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A sphere is dropped into a viscous liqui...

A sphere is dropped into a viscous liquid of viscosity `eta` from some height if the density of material and liquid are `rho` and `sigma` respectively `(rhogtsigma)` then which of the following is/are incorrect.
(a). The acceleration of the sphere just after entering the liquid is `g((rho-sigma)/(rho))`
(b). Time taken to attain terminal speed `tproprho^(0)`
(C)AT TERMINAL SPEED VISCOUS FORCE IS MAXIMUM (d). At terminal speed, the net force acting on the sphere is zero.

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The correct Answer is:
To solve the problem step by step, let's analyze each option given in the question regarding the sphere dropped into a viscous liquid. ### Step 1: Understanding the Forces Acting on the Sphere When a sphere is dropped into a viscous liquid, the forces acting on it include: 1. The gravitational force acting downwards, \( F_g = \rho V g \), where \( \rho \) is the density of the sphere, \( V \) is the volume of the sphere, and \( g \) is the acceleration due to gravity. 2. The buoyant force acting upwards, \( F_b = \sigma V g \), where \( \sigma \) is the density of the liquid. 3. The viscous drag force acting upwards, \( F_d = 6 \pi r \eta v \), where \( r \) is the radius of the sphere, \( \eta \) is the viscosity of the liquid, and \( v \) is the velocity of the sphere. ### Step 2: Analyzing Option (a) The net force acting on the sphere just after entering the liquid is given by: \[ F_{net} = F_g - F_b - F_d \] At the moment just after entering the liquid, the viscous drag force is negligible, so: \[ F_{net} = \rho V g - \sigma V g = (\rho - \sigma)V g \] Using Newton's second law, \( F = ma \): \[ (\rho - \sigma)V g = \rho V a \implies a = g \frac{\rho - \sigma}{\rho} \] Thus, option (a) is **correct**. ### Step 3: Analyzing Option (b) The time taken to attain terminal speed is not directly proportional to \( \rho^{0} \). The time to reach terminal velocity depends on various factors including the radius of the sphere and the viscosity of the liquid. Therefore, this statement is **incorrect**. ### Step 4: Analyzing Option (c) At terminal speed, the viscous force is not maximum; rather, it balances the net downward force (weight minus buoyancy). Therefore, this statement is also **incorrect**. ### Step 5: Analyzing Option (d) At terminal speed, the net force acting on the sphere is indeed zero, as the downward gravitational force is balanced by the upward buoyant force and the viscous drag force. Thus, this statement is **correct**. ### Conclusion From the analysis: - (a) is correct. - (b) is incorrect. - (c) is incorrect. - (d) is correct. Thus, the incorrect options are **(b) and (c)**. ### Final Answer The incorrect options are **(b)** and **(c)**. ---

To solve the problem step by step, let's analyze each option given in the question regarding the sphere dropped into a viscous liquid. ### Step 1: Understanding the Forces Acting on the Sphere When a sphere is dropped into a viscous liquid, the forces acting on it include: 1. The gravitational force acting downwards, \( F_g = \rho V g \), where \( \rho \) is the density of the sphere, \( V \) is the volume of the sphere, and \( g \) is the acceleration due to gravity. 2. The buoyant force acting upwards, \( F_b = \sigma V g \), where \( \sigma \) is the density of the liquid. 3. The viscous drag force acting upwards, \( F_d = 6 \pi r \eta v \), where \( r \) is the radius of the sphere, \( \eta \) is the viscosity of the liquid, and \( v \) is the velocity of the sphere. ...
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