Question 6

$$T$$ is the time period of simple pendulum on the earth's surface. Its time period becomes $$xT$$ when taken to a height $$R$$ (equal to earth's radius) above the earth's surface. Then, the value of $$x$$ will be:

We need to find the time period of a simple pendulum when taken to a height $$R$$ (equal to Earth's radius) above the Earth's surface.

Find the acceleration due to gravity at height $$h = R$$.

At height $$h$$ above the Earth's surface:

$$ g' = g \cdot \frac{R^2}{(R + h)^2} $$

At $$h = R$$:

$$ g' = g \cdot \frac{R^2}{(R + R)^2} = g \cdot \frac{R^2}{4R^2} = \frac{g}{4} $$

Find the new time period.

The time period of a simple pendulum is $$T = 2\pi\sqrt{\frac{L}{g}}$$.

At height $$R$$:

$$ T' = 2\pi\sqrt{\frac{L}{g'}} = 2\pi\sqrt{\frac{L}{g/4}} = 2\pi\sqrt{\frac{4L}{g}} = 2 \cdot 2\pi\sqrt{\frac{L}{g}} = 2T $$

Find the value of $$x$$.

Since $$T' = xT$$ and $$T' = 2T$$:

$$ x = 2 $$

The correct answer is Option B: $$2$$.

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