(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)1/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
(D) Answers of Conceptual Questions
3
−1 − 2
32. 0.01 mm
1. m kg s
33. (a) 0.005 s , (b) maximum error = 0.012 s
2.(a) 1.745×10−2 rad , (b) 2.908×10− 4 rad , (c)
34. (a) 0.11 s, (b) 0.04, (c) 4%
35.
−6
4.847×10 rad
3.(a) 10−6 ,
(b) 1.51×10 4 ( mm)2 ,
(c) 5 m ,
30 o C±1 o C
36. 7%
(d)11.3 g / cm 3 or 1.13×10 4 kg/ m 3
37. (a) (300 ±7) ohm, (b) (66.7 ± 1.8) ohm
4.(a) 107 ,
38.
(b) 10−6 ,
6.67×10−8
(c) 3.9×10 4 ,(d)
ΔZ
Δ A 1 Δ B ΔC 3 Δ D
=4
+
+
+
Z
A 3 B
C
2 D
39. 3%
6. 500
40. 0.035 mm
7. V ≈ 3×10−7 m3
41. 94.1
8. r≈ 104
42. 1 part in 1011 to 1012
9. 119 m
43. (0.25±0.08)m
10. 3.84×108 m
44.
11. ≈3×1016 m )
(1.4±0.2) m
45. (I) 13% , (II) 3.8
12.1.32 pc, 1.52″
46. (b) and (c) are wrong
13. 1.39×10 9 m
−1/2
47. m= m0 (1−(v 2 /c 2 ) )
14. ≈1 m
48. Correct
15. 1.429×10 5 km
49. (a), (c) and (e) ruled out, (b) and (d) allowed
16. 3.84×108 m
50.
17. 55.8 km
T = k √l / g
51. (b)
1 u =931.5MeV /c 2
19. 3580 km
52. (a)
ΔX
=12.5% , (b)
X
20. (a) 1.9o , (b) DEarth ≈ 4 Dmoon ,
53. dimensionless
18. 2.8×10 22 km
(c)
DSun / DEarth =100
5
21. (a) 1 parsec≈ 2×10 A.U. ,
(b) 15×10−3 min arc , (c) 30′
23.
1.7 g/ cm 3
54. (1)
√
ch
G
, (2)
√
hG
c3
56. valid
58.
tan θ=v / v rain
24. 164±3 cm 2
59. ρ Na =4.6×10 2 kg /m 3 .
25. 8.72 m 2 , 0.0855 m3
60. ρ nucleus =2.3×10 17 kg/m 3
26. (a) 2.3 kg, (b) 0.02 g
27. 311.3 m2 , 373.7 m3
28. 4.8 g/cm3
29. Clock 2
30. (a)
31. (a)
XI-I-v1-(ANQ)-15042011
X =2.8 ± 0.3
, (3)
√
hG
c5
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)2/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
Solutions:
1 kg m 2 s−2 =(1000 g )(100 cm) 2 s− 2
.
=107 g cm 2 s−2
N4. (a)
N1. From the given formula, the gravitational
(b) 1 ly = distance travelled by light (in vacuum) in
constant G may be written as
one year. Therefore,
G=
1 ly=3×108 m s−1×(365×24×3600 s) ,
F r2
m1 m2
or
Substituting SI units for force, distance and mass, we
get the following SI units for G:
or
1
ly ,
( 9.46×10−8 )
SI unit of G=m 3 kg−1 s−2 .
(c) 3 m s−2 =3( 10− 3 km)(
π
3.141
=
=1.745×10− 2 rad .
180 180
1o 1.745×10−2
1 '= =
=2.908×10− 4 rad
.
60
60
=2.91×10−4 rad
1 m =1.06×10−6 ly=10−6 ly .
1 ' 2.908×10
=
= 4.847×10−6
.
60
60
−6
= 4.85×10 rad
3
×36002 km h−2 ,
1000
or
=
or
=38880 km h −2 ,
or
=3.9×104 km h−2 .
and 1 N=1 kg m s−2 . Therefore,
1' ' =
G=6.67×10−11 m 3 kg −1 s−2 ,
or
=6.67×10−11×(100 cm) 3×(103 kg)−1×s−2
,
10−6 .
(b)Surface area = 2 π r (r+l ) ,
or
= 2×3.14×20×120( mm)2 ,
or
= 1.51×10 4 ( mm)2 .
(c) distance = speed × time
= 18(
1 − 2 −2
) h
,
3600
(d) G=6.67×10−11 N m 2 kg−2 ,
−4
N3. (a)
1m=
or
1o=
(c)
Therefore,
( kg m s−2 )( m 2 )
SI unit of G=
,
kg 2
N2. (a) Since 2 π (rad )=360 o ,
(b)
=9.46×10−8 m .
=6.67×10−11×10 6 ×(
1
) cm 3 g−1 s−2 ,
103
=6.67×10−8 cm 3 g −1 s− 2 .
N5. Let in the new system of units, the length, mass
km
1000 m
)×1 s=18×(
)×1 s ,
h
3600 s
and time units are:
m N , kg N and s N . It is
given that
= 5m .
(d) Density of water = 1 g /cm 3 . Therefore,
1 m N =β m ,
Density of lead =(relative density)×(density of water)
1 kg N =α kg ,
= 11.3 × 1 g / cm 3
1 s N =γ s .
= 11.3 g / cm 3 .
(
or Density lead = 11.3×
Therefore,
)
10−3 kg
kg
=1.13×10 4 3
−6
3
10 m
m
1 J N =1 kg N⋅m 2N⋅s−2
,
N
.
or
or
XI-I-v1-(ANQ)-15042011
=(α kg)⋅(β2 m2 )⋅( γ−2 s −2 ) ,
1 J N =(α β2 γ−2 ) kg m 2 s−2 ,
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)3/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
or
1 J N =(α β2 γ−2 ) J .
Therefore, volume of molecules of hydrogen in one
mole of gas is
Therefore,
1cal= 4.2 J =
or
V = N V 1 =6.02×10 23×4.29×10−30 m3 ,
4.2
JN ,
( αβ 2 γ−2 )
or
−1 −2 2
1 cal= 4.2 α β
γ JN .
V =2.58×10−6 .
The volume of one mole of gas
V gas= 22.4 L= 22.4×10−3 m 3 .
N6. In SI units, the speed of light is
8
c=3×10 m/s .
Therefore, the ratio molar volume to volume of
(1)
molecules is
Let the new unit of length is m N . It is given that
c=1 m N / s .
r=
(2)
1 mN =3×10 m .
V
=
22.4×10−3
=8.68×10 3≈ 104 .
2.58×10−6
This ratio is large because intermolecular separation
Comparing (1) and (2), we get
8
V gas
in a gas is much larger than the size of a molecule.
(3)
The distance between the Sun and the Earth is
d =c t=3×108 ( m/s)×500(s)=15×1010 m ,
(
10
or d =15×10 ×
N9. The parallax angle is 40o . From ΔABC,
AC = AB / tanθ
)
1
m N =500 m N .
3×10 8
= 100 / tan40o metre
= 100 / 0.8391
= 119.17
N7. Avogadro constant is
= 119 m
N A=6.022×1023 mol−1 .
The volume of one hydrogen atom is
V 1=
or
4π r 3 4×3.14×( 0.5×10−10 )3 3
=
m ,
3
3
V 1 =5.23×10−31 m 3 .
Therefore, the volume of one mole of hydrogen
atoms is
V = N A V 1 =6.02×10 23×5.23×10−31 m3 ,
or
V =3.15×10−7 m3 ≈3×10− 7 m3 .
N8. The size of hydrogen molecule is
3
−10 3
4 π r 4×3.14×(1×10
V 1=
=
3
3
)
m3 ,
V 1 =4.19×10−30 m 3 .
N10. According to parallax method
D=
b
.
θ
(1)
Given, b=1.276×107 m ,
and
θ=1o 54 '=114 ' .
Converting the degree into radians,
we get
θ=
π
×114 ,
(180×60)
or
θ=
3.141×114
180×60
or
θ=3.32×10−2 rad .
,
Substituting values of b and θ in Eq.(1), we get,
Number of molecules in one mole of hydrogen (gas)
is
D=
1.276×107
m ,
3.32×10−2
D=3.84×10 8 m .
N = N A =6.022×1023 .
XI-I-v1-(ANQ)-15042011
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)4/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
N11.
From
the
given
N13. The diameter of the Sun is
information, the distance parsec
d =Dα .
is
Substituting values, we get
D=
d =1.496×1011×1920×4.85×10−6
(b /2 )
,
θ
11
where b≈3×10
or = 1.39×10 9 m .
m , and
3.141
180×3600
= 4.85×10−6 rad
θ=1' '=
N14. In the question we are scaling up
.
the size by a factor of
Therefore,
10−5
=1010 .
10−15
1.5×1011
m
.
4.85×10−6
16
≈3×10 m
D≈
The size of an atom is ≈10−10 m .
Therefore, it will be scaled up to a size
≈10−10×1010≈1 m .
Comment:
Comment: A nucleus in an atom is as small in size as
(i) Astronomical unit is approximately equal to the
semi-major axis of the Earth's orbit.
the tip of a sharp pin placed at the centre of a sphere
of radius about a metre long.
(ii) astronomical unit is (symbol A or au)
1 A=1.49597 ..×10 11 m .
N15.The diameter of the jupitar is
(iii) One parsec (pc) is the distance at which 1A
subtends 1 sec arc.
d =Dα .
Substituting values, we get
1A
1A
1A
1 pc=
=
=
1 ' ' ( π / 648000) 4.85×10−6
=3.08567..×1016 m=3.262... ly
.
d =824.7×10 9×35.72×4.85×10−6 m
or = 1.429×10 8 m=1.429×105 km .
N12. Distance of the star
N16. If r is the radius of the lunar orbit
D = 4.29 ly. Therefore, in
around the Earth, the distance travelled by
pc unit, the distance of the
the LASER beam in going to moon and
star is
coming back to Earth is 2r. If c is the
D=
4.29
=1.32 pc .
3.262
speed of light, and t is the time taken in
covering distance 2r, then
The parallax is
2A
θ=
1.32 pc
2A
=
1.32 A/ 1 ' '
or θ=
m
)×2.56(s)
s
.
=7.68×10 8 m
2 r=c t =3×10 8 (
,
Therefore, r =
2
=1.52 ' ' .
1.32
7.68×10 8
=3.84×10 8 m .
2
N17. If d is the distance of the enemy submarine, the
speed of sound in water is v and the time-interval for
XI-I-v1-(ANQ)-15042011
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)5/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
2R
1
= rad
60 R 30
.
180o 1
=
× =1.9 o
π
30
the signal to come back to the detector is t, then
α=
2 d =v t .
It is given that v = 1450 m s–1 , and t = 77.0 s.
Therefore,
(b) Diameter of moon
2 d =1450 (
m
)×77(s)=111650 m .
s
Dmoon =(60 R)×α moon ,
or
Thus,
d=
111650
=55825 m=55.8 km .
2
N18. Let the distance of the quasar is d , speed of
(1)
Diameter of Earth
D Earth =( 60 R)×(1.9o ) .
(2)
Therefore, using (1) and (2)
D Earth
light is c and time taken is t (all in SI units). Then,
d =c t .
1 o
D moon =( 60 R)×( ) .
2
D moon
=
1/ 2o
=3.8 .
1.9 o
Thus DEarth =3.8 Dmoon ≈4 Dmoon .
Substituting values, we get
d =3.0×10 8 ( m / s)×3.0×109 ×365×24×3600(s)
.
=2.84×10 25 m
22
=2.8×10 km
(c) From the given information, we find
N19. The situation is illustrated in
Fig.N19. The diameter of the moon is
d 1= D 1 ×β .
D Sun
Here, the distance of the moon from
the Earth is D 1 =3.84×108 m and
β=1920 ' ' . Therefore,
d 1=3.84×108 ×1920× 4.85×10−6
=3.58×10 6 m
=3580 km
D moon
=
400 r
=400 .
r
Since, from calculation of (b) part of the question, we
have
D Earth
.
D moon
≈4 ,
we get
D Sun
D moon
N20. The situation is illustrated in
×
D moon
D Earth
1
=400× =100 ,
4
therefore,
Fig.N20a. The angular diameter of
D Sun
the Earth as seen from the moon is
D Earth
=100 .
N21. (a) By definition, 1 parsec is that distance at
which 1 astronomical unit (A.U.) subtends an angle
of 1 sec arc.
XI-I-v1-(ANQ)-15042011
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)6/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
1 A.U.
1 A.U.
=
1' '
4.85×10−6 ,
−5
=2.06×10 A.U.
DSun
DSun
1
=100 , and
=
D Earth
1 A.U. 2
α Mars=
r S-E . Then, by
(b) Let the Sun-Earth distance is
α Mars= 2×
Let the diameter of the Sun is D Sun . Then the
DSun
.
1A.U.
Using these values in Eq.(4) gives,
(1)
o
The angular size of sun-like star at 2 parsec distance
is
()
α Mars= 2×
or α Mars=
DSun
DSun
DSun
1
=
=
×
.
5
2 pc 2×10 A.U. 1 A.U. 2×105
() (
o
( )
1
1
1
×
=
2
200
200
,
60'
=0.3' .
200
With a magnification of 100, the angular size seen is
= 100 α Mars =30 ' .
Using the given information (Eq.(1))
o
, and
D Mars D Mars DEarth 1
1
1
=
×
= ×
=
.
D Sun DEarth DSun 2 100 200
It is given that
θ=
(5)
o
()
DSun
DSun
α=
=
.
r S-E 1 A.U.
DSun D Mars
×
.
1 A.U. DSun
Here,
DSun
1
=
1 A.U.
2
angular size of the sun is
(4)
The above may be written as
r S-E ≈1 A.U. .
θ=
(3)
D Mars
DMars
=
.
r E-M (1/ 2) A.U.
definition
α=(1/ 2)o =
.
The angular size of the Mars is
1 parsec≈ 2×105 A.U.
or
o
()
1 parsec=
)
1
1
×
=15×10−5 min arc .
2
2×10 5
With magnification of 100, the angular size of star is
This is much above the limit of resolution due to
atmospheric fluctuations. Therefore, the Mars is seen
magnified.
= 100 θ=15×10−3 min arc .
This angular size is not resolvable due to atmospheric
N22. (a) Oleic acid does not dissolve in water. It is
fluctuations. As a result stars are not seen magnified
dissolved in alcohol.
when viewed through a telescope.
(b) When lycopodium powder is spread on water, it
spreads on the entire surface. When a drop of the
(c) Let the diameters of the Earth and the Mars are,
prepared solution is dropped on water, oleic acid does
respectively, DEarth and DMars . It is given that
not dissolve in water. It spreads on the water surface
DMars 1
=
D Earth 2
pushing the lycopodium powder away to clear a
,
(1)
circular area where the drop falls. This allow
and the distance between the Earth and Mars,
r E-M=
1
A.U. .
2
(2)
It is also known that (see answer of N20)
measuring the area where oleic acid spreads.
(c)
1 mL×1
1
=
mL .
20 20
400
(d) By means of a burette and measuring cylinder and
measuring the number of drops.
XI-I-v1-(ANQ)-15042011
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)7/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
(e) If N drops of the solution make 1 mL of solution,
Thus, the area of the sheet in appropriate significant
the volume of the oleic acid in one drop will be
figures and error is
=
1
400 N
A=164±3 cm 2 .
mL .
N25. Surface Area =
N23. Density = mass / volume
4.237
g
ρ=
2.5 cm 3
g
=1.6948
cm 3
2 (l b+b h+l h ) .
Substituting values, we get
A= 2( 4.234×1.005+1.005×0.0201+4.234×0.0201)
.
or
A=2( 4.2552+0.02020+0.08510)
.
=2( 4.3605)=8.7210 m 2
Since, in multiplication or division, the final result
Retaining result up to correct significant figures, we
should retain as many significant figures as are there
find
A=8.72 m 2 .
in the original number with the least significant
figures, therefore, the density (with correct significant
The volume of the sheet is
figures) is
V =l bh= 4.234×1.005×0.0201
=0.085529 m3
ρ=1.7 g/ cm 3 .
N24. The length and breadth with the least count
errors are, respectively,
Retaining result up to correct significant figures, we
find
V =0.0855 m 3 .
L=16.2±0.1 cm , and
B=10.1±0.1 cm .
N26. (a) The addition of three masses, namely, 2.3
The area is
kg,
A= L B=16.2×10.1=163.62
Since, in multiplication or division, the final result
20.15 g and 20 .17 g, gives the following result:
should retain as many significant figures as are there
+
+
in the original number with the least significant
figures, therefore, the area up to correct significant
digits is
Since, uncertainties in subtraction or addition
combine in a different fashion (smallest number of
A=164 cm 2 .
The error is obtained from the following relation:
Δ A Δ L Δ B 0.1
0.1
=
+
=
+
'
A
L
B 16.2 10.1
or
2.3
0.02015
0.02017
2.34032
decimal places rather than the number of significant
figures in any of the number added or subtracted).
Therefore, the total mass of the box is
2.3 kg.
ΔA
=0.0062+0.0099=0.0161 .
A
(b) The difference in the masses of the pieces to
Therefore,
correct significant figures is
Δ A=164×0.0161=2.64
20.17 – 20.15 = 0.02 g
The error in the last digit of 164 is, therefore,
Δ A=3 .
N27. The number of significant figures in the
XI-I-v1-(ANQ)-15042011
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)8/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
measured length is 4. The calculated area and the
N30. The error (uncertain digit) in the measurements
volume should therefore be rounded off to 4
are as follows: (a) 0.01 mm, (b) 0.01 cm, (c) 0.01 m
significant figures.
and (d) 0.01 km. Therefore, the most precise
Surface area of the cube
measurement is (a).
=6( 7.203)2 m 2
=311.299254 m 2 '
=311.3m 2
N31. The errors in the individual measurement values
are:
Volume of the cube
(a) 5 – 4.9 = 0.1 cm
=(7.203)3 m 3
=373.7147544 m 3 .
=373.7 m 3
(b) 5 – 4.805 = 0.195 cm = 0.2 cm
(c) 5 – 5.25 = – 0.25 cm = – 0.2 cm
(d) 5 – 5.4 = – 0.4 cm
N28. Density= mass/ volume
=
=
Therefore, the most precise measurement is (a).
5.74
g /cm 3
1.2
4.783 g / cm
3
N32. Least count of vernier calliper is
.
= One division of main scale – one division
Since, in multiplication or division, the final result
of vernier scale .
should retain as many significant figures as are there
= 0.5 mm−
in the original number with the least significant
49×0.5 mm
50
figures, therefore, the density (to correct significant
= 0.5 mm −0.49 mm ,
digits) is
= 0.01 mm .
Density =
,
4.8 g/cm3 .
N33. (a) Precision tells us to what resolution or limit
N29. The range of variation over the seven days of
the quantity is measured. The precision is given by
observations is
the least count of the measuring instrument. Here the
For clock 1
least count for 20 oscillations is = 0.1 s. Therefore,
Δt1 = 142 s (12:01:30 – 11:59:08 = 00:02:22 = 142
the precision in the time- period (i.e., one oscillation)
s)
is
For clock 2
=
Δt2 = 31 s (10:15:24 – 10:14:53 = 00:00:31 = 31 s)
Although, the average reading of clock 1 is much
closer to the standard time than the average reading
of clock 2, but what we wish to measure is the time
interval. The range of error in time interval is less for
clock 2. Therefore, clock 2 should be preferred for
0.1 s
=0.005s .
20
(b) The accuracy of a measurement is a measure of
how close the measured value is to the true value of
the quantity.
The true value is taken as the mean value (in
the absence of any other information).
t mean =
measuring the time interval.
XI-I-v1-(ANQ)-15042011
39.6+39.9+39.5 119
=
s .
3
3
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)9/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
Therefore, time period,
Maximum error =
T=
119
s
60
39.9
−
= 0.012 s
∣119
60
20 ∣
0.1
=0.04 .
2.6
=
(c) Percentage error = Relative error × 100%
.
= 0.04 × 100%
= 4%
N34. The mean period of oscillation of the pendulum
N35. Temperature difference is
is
T=
or
2.63+2.56+2.42+2.71+2.80
13.12
s=
s
5
5
t 2 −t1 =50 o C−20 o C=30 o C .
The errors add
Δ t= 0.5 oC +0.5 o C=1 o C .
T =2.624 s
=2.62s
Therefore, the temperature difference and the error is
(Comment: As the periods are measured to a
written as
resolution of 0.01 s, all times are to the second
t =30 o C±1 o C .
decimal; it is proper to put this mean period also to
the second decimal.)
The absolute errors in the measurements are:
∣2.62s−2.63 s∣=0.01 s
N36.
ΔR
ΔV
ΔI
%=
%+
% .
R
V
I
Here,
ΔV
5
%=
×100=5% , and
V
100
∣2.62 s− 2.56s∣=0.06 s
ΔI
0.2
%=
×100=2% .
I
10
∣2.62 s− 2.42s∣=0.20 s
∣2.62 s− 2.71s∣=0.09 s
Therefore,
∣2.62 s− 2.80s∣=0.18 s
ΔR
%=5 %+ 2 %=7 % .
R
(a) The mean absolute error is
ΔT =
or
0.01+0.06+0.20+0.09+0.18
s
5
N37. (a) The series combination of resistances gives
0.54
ΔT =
=0.11 s .
5
R=( R1 ±Δ R 1 )+( R 2 ±Δ R 2 )
=(100±3)Ω+(200±4)Ω .
=(300± 7)Ω
Comment: The mean period of oscillation may be
reported as (without thinking over the mean absolute
(b) For parallel combination, the resistance is
error)
R '=
T =(2.62±0.11) s .
However, note that the mean absolute error is in the
From
tenth of a second, hence there is no point in reporting
=
100×200
=66.7 Ω .
300
Δ R ' Δ R1 Δ R2
= 2+ 2 ,
R '2
R1
R2
of reporting the result of measurement is
ΔT
(b) Relative error = T
,
mean
R 1+ r 2
1
1
1
= +
, we get
R ' R1 R 2
result up to hundredth of a second. The correct way
T =(2.6±0.1 ) s .)
R1 R2
or
(
)
3
4
+
×66.72
2
100
2002
.
= 4×10−4 ×4.45×103 =1.78 Ω
ΔR'=
Therefore, for parallel combinations
R '=(66.7±1.8) ohm .
XI-I-v1-(ANQ)-15042011
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)10/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
Comment: (I) Since the error is in unit place of the
N40. Observed thickness
number, the parallel combination may be better
reported as
= actual thickness × magnification
Therefore,
R '=(67±2 )ohm .
Observed thickness
magnification
.
3.5 mm
=
=0.035 mm
100
actual thickness=
(II) A different rounding convention is sometimes
followed
in
scientific
circles
when
retaining
significant digits in mean value and error. It requires
that if three highest order digits of the error lie
N41. Let the linear magnification is
between 100 and 354, we round to two significant
the area magnification is
digits. Here the three highest order digits in the error
are 178. So we round up to two significant digits and
M Linear . Then,
M Area = M 2Linear .
It is given that,
report the value as
1.55 m 2 1.55×10 4
=
1.75
.
1.75cm 2
=88.57×10 2
M Area =
R' =(66.7±1.8) ohm .
Therefore,
N38. The relative error in Z is
ΔZ
Δ A 1 Δ B ΔC 3 Δ D
=4
+
+
+
Z
A 3 B
C
2 D
M Linear = √ 88.51×10=94.1 .
.
N42. The relative error is
N39. From
T =2 π √ L/ g , we write
=
2
g=
4π L
T2
.
Since
Therefore, the relative error in g is
Δg ΔL
ΔT
=
+2
g
L
T
.
(1)
or
0.02
.
100×3.156×10 7
= 6.3×10−12 .
Thus the accuracy is 1 part in 1011 to 1012.
L=20.0 cm ,
t
n
1yr =3.156×10 7 s , we get
relative error =
it is given that,
T=
Δ t 0.02 s
=
.
t 100 yr
Δ L=0.1 cm ,
, therefore
ΔT =
Δt
n
,
N43. The value of the product is
A B= 2.5( m s−1 )×0.10(s)
.
=0.25 m
where, t = 90 s, n = 100, Δt = 1 s. Therefore,
ΔT Δt 1
= =
T
t
90
.
The relative error is
Δ ( AB) Δ A Δ B
=
+
( AB)
A
B
0.5 0.01 .
=
+
2.5 0.10
=0.3
Substituting in Eq.(1), we find
Δ g 0.1
1
=
+2×( )=0.027 .
g
20.0
90
The percentage error is
Δg
×100=0.027×100=2.7
.
g
=3%
Therefore, the product (with error) is
XI-I-v1-(ANQ)-15042011
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)11/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
= AB±Δ ( AB)
=(0.25±0.25×0.3) m
.
=(0.25±0.075) m
=(0.25±0.08) m
N46. The principle of homogeneity of dimensions
states that the final dimensions on the left hand side
of an equation should be equal to the final dimensions
on the right hand side of an equation. Further, the
N44. Let C= √ AB . Then,
argument of a trigonometric function must always be
C= √ 1.0× 2.0= √ 2=1.4
dimensionless. Therefore,
(up to the correct significant digits).
(a) In case of
The relative error in C is
y = a sin 2πt / T ,
ΔC 1 Δ A 1 Δ B
=
+
C
2 A 2 B
,
1 0.2 1 0.2
= ×
+ ×
2 1.0 2 2.0
or
we note that,
(i)
the
dimensions
of
the
argument
of
the
trigonometric function are
ΔC
=0.15 .
C
[ 2π t/T] = [ T1 / T1 ] = dimensionless,
and
Therefore, the value of C is
(ii) [y] = [L1] = [a].
C=1.4 m±1.4×0.15 m
.
=1.4 m ±0.2 m
Therefore
Comment: We have not retained uncertainties smaller
DIM LHS = DIM RHS.
than that represented by the uncertain digit in the
The relation (a) is dimensionally correct.
value 1.4 m. This is the convention of reporting error.
(b) In case of
y = a sin vt ,
N45. (I) The percentage error in P is
ΔP
Δa
Δb
1 Δc
Δd
%=3
%+2
%+
%+
%
P
a
b
2 c
d
Substituting the values we get,
trigonometric function are
[
ΔP
1
%=3×1%+2×3%+ ×4 %+1×2 % ,
P
2
or
we note that the dimensions of the argument of the
ΔP
%=13 % .
P
vt]
=
[ L T−1 T ]
≡
[L]
=not
dimensionless.
Therefore, the relation (b) is dimensionally wrong.
(c) In case of
y = (a/T) sin (t / a)
(II) The error in P is
or
Δ P =P ×0.13=3.763×0.13=0.49 ,
we note that the dimensions of the argument of the
Δ P =0.5 .
trigonometric function are
Since the uncertainty is in the first decimal place,
there is no point in retaining terms more uncertain
[ t/a ] = [ T1 / L1 ] = not dimensionless
The relation (c) is dimensionally wrong.
than this. Therefore, the value of P that should be
reported after rounding is
= 3.8
(d) in case of
y = (a √2) (sin(2πt / T) + cos(2πt / T)),
we note that,
XI-I-v1-(ANQ)-15042011
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)12/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
(i) The dimensions of the argument of the
we note that their dimensions are as follows:
[ K ]=[M L 2 T−2 ] ,
trigonometric function are
[ 2π t/T] = [ T1 / T1 ] = dimensionless
[ v]=[ L T−1 ] ,
and
[ a ]=[L T−2 ] ,
(ii) [y] = [L1] = [a].
[ m ][ M] .
Therefore,
(a) Dimensions of LHS = [M L 2 T−2 ]
DIM LHS = DIM RHS
Dimensions of RHS = [M ]2 [ L T−1 ]3 =[ M2 L 3 T−3 ]
The relation (d) is dimensionally correct.
Since
Dim LHS ≠ Dim RHS
N47. In the relation
m=
Therefore, formula (a) is dimensionally ruled out.
m0
(1−v 2 )1 /2
,
the dimensions of m and m0 are the same. Therefore,
from the principle of homogeneity of dimensions, the
denominator on RHS should be dimensionless.
Therefore, to make
(b) Dimensions of LHS = [ M L 2 T−2 ]
Dimensions of RHS = [M][L T−1 ]2 =[ M L2 T− 2 ]
Since
Dim LHS = Dim RHS
Therefore, formula (b) is dimensionally allowed.
(1−v 2 )1/2
(Comment: The validity of the numerical factor (1/2)
dimensionless, we must divide the velocity v of the
cannot be ascertained by dimensional analysis.)
particle by the velocity c of light. That is, we must
replace
2 1/ 2
by
(1−v )
2 1 /2
(1−( v/ c) )
(c) Dimensions of LHS = [M L 2 T−2 ]
.
Dimensions of RHS = [M][L T− 2 ]=[M L T−2 ]
Therefore, the correct relation should be
m=
Since
m0
1/ 2
( )
v2
1− 2
c
Dim LHS ≠ Dim RHS
.
Therefore, formula (c) is dimensionally ruled out.
(d) Dimensions of LHS = [ M L 2 T−2 ]
N48. The dimensions of LHS are
Dimensions of RHS = [M][L T−1 ]2 =[ M L2 T− 2 ]
[M] [L T–1 ]2 = [M] [ L2 T–2]
Since
= [M L2 T–2]
Dim LHS = Dim RHS
The dimensions of RHS are
Therefore, formula (d) is dimensionally allowed.
[M][L T–2] [L] = [M][L2 T–2]
(Comment: The validity of the numerical factor
= [M L2 T–2]
The dimensions of LHS and RHS are the same and
hence the equation is dimensionally correct.
N49. From the units given for the physical quantities,
(3/16)
cannot
be
ascertained
by
analysis.)
(e) Dimensions of LHS = [M L 2 T−2 ]
XI-I-v1-(ANQ)-15042011
dimensional
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)13/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
E=m c2 ,
Dimensions of first term of RHS
= [M][L T−1 ]2 =[M L2 T− 2 ]
we note that the energy equivalent of 1 u is
E=1.6605×10−27 ×(2.9979×10 8 ) kg m2 s−2
.
=14.924×10−11 J
Dimensions of second term of RHS
= [M][L T− 2 ]=[ M L T−2 ]
Since the two terms on RHS of (e) differ in
Using joule to MeV conversion, we find
E=14.924×10−11 ×
dimensions, the quantity on the right side of (e) has
no proper dimensions. (Two quantities of different
dimensions can not be added.)
1
MeV
1.6022×10−13
=931.5MeV
Therefore, we write
Therefore, formula (e) is dimensionally
ruled out.
1 u =931.5MeV
(b) One may point out that in the relation
1 u =931.5MeV
N50. Let the dependence of time period T on the
the dim of LHS are those of mass, while the dim of
quantities l, g and m as a product may be written as :
RHS are those of energy. So the above conversion
x
y
T=k l g m
z
(1)
does not exhibit correct dimensional conversion. To
where k is dimensionless constant and x, y and z are
keep the above relationship dimensionally pure, we
the exponents. By considering dimensions on both
should write it as
sides, we have
1 u=931.5 MeV /c2 .
[L0 M0 T 1 ]=[ L 1 ]x [ LT−2 ] y [ M1 ]z
=[L x+ y M z T−2y ]
(2)
Comment: If you use data with less number of
On equating the dimensions on both sides, we have
x + y = 0;
–2y = 1;
significant figures, such as
z=0
1 MeV = 1.6 ×10–13 J ;
Therefore, on solving, we find
x=
1
2
,
y=−
1 u = 1.67×10–27 kg ;
1
, z=0 .
2
you will get a different answer. Try it – but do not
Substituting these values in Eq.(1) gives,
T=k
√
l
g
.
memorize it as it represents a wrong numerical
(4)
conversion factor.
Comment: (i) Note that value of constant k can not be
obtained by the method of dimensions. Here it does
not matter if some number multiplies the right side of
this formula, because that does not affect its
dimensions.
(ii) Actually,
k = 2π
c = 3 ×108 m/s ;
(3)
√
l
so that T = 2π
g
.
N52. (a) The percentage error in the quantity X is
ΔX
Δa
Δb
5 Δc
Δd
%=2
%+3
%+
%+2
% .
X
a
b
2 c
d
Substituting values, we get
ΔX
5
%= 2×1%+3×2 %+ ×3 %+ 2×1 %
.
X
2
=12.5 %
(b) It is given that X = 2.763 . Therefore, the error in
N51. (a) Using Einstein mass energy relation,
it is
XI-I-v1-(ANQ)-15042011
Δ X =2.763×0.125=0.34 .
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)14/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
Since the error is in the first decimal place,
the value of X should be reported as 2.8. The correct
(3)
reporting (including error estimate) is
[
√ (
hG
[ML2 T−1 ][M−1 L3 T− 2 ]
]=
c5
[LT−1 ]5
2 1/2
=[T ] =[T]
1 /2
)
.
It has dimensions of time.
X =2.8±0.3
N53. The dimensions of E, m, L and G are:
[E] = [M1 L 2 T−2 ] ,
N55. From Kepler’s third law, the square of the
[m] = [M1 L 0 T 0 ] ,
period of revolution T is proportional to the cube of
the radius of the orbit r,
[L] = [M1 L 2 T−1 ] ,
T 2 ∝ r3 ,
[G] = [M−1 L3 T−2 ] .
or
Therefore, the dimensions of P are:
[ P]=[ M1 L 2 T−2 ] [M1 L 2 T−1 ]2 [ M1 ]−5 [ M−1 L3 T−2 ]−2
=[ M1+ 2−5+ 2 L 2+4−6 T−2−2+4 ]
=[ M0 L 0 T0 ]
T ∝ r 3/2 .
(1)
Let us assume that T also depends on other given
parameters, namely M, R and g. Let this dependence
is of the form
Thus, P is a dimensionless quantity.
T ∝g xRy M z .
(2)
Combining (1) and (2), we get
N54. The dimensions of c, h and G are:
T ∝ r 3/2 g x R y M z .
−1
[c]=[L T ] ,
2
Equating dimensions on the two sides, we find
−1
[h ]=[ M L T ] ,
[M0 L 0 T 1 ]=[L 3/2 ][L x T−2x ][L y ][M z ]
.
=[M z L3/ 2 + x+ y T− 2x ]
[G ]=[M−1 L3 T−2 ] .
Now consider the following combinations:
(1)
(2)
(3)
√
ch
G
√
√
hG
c3
z=0 ,
3
+ x+ y=0 ,
2
, and
−2x=1 .
Solving, we find
hG
.
c5
√ (
[LT−1 ][ML 2 T−1 ]
ch
[
]=
G
(1)
[M−1 L3 T−2 ]
=[M 2 ]1/2 =[M]
x=−
)
√ (
1
2
,
y=−1 , z = 0.
Therefore, using the above values in Eq.(3), we find
1/ 2
.
It has dimensions of mass.
(2)
(4)
Therefore,
,
Then the dimensions of these are:
[
(3)
hG
[ML2 T−1 ][M−1 L3 T−2 ]
]=
3
c
[LT−1 ]3
=[L2 ]1/ 2=[L]
or
1 /2
)
.
√
√
T∝
1 r3 ,
R g
T=
k r3 .
R g
N56. The dimensions of LHS of the formula are
[Ṽ ]=[L3 T−1 ] .
It has dimensions of length.
(1)
The dimensions of the RHS are:
XI-I-v1-(ANQ)-15042011
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)15/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
[
π P r 4 [ M L−1 T−2 ][ L 4 ]
]=
8 ηl
[M L−1 T−1 ][ L] .
=[L 3 T−1 ]
M
23
=
=3.8×10− 23 g
N A 6×1023
.
=3.8×10−26 kg
m=
(2)
Therefore, the density of sodium atom is
Thus,
dim LHS = dim RHS
ρ Na =
Therefore, the formula is dimensionally correct.
N57. A simple-minded guess suggests that the density
should be less like that of a gas ~ 1 kg/m3 or so. But
the data reveals that
The density of sodium in crystalline phase is (given)=
0.97×10 3 kg/m 3 . The orders are same, because in
the solid phase atoms are tightly packed, so the mass
density of the solid is close to the atomic mass
2.0×1030
density of the Sun =
,
( 4 π /3 )×( 7.0×10 8 )3
=
m 3.8×10−26
=
V 8.2×10−30
.
=4.6×10 3 kg/m 3
density.
(Comment: Atomic mass density is more than the
3×2.0×10 30
,
4×3.14×343×10 24
mass density of solid. Although there is a close
= 1392.7=1.4×10 3 kg/m 3 .
packing but still some empty space remains between
The mass density of the Sun is in the range of
the atoms of the solid. As a consequence, the solid
densities of liquids / solids and not gases. This high
mass density is always less than the atomic mass
density arises due to inward gravitational attraction
density.)
on outer layers due to inner layers of the Sun.
N60. The mass of a nucleus of mass number A is M =
A u (here u stands for unified atomic mass unit,
N58. The relationship
1 u =1.66×10−27 kg ). The volume of a nucleus is
tan θ=v
3
is dimensionally wrong. The LHS is dimensionless
while the RHS has dimensions [ LT−1 ] . To make
the relation dimensionally correct, one possible
V=
Therefore, density of the nucleus is
ρ Nucleus =
solution is to divide the RHS by some velocity, say
velocity of rain fall ( v rain ). then the dimensionally
correct relation is
tan θ=
v
v rain
4 π r3 4 π r 0 A .
=
3
3
It is independent of the atomic mass number, i.e., it is
constant for different nuclei.
.
The value of nuclear density is
ρ Nucleus =
N59. The diameter of the Na atom = 2.5Å.
3×1.66×10− 27
4×3.14×1.23 ×10−45 .
= 2.3×1017 kg/m 3
ρ Nucleus =
4π
3
(1.25×10−10) =8.18×10−30 m 3 .
3
The mass of one atom of Na is
3×1.66×10− 27
,
4×3.14×1.23 ×10−45
On solving we get,
Therefore, volume of the Na atom is
V=
M 3 (u )
=
V 4 π r 30 .
The nuclear density is independent of the mass
number A. Therefore, this is correct estimate of the
XI-I-v1-(ANQ)-15042011
(D) ANSWERS OF CONCEPTUAL QUESTIONS
XI-UNIT I
UNITS & MEASUREMENT
(D)16/16
Based on Lecture Notes prepared by Professor (Retd.) Sardar Singh, Mansarovar, Jaipur
density of sodium nucleus.
The density of sodium atom (calculated
earlier) is 4.6×10 3 kg/m 3 . The ratio of density of
nucleus and density of atom is
=
2.3×10 17
=0.5×10 14≈1014
3
4.6×10
Thus, nuclear density is about 1014 times larger
than the atomic density.
XI-I-v1-(ANQ)-15042011