Answer:

Explanation:
For a linear elastic material Young's modulus is a constant that is given by:

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force,
is the amount by which the length of the object changes and
is the original length of the object. In this case the force is the weight of the mass:

Replacing the given values in Young's modulus formula:

Answer:
option (B) is the correct option.
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Answer:
angle minimum θ = 41.3º
Explanation:
For this exercise let's use Newton's second law in the condition of static equilibrium
N - W = 0
N = W
The rotational equilibrium condition, where we place the axis of rotation on the wall
We assume that counterclockwise rotations are positive
fr (l sin θ) - N (l cos θ) + W (l/2 cos θ) = 0
the friction force formula is
fr = μ N
fr = μ W
we substitute
μ m g l sin θ - m g l cos θ + mg l /2 cos θ = 0
μ sin θ - cos θ + ½ cos θ= 0
μ sin θ - ½ cos θ = 0
sin θ / cos θ = 1/2 μ
tan θ = 1/2 μ
θ = tan⁻¹ (1 / 2μ)
θ = tan⁻¹ (1 (2 0.57))
θ = 41.3º