Question
Find the value of 60 W on a system having 100 g, 20 cm and 1 min as the fundamental units.
✓
Answer
Given, $n _1=60 W$, power, P is $\left[ M ^1 L^2 T^{-3}\right]$
In first system, $M _1=1 kg=1000 g, L _1=1 m=100 cm$, and $T _1=1 s$
In second system, $M _2=100 g, L _2=20 cm$ and $T _2=1 min=60 s$
In a given physical quantity, the product of its magnitude and unit is always constant.
$
\begin{aligned}
& n_1 u_1=n_2 u_2 \\
& n_2=n_1\left[\frac{M_1}{M_2}\right]^1\left[\frac{L_1}{L_2}\right]^2\left[\frac{T_1}{T_2}\right]^{-3} \\
& =60\left[\frac{1000 g}{100 g}\right]\left[\frac{100 cm}{20 cm}\right]^2\left[\frac{1 s}{60 s}\right]^{-3} \\
& =60 \times \frac{1000}{100} \times \frac{100}{20} \times \frac{100}{20} \times 60 \times 60 \times 60 \\
& =3.24 \times 10^9 \text { units }
\end{aligned}
$
In first system, $M _1=1 kg=1000 g, L _1=1 m=100 cm$, and $T _1=1 s$
In second system, $M _2=100 g, L _2=20 cm$ and $T _2=1 min=60 s$
In a given physical quantity, the product of its magnitude and unit is always constant.
$
\begin{aligned}
& n_1 u_1=n_2 u_2 \\
& n_2=n_1\left[\frac{M_1}{M_2}\right]^1\left[\frac{L_1}{L_2}\right]^2\left[\frac{T_1}{T_2}\right]^{-3} \\
& =60\left[\frac{1000 g}{100 g}\right]\left[\frac{100 cm}{20 cm}\right]^2\left[\frac{1 s}{60 s}\right]^{-3} \\
& =60 \times \frac{1000}{100} \times \frac{100}{20} \times \frac{100}{20} \times 60 \times 60 \times 60 \\
& =3.24 \times 10^9 \text { units }
\end{aligned}
$
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