In the previous problem, if two falling conductors attain velocities `v_(1) and v_(2)` respectively after falling through same height h, ratio of energy dissipated as heat to the energy dissipated in resistor per unit time for two conductors is same.
Find which of the following condition will be satisfied

Options

A. `gh+1/2 v_(1)v_(2)=0`
B. `gh+ v_(1)v_(2)=0`
C. `1/2 gh+ v_(1)v_(2)=0`
D. `gh+ (v_(1)^(2)v_(2))/(v_(1)+v_(2))=0`

Explanation

To solve the problem, we need to analyze the energy dissipated as heat and the energy dissipated in the resistor per unit time for two falling conductors. Let's denote the two conductors as conductor 1 and conductor 2, with velocities \( v_1 \) and \( v_2 \) respectively after falling through the same height \( h \).

### Step 1: Write the energy conservation equation for both conductors

For each conductor, the potential energy lost while falling through height \( h \) is converted into kinetic energy and energy dissipated in the resistor. The energy conservation equation can be written as:

For conductor 1:
\[ mgh = \frac{1}{2} mv_1^2 + E_{dissipated1} \]

For conductor 2:
\[ mgh = \frac{1}{2} mv_2^2 + E_{dissipated2} \]

Where \( E_{dissipated1} \) and \( E_{dissipated2} \) are the energies dissipated in the resistors for conductors 1 and 2 respectively.

### Step 2: Rearranging the equations

From the above equations, we can express the energies dissipated as:
\[ E_{dissipated1} = mgh - \frac{1}{2} mv_1^2 \]
\[ E_{dissipated2} = mgh - \frac{1}{2} mv_2^2 \]

### Step 3: Calculate the ratio of energies dissipated as heat to the energy dissipated in resistors per unit time

The energy dissipated in the resistor per unit time can be expressed as:
\[ P_1 = \frac{E_{dissipated1}}{t} \quad \text{and} \quad P_2 = \frac{E_{dissipated2}}{t} \]

Thus, the ratio of the energy dissipated as heat to the energy dissipated in the resistor per unit time for both conductors can be written as:
\[ \frac{E_{dissipated1}}{P_1} = \frac{mgh - \frac{1}{2} mv_1^2}{\frac{E_{dissipated1}}{t}} \quad \text{and} \quad \frac{E_{dissipated2}}{P_2} = \frac{mgh - \frac{1}{2} mv_2^2}{\frac{E_{dissipated2}}{t}} \]

### Step 4: Set the ratios equal to each other

According to the problem, these ratios are equal:
\[ \frac{mgh - \frac{1}{2} mv_1^2}{\frac{E_{dissipated1}}{t}} = \frac{mgh - \frac{1}{2} mv_2^2}{\frac{E_{dissipated2}}{t}} \]

### Step 5: Simplifying the equation

After simplification, we can equate the two expressions:
\[ mgh - \frac{1}{2} mv_1^2 = mgh - \frac{1}{2} mv_2^2 \]

This leads to:
\[ \frac{1}{2} mv_1^2 = \frac{1}{2} mv_2^2 \]

### Step 6: Finding the condition

From the above, we can conclude:
\[ v_1^2 = v_2^2 \]

Thus, the condition that will be satisfied is:
\[ v_1 = v_2 \quad \text{or} \quad v_1 = -v_2 \]

### Conclusion

The condition that will be satisfied is that the velocities of the two conductors after falling through the same height are equal.