A metal block of base area 0.10 m 2 is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless frictionless). A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.085 m / s. Find the coefficient of viscosity of the liquid. ( g =9.8 m / s 2)

Question

A metal block of base area 0.10 m 2 is connected to a 0.010 kg mass via a string that passes over an ideal pulley (considered massless frictionless). A liquid with a film thickness of 0.30 mm is placed between the block and the table. When released the block moves to the right with a constant speed of 0.085 m / s. Find the coefficient of viscosity of the liquid. ( g =9.8 m / s 2) (A) 6 × 10-2 Pa . s (B) 3.45 × 10-5 Pa.s (C) 3.45 × 10-3 Pa . s (D) 3.45 × 10-6 Pa.s

Tardigrade Admin · Accepted Answer

η=( text Shear stress / text Strain Rate ) =( F / A / v / ℓ)=( mg ℓ/ Av ) ℓ is thickness of film η=(0.01 × 9.8 × 0.3 × 10-3/0.1 × 0.085) =3.45 × 10-3 Pa - s

$6 \times 1 0^{- 2} P a . s$

$3.45 \times 1 0^{- 5}$ Pa.s

$3.45 \times 1 0^{- 3} P a . s$

$3.45 \times 1 0^{- 6}$ Pa.s

$η = \frac{Shear stress}{Strain Rate}$
$= \frac{F / A}{v / ℓ} = \frac{m g ℓ}{A v}$
$ℓ$ is thickness of film
$η = \frac{0.01 \times 9.8 \times 0.3 \times 1 0 ^{- 3}}{0.1 \times 0.085}$
$= 3.45 \times 1 0^{- 3} P a - s$

6×10−2Pa.s

3.45×10−5 Pa.s

3.45×10−3Pa.s

3.45×10−6 Pa.s

η= Strain Rate Shear stress ​ =v/ℓF/A​=Avmgℓ​ ℓ is thickness of film η=0.1×0.0850.01×9.8×0.3×10−3​ =3.45×10−3Pa−s

$6 \times 1 0^{- 2} P a . s$

$3.45 \times 1 0^{- 5}$ Pa.s

$3.45 \times 1 0^{- 3} P a . s$

$3.45 \times 1 0^{- 6}$ Pa.s

$η = \frac{Shear stress}{Strain Rate}$
$= \frac{F / A}{v / ℓ} = \frac{m g ℓ}{A v}$
$ℓ$ is thickness of film
$η = \frac{0.01 \times 9.8 \times 0.3 \times 1 0 ^{- 3}}{0.1 \times 0.085}$
$= 3.45 \times 1 0^{- 3} P a - s$