Question

The power of a motor is $1600$watt. If the unit of mass is doubled and units of length and time are halved, the power of the motor in new system is
(A) $400$ units
(B) $6400$ units
(C) $3200$ units
(D) $4800$ units

Answer 1

Hint
To solve this question, we have to convert the unit of power into the powers of the fundamental dimensions, that is, mass, length and time. Making changes in these fundamental dimensions, as given in the question will give the new unit of the power, which will give us the required answer.
The formulae used in solving this question are:
$\Rightarrow P = Fv$
$\Rightarrow F = ma$
Where the symbols have their usual meanings.

Complete step by step answer
The unit of power in SI units is watt or $W$
Let $u$ be the unit of power in the new system of units.
For getting its dimension formula, we use the formula
$\Rightarrow P = Fv$
We know that $F = ma$
So, $P = mav$
Now, writing the dimensional formula for each quantity
$\Rightarrow \left[ m \right] = \left[ {{M^1}{L^0}{T^0}} \right]$, $\left[ a \right] = \left[ {{M^0}{L^1}{T^{ - 2}}} \right]$ , $\left[ v \right] = \left[ {{M^0}{L^1}{T^{ - 1}}} \right]$
We get the dimensional formula for power as
$\Rightarrow \left[ P \right] = \left[ {{M^1}{L^0}{T^0}} \right]\left[ {{M^0}{L^1}{T^{ - 2}}} \right]\left[ {{M^0}{L^1}{T^{ - 1}}} \right]$
Simplifying, we get
$\Rightarrow \left[ P \right] = \left[ {{M^1}{L^2}{T^{ - 3}}} \right]$ (1)
Now, writing the SI units of each dimension
$\Rightarrow P = W$, $M = kg$, $L = m$, $T = s$
Substituting these in (1)
$\Rightarrow W = k{g^1}{m^2}{s^{ - 3}}$ (2)
In the new system of units, the unit of mass is doubled and units of length and time are halved, i.e.
$\Rightarrow M = 2kg$ , $L = \dfrac{m}{2}$, $T = \dfrac{s}{2}$
Substituting these in (1)
$\Rightarrow u = {\left( {2kg} \right)^1}{\left( {\dfrac{m}{2}} \right)^2}{\left( {\dfrac{s}{2}} \right)^{ - 3}}$
$\Rightarrow u = 2{\left( {\dfrac{1}{2}} \right)^2}{\left( {\dfrac{1}{2}} \right)^{ - 3}}k{g^1}{m^2}{s^{ - 3}}$
Simplifying, we get
$\Rightarrow u = 4k{g^1}{m^2}{s^{ - 3}}$
Substituting (2)
$\Rightarrow u = 4W$ (3)
Therefore, 1 unit of power in the new system is equal to 4 times watt power.
As the power given in the question is $1600W$ so multiplying both sides of (3) by $1600$
$\Rightarrow 1600u = 4(1600)W$
Dividing by four, we get
$\Rightarrow 1600W = \dfrac{{1600u}}{4}$
$\Rightarrow 1600W = 400u$
So, $1600$ watt of power is equal to $400$ units of power in the new system.
Hence, the correct answer is option (A), $400$ units.

Note
We can use any formula for the power in mechanics, to make the conversion to the basic units of mass, length and time, as per our comfort. But the preference is always given to that formula, which contains a maximum number of the fundamental dimensions on the right hand side of the formula.