Calculate the sag for a span of 200 m if the ultimate tensile strength of the conductor is 6000 Kg. Allow a factor of safety of 2. The weight of the conductor is 900 kg/km.

1.0 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2.0 m

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2.5 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.5 m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 2.0 m
Breaking strength (Ultimate strength) $= 6000 k g$

Weight of conductor, $W = 900 k g / k m$

$= 900 k g / 1000 m (∵ 1 k m = 1000 m)$

$W = 0.9 k g / m$

Working tension, $T = \frac{U l t i m a t e s t r e n g t h}{s a f e t y f a c t o r} = \frac{6000}{2}$

$T = 3000 k g$

$S a g, S = \frac{W l^{2}}{8 T}$

$= \frac{(0.9 \times 200^{2})}{8 \times 3000}$

$= 1.5 m$

Suggest Corrections

Similar questions

k g / c m^{2}

k g / c m^{2}

k g / c m^{2}

k g / c m^{2}

σ_{1} = 60 M P a .

σ_{2} = 5 M P a .

σ_{3} = - 40 M P a .

σ_{y} = 200 M P a

Join BYJU'S Learning Program