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Chapter 1: Problem 134

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

Short Answer

Expert verified
The correct dimensional formula for torque is \(\mathrm{[\tau]} = \mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\).

Step by step solution

01

Recall the formula for torque

Torque (\(\tau\)) can be defined as the cross product of force (\(\vec{F}\)) and perpendicular distance (\(\vec{r}\)) from the point of rotation. Mathematically, this is expressed as: \[\tau = |\vec{F}| \cdot |\vec{r}| \cdot sin(\theta)\]
02

Find the dimensional formula for force

According to Newton's second law, force (\(F\)) is the product of mass (\(M\)) and acceleration (\(a\)), which can be expressed as: \[F = M \cdot a\] Acceleration has the units of \(\mathrm{L} \mathrm{~T}^{-2}\), where L represents the length dimension, and T represents the time dimension. So, the dimensional formula for force (F) is given by: \[\mathrm{[F]} = \mathrm{M}^{1}\mathrm{~L}^{1}\mathrm{~T}^{-2}\]
03

Find the dimensional formula for distance

Distance (r) is a measure of length, which has the dimensional formula of L. Therefore, the dimensional formula for distance (r) is: \[\mathrm{[r]} = \mathrm{L}^{1}\]
04

Derive the dimensional formula for torque

Now, we can find the dimensional formula for torque (\(\tau\)) using the dimensional formulas for force (\(F\)) and distance (\(r\)). Since torque is the product of force and distance, we have: \[\mathrm{[\tau]} = \mathrm{[F]} \cdot \mathrm{[r]}\] Using the dimensional formulas we found earlier for force and distance: \[\mathrm{[\tau]} = \mathrm{M}^{1}\mathrm{~L}^{1}\mathrm{~T}^{-2} \cdot \mathrm{L}^{1}\] By combining the dimensions, we get: \[\mathrm{[\tau]} = \mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\]
05

Compare with the given options and find the correct answer

Now, we can compare the derived dimensional formula for torque (\(\mathrm{[\tau]} = \mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\)) with the given options: (a) \(\mathrm{M}^{2} \mathrm{~L}^{2} \mathrm{~T}^{-3}\) (b) \(\mathrm{M}^{2} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}\) We can clearly see that the derived dimensional formula for torque matches option (d). Therefore, the correct dimensional formula for torque is: \(\boxed{\mathrm{[\tau]} = \mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2}}\)

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Most popular questions from this chapter

From $\left[\mathrm{p}+\left(\mathrm{a} / \mathrm{v}^{2}\right)\right](\mathrm{v}-\mathrm{b})=$ constant equation is dimensionally correct find the dimensional formula for \(\mathrm{b}\) ? where \(\mathrm{P}=\) pressure \(\mathrm{V}=\) volume (a) \(\mathrm{M}^{0} \mathrm{~L}^{3} \mathrm{~T}^{0}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{3} \mathrm{~T}^{0}\) (c) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{3}\) (d) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-1}\)

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100 picometer \(=\ldots \ldots \ldots \ldots\) (a) \(10^{-8} \mathrm{~cm}\) (b) \(10^{-7} \mathrm{~m}\) (c) \(10 \times 10^{-6} \mu \mathrm{m}\) (d) \(10 \times 10^{-8} \mu \mathrm{m}\)

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The number of significant figures in \(0.000150\) is .............. (a) 3 (b) 5 (c) 2 (d) 4
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