For rolling without slipping, the linear velocity v is equal to the radius r times the angular velocity ω, hence v = rω.

Due to surface tension, a drop of liquid will take the shape of a sphere, as this shape has the minimum surface area for a given volume.

A viscosity of 0.5 Pa·s indicates that the fluid is relatively thick and flows less easily compared to fluids with lower viscosity.

Work Done (W) = Force * Distance = 10 N * 5 m = 50 J

Work Done (W) = Force * Distance = 10 N * 5 m = 50 J

Torque = Force × Distance × sin(θ) = 12 N × 1 m × sin(30°) = 12 N × 1 m × 0.5 = 6 Nm.

Work done = F × d × cos(θ) = 12 N × 3 m × cos(30°) = 12 N × 3 m × (√3/2) = 18 J.

Using F = ma, we find a = F/m = 12 N / 4 kg = 3 m/s².

Work done = Force × Distance = 15 N × 4 m = 60 J.

Torque = Force × Distance = 15 N × 0.4 m = 6 Nm.

Torque (τ) = F × r × sin(θ) = 15 N × 1.5 m × sin(30°) = 15 × 1.5 × 0.5 = 11.25 Nm.

Torque (τ) = F × r × sin(θ) = 15 N × 1 m × sin(30°) = 15 N × 1 m × 0.5 = 7.5 Nm.

Work done = F × d × cos(θ) = 15 N × 4 m × cos(60°) = 15 N × 4 m × 0.5 = 30 J.

Work done = F × d × cos(θ) = 15 N × 4 m × cos(60°) = 15 N × 4 m × 0.5 = 30 J.

Net force = applied force - frictional force = 15 N - 5 N = 10 N.

Net force = applied force - frictional force = 15 N - 5 N = 10 N.

Work done = Force × Distance = 15 N × 3 m = 45 J.

Work done = Force × Distance = 15 N × 4 m = 60 J.

Net force = applied force - frictional force = 30 N - 10 N = 20 N.

Work done = Force × Distance = 5 N × 4 m = 20 J.

Work done = Force × Distance × cos(θ) = 5 N × 4 m × cos(60°) = 5 N × 4 m × 0.5 = 10 J.

Work done = Force × Distance × cos(θ) = 5 N × 3 m × cos(60°) = 5 N × 3 m × 0.5 = 7.5 J.

At frequencies much lower than the natural frequency, the amplitude of the forced oscillator increases significantly.

A deflection in the galvanometer indicates that the potential difference across the galvanometer is equal to the reference voltage.

The work done by the gas during expansion against a constant external pressure is given by W = P_ext * ΔV.

Using the first law of thermodynamics, ΔU = Q - W = 100 J - 150 J = -50 J.

Using the First Law of Thermodynamics, ΔU = Q - W. Here, ΔU = 30 J - 50 J = -20 J.

Using Boyle's Law (P1V1 = P2V2), if P1 = 1 atm, V1 = 10 L, and P2 = 2 atm, then V2 = (P1V1)/P2 = (1 atm * 10 L) / 2 atm = 5 L.

Using Boyle's Law (P1V1 = P2V2), if the pressure doubles from 1 atm to 2 atm, the volume will halve from 10 L to 5 L.

The solid cylinder reaches the bottom first because it has a smaller moment of inertia compared to the hollow cylinder.