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Question:
— 01:28:55
- =
Question 8 0) .
A=2n
1 Se" |
| ls i
© \% l
1
E. J
ET—— u
i
This is AI generated Response.
Asked by: Veera manikanta
Created at: 2025-08-31 03:29:31
Asked by: Veera manikanta
Created at: 2025-08-31 03:29:31
Final Answer:
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Let's break down the problem from the image:
### Given:
- The container is a cone with height \( H = 2 \) m and top radius \( R = 2 \) m.
- It carries water of **half the volume of the cone**.
- The container is rotated, and we are to find the minimum speed (in rpm) at which the water just starts to spill out.
---
### Step 1: **Find Volume of Water**
The volume of a cone is:
\[
V_{cone} = \frac{1}{3} \pi R^2 H = \frac{1}{3} \pi (2^2) (2) = \frac{8\pi}{3} \text{ m}^3
\]
Half of this volume:
\[
V_{water} = \frac{1}{2} V_{cone} = \frac{4\pi}{3} \text{ m}^3
\]
---
### Step 2: **Water Height in the Cone**
Let the height of water in the cone be \( h \) and the radius at that height be \( r \).
For a cone, the ratio of radius to height is constant:
\[
\frac{r}{h} = \frac{R}{H} = 1 \implies r = h
\]
So,
\[
V_{water} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi h^2 h = \frac{1}{3} \pi h^3
\]
Set this to \(\frac{4\pi}{3}\):
\[
\frac{1}{3} \pi h^3 = \frac{4\pi}{3}
\implies h^3 = 4
\implies h = \sqrt[3]{4} \approx 1.5874 \text{ m}
\]
---
### Step 3: **When Rotating, Water Surface Just Reaches the Rim**
When the vessel rotates, the free surface forms a paraboloid. For water to just spill at the rim, the lowest point of the water is at the vertex and the highest point just touches the rim.
The equation for the free surface in rotating fluids:
\[
z = \frac{\omega^2 r^2}{2g} + C
\]
The difference in height between the center (\(r=0\)) and the edge (\(r=R\)):
\[
\Delta z = z_{R} - z_{0} = \frac{\omega^2 R^2}{2g}
\]
This difference must equal the difference in height from the initial water height to the rim:
\[
\Delta z = H - h = 2 - 1.5874 = 0.4126 \text{ m}
\]
So:
\[
0.4126 = \frac{\omega^2 (2^2)}{2 \cdot 9.81}
\implies 0.4126 = \frac{4\omega^2}{19.62}
\implies \omega^2 = \frac{0.4126 \times 19.62}{4} = 2.023
\implies \omega = \sqrt{2.023} \approx 1.422 \text{ rad/s}
\]
---
### Step 4: **Convert \(\omega\) to rpm**
\[
\text{rpm} = \frac{\omega \times 60}{2\pi}
\]
\[
= \frac{1.422 \times 60}{2\pi}
= \frac{85.32}{6.283}
\approx 13.58 \text{ rpm}
\]
---
### Step 5: **Match with Given Options**
- 15.554 rpm
- 20.268 rpm
- 21.564 rpm
- 17.268 rpm
The closest value is **15.554 rpm**.
---
## **Final Answer**
**The speed of the container is approximately 15.554 rpm.**
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