19.4

Yes, for example, a plane angle has no dimensions but has unit like radian for its measurement.

Mass = 5.64kg
radius of gyration = 30cm = 0.3m
SI unit of a mass is kg
SI unit of the radius (gyration) = m² = 0.9
Moment of inertia (I) = mass\(\times\)radius of gyration
= 5.64kg\(\times\)0.9m
= 1.5228 kgm²

All units in atomic standards are more useful because they never change with time.

\(Average\ length=\frac{2.48+1.46+2.49+2.50+2.48}{5}=\frac{12.41}{5}=2.48\ m\)
\(Mean\ absolute\ error=\frac{0.00+0.02+0.01+0.02+0.00}{5}=\frac{0.05}{5}=0.01\ m\)
\(Percentage\ error=\frac{0.01}{2.48}\times100\%=0.04\times100\%=0.40\%\)
Correct length = (2.48 ± 0.01)m
Correct length = (2.48m ± 0.40%)

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

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]

10^-6m equivalent of 1 \(\mu\) m
1 metre is equivalent to \({{1}\over{{10}^{-6}}}={10}^{6}\mu\) m
In one metre 10⁶ microns are present

1.496\(\times\)10¹¹m is equivalent to 1 AU.
1 metre is equivalent to \({{1}\over{1.496\times{10}^{11}}}\)
= 0.6884\(\times\)10^-11
= 6.68\(\times\)10^-12 AU
In one more, 6.68\(\times\)10^-12 astronomical units are present.

Angle \(\theta\) = 2' = 2\(\times\)60" = 120" = 120\(\times\)4.85\(\times\)10^-6 rad
\(\theta\) = 5.82\(\times\)10^-4 rad;
d = 2 x r
d = 2 x 6400 = 12800 x 10³ m
The distance of heavenly body
\(D=\frac{d}{\theta}=\frac{12800\times10^3}{5.82\times10^{-4}}\)
D = 2.19\(\times\)10¹⁰m.

\(Average \ length ={3.48+3.46+3.49+3.50+3.48\over 5}={17.41\over5}\)
= 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 =\({0.01\over 3.48}\times 100=0.29\%\)