Q.12 A horizontal force of 0.5N is required to move a metal plate of area 10⁻² m² with a velocity of 3×10⁻² m/s, when it rests on 0.5×10⁻³ m thick layer of glycerin. Find the coefficient of viscosity of glycerin.

Answer: η ≈ 0.833 N s/m ² (Pa s) Given: F = 0.5 N, A = 10⁻² m ² , dv = 3×10⁻² m/s, dx = 0.5×10⁻³ m. Newton's law of viscosity: F = η Adv/dx , so η = F dx/A dv = dfrac0.5×0.5×10⁻³10⁻²×3×10⁻² = dfrac2.5×10⁻⁴3×10⁻⁴ = 0.833 N s/m ² .

Chapter: Mechanical Properties of Fluids· Physics

A horizontal force of 0.5N is required to move a metal plate of area \(10^{-2}\,\text{m}^2\) with a velocity of \(3\times10^{-2}\) m/s, when it rests on \(0.5\times10^{-3}\) m thick layer of glycerin. Find the coefficient of viscosity of glycerin.

Q.12 A horizontal force of 0.5N is required to move a metal plate of area \(10^{-2}\,\text{m}^2\) with a velocity of \(3\times10^{-2}\) m/s, when it rests on \(0.5\times10^{-3}\) m thick layer of glycerin. Find the coefficient of viscosity of glycerin.

Doubtly Asked 29 Oct 2026 Board exam · March 2024 hsc 12th Save WhatsApp Report

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Answer: \(\eta \approx 0.833\) N s/m\(^2\) (Pa s)

Given: \(F = 0.5\) N, \(A = 10^{-2}\) m\(^2\), \(dv = 3\times10^{-2}\) m/s, \(dx = 0.5\times10^{-3}\) m.

Newton's law of viscosity: \(F = \eta A\dfrac{dv}{dx}\), so

\(\eta = \dfrac{F\,dx}{A\,dv} = \dfrac{0.5\times0.5\times10^{-3}}{10^{-2}\times3\times10^{-2}} = \dfrac{2.5\times10^{-4}}{3\times10^{-4}} = 0.833\) N s/m\(^2\).

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A horizontal force of 0.5N is required to move a metal plate of area \(10^{-2}\,\text{m}^2\) with a velocity of \(3\times10^{-2}\) m/s, when it rests on \(0.5\times10^{-3}\) m thick layer of glycerin. Find the coefficient of viscosity of glycerin.

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