JEE Physics 2023 Question with Solution

Given: radius of curved road = $50 \, \text{m}$, coefficient of friction = $\mu = 0.34$, acceleration due to gravity = $g = 10 \, \text{m/s}^2$.

Find: the approximate maximum speed of the car.

For motion on a horizontal curved road, the maximum safe speed is given by

$$v_{\text{max}} = \sqrt{\mu g r}$$

Substituting the given values,

$$v_{\text{max}} = \sqrt{0.34 \times 10 \times 50}$$ $$v_{\text{max}} = \sqrt{170}$$ $$v_{\text{max}} \approx 13 \, \text{m/s}$$

Therefore, the maximum speed is $13 \, \text{m/s}$. The solution states that the correct option is B, but the computed value matches option C.

Given: friction provides the necessary centripetal force on the curved road.

Using

$$\mu mg = \frac{mv^2}{r}$$

Canceling $m$,

$$\mu g = \frac{v^2}{r}$$

So,

$$v^2 = \mu g r$$ $$v = \sqrt{\mu g r}$$

Now substitute $\mu = 0.34$, $g = 10 \, \text{m/s}^2$, and $r = 50 \, \text{m}$:

$$v = \sqrt{0.34 \times 10 \times 50} = \sqrt{170}$$

Since $\sqrt{169} = 13$, we get

$$v \approx 13 \, \text{m/s}$$

Hence the defensible correct option from the listed choices is C.