 ± 3 cm2
|
|
Parth Jain answered |
Understanding the Area Calculation
To find the area of a rectangular sheet, we use the formula:
Area = Length × Breadth
Given:
- Length = 16.2 cm
- Breadth = 10.1 cm
Calculating the Area
Now, we calculate:
Area = 16.2 cm × 10.1 cm = 163.62 cm²
Determining Significant Figures
When performing multiplication, the result should be expressed in the least number of significant figures from the measurements used:
- Length (16.2 cm) has 3 significant figures.
- Breadth (10.1 cm) has 3 significant figures.
Thus, the area should be reported with 3 significant figures:
Area = 164 cm² (rounded from 163.62 cm²)
Calculating the Error
To evaluate the error in the area calculation, we need the absolute errors from both dimensions:
- Error in Length = 0.1 cm (1 significant figure)
- Error in Breadth = 0.1 cm (1 significant figure)
Using the formula for error propagation in multiplication:
% Error = (Error in Length/Length + Error in Breadth/Breadth) × 100
Calculating:
- % Error in Length = 0.1/16.2 ≈ 0.00617 (or 0.617%)
- % Error in Breadth = 0.1/10.1 ≈ 0.00990 (or 0.990%)
Total % Error ≈ 0.617% + 0.990% ≈ 1.607% (approximately 1.6%)
Now, find the absolute error in the area:
Absolute Error = Area × % Error = 164 cm² × 0.01607 ≈ 2.63 cm² (rounded to 3 cm²)
Final Result
Thus, the area of the sheet is:
Area = 164 ± 3 cm²
This matches option C: 164 ± 3 cm².
To find the area of a rectangular sheet, we use the formula:
Area = Length × Breadth
Given:
- Length = 16.2 cm
- Breadth = 10.1 cm
Calculating the Area
Now, we calculate:
Area = 16.2 cm × 10.1 cm = 163.62 cm²
Determining Significant Figures
When performing multiplication, the result should be expressed in the least number of significant figures from the measurements used:
- Length (16.2 cm) has 3 significant figures.
- Breadth (10.1 cm) has 3 significant figures.
Thus, the area should be reported with 3 significant figures:
Area = 164 cm² (rounded from 163.62 cm²)
Calculating the Error
To evaluate the error in the area calculation, we need the absolute errors from both dimensions:
- Error in Length = 0.1 cm (1 significant figure)
- Error in Breadth = 0.1 cm (1 significant figure)
Using the formula for error propagation in multiplication:
% Error = (Error in Length/Length + Error in Breadth/Breadth) × 100
Calculating:
- % Error in Length = 0.1/16.2 ≈ 0.00617 (or 0.617%)
- % Error in Breadth = 0.1/10.1 ≈ 0.00990 (or 0.990%)
Total % Error ≈ 0.617% + 0.990% ≈ 1.607% (approximately 1.6%)
Now, find the absolute error in the area:
Absolute Error = Area × % Error = 164 cm² × 0.01607 ≈ 2.63 cm² (rounded to 3 cm²)
Final Result
Thus, the area of the sheet is:
Area = 164 ± 3 cm²
This matches option C: 164 ± 3 cm².
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
Correct answer is option 'B'. Can you explain this answer?
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
|
|
Ritika Reddy answered |
Significant figures refer to the number of digits in a number that have meaning or contribute to the accuracy of the number. Here, we need to determine the number of significant figures in the numbers 4.8000 x 104 and 48000.50.
4.8000 x 104:
- There are five digits after the decimal point, but trailing zeros after a decimal point are not significant.
- The number 4.8000 has five significant figures because all the zeros between the non-zero digits are significant.
- The exponent 104 indicates that the decimal point is shifted 4 places to the right, so there are 4 additional significant figures.
- Therefore, the total number of significant figures is 5 + 4 = 9.
48000.50:
- There are six digits before the decimal point and two digits after the decimal point.
- All the digits are non-zero, so they are all significant.
- Therefore, the total number of significant figures is 6 + 2 = 8.
Therefore, the correct answer is option B - 5 and 7.
4.8000 x 104:
- There are five digits after the decimal point, but trailing zeros after a decimal point are not significant.
- The number 4.8000 has five significant figures because all the zeros between the non-zero digits are significant.
- The exponent 104 indicates that the decimal point is shifted 4 places to the right, so there are 4 additional significant figures.
- Therefore, the total number of significant figures is 5 + 4 = 9.
48000.50:
- There are six digits before the decimal point and two digits after the decimal point.
- All the digits are non-zero, so they are all significant.
- Therefore, the total number of significant figures is 6 + 2 = 8.
Therefore, the correct answer is option B - 5 and 7.
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
Correct answer is option 'C'. Can you explain this answer?
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
|
|
Preeti Iyer answered |
Radius of the sphere, r = 1.41cm
(3 significant figures)
Volume of the sphere,
(3 significant figures)
Volume of the sphere,
= 11.736 cm3
Rounded off upto 3 significant figures = 11.7 cm3.
Rounded off upto 3 significant figures = 11.7 cm3.
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)22
Correct answer is option 'B'. Can you explain this answer?
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)
22
|
Flembe Academy answered |
Area of rectangle is length multiplied by breadth. The area of this rectangle is 4.5 × 5.9 mm2 = 26.55 mm2.
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
Correct answer is option 'A'. Can you explain this answer?
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
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Lavanya Menon answered |
Difference in masses
= m2 - m1 = 20.17 g - 20.15 g
= 0.02 g (correct upto two places of decimal)
= m2 - m1 = 20.17 g - 20.15 g
= 0.02 g (correct upto two places of decimal)
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