Answer is (4)
V = K (h)a (I)b (G)c (C)d (V is voltage)
we know [h] = ML2T–1
[I] = A
[G] = M–1 L3 T–2
[C] = L T–1
[V] = M L2 T–3 A–1
M L2 T–3 A–1 = (M L2 T–1)a (A)b (M–1L3T-2)c (LT–1)d
ML2T–3A–1 = Ma–c L2a+3c+d T–a–2c–d Ab
a – c = 1………………(1)
2a + 3c + d = 2………………(2)
–a –2c –d = –3 ………………(3)
b = –1………………(4)
on solving
c = –1
a = 0
d = 5,
b = –1
V = K (h)° (I)–1 (G)–1 (C) 5