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Step 1: Identify the Given Data
• Diameter of narrow bore 1, $d_1 = 5.0\,\text{mm}$, so radius $r_1 = \frac{d_1}{2} = 2.5\,\text{mm} = 2.5\times 10^{-3}\,\text{m}$.
• Diameter of narrow bore 2, $d_2 = 8.0\,\text{mm}$, so radius $r_2 = \frac{d_2}{2} = 4.0\,\text{mm} = 4.0\times 10^{-3}\,\text{m}$.
• Surface tension of water, $T = 7.3\times 10^{-2}\,\text{N\,m}^{-1}$.
• Angle of contact, $\theta = 0$.
• Acceleration due to gravity, $g = 10\,\text{m\,s}^{-2}$.
• Density of water, $\rho = 1.0\times 10^{3}\,\text{kg\,m}^{-3}$.
We want to find the difference in height, $\Delta h$, between the water levels in the two limbs.
Step 2: Write the Pressure Balance Equation
Consider a common horizontal level in the two limbs of the U-tube. Let the left limb (with radius $r_1$) have its fluid level depressed by an amount $(x + \Delta h)$ relative to that horizontal line, and the right limb (with radius $r_2$) have its fluid level depressed by $x$. The pressure at the same level in each limb must be equal. Symbolically:
$P_{\mathrm{atm}} - \frac{2T}{r_1} + \rho\,g\,(x + \Delta h) \;=\; P_{\mathrm{atm}} - \frac{2T}{r_2} + \rho\,g\,x$.
Step 3: Simplify the Pressure Balance
Subtract $P_{\mathrm{atm}} + \rho\,g\,x$ from both sides to isolate $\Delta h$:
$-\frac{2T}{r_1} + \rho\,g\,\Delta h \;=\; -\frac{2T}{r_2}$.
Rearrange to get an expression for $\Delta h$:
$\rho\,g\,\Delta h \;=\; 2T \left(\frac{1}{r_1} - \frac{1}{r_2}\right)$
$\Delta h \;=\; \frac{2T}{\rho\,g} \;\left(\frac{1}{r_1} - \frac{1}{r_2}\right).$
Step 4: Plug in the Numerical Values
Substitute $T = 7.3 \times 10^{-2}\,\text{N\,m}^{-1}$, $r_1 = 2.5 \times 10^{-3}\,\text{m}$, $r_2 = 4.0 \times 10^{-3}\,\text{m}$, $\rho = 10^{3}\,\text{kg\,m}^{-3}$, and $g = 10\,\text{m\,s}^{-2}$:
$\Delta h
= \frac{2 \times 7.3 \times 10^{-2}}{(10^{3} \times 10)}
\Bigl(\frac{1}{2.5 \times 10^{-3}} - \frac{1}{4.0 \times 10^{-3}}\Bigr).$
Step 5: Calculate Step by Step
Compute the term inside the brackets:
$\frac{1}{2.5 \times 10^{-3}} = 400 \,\text{m}^{-1}, \quad
\frac{1}{4.0 \times 10^{-3}} = 250 \,\text{m}^{-1}.$
So, $\left(\frac{1}{r_1} - \frac{1}{r_2}\right) = 400 - 250 = 150\,\text{m}^{-1}.$
Multiply by $2 \times 7.3 \times 10^{-2} = 1.46 \times 10^{-1}.$
Then $1.46 \times 10^{-1} \times 150 = 21.9.$
Divide by $\rho g = (10^{3} \times 10) = 10^{4}.$
Thus, $\Delta h = \frac{21.9}{10^{4}} = 2.19 \times 10^{-3}\,\text{m}.$
Therefore,
$\Delta h = 2.19 \times 10^{-3}\,\text{m} = 2.19\,\text{mm}.$
Step 6: State the Final Answer
Hence, the difference in the levels of the water in the two limbs of the U-tube is
$2.19\,\text{mm}.$
