Fluid Mechanics

JEE Main level, numerical answer, medium

177 of 192

Question 177 of 192

A metal block of base area 0.20m20.20{m}^{2} is placed on a table, as shown in figure. A liquid film of thickness 0.25mm0.25mm is inserted between the block and the table. The block is pushed by a horizontal force of 0.1N0.1N and moves with a constant speed. If the viscosity of the liquid is 5.0×10–3Pl5.0\times {10}^{–3}Pl, the speed of block is ______ ×10−3ms−1\times {10}^{-3}m{s}^{-1}.

Show answer and solution

Answer: 25

Viscous force is given by,
∣F∣=ηAΔvΔh|F|=\eta A\dfrac{\Delta v}{\Delta h}
⇒0.1=5×10−3×0.2×v0.25×10−3\Rightarrow 0.1=5\times {10}^{-3}\times 0.2\times \dfrac{v}{0.25\times {10}^{-3}}
⇒v=0.025ms−1\Rightarrow v=0.025m{s}^{-1}
⇒v=25×10−3ms−1\Rightarrow v=25\times {10}^{-3}m{s}^{-1}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Fluid mechanics simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Fluid Mechanics (Viscosity), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.