JEE Physics 2023 Question with Solution

Given: A body moves with constant speed in a circle of radius $R = 10 \, \text{m}$ and time period $T = 4 \, \text{s}$.

Find: The displacement from the starting point at the end of $3 \, \text{s}$.

From the given figure above, speed is constant, radius is $R = 10 \, \text{m}$, and $T = 4 \, \text{s}$. At the end of **$3$**rd second, the particle will be at $D$ starting from $A$.

The displacement is the straight-line distance from $A$ to $D$.

Therefore, the displacement is $10\sqrt{2} \, \text{m}$. The correct option is D. The solution labels it as B, but the computed value matches option D in the given options.

Given: Radius of the circle is $R = 10 \, \text{m}$, speed is constant, and one complete revolution takes $4 \, \text{s}$.

Find: Displacement after $3 \, \text{s}$.

Total time of $4 \, \text{s}$ is evenly distributed over the circular path, so each quarter of the circle takes $1 \, \text{s}$.

• From $A$ to $B$: $1 \, \text{s}$
• From $B$ to $C$: $1 \, \text{s}$
• From $C$ to $D$: $1 \, \text{s}$
• From $D$ to $A$: $1 \, \text{s}$

So, at the end of the **$3$**rd second, the body is at $D$.

Now $AOD$ is a right-angled triangle with

Applying Pythagoras theorem,

Therefore, at the end of the **$3$**rd second, the displacement of the body from its starting point is $10\sqrt{2} \, \text{m}$.