The power of a motor is 200W. If the unit of length is halved, that of mass is doubled and that of time is aiso doubled, then the power of the motor in the new system is

Options

A. 3200 W
B. 3200 new units
C. 12.5 new units
D. 12.5 W

Answer

To solve the problem step by step, we need to understand how the changes in the units affect the power of the motor.

### Step 1: Understand the formula for power
Power (P) is defined as work done (W) per unit time (t). In terms of units, power can be expressed as:
\[ P = \frac{W}{t} = \frac{F \cdot d}{t} \]
where \( F \) is force and \( d \) is distance. The SI unit of power is Watts (W), which can be expressed in terms of base units as:
\[ 1 \text{ W} = 1 \frac{kg \cdot m^2}{s^3} \]

### Step 2: Identify the initial unit of power
The initial power of the motor is given as 200 W. In terms of base units, this can be expressed as:
\[ P = 200 \frac{kg \cdot m^2}{s^3} \]

### Step 3: Apply the changes in units
According to the problem:
- The unit of length (m) is halved: \( L' = \frac{L}{2} \)
- The unit of mass (kg) is doubled: \( M' = 2M \)
- The unit of time (s) is doubled: \( T' = 2T \)

### Step 4: Substitute the new units into the power formula
Now, substituting the new units into the power formula:
\[ P' = \frac{F' \cdot d'}{t'} \]
Where:
- Force \( F' = M' \cdot a' \)
- Acceleration \( a' \) can be expressed as \( \frac{d'}{(t')^2} \)

Given that \( d' = \frac{d}{2} \) and \( t' = 2t \), we can express the new force as:
\[ F' = (2M) \cdot \left(\frac{d/2}{(2t)^2}\right) \]
\[ F' = (2M) \cdot \left(\frac{d/2}{4t^2}\right) \]
\[ F' = \frac{2M \cdot d}{8t^2} = \frac{M \cdot d}{4t^2} \]

### Step 5: Calculate the new power
Substituting \( F' \) into the power formula:
\[ P' = \frac{F' \cdot d'}{t'} = \frac{\left(\frac{M \cdot d}{4t^2}\right) \cdot \left(\frac{d}{2}\right)}{2t} \]
\[ P' = \frac{M \cdot d^2}{8t^3} \]

### Step 6: Relate the new power to the original power
Now we can relate the new power \( P' \) to the original power \( P \):
\[ P' = \frac{1}{8} \cdot P \]
Since \( P = 200 \text{ W} \):
\[ P' = \frac{1}{8} \cdot 200 = 25 \text{ W} \]

### Step 7: Final calculation
Now, we need to adjust for the changes in units:
- The new unit of power is \( P' = 200 \cdot \frac{2^2}{(1/2)^2} = 200 \cdot \frac{4}{1/4} = 200 \cdot 16 = 3200 \text{ W} \)

### Conclusion
Thus, the power of the motor in the new system is:
\[ \text{Power} = 3200 \text{ W} \]