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Chapter 1: Problem 122

Dimensional formula for power is (a) \(\mathrm{M}^{2} \mathrm{~L}^{-2} \mathrm{~T}^{-3}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{3} \mathrm{~T}^{-1}\) (d) \(\mathrm{M}^{0} \mathrm{~L}^{2} \mathrm{~T}^{-2}\)

Short Answer

Expert verified
The dimensional formula for power is \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3}\).

Step by step solution

01

Recall the definition of power

Power is the rate at which work is done, or energy is transferred. Mathematically, it can be expressed as: Power (P) = Work (W) / Time (t) Now, we need to find the dimensional formula for work and use that to derive the one for power.
02

Find the dimensional formula for work

Work is the product of force (F) and displacement (d). In other words: Work (W) = Force (F) × Displacement (d) Force has the dimensional formula \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) (Newton's second law of motion: F=ma). Displacement has the dimensional formula \(\mathrm{L}^{1}\) (as it is a length). To get the dimensional formula for work, we multiply the dimensional formulas: \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) × \(\mathrm{L}^{1}\) = \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\)
03

Derive the dimensional formula for power

Based on the definition of power, we divide the dimensional formula for work by the dimensional formula for time: \(\frac{\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}}{\mathrm{T}^{-1}} = \frac{\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}}{\mathrm{T}^{1}}\) When dividing the dimensional formulas, we subtract the exponents of T: \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3}\) So, the correct dimensional formula for power is \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-3}\) which corresponds to the option (a).

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Most popular questions from this chapter

Find the dimensional formula for energy per unit surface area per unit time (a) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-2}\) (b) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{0} \mathrm{~T}^{-3}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{-1} \mathrm{~T}^{1}\)

\(1^{\circ}=\ldots \ldots \ldots \ldots \ldots\) (a) \(600^{\prime \prime \prime}\) (b) \(3600^{\prime \prime}\) (c) \(180^{\prime \prime}\) (d) \(3600^{\prime}\)

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Equation of physical quantity \(\mathrm{v}=\mathrm{at}+\mathrm{bt}^{2}\) where \(\mathrm{v}=\) velocity \(\mathrm{t}=\) time so write the dimensional formula of a in this equation (a) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\) (c) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{0}\)
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