Question

The length of an offset is 15 m. The maximum error is its length is 7 cm and scale used is 1 cm = 20 m. What is the maximum permissible error in the laying of the direction of the offset so that the maximum displacement does not exceed 0.75 mm?

• A. $3^{\circ }{12}^{\prime }$
• B. $5^{\circ }{43}^{\prime }$
• C. $4^{\circ }{28}^{\prime }$
• D. $2^{\circ }{32}^{\prime }$

Answer 1

The correct option is B $5^{\circ }{43}^{\prime }$

Length of offset, l = 15 m

e = 7 cm = 0.07 m

Scale, 1 cm = 20 m

Maximum displacement = 0.75 mm = 0.075 cm

Displacement of point due to incorrect direction,

$P_2{P}_1=l\sin \alpha =15\sin \alpha$

Maximum error in the length of the offset
$=P{P}_1=0.07m$

Maximum displacement due to both errors,

$P{P}_2=\sqrt{(15\sin \alpha {)}^2+(0.07{)}^2}meter$

Maximum displacement on paper,

$=\frac{\sqrt{(15\sin \alpha {)}^2+(0.07{)}^2}}{20}cm=0.075\left(given\right)$

$\therefore (15\sin \alpha {)}^2+(0.07{)}^2=(0.075\times 20{)}^2$

$\Rightarrow 225{\sin }^2\alpha =2.2451$

$\Rightarrow \alpha ={5.7329}^{\circ }=5^{\circ }{43}^{\prime }{58.4}^{\prime }\simeq 5^{\circ }{43}^{\prime }$