Question

If Momentum p and velocity V are related as P is equals to root a upon B + v square then the dimensional formula of area ab-2 is

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Answer 1

Momentum p and velocity v are related as P = $\frac{\sqrt{a}}{b+ v^2}$

from dimensional analysis,

dimension of b = dimension of v²....(1)

and dimension of (√a/v²) = dimension of p.....(2)

we know,

dimensional formula of momentum, P = [MLT-¹]

dimensional formula of velocity, v = [LT-¹]

now,

from equation (1),

dimension of b = [LT-¹]² = [L²T-²]

from equation (2),

dimension of √a/dimension of v² = dimension of p

or, dimension of √a = dimension of p × dimension of v²

= [MLT-¹] × [L²T-²]

= [ML³T-³]

so, dimension of a = [M²L^6T^-6]

now, dimension of ab^-2 = [M²L^6T^-6]/[L⁴T-⁴] = [M²L²T-²]

Answer 2

Answer:

solve via dimensional analysis!!

Explanation:

Momentum p and velocity v are related as P = \frac{\sqrt{a}}{b+ v^2}b+v2a

from dimensional analysis,

dimension of b = dimension of v²....(1)

and dimension of (√a/v²) = dimension of p.....(2)

we know,

dimensional formula of momentum, P = [MLT-¹]

dimensional formula of velocity, v = [LT-¹]

now,

from equation (1),

dimension of b = [LT-¹]² = [L²T-²]

from equation (2),

dimension of √a/dimension of v²=dimension of p

or, dimension of √a = dimension of p × dimension of v²

= [MLT-¹] × [L²T-²]

= [ML³T-³]

so, dimension of a = [M²L^6T^-6]

now, dimension of ab^-2 = [M²L^6T^-6]/[L⁴T-⁴] = [M²L²T-²]

Hope it helps!

cheers!