Answer:
To determine the relation for velocity using the method of dimensions, we'll consider the variables involved: weight of the ball \( mg \), coefficient of viscosity \( n \), and radius of the ball \( r \). We'll express the velocity \( v \) as a function of these variables.
Let's denote:
- \( v \) as velocity
- \( m \) as mass of the ball
- \( g \) as acceleration due to gravity
- \( \eta \) as coefficient of viscosity
- \( r \) as radius of the ball
We can express the velocity \( v \) as a function of these variables using dimensional analysis:
\[ v = k \cdot m^a \cdot g^b \cdot \eta^c \cdot r^d \]
where \( k \) is a dimensionless constant, and \( a, b, c, \) and \( d \) are the exponents to be determined.
Now, let's analyze the dimensions of each variable:
- Dimension of velocity \( [v] = [LT^{-1}] \) (where L represents length, and T represents time)
- Dimension of mass \( [m] = [M] \) (where M represents mass)
- Dimension of acceleration due to gravity \( [g] = [LT^{-2}] \)
- Dimension of coefficient of viscosity \( [\eta] = [ML^{-1}T^{-1}] \)
- Dimension of radius \( [r] = [L] \)
Now, let's substitute these dimensions into the equation:
\[ [LT^{-1}] = k \cdot [M]^a \cdot [LT^{-2}]^b \cdot [ML^{-1}T^{-1}]^c \cdot [L]^d \]
Equating dimensions on both sides, we get:
\[ L^1 T^{-1} = k \cdot M^a \cdot L^{b+c} T^{-2b-c} \]
Now, equating the powers of each dimension, we get:
For length dimension:
\[ 1 = b + c + d \]
For time dimension:
\[ -1 = -2b - c \]
For mass dimension:
\[ 0 = a + c \]
Solving these equations, we find:
\[ b = -\frac{1}{2}, \, c = \frac{1}{2}, \, d = 1 \]
Now, substituting these values back into the equation for velocity:
\[ v = k \cdot m^a \cdot g^b \cdot \eta^c \cdot r^d \]
\[ v = k \cdot m^a \cdot g^{-\frac{1}{2}} \cdot \eta^{\frac{1}{2}} \cdot r \]
As \( g \) is constant and \( a = 0 \) (since velocity does not depend on mass), we can rewrite the equation as:
\[ v = k \cdot \frac{\sqrt{\eta}}{\sqrt{g}} \cdot r \]
This is the relation for velocity in terms of the given variables.
Explanation:
Mark me as brainleist
