Question

When a load is attached at the midpoint of a steel bar A of length \(80\,\text{cm}\), breadth \(2.5\,\text{cm}\) and thickness \(2\,\text{mm}\), it sags by \(1.2\,\text{mm}\). If the same load is attached at the midpoint of another steel bar B of length \(120\,\text{cm}\), breadth \(3\,\text{cm}\) and thickness \(3\,\text{mm}\), then the bar B sags by \[ (\text{In both the cases, the bars are supported at the two ends}) \]

• A. \(0.18\,\text{cm}\)
• B. \(0.10\,\text{cm}\)
• C. \(0.12\,\text{cm}\)
• D. \(0.15\,\text{cm}\)

Hint:

For a simply supported rectangular beam carrying a central load, \[ \boxed{ y\propto\frac{L^3}{bd^3}. } \] The sag increases rapidly with the length of the beam and decreases with the cube of its thickness.

Answer 1

The Correct Option is B

Step 1: Use the relation for sag of a simply supported beam. For a beam supported at both ends and loaded at its centre, \[ y\propto\frac{L^3}{bd^3}, \] where \[ L=\text{length}, \] \[ b=\text{breadth}, \] and \[ d=\text{thickness}. \] Hence, \[ \frac{y_2}{y_1} = \frac{L_2^3}{L_1^3} \cdot \frac{b_1}{b_2} \cdot \frac{d_1^3}{d_2^3}. \]

Step 2: Substitute the given values. Given, \[ L_1=80\,\text{cm}, \qquad L_2=120\,\text{cm}, \] \[ b_1=2.5\,\text{cm}, \qquad b_2=3\,\text{cm}, \] \[ d_1=2\,\text{mm}, \qquad d_2=3\,\text{mm}, \] and \[ y_1=1.2\,\text{mm}=0.12\,\text{cm}. \] Therefore, \[ y_2 = 0.12 \left(\frac{120}{80}\right)^3 \left(\frac{2.5}{3}\right) \left(\frac{2}{3}\right)^3. \]

Step 3: Simplify. Now, \[ \left(\frac{120}{80}\right)^3 = \left(\frac32\right)^3 = \frac{27}{8}, \] and \[ \left(\frac23\right)^3 = \frac{8}{27}. \] Hence, \[ y_2 = 0.12 \times \frac{27}{8} \times \frac56 \times \frac{8}{27} = 0.12\times\frac56 = 0.10\,\text{cm}. \] Thus, \[ \boxed{y_2=0.10\,\text{cm}.} \] Therefore, the correct option is \(\boxed{(B)}\).