# Hydrometer Construction and Calculations ## Given Data - **Weight (empty):** \( 9 \) gf - **Cylinder 1:** Diameter \( d_1 = 6 \) mm, Length \( L_1 = 100 \) mm - **Cylinder 2:** Diameter \( d_2 = 24 \) mm, Length \( L_2 = 40 \) mm - **Total length:** \( 100 + 40 = 140 \) mm - **Immersion in water:** \( h = 71 \) mm - **Scale range:** \( \pm 30 \) mm from the water immersion point (i.e., total range \( 71 - 30 = 41 \) mm to \( 71 + 30 = 101 \) mm) - **Density of water at \( 4^\circ \mathrm{C} \):** \( \rho_\text{water} = 1 \) g/cm³ --- ## Part (a): Mass of Steel Balls Needed ### **Step 1: Buoyant Force Must Balance Weight** At equilibrium (when floating in water, vertical), the **weight of the hydrometer** (including steel balls) equals the **buoyant force**: \[ W_\text{total} = \text{Buoyant force} = \rho_\text{water} \cdot V_\text{immersed} \cdot g \] *Since we are using grams and cm³, and \( g \) cancels out, we can use mass directly:* \[ \text{Total mass (g)} = \text{Volume immersed (cm}^3\text{)} \times \rho_\text{water} (\text{g/cm}^3) \] --- ### **Step 2: Calculate Volume Immersed** The *immersed length* is \( 71 \) mm. We need to find the volume of the hydrometer submerged up to 71 mm. - The bottom cylinder has \( d_2 = 24 \) mm, length \( L_2 = 40 \) mm. - The upper cylinder has \( d_1 = 6 \) mm, length \( L_1 = 100 \) mm. - The 71 mm will be composed of all 40 mm of the bottom (wide) section + 31 mm of the upper (narrow) section. #### **1. Volume of lower (wide) section (fully submerged):** \[ V_2 = \pi \left( \frac{d_2}{2} \right)^2 L_2 \] \[ = \pi \left( \frac{24\, \text{mm}}{2} \right)^2 \times 40\, \text{mm} = \pi (12\, \text{mm})^2 \times 40\, \text{mm} = \pi \cdot 144\, \text{mm}^2 \times 40\, \text{mm} = \pi \cdot 576\, \text{mm}^3 \] #### **2. Volume of upper (narrow) section (partially submerged):** \[ \text{Submerged length: } 31\, \text{mm} \] \[ V_1 = \pi \left( \frac{d_1}{2} \right)^2 \times \text{(submerged length)} = \pi (3\, \text{mm})^2 \times 31\, \text{mm} = \pi \cdot 9\, \text{mm}^2 \times 31\, \text{mm} = \pi \cdot 279\, \text{mm}^3 \] #### **3. Total Submerged Volume** \[ V_\text{total} = \pi (576 + 279)\, \text{mm}^3 = \pi \times 6039\, \text{mm}^3 \] #### **Convert mm³ to cm³:** \[ 1\, \text{cm}^3 = 100\, \text{mm}^3 \] \[ V_\text{total} = \pi \times 6039\, \text{mm}^3 \times \frac{1\, \text{cm}^3}{100\, \text{mm}^3} = \pi \times 6.039\, \text{cm}^3 \] \[ = 3.1416 \times 6.039 \approx 18.98\, \text{cm}^3 \] --- ### **Step 3: Total Mass for Buoyancy** \[ \text{Required total mass} = V_\text{total} \times \rho_\text{water} = 18.98\, \text{g} \] - The hydrometer (without steel balls) weighs \( 9 \) g. - Therefore, mass of steel balls required: \[ m_\text{balls} = 18.98\, \text{g} - 9\, \text{g} = 9.98\, \text{g} \] --- ### **Step 4: Final Answer for (a)** **The mass of steel balls that must be placed in the lower section is:** \[ \boxed{9.98~\text{g}} \] --- ## Part (b): Range of Specific Gravities ### **Step 1: Immersion Range** - The scale extends \(30\) mm above and below the \(71\) mm point. - So, the **submerged length** can vary from \(41\) mm to \(101\) mm. ### **Step 2: Calculate Submerged Volume at Each Limit** #### **Case 1: Minimum immersion (41 mm)** - All 40 mm of the wide section is submerged. - Remaining \(1\) mm is from the narrow section. \[ V_{41} = \pi (12^2 \times 40 + 3^2 \times 1) = \pi (576 + 9) = \pi \times 5769\, \text{mm}^3 = 3.1416 \times 5.769\, \text{cm}^3 \approx 18.12\, \text{cm}^3 \] #### **Case 2: Maximum immersion (101 mm)** - Both cylinders fully submerged (\(40 + 61 = 101\) mm, all 100 mm of upper section, and 1 mm more, but the upper is only 100 mm long). - So, maximum is all 40 mm of lower, all 100 mm of upper: 140 mm total, but only 101 mm is possible, so all 40 mm of lower + 61 mm of upper. \[ \text{Length in upper section: } 101 - 40 = 61\, \text{mm} \] \[ V_{101} = \pi (12^2 \times 40 + 3^2 \times 61) = \pi (576 + 549) = \pi \times 6309\, \text{mm}^3 = 3.1416 \times 6.309\, \text{cm}^3 \approx 19.82\, \text{cm}^3 \] --- ### **Step 3: Specific Gravity Calculation** The **specific gravity** (\(s\)) of a liquid is: \[ s = \frac{\text{weight of displaced water}}{\text{weight of the instrument}} \] But since the weight is constant, the specific gravity is: \[ s = \frac{W_\text{hydrometer}}{V_\text{immersed} \cdot \rho_\text{water}} \] Where \(W_\text{hydrometer} = 18.98\, \text{g}\) (from part a). #### **Minimum specific gravity (maximum immersion):** \[ s_\text{min} = \frac{18.98}{19.82} = .958 \] #### **Maximum specific gravity (minimum immersion):** \[ s_\text{max} = \frac{18.98}{18.12} = 1.047 \] --- ### **Step 4: Final Answer for (b)** **The specific gravity range the apparatus can read is:** \[ \boxed{.958 \leq s \leq 1.047} \] --- ## **Summary** ### **(a) Mass of steel balls required** \[ \boxed{9.98~\text{g}} \] ### **(b) Range of specific gravities** \[ \boxed{.958 \leq s \leq 1.047} \] --- **Note:** - All calculations use \(\pi \approx 3.1416\). - Volumes calculated assuming perfect cylinders. - The range covers typical liquid densities near water.