The number of significant figures in 0.06900 is
5
4
2
3
Get Answer
(b) In , the two zeroes before six are not siginificant and two zeroes on right side of 9 are significant figures. Hence, number of significant figures are four (6900).
The sum of the numbers and 0.301 in appropriate significant figures is
663.821
664
663.8
663.82
(b) In addition the result will be in least number of places after decimal.and minimum number of significant figure. The sum of the given numbers can be calculated as 663.821 . The number with least decimal places is 227.2 is correct to only one decimal place but in addition of numbers, the final result should be rounded off to one decimal place i.e., 664.
The mass and volume of a body are and , respectively. The density of the material of the body in correct significant figures is
(c) As we know that in multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures.
The significant figure in given numbers and 2.5 are four and two respectivey so, density should be reported to two significant figures.
Density
As rounding off the number upto 2 significant figures, we get density .
The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give
2.75 and 2.74
2.74 and 2.73
2.75 and 2.73
2.74 and 2.74
(d) Rounding off 2.745 upto 3 significant figures here IVth digit is 5 and its preceding is even, so no change in 4. Thus answer would be 2.74 . Rounding off 2.735 upto 3 significant figures, here IV digit is 5 and its preceding digit is 3 (odd). So 3 is increased by 1 answer become would be 2.74 .
The length and breadth of a rectangular sheet are and , respectively. The area of the sheet in appropriate significant figures and error is
(a) If is error in a physical quantity, then relative error is calculated as .
Given that,
Length
Breadth Area
In significant figure rounding off to three significant digits, area
So,
Now rounding off up to one significant figure .
So, Area .
Which of the following pairs of physical quantities does not have same dimensional formula?
Work and torque
Angular momentum and Planck's constant
Tension and surface tension
Impulse and linear momentum
(c) Tension force
Surface tension
. The value of will be
(a) By applying the Rule of significant figure in multiplication and addition.
As given that,
(consider only significant figure value)
(rounding off to two significant figures.)
You measure two quantities as . We should report correct value for as
(d) As given that,
Rounding off upto two significant digit
(Let)
Rounding off upto one significant digit
So, correct value of
Young's modulus of steel is . When expressed in CGS units of dyne , it will be equal to (1N dyne, )
(c) It is given that Young's modulus is,
dyne
So, dyne
Convert meter to centimeter
If momentum , area and time are taken to be fundamental quantities, then energy has the dimensional formula
(d) Given that fundamental quantities are momentum , area and time .
Let us consider the dimensional formula for
where is dimensionless constant of proportionality.
Dimensions of energy and Dimension of
momentum
Dimension of Area
Dimension of Time
Dimension of energy
Putting all the dimensions, value
By principle of homogeneity of dimensions,
So, Dimensional formula (of energy)
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