Given, Angular diameter, \(\theta =1920"\)
Distance of the sun from earth, \(r=1.496\times { 10 }^{ 11 }m\)
The diameter of the sun, d = ?
\(\theta =1920"=1920\times 4.85\times { 10 }^{ -6 }rad\)
= \(9.31\times { 10 }^{ -3 }\) rad
We know that, \(d=\theta r\)
\(d=9.31\times { 10 }^{ -3 }\times 1.496\times { 10 }^{ 11 }=1.39\times { 10 }^{ 9 }m\)
Thus, the diameter of the sun is \(1.39\times { 10 }^{ 9 }m\) 

Fractional error,
\({ \delta }_{ \mu }=\frac { \triangle \mu }{ { \mu }_{ m } } =\frac { 0.02 }{ 1.33 } =0.015=0.02\)

Two

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

Here \(T=2\pi \sqrt { \frac { l }{ g } } \)
\(\\ { T }^{ 2 }=4{ \pi }^{ 2 }\frac { l }{ g } \)
\(\\ g=4{ \pi }^{ 2 }\frac { l }{ { T }^{ 2 } } \)
Given l = 100 cm, T = \(\frac{40s}{20}\) = 2s
g =  \(4\times { (3.14) }^{ 2 }\times \frac { 1000cm }{ { (2s) }^{ 2 } } =\frac { 4\times 9.8596 \times 100 }{ 4 } cm{ s }^{ -2 }\)
= 985.9 cms^-2
Let us now calculate the maximum error
\(g=4{ \pi }^{ 2 }\times \frac { 1 }{ { T }^{ 2 } } =4{ \pi }^{ 2 }\frac { 1 }{ { \left( \frac { t }{ 20 } \right) }^{ 2 } } \) (taking T=\(\frac{t}{20}\))
or \(g=\frac { 4{ \pi }^{ 2 }l\times ({ 20) }^{ 2 } }{ { t }^{ 2 } } \)
Taking log on both sides, we get
log g log 4 + 2 log rt + log I + 2 log 20 - 2 log t [differentiating both the sides,]
\(\frac { \triangle g }{ g } =\frac { \triangle l }{ l } -2\frac { \triangle t }{ t } \)
Given l =100 cm
\(\triangle\)l = 0.1 cm (least count of the metre scale)
t 40 s,  \(\triangle\)t = 0.1 5 (least count of a stop watch)
\(\therefore\) Maximum error in g = \(\frac { 0.1 }{ 100 } +2\times \frac { 0.1 }{ 40 } \)
= 0.001 + 0.005 = 0.006
= 0.006 x 100% = 0.6%
Here 0.1 % is the error in the measurement of length, and 0.5% is the error in the measurement of time. Therefore, time needs more careful measurement.

\(Fractional\quad error=\frac { \triangle { D }_{ mean } }{ D } =\pm \frac { 0.004 }{ 1.330 } =\pm 0.003\).

Mean absolute error,
\(\triangle { D }_{ mean }=\frac { |{ D }_{ 1 }|+|{ D }_{ 2 }|+|{ D }_{ 3 }|+|{ D }_{ 4 }|+|{ D }_{ 5 }|+|{ D }_{ 6 }| }{ 6 } \) 
\(=\frac { 0.002+0+0.005+0.004+0.004+0.006 }{ 6 } \) 
\(=\frac { 0.021 }{ 6 } =0.0035=0.004\)
(rounding off to 3 decimal places)

It is given that viscous force F depends on (i) radius r of steel ball, (ii) coefficient of viscosity \(\eta \) of viscous liquid (iii) Speed v of the ball
i.e., F = kr^a\(\eta \)^bv^c, where k is dimensionless constant Dimensional formula of force
F = [MLT^-2], r=[L]
\(\eta \) = [M¹L^-1T^-1] and v=[LT^-1], we have
[MLT^-2] = [L]^a[M¹L^-1T^-1]^b[LT^-1]^c
= [M^aL^a-b+cT^-b-c]
Comparing powers of M,L and T on either side of equation, we get
a =1
a - b + c = 1
-b - c = -2
On solving, these above equations, we get
a = 1, b = 1 and c = 1
Hence, the relation becomes
F = kr\(\eta \)v

Absolute error in different measurements are
\({ \triangle \mu }_{ 1 }\)= 1.33-1.29=0.04
\({ \triangle \mu }_{ 2 }\)= 1.33-1.33=0.00
\(\triangle { \mu }_{ 3 }\)= 1.33-1.34=-0.01
\(\triangle { \mu }_{ 4 }\)= 1.33-1.35=-0.02
\(\triangle { \mu }_{ 5 }\)= 1.33-1.32=0.01
\(\triangle { { \mu } }_{ 6 }\)= 1.33-1.36=-0.03
\(\triangle { { \mu } }_{ 7 }\)= 1.33-1.30=0.03
\(\triangle { { \mu } }_{ 8 }\)= 1.33-1.33=0.00
Mean absolute error,
\(\triangle { { \mu } }=\frac { \sum { \left| \triangle { \mu }_{ i } \right| } }{ n }\)
\( \\ \triangle { { \mu } }=\frac { 0.04+0.00+0.01+0.02+0.01+0.03+0.03+0.00 }{ 8 } \)
\(=\frac { 0.14 }{ 8 } =0.0175\approx 0.02\)

Given, Radius of olive oil, r = 0.25 mm = 0.025 cm
Radius of circular film, R = 10 cm and Molecular size, t = ?
We know that,
\(t=\frac { Volume\ of\ oil\ drop }{ Area\ of\ film } =\frac { \frac { 4 }{ 3 } \pi { r }^{ 3 } }{ \pi { R }^{ 2 } }\)
\( \\ t=\frac { \frac { 4 }{ 3 } \times \pi \times { 0.025 }^{ 3 } }{ \pi { \left( 10 \right) }^{ 2 } }\)
\( \\ =\frac { 4 }{ 3 } \times { \left( 25 \right) }^{ 3 }\times { 10 }^{ -11 }\)
\(\\ =2.08\times { 10 }^{ -7 }cm\)
Thus, the molecular size of olive oil is \(2.08\times { 10 }^{ -7 }cm\)