Question

following observations were taken during a reciprocal levelling .
If reduced level P is 140 .815 m,
then the reduced level of Q is

Instrument near	P	Q
Staff reading at P(in m)	1.824	0.928
Staff reading at Q (in m)	2.748	1.606

• A. 140.014 m
• B. 139.616 m
• C. 141.616 m
• D. 138.014 m

Answer 1

The correct option is A 140.014 m

$h_P=1.824m;h_Q=2.748m$

$h_P^{\prime }=0.928m;h_Q^{\prime }=1.606m$

Correct difference of elevation between P and Q is

$h=\frac{(h_Q^{\prime }-h_P^{\prime })+(h_Q-h_p)}{2}$

$=\frac{(1.606-0.928)+(2.748-1.824)}{2}$

= 0.801 m > 0 and thus Q is at lower elevation than P.

$\therefore$ The reduced level of Q = (140.815 - 0.801)
= 140.014 m