Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The velocity $v$ (in $\text{cm/sec}$ ) of a particle is given in terms of time $t$ (in $\text{sec}$ ) by the relation $v = at + \frac{b}{t+c}$ ; the dimensions of $a$ , $b$ and $c$ are

$a = L^2, b = T, c = L T^2$

$a = L T^2, b = L T, c = L$

$a = L T^{-2}, b = L, c = T$

$a = L, b = L T, c = T^2$

Step-by-Step Solution

According to the principle of homogeneity of dimensions, only quantities with the same dimensions can be added, subtracted, or equated. From the given equation $v = at + \frac{b}{t+c}$ :

In the denominator, $c$ is added to time $t$ . Therefore, the dimension of $c$ must be the same as that of time $t$ . So, $[c] = [T]$ .
The dimension of the term $at$ must be equal to the dimension of velocity $v$ . So, $[a][T] = [L T^{-1}] \implies [a] = [L T^{-2}]$ .
The dimension of the term $\frac{b}{t+c}$ must also be equal to the dimension of $v$ . So, $\frac{[b]}{[T]} = [L T^{-1}] \implies [b] = [L]$ . Therefore, the dimensions of $a$ , $b$ , and $c$ are $[L T^{-2}]$ , $[L]$ , and $[T]$ respectively .

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