Question

force F and density d are related as F is equal to A divided by B plus root d obtain the dimensions of A and B

Answer 1

F = A/(B+√d)

then B has same dimensions as √d

= M^(1/2)L^(-3/2)

F => MLT-²

so

A => MLT-² (M^0.5L^-1.5)

= (M^3/2)(L^-1/2)T-²

Like

Answer 2

The dimensions of A and B is

${M}^{ \frac{1}{1} } {L}^{ \frac{ 1}{1} } {T}^{ - 2} and \: {M}^{ \frac{1}{2} } {L}^{ \frac{ - 3}{2} } {T}^{0}$ respectively.

Given:

A force F and density d are related as F are equal to A divided by B plus root d.

To Find:

The dimensions of A and B.

Solution:

To find the dimensions of A and B will follow the following steps:

As given:

$F = \frac{A}{ B+ \sqrt{d} }$

Now,

Dimensions of B will be equal to the dimensions of the square root of d.

B = √d

Density unit =

$kg {m}^{ - 3}$

B =

$\sqrt{kg {m}^{ - 3} }$

B =

${kg}^{ \frac{1}{2} } { {m}^{ \frac{ - 3}{2} } }$

Unit of force F = Kgms-²

Unit of A = unit of F = Kgms-²

Dimensions of A=

${M}^{ \frac{1}{1} } {L}^{ \frac{ 1}{1} } {T}^{ - 2}$

Dimensions of B =

${M}^{ \frac{1}{2} } {L}^{ \frac{ - 3}{2} } {T}^{0}$

Henceforth, the dimensions of A and B are

${M}^{ \frac{1}{1} } {L}^{ \frac{ 1}{1} } {T}^{ - 2} and \: {M}^{ \frac{1}{2} } {L}^{ \frac{ - 3}{2} } {T}^{0}$

${M}^{ \frac{1}{1} } {L}^{ \frac{ 1}{1} } {T}^{ - 2} and \: {M}^{ \frac{1}{2} } {L}^{ \frac{ - 3}{2} } {T}^{0}$Respectively.

#SPJ2