A piece of silver of mass 2400 g has a volume of 40 cm3. Calculate the density of copper in CGS unit and in SI unit
Answers
Given:
\mathbf{A\ quantity\ =1.4\ \ \dfrac{kJ}{s.m^{2}}}A quantity =1.4
s.m
2
kJ
To Find:
Express in some other unit.
Solution:
We know some standard conversion which are given that:
⇒ \mathbf{1\ \dfrac{J}{s}= 1\ W}1
s
J
=1 W
⇒ \mathbf{1\ \dfrac{kJ}{s}= 1\ kW=10^{3}\ W}1
s
kJ
=1 kW=10
3
W
⇒ \mathbf{1\ m^{2}=10^{4}\ cm^{2}}1 m
2
=10
4
cm
2
Now come to question:
\mathbf{Quantity\ =1.4\ \ \dfrac{kJ}{s.m^{2}}}Quantity =1.4
s.m
2
kJ
Above can be written as:
\mathbf{Quantity\ =1.4\ \ \dfrac{kW}{m^{2}}}Quantity =1.4
m
2
kW
Above can be written as:
\mathbf{Quantity\ =1.4\times 10^{3} \ \dfrac{W}{m^{2}}}Quantity =1.4×10
3
m
2
W
(Which is same as option A)
\mathbf{Quantity\ =\dfrac{1.4\times 10^{3}}{10^{4}} \ \dfrac{W}{cm^{2}}}Quantity =
10
4
1.4×10
3
cm
2
W
On simplify:
\mathbf{Quantity\ =0.14 \ \dfrac{W}{cm^{2}}}Quantity =0.14
cm
2
W
(Which is same as option B)
Again consider the term:
\mathbf{Quantity\ =1.4\ \ \dfrac{kW}{m^{2}}}Quantity =1.4
m
2
kW
Above can be written as:
\mathbf{Quantity\ =\dfrac{1.4}{10^{4}}\ \ \dfrac{kW}{cm^{2}}}Quantity =
10
4
1.4
cm
2
kW
On simplify above:
\mathbf{Quantity\ =1.4\times 10^{-4}\ \ \dfrac{kW}{cm^{2}}}Quantity =1.4×10
−4
cm
2
kW
(Which is same as option C)
Means:
\mathbf{1.4\ \dfrac{kJ}{s.m^{2}}}1.4
s.m
2
kJ
is not equal to option D.