By the method of dimensions, test the accuracy of the equation : `delta = (mgl^3)/(4bd^3Y)` where `delta` is depression in the middle of a bar of length I, breadth b, depth d, when it is loaded in the middle with mass m. Y is Young's modulus of material of the bar.

Options

A. `(Wl^(3))/(4bd^(3)Y)`
B. `(Wb^(3))/(4dl^(3)Y)`
C. `(Wd^(3))/(4lb^(3)Y)`
D. `(Wl^(3))/(bd^(3)Y)`

Explanation

To test the accuracy of the equation \(\delta = \frac{mgl^3}{4bd^3Y}\) using the method of dimensions, we will analyze the dimensions of both sides of the equation step by step.

### Step 1: Identify the dimensions of each variable
1. **Depression (\(\delta\))**: This is a length, so its dimension is:
\[ [\delta] = L \]

2. **Mass (\(m\))**: The dimension of mass is:
\[ [m] = M \]

3. **Acceleration due to gravity (\(g\))**: The dimension of acceleration is:
\[ [g] = LT^{-2} \]

4. **Length (\(l\))**: The dimension of length is:
\[ [l] = L \]

5. **Breadth (\(b\))**: The dimension of breadth is:
\[ [b] = L \]

6. **Depth (\(d\))**: The dimension of depth is:
\[ [d] = L \]

7. **Young's modulus (\(Y\))**: Young's modulus is defined as stress divided by strain. The dimension of stress (force per unit area) is:
\[ [\text{Stress}] = \frac{[F]}{[A]} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \]
The dimension of strain is dimensionless, so:
\[ [Y] = ML^{-1}T^{-2} \]

### Step 2: Analyze the right-hand side of the equation
Now, we will analyze the right-hand side of the equation:
\[ \text{RHS} = \frac{mgl^3}{4bd^3Y} \]

1. **Substituting the dimensions**:
- The dimension of \(m\) is \(M\).
- The dimension of \(g\) is \(LT^{-2}\).
- The dimension of \(l^3\) is \(L^3\).
- The dimension of \(b\) is \(L\).
- The dimension of \(d^3\) is \(L^3\).
- The dimension of \(Y\) is \(ML^{-1}T^{-2}\).

2. **Combining the dimensions**:
\[ \text{RHS} = \frac{M \cdot (LT^{-2}) \cdot (L^3)}{L \cdot (L^3) \cdot (ML^{-1}T^{-2})} \]
Simplifying the right-hand side:
\[ = \frac{M \cdot L^4 T^{-2}}{M \cdot L^4 T^{-2}} = 1 \]

### Step 3: Compare dimensions of both sides
The left-hand side has the dimension:
\[ [\delta] = L \]

The right-hand side simplifies to:
\[ [\text{RHS}] = L \]

### Conclusion
Since both sides of the equation have the same dimension, we conclude that the equation is dimensionally consistent and hence is accurate.