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JEE Main 201511 AprilOnline
Exam
JEE Main
Year
2015
Date
11 April
Mode
Online
Subject
Physics
Chapter
Mathematics in Physics

JEE Main 2015 · 11 April · Online · Physics · Mathematics in Physics

A vector → A is rotated by a small angle ∆ θ radians (∆ θ ≪ 1) to...

A vector A→\overset{\to}{A} is rotated by a small angle ∆θ∆ \theta radians \left(∆ \theta ≪ 1\right)togetanewvectorto get a new vector\overset{\to}{B}.Inthatcase∣B→−A→∣. In that case \left|\overset{\to}{B} - \overset{\to}{A}\right| is :A vector → A is rotated by a small angle ∆ θ radians (∆ θ ≪ 1) to get a new vector → B . In that case | → B - → A | is :

Options

  1. A∣A→∣[1−(∆θ)22]\left|\overset{\to}{A}\right|\left[1 - \dfrac{{\left(∆ \theta\right)}^{2}}{2}\right]
  2. B00
  3. C∣A→∣∆θ\left|\overset{\to}{A}\right| ∆ \theta
  4. D∣B→∣∆θ−∣A→∣\left|\overset{\to}{B}\right| ∆ \theta - \left|\overset{\to}{A}\right|

Correct answer

C. ∣A→∣∆θ\left|\overset{\to}{A}\right| ∆ \theta

Step-by-step solution

A vector → A is rotated by a small angle ∆ θ radians (∆ θ ≪ 1) to get a new vector → B . In that case | → B - → A | is :

∣B→∣=∣A→∣\left|\overset{\to}{B}\right| = \left|\overset{\to}{A}\right|

\left|\overset{\to}{B} - \overset{\to}{A}\right| = \sqrt{{\left|\overset{\to}{A}\right|}^{2} + {\left|\overset{\to}{A}\right|}^{2} + 2\left|\overset{\to}{A}\right|\left|\overset{\to}{A}\right|\cos \left(180 ° - ∆ \theta\right)}</p><p></p><p>= \sqrt{{\left|\overset{\to}{A}\right|}^{2} + {\left|\overset{\to}{A}\right|}^{2} - 2{\left|\overset{\to}{A}\right|}^{2}cos ∆ \theta}</p><p></p><p>= \sqrt{2{\left|\overset{\to}{A}\right|}^{2} - 2{\left|\overset{\to}{A}\right|}^{2}cos ∆ \theta}</p><p></p><p>= \sqrt{2{\left|\overset{\to}{A}\right|}^{2}\left(1 - cos ∆ \theta\right)}</p><p></p><p>∵ 1 - cos ∆ \theta = 2\sin^{2}\left(\dfrac{∆ \theta}{2}\right)</p><p></p><p>= \sqrt{2{\left|\overset{\to}{A}\right|}^{2} \times 2\sin^{2}\left(\dfrac{∆ \theta}{2}\right)}</p><p></p><p>= 2\left|\overset{\to}{A}\right|\sin \left(\dfrac{∆ \theta}{2}\right)</p><p>Forsmall</p><p>For small∆ \theta, \sin \left(\dfrac{∆ \theta}{2}\right) ≃ \left(\dfrac{∆ \theta}{2}\right)</p><p>∣B→−A→∣=2∣A→∣(∆θ2)</p><p>\left|\overset{\to}{B} - \overset{\to}{A}\right| = 2\left|\overset{\to}{A}\right|\left(\dfrac{∆ \theta}{2}\right)

=∣A→∣∆θ= \left|\overset{\to}{A}\right| ∆ \theta

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