JEE MAIN - Physics (2002 - No. 66)
The minimum velocity (in $$m{s^{ - 1}}$$) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction $$0.6$$ to avoid skidding is
$$60$$
$$30$$
$$15$$
$$25$$
Обяснение
For no skidding along curved track,
The maximum velocity possible $${v_{\max }} = \sqrt {\mu rg} $$
Here $$\mu = 0.6,\,r = 150m,\,g = 9.8$$
$$\therefore$$ $${v_{\max }} = \sqrt {0.6 \times 150 \times 9.8} \simeq 30m/s$$
The maximum velocity possible $${v_{\max }} = \sqrt {\mu rg} $$
Here $$\mu = 0.6,\,r = 150m,\,g = 9.8$$
$$\therefore$$ $${v_{\max }} = \sqrt {0.6 \times 150 \times 9.8} \simeq 30m/s$$