A rod with circular cross-section of radius 5 mm is stretched such that its cross-section remains circular. If after stretching the rod its length becomes four times the original length, then what is the radius of the new cross-section ?
Let the original length of the rod be $L_1$.
The original volume of the rod (assuming it’s a cylinder) is $V_1 = \pi r_1^2 L_1 = \pi (5)^2 L_1 = 25\pi L_1$.
The rod is stretched such that its cross-section remains circular. This implies the stretched rod is still a cylinder.
After stretching, the new length becomes four times the original length: $L_2 = 4 L_1$.
Let the new radius of the circular cross-section be $r_2$.
The new volume of the rod is $V_2 = \pi r_2^2 L_2 = \pi r_2^2 (4L_1)$.
Assuming the material of the rod is incompressible (a common assumption in such problems unless otherwise stated, meaning the density remains constant), the volume of the rod remains constant during stretching.
Therefore, $V_1 = V_2$.
$25\pi L_1 = \pi r_2^2 (4L_1)$.
We can cancel $\pi$ and $L_1$ from both sides (assuming $L_1 > 0$):
$25 = 4 r_2^2$.
Solve for $r_2^2$:
$r_2^2 = \frac{25}{4}$.
Solve for $r_2$:
$r_2 = \sqrt{\frac{25}{4}} = \frac{\sqrt{25}}{\sqrt{4}} = \frac{5}{2}$.
$r_2 = 2.5$ mm.
The radius of the new cross-section is 2.5 mm.
– The shape of the rod is a cylinder, so its volume is given by $\pi r^2 L$.
– Set the initial volume equal to the final volume and solve for the unknown radius.