/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 The ratio of the dimension of Pl... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 55

The ratio of the dimension of Planck's constant and that of moment of inertia is the dimension of (A) Time (B) Frequency (C) Angular momentum (D) Velocity

Short Answer

Expert verified
The dimension of the ratio of Planck's constant and moment of inertia is Frequency, hence the correct option is (B) Frequency.

Step by step solution

01

Dimensions of Planck's constant

Planck鈥檚 constant (h) is associated with the energy of a photon and is given as \(E = h\nu\), where \(\nu\) is the frequency of the photon. From this, we obtain the dimensions of Planck鈥檚 constant, which would be \([M][L]^2[T]^{-1}\). This can be interpreted as mass times distance squared per unit time.
02

Dimensions of moment of inertia

Moment of Inertia (I), is defined as \(I=mr^2\), where m is mass and r is radius. Hence, the dimensions of moment of inertia are \([M][L]^2\), which can be interpreted as mass times distance squared.
03

The ratio of dimensions

Now, we take the ratio of the dimensions of Planck鈥檚 constant to the dimensions of the moment of inertia. The mass and length squared dimensions cancel out, leaving a time inverse \([T]^{-1}\) remaining. The dimension \([T]^{-1}\) signifies frequency.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The vector sum of two forces is perpendicular to their vector differences. In that case, the forces (A) are not equal to each other in magnitude. (B) cannot be predicted. (C) are equal to each other. (D) are equal to each other in magnitude.

\(\int \frac{d x}{\sqrt{a^{2}-x^{2}}}=\frac{1}{a} \sin ^{-1} \frac{a}{x}\) (A) is dimensionally correct. (B) dimensionally incorrect. (C) such mathematical relations cannot be tested. (D) cannot say.

Which one of the following represents the correct dimensions of the coefficient of viscosity? (A) \(\left[M L^{-1} T^{-2}\right]\) (B) \(\left[M L T^{-1}\right]\) (C) \(\left[M L^{-1} T^{-1}\right]\) (D) \(\left[M L^{-2} T^{-2}\right]\)

Which of the following cannot be in equilibrium? (A) \(10 \mathrm{~N}, 10 \mathrm{~N}, 5 \mathrm{~N}\) (B) \(5 \mathrm{~N}, 7 \mathrm{~N}, 9 \mathrm{~N}\) (C) \(8 \mathrm{~N}, 4 \mathrm{~N}, 13 \mathrm{~N}\) (D) \(9 \mathrm{~N}, 6 \mathrm{~N}, 5 \mathrm{~N}\)

What are the dimensions of electrical permittivity? (A) \(M L^{-2} T^{-2} Q^{-2}\) (B) \(M^{-1} L^{2} T^{-3} Q^{-1}\) (C) \(M^{-1} L^{-3} T^{2} Q^{2}\) (D) \(M^{-1} L^{3} T^{-2} Q^{-2}\)
See all solutions

Recommended explanations on Physics Textbooks

Torque and Rotational Motion

Read Explanation

Rotational Dynamics

Read Explanation

Space Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Fundamentals of Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.