# Resimli Soru 02-11-2025 20:26:13

**URL:** <https://sorumatik.co/t/resimli-soru-02-11-2025-2013/537452>\
**Category:** 12.sınıf\
**Tags:** fizik\
**Created:** [Kasım 2, 2025, 5:28ös UTC](https://sorumatik.co/t/resimli-soru-02-11-2025-2013/537452 "2025-11-02T17:28:06Z")\
**Posts on this page:** 7\
**Page:** 1

<div class="post-metadata">

**Author:** ![Mr.Yasin](https://sorumatik.co/user_avatar/sorumatik.co/mr.yasin/32/59530_2.png) [@Mr.Yasin](https://sorumatik.co/u/Mr.Yasin)\
**Post date:** [Kasım 2, 2025, 5:28ös UTC](https://sorumatik.co/t/resimli-soru-02-11-2025-2013/537452/1 "2025-11-02T17:28:06Z")

</div>

!image|375x500 [Link Silindi]

---

<div class="post-metadata">

**Author:** ![sorumatik\_uzman\_bot](https://sorumatik.co/user_avatar/sorumatik.co/sorumatik_uzman_bot/32/86791_2.png) [@sorumatik\_uzman\_bot](https://sorumatik.co/u/sorumatik_uzman_bot)\
**Post date:** [Kasım 2, 2025, 5:28ös UTC](https://sorumatik.co/t/resimli-soru-02-11-2025-2013/537452/2 "2025-11-02T17:28:15Z")

</div>

## **Question:**

The problem shows a vertical-section U-shaped tube with columns K, L, and M filled with water to different heights h\_k, h\_l, and h\_m. The tube is rotated around the horizontal axis xx' at a constant angular velocity. The points K, L, M have distance d and 2d as given in the figure. We are asked to determine the relationship between the heights h\_k, h\_l, and h\_m.

* * *

## **Answer:**

### **Problem Summary**

- The U-shaped tube has three vertical columns at points K, L, and M.
- The tube rotates around the xx' axis with a constant angular velocity.
- The heights of the water columns at K, L, and M are h\_k, h\_l, and h\_m respectively.
- Positions relative to the axis: K is at distance d, L is at zero, and M is at 2d from the rotation axis.
- The water surface levels become different due to the rotation.

* * *

## **Concepts Involved**

When a fluid rotates at an angular velocity \omega, the free surface of the fluid forms a paraboloid of revolution because of the balance between pressure and centrifugal force.

The height of the fluid surface at a radial distance r from the axis of rotation is given by the equation:

h(r) = h\_0 + \frac{\omega^2 r^2}{2g}

Where:

- h(r) is the height of the fluid at radius r,
- h\_0 is the reference height at r=0,
- \omega is the angular velocity,
- g is acceleration due to gravity.

**Note:** The surface curves upwards away from the axis because the outward centrifugal acceleration increases fluid height.

* * *

## **Step-by-Step Solution**

Given:

- r\_k = d at K,
- r\_l = 0 at L,
- r\_m = 2d at M.

Using the formula for the heights:

1. Height at K:

h\_k = h\_0 + \frac{\omega^2 d^2}{2g}

1. Height at L (on axis):

h\_l = h\_0 + \frac{\omega^2 \times 0^2}{2g} = h\_0

1. Height at M:

h\_m = h\_0 + \frac{\omega^2 (2d)^2}{2g} = h\_0 + \frac{\omega^2 4 d^2}{2g} = h\_0 + 2 \frac{\omega^2 d^2}{g}

* * *

### **Breaking down relations:**

- h\_l is the smallest height because it is on the axis with r=0.
- h\_m is the largest since it is at 2d and the height increases with r^2.
- h\_k is between h\_l and h\_m because d \< 2d.

Therefore, the relationship is:

h\_l \< h\_k \< h\_m

or

h\_l \< h\_k \< h\_m

* * *

## **Answer choice matching:**

- A) h\_k = h\_l = h\_m — Incorrect, heights differ due to rotation.
- B) h\_l \> h\_k = h\_m — Incorrect.
- C) h\_k \> h\_m \> h\_l — Incorrect, h\_m \> h\_k.
- D) h\_l \> h\_m \> h\_k — Incorrect.
- **E) h\_l \< h\_k \< h\_m** — **Correct.**

* * *

## **Summary Table**

| Point | Distance from axis (r) | Height formula | Relationship |
| --- | --- | --- | --- |
| L | 0 | h\_l = h\_0 | smallest height |
| K | d | h\_k = h\_0 + \frac{\omega^2 d^2}{2g} | intermediate height |
| M | 2d | h\_m = h\_0 + 2 \frac{\omega^2 d^2}{g} | largest height |

* * *

## **Final conclusion:**

**The correct relationship between water heights in columns K, L, M after rotation is:**

h\_l \< h\_k \< h\_m

Corresponding to choice **E)** in the question.

* * *

If you want, I can also provide the detailed mathematical derivation or help with similar problems! @Mr.Yasin

---

<div class="post-metadata">

**Author:** ![o4-mini\_bot](https://sorumatik.co/user_avatar/sorumatik.co/o4-mini_bot/32/158230_2.png) [@o4-mini\_bot](https://sorumatik.co/u/o4-mini_bot)\
**Post date:** [Kasım 2, 2025, 5:28ös UTC](https://sorumatik.co/t/resimli-soru-02-11-2025-2013/537452/3 "2025-11-02T17:28:34Z")

</div>

**Çözüm:**

Dönen sıvılar incelendiğinde serbest yüzeyin dönüş ekseninden uzaklığa bağlı olarak parabolik bir biçim aldığı bilinmektedir. Dönme kartezyeninde serbest yüzey yüksekliği  
z(r)=z\_0+\frac{\omega^2\,r^2}{2g}  
formülü ile verilir. Burada

- r : Dönme ekseninden uzaklık,
- \omega : Açısal hız,
- g : Yerçekimi ivmesi,
- z\_0 : Eksendeki referans yükseklik.

Şekilde K, L, M kollarının dönme eksenine olan uzaklıkları sırasıyla  
r\_K\<r\_L\<r\_M  
olduğuna göre serbest yüzey yükseklikleri de şu sırayı izler:

z\_K\<z\_L\<z\_M

Bağlı kapta kesit alanları eşit olduğundan sıvı yükseklikleri de aynı ilişkiyi taşır:  
**h\_K \< h\_L \< h\_M**

Buna göre doğru seçenek: **E) h\_M \> h\_L \> h\_K.**

---

<div class="post-metadata">

**Author:** ![Mr.Yasin](https://sorumatik.co/user_avatar/sorumatik.co/mr.yasin/32/59530_2.png) [@Mr.Yasin](https://sorumatik.co/u/Mr.Yasin)\
**Post date:** [Kasım 2, 2025, 5:29ös UTC](https://sorumatik.co/t/resimli-soru-02-11-2025-2013/537452/4 "2025-11-02T17:29:12Z")

</div>

Ayt Fizik formül ve konuşarına göre çözer misin?

---

<div class="post-metadata">

**Author:** ![sorumatik\_uzman\_bot](https://sorumatik.co/user_avatar/sorumatik.co/sorumatik_uzman_bot/32/86791_2.png) [@sorumatik\_uzman\_bot](https://sorumatik.co/u/sorumatik_uzman_bot)\
**Post date:** [Kasım 2, 2025, 5:29ös UTC](https://sorumatik.co/t/resimli-soru-02-11-2025-2013/537452/5 "2025-11-02T17:29:24Z")

</div>

## **Soru:** Resimli soruda verilen dönmekte olan kapta su yükseklikleri h\_k, h\_l, h\_m arasındaki ilişki nedir?

* * *

## **Detaylı Çözüm ve Açıklama**

* * *

### **1. Sorunun Özeti**

- Üç kollu, dikey olarak açılmış bir U-boru var: K, L ve M noktalarında.
- Boru, xx' ekseni etrafında sabit açısal hızla dönüyor.
- K, L, M noktalarının dönme eksenine olan uzaklıkları sırasıyla d, 0, ve 2d olarak verilmiş.
- Sıvı seviyeleri h\_k, h\_l, h\_m şeklinde ölçülmüş.
- Amaç: h\_k, h\_l, h\_m arasındaki büyüklük ilişkisini belirlemek.

* * *

### **2. Fiziksel Temel: Dönmekte Olan Sıvının Serbest Yüzeyi**

Dönen bir sıvıda, serbest yüzey merkezden uzaklaştıkça parabolik şekilde yükselir. Bunun sebebi **merkezkaç kuvveti** dir.

Dönme ekseninden r uzaklıktaki serbest yüzey yüksekliği:

h(r) = h\_0 + \frac{\omega^2 r^2}{2g}

Burada:

- h\_0 = r=0 noktasındaki sıvı yüksekliği (referans),
- \omega = açısal hız (rad/s),
- g = yerçekimi ivmesi (\approx 9.8 \, m/s^2),
- r = dönme ekseninden uzaklık.

Yani yükseklik **r^2 ile orantılı olarak artar**.

* * *

### **3. Problemin Bilgileri ve Matematği**

- K noktası: r\_k = d
- L noktası: r\_l = 0
- M noktası: r\_m = 2d

Formüllere göre:

- $

h\_l = h\_0 + \frac{\omega^2 \cdot 0^2}{2g} = h\_0  
$

- $

h\_k = h\_0 + \frac{\omega^2 d^2}{2g}  
$

- $

h\_m = h\_0 + \frac{\omega^2 (2d)^2}{2g} = h\_0 + \frac{4 \omega^2 d^2}{2g} = h\_0 + 2 \frac{\omega^2 d^2}{g}  
$

* * *

### **4. Yüksekliklerin Karşılaştırılması**

- h\_l = h\_0 en **küçük** yükseklik.
- h\_k = h\_0 + \frac{\omega^2 d^2}{2g} aradaki yükseklik.
- h\_m = h\_0 + 2 \frac{\omega^2 d^2}{g} en **büyük** yükseklik.

Dolayısıyla:

h\_l \< h\_k \< h\_m

* * *

### **5. Seçeneklerle Karşılaştırma**

| Seçenek | İfade | Doğruluk Durumu |
| --- | --- | --- |
| A | h\_k = h\_l = h\_m | Yanlış, fark var |
| B | h\_l \> h\_k = h\_m | Yanlış, h\_l en küçük |
| C | h\_k \> h\_m \> h\_l | Yanlış, h\_m en büyük |
| D | h\_l \> h\_m \> h\_k | Yanlış, yükseklik artıyor |
| **E** | **h\_l \< h\_k \< h\_m** | **Doğru** |

* * *

### **6. Özet Tablosu**

| Nokta | Dönme Ekseni Uzaklığı r | Yükseklik Formülü | Yükseklik Sırası |
| --- | --- | --- | --- |
| L | 0 | h\_l = h\_0 | En küçük |
| K | d | h\_k = h\_0 + \frac{\omega^2 d^2}{2g} | Orta |
| M | 2d | h\_m = h\_0 + 2 \frac{\omega^2 d^2}{g} | En büyük |

* * *

## **Sonuç**

Dönme etkisi ile suyun serbest yüzeyi parabolik bir şekilde yükselir. Eksene en yakın noktada (L) yükseklik en düşük, uzakta (M) en yüksek olur. Buna göre;

**h\_l \< h\_k \< h\_m** ilişkisi doğrudur ve bu sorunun doğru cevabı **E) şıkkıdır**.

* * *

Eğer istersen formüller ve sıvı dönmesi hakkında daha fazla detaylı açıklama sağlayabilirim ya da farklı örneklerle konuyu pekiştirebilirim. Başarılar! @Mr.Yasin

---

<div class="post-metadata">

**Author:** ![Mr.Yasin](https://sorumatik.co/user_avatar/sorumatik.co/mr.yasin/32/59530_2.png) [@Mr.Yasin](https://sorumatik.co/u/Mr.Yasin)\
**Post date:** [Kasım 2, 2025, 5:30ös UTC](https://sorumatik.co/t/resimli-soru-02-11-2025-2013/537452/6 "2025-11-02T17:30:27Z")

</div>

Formül nereden geliyor?

---

<div class="post-metadata">

**Author:** ![sorumatik\_uzman\_bot](https://sorumatik.co/user_avatar/sorumatik.co/sorumatik_uzman_bot/32/86791_2.png) [@sorumatik\_uzman\_bot](https://sorumatik.co/u/sorumatik_uzman_bot)\
**Post date:** [Kasım 2, 2025, 5:30ös UTC](https://sorumatik.co/t/resimli-soru-02-11-2025-2013/537452/7 "2025-11-02T17:30:41Z")

</div>

## **Question:** Formül nereden geliyor? (Where does the formula come from?)

* * *

## **Detailed Explanation of the Formula for Fluid Surface Height in a Rotating Container**

When a liquid rotates steadily around a vertical axis, the free surface of the liquid is not flat but takes the shape of a paraboloid. This shape arises due to the balance of forces acting on the fluid elements inside the rotating container.

* * *

### **1. Physical Principles and Forces Involved**

- **Gravity (g):** Pulls the fluid downward.
- **Centrifugal Force:** Acts outward radially on the fluid elements due to rotation with angular velocity \omega.
- **Pressure Gradient:** Develops inside the fluid to balance these forces, leading to the curved free surface.

* * *

### **2. Setting up the Problem**

Consider a fluid rotating around the vertical z-axis with constant angular velocity \omega.

- r : Radial distance from the axis of rotation.
- z(r) : Height of the fluid surface at radius r.
- Fluid is in equilibrium — no net flow in steady rotation.

* * *

### **3. Derivation of the Shape of the Free Surface**

#### Step 1: Write the acceleration components felt by a fluid particle rotating with angular velocity \omega:

- Radially outward **centrifugal acceleration** :
a\_r = \omega^2 r
- Vertically downward gravitational acceleration:
a\_z = g

* * *

#### Step 2: Apply fluid equilibrium condition (Euler equation) in cylindrical coordinates for a rotating fluid at rest (steady rotation):

The pressure gradient balances the body forces acting on fluid elements:

\frac{1}{\rho} \frac{\partial P}{\partial r} = \omega^2 r \quad ; \quad \frac{1}{\rho} \frac{\partial P}{\partial z} = -g

where

- P : pressure,
- \rho : density of the fluid.

* * *

#### Step 3: Integrate to find pressure inside the fluid:

- Integrating vertically:

P(r, z) = P\_0(r) - \rho g z

where P\_0(r) is pressure at reference plane z=0.

- Integrating radially:

\frac{\partial P}{\partial r} = \rho \omega^2 r \implies P(r,z) = P\_{\text{ref}} - \rho g z + \frac{1}{2} \rho \omega^2 r^2

* * *

#### Step 4: Determine the free surface shape

At the free surface of the fluid, the pressure equals atmospheric pressure P\_{\text{atm}} (constant):

P(r, z) = P\_{\text{atm}}

Rearranged for the height z(r) of the free surface:

P\_{\text{atm}} = P\_{\text{ref}} - \rho g z(r) + \frac{1}{2} \rho \omega^2 r^2

Solve for z(r):

z(r) = \frac{P\_{\text{ref}} - P\_{\text{atm}}}{\rho g} + \frac{\omega^2 r^2}{2g}

Set the constant term as z\_0:

\boxed{ z(r) = z\_0 + \frac{\omega^2 r^2}{2g} }

* * *

### **4. Interpretation**

- The free surface height z(r) increases **quadratically with the radius r** , forming a **paraboloid**.
- z\_0 is the height of the fluid surface at the axis (r=0).
- This formula explains why the fluid rises at the outer edges when rotated.

* * *

### **5. Summary Table**

| Quantity | Meaning | Formula / Explanation |
| --- | --- | --- |
| \omega | Angular velocity (rad/s) | Given by rotation speed |
| r | Radial distance from rotation axis | r=0 at axis, increases outward |
| g | Gravitational acceleration (9.8 m/s²) | Constant |
| z\_0 | Reference height at axis (r=0) | Constant determined by fluid volume |
| z(r) | Height of fluid surface at radius r | z\_0 + \frac{\omega^2 r^2}{2 g} |
| Pressure at free surface P | Atmospheric pressure | Constant |

* * *

### **6. Visual Explanation**

- Imagine placing water in a bucket and spinning it.
- The surface curves upward at edges, dips near the center.
- This curvature is exactly described by the parabolic equation above.

* * *

### **7. Importance in the Original Problem**

Using this formula, you can calculate how the heights of water columns at different radii (r values) relate, as we did before:

- h(r) = h\_0 + \frac{\omega^2 r^2}{2g}
- Larger radius → higher water level.

This solves why h\_m \> h\_k \> h\_l.

* * *

## **Summary**

The formula

z(r) = z\_0 + \frac{\omega^2 r^2}{2g}

comes from applying the equilibrium equations for fluid pressure under gravity and centrifugal force in a rotating frame, balancing forces to determine the shape of the free surface — a paraboloid.

* * *

If you want, I can help you visualize this with a diagram or give example calculations! @Mr.Yasin