Question

write the dimension of F= X +bt^2​

Answer 1

Answer:

it is on the graph . please try to do it yourself

Answer 2

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

X =$ML{T}^{-2}$, B =$ML{T}^{-4}$

According to principal of homogeneity,

The dimension of all the terms in an equation must be identical.

One can add or subtract or compare the same physical quantity.

The dimension of force is :- $\bold{</strong><strong>ml {t}^{ - 2} </strong><strong> }$

F = X + $\bold{</strong><strong>b{t}^{2}</strong><strong>}$

F = X

Then, dimension of X is equal to F.

Therefore, X = $\bold{</strong><strong>ml {t}^{ - 2} </strong><strong> }$

Now,

F = $\bold{</strong><strong>b</strong><strong>{</strong><strong>t</strong><strong>}</strong><strong>^</strong><strong>{</strong><strong>2</strong><strong>}</strong><strong>}$

$\bold{</strong><strong>b</strong><strong> </strong><strong>=</strong><strong>\frac{ml {t}^{ - 2} }{ {t}^{2} } </strong><strong> </strong><strong> }$

$\bold{</strong><strong>b</strong><strong> </strong><strong>=</strong><strong> </strong><strong>ml {t}^{ - 4} </strong><strong>}$