Let the height of the pole be h
Solution \(\frac { x+45 }{ h } \)  = cot 30^o ⇒ h =\(\frac { x+45 }{ cot\quad { 30 }^{ o } } \)
\(\frac { x }{ h } \) = cot 30^o ⇒ x = h cot 60^o
Substituting the values of x in the above equation
h = \(\frac { h\quad cot \ { 60 }^{ o }+45 }{ cot \ { 30 }^{ o } } \)
\(h \cot 30^{\circ} =h \cot 60^{\circ}+45  \)
\(h\left(\cot 30^{\circ}-\cot 60^{\circ}\right) =45  \)
\(h =\frac{45}{\cot 30^{\circ}-\cot 60^{\circ}}=\frac{45}{\sqrt{3}-\frac{1}{\sqrt{3}}}=38.97 \mathrm{~m}\)

The given readings are 2.36 cm, 2.27 cm, 2.26 cm, 2.28 cm, 2.31 cm, 2.28 cm and 2.29 cm
The mass value \(\overrightarrow { l } =\frac { 2.36+2.27+2.26+2.28+2.31+2.28+2.29 }{ 7 } =\frac { 16.08 }{ 7 } \)  = 2.29 cm
Absolute errors in the measurements are
\({ \Delta }l_{ _{ 1 } }\) = 2.29 - 2.36 = -0.07
\({ \Delta }l_{ _{ 2 } }\) = 2.29 - 2.27 = 0.02
\({ \Delta }l_{ _{ 3 } }\)= 2.29 - 2.26 = 0.03
\({ \Delta }l_{ _{ 4 } }\)= 2.29 - 2.28 = 0.01
\({ \Delta }l_{ _{ 5 } }\)=2.29 - 2.31 = -0.02
\({ \Delta }l_{ _{ 6 } }\)= 2.29 - 2.28 = 0.01
\({ \Delta }l_{ _{ 7 } }\)= 2.29 - 2.29 = 0.00
Mean Absolute error
Δl_mean =\(\frac { 0.7+0.02+0.03+0.01+0.02+0.01+0.00 }{ 7 } =\frac { 0.16 }{ 7 } \)=0.02
Relative error =  \(\frac { { \Delta l }_{ mean } }{ \overline { l } } =\pm \frac { 0.02 }{ 2.29 } =\pm 8.7\times { 10 }^{ -3 }\)
Percentage error = 8.7\(\times\)10^-3\(\times\)100 = 0.87%\(\times\)100 = ± (8.7\(\times\)10^-1) = 0.9%

The equivalent resistance of a parallel combination
\(R'=\frac{R_1R_2}{R_1+R_2}=\frac{150\times220}{150+220}=\frac{33000}{370}=89.1\ Ohm\)
We know that, \(\frac{1}{R'}=\frac{1}{R_1}+\frac{1}{R_2}\)
\(\frac { \triangle { R }^{ ' } }{ \left( { R }^{ ' } \right) ^{ 2 } } =\frac { \triangle { R }_{ 1 } }{ { R }_{ 1 }^{ 2 } } +\frac { \triangle { R }_{ 2 } }{ { R }_{ 2 }^{ 2 } } \)
\(\triangle { R }^{ ' }=\left( { R }^{ ' } \right) ^{ 2 }\frac { \triangle { R }_{ 1 } }{ { R }_{ 1 }^{ 2 } } +\left( { R }^{ ' } \right) ^{ 2 }\frac { \triangle { R }_{ 2 } }{ { R }_{ 2 }^{ 2 } } =\left( \frac { { R }^{ ' } }{ { R }_{ 1 } } \right) ^{ 2 }\triangle { R }_{ 1 }+\left( \frac { { R }^{ ' } }{ { R }_{ 2 } } \right) ^{ 2 }\triangle { R }_{ 2 }\)
Substituting the value,
\(\triangle { R }^{ ' }=\left[ \frac { 89.1 }{ 150 } \right] ^{ 2 }\times 2+\left[ \frac { 89.1 }{ 220 } \right] ^{ 2 }\times 6=0.070+0.098=0.168\)
R' = 89.1 ± 0.168 Ohm.

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

T ∝ m^a l^b g^c;
T = k.m^a l^b g^c
Here k is the dimensionless constant. Rewriting the above equation with dimensions
[T¹] = [M^a] [L^b] [LT^-2]^c
[M⁰L⁰T¹] = [M^aL^b+cT^-2c]
Comparing the powers of M, L and T on both sides, a = 0, b + c = 0, -2c = 1
Solving for a, b and c a = 0, b = 1/2, and c = -1/2
From the above equation T = k.m⁰ l^1/2 g^-1/2
T= k\(\left( \frac { 1 }{ g } \right) ^{ 1/2 }=k\sqrt { \frac { l }{ g } } \)
Experimentally k = 2π hence
T = \(2\pi \sqrt { l/g } \)