Dimensions of impulse are. (a) \(\mathrm{M}^{-1} \mathrm{~L}^{-1} \mathrm{~T}^{1}\) (b) $\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}$ (c) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{1}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\)

The formula for impulse (I) is given by \(I = F \cdot \Delta t\) where \(F\) is the force and \(\Delta t\) is the time interval.

The dimensions of Force (F) are given by: \(\mathrm{[F]} = \mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) since force is a product of mass and acceleration, and acceleration has dimensions of \(\mathrm{L}^{1} \mathrm{~T}^{-2}\). The dimensions of Time (\(\Delta t\)) are: \(\mathrm{[\Delta t]} = \mathrm{T}^{1}\)

Since impulse is a product of force and time, we multiply the dimensions of force by the dimensions of time: \(\mathrm{[I]} = \mathrm{[F]} \times \mathrm{[\Delta t]} \) \(\mathrm{[I]} = (\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}) \times (\mathrm{T}^{1})\) By multiplying the exponents with similar bases, we find the dimensions of impulse: \(\mathrm{[I]} = \mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) Now, let's compare the found dimensions with the options given in the exercise: (a) \(\mathrm{M}^{-1} \mathrm{~L}^{-1} \mathrm{~T}^{1}\) (b) \(\boxed{\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{1}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\) The correct answer is (b) because it matches the dimensions we found for impulse: \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\).