Answer
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Hint: It is a problem in which we check correctness of units for a physical quantity. Use the formula and definition of energy for comparing the left hand side and right hand side of the relation. In case of correct units, both the sides will produce the same results.
Formula used:
$E = F.s$
Where, $E$ is energy of an object, $F$ is applied force on the object, $s$ is total displacement of the object.
Complete step by step answer:
The S.I. unit for energy is $Joule(J)$, for force it is $Newton(N)$ or $kgm{s^{ - 2}}$ and for displacement it is $m$.
Now if we put the formula in terms of units of respective items, we get $J = kg{m^2}{s^{ - 2}}$ .
Similarly, the c.g.s unit of energy is $erg(gc{m^2}{s^{ - 2}})$
Let’s start analysing the options.
For option A, we get
$Js = kg{m^2}{s^{ - 1}}$
For option B, we get
$Ws = kg{m^2}{s^{ - 2}}$, because if we expand $W$ (watts) like we have done energy, we will get its S.I. unit to be $kg{m^2}{s^{ - 3}}$.
For option C, we get
$kWh = kg{m^2}{s^{ - 2}}$ in the same way as option B.
For option D,
$erg$ is the c.g.s unit for energy.
So the correct answer is option A, because it does not correspond with units (S.I. or c.g.s. system).
Additional Information:
Energy is the ability of a system to do work. Dimensionally it is similar to work.
The SI unit of power $(P)$ is watt $(W)$
Formula for power: $P = \dfrac{{dW}}{{dt}}$
Where, $dW$ is the rate at which work is done by a force.
$dt$ is the change in time.
Note: In these types of questions where we are asked to check the correctness of a unit for a physical quantity, we must analyse each option separately. We can also check the units by analysing dimensions of both sides of the relation corresponding to the definition of that physical quantity, but it is a lengthier procedure. While analysing units, be sure to do that in a single system of units, if the systems are different, convert them. One should ignore the numerical coefficients written before the units if checking for correctness of units only.
Formula used:
$E = F.s$
Where, $E$ is energy of an object, $F$ is applied force on the object, $s$ is total displacement of the object.
Complete step by step answer:
The S.I. unit for energy is $Joule(J)$, for force it is $Newton(N)$ or $kgm{s^{ - 2}}$ and for displacement it is $m$.
Now if we put the formula in terms of units of respective items, we get $J = kg{m^2}{s^{ - 2}}$ .
Similarly, the c.g.s unit of energy is $erg(gc{m^2}{s^{ - 2}})$
Let’s start analysing the options.
For option A, we get
$Js = kg{m^2}{s^{ - 1}}$
For option B, we get
$Ws = kg{m^2}{s^{ - 2}}$, because if we expand $W$ (watts) like we have done energy, we will get its S.I. unit to be $kg{m^2}{s^{ - 3}}$.
For option C, we get
$kWh = kg{m^2}{s^{ - 2}}$ in the same way as option B.
For option D,
$erg$ is the c.g.s unit for energy.
So the correct answer is option A, because it does not correspond with units (S.I. or c.g.s. system).
Additional Information:
Energy is the ability of a system to do work. Dimensionally it is similar to work.
The SI unit of power $(P)$ is watt $(W)$
Formula for power: $P = \dfrac{{dW}}{{dt}}$
Where, $dW$ is the rate at which work is done by a force.
$dt$ is the change in time.
Note: In these types of questions where we are asked to check the correctness of a unit for a physical quantity, we must analyse each option separately. We can also check the units by analysing dimensions of both sides of the relation corresponding to the definition of that physical quantity, but it is a lengthier procedure. While analysing units, be sure to do that in a single system of units, if the systems are different, convert them. One should ignore the numerical coefficients written before the units if checking for correctness of units only.
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