A magnetic needle of magnetic moment $$6.7 \times 10^{-2}$$ A m$$^{2}$$ and moment of inertia $$7.5 \times 10^{-6}$$ kg m$$^{2}$$ is performing simple harmonic oscillations in a magnetic field of 0.01 T. Time taken for 10 complete oscillations is:

We know that a magnetic needle of magnetic moment $$M$$ placed in a uniform magnetic field $$B$$ executes torsional simple harmonic motion. The restoring torque on the needle is $$\tau = -MB\sin\theta$$. For very small angular displacements we can use the small-angle approximation $$\sin\theta \approx \theta$$, so the torque becomes $$\tau = -MB\,\theta$$.

Newton’s rotational equation is $$I\,\dfrac{d^{2}\theta}{dt^{2}} = \tau$$, where $$I$$ is the moment of inertia. Substituting the expression for $$\tau$$, we get

$$I\,\dfrac{d^{2}\theta}{dt^{2}} = -MB\,\theta.$$

This is the differential equation of a simple harmonic oscillator of the form $$\dfrac{d^{2}\theta}{dt^{2}} + \omega^{2}\theta = 0$$, where the square of the angular frequency is identified as

$$\omega^{2} = \dfrac{MB}{I}.$$

Therefore, the angular frequency is

$$\omega = \sqrt{\dfrac{MB}{I}}.$$

The time period $$T$$ of one complete oscillation is related to $$\omega$$ by the standard formula for SHM:

$$T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{I}{MB}}.$$

Now we substitute the given numerical values. We have

$$M = 6.7 \times 10^{-2}\ \text{A m}^{2},\quad B = 0.01\ \text{T},\quad I = 7.5 \times 10^{-6}\ \text{kg m}^{2}.$$

First compute the product $$MB$$:

$$MB = \left(6.7 \times 10^{-2}\right)\left(0.01\right) = 6.7 \times 10^{-4}.$$

Next form the ratio $$\dfrac{I}{MB}$$:

$$\dfrac{I}{MB} = \dfrac{7.5 \times 10^{-6}}{6.7 \times 10^{-4}} = \dfrac{7.5}{6.7}\times 10^{-6+4} = 1.1194 \times 10^{-2}.$$

Taking the square root gives

$$\sqrt{\dfrac{I}{MB}} = \sqrt{1.1194 \times 10^{-2}} \approx 0.1058.$$

Now insert this value into the expression for the period:

$$T = 2\pi \left(0.1058\right) = 6.283 \times 0.1058 \approx 0.6647\ \text{s}.$$

This is the time for one complete oscillation. The question asks for the time taken for 10 complete oscillations, so we multiply by 10:

$$t_{10} = 10T = 10 \times 0.6647 \approx 6.647\ \text{s}.$$

Rounding to three significant figures, we obtain $$t_{10} \approx 6.65\ \text{s}.$$ Hence, the correct answer is Option B.