The distance travelled by a particle in `n^(th)` second is `S_n =u+a/2 (2n-1)` where u is the velocity and a is the acceleration. The equation is 

Options

A. dimensionally true
B. dimensionally false
C. numerically may be true or false
D. 1 and 3 are correct

Answer

To solve the problem, we need to analyze the equation given for the distance traveled by a particle in the nth second, which is:

\[ S_n = u + \frac{a}{2} (2n - 1) \]

where \( u \) is the initial velocity and \( a \) is the acceleration. We need to determine if this equation is dimensionally true or false, and also check the numerical validity of the equation.

### Step 1: Understand the equation

The equation states that the distance traveled by a particle in the nth second is a function of the initial velocity \( u \) and the acceleration \( a \). The term \( (2n - 1) \) suggests that the distance depends on the time interval \( n \).

### Step 2: Use the equations of motion

We can relate this equation to the equations of motion. The distance traveled in the nth second can be derived from the total distance traveled in time \( n \) and \( n-1 \):

\[ S_n = S(n) - S(n-1) \]

Where \( S(t) = ut + \frac{1}{2} a t^2 \).

### Step 3: Calculate \( S(n) \) and \( S(n-1) \)

1. For \( S(n) \):
\[ S(n) = u \cdot n + \frac{1}{2} a n^2 \]

2. For \( S(n-1) \):
\[ S(n-1) = u \cdot (n-1) + \frac{1}{2} a (n-1)^2 \]

Expanding \( S(n-1) \):
\[ S(n-1) = u(n-1) + \frac{1}{2} a (n^2 - 2n + 1) = un - u + \frac{1}{2} a n^2 - an + \frac{1}{2} a \]

### Step 4: Find \( S_n \)

Now, substituting \( S(n) \) and \( S(n-1) \) into the equation for \( S_n \):
\[ S_n = \left( un + \frac{1}{2} a n^2 \right) - \left( un - u + \frac{1}{2} a n^2 - an + \frac{1}{2} a \right) \]

Simplifying this:
\[ S_n = un + \frac{1}{2} a n^2 - un + u - \frac{1}{2} a n^2 + an - \frac{1}{2} a \]
\[ S_n = u + \frac{1}{2} a (2n - 1) \]

### Step 5: Check the dimensional correctness

To check if the equation is dimensionally true, we need to analyze the dimensions of each term:

- The dimension of \( S_n \) (distance) is \([L]\).
- The dimension of \( u \) (velocity) is \([L][T^{-1}]\), and when multiplied by time \( n \) (which has dimension \([T]\)), it gives \([L]\).
- The dimension of \( a \) (acceleration) is \([L][T^{-2}]\), and when multiplied by \( n^2 \), it gives \([L][T^{-2}][T^2] = [L]\).

Thus, both terms on the right side of the equation have the dimension of length, confirming that the equation is dimensionally correct.

### Conclusion

1. The equation is dimensionally true.
2. The numerical correctness can vary based on the values of \( u \) and \( a \), hence it may be true or false depending on the context.

### Final Answer

The correct options are:
- Option 1: Dimensionally true
- Option 3: Numerically may be true or false
- Option 4: Both 1 and 3 are correct