Question

If the force is given by $F=at+b{t}^2$ with $t$ as time. The dimensions of $a$ and $b$ are

• A. $\left[ML{T}^{-4}\right],\left[ML{T}^{-4}\right]$
• B. $\left[ML{T}^{-3}\right],\left[ML{T}^{-4}\right]$
• C. $\left[M{L}^2{T}^{-3}\right],\left[M{L}^2{T}^{-2}\right]$
• D. $\left[M{L}^2{T}^{-3}\right],\left[M{L}^3{T}^{-4}\right]$

Answer 1

The correct option is B

$\left[ML{T}^{-3}\right],\left[ML{T}^{-4}\right]$

Step 1: Given Equation

$F=at+b{t}^2$

$t$ is the time and $F$ is the force.

Step 2: Formula Used

The dimension of Force is $[M{]}^1[L{]}^1[T{]}^{-2}$ and that of time is ${\left[M\right]}^0{\left[L\right]}^0{\left[T\right]}^1$.

The dimension of $at$ and $b{t}^2$ has to be the same to find their dimension.

This is because objects with the same dimensions can only be equated, added, or subtracted.

Step 3: Calculate the dimensions of $a$

Here in the question, $F$ =$at+b{t}^2$, where $t$ is time and has the dimension: $[M{]}^0[L{]}^0[T{]}^1$
As we discussed above,$F$,$at$ and $b{t}^2$ have the same dimension.
The dimension of $F$ is $[M{]}^1[L{]}^1[T{]}^{-2}$.
So the dimension of $at$ should also be $[M{]}^1[L{]}^1[T{]}^{-2}$
Substituting the above dimension of $t$ as $[M{]}^0[L{]}^0[T{]}^1$, we get
$[a]\times [M{]}^0[L{]}^0[T{]}^1=[M{]}^1[L{]}^1[T{]}^{-2}$
So we get $[a]=[M{]}^1[L{]}^1[T{]}^{-3}$
Step 4: Calculate the dimensions of $b$
The dimension of $b{t}^2$ should also be $[M{]}^1[L{]}^1[T{]}^{-2}$.
$[b]\times {\left({[M]}^0{[L]}^0{[T]}^1\right)}^2=[M{]}^1[L{]}^1[T{]}^{-2}$
$[b]\times [M{]}^0[L{]}^0[T{]}^2=[M{]}^1[L{]}^1[T{]}^{-2}$
So, we get $[b]=[M{]}^1[L{]}^1[T{]}^{-4}$
Hence option B is the correct answer.