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Question

Question: Show that 1 Joule = \({10^7}erg\)....

Show that 1 Joule = 107erg{10^7}erg.

Explanation

Solution

Hint: In this question use the concept that Joule is a unit in the SI system whereas erg is a unit in the C.G.S system of measurement. C.G.S refers to the unit of centimeter, gram and second. So 1 erg = 1 g cm2s2\dfrac{cm^2} {s^2}. This will help getting the right conversion for the given problem statement.

Complete step-by-step answer:
Proof –
As we know that both Joule and erg are the unit of energy.
Where, Joule is a S.I unit and erg is a C.G.S unit.
S.I = system of international unit, where as
C.G.S = centimeter gram second.
Now as we know that 1 joule = 1 Kg m2s2\dfrac{m^2} {s^2}.
And 1 erg = 1 g cm2s2\dfrac{cm^2} {s^2}.
Now first convert kilogram into gram we have,
As we know that 1Kilograms = 1000grams....................... (1)
Now convert meter into centimeter we have,
As we know that 1m = 100cm
Now square on both sides we have,
1 square meter = 10000 square centimeter.......................... (2)
Now multiply equation (1) and (2) we have,
1kg m2m^2 = 1000(10000) gm cm2cm^2.
Now divide by second square on both sides we have,
1kg m2s2\dfrac{m^2} {s^2} = 1000(10000) gm cm2s2\dfrac{cm^2} {s^2}.
Therefore,
1kg m2s2\dfrac{m^2} {s^2} = 107{10^7} gm cm2s2\dfrac{cm^2} {s^2}.
Therefore, 1 joule = 107{10^7} erg.
So this is the required answer.
Hence proved.

Note – SI units is known as international systems of units, it is a system of physical units based upon the meter, kilogram, second, ampere, kelvin, candela and mole together with a set of prefixes to indicate the multiplication or division by a power of 10.