\(\pi=3.14 \)
\(\pi^{2} =3.14 \times 3.14 \)
\(=9.8596=9.86 \)

\(\text { Error in radius }=2 \%\)
\(\Delta r=\frac{2}{100}=0.02\)
\(\text {Volume of the sphere }=\frac{4}{3} \pi r^{3}\)
\(V =\frac{4}{3} \pi r^{3} \)
\(\frac{d V}{V} =\frac{4}{3} \pi \times 3 r^{2} d r \)
\(=3 d r=3(2 \%)=6 \%\)

Unit of a (Planck's constant) - Js
Unit of c (Velocity of light) - ms^-1
Unit of G (Gravitational Constant) - \(\frac{\mathrm{Nm}^{2}}{\mathrm{Kg}^{2}}\)
\(\therefore \text { Unit of } \frac{h c}{G} \text { is }=\frac{J s \times m s^{-1}}{N m^{2} / k g^{2}} \)
\(=\frac{N m s \times m s^{-1} \times k g^{2}}{N m^{2}}[J=N m] =\mathrm{kg}^{2}\)

Let G_SI be the gravitational constant in the SI system and G_cgs in the cgs system. Then 
G_SI = 6.6 10^-11 Nm² kg^-2;
G_cgs = ?
n₂ =\(n_1{ \left[ \frac { { M }_{ 1 } }{ { M }_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { L_{ 1 } }{ L_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { T_{ 1 } }{ T_{ 2 } } \right] }_{ }^{ c }\)
G_cgs = G_SI \({ \left[ \frac { { M }_{ 1 } }{ { M }_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { L_{ 1 } }{ L_{ 2 } } \right] }_{ }^{ a }{ \left[ \frac { T_{ 1 } }{ T_{ 2 } } \right] }_{ }^{ c }\)
M₁= 1 kg L₁ = 1 m T₁= 1s
M₂= 1 kg L₂ = 1 m T₂= 1s
The dimensional formula for G is M^-1L³ T^-2
a = -1 b = 3 and c =-2
G_cgs = 6.6\(\times\)10^-11 \(\left[ \frac { 1kg }{ 1g } \right] ^{ -1 }\left[ \frac { 1m }{ 1cm } \right] ^{ 3 }\left[ \frac { 1s }{ 1s } \right] ^{ -2 }\)
=  6.6\(\times\)10^-11 \(\left[ \frac { 1kg }{ { 10 }^{ -3 }kg } \right] ^{ -1 }\left[ \frac { 1m }{ { 10 }^{ -2 }m } \right] ^{ 3 }\left[ \frac { 1s }{ 1s } \right] ^{ -2 }\)
= 6.6\(\times\)10^-11\(\times\)10^-3\(\times\)10⁶\(\times\)1
G_cgs = 6.6\(\times\)10^-8 dyne cm² g^-2

\(Average \ length ={3.48+3.46+3.49+3.50+3.48\over 5}={17.41\over5}\)
= 3.482 m = 3.48 m
(Round off to 2 places of decimal point)
The absolute errors in the different measurements are
\(\triangle\) L₁= 3.48 - 3.48 = 0.00 m
\(\triangle\) L₂= 3.48 - 3.46 = 0.02 m
\(\triangle\) L₃=3.48 - 3.49 = - 0.01 m
\(\triangle\) L₄= 3.48 - 3.50 = - 0.02 m
\(\triangle\) L₅= 3.48 - 3.48 = 0.00 m .
The absolute error =\({\sum |\triangle L_i|\over 5}\)
=\(0.00+0.02+0.01+0.02+0.00\over 5\)
=\({0.05\over5}=0.01m\)
\(\therefore\) Correct length = 3.48 ± 0.01m
Percentage error =\({0.01\over 3.48}\times 100=0.29\%\)

Propagation of errors
A number of measured quantities may be involved in the final calculation of an experiment. Different types of instruments might have been used for taking readings. Then we may have to look at the errors in measuring various quantities, collectively.
The error in the final result depends on
(i) The errors in the individual measurements
(ii) On the nature of mathematical operations performed to get the final result. So we should know the rules to combine the errors.
The various possibilities of the propagation or combination of errors in different mathematical operations are discussed below:
(i) Error in the sum of two quantities:
Let A\(\triangle\) and \(\triangle\)B be the absolute errors in the two quantities A and B respectively. Then, 
Measured value of A = A \(\pm\triangle\) A
Measured value of B = B \(\pm\triangle\) B
Consider the sum, Z = A + B
The error \(\triangle\) Z in Z is the given by
Z  \(\pm\triangle\) Z = (A \(\pm\triangle\)A) + ( B \(\pm\triangle\) B)
= ( A + B ) \(\pm\) (\(\triangle\)A+ \(\triangle\) B)
= Z \(\pm\) ( \(\triangle\) A + \(\triangle\) B )
(or) \(\triangle\)Z = \(\triangle\) A+ \(\triangle\) B
The maximum possible error in the sum of two quantities is equal to the sum of the absolute errors in the individual quantities.
(ii) Error in the difference of two quantities:
Let \(\triangle\)A and \(\triangle\)B be the absolute errors in the two quantities, A and B, respectively. Consider the product Z = AB
Let \(\triangle\)A and \(\triangle\)B be the absolute errors in the two quantities, A and B, respectively. Consider the product Z = AB
The error \(\triangle\)Z in Z is given by \(Z \pm \Delta Z=(A \pm \Delta A)(B \pm \Delta B)\)
\(=(A B) \pm(A \Delta) \pm(B \Delta A) \pm(\Delta A . \Delta B)\)
Dividing L.H.S by Z and R.H.S by AB, we get,
\(1 \pm \frac{\Delta Z}{Z} \cdot 1 \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \pm \frac{\Delta A}{A} \cdot \frac{\Delta B}{B}\)
As \(\triangle\)A/A, \(\triangle\)B/B are both small quantities, their product term \(\frac{\Delta A}{A} \cdot \frac{\Delta B}{B}\) can be neglected. The maximum fractional error in Z is
\(\frac{\Delta Z}{Z}=\pm\left(\frac{\Delta A}{A}+\frac{\Delta B}{B}\right)\)
The maximum error in difference of two quantities is equal to the sum of the absolute errors in the individual quantities.
(iii) Error in the division or quotient of two quantities
Let \(\triangle\)A and \(\triangle\)B be the absolute errors in the two quantities A and B respectively.
Consider the quotient, \(\mathrm{Z}=\frac{A}{B}\)
The error \(\triangle\)Z in Z is given by
\(Z \pm \Delta Z =\frac{A \pm \Delta A}{B \pm \Delta B}=\frac{A\left(1 \pm \frac{\Delta A}{A}\right)}{B\left(1 \pm \frac{\Delta B}{B}\right)} \)
\(=\frac{A}{B}\left(1 \pm \frac{\Delta A}{A}\right)\left(1 \pm \frac{\Delta B}{B}\right)^{-1}\)
or \(Z \pm \Delta Z=Z\left(1 \pm \frac{\Delta A}{A}\right)\left(1 \pm \frac{\Delta B}{B}\right)\)
[using (1+x) n \(\approx\) 1+n x, when x<1]
Dividing both sides by Z, we get,
\(1 \pm \frac{\Delta Z}{Z} =\left(1 \pm \frac{\Delta A}{A}\right)\left(1 \pm \frac{\Delta B}{B}\right)  \)
\(=1 \pm \frac{\Delta Z}{Z} \pm \frac{\Delta B}{B} \pm \frac{\Delta A}{A} \cdot \frac{\Delta B}{B}\)
As the terms \(\triangle\)A /A and \(\triangle\)B/ B are small, their product term can be neglected. The maximum fractional error in Z is given by
\(\frac{\Delta Z}{Z}=\left(\frac{\Delta A}{A}+\frac{\Delta B}{B}\right)\)
The maximum fractional error in the quotient of two quantities is equal to the sum of their individual fractional errors.

(i) Physics is a branch of science.
(ii) The word comes from a Greek word meaning 'nature'.
(iii) It deals with the study of nature and natural phenomena.

Length of a simple pendulum \(l=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}\)
\(\text {Accuracy }=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\)
Time for one oscillation \(=\frac{40}{50}=0.88\)
\(T =2 \pi \sqrt{\frac{l}{g}} ; T^{2}=4 \pi^{2}\left(\frac{l}{g}\right) \)
\(g =\frac{4 \pi^{2} l}{T^{2}}\)
\(\frac{\Delta g}{g} \times 100 ==\frac{\Delta l}{l} \times 100+2 \frac{\Delta T}{T} \times 100  \)
\(l =20 \mathrm{~cm} \Delta l=2 \mathrm{~mm}=0.2 \mathrm{~cm} \Delta T=1 \mathrm{~s}  \)
\(\frac{\Delta g}{g} =\frac{\Delta l}{l}+2 \frac{\Delta T}{T}  \)
\(=\frac{0.2}{20}+2 \times \frac{1}{40} \)
= 0.01 + 0.05 = 0.06
\(\therefore\) Percentage accuracy in the determination of g=0.06 x 100=6 %

Radius of the circle r = 3.12 m
Area of the circle A = \(\pi\) r²
= 3.14\(\times\)3.12\(\times\)3.12
= 30.566016 m²
According to the rule of significant 
A = 30.6 m² 

Dimension of Pressure = [ML^-1T^-2]
Dimension of density = [ML^-3]
Dimension of Energy = [ML²T^-2]
By substituting in the given equation,
\(=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]^{-5 / 6}\left[\mathrm{ML}^{-3}\right]^{1 / 2}\left[\mathrm{ML}^{2} \mathrm{~T}^{-2}\right]^{1 / 3} \)
\(=\mathrm{M}^{-5 / 6+1 / 2+1 / 3} \mathrm{~L}^{5 / 6-3 / 2+2 / 3} \mathrm{~T}^{5 / 3-2 / 3}\)
=M⁰L⁰T¹=[T]

By the principle of homogeneity, a / V² is of the dimensions of pressure and b is of the dimensions of volume.
[a] = [pressure] [V²] = [ML⁻¹T⁻²] [L⁶]
= [ML⁵T^-2]
[b] = [V] = L³

The SI unit for Luminous intensity is candela. Its symbol is cd.
Definition:
One candela is the luminous intensity in a given direction, of a source that emits monochromatic radiation of frequency 5.4\(\times\) 10⁴ Hz and that has a radiant intensity of \({{1}\over{638}}\) watt/steradian in that direction.

The SI unit for electric current is ampere (A)
Definition:
One ampere is the constant current, which when maintained in each of the two straight parallel conductors of infinite length and negligible cross section, held one metre apart in vacuum shall produce a force per unit length of 2\(\times\)10^-7N/m between them.

(i) Systematic observation
(ii) Controlled experimentation
(iii) Reasoning (qualitative and quantitative)
(iv) Modelling (Mathematical)
(v) Prediction and verification (theories)

Time becomes irrelevant. Because day and year based on spinning and revolution of the planet. So each planet has its own year length.
Eg: Uranus and Neptune move too slow.

Frequency of a vibrating body \(\gamma \alpha \frac{1}{l} \sqrt{\frac{F}{M}}\)
\( a\text { Dimension of frequency } =\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}  \)
\(\text {Dimension of length } =\mathrm{L}=\mathrm{M}^{0} \mathrm{LT}^{0}  \)
\(\text {Dimension of Force } =\mathrm{MLT}^{-2}  \)
\(\text {Dimension of Mass } =\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{0}  \)
\(\text {Frequency } \gamma =x  \)
\(\gamma =\mathrm{K}\left[\mathrm{F}^{x}\right][\mathrm{M}]^{y}[\mathrm{~L}]^{z}\)
Using dimensions we get
\(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}=\left[\mathrm{MLT}^{-2}\right]^{x}[\mathrm{M}]^{y}[\mathrm{~L}]^{z}  \)
\(\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{-1}=\mathrm{M}^{x+y} \mathrm{~L}^{x+z} \mathrm{~T}^{-2 x}\)
Comparing the powers we get
x + y = 0  z = -x = \(\frac{1}{2} \)
x + z = 0  y = -x = \(\frac{-1}{2} \)
-2 x = -1 
\(\therefore x =\frac{-1}{-2}=\frac{1}{2}  \)
\(\therefore \gamma =1 \times[\mathrm{F}]^{1 / 2}[\mathrm{M}]^{-1 / 2}[\mathrm{~L}]^{-1 / 2}  \)
\(\gamma =\frac{1}{l} \sqrt{\frac{F}{m}}\)