JEE MAIN - Physics (2024 - 27th January Morning Shift - No. 6)
A train is moving with a speed of $$12 \mathrm{~m} / \mathrm{s}$$ on rails which are $$1.5 \mathrm{~m}$$ apart. To negotiate a curve radius $$400 \mathrm{~m}$$, the height by which the outer rail should be raised with respect to the inner rail is (Given, $$g=10 \mathrm{~m} / \mathrm{s}^2)$$ :
6.0 cm
5.4 cm
4.8 cm
4.2 cm
Explanation
$$\tan \theta=\frac{v^2}{R g}=\frac{12 \times 12}{10 \times 400}$$
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$$\begin{aligned} & \tan \theta=\frac{\mathrm{h}}{1.5} \\\\ & \Rightarrow \frac{\mathrm{h}}{1.5}=\frac{144}{4000} \\\\ & \mathrm{~h}=5.4 \mathrm{~cm} \end{aligned}$$
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