Given that `Y = a sin omegat + bt + ct^(2) cos omegat`. The unit of `abc` is same as that of

Options

A. `y`
B. `y//t`
C. `( y //t)^(2)`
D. `(y//t)^(3)`

Explanation

To solve the problem, we need to analyze the given expression for \( Y \) and determine the relationship between the units of \( a \), \( b \), and \( c \) with respect to \( Y \).

### Step 1: Identify the expression for \( Y \)
The expression given is:
\[ Y = a \sin(\omega t) + bt + ct^2 \cos(\omega t) \]

### Step 2: Determine the dimensions of \( Y \)
Since \( Y \) is a function of time, we can assume that it has dimensions of length (L), which is a common assumption in physics unless stated otherwise. Therefore:
\[ [Y] = L \]

### Step 3: Analyze the first term \( a \sin(\omega t) \)
The term \( \sin(\omega t) \) is dimensionless because trigonometric functions do not have dimensions. Thus, the dimension of \( a \) must be the same as that of \( Y \):
\[ [a] = [Y] = L \]

### Step 4: Analyze the second term \( bt \)
In this term, \( t \) has dimensions of time (T). Therefore, for the term \( bt \) to have dimensions of length (L), the dimension of \( b \) must be:
\[ [b] = \frac{[Y]}{[t]} = \frac{L}{T} \]

### Step 5: Analyze the third term \( ct^2 \cos(\omega t) \)
Similar to the first term, \( \cos(\omega t) \) is also dimensionless. Therefore, for the term \( ct^2 \) to have dimensions of length (L), we have:
\[ [ct^2] = L \implies [c][t^2] = L \]
Since \( t^2 \) has dimensions of \( T^2 \), we can write:
\[ [c][T^2] = L \implies [c] = \frac{L}{T^2} \]

### Step 6: Calculate the product \( abc \)
Now, we can find the product \( abc \):
\[ [abc] = [a][b][c] = (L) \left(\frac{L}{T}\right) \left(\frac{L}{T^2}\right) \]
Calculating this gives:
\[ [abc] = L \cdot \frac{L}{T} \cdot \frac{L}{T^2} = \frac{L^3}{T^3} \]

### Conclusion
The units of \( abc \) are the same as \( \frac{L^3}{T^3} \).