Balakumar

Water pressure at the bottom, p = 1.1 × 10⁸ Pa
Initial volume of the steel ball, V = 0.32 m³
Bulk modulus of steel, B = 1.6 × 10¹¹ Nm^–2
The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface.
Let the change in the volume of the ball on reaching the bottom of the trench be ΔV.
Bulk modulus, B = p / (?V/V)
?V  =  pV/B
= 1.1 × 10⁸ × 0.32 / (1.6 × 10¹¹ )  =  2.2 × 10^-4 m³
Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 × 10^–4 m³.

Cross-sectional area of wire A, a₁ = 1.0 mm² = 1.0 × 10^–6 m²
Cross-sectional area of wire B, a₂ = 2.0 mm² = 2.0 × 10^–6 m²
Young’s modulus for steel, Y₁ = 2 × 10¹¹ Nm^–2
Young’s modulus for aluminium, Y₂ = 7.0 ×10¹⁰Nm^–2

(a) Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
Stress in the wire = Force / Area  =  F / a
If the two wires have equal stresses, then:
F₁ / a₁  =  F₂ / a₂
Where,
F₁ = Force exerted on the steel wire
F₂ = Force exerted on the aluminum wire
F₁ / F₂ = a₁ / a₂  =  1 / 2    ….(i)
The situation is shown in the following figure:

Taking torque about the point of suspension, we have:
F₁y = F₂ (1.05 – y)
F₁ / F₂ = (1.05 – y) / y    ……(ii)
Using equations (i) and (ii), we can write:
(1.05 – y) / y  = 1 / 2
2(1.05 – y)  =  y
y = 0.7 m
In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.

(b) Young’s modulus = Stress / Strain
Strain = Stress / Young’s modulus  =  (F/a) / Y
If the strain in the two wires is equal, then:
(F₁/a₁) / Y₁  =  (F₂/a₂) / Y₂
F₁ / F₂ = a₁Y₁ / a₂Y₂
a₁ / a₂ = 1/2
F₁ / F₂ = (1 / 2) (2 × 10¹¹ / 7 × 10¹⁰)  =  10 / 7      …….(iii)
Taking torque about the point where mass m, is suspended at a distance y₁ from the side where wire A attached, we get:
F₁y₁ = F₂ (1.05 – y₁)
F₁ / F₂  =  (1.05 – y₁) / y₁    ….(iii)
Using equations (iii) and (iv), we get:
(1.05 – y₁) / y₁  =  10 / 7
7(1.05 – y₁)  =  10y₁
y₁ = 0.432 m
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached

Radius of the steel cable, r = 1.5 cm = 0.015 m
Maximum allowable stress = 10⁸ N m^–2
Maximum stress = Maximum force / Area of cross-section
∴ Maximum force = Maximum stress × Area of cross-section
= 10⁸ × π (0.015)²
= 7.065 × 10⁴ N
Hence, the cable can support the maximum load of 7.065 × 10⁴ N.

Mass of the shell, m = 0.020 kg
Muzzle speed of the shell, v = 80 m/s
Recoil speed of the gun = V
Both the gun and the shell are at rest initially.
Initial momentum of the system = 0
Final momentum of the system = mv – MV
Here, the negative sign appears because the directions of the shell and the gun are opposite to each other.
According to the law of conservation of momentum:
Final momentum = Initial momentum
mv – MV = 0
∴ V = mv / M
= 0.020 × 80 / (100 × 1000) = 0.016 m/s