Physics Q29 – JEE-MAIN 2025

Home Physics Rotational MotionQuestion #452

A rod of length

5 L

is bent right angle keeping one side length as

2 L

. The position of the centre of mass of the system: (Consider

L = 10

cm)

Clue 1 of 3: Divide the System

Treat the bent rod as two separate uniform rod segments, each with its own mass and center of mass.

🥷

Ninja Strategy•X-coordinate Elimination

Calculate only the x-coordinate of the center of mass, $X_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} = \frac{(λ \cdot 2 L) \cdot L + (λ \cdot 3 L) \cdot 0}{5 λ L} = \frac{2 L}{5}$ . With $L = 10$ cm, $X_{C M} = 4$ cm. Only one option matches this x-coordinate.

🧠 Full Solution Path

Step 1: Identify Segments and Their Properties✦ Active

The rod of total length $5 L$ is bent into two segments at a right angle. One segment has length $2 L$ , so the other segment must have length $5 L - 2 L = 3 L$ . Let the bent corner be at the origin $(0, 0)$ . The horizontal segment (length $2 L$ ) lies along the x-axis, and the vertical segment (length $3 L$ ) lies along the y-axis.

Assuming uniform linear mass density $λ$ :

m_{1} = λ (2 L) (mass of horizontal segment)

C M_{1} = (L, 0) (center of mass of horizontal segment)

m_{2} = λ (3 L) (mass of vertical segment)

C M_{2} = (0, \frac{3 L}{2}) (center of mass of vertical segment)

Step 2: Calculate the Overall Center of Mass○ Expand

The coordinates of the center of mass of the composite system are given by:

X_{C M} = \frac{m_{1} x_{1} + m_{2} x_{2}}{m_{1} + m_{2}} = \frac{(λ \cdot 2 L) \cdot L + (λ \cdot 3 L) \cdot 0}{λ \cdot 2 L + λ \cdot 3 L} = \frac{2 λ L^{2}}{5 λ L} = \frac{2 L}{5}

Y_{C M} = \frac{m_{1} y_{1} + m_{2} y_{2}}{m_{1} + m_{2}} = \frac{(λ \cdot 2 L) \cdot 0 + (λ \cdot 3 L) \cdot \frac{3 L}{2}}{λ \cdot 2 L + λ \cdot 3 L} = \frac{\frac{9}{2} λ L^{2}}{5 λ L} = \frac{9 L}{10}

💡 Teacher's Secret Hint

Ensure correct substitution of individual masses and CM coordinates into the formula.

Step 3: Substitute the Value of L and Determine the Position Vector○ Expand

Given $L = 10$ cm, substitute this value into the $X_{C M}$ and $Y_{C M}$ expressions:

X_{C M} = \frac{2 (10 cm)}{5} = 4 cm

Y_{C M} = \frac{9 (10 cm)}{10} = 9 cm

The position of the center of mass is $4 \hat{i} + 9 \hat{j}$ cm.

💡 Teacher's Secret Hint

Always double-check units and final vector notation.

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Physics Question 29 – JEE-MAIN 2025

Clue 1 of 3: Divide the System

🧠 Full Solution Path