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Chapter 2: Problem 168

To locate the position of the particle we need ... (A) a frame of reference (B) direction of the particle (C) size of the particle (D) mass of the particle

Short Answer

Expert verified
To locate the position of a particle, we need \(A\) a frame of reference, as it provides a background against which we can measure the position and movement of the particle.

Step by step solution

01

Option A: A Frame of Reference

A frame of reference is necessary to locate a position of a particle, as it provides a background against which we can measure the position and movement of the particle. Therefore, this option seems to be correct.
02

Option B: Direction of the Particle

The direction of the particle indicates where it is moving, but it does not give us the location of the particle. In other words, knowing the direction of the particle alone is not enough to determine its position. Thus, this option is not correct.
03

Option C: Size of the Particle

The size of the particle tells us its dimensions, but it does not help in determining the position of the particle. Therefore, this option is not correct as well.
04

Option D: Mass of the Particle

Mass is a measure of the amount of matter in a particle. However, it is not related to the location of the particle. Hence, this option is not correct. Since option A is correct, the answer is: (A) A frame of reference

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Most popular questions from this chapter

\(\mathrm{A}=+\mathrm{i} \wedge+\mathrm{j} \wedge-2 \mathrm{k} \wedge\) and $\mathrm{B} \overrightarrow{\mathrm{i}} \wedge-\mathrm{j} \wedge+\mathrm{k} \wedge$ Find the unit vector in direction of \(\mathrm{A} \rightarrow \times \mathrm{B}^{\rightarrow}\) (A) $[1 / \sqrt{(23)}](-\mathrm{i} \wedge-5 \mathrm{j} \wedge-2 \mathrm{k} \wedge)$ (B) $[1 / \sqrt{(35)]}(-\mathrm{i} \wedge-5 \mathrm{j} \wedge-3 \mathrm{k} \wedge)$ (C) \([1 / \sqrt{(29})](-i \wedge-5 j \wedge-3 k \wedge)\) (D) \([1 / \sqrt{(35)]}(-\mathrm{i} \wedge-5 j \wedge-3 \mathrm{k} \wedge)\)

A particle is thrown in upward direction with initial velocity of $60 \mathrm{~m} / \mathrm{s}$. Find average speed and average velocity after 10 seconds. \(\left[\mathrm{g}=10 \mathrm{~ms}^{-2}\right]\) (A) \(26 \mathrm{~ms}^{-1}, 16 \mathrm{~ms}^{-1}\) (B) \(26 \mathrm{~ms}^{-1}, 10 \mathrm{~ms}^{-1}\) (C) \(20 \mathrm{~ms}^{-1}, 10 \mathrm{~ms}^{-1}\) (D) \(15 \mathrm{~ms}^{-1}, 25 \mathrm{~ms}^{-1}\)

Comprehensions type questions. A particle is moving in a circle of radius \(R\) with constant speed. The time period of the particle is \(\mathrm{T}\) Now after time \(\mathrm{t}=(\mathrm{T} / 6)\) Range of a projectile is \(R\) and maximum height is \(\mathrm{H}\). Find the area covered by the path of the projectile and horizontal line. (A) \((2 / 3) \mathrm{RH}\) (B) \((5 / 3) \mathrm{RH}\) (C) \((3 / 5) \mathrm{RH}\) (D) \((6 / 5) \mathrm{RH}\)

If \(\mathrm{A}^{\rightarrow} \cdot \mathrm{B}^{\rightarrow}=0\) then (A) \(\left|\mathrm{A}^{\rightarrow}\right|\) must be zero (B) \(\mathrm{B}^{\rightarrow} \mid\) must be zero (C) either \(\mathrm{A}^{\rightarrow}=0, \mathrm{~B}^{\rightarrow}=0\) or \(\theta=0\) (D) either \(\mathrm{A}^{\rightarrow}=0, \mathrm{~B}^{\rightarrow}=0\) or \(\theta=(\pi / 2)\)

A particle moves in one direction with acceleration \(2 \mathrm{~ms}^{-2}\) and initial velocity \(3 \mathrm{~ms}^{-1}\).After what time its displacement will be \(10 \mathrm{~m}\) ? (A) \(1 \mathrm{~s}\) (B) \(2 \mathrm{~s}\) (C) \(3 \mathrm{~s}\) (D) \(4 \mathrm{~s}\)
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