Number of protons = \(\frac { Total \ mass \ }{ Mass \ of \ each \ proton } \)
\(\\ =\frac { { 10 }^{ -3 } }{ 1.67\times { 10 }^{ -27 } }\)
\( \\ =5.99\times { 10 }^{ 23 }\)

\(\frac { \triangle x }{ x } =\frac { 0.01 }{ 40.00 } =0.00025\)

\(\frac { \triangle x }{ x } =\frac { 0.001 }{ 0.004 } =0.25\)

For measuring large time intervals, we use the technique of radioactive dating. Large time intervals are measured by studying the ratio of number of radioactive atoms decayed to the number of surviving atoms in the specimen.

Given, Area of object = 1.75 x 10^-4 m²
Area of image = 1.55 m²
Areal magnification = \(\frac { Area\ of\ image }{ Area\ of\ object } \)
\(\frac { 1.55 }{ 1.74X{ 10 }^{ -4 } } \approx 8857\)
Linear magnification = \(\sqrt { \text{A real maginification} } \)
\(=\sqrt { 8857 } \)
= 94.1

Five

Three

\(D=-N\left( \frac { { x }_{ 2 }-{ x }_{ 1 } }{ { n }_{ 2 }-{ n }_{ 1 } } \right) \)
\(\\ [N]=\frac { { N }_{ 0 } }{ { [L }^{ 2 }T] } =[{ L }^{ -2 }{ T }^{ -1 }]\)
\(\\ [D]=\frac { [{ L }^{ -2 }{ T }^{ -1 }L] }{ [{ L }^{ -3 }] } =[{ L }^{ 2 }{ T }^{ -1 }]\)
\(\\ [{ x }_{ 2 }]=[{ x }_{ 1 }]=[L]\ \)
\(\ [{ n }_{ 2 }]=[{ n }_{ 1 }]=\frac { { N }_{ 0 } }{ [{ L }^{ 3 }] } =[{ L }^{ -3 }]\) 

In mechanics length, mass and time are chosen as base quantities because
(i) there is nothing simpler to length, mass and time.
(ii) all other quantities in mechanics can be expressed in terms of length, mass and time.
(iii) length, mass and time cannot be derived from one another.

187.2kg - 63.54kg = 123.66kg = 123.7kg
[rounded off into first decimal place]

Given, x = 4.29 light years = 4.29 x 9.46 x 10¹⁵ m
= \(\frac { 4.29\times9.46\times{ 10 }^{ 15 } }{ 3.08\times{ 10 }^{ 16 } } par\ sec\ =1.317par\ sec\)
\(\\ \theta =\frac { l }{ r } =\frac { 2AU }{ x } \)
Radius of the Earth orbit = 1 AU
= 1.496 x 10¹¹m
\(=\frac { 2\times1.496\times{ 10 }^{ 11 } }{ 4.29\times9.46\times{ 10 }^{ 15 } } rad\)
\(\\ =7.39\times{ 10 }^{ -6 }rad\)
\(\\ =.39\times{ 10 }^{ -6 }\times\frac { 180\times60\times60 }{ \pi } s=1.52s\) 

The number of significant figures in the given quantities are given below.
(i) In 0.007, the number of significant figures is 1 because in a number less than 1, the zero's on the right of the decimal point but to the left of the first non-zero digit are not significant.
(ii) In 2.64 × 10²⁴ kg, the number of significant figures is 3 because all non-zero digits are significant, power of 10 are not taken in significant figure.
(iii) In 0.2370, the number of significant figures is 4, as all non-zero digits left to decimal and trailing zero are significant.
(iv) In 6.320, the number of significant figures is 4, as all non-zero digits left to decimal and trailing zero are significant.
(v) In 6.032, the number of significant figures is 4, as all non-zero digits left to decimal and trailing zero are significant.
(vi) In 0.0006032, the number of significant figures is 4, because in a number less than 1, the zero's on the right of the decimal point but to the left of the first non-zero digit are not significant.

Obtain dimensionally an expression for n_R" Given, n_R is directly proportional to r.
Let n_R = p^x n^y v^z r
Note in Eqn. (1) we have used the information that n_R is directly proportional to r. If this information was not available there will be four unknowns. By equating powers ofM, L and T only three independent equations will be obtained and they cannot give values of the four unknowns. Now
[n_R] = M⁰L⁰T⁰
[P] = M L^-3
[n] = M L^-1T^-1
[r] = L
Substituting dimensions of parameters involved in Eqn. (1), we have
M⁰L⁰T⁰ = (M L^-3)^x (M L^-1 T^-1)^y (L T^-1)^z L⁺¹
= M^x+y L -3x -y +z + 1 T -^{Y -Z} 
By the principle of homogeneity of dimensions
x + y a
-3x - y + z + 1 = a
-y-z = a
Solving these equations, we get
x = 1, Y =-1, Z =1
Hence, n_R = K r p¹n^-1v¹
Or  n_R = \(\frac { Krpv }{ n } \)

Let the distance covered is S,
Then S = K u^a  f^b t^c ; where k is a constant. Writing dimensions on both the sides, we have
[L] = [LT^-1]^a[LT^-2]^b[T]^c
or [L] = [L^a+bT^-a-2b+c]
Comparing powers on both sides, we get
1 = a + b
and 0 = - a - 2b + C
We have only two equations with three unknowns, therefore, we split the problem into two parts.
(a) Let the body have no acceleration,
then S = k₁u^at^b
or [L] = [LT^-1]^a[T]^b
= [L^aT^-a+b] 
or a = 1
-a + b = 0 or b = 1
S = k₁ut  ------------------(i)
(b) Suppose the body has no initial velocity
then S=k₂f^at^b
[L] = k₂[LT^-2]^a[T]^b
or [L] =[L^aT^-2a+b] or a = 1
- 2a + bOor b = 2a = 2
S = k₂ft² ------------------(ii)
If the body has both the initial velocity and acceleration comparing (i) and (ii), we get,
S = k₁ ut + k₂ ft²
This is the required equation
If we put k₁= 1, k₂ =\(\frac{1}{2}\), we get
S = ut +\(\frac{1}{2}\)ft²