Equation of simple harmonic progressive wave is given by $y=\frac{1}{\sqrt{a}}\sin(\omega t)\pm \frac{1}{\sqrt{b}}\cos(\omega t)$. Then the resultant amplitude of the wave is ($\cos 90^{\circ}=0$):
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Logic Tip: Orthogonal wave components (sine and cosine) combine using the Pythagorean theorem for the resultant amplitude.
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 $\Delta V \propto V$.