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Question:
Solve correct option
= ea
=
= >
=
EE = = —————
_—----
_—
= ——
=
-_-
= ——— — —— — — =
Ee =
-——
= — ~~ =
_ ————
= — — —— —
ss = =
= = — =
== — — — ==
——— 1 I= = — —— ——
—=—=——— -— =a
ra es — oan
J cylnanea messing nsuments being designed to tu the Geng he is proposed tht th instrument shoud fost ith some part above the weer & hGH
— —
efgueteon - -—— — =
— - = -—-— —— — =
ss = - —
= aa — — — — — — ===
— = = =
= — -— — — = =
= == —————— — —
= == @ 9 — 9 —
EE — -.s —
diay —— = == ——
===) — =
EF =
Conse ge certo te cdl sumer tel 20 bo mand height 15 m deine he minum metro he onic sument HE
Boston ns sie equim (Dens of water S000 g/m) = =
- = _—
oo Eoneatiri Sa EE =
This is AI generated Response.
Asked by: Sachin hanjagi
Created at: 2025-10-11 11:27:29
Asked by: Sachin hanjagi
Created at: 2025-10-11 11:27:29
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## Problem Restatement
A cylindrical instrument (density = 920 kg/m³, height \( H = 1.5 \) m) is to float upright in water (density = 100 kg/m³). You are asked **the minimum diameter \( D \)** for stable equilibrium.
### Options:
- 1.152 m
- 1.023 m
- .321 m
- .962 m
---
## Step 1: Floating Condition (Buoyancy)
Let \( D \) be the diameter, \( H \) the height, and \( h \) the submerged height.
**Volume of cylinder:**
\( V = \frac{\pi}{4} D^2 H \)
**Weight of cylinder:**
\( W = \rho_c V g = 920 \cdot \frac{\pi}{4} D^2 H g \)
**Buoyant force (equals weight of displaced water):**
\( F_B = \rho_w \cdot (\frac{\pi}{4} D^2 h) g = 100 \cdot \frac{\pi}{4} D^2 h g \)
Set \( F_B = W \):
\[
920 \cdot \frac{\pi}{4} D^2 H = 100 \cdot \frac{\pi}{4} D^2 h
\]
\[
920 H = 100 h
\]
\[
h = .92H
\]
\[
h = .92 \times 1.5 = 1.38 \text{ m}
\]
---
## Step 2: Stability Condition (Metacentric Height)
For stable floating, the **metacentric height** (\( GM \)) must be positive:
\[
GM = BM - BG >
\]
Where:
- \( BM = \frac{I}{V_{disp}} \), \( I = \) second moment of area \( = \frac{\pi}{64} D^4 \)
- \( V_{disp} = \frac{\pi}{4} D^2 h \)
- \( BG = \) distance from center of buoyancy to center of gravity \( = \frac{H - h}{2} \)
So,
\[
BM = \frac{\frac{\pi}{64} D^4}{\frac{\pi}{4} D^2 h} = \frac{D^2}{16 h}
\]
\[
BG = \frac{H - h}{2} = \frac{1.5 - 1.38}{2} = .06 \text{ m}
\]
For stability: \( BM > BG \)
\[
\frac{D^2}{16h} > .06
\]
\[
D^2 > .06 \times 16h = .96 \times 1.38 = 1.3248
\]
\[
D > \sqrt{1.3248} \approx 1.151 \text{ m}
\]
---
## Step 3: Choose Correct Option
Closest to our answer: **1.152 m**
---
## **Final Answer**
\[
\boxed{1.152 \text{ m}}
\]
This is the minimum diameter for stable equilibrium.
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