The Question is from neet-physics , and it was created by the author P.NARASIMHA

Question

The viscous force $F$ acting on a sphere moving through a liquid depends on the radius $r$ of the sphere, the velocity $v$ of the sphere, and the coefficient of viscosity $η$ of the liquid.  If $F = kη^arv^b$, where $k$ is a dimensionless constant, using dimensional analysis, find the values of $a$, $b$.

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: Option A
## CONCEPT FRAMEWORK
**Primary Concept**: Dimensional Analysis
**Related Topics**: Units and Dimensions, Viscosity
**Prerequisites**: Knowledge of basic physical quantities and their dimensions
## SOLUTION PATHWAY
**Given Information**:
    *   $F = k\eta^arv^b$
    *   $k$ is dimensionless
    *   $F$ is force, $r$ is radius, $v$ is velocity, $\eta$ is coefficient of viscosity
**Method & Steps**:
    1.  Write the dimensions of each quantity:
        Reasoning: Express each physical quantity in terms of fundamental dimensions (Mass (M), Length (L), Time (T)).
        Formula/Rule: 
            *   $[F] = MLT^{-2}$ (Force)
            *   $[r] = L$ (Length)
            *   $[v] = LT^{-1}$ (Velocity)
            *   $[\eta] = ML^{-1}T^{-1}$ (Coefficient of Viscosity)
    2.  Substitute the dimensions into the given equation:
        Working: $[MLT^{-2}] = [ML^{-1}T^{-1}]^a [L] [LT^{-1}]^b$
        Key Point: The dimensions on both sides of the equation must be equal.
    3.  Simplify and equate the powers of M, L, and T:
        Working: $[MLT^{-2}] = M^a L^{-a}T^{-a} L^1 L^b T^{-b} = M^a L^{-a+1+b} T^{-a-b}$
        *   Comparing powers of M: $1 = a$
        *   Comparing powers of L: $1 = -a + 1 + b$
        *   Comparing powers of T: $-2 = -a - b$
    4.  Solve the system of equations:
        Working: From M, $a=1$. Substituting $a=1$ in the equation for L, we have $1 = -1 + 1 + b$, so $b=1$. We can verify this with T: $-2 = -1 - 1$ which is true.
        Result: $a=1$, $b=1$
    5.  Express the final result:
        Result: $a=1$ and $b=1$
        Verification: The derived values satisfy all the dimensional constraints.
 ## OPTION ANALYSIS
[A] ✓: $a=1$ and $b=1$ which is the correct solution.
[B] :x: $a=2$ and $b=1$ does not satisfy the dimensional analysis equations.
[C] :x: $a=1$ and $b=2$ does not satisfy the dimensional analysis equations.
[D] :x: $a=2$ and $b=2$ does not satisfy the dimensional analysis equations.
 ## LEARNING AIDS
**Common Mistakes**:
    1.  Incorrect dimensions for physical quantities: Ensure the correct dimensions are used for force, velocity, viscosity, etc.
    2.  Errors in equating powers: Pay close attention to the exponents of each dimension and ensure correct algebraic manipulation.
**Quick Tips**:
    1.  Memorize basic dimensions: Knowing the dimensions of force, velocity, acceleration, etc., can speed up the process.
    2.  Double-check equations: Always verify that the dimensions on both sides of the equation are equal after each step.

P.NARASIMHA · Answer

The correct Option is "$a=1, b=1$"

P.NARASIMHA · Answer

$a=2, b=1$

P.NARASIMHA · Answer

$a=1, b=2$

P.NARASIMHA · Answer

$a=2, b=2$

NEET Physics MCQs

Select Chapter

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company