In melde's experiment in perpendicular position , 6 loops are obtained when a mass of 6gm is put in the pan. When the arrangement is changed to parallel position without disturbing other setup, a mass of 14.75 gm is to be put in the pan to obtain 2 loops . then the mass of the pan will be

Options

A. 2gm
B. 1gm
C. 1.8gm
D. 0gm

Explanation

To solve the problem step by step, we will use the principles of Melde's experiment and the relationships between frequency, mass, and tension in the string.

### Step 1: Understand the Setup
In Melde's experiment, the number of loops formed is related to the frequency of the vibrating string. The frequency \( f \) of the string can be expressed as:
\[ f = \frac{N}{2L} \sqrt{\frac{T}{\mu}} \]
where:
- \( N \) = number of loops
- \( L \) = length of the string
- \( T \) = tension in the string
- \( \mu \) = mass per unit length of the string

### Step 2: Analyze the Perpendicular Position
When 6 loops are formed with a mass of 6 g in the pan:
- The tension \( T \) is given by \( T = (m + M)g \), where \( m \) is the mass in the pan (6 g) and \( M \) is the mass of the pan.
- The frequency for 6 loops is:
\[ f_1 = \frac{6}{2L} \sqrt{\frac{(6 + M)g}{\mu}} \]

### Step 3: Analyze the Parallel Position
When the setup is changed to the parallel position, we need 14.75 g in the pan to obtain 2 loops:
- The new tension is \( T = (14.75 + M)g \).
- The frequency for 2 loops is:
\[ f_2 = \frac{2}{2L} \sqrt{\frac{(14.75 + M)g}{\mu}} \]

### Step 4: Relate the Frequencies
In the parallel position, the frequency of the fork is twice that of the string:
\[ f_{fork} = 2f_2 \]
Thus, we can set up the equation:
\[ \frac{6}{2L} \sqrt{\frac{(6 + M)g}{\mu}} = 2 \cdot \frac{2}{2L} \sqrt{\frac{(14.75 + M)g}{\mu}} \]

### Step 5: Simplify the Equation
Cancel \( \frac{1}{2L} \) and \( g \) from both sides:
\[ 6 \sqrt{6 + M} = 4 \sqrt{14.75 + M} \]

### Step 6: Square Both Sides
Squaring both sides gives:
\[ 36(6 + M) = 16(14.75 + M) \]

### Step 7: Expand and Rearrange
Expanding both sides:
\[ 216 + 36M = 236 + 16M \]
Rearranging gives:
\[ 36M - 16M = 236 - 216 \]
\[ 20M = 20 \]

### Step 8: Solve for M
Dividing both sides by 20:
\[ M = 1 \text{ g} \]

### Conclusion
The mass of the pan is \( 1 \text{ g} \).

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