Two strings (A, B) having linear densitiesmu_A = 2 times 10^-4text kg/m$\mu_{A} = 2 \times 10^{-4}\text{ kg/m}$ and mu_B = 4 times 10^-4text kg/m$\mu_{B} = 4 \times 10^{-4}\text{ kg/m}$ and lengths L_A = 2.5text m$L_{A} = 2.5\text{ m}$ and L_B = 1.5text m$L_{B} = 1.5\text{ m}$ 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_1$t_{1}$ and t_2$t_{2}$ , respectively, to reach the joint. The ratio t_1/t_2$t_{1}/t_{2}$ is :
A.1.08
B.1.90
C.1.67
D.1.18
Solution & Explanation
### Related Formula
v = sqrtfracTmu$$v = \sqrt{\frac{T}{\mu}}$$t = fracLv$$t = \frac{L}{v}$$
### Core Logic
Given L_A = 2.5text m$L_{A} = 2.5\text{ m}$, L_B = 1.5text m$L_{B} = 1.5\text{ m}$, T = 500text N$T = 500\text{ N}$.
Velocity in string A:
v_A = sqrtfracTmu_A = sqrtfrac5002 times 10^-4 = sqrt2500000 = 5 sqrt10 times 10^2text m/s$$v_A = \sqrt{\frac{T}{\mu_A}} = \sqrt{\frac{500}{2 \times 10^{-4}}} = \sqrt{2500000} = 5 \sqrt{10} \times 10^{2}\text{ m/s}$$
Velocity in string B:
v_B = sqrtfracTmu_B = sqrtfrac5004 times 10^-4 = sqrt1250000 = 5 sqrt5 times 10^2text m/s$$v_B = \sqrt{\frac{T}{\mu_B}} = \sqrt{\frac{500}{4 \times 10^{-4}}} = \sqrt{1250000} = 5 \sqrt{5} \times 10^{2}\text{ m/s}$$
### Step 1: Calculate Ratio of Times
Time taken to reach joint:
t_1 = fracL_Av_A = frac2.55 sqrt10 times 10^2$$t_{1} = \frac{L_{A}}{v_{A}} = \frac{2.5}{5 \sqrt{10} \times 10^{2}}$$t_2 = fracL_Bv_B = frac1.55 sqrt5 times 10^2$$t_{2} = \frac{L_{B}}{v_{B}} = \frac{1.5}{5 \sqrt{5} \times 10^{2}}$$
Ratio:
fract_1t_2 = frac2.55sqrt10 times frac5sqrt51.5 = frac2.51.5 times fracsqrt5sqrt10 = frac53 times frac1sqrt2$$\frac{t_{1}}{t_{2}} = \frac{2.5}{5\sqrt{10}} \times \frac{5\sqrt{5}}{1.5} = \frac{2.5}{1.5} \times \frac{\sqrt{5}}{\sqrt{10}} = \frac{5}{3} \times \frac{1}{\sqrt{2}}$$fract_1t_2 = frac1.6661.414 approx 1.18$$\frac{t_{1}}{t_{2}} = \frac{1.666}{1.414} \approx 1.18$$
### Pattern Recognition
Time ratio is proportional to fracLsqrtT/mu = L sqrtmu/T$\frac{L}{\sqrt{T/\mu}} = L \sqrt{\mu/T}$. Thus t_1/t_2 = (L_1/L_2) sqrtmu_1/mu_2$t_1/t_2 = (L_1/L_2) \sqrt{\mu_1/\mu_2}$ directly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
Keywords:#Two strings (A, B) having linear densities#JEE Main 2026 Morning Q43#Waves JEE Main 2026#Wave on a String JEE Main 2026
More Waves Previous-Year Questions — Page 4
Q45jee_main_2024_31_jan_morningOrgan Pipes
The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If length of the open pipe is 60mathrm~cm$60\mathrm{~cm}$, the length of the closed pipe will be:
A.60mathrm~cm$60\mathrm{~cm}$
B.45mathrm~cm$45\mathrm{~cm}$
C.30mathrm~cm$30\mathrm{~cm}$
D.15mathrm~cm$15\mathrm{~cm}$
Solution
### Related Formula
f_textclosed, fundamental = fracv4L_c$$f_{\text{closed, fundamental}} = \frac{v}{4L_c}$$f_textopen, 1st overtone = frac2v2L_o$$f_{\text{open, 1st overtone}} = \frac{2v}{2L_o}$$
### Core Logic
Organ Pipes diagram for Q45 - JEE Main 2024 MorningOrgan Pipes diagram for Q45 - JEE Main 2024 Morning
For a closed organ pipe, the fundamental frequency (1st harmonic) is:
f_1 = fracvlambda = fracv4L_1$$f_1 = \frac{v}{\lambda} = \frac{v}{4L_1}$$
where L_1$L_1$ is the length of the closed pipe.
For an open organ pipe, the first overtone (2nd harmonic) is:
f_2 = frac2v2L_2 = fracvL_2$$f_2 = \frac{2v}{2L_2} = \frac{v}{L_2}$$
where L_2$L_2$ is the length of the open pipe (L_2 = 60mathrm\,cm$L_2 = 60\mathrm{\,cm}$).
### Step 2: Equating Frequencies
Given f_1 = f_2$f_1 = f_2$:
fracv4L_1 = fracvL_2$$\frac{v}{4L_1} = \frac{v}{L_2}$$L_2 = 4L_1$L_2 = 4L_1$60 = 4 times L_1$$60 = 4 \times L_1$$L_1 = 15mathrm\,cm$$L_1 = 15\mathrm{\,cm}$$
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Waves
More Waves Questions — jee_main_2026_21_jan_morning
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