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Chapter 1: Problem 139

Write the dimensional formula of the ratio of linear momentum to angular momentum. (a) \(\mathrm{M}^{0} \mathrm{~L}^{-1} \mathrm{~T}^{0}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{0}\) (c) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{0}\) (d) \(\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{1}\)

Short Answer

Expert verified
The short answer is: The dimensional formula of the ratio of linear momentum to angular momentum is (a) \(\mathrm{M}^{0} \mathrm{~L}^{-1} \mathrm{~T}^{0}\).

Step by step solution

01

Write the formula for linear momentum

Linear momentum is calculated by the product of the mass and velocity of the object: P = m*v, where P is the linear momentum, m is the mass, and v is the velocity.
02

Find the dimensional formula for linear momentum

To find the dimensional formula for linear momentum, we need to determine the dimensions of mass and velocity. Mass has the dimension M and velocity has dimensions of length L and time T (L/T). Therefore, the dimensional formula for linear momentum is: \[ [P] = [m] \times [v] = M \times \frac{L}{T} = M L T^{-1} \]
03

Write the formula for angular momentum

Angular momentum is calculated by the product of the moment of inertia and angular velocity of the object: L = I*ω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
04

Find the dimensional formula for angular momentum

To find the dimensional formula for angular momentum, we need to determine the dimensions of the moment of inertia and angular velocity. The moment of inertia has dimensions of mass and length (ML²) and angular velocity has dimensions of time (T^(-1)). Therefore, the dimensional formula for angular momentum is: \[ [L] = [I] \times [\omega] = (M L^2) \times (T^{-1}) = M L^2 T^{-1} \]
05

Determine the dimensional formula for the ratio of linear momentum to angular momentum

To find the dimensional formula for the ratio of linear momentum to angular momentum, we need to divide the dimension of linear momentum by the dimension of angular momentum. \[ \frac{[P]}{[L]} = \frac{M L T^{-1}}{M L^2 T^{-1}} = M^0 L^{-1} T^0 \]
06

Match the result to one of the answer options

Comparing the result (\(M^0 L^{-1} T^0\)) to the options given, we can see that it matches with option (a). Therefore, the correct answer is (a) \(\mathrm{M}^{0} \mathrm{~L}^{-1} \mathrm{~T}^{0}\).

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Most popular questions from this chapter

If $\mathrm{P}=\left[\left(\mathrm{A}^{2} \mathrm{~B}\right) /\left(\mathrm{C}^{3}\right)\right]\( where percentage error in \)\mathrm{A}, \mathrm{B}\( and \)\mathrm{C}$ are respectively \(\pm 2 \% \pm 3 \%\) and \(\pm 5 \%\) then total percentage error in measurement of \(\mathrm{p}\) (a) \(18 \%\) (b) \(14 \%\) (c) \(21 \%\) (d) \(12 \%\)

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) \(10 \times 10^{-8} \mu \mathrm{m}\)

\([(1\) femtometer \() /(100\) nanometer \()]=\ldots \ldots \ldots \ldots \ldots\) (a) \(10^{-6}\) (b) \(10^{-8}\) (c) \(10^{24}\) (d) \(10^{-24}\)

The period of oscillation of a simple pendulum is given by $\mathrm{T}=2 \pi \sqrt{(} \ell / \mathrm{g})$ what is the equation of the relative error \(\Delta \mathrm{T} / \mathrm{T}\) in measurement of period \(\mathrm{T}\) ? (a) \((1 / 2)(\Delta \ell / \ell)\) (b) \(2(\Delta \ell / \ell)\) (c) \((1 / 4)(\Delta \ell / \ell)\) (d) \(4(\Delta \ell / \ell)\)

The radius of circle is \(1.26 \mathrm{~cm}\). According to the concept of significant figures area of it can be represented as - (a) \(4.9850 \mathrm{~cm}^{2}\) (b) \(4.985 \mathrm{~cm}^{2}\) (c) \(4.98 \mathrm{~cm}^{2}\) (d) \(9.98 \mathrm{~cm}^{2}\)
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