The density of a gas filled in electric lamp is `0.75kg//m^(3)` . After the lamp has been switched on, the pressure in it increases from `4xx10^(4)` Pa to `9xx10^(4)` Pa. What is increases in `u_("rms")` ?

Options

A. 100
B. 200
C. 300
D. None of these

Explanation

To solve the problem of finding the increase in the root mean square speed (u_rms) of a gas in an electric lamp after the pressure changes, we will follow these steps:

### Step 1: Understand the formula for u_rms
The root mean square speed (u_rms) of a gas is given by the formula:
\[ u_{rms} = \sqrt{\frac{3P}{\rho}} \]
where \( P \) is the pressure and \( \rho \) is the density of the gas.

### Step 2: Identify the initial and final pressures
From the problem, we have:
- Initial pressure \( P_1 = 4 \times 10^4 \) Pa
- Final pressure \( P_2 = 9 \times 10^4 \) Pa

### Step 3: Use the density of the gas
The density of the gas is given as:
\[ \rho = 0.75 \, \text{kg/m}^3 \]

### Step 4: Calculate the initial u_rms
Using the formula for u_rms, we can calculate the initial u_rms:
\[ u_{rms1} = \sqrt{\frac{3P_1}{\rho}} = \sqrt{\frac{3 \times (4 \times 10^4)}{0.75}} \]

Calculating the value:
\[ u_{rms1} = \sqrt{\frac{12 \times 10^4}{0.75}} = \sqrt{16 \times 10^4} = 4 \times 10^2 = 400 \, \text{m/s} \]

### Step 5: Calculate the final u_rms
Now, we calculate the final u_rms using the final pressure:
\[ u_{rms2} = \sqrt{\frac{3P_2}{\rho}} = \sqrt{\frac{3 \times (9 \times 10^4)}{0.75}} \]

Calculating the value:
\[ u_{rms2} = \sqrt{\frac{27 \times 10^4}{0.75}} = \sqrt{36 \times 10^4} = 6 \times 10^2 = 600 \, \text{m/s} \]

### Step 6: Calculate the increase in u_rms
The increase in u_rms is given by:
\[ \Delta u_{rms} = u_{rms2} - u_{rms1} = 600 \, \text{m/s} - 400 \, \text{m/s} = 200 \, \text{m/s} \]

### Final Answer
The increase in u_rms is:
\[ \Delta u_{rms} = 200 \, \text{m/s} \]

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