Question

A force F is given by $F = at + bt^{2},$ where $t$ is time. What are the dimensions of a and b

• A. $MLT^{- 3}$and$ML^{2}T^{- 4}$
• B. $MLT^{- 3}$ and $MLT^{- 4}$
• C. $MLT^{- 1}$and$MLT^{0}$
• D. $MLT^{- 4}$ and $MLT^{1}$

Answer 1

Answer: (B)

$MLT^{- 3}$ and $MLT^{- 4}$

Answer 2

From the principle of dimensional homogenity $\lbrack F\rbrack = \lbrack at\rbrack$ $\therefore\lbrack a\rbrack = \left\lbrack \frac{F}{t} \right\rbrack = \left\lbrack \frac{MLT^{- 2}}{T} \right\rbrack = \lbrack MLT^{- 3}\rbrack$

Similarly $\lbrack F\rbrack = \lbrack bt^{2}\rbrack$ $\therefore\lbrack b\rbrack = \left\lbrack \frac{F}{t^{2}} \right\rbrack = \left\lbrack \frac{MLT^{- 2}}{T^{2}} \right\rbrack = \lbrack MLT^{- 4}\rbrack$.