Answer

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**Hint:-**

Force can write as a product of mass and acceleration.

The unit of force is Newton (\[N\]) or \[kgm{s^{ - 2}}\]

Work is the scalar product of force and displacement

Weight is related to gravitational force.

**Complete step by step solution:-**

Force is given by Newton’s second law,

‘Rate of change of momentum is directly proportional to the applied force’.

\[f \propto \dfrac{{dp}}{{dt}}\]

Proportionality constant is unity, so

\[f = \dfrac{{dp}}{{dt}}\]

Where,

\[f\] is the force

\[p\] is the momentum, \[dp\] is the change in momentum.

\[dt\] is a change in time.

Momentum \[p = mv\]

\[m\] is the mass, \[v\] is the velocity.

Unit of velocity\[ = m{s^{ - 1}}\]

Unit of momentum \[ = kg \times m{s^{ - 1}} = kgm{s^{ - 1}}\]

So

Unit of force \[ = kgm{s^{ - 1}} \times {s^{ - 1}} = kgm{s^{ - 2}}\]

Dimension of force \[f \equiv ML{T^{ - 2}}\]

Or

If mass is constant f is the product of mass (\[m\]) and acceleration (\[a\]).

\[f = ma\]

Dimension of \[a \equiv L{T^{ - 2}}\].

So Dimension of force \[f \equiv ML{T^{ - 2}}\]

\[M\] Represent the mass, \[L\]represent the length, and \[T\] represent the time.

Weight is given by force applied by the gravitation on an object.

Weight is given by \[{f_w} = mg\]

\[g\] is the acceleration due to gravity. Which have the same dimension of acceleration.

Dimension of force \[{f_w} \equiv ML{T^{ - 2}}\]

So, Weight and force have the same dimension.

Rate of change of Momentum.

By the definition we already understood that the rate of change of momentum is equal to the force.

So, the rate of change of momentum also has the same dimension of force.

Work per unit length.

Work is defined as a scalar product of mass and the displacement.

Work

\[S\] is the displacement.

Work per unit length \[w = W/S = F\]

So work per unit length also has the same dimension of force.

Work done per unit time.

Work already defined.

Work

So, Dimension of work

\[W \equiv ML{T^{ - 2}} \times L \equiv M{L^2}{T^{ - 2}}\]

So work per unit time

\[W/t \equiv M{L^2}{T^{ - 2}} \times {T^{ - 1}} \equiv M{L^2}{T^{ - 2}}\]

Work per unit time does not have a dimension of force.

**So the answer is (D) Work done per unit time.**

**Note:-**

Work per unit time is called power.

Unit of power is Watt.

\[f = ma\]is correct only if mass is a constant quantity.

Work and energy have the same dimensional formula.

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