Problem 21 - Master Resource Book in JEE Main Physics

Question

What will be the unit of time in that system in which the unit of length is metre, unit of mass is \(\mathrm{kg}\) and unit of force is kg-wt? (a) \((9.8)^{2} \mathrm{~s}\) (b) \(9.8 \mathrm{~s}\) (c) \(\sqrt{9.8} \mathrm{~s}\) (d) \(\frac{1}{\sqrt{9.8}} 5\)

Step-by-Step Solution

Answer

The time unit is \( \frac{1}{\sqrt{9.8}} \) seconds, option (d).

Step 1: Understanding the Units

In this problem, we are under a system where the unit of length is meters, the unit of mass is kilograms, and the unit of force is "kg-wt". Here, "kg-wt" represents a force equivalent to the gravitational force acting on a mass of one kilogram at the Earth's surface.

Step 2: Relating Units to the Gravitational Constant

The kg-wt is related to the newton (N) by the gravitational constant (g = 9.8 m/s²). Hence, 1 kg-wt = 9.8 N. In this special system, 1 kg-wt = 1 force unit which means 1 force unit = 9.8 kg·m/s².

Step 3: Expressing Force Using the Given System

Using the units from our system, the formula for force is given by \[ F = m imes a \] where the force is 1 kg-wt, mass is in kg, and acceleration is in the units we'll determine. Here, acceleration will relate to the unit of time we need to find.

Step 4: Applying Dimensional Analysis

From the force relation using newton, \[ 1 ext{ force unit} = 1 ext{ kg} imes a \]In terms of dimensions, \[ 1 imes 9.8 ext{ m/s² } = 1 ext{ kg} imes a \] which implies \[ a = 9.8 ext{ m/s² } \]This factor must be adjusted to determine the time unit's dimension.

Step 5: Determining the Time Unit from Force Formula

Since \[ F = m imes a \], and substituting in terms of our force unit definition: \[ 1 ext{ force unit} = 1 ext{ kg} imes rac{m}{t^2} = 9.8 ext{ kg} imes rac{m}{t^2}\] equating dimensions for time gives: \[ t^2 = rac{m}{9.8} \] Therefore, \[ t = rac{1}{ ext{√9.8}} \] seconds.

Key Concepts

Unit of Force

In physics, the unit of force is a fundamental aspect used to measure a force applied on an object. The standard unit of force in the International System of Units (SI) is the newton (N). However, in this particular exercise, we are looking at a different system: using "kg-wt" as the unit of force.
"Kg-wt", or kilogram-weight, describes the gravitational force on a mass of one kilogram at Earth's surface. In more everyday terms, it's the force needed to hold a 1 kg mass still against gravity.

• This relationship is essential in understanding customized systems of units.
• 1 kg-wt is defined as 9.8 newtons due to Earth's gravity.

Understanding how units like kg-wt relate to the newton allows us to analyze situations where gravitational forces are involved, such as when different unit systems are used.

Gravitational Constant

The gravitational constant, usually denoted as "g", is a crucial quantity in physics that represents the acceleration due to Earth's gravity. At Earth's surface, the standard value is approximately 9.8 meters per second squared (\( g = 9.8 \, \text{m/s}^2 \)).
This constant helps us to understand how the force of gravity operates over various objects near Earth's surface:

• It's the rate at which objects accelerate when dropped,
• which directly affects the force measured in units like the newton or kg-wt.

Recognizing the gravitational constant's role allows us to make connections between force and motion, enabling calculations like determining unknown acceleration or time, crucial in dimensional analysis.

Dimensional Formula

A dimensional formula expresses a physical quantity in terms of the basic dimensions of mass \([M]\), length \([L]\), and time \([T]\). It helps us derive relationships between different quantities by ensuring consistent dimensions.

In this problem, dimensional analysis is used to determine the appropriate unit of time. Using the expression for force, \( F = m \times a \), we can substitute the known force formula in our system:

• Given 1 kg-wt = 1 force unit = 9.8 kg·m/s²,
• Means \( F = M \times \frac{L}{T^2} \).

From this relationship, the formula provides dimensional clarity, allowing one to adjust units consistently and derive the equation:\[ \frac{m}{t^2} = 9.8 \]. This equation ultimately guides us in calculating the time unit in a non-standard force unit.

Acceleration and Units

Acceleration, in physics, is defined as the rate of change of velocity of an object with respect to time. It is a vector quantity, which means it has both magnitude and direction, and its standard unit in the SI system is meters per second squared (\( \text{m/s}^2 \)).
In this curious unit system where the force unit is "kg-wt", acceleration is a derived unit based on the standard gravitational pull:

• Since force \( F = m \times a \), through dimensional analysis, we have acceleration related to gravitational constant: \( a = 9.8 \text{ m/s}^2 \).
• It connects back to finding the unit of time by correlating the force equation with acceleration dimensions.

Acceleration's unit role in determining the time demonstrates the intricate linkage between different units and explores how they can be interdependent when analyzing physics problems.