Check by the method of dimensions, the formula ` upsilo = (1)/(lambda)sqrt((K)/(d)),` where `upsilon` is velocity of longitudinal waves, `lambdas` is wavelength of wave, K is coefficient of volume elasticity and d is density of the medium.

Answer

To check the formula \( \upsilon = \frac{1}{\lambda} \sqrt{\frac{K}{d}} \) using the method of dimensions, we will follow these steps:

### Step 1: Identify the dimensions of each term in the formula.

1. **Velocity of longitudinal waves (\( \upsilon \))**:
- The dimension of velocity is given by:
\[ [\upsilon] = LT^{-1} \]

2. **Wavelength (\( \lambda \))**:
- The dimension of wavelength is:
\[ [\lambda] = L \]

3. **Coefficient of volume elasticity (\( K \))**:
- The coefficient of volume elasticity is defined as volumetric stress divided by volumetric strain.
- The dimension of stress (pressure) is:
\[ [\text{Stress}] = \frac{[\text{Force}]}{[\text{Area}]} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \]
- Since strain is dimensionless, we have:
\[ [K] = ML^{-1}T^{-2} \]

4. **Density (\( d \))**:
- The dimension of density is:
\[ [d] = \frac{[\text{Mass}]}{[\text{Volume}]} = \frac{M}{L^3} = ML^{-3} \]

### Step 2: Substitute the dimensions into the right-hand side (RHS) of the formula.

The RHS of the formula is:
\[ \frac{1}{\lambda} \sqrt{\frac{K}{d}} \]

1. Substitute the dimensions:
- The dimension of \( \frac{1}{\lambda} \) is:
\[ \left[\frac{1}{\lambda}\right] = L^{-1} \]
- Now, calculate the dimensions of \( \frac{K}{d} \):
\[ \frac{K}{d} = \frac{ML^{-1}T^{-2}}{ML^{-3}} = \frac{L^{3}}{L^{-1}}T^{-2} = L^{2}T^{-2} \]
- Taking the square root:
\[ \sqrt{\frac{K}{d}} = \sqrt{L^{2}T^{-2}} = LT^{-1} \]

2. Combine the dimensions:
- Now, multiply \( \frac{1}{\lambda} \) and \( \sqrt{\frac{K}{d}} \):
\[ \left[\frac{1}{\lambda} \sqrt{\frac{K}{d}}\right] = L^{-1} \cdot LT^{-1} = T^{-1} \]

### Step 3: Compare the dimensions of the left-hand side (LHS) and the right-hand side (RHS).

- The dimension of the LHS (\( \upsilon \)):
\[ [\upsilon] = LT^{-1} \]

- The dimension of the RHS:
\[ \left[\frac{1}{\lambda} \sqrt{\frac{K}{d}}\right] = LT^{-1} \]

### Conclusion:

Since the dimensions of the LHS and RHS are different, the formula \( \upsilon = \frac{1}{\lambda} \sqrt{\frac{K}{d}} \) is dimensionally incorrect.