Q1: The sag of 50 m tape weighing 4 kg under 5 kg tension is roughly

ANS:E - 0.083 m

.083 m is not the correct answer for given data.

Here, Given data will be same except W= 1kg
So, ( 1kg)^2 * (50 m) /[ 24 * ( 5kg)^2]= .083 m. Correction for sag = W^2*L / 24*P^2 , where w is in kg,N,kN etc.

Correction for sag = w^2*L^3 / 24*P^2 , where w is in kg/m, N/m etc. ((4/50)^2 * 50)/(24*5^2) = 0.0833m.

Y (4/50)----->w is the weight per meter length. (W)^2=(4kg )^2 = 16kg.
L=50.
p=5^2=25.

Sag correction = 16*50/24*25.
800/600 = -1.33ft.

Note: Sag correction is Always negative. The correct steps:

(W^2 * L)÷(24 *P^2).

W=weight of tape in kg/m,
L=length of tape in m,
P= pull applied or tension in kg,
So, W= 5kg/50m= 0.1kg/m.
L=50m and P=5kg.