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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Physics
Chapter
Units and Dimensions

JEE Main 2019 · 11 January · Evening Shift · Physics · Units and Dimensions

If speed (V), acceleration (A) and force (F) are considered as...

If speed (V), acceleration (A) and force (F) are considered as fundamental units, the dimension of Young's modulus will be :

Options

  1. AV−2 A2 F−2\mathrm{V}^{-2} \mathrm{~A}^{2} \mathrm{~F}^{-2}
  2. BV−2 A2 F2\mathrm{V}^{-2} \mathrm{~A}^{2} \mathrm{~F}^{2}
  3. CV−4 A−2 F\mathrm{V}^{-4} \mathrm{~A}^{-2} \mathrm{~F}
  4. DV−4 A2 F\mathrm{V}^{-4} \mathrm{~A}^{2} \mathrm{~F}

Correct answer

D. V−4 A2 F\mathrm{V}^{-4} \mathrm{~A}^{2} \mathrm{~F}

Step-by-step solution

Let [Y] = [V] a[ F]b[A]c^{a}[\mathrm{~F}]^{\mathrm{b}}[\mathrm{A}]^{\mathrm{c}} [ML−1 T−2]=[LT−1]a[MLT−2]b[LT−2]c\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]=\left[\mathrm{LT}^{-1}\right]^{\mathrm{a}}\left[\mathrm{MLT}^{-2}\right]^{\mathrm{b}}\left[\mathrm{LT}^{-2}\right]^{\mathrm{c}} [ML−1 T−2]=[Mb La+b+cT−a−2 b−2c]\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^{b} \mathrm{~L}^{\mathrm{a}+b+\mathrm{c}} \mathrm{T}^{-\mathrm{a}-2 \mathrm{~b}-2 \mathrm{c}}\right] Comparing power both side of similar terms we get, b=1,a+b+c=−1,−a−2b−2c=−2b=1, a+b+c=-1,-a-2 b-2 c=-2 solving above equations we get a=−4,b=1,c=2a=-4, b=1, c=2 so $[\mathrm{Y}]=\left[\mathrm{V}^{-4} \mathrm{FA}^{2}\right]=\left[\mathrm{V}^{-4} \mathrm{~A}^{2} \mathrm{~F}\right]
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