To solve the problem, we will follow these steps:
### Step 1: Understand the System
We have a spherical tank of radius \( R = 0.35 \, \text{m} \) that is half-filled with oil of relative density \( \rho_r = 0.8 \). The tank is accelerating horizontally with \( a = 10 \, \text{m/s}^2 \).
### Step 2: Calculate the Density of the Oil
The density of the oil can be calculated using the formula:
\[
\rho = \rho_r \times \rho_{\text{water}}
\]
where \( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \). Thus,
\[
\rho = 0.8 \times 1000 = 800 \, \text{kg/m}^3
\]
### Step 3: Determine the Effective Gravitational Force
When the tank accelerates, the effective gravitational force acting on the oil is the vector sum of gravitational force and the pseudo force due to acceleration. The effective gravitational acceleration \( g_{\text{effective}} \) can be calculated as:
\[
g_{\text{effective}} = \sqrt{g^2 + a^2}
\]
Substituting \( g = 10 \, \text{m/s}^2 \) and \( a = 10 \, \text{m/s}^2 \):
\[
g_{\text{effective}} = \sqrt{10^2 + 10^2} = \sqrt{200} = 10\sqrt{2} \, \text{m/s}^2
\]
### Step 4: Calculate the Maximum Depth of Oil
The maximum depth of the oil occurs at the bottom of the tank. Since the tank is half-filled, the depth \( h \) of the oil is:
\[
h = \frac{R}{2} = \frac{0.35}{2} = 0.175 \, \text{m}
\]
### Step 5: Calculate the Maximum Gauge Pressure
The maximum gauge pressure \( P \) at the bottom of the tank can be calculated using the formula:
\[
P = \rho \cdot g_{\text{effective}} \cdot h
\]
Substituting the values:
\[
P = 800 \, \text{kg/m}^3 \cdot (10\sqrt{2}) \, \text{m/s}^2 \cdot 0.175 \, \text{m}
\]
Calculating this:
\[
P = 800 \cdot 10\sqrt{2} \cdot 0.175
\]
\[
P = 800 \cdot 1.414 \cdot 1.75 \approx 800 \cdot 2.449 \approx 1959.2 \, \text{Pa}
\]
### Step 6: Convert to Gauge Pressure in the Required Form
The problem states that the maximum gauge pressure can be expressed as \( n \times 10^3 \, \text{Pa} \). Thus,
\[
P \approx 1959.2 \, \text{Pa} \approx 1.9592 \times 10^3 \, \text{Pa}
\]
So, \( n \approx 1.96 \).
### Final Answer
Since we need to approximate \( n \):
\[
n \approx 2
\]
