
$ML{T^{ - 1}}$ represents the dimensional formula of
a. Power
b. Momentum
c. Force
d. Couple
Answer
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Hint: We can find out the dimensional formula of all the quantities given in the question by their simple formulas. We can break the quantities into others and then find out the dimensional formula for each of them in terms of the units of fundamental units.
Complete step by step answer:
The dimensional formula of power:
As we know that power is work done per unit time, hence $P = \dfrac{W}{t}$, where $P$ is the power, $W$ is the work done and $t$ is the time taken.
We know that work is the product of force and displacement in the direction of applied force, therefore the dimensional formula for work will be $[{M^1}{L^2}{T^{ - 2}}]$.
Therefore, the dimensional formula for power will be $P = \dfrac{{[{M^1}{L^2}{T^{ - 2}}]}}{{[{T^1}]}} = [{M^1}{L^2}{T^{ - 2}}][{T^{ - 1}}]$
$\Rightarrow P = [{M^1}{L^2}{T^{ - 3}}]$
For momentum, we know that momentum $M = m \times v$, where $m$ is the mass and $v$ is the velocity.
Therefore, the dimensional formula of momentum will be $M = [{M^1}{L^0}{T^0}] \times [{M^0}{L^1}{T^{ - 1}}]$
$\Rightarrow M = [{M^1}{L^1}{T^{ - 1}}]$
For force, we know that force $F = m \times a$, where $m$ is the mass and $a$ is the acceleration.
Therefore, the dimensional formula of force will be $F = [{M^1}{L^0}{T^0}] \times [{M^0}{L^1}{T^{ - 2}}]$
$\Rightarrow F = [{M^1}{L^1}{T^{ - 2}}]$
We know that a couple is a moment generated by two anti-parallel forces and is the product of force and perpendicular distance between the two forces.
Therefore a couple $C = F \times d$, $F$ is the force and $d$ is the distance.
The dimensional formula of couple $C = [{M^1}{L^1}{T^{ - 2}}] \times [{M^0}{L^1}{T^0}]$
$\Rightarrow C = [{M^1}{L^2}{T^{ - 2}}]$
Hence, the correct answer is option (B).
Note: Momentum is a vector quantity and can be expressed in terms of the fundamental quantities and therefore it is known as a derived unit. There are seven fundamental quantities (mass, length, time, current, temperature, intensity of light and quantity of substance) from which the units of the derived quantities can be found out.
Complete step by step answer:
The dimensional formula of power:
As we know that power is work done per unit time, hence $P = \dfrac{W}{t}$, where $P$ is the power, $W$ is the work done and $t$ is the time taken.
We know that work is the product of force and displacement in the direction of applied force, therefore the dimensional formula for work will be $[{M^1}{L^2}{T^{ - 2}}]$.
Therefore, the dimensional formula for power will be $P = \dfrac{{[{M^1}{L^2}{T^{ - 2}}]}}{{[{T^1}]}} = [{M^1}{L^2}{T^{ - 2}}][{T^{ - 1}}]$
$\Rightarrow P = [{M^1}{L^2}{T^{ - 3}}]$
For momentum, we know that momentum $M = m \times v$, where $m$ is the mass and $v$ is the velocity.
Therefore, the dimensional formula of momentum will be $M = [{M^1}{L^0}{T^0}] \times [{M^0}{L^1}{T^{ - 1}}]$
$\Rightarrow M = [{M^1}{L^1}{T^{ - 1}}]$
For force, we know that force $F = m \times a$, where $m$ is the mass and $a$ is the acceleration.
Therefore, the dimensional formula of force will be $F = [{M^1}{L^0}{T^0}] \times [{M^0}{L^1}{T^{ - 2}}]$
$\Rightarrow F = [{M^1}{L^1}{T^{ - 2}}]$
We know that a couple is a moment generated by two anti-parallel forces and is the product of force and perpendicular distance between the two forces.
Therefore a couple $C = F \times d$, $F$ is the force and $d$ is the distance.
The dimensional formula of couple $C = [{M^1}{L^1}{T^{ - 2}}] \times [{M^0}{L^1}{T^0}]$
$\Rightarrow C = [{M^1}{L^2}{T^{ - 2}}]$
Hence, the correct answer is option (B).
Note: Momentum is a vector quantity and can be expressed in terms of the fundamental quantities and therefore it is known as a derived unit. There are seven fundamental quantities (mass, length, time, current, temperature, intensity of light and quantity of substance) from which the units of the derived quantities can be found out.
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