A metal block of base area 0.20 m² is placed on a table, as shown in figure. A liquid film of thickness 0.25 mm is inserted between the block and the table. The block is pushed by a horizontal force of 0.1 N and moves with a constant speed. IF the viscosity of the liquid is 5.0×10⁻³ Pl, the speed of block is ____________ ×10⁻³ m/s.

Solution

Physics sir · 4 steps

Constant speed means the push exactly balances the viscous drag, so set 0.1 N equal to η A v/d.

1. Forces balance. The block moves at constant speed, so its acceleration is zero. The applied 0.1 N is exactly cancelled by the viscous force from the liquid film.
2. Viscous force. The film touching the table is at rest and the film touching the block moves at speed v. The velocity gradient across the thin film is v/d, so Newton's law of viscosity gives the drag below.

   F = η A v/d ⇒ v = (F d)/(η A)

3. Put in numbers. With d = 0.25 mm = 2.5 × 10⁻⁴ m: v = (0.1 × 2.5 × 10⁻⁴)/(5.0 × 10⁻³ × 0.20) = (2.5 × 10⁻⁵)/(1.0 × 10⁻³) = 0.025 m s⁻¹.
4. So. v = 25 × 10⁻³ m s⁻¹, so the blank is 25.

The common slip: 0.025. Writing the speed straight in m/s gives 0.025. The blank already carries × 10⁻³, so the entry is 25.

The idea to keep. For a plate sliding on a thin liquid film, the velocity changes linearly across the film, so the drag is η A v/d. Constant speed tells you this drag equals the applied force.

Answer from the NTA final answer key. Solution by PrepWiser.