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A force is given by $$F = at + b{t^2}$$, where $$t$$ is time, the dimensions of $$a$$ and $$b$$ are respectively :（ ）
A. $$\left[ {ML{T^{ - 4}}} \right]$$ and $$\left[ {ML{T^{ - 1}}} \right]$$
B. $$\left[ {ML{T^{ - 1}}} \right]$$ and $$\left[ {ML{T^0}} \right]$$
C. $$\left[ {ML{T^{ - 3}}} \right]$$ and $$\left[ {ML{T^{ - 4}}} \right]$$
D. $$\left[ {ML{T^{ - 3}}} \right]$$ and $$\left[ {ML{T^0}} \right]$$
Speed
00:00
04:14

## QuestionPhysicsClass 11

A force is given by $$F = at + b{t^2}$$, where $$t$$ is time, the dimensions of $$a$$ and $$b$$ are respectively :（ ）
A. $$\left[ {ML{T^{ - 4}}} \right]$$ and $$\left[ {ML{T^{ - 1}}} \right]$$
B. $$\left[ {ML{T^{ - 1}}} \right]$$ and $$\left[ {ML{T^0}} \right]$$
C. $$\left[ {ML{T^{ - 3}}} \right]$$ and $$\left[ {ML{T^{ - 4}}} \right]$$
D. $$\left[ {ML{T^{ - 3}}} \right]$$ and $$\left[ {ML{T^0}} \right]$$

C
4.6
4.6

## Solution

Since $$F=at+bt^2---(1)$$
$$\therefore$$ Dimension of $$at$$ and $$bt^2$$ must be equal to force only.
Hence $$[F]=[M^{1} L^{1} T^{-2}]---(2)$$
from $$(1)$$ and $$(2)$$
$$\Rightarrow \ [at]=a[T] =[F]$$
$$\therefore \ a[T] =[M^{1} L^{1} T^{-2}]$$
$$\therefore \ [a]=[M^{1} L^{1} T^{-3}]$$
$$bt^2=[b][T^{2}]=[a][t]=[M^{1} L^{1} T^{-2}]$$
$$\therefore \ \boxed {[b]= [M^{1} L^{1} T^{-4}]}$$
Hence, option $$(C)$$ is correct.