To solve the problem, we need to convert the units from the TK system to the SI and MKS systems for acceleration, kinetic energy, pressure, and work. Let's break down each part step by step.
### Step 1: One Unit of Acceleration in TK System
1. **Understanding Acceleration**:
- Acceleration is defined as the change in velocity per unit time. Its dimensions are given by \( L T^{-2} \) (length per time squared).
2. **Units in TK System**:
- Length (Akshay) = 1 km = 1000 m
- Time (Aamir) = 1 minute = 60 seconds
3. **Calculating Acceleration**:
\[
\text{Acceleration} = \frac{\text{Length}}{\text{Time}^2} = \frac{1000 \, \text{m}}{(60 \, \text{s})^2} = \frac{1000}{3600} \, \text{m/s}^2
\]
\[
= \frac{1000}{36 \times 100} = \frac{10}{36} \, \text{m/s}^2 = \frac{5}{18} \times 10^{-4} \, \text{SI unit}
\]
4. **Conclusion for Acceleration**:
- Therefore, one unit of acceleration in the TK system corresponds to \( \frac{5}{18} \times 10^{-4} \) SI units.
### Step 2: One Unit of Kinetic Energy in TK System
1. **Understanding Kinetic Energy**:
- Kinetic energy is given by the formula \( KE = \frac{1}{2} mv^2 \).
- Its dimensions are \( M L^2 T^{-2} \).
2. **Calculating Kinetic Energy**:
- Mass (Shahrukh) = 1 Quintal = 100 kg
- Length (Akshay) = 1 km = 1000 m
- Time (Aamir) = 1 minute = 60 seconds
3. **Substituting Values**:
\[
KE = \frac{1}{2} \times 100 \, \text{kg} \times \left(\frac{1000 \, \text{m}}{60 \, \text{s}}\right)^2
\]
\[
= \frac{1}{2} \times 100 \times \frac{1000000}{3600} = 50 \times \frac{1000000}{3600} = 50 \times \frac{10000}{36} = \frac{500000}{36} = \frac{5}{18} \times 10^5 \, \text{MKS unit}
\]
4. **Conclusion for Kinetic Energy**:
- Therefore, one unit of kinetic energy in the TK system corresponds to \( \frac{5}{18} \times 10^{5} \) MKS units.
### Step 3: One Unit of Pressure in TK System
1. **Understanding Pressure**:
- Pressure is defined as force per unit area. Its dimensions are \( M L^{-1} T^{-2} \).
2. **Calculating Pressure**:
- Force = Mass × Acceleration
\[
\text{Force} = 100 \, \text{kg} \times \frac{1000 \, \text{m}}{(60 \, \text{s})^2} = 100 \times \frac{1000}{3600} = \frac{100000}{3600} = \frac{1000}{36} \, \text{N}
\]
- Area = Length² = \( (1000 \, \text{m})^2 = 1000000 \, \text{m}^2 \)
3. **Substituting Values**:
\[
\text{Pressure} = \frac{\text{Force}}{\text{Area}} = \frac{\frac{1000}{36}}{1000000} = \frac{1000}{36000000} = \frac{5}{18} \times 10^{-4} \, \text{SI unit}
\]
4. **Conclusion for Pressure**:
- Therefore, one unit of pressure in the TK system corresponds to \( \frac{5}{18} \times 10^{-4} \) SI units.
### Step 4: One Unit of Work in TK System
1. **Understanding Work**:
- Work is defined as force times distance. Its dimensions are \( M L^2 T^{-2} \).
2. **Calculating Work**:
- Using the force calculated earlier:
\[
\text{Work} = \text{Force} \times \text{Distance} = \left(\frac{1000}{36} \, \text{N}\right) \times (1000 \, \text{m}) = \frac{1000000}{36} \, \text{J}
\]
3. **Conclusion for Work**:
- Therefore, one unit of work in the TK system corresponds to \( \frac{5}{18} \times 10^{5} \) SI units.
### Final Answers
- (A) Acceleration: \( \frac{5}{18} \times 10^{-4} \) SI unit (P)
- (B) Kinetic Energy: \( \frac{5}{18} \times 10^{5} \) MKS unit (Q)
- (C) Pressure: \( \frac{5}{18} \times 10^{-4} \) SI unit (R)
- (D) Work: \( \frac{5}{18} \times 10^{5} \) SI unit (S)
