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

To solve the problem of finding \frac{b+c}{a+d}, we need to use the principles of dimensional analysis. Let's outline the steps:

1. Identify the physical quantities involved and their dimensions:
   • The time \(t\) has dimensions [T].
   • The density of the material \(\rho\) has dimensions [M][L]^-3.
   • The density of the liquid \(\sigma\) also has dimensions [M][L]^-3.
   • The radius \(r\) has dimensions [L].
   • The viscosity \(\eta\) has dimensions [M][L]^-1[T]^-1.
2. Express the equation for \(t\):

   \(t = k\rho^a r^b \eta^c \sigma^d\)

3. Write down the dimensional equation for the right-hand side:

   \([T] = [M]^a[L]^{-3a} \times [L]^b \times [M]^c[L]^{-c}[T]^{-c} \times [M]^d[L]^{-3d}\)

   Combine the dimensions:

   \([T] = [M]^{a+c+d} \times [L]^{-3a+b-c-3d} \times [T]^{-c}\)

4. Equate the dimensions on both sides:
   • For [M]: \(a + c + d = 0\)
   • For [L]: \(-3a + b - c - 3d = 0\)
   • For [T]: \(-c = 1\)
5. Solve these equations:
   • From \(-c = 1\), we get \(c = -1\).
   • Substitute \(c = -1\) in \(a + c + d = 0\):
     \(a - 1 + d = 0\), which gives \(a + d = 1\).
   • Substitute \(c = -1\) in \(-3a + b - c - 3d = 0\):
     \(-3a + b + 1 - 3d = 0\), simplify to \(-3a + b - 3d = -1\).
6. Find \(b\) using \(a + d = 1\):
   • Substitute \(a = 1 - d\) into \(-3(1 - d) + b - 3d = -1\):
     \(-3 + 3d + b - 3d = -1\), simplifies to \(b = 2\).
7. Calculate \(\frac{b+c}{a+d}\):

   Substitution gives \(\frac{b+c}{a+d} = \frac{2-1}{1} = 1\).

8. Conclusion: The value of \(\frac{b+c}{a+d}\) is 1.

Thus, the correct answer is option 1.