Answer :
To determine which formula for kinetic energy (K) is ruled out based on dimensional arguments, let's analyze each formula in terms of dimensions.
The dimensional formula for kinetic energy (K) can be derived from its definition as the work needed to accelerate a body of mass \( m \) from rest to a velocity \( v \):
\[ K = \frac{1}{2} mv^2 \]
Let's check the dimensions of kinetic energy (K) using the formula \( [K] \):
\[ [K] = [ \frac{1}{2} mv^2 ] \]
\[ [K] = [m] \cdot [v^2] \]
\[ [K] = M \cdot (LT^{-1})^2 \]
\[ [K] = M \cdot L^2 \cdot T^{-2} \]
Where:
- \( M \) denotes mass (dimension \( [M] = M^1 \)),
- \( L \) denotes length (dimension \( [L] = L^1 \)),
- \( T \) denotes time (dimension \( [T] = T^1 \)).
Therefore, the dimensional formula for kinetic energy \( K \) is \( ML^2T^{-2} \).
Now, let's analyze each given option:
**(a) \( k = m^2v^3 \)**
- Dimensions of \( m^2v^3 \):
\[ [m^2v^3] = M^2 \cdot L^3 \cdot T^{-3} \]
- Kinetic energy dimension:
\[ [K] = M \cdot L^2 \cdot T^{-2} \]
Clearly, \( m^2v^3 \) does not match the dimensions of kinetic energy \( ML^2T^{-2} \). Hence, option (a) is ruled out because its dimensional formula does not correspond to the correct dimensions of kinetic energy.
**(b) \( k = mv^2 \)**
- Dimensions of \( mv^2 \):
\[ [mv^2] = M^1 \cdot L^2 \cdot T^{-2} \]
- Kinetic energy dimension:
\[ [K] = M \cdot L^2 \cdot T^{-2} \]
Option (b) matches the correct dimensional formula for kinetic energy \( ML^2T^{-2} \).
**(c) \( k = \frac{1}{2} mv^2 \)**
- Dimensions of \( \frac{1}{2} mv^2 \):
\[ [\frac{1}{2} mv^2] = M^1 \cdot L^2 \cdot T^{-2} \]
- Kinetic energy dimension:
\[ [K] = M \cdot L^2 \cdot T^{-2} \]
Option (c) also matches the correct dimensional formula for kinetic energy \( ML^2T^{-2} \).
**(d) \( k = \frac{3}{6} mv^2 \)**
- Simplifying \( \frac{3}{6} mv^2 = \frac{1}{2} mv^2 \),
- Dimensions of \( \frac{1}{2} mv^2 \):
\[ [\frac{1}{2} mv^2] = M^1 \cdot L^2 \cdot T^{-2} \]
- Kinetic energy dimension:
\[ [K] = M \cdot L^2 \cdot T^{-2} \]
Option (d) simplifies to \( \frac{1}{2} mv^2 \), which matches the correct dimensional formula for kinetic energy \( ML^2T^{-2} \).
### Conclusion:
Based on dimensional arguments, option (a) \( k = m^2v^3 \) is ruled out because its dimensional formula \( M^2 \cdot L^3 \cdot T^{-3} \) does not match the dimensional formula of kinetic energy \( ML^2T^{-2} \). The correct formula for kinetic energy is \( k = \frac{1}{2} mv^2 \), where \( [K] = ML^2T^{-2} \).
The dimensional formula for kinetic energy (K) can be derived from its definition as the work needed to accelerate a body of mass \( m \) from rest to a velocity \( v \):
\[ K = \frac{1}{2} mv^2 \]
Let's check the dimensions of kinetic energy (K) using the formula \( [K] \):
\[ [K] = [ \frac{1}{2} mv^2 ] \]
\[ [K] = [m] \cdot [v^2] \]
\[ [K] = M \cdot (LT^{-1})^2 \]
\[ [K] = M \cdot L^2 \cdot T^{-2} \]
Where:
- \( M \) denotes mass (dimension \( [M] = M^1 \)),
- \( L \) denotes length (dimension \( [L] = L^1 \)),
- \( T \) denotes time (dimension \( [T] = T^1 \)).
Therefore, the dimensional formula for kinetic energy \( K \) is \( ML^2T^{-2} \).
Now, let's analyze each given option:
**(a) \( k = m^2v^3 \)**
- Dimensions of \( m^2v^3 \):
\[ [m^2v^3] = M^2 \cdot L^3 \cdot T^{-3} \]
- Kinetic energy dimension:
\[ [K] = M \cdot L^2 \cdot T^{-2} \]
Clearly, \( m^2v^3 \) does not match the dimensions of kinetic energy \( ML^2T^{-2} \). Hence, option (a) is ruled out because its dimensional formula does not correspond to the correct dimensions of kinetic energy.
**(b) \( k = mv^2 \)**
- Dimensions of \( mv^2 \):
\[ [mv^2] = M^1 \cdot L^2 \cdot T^{-2} \]
- Kinetic energy dimension:
\[ [K] = M \cdot L^2 \cdot T^{-2} \]
Option (b) matches the correct dimensional formula for kinetic energy \( ML^2T^{-2} \).
**(c) \( k = \frac{1}{2} mv^2 \)**
- Dimensions of \( \frac{1}{2} mv^2 \):
\[ [\frac{1}{2} mv^2] = M^1 \cdot L^2 \cdot T^{-2} \]
- Kinetic energy dimension:
\[ [K] = M \cdot L^2 \cdot T^{-2} \]
Option (c) also matches the correct dimensional formula for kinetic energy \( ML^2T^{-2} \).
**(d) \( k = \frac{3}{6} mv^2 \)**
- Simplifying \( \frac{3}{6} mv^2 = \frac{1}{2} mv^2 \),
- Dimensions of \( \frac{1}{2} mv^2 \):
\[ [\frac{1}{2} mv^2] = M^1 \cdot L^2 \cdot T^{-2} \]
- Kinetic energy dimension:
\[ [K] = M \cdot L^2 \cdot T^{-2} \]
Option (d) simplifies to \( \frac{1}{2} mv^2 \), which matches the correct dimensional formula for kinetic energy \( ML^2T^{-2} \).
### Conclusion:
Based on dimensional arguments, option (a) \( k = m^2v^3 \) is ruled out because its dimensional formula \( M^2 \cdot L^3 \cdot T^{-3} \) does not match the dimensional formula of kinetic energy \( ML^2T^{-2} \). The correct formula for kinetic energy is \( k = \frac{1}{2} mv^2 \), where \( [K] = ML^2T^{-2} \).