Question

Time taken to achieve terminal velocity by a body depends on density of material (\(\rho\)), density of liquid (\(\sigma\)), radius of material (r) and viscosity of liquid (\(\eta\)) as \(t = k\rho^a r^b \eta^c \sigma^d\). Find \(\frac{b+c}{a+d}\) ?

• A. 1
• B. \(\frac{1}{2}\)
• C. 3
• D. 2

Hint:

In dimensional analysis problems, always write down the dimensions of all physical quantities carefully.
Look for opportunities to simplify the system of equations. In this case, solving for \(c\) and the sum \(a+d\) first made finding \(b\) much easier.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a formula for the time (\(t\)) to achieve terminal velocity, expressed in terms of several physical quantities. We need to find the value of an expression involving the exponents (\(a, b, c, d\)) using the principle of dimensional homogeneity.
Step 2: Key Formula or Approach:
The principle of dimensional homogeneity states that an equation is dimensionally correct if the dimensions of all the terms on both sides of the equation are the same. We will write the dimensions of each quantity and equate the powers of mass (M), length (L), and time (T).
Step 3: Detailed Explanation:
First, let's list the dimensions of each quantity involved:
- Time (\(t\)): \([T]\)
- Density (\(\rho, \sigma\)): \([ML^{-3}]\)
- Radius (\(r\)): \([L]\)
- Viscosity (\(\eta\)): \([ML^{-1}T^{-1}]\)
- \(k\) is a dimensionless constant.
Now, we write the dimensional equation for the given relation \(t = k\rho^a r^b \eta^c \sigma^d\):
\[ [M^0 L^0 T^1] = [ML^{-3}]^a [L]^b [ML^{-1}T^{-1}]^c [ML^{-3}]^d \] Combine the powers of M, L, and T on the right side:
\[ [M^0 L^0 T^1] = [M^{a+c+d}] [L^{-3a+b-c-3d}] [T^{-c}] \] Now, we equate the exponents of M, L, and T from both sides:
1. For T: \( 1 = -c \implies c = -1 \)
2. For M: \( 0 = a+c+d \implies a + (-1) + d = 0 \implies a+d = 1 \)
3. For L: \( 0 = -3a+b-c-3d \). We can rearrange this as \( 0 = -3(a+d) + b - c \).
We have the values for \(c\) and \(a+d\). Let's substitute them into the equation for L to find \(b\):
\[ 0 = -3(1) + b - (-1) \] \[ 0 = -3 + b + 1 \] \[ 0 = b - 2 \implies b = 2 \] Step 4: Final Answer:
We need to find the value of the expression \(\frac{b+c}{a+d}\).
We found \(b=2\), \(c=-1\), and \(a+d=1\).
\[ \frac{b+c}{a+d} = \frac{2 + (-1)}{1} = \frac{1}{1} = 1 \]