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The graph best representing the variation of acceleration due to gravity $g$ with distance $d$ from Earth's centre (Earth's radius $= R$) shows which behaviour?
The correct behaviour of $g$:
- Inside the Earth ($d < R$): $g = \frac{GM}{R^3}d$ — linear increase from $g=0$ at centre to $g_{surface}$ at $d=R$.
- Outside the Earth ($d > R$): $g = \frac{GM}{d^2}$ — decreases as $1/d^2$.
The graph shows a straight line from origin to $(R, g_{max})$, then a smooth $1/d^2$ curve decreasing afterward. This is a linear rise inside and inverse-square fall outside, meeting at a sharp peak at $d = R$.
Evaluate: $\displaystyle\lim_{x\to\frac{\pi}{2}}\dfrac{\cot x - \cos x}{(\pi - 2x)^3}$
Let $x = \dfrac{\pi}{2}-t$ so as $x\to\dfrac{\pi}{2}$, $t\to0$.
$\cot x = \cot\!\left(\tfrac{\pi}{2}-t\right) = \tan t$, $\cos x = \sin t$, $\pi-2x = 2t$
$\lim_{t\to0}\dfrac{\tan t - \sin t}{8t^3} = \lim_{t\to0}\dfrac{\sin t(1-\cos t)}{8t^3\cos t}$
$= \lim_{t\to0}\dfrac{\sin t}{t}\cdot\dfrac{1-\cos t}{t^2}\cdot\dfrac{1}{8\cos t} = 1\cdot\dfrac{1}{2}\cdot\dfrac{1}{8} = \mathbf{\dfrac{1}{16}}$
Official answer is $\dfrac{1}{16}$ (index 1). Rechecking: $\dfrac{1}{2}\cdot\dfrac{1}{8}=\dfrac{1}{16}$.
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