Question

A force, F= ax + bx^2, where x is displacement. Find the dimension of a and b.​

Answer 1

Answer:

DIMENSION OF a = $M^1L^0T^{-2}$ AND DIMENSION OF b = $M^1L^{-1}T^{-2}$

Explanation:

$F=ax+bx^{2}$   (given)

now, we know that two values can only be added or subtracted if their dimensions are the same.

Hence, dimension of = $F$ dimension of $ax$ =dimension of $bx^{2}$

⇒$M^{1} L^1T^{-2}=$ dimension of $ax$   {dimension of force = $M^{1} L^{1} T^{-2}$}

⇒$M^1L^1T^{-2}=L^1$ × dimension of a   {dimension of displacement = $L^{1}$}

⇒dimension of a = $\frac{M^1L^1T^{-2}}{L^1}$

⇒dimension of a =$M^1L^0T^{-2}$

Similarly,  

dimension of F = dimension of $bx^{2}$

⇒$M^{1} L^{1} T^{-2} = {(L^1)}^2$× dimension of b

⇒dimension of b = $\frac{M^1L^1T^{-2}}{L^2}$

⇒dimension of b = $M^1L^{-1}T^{-2}$