Units & Dimensions

JEE Main level, single correct, medium

119 of 149

Question 119 of 149

The dimension of stopping potential V0{V}_{0} in photoelectric effect in units of Planck’s constant ‘ hh ’, speed of light ‘ cc ’ and Gravitational constant ‘ GG ’ and ampere AA is:

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Answer: (B) h0c5G−1A−1{h}^{0}{c}^{5}{G}^{-1}{A}^{-1}

V=K(h)a(I)b(G)c(C)dV=K{(h)}^{a}{(I)}^{b}{(G)}^{c}{(C)}^{d} (V is voltage)
We know [h]=ML2T−1[h]=M{L}^{2}{T}^{-1}
[I]=A[I]=A
[G]=M−1L3T−2[G]={M}^{-1}{L}^{3}{T}^{-2}
[C]=LT−1[C]=L{T}^{-1}
[V]=ML2T−3A−1[V]=M{L}^{2}{T}^{-3}{A}^{-1}
ML2T−3A−1=(ML2T−1)a(A)b(M−1L3T−2)c(LT−1)dM{L}^{2}{T}^{-3}{A}^{-1}={(M{L}^{2}{T}^{-1})}^{a}{(A)}^{b}{({M}^{-1}{L}^{3}{T}^{-2})}^{c}{(L{T}^{-1})}^{d}
ML2T−3A−1=Ma−cL2a+3c+dT−a−2c−dAbM{L}^{2}{T}^{-3}{A}^{-1}={M}^{a-c}{L}^{2a+3c+d}{T}^{-a-2c-d}{A}^{b}
a−c=1.....(1)a-c=1.....(1)
2a+3c+d=2.....(2)2a+3c+d=2.....(2)
−a−2c−d=−3.....(3)-a-2c-d=-3.....(3)
b=−1...(4)b=-1...(4)
On solving
c=−1c=-1
a=0a=0
d=5,b=−1d=5,b=-1
V=K(h)0(A)−1(G)−1(C)5V=K{(h)}^{0}{(A)}^{-1}{(G)}^{-1}{(C)}^{5}

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