M -> V, d (density) and g
 
    Let M (proportional to) Vadbgc
Now Hence M = k Vadbgc  where k is constant of proportionality
[M] = [V]a[d]b[g]c
M1L0T0 = (L1T-1)a(M1L-3)b(L1T-2)c
M1L0T0 = Mb La - 3b + c  T-a - 2c
Comparing the powers,
 
b = 1
a - 3b + c = 0
Hence,
 
a - 3(1) + c = 0
a + c = 3  ...........1
-a - 2c = 0 ............2
 
Solving equations 1 and 2,
a + c =  3
-a - 2c = 0
---------------
     -c = 3
Hence, c = - 3.
Substituting c = -3 in equation 1, we get a = 6.
Thus, M = k V6dg
hence, it can be said that M varies with 6th power of velocity.
Hint: In any type of problems of dimensional analysis where it is given that This quantity depends on so and so... solve by this method always. It would be easier.
 
						

							
M is directly proportional to vˆx
M is directly proportional to dˆy
M is directly proportional to gˆz
m=k*v^x*d^y*g^z
m=(l*t^-1)^x*(m*l)^y*(lt^-2)^z
y=1
-x-2*z=0
z=-x/2
x-3*y-x/2=0
x=6
m=v^6
-3
						

							
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M is directly proportional to vˆx
M is directly proportional to dˆy
M is directly proportional to gˆz
m=k*v^x*d^y*g^z
m=(l*t^-1)^x*(m*l)^y*(lt^-2)^z
y=1
-x-2*z=0
z=-x/2
x-3*y-x/2=0
x=6
m=v^6
-3