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

To solve this problem, we need to determine the time taken \(t_A\) and \(t_B\) for pulses to travel through strings A and B, and then find the ratio \(t_A/t_B\).

The speed of a wave on a string is given by the formula:

v = \sqrt{\frac{T}{\mu}}

where \(T\) is the tension in the string and \(\mu\) is the linear density of the string.

Given:

• For string A, \(\mu_A = 2 \times 10^{-4}\) kg/m and \(L_A = 2.5\) m
• For string B, \(\mu_B = 4 \times 10^{-4}\) kg/m and \(L_B = 1.5\) m
• Tension \(T = 500\) N

First, calculate the speed of the wave on string A:

v_A = \sqrt{\frac{T}{\mu_A}} = \sqrt{\frac{500}{2 \times 10^{-4}}} = \sqrt{2.5 \times 10^6} = 1581.14 \text{ m/s}

Next, calculate the speed of the wave on string B:

v_B = \sqrt{\frac{T}{\mu_B}} = \sqrt{\frac{500}{4 \times 10^{-4}}} = \sqrt{1.25 \times 10^6} = 1118.03 \text{ m/s}

Now, calculate the time \(t_A\) for the pulse to travel through string A:

t_A = \frac{L_A}{v_A} = \frac{2.5}{1581.14} \approx 0.00158 \text{ s}

Calculate the time \(t_B\) for the pulse to travel through string B:

t_B = \frac{L_B}{v_B} = \frac{1.5}{1118.03} \approx 0.00134 \text{ s}

Finally, find the ratio \(t_A/t_B\):

\frac{t_A}{t_B} = \frac{0.00158}{0.00134} \approx 1.18

Therefore, the correct answer is 1.18, which matches the provided correct answer.