Physical World Units and Measurements - Result Question 18
18. The velocity $v$ of a particle at time $t$ is given by $v=a t+\frac{b}{t+c}$, where $a, b$ and $c$ are constant. The dimensions of $a, b$ and $c$ are respectively
[2006]
(a) $[L^{2}, T.$ and $.LT^{2}]$
(c) $[L, LT.$ and $.T^{2}]$
(b) $[LT^{2}, LT.$ and $.L]$
(d) $[LT^{-2}, L.$ and $.T]$
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Answer:
Correct Answer: 18. (d)
Solution:
- (d) Dimension of $a \cdot t=$ dimension of velocity
a.t $=[LT^{-1}] \Rightarrow[a=LT^{-2}]$
Dimension of $c=$ dimension of $t$
(two physical quantity of same dimension can only be added)
So, dimension of $c=[T]$
Dimension of $\frac{b}{t+c}=$ Dimension of velocity
$\frac{b}{T+T}=[LT^{-1}][LT^{-1}] \Rightarrow[b \cdot T^{-1}]=[LT^{-1}]$
$\Rightarrow b=[L]$
So, answer is $[LT^{-2}]$, $[L]$ and $[T]$
A dimensionally correct equation may or may not be physically correct. In a dimensionally correct equation, the dimensions of each term on both sides of an equation must be the same.
