JEE MAIN - Physics (2023 - 29th January Morning Shift - No. 10)
A car is moving on a horizontal curved road with radius 50 m. The approximate maximum speed of car will be, if friction between tyres and road is 0.34. [take g = 10 ms$$^{-2}$$]
3.4 ms$$^{-1}$$
13 ms$$^{-1}$$
22.4 ms$$^{-1}$$
17 ms$$^{-1}$$
Explanation
$f_{s}=\frac{m v^{2}}{r}$
For maximum speed in safe turning,
$\mathrm{f}_{\mathrm{s}}=\mathrm{f}_{\mathrm{s}} \max =\mu \mathrm{mg}$
$\mathrm{v}_{\max }$ (for safe turning) $=\sqrt{\mu \mathrm{rg}}$
$=\sqrt{0.34 \times 50 \times 10} \approx 13 \mathrm{~m} / \mathrm{s}$
For maximum speed in safe turning,
$\mathrm{f}_{\mathrm{s}}=\mathrm{f}_{\mathrm{s}} \max =\mu \mathrm{mg}$
$\mathrm{v}_{\max }$ (for safe turning) $=\sqrt{\mu \mathrm{rg}}$
$=\sqrt{0.34 \times 50 \times 10} \approx 13 \mathrm{~m} / \mathrm{s}$
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