Q. The human body can safely stand an acceleration $9$ times that due to gravity which is $10 \,m / s ^{2}$. The minimum radius of curvature with which a pilot may safely turn a plane vertically upward at the end of a dive, when the plane's speed is $720 \, km / hr$ is:

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A

$500 \,m$

B

$612 \, m$

C

$475 \, m$

D

$323 \,m$

Solution:

Speed of plane $=720\, km / h =200\, m / s$

$a=\frac{v^{2}}{r}$
$r=\frac{v^{2}}{a}=\frac{(200)^{2}}{9 g-g}=\frac{40000}{8 \times 10}$
$r=500\, m$

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