Question

Convert 1 joule into erg.​

Answer 1

Converting 1 joule into 1 erg.

So,

$\boxed{\red{n_{1}[M_{1}^{a}\ L_{1}^{b}\ T_{1}^{c}]=n_{2}[M_{2}^{a}\ L_{2}^{b}\ T_{2}^{c}]}}$

Here,

System 1(The value of known) [S.I unit]

• n₁ = 1
• M₁ = 1 kg
• L₁ = 1 m
• T₁ = 1 sec

Now,

System 2(The value of unknown) [C.G.S unit]

• n₂ =?
• M₂ = 1 g
• L₂ = 1 cm
• T₂ = 1 sec

So,

Unit of work - [$\green{M^{1}L^{2}T^{-2}}$ ]

So, a = 1, b = 2, c = (-2)

$\boxed{\implies n_{2}=n_{1}[\frac{M_{1}}{M_{2}}]^{a}\ [\frac{L_{1}}{L_{2}}]^{b}\ [\frac{T_{1}}{T_{2}}]^{c}}$

$\implies 1[\frac{1kg}{1g}]^{1}\ [\frac{1m}{1cm}]^{2}\ [\frac{1sec}{1sec}]^{-2}$

$\implies 1[\frac{1000g}{1g}]^{1}\ [\frac{100cm}{1cm}]^{2}\ [\frac{\not1sec}{\not1sec}]^{-2}$

$\implies 1[\frac{1000\not{g}}{1\not{g}}]^{1}\ [\frac{100\not{cm}}{1\not{cm}}]^{2}$

$\implies 1000 \times (100)^{2}$

$\implies 1000 \times 10000$

$\implies 10^{3} \times 10^{4}$

$\implies 10^{3+4}$

[As $a^{n} \times a^{m} = a^{n+m}$ ]

$\boxed{\purple{\sf{=10^{7}erg}}}$

Hence,

1 joule is 10⁷ erg.

Answer 2

$\bold{\huge\blue{\boxed{{{Question}}}}}$

Convert 1 joule into erg.

$\bold{\huge\blue{\boxed{{{Answer}}}}}$

Unit of energy is joule = kg*m²/s²

Joule is a MKS unit and erg is a CGS unit

MKS = CGS

1 m = 100 cm

1 s = 1 s

1 kg = 1000 gm

From above :

1 joule = 1 kg*m²/s²

= 1000 * 10000 / 1 cm*m²/s²

= 10^7 gm*cm²/s²

= 10^7 erg

So, 1 joule = 10^7 erg