Two solid spheres of same metal but of mass \(\mathrm{M}\) and \(8 \mathrm{M}\) full simultaneously on a viscous liquid and their terminal velocity are \(\mathrm{V}\) and ' \(\mathrm{nV}^{\prime}\) then value of 'n' is (A) 16 (B) 8 (C) 4 (D) 2

Short Answer

Expert verified
The value of 'n' is 4. Option (C) is correct.

Step by step solution

01

Understand the terminal velocity formula.

The formula for terminal velocity is given by: \(v_t = \dfrac{2}{9}\dfrac{(\rho_p - \rho_f)r^2g}{\eta}\) Where: - \(v_t\) is the terminal velocity - \(\rho_p\) is the density of the sphere - \(\rho_f\) is the density of the fluid - \(r\) is the radius of the sphere - \(g\) is the acceleration due to gravity - \(\eta\) is the dynamic viscosity of the fluid
02

Set up equations for both spheres.

Since both the spheres have the same metal and are falling through the same fluid, their densities and fluid density will be the same. So, we can set up the equations for both spheres as follows: Sphere 1: \(V = \dfrac{2}{9}\dfrac{(\rho_p - \rho_f)r_1^2g}{\eta}\) Sphere 2: \(nV = \dfrac{2}{9}\dfrac{(\rho_p - \rho_f)r_2^2g}{\eta}\) We know that the mass of sphere 2 is 8 times sphere 1. So, \(M_2 = 8M_1\) Since both spheres are made of the same material, their volume ratio will also be the same as their mass ratio. And from the mass ratio, we can find the ratio of their radii: \(\dfrac{M_2}{M_1} = \dfrac{8}{1} = \dfrac{4 \PageIndex{3} r_2^3}{4/3 r_1^3}\) This simplifies to: \(\dfrac{r_2}{r_1} = 2\)
03

Solve for 'n'.

We can now substitute the radius ratio in both sphere equations: \(V = \dfrac{2}{9}\dfrac{(\rho_p - \rho_f)r_1^2g}{\eta}\) \(nV = \dfrac{2}{9}\dfrac{(\rho_p - \rho_f)(2r_1)^2g}{\eta}\) Now, we can divide the second equation by the first one to solve for 'n': \(\dfrac{nV}{V} = \dfrac{\dfrac{2}{9}\dfrac{(\rho_p - \rho_f)(2r_1)^2g}{\eta}}{\dfrac{2}{9}\dfrac{(\rho_p - \rho_f)r_1^2g}{\eta}}\) Cancelling out all the common terms, we get: \(n = \dfrac{(2r_1)^2}{r_1^2} = 2^2 = 4\) So, the value of 'n' is 4. The correct option is (C) 4.

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