1
Units and Measurements
Multiple Choice Questions (MCQs)
Q. 1 The number of significant figures in 0.06900 is
(a) 5 (b) 4 (c) 2 (d) 3
K Thinking Process
If the number is less than 1, the zero(s) on the right of decimal point and before the
first non-zero digit are not significant.
Ans. (b) In 0.06900, the underlined zeroes are not significant. Hence, number of significant
figures are four (6900).
Q. 2 The sum of the numbers 436.32, 227.2 and 0.301 in appropriate
significant figures is
(a) 663.821 (b) 664 (c) 663.8 (d) 663.82
Ans. (b) The sum of the numbers can be calculated as 663.821 arithmetically. The number with
least decimal places is 227.2 is correct to only one decimal place.
The final result should, therefore be rounded off to one decimal place i.e., 664.
Note In calculating the sum, we should not confuse with the number of decimal places
and significant figures. The result should have least number of decimal places.
Q. 3 The mass and volume of a body are 4.237 g and 2.5 cm 3 , respectively. The
density of the material of the body in correct significant figures is
(a) 1.6048 g cm-3 (b) 1.69 g cm-3 (c) 1.7 g cm-3 (d) 1.695 g cm-3
K Thinking Process
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.
Ans. (c) In this question, density should be reported to two significant figures.
4.237g
Density = = 1.6948
2.5 cm3
As rounding off the number, we get density = 17
.
,Q. 4 The numbers 2.745 and 2.735 on rounding off to 3 significant figures will
give
(a) 2.75 and 2.74 (b) 2.74 and 2.73 (c) 2.75 and 2.73 (d) 2.74 and 2.74
Ans. (d) Rounding off 2.745 to 3 significant figures it would be 2.74. Rounding off 2.735 to
3 significant figures it would be 2.74.
Q. 5 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 is
(a) 164 ± 3 cm2 (b) 163.62 ± 2.6 cm2
(c) 163.6 ± 2.6 cm2 (d) 163.62 ± 3 cm2
K Thinking Process
Dx
If D x is error in a physical quantity, then relative error is calculated as .
x
Ans. (a) Given, length l = (16.2 ± 0.1) cm
Breadth b = (10.1 ± 0.1) cm
Area A = l ´ b
= (16.2 cm) ´ (10.1 cm) = 163.62 cm2
Rounding off to three significant digits, area A = 164 cm2
D A Dl Db 0.1 0.1
= + = +
A l b 16.2 10.1
1.01 + 1.62 2.63
= =
16.2 ´ 10.1 163.62
2.63 2.63
Þ DA = A ´ = 163.62 ´ = 2.63 cm2
163.62 163.62
DA = 3 cm2 (By rounding off to one significant figure)
\ Area, A = A ± DA = (164 ± 3) cm2 .
Q. 6 Which of the following pairs of physical quantities does not have same
dimensional formula?
(a) Work and torque (b) Angular momentum and Planck’s constant
(c) Tension and surface tension (d) Impulse and linear momentum
Ans. (c) (a) Work =force ´ distance= [MLT-2 ][L] [ML2 T-2 ]
Torque=force ´ distance=[ML2 T-2 ]
(b) Angular momentum =mvr=[M][LT1][L] = [ML2 T-1]
E [ML2 T-2 ]
Planck’s constant = = =[ML2 T-1 ]
V [T-1 ]
(c) Tension=force=[MLT-2 ]
force [MLT-2 ]
Surface tension = = =[ML0 T-2 ]
length [L]
(d) Impulse =force ´ time=[MLT-2 ][T] = [MLT-1 ]
Momentum = mass ´ velocity =[M][LT-1 ] = [MLT-1 ]
Note One should not be confused with the similar form tension in both the physical
quantities-surface tension and tension. Dimensional formula for both of them is
not same.
,Q. 7 Measure of two quantities along with the precision of respective
measuring instrument is
A = 2.5 ms -1 ± 0.5 ms -1 , B = 0.10 s ± 0.01 s. The value of AB will be
(a) (0.25 ± 0.08) m (b) (0.25 ± 0.5) m (c) (0.25 ± 0.05) m (d) (0.25 ± 0.135) m
-1
Ans. (a) Given, A = 2.5 ms ± 0.5 ms -1, B = 0.10 s ± 0.01 s
x = AB = (2.5)(010
. ) = 0.25m
Dx DA DB
= +
x A B
0.5 0.01 0.05 + 0.025 0.075
= + = =
2.5 0.10 0.25 0.25
Dx = 0.075 = 0.08 m, rounding off to two significant figures.
AB = (0.25 ± 0.08) m
Q. 8 You measure two quantities as A = 1.0 m ± 0.2 m, B = 2 .0 m ± 0.2 m. We
should report correct value for AB as
(a) 1.4 m ± 0.4 m (b) 1.41 m ± 0.15 m (c) 1.4 m ± 0.3 m (d) 1.4 m ± 0.2 m
Ans. (d) Given, A = 10. m ± 0.2 m, B = 2 . 0 m ± 0.2 m
Let, Y = AB = (10 . )(2.0) = 1414
. m
Rounding off to two significant digit Y = 14. m
DY 1 é DA DB ù 1 é 0.2 0.2 ù 0.6
= + = + =
Y ê
2ë A ú ê ú
B û 2 ë 1.0 2.0 û 2 ´ 2.0
0.6Y 0.6 ´ 14.
Þ DY = = = 0.212
2 ´ 2.0 2 ´ 2.0
Rounding off to one significant digit DY = 0.2 m
Thus, correct value for AB = r + D r = 1.4 ± 0.2m
Q. 9 Which of the following measurement is most precise?
(a) 5.00 mm (b) 5.00 cm (c) 5.00 m (d) 5.00 km
Ans. (a) All given measurements are correct upto two decimal places. As here 5.00 mm has the
smallest unit and the error in 5.00 mm is least (commonly taken as 0.01 mm if not
specified), hence, 5.00 mm is most precise.
Note In solving these type of questions, we should be careful about units although
their magnitude is same.
Q. 10 The mean length of an object is 5 cm. Which of the following
measurements is most accurate?
(a) 4.9 cm (b) 4.805 cm (c) 5.25 cm (d) 5.4 cm
Ans. (a) Given length l = 5 cm
Now, checking the errors with each options one by one, we get
D l 1 = 5 - 4.9 = 0.1cm
D l 2 = 5 - 4.805 = 0.195 cm
D l 3 = 5.25 - 5 = 0.25 cm
D l 4 = 5. 4 - 5 = 0. 4 cm
Error D l 1 is least.
Hence, 4.9 cm is most precise.
, Q. 11 Young’s modulus of steel is 1.9 ´ 10 11 N / m2 . When expressed in CGS
units of dyne/cm2 , it will be equal to (1N = 10 5 dyne, 1 m2 = 10 4 cm2 )
(a) 1.9 ´ 10 10 (b) 1.9 ´ 10 11 (c) 1.9 ´ 10 12 (d) 1.9 ´ 10 13
11 2
Ans. (c) Given, Young’s modulus Y = 19
. ´ 10 N/ m
1N = 105 dyne
Hence, . ´ 1011 ´ 105 dyne/ m2
Y = 19
We know that 1m = 100cm
\ . ´ 1011 ´ 105 dyne/(100)2 cm2
Y = 19
= 1.9 ´ 1016 - 4 dyne/cm2
. ´ 1012 dyne/cm2
Y = 19
Note While we are going through units conversion, we should keep in mind that proper
relation between units are mentioned.
Q. 12 If momentum (p), area (A) and time (T ) are taken to be fundamental
quantities, then energy has the dimensional formula
(a) [pA -1T 1] (b) [p 2AT] (c) [p A -1/2T] (d) [ pA1/2T]
Ans. (d) Given, fundamental quantities are momentum (p), area ( A) and time (T ).
We can write energy E as
E µ pa A bTc
E = kpa A A Tc
where k is dimensionless constant of proportionality.
Dimensions of E = [E ] = [ML2 T-2 ] and [ p] = [MLT-1 ]
[A] = [L2 ]
[T] = [T]
[ ] a[A] b[T]c
[E] = [K]p
Putting all the dimensions, we get
ML2 T-2 = [MLT-1 ] a[L2 ] b[T]c
= M aL 2 b + a T - a + c
By principle of homogeneity of dimensions,
a = 1, 2 b + a = 2
Þ 2b + 1 = 2
Þ b = 1/2 - a + c = - 2
Þ c = -2 + a = -2 + 1= -1
Hence, E = pA1/ 2T -1
Units and Measurements
Multiple Choice Questions (MCQs)
Q. 1 The number of significant figures in 0.06900 is
(a) 5 (b) 4 (c) 2 (d) 3
K Thinking Process
If the number is less than 1, the zero(s) on the right of decimal point and before the
first non-zero digit are not significant.
Ans. (b) In 0.06900, the underlined zeroes are not significant. Hence, number of significant
figures are four (6900).
Q. 2 The sum of the numbers 436.32, 227.2 and 0.301 in appropriate
significant figures is
(a) 663.821 (b) 664 (c) 663.8 (d) 663.82
Ans. (b) The sum of the numbers can be calculated as 663.821 arithmetically. The number with
least decimal places is 227.2 is correct to only one decimal place.
The final result should, therefore be rounded off to one decimal place i.e., 664.
Note In calculating the sum, we should not confuse with the number of decimal places
and significant figures. The result should have least number of decimal places.
Q. 3 The mass and volume of a body are 4.237 g and 2.5 cm 3 , respectively. The
density of the material of the body in correct significant figures is
(a) 1.6048 g cm-3 (b) 1.69 g cm-3 (c) 1.7 g cm-3 (d) 1.695 g cm-3
K Thinking Process
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.
Ans. (c) In this question, density should be reported to two significant figures.
4.237g
Density = = 1.6948
2.5 cm3
As rounding off the number, we get density = 17
.
,Q. 4 The numbers 2.745 and 2.735 on rounding off to 3 significant figures will
give
(a) 2.75 and 2.74 (b) 2.74 and 2.73 (c) 2.75 and 2.73 (d) 2.74 and 2.74
Ans. (d) Rounding off 2.745 to 3 significant figures it would be 2.74. Rounding off 2.735 to
3 significant figures it would be 2.74.
Q. 5 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 is
(a) 164 ± 3 cm2 (b) 163.62 ± 2.6 cm2
(c) 163.6 ± 2.6 cm2 (d) 163.62 ± 3 cm2
K Thinking Process
Dx
If D x is error in a physical quantity, then relative error is calculated as .
x
Ans. (a) Given, length l = (16.2 ± 0.1) cm
Breadth b = (10.1 ± 0.1) cm
Area A = l ´ b
= (16.2 cm) ´ (10.1 cm) = 163.62 cm2
Rounding off to three significant digits, area A = 164 cm2
D A Dl Db 0.1 0.1
= + = +
A l b 16.2 10.1
1.01 + 1.62 2.63
= =
16.2 ´ 10.1 163.62
2.63 2.63
Þ DA = A ´ = 163.62 ´ = 2.63 cm2
163.62 163.62
DA = 3 cm2 (By rounding off to one significant figure)
\ Area, A = A ± DA = (164 ± 3) cm2 .
Q. 6 Which of the following pairs of physical quantities does not have same
dimensional formula?
(a) Work and torque (b) Angular momentum and Planck’s constant
(c) Tension and surface tension (d) Impulse and linear momentum
Ans. (c) (a) Work =force ´ distance= [MLT-2 ][L] [ML2 T-2 ]
Torque=force ´ distance=[ML2 T-2 ]
(b) Angular momentum =mvr=[M][LT1][L] = [ML2 T-1]
E [ML2 T-2 ]
Planck’s constant = = =[ML2 T-1 ]
V [T-1 ]
(c) Tension=force=[MLT-2 ]
force [MLT-2 ]
Surface tension = = =[ML0 T-2 ]
length [L]
(d) Impulse =force ´ time=[MLT-2 ][T] = [MLT-1 ]
Momentum = mass ´ velocity =[M][LT-1 ] = [MLT-1 ]
Note One should not be confused with the similar form tension in both the physical
quantities-surface tension and tension. Dimensional formula for both of them is
not same.
,Q. 7 Measure of two quantities along with the precision of respective
measuring instrument is
A = 2.5 ms -1 ± 0.5 ms -1 , B = 0.10 s ± 0.01 s. The value of AB will be
(a) (0.25 ± 0.08) m (b) (0.25 ± 0.5) m (c) (0.25 ± 0.05) m (d) (0.25 ± 0.135) m
-1
Ans. (a) Given, A = 2.5 ms ± 0.5 ms -1, B = 0.10 s ± 0.01 s
x = AB = (2.5)(010
. ) = 0.25m
Dx DA DB
= +
x A B
0.5 0.01 0.05 + 0.025 0.075
= + = =
2.5 0.10 0.25 0.25
Dx = 0.075 = 0.08 m, rounding off to two significant figures.
AB = (0.25 ± 0.08) m
Q. 8 You measure two quantities as A = 1.0 m ± 0.2 m, B = 2 .0 m ± 0.2 m. We
should report correct value for AB as
(a) 1.4 m ± 0.4 m (b) 1.41 m ± 0.15 m (c) 1.4 m ± 0.3 m (d) 1.4 m ± 0.2 m
Ans. (d) Given, A = 10. m ± 0.2 m, B = 2 . 0 m ± 0.2 m
Let, Y = AB = (10 . )(2.0) = 1414
. m
Rounding off to two significant digit Y = 14. m
DY 1 é DA DB ù 1 é 0.2 0.2 ù 0.6
= + = + =
Y ê
2ë A ú ê ú
B û 2 ë 1.0 2.0 û 2 ´ 2.0
0.6Y 0.6 ´ 14.
Þ DY = = = 0.212
2 ´ 2.0 2 ´ 2.0
Rounding off to one significant digit DY = 0.2 m
Thus, correct value for AB = r + D r = 1.4 ± 0.2m
Q. 9 Which of the following measurement is most precise?
(a) 5.00 mm (b) 5.00 cm (c) 5.00 m (d) 5.00 km
Ans. (a) All given measurements are correct upto two decimal places. As here 5.00 mm has the
smallest unit and the error in 5.00 mm is least (commonly taken as 0.01 mm if not
specified), hence, 5.00 mm is most precise.
Note In solving these type of questions, we should be careful about units although
their magnitude is same.
Q. 10 The mean length of an object is 5 cm. Which of the following
measurements is most accurate?
(a) 4.9 cm (b) 4.805 cm (c) 5.25 cm (d) 5.4 cm
Ans. (a) Given length l = 5 cm
Now, checking the errors with each options one by one, we get
D l 1 = 5 - 4.9 = 0.1cm
D l 2 = 5 - 4.805 = 0.195 cm
D l 3 = 5.25 - 5 = 0.25 cm
D l 4 = 5. 4 - 5 = 0. 4 cm
Error D l 1 is least.
Hence, 4.9 cm is most precise.
, Q. 11 Young’s modulus of steel is 1.9 ´ 10 11 N / m2 . When expressed in CGS
units of dyne/cm2 , it will be equal to (1N = 10 5 dyne, 1 m2 = 10 4 cm2 )
(a) 1.9 ´ 10 10 (b) 1.9 ´ 10 11 (c) 1.9 ´ 10 12 (d) 1.9 ´ 10 13
11 2
Ans. (c) Given, Young’s modulus Y = 19
. ´ 10 N/ m
1N = 105 dyne
Hence, . ´ 1011 ´ 105 dyne/ m2
Y = 19
We know that 1m = 100cm
\ . ´ 1011 ´ 105 dyne/(100)2 cm2
Y = 19
= 1.9 ´ 1016 - 4 dyne/cm2
. ´ 1012 dyne/cm2
Y = 19
Note While we are going through units conversion, we should keep in mind that proper
relation between units are mentioned.
Q. 12 If momentum (p), area (A) and time (T ) are taken to be fundamental
quantities, then energy has the dimensional formula
(a) [pA -1T 1] (b) [p 2AT] (c) [p A -1/2T] (d) [ pA1/2T]
Ans. (d) Given, fundamental quantities are momentum (p), area ( A) and time (T ).
We can write energy E as
E µ pa A bTc
E = kpa A A Tc
where k is dimensionless constant of proportionality.
Dimensions of E = [E ] = [ML2 T-2 ] and [ p] = [MLT-1 ]
[A] = [L2 ]
[T] = [T]
[ ] a[A] b[T]c
[E] = [K]p
Putting all the dimensions, we get
ML2 T-2 = [MLT-1 ] a[L2 ] b[T]c
= M aL 2 b + a T - a + c
By principle of homogeneity of dimensions,
a = 1, 2 b + a = 2
Þ 2b + 1 = 2
Þ b = 1/2 - a + c = - 2
Þ c = -2 + a = -2 + 1= -1
Hence, E = pA1/ 2T -1
