Question 7 A thick uniform rubber rope of density 0.7 gcm-3 and Young Modulus 5 x 106 Nm 2 has a length 6.7 m. when
hung from the ceiling of the room, calculate the increase in length due to its own weight ?

O A. 35.79 x 10-2 m
O B. 43.81 x 10-2 m
OC.52.34 x 10-2 m
O D.61.65 x 10-2m

Increase in length, e = 61.59 * 10⁻²m

The closest value from the options is option D. 61.65 x 10-2m

Explanation:

Young Modulus = stress/strain = (F/A)(/e/L) = F * L/A * e

where F = force; L = original length; A = area; e = increase in length

making e subject of formula;  

Also force, F =  mass * acceleration due to gravity;

mass = density * volume = density * area * length = ρ * A * L

Therefore, force, F = ρ * A * L * g

e = ρ *A * L* L * g/Y * A = ρgL²/Y

density = 0.7 gcm⁻³ = 0.7 * 10³ kgm⁻³; g = 9.8 ms⁻²

e = {0.7 * 10³ kgm⁻³ * (6.7 m)² * 9.8 ms⁻²}/5 * 10⁶ Nm²

e = 61.59 * 10⁻² m

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

pretty sure it's false