/*! 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 13 The dimensions of physical quant... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The dimensions of physical quantity \(\mathrm{X}\) in the equation Force \(=\frac{X}{\text { Density }}\) is given by (a) \(M^{1} L^{4} T^{-2}\) (b) \(M^{2} L^{-2} T^{-1}\) (c) \(M^{2} L^{-2} T^{-2}\) (d) \(M^{1} L^{-2} T^{-1}\)

Short Answer

Expert verified
The correct option is (c) \(M^{2} L^{-2} T^{-2}\)

Step by step solution

01

List the dimensions of force and density

The dimensions of force are given by \(M^{1}L^{1}T^{-2}\) and the dimensions of density are \(M^{1}L^{-3}\).
02

Calculate the dimensions of X

According to given equation, force equals to X divided by density, which means X= force * density. Hence dimensions of X would be obtained by multiplying dimensions of force and density. We get, X= \( M^{1}L^{1}T^{-2} * M^{1}L^{-3} = M^{2}L^{-2}T^{-2}\)
03

Match calculated dimensions to options

Upon examination of the options given, we find that the calculated dimensions \(M^{2}L^{-2}T^{-2}\) are same as option (c).

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Physical Quantities
Physical quantities are the building blocks of physics. They describe the characteristics of the world in a quantifiable manner. Essentially, these are anything you can measure, ranging from length, time, and mass, to temperature, current, and more complex derivations like speed and force.
\(\)
Each physical quantity has two main components:
  • Numerical Magnitude: This tells us "how much" of the quantity we have.
  • Unit: This specifies "what" we are measuring, such as meters for length or seconds for time.
Because a physical quantity can be expressed as a number multiplied by a unit, scientists developed a systematic method to standardize these units, known as the International System of Units (SI units). This allows for consistent measurement and communication of scientific information across the world.
Dimensions of Force
Force is a fundamental concept in physics. It represents an influence that can change the motion of an object, either accelerating it, decelerating it, or changing its direction.
\(\)
The dimensions of force are derived from Newton's Second Law of Motion: force equals mass times acceleration \( F = ma \). Given the dimensions for mass \( (M) \) and acceleration (which is the rate of change of velocity), we express acceleration as the change in velocity \( (L/T) \) over time \( (T) \). Thus, acceleration has dimensions of \( L T^{-2} \).
\(\)
By multiplying the dimensions of mass \( M \) with acceleration \( L T^{-2} \), we find that the dimensions of force are \( M L T^{-2} \). These dimensions help describe the fundamental physical influence that modifies an object's motion, allowing us to solve problems involving forces in various physical contexts.
Density in Physics
Density is another vital physical characteristic, describing how compact a substance is. It's a measure of how much mass is contained in a given volume. Mathematically, it is expressed as density equals mass divided by volume \( \rho = \frac{m}{V} \).
\(\)
Analyzing its dimensions helps to understand its role in physical phenomena:
  • Mass: This is typically measured in kilograms \( (M) \)
  • Volume: Given volume has dimensions \( L^{3} \), since it can be expressed as length times width times height.
When combining these units, the formula for density allows us to express its dimensions as \( M L^{-3} \). This equation shows that a small mass in a large volume results in a low density, while a large mass in a small volume indicates a high density. This characteristic is critical in fields spanning from material science to astrophysics, influencing how we understand phenomena from buoyant forces to the composition of distant planets.

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

To determine the Young's modulus of a wire, the formula is \(Y=\frac{F}{A} \cdot \frac{L}{\Delta l} ;\) where \(L=\) length, \(\mathrm{A}=\) area of cross- section of the wire, \(\Delta L=\) Change in length of the wire when stretched with a force \(F\). The conversion factor to change it from CGS to MKS system is (a) 1 (b) 10 (c) \(0.1\) (d) \(0.01\)

The SI unit of universal gas constant \((R)\) is (a) Watt \(\mathrm{K}_{1}^{-1} \mathrm{~mol}^{-1}\) (b) Newton \(\mathrm{K}^{-1} \mathrm{~mol}^{-1}\) (c) Joule \(\mathrm{K}^{-1} \mathrm{~mol}^{-1}\) (d) \(\operatorname{Erg} K^{-1} \mathrm{~mol}^{-1}\)

In an experiment, the following observation's were recorded : \(L=2.820 \mathrm{~m}, M=3.00 \mathrm{~kg}, l=0.087 \mathrm{~cm}\), Diameter \(D=0.041 \mathrm{~cm}\) Taking \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\) using the formula, \(Y=\frac{4 M g}{\pi D^{2} l}\), the maximum permissible error in \(Y\) is (a) \(7.96 \%\) (b) \(4.56 \%\) (c) \(6.50 \%\) (d) \(8.42 \%\)

If the present units of length, time and mass \((m, s, k g)\) are changed to \(100 m, 100 s\), and \(\frac{1}{10} k g\) then (a) The new unit of velocity is increased 10 times (b) The new unit of force is decreased \(\frac{1}{1000}\) times (c) The new unit of energy is increased 10 times (d) The new unit of pressure is increased 1000 times

Suppose we employ a system in which the unit of mass equals \(100 \mathrm{~kg}\), the unit of length equals \(1 \mathrm{~km}\) and the unit of time \(100 \mathrm{~s}\) and call the unit of energy eluoj (joule written in reverse order), then (a) 1 eluoj \(=10^{4}\) joule (b) 1 eluoj \(=10^{-3}\) joule (c) 1 eluoj \(=10^{-4}\) joule (d) 1 joule \(=10^{3}\) eluoj

See all solutions

Recommended explanations on Physics Textbooks

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.