Units & Dimensions

JEE Main level, single correct, medium

64 of 149

Question 64 of 149

If mass is written as m=kcPG−1/2h1/2m=k{c}^{P}{G}^{-1/2}{h}^{1/2}, then the value of PP will be : (Constants have their usual meaning with kk a dimensionless constant)

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Answer: (A) 12\dfrac{1}{2}

Given:
m=kcPG−1/2h1/2m=k{c}^{P}{G}^{-1/2}{h}^{1/2}
Writing dimensions on both sides, we get
[M1L0T0]=[LT−1]P[M−1L3T−2]−1/2[ML2T−1]1/2\left[{M}^{1}{L}^{0}{T}^{0}\right]={\left[{LT}^{-1}\right]}^{P}{\left[{M}^{-1}{L}^{3}{T}^{-2}\right]}^{-1/2}{\left[{ML}^{2}{T}^{-1}\right]}^{1/2}
By comparing exponent of LL on both the sides, we get
0=P−32+1⇒P=120=P-\dfrac{3}{2}+1 \Rightarrow P=\dfrac{1}{2}

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