NEET 2022 Physics Graphs MCQ Question

Chapter: Motion in a Straight Line

Class: 11 Physics Topic: Displacement–Time Graph Difficulty: 🟡 Medium

✅ Correct Answer: B — $1:\sqrt3$

In a displacement–time graph, the slope gives velocity:

$$v=\frac{\Delta x}{\Delta t}=\tan\theta$$

For the two particles:

$$v_1=\tan30^\circ=\frac{1}{\sqrt3}$$ $$v_2=\tan45^\circ=1$$

Therefore,

$$v_1:v_2=\frac{1}{\sqrt3}:1$$

Multiplying by $\sqrt3$:

$$\boxed{v_1:v_2=1:\sqrt3}$$

❌ Why other options are wrong?

• A: $1:2$ ❌ $\tan30^\circ eq \frac12$
• B: $1:\sqrt3$ ✅ Correct
• C: $\sqrt3:1$ ❌ This is the inverse ratio.
• D: $1:1$ ❌ Different slopes mean different velocities.

📌 NCERT Concept

The slope of the position/displacement–time graph represents velocity. Therefore, the angle made with the time axis directly determines velocity through $v=\tan\theta$ (with appropriate graph-axis scales).

🧠 NEET Trick

For an $x-t$ graph:

$$\boxed{\text{Steeper line} \Rightarrow \text{greater velocity}}$$

Here:

$$\tan30^\circ:\tan45^\circ =\frac1{\sqrt3}:1 =\boxed{1:\sqrt3}$$

So the screenshot's marked A is incorrect, and D ($1:1$) is also incorrect. Correct option = B.