Question

Convert 5 joule work in ergs using dimension

Answer 1

Heya!
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◾➖◾Convert 5J into Ergs Using Dimensions .
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S. I unit -----------------------> C.G.S unit


We Have
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◾M1 = 1Kg , L1 = 1m and T1 = 1 sec.

◾M2= 1g , L2= 1cm and T2 = 1sec .


◾➖◾Dimensions Of Energy => [ ML²T^-2 ]


Applying in formula ,


N1 × [ M1] [ L1] [ T1] = N2 [ M2] [ L2] [ T2]


=> 5 × [ 1Kg ] [ 1 m ] [ 1sec ] = N2 [ 1g ] [ 1cm ] [ 1sec]


$= > 5 \times ( \frac{1kg}{1g} )( \frac{1m}{1cm} ) {}^{2} ( \frac{1sec}{1sec} ) {}^{ - 2} = n2 \\ \\ = > 5 \times 1000 \times 100 \times 100 = n2 \\ \\ = > n2 = 5 \times 10 {}^{7} ergs$



◾➖◾Alternative Way :- We Know 1 J = 10^7 Ergs.

=> Hence 5 J = 5 × 10^7 Ergs.




➖◾➖Answer => 5 × 10^7 Ergs.


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Answer 2

$\underline{\bold{Solution:-}}$

First of all we have to write the dimensional formula of joule.

Dimensional formula of joule is ${M}^{1} {L}^{2} {T}^{ - 2}$

n1 = 1

M1 = 1 kg

M2 = 1 gm

L1 = 1 m

L2 = 1 cm

T1 = 1 sec

T2 = 1 sec

Let n1 be the number of joule and n2 be the number of erg.

n1 [M₁¹L₁²T₁⁻²] = n2 [M₂¹L₂²T₂⁻²]

On dividing both sides by [M₂¹L₂²T₂⁻²].

n1 [M₁¹L₁²T₁⁻²] / [M₂¹L₂²T₂⁻²] = n2

n2 = n1 [M₁¹L₁²T₁⁻²] / [M₂¹L₂²T₂⁻²]

On Separating same term

$n_{2} \: = \: n_{1} \times { [\frac{M_{1} }{M_{2} }]}^{1} \times { [\frac{L_{1} }{L_{2} }]}^{2} \times { [\frac{T_{1} }{T_{2} }]}^{ - 2} \\ \\on \: putting \: there \: values \\ \\ n_{2} \: = \: n_{1} \times { [\frac{1 \: kg}{1 \: gm} ]}^{1} \times {[ \frac{1 \: m}{1 \: cm} ]}^{2} \times { [\frac{1 \: sec}{1 \: sec} ]}^{ - 2} \\ \\ n_{2} \: = \: n_{1} \times {[ \frac{1000 \: gm}{1 \: gm}]}^{1} \times { [\frac{100 \: cm}{1 \: cm} ]}^{2} \times { 1 }^{ - 2} \\ \\ n_{2} \: = \: 1 \times {10}^{7} \\ \\ n_{2} \: = \: {10}^{7} \\ \\ so \: 1 \: joule \: = \: {10}^{7} \: erg$