Question

GOOD MORNING.
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Q. 7
GIVE ME THE WELL EXPLAINED SOLUTION. ​

Answer 1

ANSWER

Let L=cagbpc -----1

[L]= [L]

[c]=[LT-1]

[g]=[LT-2]

[p]=[ML-1T-2]

[L]=[LT-1]a[LT-2]b[ML-1T-2]c

[L]=[McLa+b-cT-a-2b-2c]

COMPARING BOTH SIDES,

c=0, a+b-c=1

a+b+0=1

a+b=1 -------2

-a-2b-2c=0

-a-2b=0

a=-2b

PUTTING a=-2b , in equation 2

-2b+b=1

b=-1

a+b=1 (eqn. 2)

a-1=1

a=2

PUTTING VALUES IN EQN. 1

L=c2g-1p0

L=c2g-1

Answer 2

Units of...

c = m/s

g = m/s^2

P = (kgm/s^2) / m^2 = kg / (ms^2)

Dimensional formula of...

c = [LT^-1]

g = [LT^-2]

P = [ML^-1 T^-2]

c^a g^b P^c = [L] = [LT^-1]^a [LT^-2]^b [ML^-1 T^-2]^c

[L] = [L]^(a + b - c) [T]^(-a - 2b - 2c) [M]^c

From above equation

1 = a + b - c ——(1)

0 = -a - 2b - 2c —-(2)

0 = c ——(3)

Substitute c = 0 in equation (1) and (2)

1 = a + b ; a = -2b

From above equations

b = -1 and a = 2

Finally we got

a = 2

b = -1

c = 0

The dimension of length is therefore

c^a g^b P^c = c^2 g^-1 P^0 = (c^2) / g