Question

Please help me out with that question as an step by step attachment​

Answer 1

• Area of the plate, A = 100 cm²

$1 \: cm \: = {10}^{ - 2 } \: m \\ \\ 1 \: {cm}^{2} = { ({10}^{ - 2}) }^{2} \: {m}^{2} \\ \\ 1 \: {cm}^{2} = {10}^{ - 4} \: {m}^{2} \\ \\ 100\: {cm}^{2} =100 \times {10}^{ - 4} \: {m}^{2} \\ \\ = {10}^{2} \times {10}^{ - 4} \: {m}^{2} \\ \\ A = {10}^{ - 2} \: {m}^{2}$

• Thickness of castor oil, dx = 2 mm

$1 \: mm = {10}^{ - 3} \: m \\ \\dx = 2 \: mm \: = 2 \times {10}^{ - 3} \: m$

• Coefficient of viscosity, η = 15.5 poise

1 poise = 10 deca poise

15.5 poise = 15.5 × 10 deca poise

η= 1.55 deca poise

• Velocity, dv = 3 cm/s

$1 \: \dfrac{cm}{s} \: = {10}^{ - 2} \: \dfrac{m}{s} \\ \\ dv = 3\: \dfrac{cm}{s} \: = 3 \times {10}^{ - 2} \: \dfrac{m}{s}$

Viscous force, f = ηAdv/dx

$f = \dfrac{1.55 \times {10}^{ - 2} \times 3 \times {10}^{ - 2} }{2 \times {10}^{ - 3} } \\ \\ = \dfrac{1.55 \times 3}{2} \times \dfrac{ {10}^{ - 4} }{ {10}^{ - 3} } \\ \\ = 1.55 \times 1.5 \times {10}^{ - 1} \\ \\ = 1.55 \times 1.5 \times \dfrac{1}{10} \\ \\ = 1.55 \times 0.3 \times \dfrac{1}{2} \\ \\ f = 0.2325 \: newton$

The viscous force is 0.2425 N.