The distance travelled by a particle in $n^{t h}$ second is $S_{n} = u + \frac{a}{2} (2 n - 1)$ where u is the velocity and a is the acceleration. The equation is

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Dimensionally true

A and C are correct

Dimensionally false

Numerically may be true or false

answer is A.

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Detailed Solution

$S_{n} = S_{1} - S_{2}$

$S_{n} = u n + \frac{1}{2} a n^{2} - (u (n - 1) + \frac{1}{2} a {(n - 1)}^{2})$ $S_{n} = u n + \frac{1}{2} a n^{2} - (u n - u + \frac{1}{2} a (n^{2} + 1.1 - 2 n (1)))$ $S_{n} = u n + \frac{1}{2} a n^{2} - u n + u . 1 - \frac{1}{2} a n^{2} - \frac{1}{2} a . 1.1 + \frac{1}{2} a . 2 . n (1)$ $S_{n} = u . 1 + \frac{a}{2} (2 n . 1 - 1.1) H e r e n a n d 1 a r e t h e m a g n i t u d e s o f t i m e a n d m e a s u r e d i n s e c o n d s .$ $D . F o f L . H . S [S_{n}] = [L]$

$R . H . S D . F o f u . 1 = [L T^{- 1} . T] = [L]$ $D . F o f \frac{a}{2} (2 n . 1 - 1.1) = [L T^{- 2}] [T . T - T . T] = [L T^{- 2}] [T^{2}] = [L]$ $∴ R . H . S = [L + L] = [L]$