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A spherical ball of diameter 1 cm and de...

A spherical ball of diameter 1 cm and density `5xx10^(3)kg m^(-3)` is dropped gently in a large tank containing viscous liquid of density `3xx10^(3)kgm^(-3)` and coefficient of viscosity `0.1 Nsm^(-2)`. The distance that the ball moves n 1 s after attaining terminal velocity is `(g=10ms^(-2))`

A

`(10)/(2)m`

B

`(2)/(3)m`

C

`(4)/(9)m`

D

`(4)/(5)m`

Text Solution

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The correct Answer is:
To solve the problem, we need to find the distance that a spherical ball moves in 1 second after attaining terminal velocity when dropped in a viscous liquid. Here’s a step-by-step solution: ### Step 1: Identify the given data - Diameter of the ball, \( d = 1 \, \text{cm} = 0.01 \, \text{m} \) - Radius of the ball, \( r = \frac{d}{2} = \frac{0.01}{2} = 0.005 \, \text{m} \) - Density of the ball, \( \rho_{\text{ball}} = 5 \times 10^3 \, \text{kg/m}^3 \) - Density of the liquid, \( \rho_{\text{liquid}} = 3 \times 10^3 \, \text{kg/m}^3 \) - Coefficient of viscosity, \( \eta = 0.1 \, \text{Ns/m}^2 \) - Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \) ### Step 2: Calculate the terminal velocity The formula for terminal velocity \( v_t \) of a sphere falling through a viscous fluid is given by: \[ v_t = \frac{2}{9} \cdot \frac{g r^2 (\rho_{\text{ball}} - \rho_{\text{liquid}})}{\eta} \] Substituting the known values into the formula: \[ v_t = \frac{2}{9} \cdot \frac{10 \cdot (0.005)^2 \cdot (5 \times 10^3 - 3 \times 10^3)}{0.1} \] ### Step 3: Simplify the expression Calculating \( \rho_{\text{ball}} - \rho_{\text{liquid}} \): \[ \rho_{\text{ball}} - \rho_{\text{liquid}} = 5 \times 10^3 - 3 \times 10^3 = 2 \times 10^3 \, \text{kg/m}^3 \] Now substituting this back into the terminal velocity equation: \[ v_t = \frac{2}{9} \cdot \frac{10 \cdot (0.005)^2 \cdot (2 \times 10^3)}{0.1} \] Calculating \( (0.005)^2 = 2.5 \times 10^{-5} \): \[ v_t = \frac{2}{9} \cdot \frac{10 \cdot 2.5 \times 10^{-5} \cdot 2 \times 10^3}{0.1} \] Calculating the numerator: \[ 10 \cdot 2.5 \times 10^{-5} \cdot 2 \times 10^3 = 5 \times 10^{-2} \] Now substituting this back: \[ v_t = \frac{2}{9} \cdot \frac{5 \times 10^{-2}}{0.1} \] This simplifies to: \[ v_t = \frac{2}{9} \cdot 0.5 = \frac{1}{9} \, \text{m/s} \] ### Step 4: Calculate the distance moved in 1 second After attaining terminal velocity, the distance \( d \) moved in 1 second is given by: \[ d = v_t \cdot t \] Substituting \( t = 1 \, \text{s} \): \[ d = \frac{1}{9} \cdot 1 = \frac{10}{9} \, \text{m} \] ### Final Answer The distance that the ball moves in 1 second after attaining terminal velocity is: \[ \frac{10}{9} \, \text{m} \]
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