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

Explanation

To solve the problem, we start with the given equation for the time period of oscillation:

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

Where:
- \( T \) is the time period,
- \( m \) is the mass,
- \( g \) is the acceleration due to gravity,
- \( A \) is an unknown quantity,
- \( K \) is the kinetic energy.

### Step 1: Identify the dimensions of each variable
- The dimension of mass \( m \) is \( [M] \).
- The dimension of acceleration due to gravity \( g \) is \( [L T^{-2}] \).
- The dimension of kinetic energy \( K \) is given by the formula for kinetic energy, which is \( \frac{1}{2} mv^2 \). Thus, the dimension of kinetic energy is \( [M L^2 T^{-2}] \).

### Step 2: Rewrite the equation in terms of dimensions
We can express the dimensions of the right-hand side of the equation:

\[ T^2 = \frac{mgA}{K} \]

This implies:

\[ [T^2] = \frac{[M][L T^{-2}][A]}{[M L^2 T^{-2}]} \]

### Step 3: Simplify the dimensions
Now, substituting the dimensions into the equation:

\[ [T^2] = \frac{[M][L T^{-2}][A]}{[M L^2 T^{-2}]} \]

The \( [M] \) and \( [T^{-2}] \) terms cancel out:

\[ [T^2] = \frac{[L][A]}{[L^2]} \]

This simplifies to:

\[ [T^2] = [L^{-1}][A] \]

### Step 4: Solve for the dimension of \( A \)
Rearranging the equation gives us:

\[ [A] = [T^2][L] \]

### Step 5: Write the dimensions of \( A \)
Thus, we have:

\[ [A] = [L^1][T^2] \]

### Step 6: Identify the powers of M, L, and T
From the dimension of \( A \):

- The power of \( M \) is \( 0 \) (since \( M \) does not appear),
- The power of \( L \) is \( 1 \),
- The power of \( T \) is \( 2 \).

This means:

- \( x = 0 \)
- \( y = 1 \)
- \( z = 2 \)

### Step 7: Calculate \( x + y + z \)
Now, we can find:

\[ x + y + z = 0 + 1 + 2 = 3 \]

Thus, the final answer is:

**Answer: 3**

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