STATEMENT-1: In the expression `y=A sin (kx-omegat)`, dimensions of omega must be reciprocal of that of t.
STATEMENT 2 The expression (kx-omegat) must be dimensionless.

Options

A. Statement-1 is True , Statement-2 is True ,Statement-2 is a correct explanation for Statement-1
B. Statement-1 is True , Statement-2 is True ,Statement-2 is NOT a correct explanation for Statement-1
C. Statement-1 is True , Statement-2 is False
D. Statement-1 is False, Statement-2 is True

Answer

To solve the question, we need to analyze both statements provided in the question regarding the expression \( y = A \sin(kx - \omega t) \).

### Step 1: Analyze Statement 1
The first statement claims that in the expression \( y = A \sin(kx - \omega t) \), the dimensions of \( \omega \) must be the reciprocal of that of \( t \).

- In the expression \( kx - \omega t \), both \( kx \) and \( \omega t \) must have the same dimensions because they are being subtracted.
- The term \( t \) represents time, which has the dimension \( [T] \).
- Therefore, if \( \omega \) is angular frequency, its dimensions must be such that \( \omega t \) is dimensionless.
- The dimension of \( \omega \) can be derived as follows:
\[ [\omega] = \frac{[1]}{[T]} = [T^{-1}] \]
- This confirms that the dimensions of \( \omega \) are indeed the reciprocal of the dimensions of \( t \).

**Conclusion for Statement 1**: True.

### Step 2: Analyze Statement 2
The second statement asserts that the expression \( kx - \omega t \) must be dimensionless.

- For any argument of a trigonometric function (like sine), it must be dimensionless. Thus, \( kx - \omega t \) must have no dimensions.
- Since \( k \) is the wave number, it has dimensions of \( [L^{-1}] \) (inverse length).
- Therefore, the term \( kx \) has dimensions:
\[ [kx] = [L^{-1}][L] = [1] \quad \text{(dimensionless)} \]
- Similarly, since \( \omega t \) has dimensions:
\[ [\omega t] = [T^{-1}][T] = [1] \quad \text{(dimensionless)} \]
- Both terms \( kx \) and \( \omega t \) are dimensionless, confirming that their difference \( kx - \omega t \) is also dimensionless.

**Conclusion for Statement 2**: True.

### Final Conclusion
Both statements are true, and Statement 2 provides the correct explanation for Statement 1.