Problem 34 - Master Resource Book in JEE Main Physics

Question

The dimensions of distance travelled in \(n\)th second are (a) \(\left[\mathrm{M}^{0} \mathrm{LT}\right]\) (b) \(\left[\mathrm{M}^{\mathrm{O}} \mathrm{L}^{0} \mathrm{~T}^{0}\right]\) (c) \(\left[\mathrm{M}^{0} \mathrm{LT}^{-1}\right]\) (d) \(\left[\mathrm{M}^{0} \mathrm{LT}^{0}\right]\)

Step-by-Step Solution

Answer

The correct choice is (d) \(\left[\mathrm{M}^{0} \mathrm{LT}^{0}\right]\).

Step 1: Understand the variables involved

The problem involves the concept of distance and time. We are specifically asked about the distance traveled during the \(n\)th second, which means we need to consider both distance \(s\) and time \(t\).

Step 2: Identify the formula for distance in nth second

The formula for distance traveled in the \(n\)th second is given by: \( s_n = u + \frac{a}{2}(2n-1) \) where \(u\) is the initial velocity and \(a\) is the acceleration. This formula involves both time and velocity terms.

Step 3: Analyze the formula dimensional components

Breaking down the formula: the term \(u\) has dimensions of velocity, which is \([LT^{-1}]\), and the term \(\frac{a}{2}(2n-1)\) simplifies to acceleration \(a\), which has dimensions \([LT^{-2}]\) times a dimensionless number \((2n-1)\). Therefore, the dimensional analysis of the \(n\)th second distance is maintained by the dominant dimension, which is distance \(L\).

Step 4: Determine the dimensional formula

The dimensions of distance traveled in the \(n\)th second are thus simply the dimensions of distance \([L]\), with no influence from mass \([M^0]\) or components of time \([T^0]\) except as they factor into the expression already as constants associated with time units, yielding \([L]\) remaining.

Step 5: Identify the correct dimensional formula from the options

The dimension of distance \([L]\) as deduced from the dimensional analysis corresponds to the choice \(\left[\mathrm{M}^{0} \mathrm{LT}^{0}\right]\) as the mass is not involved and the influence of time is effectively neutralized in the sense of distinct dimensions here.

Key Concepts

Distance Formula

The distance formula is an essential concept in physics, providing us with a way to calculate the distance an object travels during a particular period of motion, such as the nth second. In kinematics, this becomes especially important when understanding motion with uniform acceleration. To find the distance traveled by an object in the nth second, we use the formula:\[ s_n = u + \frac{a}{2}(2n-1) \]where:

• \(s_n\) is the distance traveled in the nth second
• \(u\) is the initial velocity
• \(a\) is the acceleration

This formula helps us analyze motion by breaking down into understandable parts; the initial velocity term \(u\) represents the object's speed at the start, and the expression \(\frac{a}{2}(2n-1)\) accounts for the continuous change in speed due to acceleration over the given time unit.

Kinematics

Kinematics is the branch of physics that studies the motion of objects without considering the forces causing the motion. The equations used in kinematics are invaluable for calculating distances, velocities, and accelerations of objects in motion. In the context of the distance traveled during the nth second, kinematics provides a systematic approach to measure how an object moves over time under constant acceleration. It allows us to:

• Break down motion into initial velocity and acceleration components
• Utilize equations like the distance formula to measure movement in specific time frames
• Analyze the progression of an object's speed and position over intervals

These kinematic principles help in understanding trajectory, speed changes, and time intervals, which are critical in solving problems related to motion.

Dimensional Formula

The dimensional formula is a way to express a quantity in terms of its basic dimensions: mass ([M]), length ([L]), and time ([T]). This approach is crucial when performing dimensional analysis, which verifies the consistency of an equation.When analyzing the distance traveled within a specific time frame, such as the nth second, the dimensional formula guides us in determining the correct dimensions involved. In this scenario:

• The dimension of initial velocity \(u\) is \([LT^{-1}]\)
• The dimension of acceleration \(a\) is \([LT^{-2}]\)
• The overall dimensional formula simplifies, highlighting that the distance \(L\) indeed dominates

Therefore, the dimensional formula for the distance traveled in the nth second reflects just the length dimension \([L]\). This indicates no direct involvement of mass \([M^0]\) or time alterations beyond consistent units, which underscores the neutralized impact of time's dimensional factors.