Question

Two uniform brass rods A and B of length 'l' and '2l' and their radii '2r' and 'r' respectively are heated to same temperature. The ratio of the increase in the volume of rod A to that of rod B is

• A. 1:1
• B. 1:2
• C. 2:1
• D. 1:4

Hint:

Logic Tip: Increasing the radius has a greater impact on volume than increasing the length because the radius is squared in the volume formula.

Answer 1

The Correct Option is C

Concept:
When a solid body is heated, its increase in volume is given by: \[ \Delta V=\beta V \Delta T \] where:

• $\Delta V$ = increase in volume
• $\beta$ = coefficient of volumetric expansion
• $V$ = initial volume
• $\Delta T$ = rise in temperature

If two bodies are made of the same material and heated through the same temperature rise, then $\beta$ and $\Delta T$ remain same. Hence: \[ \Delta V \propto V \] So, increase in volume depends directly on initial volume.
Step 1: Find volume of cylinder A
For a cylinder: \[ V=\pi r^2 h \] Cylinder A has radius $2r$ and length $l$. \[ V_A=\pi (2r)^2 l \] \[ V_A=\pi (4r^2)l \] \[ V_A=4\pi r^2 l \]
Step 2: Find volume of cylinder B
Cylinder B has radius $r$ and length $2l$. \[ V_B=\pi r^2 (2l) \] \[ V_B=2\pi r^2 l \]
Step 3: Use volumetric expansion relation
Since both cylinders are heated through same temperature rise and made of same material: \[ \frac{\Delta V_A}{\Delta V_B}=\frac{V_A}{V_B} \] Substitute values: \[ \frac{\Delta V_A}{\Delta V_B}= \frac{4\pi r^2 l}{2\pi r^2 l} \] Cancel common terms: \[ \frac{\Delta V_A}{\Delta V_B}=2 \]
Step 4: Final Answer
The ratio of increase in volume of cylinders A and B is: \[ \boxed{2:1} \] Quick Tip:
For same material and same heating, compare only original volumes because $\Delta V \propto V$.