Name: Conversion of Units Using Dimensional Analysis
Uploaded: 2019-07-19
Duration: 6 min 14 s
Description: How do you solve problems if the units of quantity are different? Is there any conversion method? Let us discuss the conversion using dimensional analysis in this video.

Question

The value of the gravitational constant

G

in MKS system is

6.67 \times 10^{- 11} N m^{2} {kg}^{- 2}

. What will be its value in CGS system?

Accepted Answer

According to the question, we have to convert the value of quantity in M.K.S system to C.G.S system.

So,

M.K.S system.
C.G.S system.

$n_{1} = 6.67 \times 10^{- 11} \frac{N m^{2}}{{kg}^{2}}$
$n_{2} = ?$

$l_{1} = 1 m$ $l_{2} = 1 cm$

$m_{1} = 1 kg$ $m_{2} = 1 gm$

$t_{1} = 1 s$ $t_{2} = 1 s$

The dimensional formula for gravitational constant $= [M^{- 1} L^{3} T^{- 2}]$

Here, $x = - 1$ , $y = 3$ and $z = - 2$

From $n_{2} = n_{1} [{(\frac{m_{1}}{m_{2}})}^{x} . {(\frac{l_{1}}{l_{2}})}^{y} . {(\frac{t_{1}}{t_{2}})}^{z}]$

Now substituting all the value in the above formula, we get,

$n_{2} = 6.67 \times 10^{- 11} [{(\frac{1 kg}{1 gm})}^{- 1} . {(\frac{1 m}{1 cm})}^{3} . {(\frac{1 s}{1 s})}^{- 2}]$

$= 6.67 \times 10^{- 11} [{(\frac{10^{3} gm}{1 gm})}^{- 1} . {(\frac{10^{2} cm}{1 cm})}^{3} . {(\frac{1 s}{1 s})}^{- 2}]$

$∴$ $n_{2} = 6.67 \times 10^{- 8} \frac{dyne \times {cm}^{2}}{{gm}^{2}}$

Hence, the value of $G$ in C.G.S system is, $6.67 \times 10^{- 8} \frac{dyne \times {cm}^{2}}{{gm}^{2}}$ .

$(A)$	$(B)$
(a) Base unit	(i) $N$
(b) Derived SI unit	(ii) $h p$
(c) Improper unit	(iii) $k g - w a t$
(d) Practical unit	(iv) $r a d$
(e) Supplmentary unit	(v) $k g$

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