The time period of oscillation of a body is given by `T=2pisqrt((mgA)/(K))`
K: Represents the kinetic energy, m mass, g acceleration due to gravity and A is unknown If `[A]=M^(x)L^(y)T^(z)`, then what is the value of x+y+z?

Options

A. 3
B. 2
C. 1
D. 5

Explanation

To solve the problem, we need to analyze the given equation for the time period of oscillation:

\[ T = 2\pi \sqrt{\frac{mgA}{K}} \]

Where:
- \( K \) is the kinetic energy, given by \( K = \frac{1}{2} mv^2 \)
- \( m \) is the mass
- \( g \) is the acceleration due to gravity
- \( A \) is an unknown quantity with dimensions represented as \( [A] = M^x L^y T^z \)

### Step 1: Identify the dimensions of each variable
1. **Dimensions of \( T \)** (Time period):
\[ [T] = T^1 \]

2. **Dimensions of \( m \)** (Mass):
\[ [m] = M^1 \]

3. **Dimensions of \( g \)** (Acceleration due to gravity):
\[ [g] = L^1 T^{-2} \]

4. **Dimensions of \( K \)** (Kinetic energy):
\[ [K] = [\frac{1}{2} mv^2] = M^1 L^2 T^{-2} \]

### Step 2: Substitute the dimensions into the equation
The right-hand side of the equation can be expressed as:
\[ [T^2] = \frac{[m][g][A]}{[K]} \]

Substituting the dimensions we have:
\[ [T^2] = \frac{(M^1)(L^1 T^{-2})([A])}{(M^1 L^2 T^{-2})} \]

### Step 3: Simplify the dimensions
Now, simplifying the right-hand side:
\[ [T^2] = \frac{M^1 L^1 T^{-2} [A]}{M^1 L^2 T^{-2}} \]

The \( M^1 \) and \( T^{-2} \) cancel out:
\[ [T^2] = \frac{L^1 [A]}{L^2} \]
\[ [T^2] = L^{-1} [A] \]

### Step 4: Rearranging to find dimensions of \( A \)
Rearranging gives:
\[ [A] = L^2 [T^2] \]

Since \( [T^2] = T^2 \), we can express this as:
\[ [A] = L^2 T^2 \]

### Step 5: Identify the dimensions of \( A \)
From the expression \( [A] = L^2 T^2 \), we can identify:
- \( x = 0 \) (no mass dimension)
- \( y = 2 \) (length dimension)
- \( z = 2 \) (time dimension)

### Step 6: Calculate \( x + y + z \)
Now, we can find:
\[ x + y + z = 0 + 2 + 2 = 4 \]

### Final Answer
Thus, the value of \( x + y + z \) is:
\[ \boxed{4} \]