JEE Physics 2025 Question with Solution

Given: Mass of the object is $500 \, \text{g}$ and speed is $v = 4\sqrt{x}$.

Find: The force acting on the object.

When velocity is given as a function of position, use

$$a = v\frac{dv}{dx}$$

First convert mass into SI units:

$$m = 0.5 \, \text{kg}$$

Now differentiate the given velocity:

$$\frac{dv}{dx} = \frac{d}{dx}(4\sqrt{x}) = \frac{4}{2\sqrt{x}} = \frac{2}{\sqrt{x}}$$

Then acceleration is

$$a = \frac{dv}{dt} = \frac{dv}{dx}\cdot\frac{dx}{dt} = \frac{dv}{dx}\cdot v = \left(\frac{2}{\sqrt{x}}\right)(4\sqrt{x}) = 8$$

So,

$$a = 8 \, \text{m/s}^2$$

Using Newton's second law,

$$F = ma = 0.5 \times 8 = 4 \, \text{N}$$

Therefore, the force acting on the object is $4 \, \text{N}$. The correct option is D.

Given: $m = 500 \, \text{g}$ and $v = 4\sqrt{x}$.

Find: Force on the object.

1. Convert mass to SI unit:

$$m = 500 \, \text{g} = 0.5 \, \text{kg}$$

2. Use the chain rule for acceleration:

$$a = \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}$$

3. Differentiate velocity with respect to position:

$$\frac{dv}{dx} = \frac{d}{dx}(4\sqrt{x}) = 4\cdot\frac{1}{2\sqrt{x}} = \frac{2}{\sqrt{x}}$$

4. Substitute in the acceleration expression:

$$a = (4\sqrt{x})\left(\frac{2}{\sqrt{x}}\right) = 8 \, \text{m/s}^2$$

The acceleration is constant along the x-axis.

5. Apply Newton's second law:

$$F = m\cdot a = (0.5 \, \text{kg})(8 \, \text{m/s}^2) = 4 \, \text{N}$$

Therefore, the force acting on the object is $4 \, \text{N}$.