Root mean square speed of a gas is `5 ms^(-1)` If some molecules out of 10 molecules in all are moving with `7ms^(-1)` and rest all the molecules moving with `3 m sec^(-1)` then number of molecules moving with higher speed is ____________.

Explanation

To find the number of molecules moving with a higher speed of `7 m/s`, we can follow these steps:

### Step 1: Understand the Problem
We know the root mean square (RMS) speed of a gas is given as `5 m/s`. There are a total of `10 molecules`, some of which are moving at `7 m/s` and the rest at `3 m/s`. We need to determine how many molecules are moving at `7 m/s`.

### Step 2: Set Up Variables
Let:
- \( n_1 \) = number of molecules moving at `7 m/s` = \( x \)
- \( n_2 \) = number of molecules moving at `3 m/s` = \( 10 - x \)

### Step 3: Write the RMS Speed Formula
The formula for the root mean square speed is:
\[ RMS = \sqrt{\frac{n_1 u_1^2 + n_2 u_2^2}{n_1 + n_2}} \]
Where:
- \( u_1 = 7 \, m/s \)
- \( u_2 = 3 \, m/s \)

### Step 4: Substitute Values into the Formula
Substituting the known values into the RMS formula:
\[ 5 = \sqrt{\frac{x \cdot (7)^2 + (10 - x) \cdot (3)^2}{10}} \]
This simplifies to:
\[ 5 = \sqrt{\frac{49x + 9(10 - x)}{10}} \]

### Step 5: Simplify the Equation
Now, simplify the equation:
\[ 5 = \sqrt{\frac{49x + 90 - 9x}{10}} \]
\[ 5 = \sqrt{\frac{40x + 90}{10}} \]

### Step 6: Square Both Sides
To eliminate the square root, square both sides:
\[ 25 = \frac{40x + 90}{10} \]

### Step 7: Multiply Through by 10
Multiply both sides by `10` to remove the denominator:
\[ 250 = 40x + 90 \]

### Step 8: Solve for \( x \)
Now, solve for \( x \):
\[ 250 - 90 = 40x \]
\[ 160 = 40x \]
\[ x = \frac{160}{40} = 4 \]

### Step 9: Conclusion
Thus, the number of molecules moving with the higher speed of `7 m/s` is \( \boxed{4} \).

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