JEE Physics 2025 Question with Solution

Given: Radius at point (1) is $r_1 = 2 \, \text{cm}$, radius at point (2) is $r_2 = 1 \, \text{cm}$, and entry velocity is $v_1 = 2 \, \text{m/s}$.

Find: Velocity of water at point (2), $v_2$.

For an incompressible fluid, the continuity equation applies:

$$A_1 v_1 = A_2 v_2$$

The cross-sectional area is

$$A = \pi r^2$$

So,

$$\pi r_1^2 v_1 = \pi r_2^2 v_2$$

Simplifying,

$$r_1^2 v_1 = r_2^2 v_2$$

Substituting $r_1 = 2 \, \text{cm}$, $r_2 = 1 \, \text{cm}$, and $v_1 = 2 \, \text{m/s}$:

$$(2^2)(2) = (1^2)(v_2)$$ $$8 = v_2$$

Thus, the velocity of water leaving point (2) is $8 \, \text{m/s}$.

Therefore, the correct option is C.

Given: The tube carries an incompressible fluid. The radii are $r_1 = 2 \, \text{cm}$ and $r_2 = 1 \, \text{cm}$, with $v_1 = 2 \, \text{m/s}$.

Find: The exit velocity $v_2$.

Step 1: Use the principle of continuity.

$$A_1 v_1 = A_2 v_2$$

where $A$ is cross-sectional area and $v$ is fluid speed.

Step 2: Express area in terms of radius.

$$A = \pi r^2$$

Hence,

$$\pi r_1^2 v_1 = \pi r_2^2 v_2$$

Cancelling $\pi$,

$$r_1^2 v_1 = r_2^2 v_2$$

Step 3: Substitute the given values.

$$(2)^2(2) = (1)^2 v_2$$ $$8 = v_2$$

Step 4: Final result.

$$v_2 = 8 \, \text{m/s}$$

Therefore, the speed of water leaving point (2) is $8 \, \text{m/s}$, so the correct option is C.