QJEEM23Ph8e64: JEE Main 2023 Physics - Class 11 JEEM23Ph8e64 Solution: A body of mass moves along -axis such that it's velocity varies …

Given that the velocity of the body varies with displacement x according to the relation:

$v = 10\sqrt{x}\,\mathrm{ms}^{-1}$

To find the force acting on the body, we first need to find its acceleration, which can be obtained by differentiating the velocity with respect to time. However, we don't have the velocity expressed as a function of time, but rather as a function of displacement. To work around this, we will use the chain rule:

$\frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$

Now, differentiate the velocity with respect to displacement:

$\frac{dv}{dx} = \frac{1}{2} \cdot 10 \cdot x^{-1/2} = 5x^{-1/2}$

Recall that $\frac{dx}{dt}$ is the velocity, so we have:

$\frac{dv}{dt} = 5x^{-1/2} \cdot 10\sqrt{x} = 50$

Thus, the acceleration is constant and equal to 50 m/s². Now we can find the force acting on the body using Newton's second law:

$F = ma$

First, convert the mass from grams to kilograms:

$m = \frac{500\,\mathrm{g}}{1000} = 0.5\,\mathrm{kg}$

Now, calculate the force:

$F = (0.5\,\mathrm{kg})(50\,\mathrm{ms}^{-2}) = 25\,\mathrm{N}$

The force acting on the body is 25 N.