Modern family: Addition

State level Olympiad (3)

Despite all these thoughts going on in my mind; I resisted the urge to call my team from hotel telephone. Afterall during our conversation there will be some sensitive information; so even if their is very little chance for someone to evasdrop at our conversation it's not worth the risk. Even more so, I'm not in mood to talk about these things in middle of exam period.

Apart from taking an exam in San francisco; there are few iconic places that I would like to visit.

Eg. Golden gate bridge : Its iconic orange-red hue contrasted against the azure sky, creating a picturesque scene that would forever be etched in your memory. The bridge spanned gracefully across the vast expanse of the Pacific Ocean, connecting the city to the wonders that lay beyond.

The rolling hills of the Bay Area :  Adorned with vibrant wildflowers, their colors dancing in harmony with the gentle breeze. Majestic redwood trees towered above, their branches reaching towards the sky, providing shelter to an array of wildlife.

Iconic Transamerican pyramid :
48 story modernist skyscraper, second tallest building in San Francisco skyline. Opened in 1972  and was tallest building in San Francisco until 2017 in my previous life where it was surpassed by newly constructed Salesforce tower. Building had changed many hands since it's inception; and I also might consider owning it if necessary conditions arose as such.

Well, San Francisco is famous for it's classic and modern mixed type architecture which makes this City quite touristic in nature.
Nonetheless, these travel plans will be halted until the all three exams get completed.

Same as previous one, I along with my dad reached the exam centre on exam date. After necessary checking and verification, I entered in exam hall. Papers were distributed and I started reading it.

Q 1 : A simple gun can be made from a uniform cylinder of length L_0 and inside radius r_c. One
end of the cylinder is sealed with a moveable plunger and the other end is plugged with a
cylindrical cork bullet. The bullet is held in place by friction with the walls of the cylinder.
The pressure outside the cylinder is atmospheric pressure, P_0. The bullet will just start to slide
out of the cylinder if the pressure inside the cylinder exceeds
P_cr.

a. There are two ways to launch the bullet: either by heating the gas inside the cylinder and
keeping the plunger fixed, or by suddenly pushing the plunger into the cylinder. In either case,
assume that an ideal monatomic gas is inside the cylinder, and that originally the gas is at
temperature T_0, the pressure inside the cylinder is P_0 , and the length of the cylinder is L_0.

(I). Assume that we launch the bullet by heating the gas without moving the plunger.
Find the minimum temperature of the gas necessary to launch the bullet. Express
your answer in terms of any or all of the variables: r_c, T_0, L_0, P_0 and P_cr.

(II). Assume, instead that we launch the bullet by pushing in the plunger, and that we
do so quickly enough so that no heat is transferred into or out of the gas. Find the length of the gas column inside the cylinder when the bullet just starts to move.
Express your answer in terms of any or all of the variables: r_c, T_0, L_0, P_0 and P_cr.

b. It is necessary to squeeze the bullet to get it into the cylinder in the first place. The
bullet normally has a radius r_b that is slightly larger than the inside radius of the cylinder;
r_b - r_c = ∆r,  ∆r is small compared to r_c . The bullet has a length h<< L_0. The walls of the cylinder apply a pressure to the cork bullet. When a pressure P is applied to the bullet along a given
direction, the bullet’s dimensions in that direction change by

(∆x/x) = - (P/E)

for a constant E known as Young’s modulus. You may assume that compression along one
direction does not cause expansion in any other direction. (This is true if the so-called Poisson ratio is close to zero, which is the case for cork.)
If the coefficient of static friction between the cork and the cylinder is μ , find an expression
for P_cr . Express your answer in terms of any or all of the variables: P_0, h, E, ∆r, μ and r_c .

Solution 1 :
a. Two parts, solved individually.
i. Isochoric compression:
P_f / P_i = T_f / T_i.             (Q 1-1)

so  T = (P_cr / P_0) T_0.      (Q 1-2)

ii. Adiabatic compression:

(P V)^ γ = const,                 (Q 1-3)

where γ = C_P /C_V = (C_V + 1)/C_V = 5/3 for a monatomic gas. Consequently

(P_0 L_0)^γ  = (P_cr L )^γ,  (Q 1-4)
and then

L = L_0 (P_0/P_cr)^ 5/3.    (Q 1-5)

b. The normal pressure on the bullet comes from
P = (∆r / r_c)E.                 (Q 1-6)

Therefore, the normal force on the bullet is
F_N=(∆r /r_c)E×(2π r_c h), (Q 1-7)
and finally the force of friction is µF_N . The force due to the pressure difference between the
inside of the barrel and the outside must equal the normal force, so
(π(r_c)^2)(P_cr − P_0) = 2πhµ ∆r E,                    ( Q 1-8)

and then
P_cr = P_0 + (2µEh ∆r/ (r_c)^2).
(Q 1-9)

Q 2. A volume V_f of fluid with uniform charge density ρ is sprayed into a room, forming
spherical drops. As they float around the room, the drops may break apart into smaller drops
or coalesce into larger ones. Suppose that all of the drops have radius R. Ignore inter-drop
forces and assume that V_f >> R^3

a. Calculate the electrostatic potential energy of a single drop. (Hint: suppose the
sphere has radius r. How much work is required to increase the radius by dr?).

b. What is the total electrostatic energy of the drops?

Your answer to (b) should indicate that the total energy increases with R. In the absence of surface tension, then, the fluid would break apart into infinitesimally small drops. Suppose, however, that the fluid has a surface tension γ . (This value is the potential energy per unit surface area, and is positive.)

c. What is the total energy of the drops due to surface tension?

d. What is the equilibrium radius of the drops?

Solution 2 :
a. If the sphere has radius r, it has charge q = (4 /3 ) πρ r^3    (Q 2-1)

and thus its surface is at electrostatic potential
V = q /(4π€_0 r) = (ρ r^2)/(3 €_0)
(Q 2-2)

To increase the radius by dr, an additional charge dq = 4πr^2dr must be brought in from
infinity, requiring work
dU = V dq = {(4π r^4 ρ^2)/(3 €_0)}dr               (Q 2-3)

Thus to grow the sphere from r = 0 to r = R requires

U = Integral ( 0 to R) {(4π r^4 ρ^2)/(3 €_0)} dr = (4π R^5 ρ^2)/(150 €_0)
(Q 2-4)

b. Each drop has volume
V_d = (4 /3) π R^3 ,
so the number of drops is

n = (V_f /V_d) = V_f /{(4 /3) π R^3}
(Q 2-5)

Since we are ignoring inter-drop forces, the total energy of the drops is simply the sum of the
energies of each individual drop:
U_e,tot = nU = [V_f/{(4/3)π R^3}]×
[(4π R^5 ρ^2)/(150 €_0)]
= [(R^2 ρ^2)/(5 €_0)]× V_f    (Q 2-6)

c. Each drop has surface area
(4π R^2) and thus surface tension energy (4π R^2 γ). As before, the total energy due to surface tension is just the sum of the energies of the individual drops:
U_s,tot = 4πR^2 γ n
= (4π R^2 γ) ×  (V_f)/{(4/ 3)π R^3}
= 3γV_f  / R              (Q 2-7)

d. The total potential energy from both sources is
U_tot = [{(R^2 ρ^2)/(5 €_0)} +
{3γ/R}] × V_f            (Q 2-8)

Equilibrium is reached when the total energy is a minimum; since U → ∞ at both R → 0 and R → ∞, it must have an interior minimum.
d/dR (U_tot) = [{(2R ρ^2)/(5 €_0)}
− {3γ/ R^2}]× V_f       (Q 2-9)

Setting this equal to zero,
(2R ρ^2)/(5 €_0) = 3γ/R^2  (Q 2-10)

R^3 = (15 γ €_0)/(2ρ^2).    (Q 2-11)

R = {(15γ €_0)/(2 ρ^2)}^ (1/3)
(Q 2-12)