Two strings (A, B) having linear densitiesμA = 2 × 10⁻⁴ kg/m$\mu_{A} = 2 \times 10^{-4}\text{ kg/m}$ and μB = 4 × 10⁻⁴ kg/m$\mu_{B} = 4 \times 10^{-4}\text{ kg/m}$ and lengths LA = 2.5 m$L_{A} = 2.5\text{ m}$ and LB = 1.5 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 t₁$t_{1}$ and t₂$t_{2}$ , respectively, to reach the joint. The ratio t₁/t₂$t_{1}/t_{2}$ is :
A.1.08
B.1.90
C.1.67
D.1.18
Solution & Explanation
Related Formula
v = √((T)/(μ))$$v = \sqrt{\frac{T}{\mu}}$$t = (L)/(v)$$t = \frac{L}{v}$$
Core Logic
Given LA = 2.5 m$L_{A} = 2.5\text{ m}$, LB = 1.5 m$L_{B} = 1.5\text{ m}$, T = 500 N$T = 500\text{ N}$.
Velocity in string A:
Time ratio is proportional to L√(T/μ) = L √(μ/T)$\frac{L}{\sqrt{T/\mu}} = L \sqrt{\mu/T}$. Thus t₁/t₂ = (L₁/L₂) √(μ₁/μ₂)$t_1/t_2 = (L_1/L_2) \sqrt{\mu_1/\mu_2}$ directly.
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
Q46jee_main_2026_22_january_morningInterference of Sound Waves
Two loudspeakers L₁$L_{1}$ and L₂$L_{2}$ are placed with a separation of 10 m, as shown in figure. Both speakers are fed with an audio input signal of same frequency with constant volume. A voice recorder, initially at point A, at equidistance to both loud speakers, is moved by 25 m along the line AB while monitoring the audio signal. The measured signal was found to undergo 10 cycles of minima and maxima during the movement. The frequency of the input signal is \_\_\_\_ Hz
(Speed of sound in air is 324~m/s$324~\mathrm{m/s}$ and √(5)=2.23$\sqrt{5}=2.23$)
Two loudspeakers setup with path difference for sound interference.
Numerical Answer.Answer: 600 to 600
Solution
Related Formula
Δ x = nλ, v = fλ$$\Delta x = n\lambda, \quad v = f\lambda$$
Core Logic
Two loudspeakers setup with path difference for sound interference.
Sees: Loudspeaker interference cycles along displacement line.
Shortcut: Compute path difference at endpoint B, relate to total interference cycles to find wavelength, then frequency.
Check: Numerical answer is 600 Hz. ✓
In an open organ pipe v₃$v_{3}$ and v₆$v_{6}$ are 3rd$3^{\mathrm{rd}}$ and 6th$6^{\mathrm{th}}$ harmonic frequencies, respectively. If v₆ - v₃ = 2200$v_{6} - v_{3} = 2200$ Hz then length of the pipe is ____ mm.
(Take velocity of sound in air is 330 m/s.)
A.275$275$
B.225$225$
C.200$200$
D.250$250$
Solution
Related Formula
fₙ = n ((v)/(2L))$$f_n = n \left(\frac{v}{2L}\right)$$
Core Logic
For an open organ pipe, harmonic frequency vₙ = n · f₀ = n ((v)/(2L))$v_n = n \cdot f_0 = n \left(\frac{v}{2L}\right)$.
The speed of sound in an ideal gas is directly proportional to the square root of its absolute temperature in Kelvin.
Initial state: T₁ = 0°C = 273 K$T_1 = 0^{\circ}\mathrm{C} = 273 \, \mathrm{K}$, V₁ = V₀$V_1 = V_0$
Final state: T₂ = α°C = (α + 273) K$T_2 = \alpha^{\circ}\mathrm{C} = (\alpha + 273) \, \mathrm{K}$, V₂ = 2V₀$V_2 = 2V_0$
If velocity doubles, absolute temperature must quadruple (2² = 4$2^2 = 4$). 4 × 273 = 1092 K$4 \times 273 = 1092 \, \mathrm{K}$, which translates perfectly to 819°C$819^{\circ}\mathrm{C}$.
Chapter Mix
Class 11 Physics: Waves
Class 11 Physics: Kinetic Theory of Gases
Q34jee_main_2026_24_january_eveningOrgan Pipes and Resonance
The fifth harmonic of a closed organ pipe is found to be in unison with the first harmonic of an open pipe. The ratio of lengths of closed pipe to that of the open pipe is 5/x. The value of x is ____.
A.4$4$
B.2$2$
C.1$1$
D.3$3$
Solution
Related Formula
fn, closed = n v4Lclosed$$f_{n, \text{closed}} = \frac{n v}{4L_{\text{closed}}}$$ (where n$n$ is an odd integer)
fn, open = n v2Lₒₚₑₙ$$f_{n, \text{open}} = \frac{n v}{2L_{\text{open}}}$$ (where n$n$ is any integer)
Core Logic
We are given that the fifth harmonic of the closed organ pipe equals the first harmonic of the open pipe:
For a closed pipe, the fundamental is v/4L$v/4L$. For an open pipe, it's v/2L$v/2L$. If an odd harmonic n$n$ of a closed pipe matches the m$m$-th harmonic of an open pipe, Lc / Lₒ = n/2m$L_c / L_o = n/2m$.
Chapter Mix
Class 11 Physics: Waves
More Waves Questions — jee_main_2026_21_jan_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
