A vertical spring supports a cylindrical object with an outer diameter of $10 \text{ cm}$ , a height of $20 \text{ cm}$ , and a density of $8000 \text{ kg m}^{-3}$ . The object rests half dipped in water in its equilibrium state, as shown in the figure. Determine the elongation of the spring. The spring constant is $500 \text{ N m}^{-1}$ .

Question

A vertical spring supports a cylindrical object with an outer diameter of

10 \text{ cm}

, a height of

20 \text{ cm}

, and a density of

8000 \text{ kg m}^{-3}

. The object rests half dipped in water in its equilibrium state, as shown in the figure. Determine the elongation of the spring. The spring constant is

500 \text{ N m}^{-1}

.

A vertical spring supports a cylindrical object with an outer diameter of 10 cm , a height of 20 cm , and a density of

Quantrex Academy · Accepted Answer

Correct answer: D. 23.5 cm

Volume of the cylinder, V = π r^2 h = π ((0.1)/(2))^2 (0.2) = 0.0005 π m^3 Mass of the cylinder, m = ho V = 8000 × 0.0005 π = 4 π kg Weight of the cylinder, W = mg = 4 π × 10 = 40 π N Since the cylinder is half dipped in water, the buoyant force is: F_B = (V)/(2) ho_w g = (0.0005 π)/(2) × 1000 × 10 = 2.5 π N At equilibrium, the upward forces balance the downward weight: kx + F_B = W 500 x + 2.5 π = 40 π 500 x = 37.5 π x = (37.5 π)/(500) m = 0.075 π m x = 7.5 π cm 7.5 × 3.14 cm = 23.55 cm 23.5 cm Answer: 23.5 cm

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