QJEEM22Pha4d2: JEE Main 2022 Physics - Class 11 JEEM22Pha4d2 Solution: A curved in a level road has a radius 75 m. The maximum speed…

For a car to move safely around a curved road without skidding, the centripetal force needed is provided by the frictional force between the tires and the road. The relationship governing this scenario can be expressed as:

$f_{\text{friction}} = f_{\text{centripetal}}$

The frictional force is given by:

$f_{\text{friction}} = \mu \cdot m \cdot g$

And the centripetal force is given by:

$f_{\text{centripetal}} = \frac{m \cdot v^2}{r}$

Where:

$\mu$ is the coefficient of friction,

$m$ is the mass of the car,

$g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$ ),

$v$ is the speed of the car,

$r$ is the radius of the curve.

Since the frictional force equals the centripetal force, we have:

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

Cancelling $m$ from both sides (assuming $m \neq 0$ ) gives:

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

Rewriting for $v$ gives the speed:

$v = \sqrt{\mu \cdot g \cdot r}$

For the first scenario with $r = 75 \, \text{m}$ and $v = 30 \, \text{m/s}$ :

$v_1^2 = \mu \cdot g \cdot r_1$

Thus,

$\mu \cdot g = \frac{v_1^2}{r_1} = \frac{30^2}{75} = \frac{900}{75} = 12$

For the second scenario with $r_2 = 48 \, \text{m}$ , the new speed $v_2$ is:

$v_2 = \sqrt{\mu \cdot g \cdot r_2} = \sqrt{12 \times 48}$

Calculating inside the square root:

$12 \times 48 = 576$

So,

$v_2 = \sqrt{576} = 24$

Hence, the maximum allowed speed for a curve with a radius of $48 \, \text{m}$ is 24 m/s.