1) D
2) D.) Greater than 
Explanation:
1)
The phenomenon of total internal reflection occurs when a ray of light hitting the interface between two mediums is totally reflected back into the original medium, therefore no refraction into the second medium occurs.
This phenomenon occurs only if two conditions are satisfied:
- The index of refraction of the first medium is larger than the index of refraction of the 2nd medium
- The angle of incidence is greater than a certain angle called critical angle
In picture 1, we have 4 different diagrams. In the diagrams:
- The red arrow represents the incident ray
- The green arrow represents the refracted ray
- The blue arrow represents the reflected ray
Total internal reflection occurs when there is no refraction, therefore when there is no green arrow: this occurs only in figure D, so this is the correct option. (in figure C, there is a refracted ray but it is parallel to the interface: this condition occurs when the angle of incidence is exactly equal to the critical angle, however in this problem, the angle of incidence is greater than the critical angle, so the correct option is D)
2)
As we stated in problem 1), total internal reflection occurs when the angle of incidence is equal or greater than the critical angle. Therefore in this case, the angle of incidence must be
D.) Greater than
Answer
given,
copper rod length = 50 cm
density of the copper = 8.92 g/cm³
iron rod length = 50 cm
density of iron = 7.86 g/cm³
mass of iron = density × volume
m_i = 7.86 × A × l/2
m_c = 8.96 × A × l/2
taking the intersection of copper and iron rod be starting point cooper side is taken as positive side and iron side length is taken to be -ve side.
center of mass
= 
= 
= 
= 
= 0.015793 m
= 1.579 m (+ve)
center of mass shift to cooper because cooper is heavy.
To solve this problem we will apply the principle of conservation of energy, for which the initial potential and kinetic energy must be equal to the final one. The final kinetic energy will be transformed into rotational and translational energy, so the mathematical expression that approximates this deduction is
KE_i+PE_i = KE_{trans}+KE_{rot} +PE_f
, since initially cylinder was at rest
since at the ground potential energy is zero
The mathematical values are,
Here,
m = mass
g= Gravity
h = Height
V = Velocity
moment of Inertia in terms of its mass and radius
Angular velocity in terms of tangential velocity and its radius
Replacing the values we have that
mgh = \frac{1}{2} mv^2 +\frac{1}{2} (\frac{mr^2}{2})(\frac{v}{r})^2
gh = \frac{v^2}{2}+\frac{v^2}{4}
v = \sqrt{\frac{4gh}{3}}
From trigonometry the vertical height of inclined plane is the length of this plane for 
, then
Replacing,
Therefore the cylinder's speedat the bottom of the ramp is 3.32m/s
The answer is 7.93 pounds
