A magnetic needle of magnetic moment `6*7xx10^-2Am^2` and moment of inertia `7*5xx10^-6kgm^2` is performing simple harmonic oscillations in a magnetic field of `0*01T`. Time taken for 10 complete oscillation is

Options

A. 6.98 s
B. 8.76 s
C. 6.65 s
D. 8.89 s

Answer

To solve the problem, we need to find the time taken for 10 complete oscillations of a magnetic needle performing simple harmonic motion in a magnetic field. The formula for the time period \( T \) of oscillation is given by:

\[ T = 2\pi \sqrt{\frac{I}{mB}} \]

where:
- \( I \) is the moment of inertia,
- \( m \) is the magnetic moment,
- \( B \) is the magnetic field.

### Step 1: Identify the given values
- Magnetic moment \( m = 6.7 \times 10^{-2} \, \text{Am}^2 \)
- Moment of inertia \( I = 7.5 \times 10^{-6} \, \text{kg m}^2 \)
- Magnetic field \( B = 0.01 \, \text{T} \)

### Step 2: Substitute the values into the formula for the time period
Substituting the values into the formula:

\[ T = 2\pi \sqrt{\frac{7.5 \times 10^{-6}}{6.7 \times 10^{-2} \times 0.01}} \]

### Step 3: Calculate the denominator
Calculate the denominator:

\[ 6.7 \times 10^{-2} \times 0.01 = 6.7 \times 10^{-4} \]

### Step 4: Calculate the fraction inside the square root
Now calculate the fraction:

\[ \frac{7.5 \times 10^{-6}}{6.7 \times 10^{-4}} = \frac{7.5}{6.7} \times 10^{-2} \]

### Step 5: Calculate the square root
Now we can calculate:

\[ \sqrt{\frac{7.5}{6.7} \times 10^{-2}} = \sqrt{\frac{7.5}{6.7}} \times 10^{-1} \]

### Step 6: Calculate the numerical values
Calculating \( \frac{7.5}{6.7} \):

\[ \frac{7.5}{6.7} \approx 1.1194 \]

So,

\[ \sqrt{1.1194} \approx 1.058 \]

Thus,

\[ \sqrt{\frac{7.5}{6.7} \times 10^{-2}} \approx 1.058 \times 10^{-1} = 0.1058 \]

### Step 7: Calculate the time period
Now substituting back into the time period formula:

\[ T \approx 2\pi \times 0.1058 \approx 0.665 \, \text{s} \]

### Step 8: Calculate the time for 10 oscillations
To find the time for 10 complete oscillations:

\[ \text{Time for 10 oscillations} = 10 \times T = 10 \times 0.665 = 6.65 \, \text{s} \]

### Final Answer
The time taken for 10 complete oscillations is:

\[ \boxed{6.65 \, \text{s}} \]