Question

Two strings (A, B) having linear densities \(\mu_A = 2 \times 10^{-4}\) kg/m and, \(\mu_B = 4 \times 10^{-4}\) kg/m and lengths \(L_A = 2.5\) m and \(L_B = 1.5\) m respectively are joined. Free ends of A and B are tied to two rigid supports C and D, respectively creating a tension of 500 N in the wire. Two identical pulses, sent from C and D ends, take time \(t_A\) and \(t_B\), respectively, to reach the joint. The ratio \(t_A/t_B\) is:

• A. 1.08
• B. 1.90
• C. 1.18
• D. 1.67

Hint:

In ratio problems, always write out the full expressions for each quantity before substituting numbers. Often, common terms like the tension (T) in this case will cancel, simplifying the calculation significantly.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We have two strings of different lengths and linear mass densities joined together and kept under a constant tension. We need to find the ratio of the times it takes for a pulse to travel the length of each string.
Step 2: Key Formula or Approach:
The speed of a transverse wave on a string is given by:
\[ v = \sqrt{\frac{T}{\mu}} \] where T is the tension and \(\mu\) is the linear mass density.
The time taken to travel a length L is:
\[ t = \frac{L}{v} = \frac{L}{\sqrt{T/\mu}} = L \sqrt{\frac{\mu}{T}} \] Step 3: Detailed Explanation:
We need to find the ratio \( \frac{t_A}{t_B} \).
Using the formula for time:
\[ t_A = L_A \sqrt{\frac{\mu_A}{T}} \] \[ t_B = L_B \sqrt{\frac{\mu_B}{T}} \] The tension T is the same for both strings. So, when we take the ratio, T will cancel out.
\[ \frac{t_A}{t_B} = \frac{L_A \sqrt{\mu_A/T}}{L_B \sqrt{\mu_B/T}} = \frac{L_A}{L_B} \sqrt{\frac{\mu_A}{\mu_B}} \] Now, substitute the given values:
\(L_A = 2.5\) m
\(L_B = 1.5\) m
\(\mu_A = 2 \times 10^{-4}\) kg/m
\(\mu_B = 4 \times 10^{-4}\) kg/m
\[ \frac{t_A}{t_B} = \frac{2.5}{1.5} \sqrt{\frac{2 \times 10^{-4}}{4 \times 10^{-4}}} = \frac{25}{15} \sqrt{\frac{2}{4}} = \frac{5}{3} \sqrt{\frac{1}{2}} \] \[ \frac{t_A}{t_B} = \frac{5}{3} \frac{1}{\sqrt{2}} = \frac{5}{3\sqrt{2}} \] To get a numerical value, use \( \sqrt{2} \approx 1.414 \):
\[ \frac{t_A}{t_B} \approx \frac{5}{3 \times 1.414} = \frac{5}{4.242} \approx 1.1786 \] Step 4: Final Answer:
The ratio \(t_A/t_B\) is approximately 1.18. This corresponds to option (C).