Units & Dimensions

JEE Main level, single correct, medium

124 of 149

Question 124 of 149

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

as fundamental units, the dimension of Young's modulus

will be :

Show answer and solution

Answer: (D) V−4 A2 F\mathrm{V}^{-4} \mathrm{~A}^{2} \mathrm{~F}

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 [Y]=[V−4FA2]=[V−4 A2 F][\mathrm{Y}]=\left[\mathrm{V}^{-4} \mathrm{FA}^{2}\right]=\left[\mathrm{V}^{-4} \mathrm{~A}^{2} \mathrm{~F}\right]

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