The number of significant figures of 6.3200 J is 5

\(\Delta l =0.01 \)
\(l =3.51 \)
\(\% \text { error } =\frac{\Delta l}{l} \times 100=\frac{0.01}{3.51} \times 100 \)
\(=0.00284 \times 100 \)
\(=0.284=0.28 \% \)

\(T =2 \pi \sqrt{\frac{l}{g}} \)
\(g =4 \pi^{2} l / T^{2} \)
\(\frac{d g}{g} =\frac{d l}{l}-\frac{2 d T}{T} \)
\(\frac{d g}{g} \% =\left(\frac{d l}{l}\right) \%-2\left(\frac{d T}{T}\right) \% \)
\(=1 \%-2 \times(-3 \%) \)
\(=1+6=7 \% \)

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

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

The distance of the planet from the Earth
d = 6.3\(\times\)10¹⁰ m
Time t = 7 minutes = 7\(\times\)60 s.
The speed of signal v = ?
The speed of signal v = \({2d\over t}={2\times 6.3 \times 10^8\over 7\times 60}=3\times 10^8 ms^{-1}\)

Distance, D = 824.7\(\times\)10⁶ km
\(\theta=35.72"={{35.72}\over{60\times 60}}\times{{\pi}\over{180}}\) rad
\(\theta \ \rightarrow\) is angular diameter
Diameter, d = ?
\(d=D\times\theta\)
\(=824.7\times{10}^{6}\times{{35.72}\over{60\times 60}}\times{{\pi}\over{180}}\) km 
\(=824.7\times{10}^{6}\times{{35.72}\over{3600}}\times{{\pi}\over{180}}\) km
\(=824.7\times{10}^{6}\times{{35.72\times3.14}\over{3600\times180}}\)km
Diameter of Jupiter = 1.427\(\times\)10⁵ km

t₁ = 2.63s, t₂ = 2.56s,
t₃ = 2.42s, t₄ = 2.71s, t₅ = 2.80s
(i) T_m =\({t_1+t_2+t_3+t_4+t_5\over 5}={2.63+2.56+2.42+2.71+2.80\over 5}\)
T_m = 2.62s (Rounded off to 2^nd decimal place)
(ii) Absolute error
\(\triangle\)T = Tm - t
\(\triangle\)T_I = 2.62 - 2.63 = +0.01s
\(\triangle\)T₂ = 2.62 - 2.56 = +0.06s
\(\triangle\)T₃ = 2.62 - 2.42 = +0.20s
\(\triangle\)T₄ = 2.62 - 2.71 = +0.09s
\(\triangle\)T₅ = 2.62 - 2.80 = +0.18s
(iii) Mean absolute error = \({\sum |\triangle T_1|\over n}\)
\(\triangle\)T_m = \({0.01+0.06+0.20+0.09+0.18\over 5}\)
\(\triangle\)T_m​​​​=\({0.54\over 5}=0.108s=0.11s\) (Rounded off to 2^nd decimal place)
(iv) Relative error: S_T =\({\triangle T_m\over T_m}={0.11\over 2.62}=0.0419\)
S_T = 0.04
(v) Percentage error in T = 0.04\(\times\)100% = 4%
(vi) Time period of simple pendulum = T = (2.62 ± 4%)s

Angle \(\theta\) = 1° 55'= 115' = (115\(\times\)60)"\(\times\) (4.85\(\times\) 10^-6) rad = 3.34\(\times\)10^-2 rad
Since 1" = 4.85\(\times\)10^-6 rad
Radius of the Earth = 6.4\(\times\)10⁶m
From the Figure AB is the diameter of the Earth (b) = 2\(\times\)6.4\(\times\)10⁶m
Distance of the Moon from the Earth x = ?
\(x={b\over \theta}={2\times 6.4 \times 10^6\over 3.34\times 10^{-2}}\)
x = 3.83\(\times\)10⁸m

Angle \(\theta\) = 60 ^o
The distance between the tree and a point
x = 50m
Height of the tree (h)=?
For triangulation method tan\(\theta\) =\({h\over x}\)
h= x tan \(\theta\) = 50\(\times\)tan 60⁰ = 50\(\times\)1.732
h = 86.6m
The height of the tree is 86.6m.