Question

Locations \(P\) and \(Q\) are separated by a wide valley. The difference in levels of locations \(P\) and \(Q\) measured by a levelling instrument stationed near \(P\) is 3.0 m. The same instrument stationed near \(Q\) measures the difference in levels of locations \(P\) and \(Q\) as \(-1.0\) m. Assume that the atmospheric refraction is the same during the measurements. The true difference in levels (in m) of locations \(P\) and \(Q\) is

• A. 1.0
• B. 1.5
• C. 2.0
• D. 4.0

Hint:

Use the reciprocal levelling rule: true difference = mean of the two apparent (reciprocal) observations.

Answer 1

The Correct Option is A

Step 1: Set up staff-reading notation.
Let $p_1$ and $q_1$ be the staff readings on $P$ and $Q$ when the level is set up near $P$, and let $p_2$ and $q_2$ be the staff readings on $P$ and $Q$ when the level is set up near $Q$. Because the line of sight bends and the earth curves away over a long distance, every reading carries a small systematic error that grows with the sight length; call this combined curvature-and-refraction correction $c$.

Step 2: Write the apparent difference at each set-up.
Near $P$: the sight to $P$ is short, so its reading is free of the correction, while the sight to $Q$ is long and picks up $+c$. The apparent difference from $P$ to $Q$ is
\[ d_1 = 3.0 \text{ m} \]
Near $Q$: now the sight to $Q$ is short (correction-free) and the sight to $P$ is long, so this time the correction $c$ shows up with the opposite sign in the computed difference:
\[ d_2 = -1.0 \text{ m} \]

Step 3: Write both readings in terms of the true difference $H$ and the error.
\[ d_1 = H + c \]
\[ d_2 = H - c \]
This is because the same physical error $c$ always attaches to the long sight, and swapping which point the instrument is near also swaps which sight is long, flipping the sign of $c$ while $H$ (the real level difference) stays fixed.

Step 4: Eliminate $c$ by adding the two equations.
\[ d_1 + d_2 = 2H \]
\[ H = \frac{d_1 + d_2}{2} \]
Notice $c$ never has to be known numerically. Adding the equations removes it automatically; that is the whole point of taking readings reciprocally instead of from a single set-up.

Step 5: Plug in numbers.
\[ H = \frac{3.0 + (-1.0)}{2} = \frac{2.0}{2} = 1.0 \text{ m} \]
So the level difference free of curvature, refraction and residual collimation error is 1.0 m.

Final Answer:
\[ \boxed{H = 1.0 \text{ m, option (A)}} \]