Two strings (A, B) having linear densities muA = 2 x 10-4 kg/m and muB = 4 x 10-4 kg/m and lengths LA = 2.5 m and LB = 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 t1 and t2 , respectively, to reach the joint. The ratio t1/t2 is :

Given LA = 2.5 m, LB = 1.5 m, T = 500 N. Velocity in string A: Velocity in string B: Time taken to reach joint: Ratio: Time ratio is proportional to (L/sqrt(T/mu)) = L sqrt(mu/T). Thus t1/t2 = (L1/L2) sqrt(mu1/mu2) directly. Class 11 Physics: Waves

Two strings (A, B) having linear densities muA = 2 x 10-4 kg/m and muB = 4 x 10-4 kg/m and lengths LA = 2.5 m and LB = 1.5 m respectively are joined. Free ends of A and B are tied to two rigid support

Given LA = 2.5 m, LB = 1.5 m, T = 500 N. Velocity in string A: Velocity in string B: Time taken to reach joint: Ratio: Time ratio is proportional to (L/sqrt(T/mu)) = L sqrt(mu/T). Thus t1/t2 = (L1/L2) sqrt(mu1/mu2) directly. Class 11 Physics: Waves

Two strings (A, B) having linear densities muA = 2 x 10-4 kg/m and muB = 4 x 10-4 kg/m and lengths LA = 2.5 m and LB = 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 t1 and t2 , respectively, to reach the joint. The ratio t1/t2 is :

Given LA = 2.5 m, LB = 1.5 m, T = 500 N. Velocity in string A: Velocity in string B: Time taken to reach joint: Ratio: Time ratio is proportional to (L/sqrt(T/mu)) = L sqrt(mu/T). Thus t1/t2 = (L1/L2) sqrt(mu1/mu2) directly. Class 11 Physics: Waves

Two strings (A, B) having linear densities muA = 2 x 10-4 kg/m and muB = 4 x 10-4 kg/m and lengths LA = 2.5 m and LB = 1.5 m respectively are joined. Free ends of A and B are tied to two rigid support

Given LA = 2.5 m, LB = 1.5 m, T = 500 N. Velocity in string A: Velocity in string B: Time taken to reach joint: Ratio: Time ratio is proportional to (L/sqrt(T/mu)) = L sqrt(mu/T). Thus t1/t2 = (L1/L2) sqrt(mu1/mu2) directly. Class 11 Physics: Waves

Solution & Explanation

### Related Formula

v = sqrtfracTmu

t = fracLv

### Core Logic Given

L_A = 2.5text m

L_B = 1.5text m

T = 500text N

. Velocity in string A:

v_A = sqrtfracTmu_A = sqrtfrac5002 times 10^-4 = sqrt2500000 = 5 sqrt10 times 10^2text m/s

Velocity in string B:

v_B = sqrtfracTmu_B = sqrtfrac5004 times 10^-4 = sqrt1250000 = 5 sqrt5 times 10^2text 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_2 = fracL_Bv_B = frac1.55 sqrt5 times 10^2

Ratio:

fract_1t_2 = frac2.55sqrt10 times frac5sqrt51.5 = frac2.51.5 times fracsqrt5sqrt10 = frac53 times frac1sqrt2

fract_1t_2 = frac1.6661.414 approx 1.18

### Pattern Recognition Time ratio is proportional to

fracLsqrtT/mu = L sqrtmu/T

. Thus

t_1/t_2 = (L_1/L_2) sqrtmu_1/mu_2

directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

More Waves Previous-Year Questions — Page 2

Q17 jee_main_2025_08_april_evening Superposition of Waves

The amplitude and phase of a wave that is formed by the superposition of two harmonic travelling waves,

y_1(x,t) = 4sin(kx - omega t)

and

y_2(x,t) = 2sinleft(kx - omega t + frac2pi3 ight)

, are: (Take the angular frequency of initial waves same as

omega

)

A. $left[6, frac2pi3right]$ $\left[6, \frac{2\pi}{3}\right]$
B. $left[6, fracpi3right]$ $\left[6, \frac{\pi}{3}\right]$
C. $left[sqrt3, fracpi6right]$ $\left[\sqrt{3}, \frac{\pi}{6}\right]$
D. $left[2sqrt3, fracpi6right]$ $\left[2\sqrt{3}, \frac{\pi}{6}\right]$

Solution

### Related Formula

A_textres = sqrtA_1^2 + A_2^2 + 2 A_1 A_2 cosphi

tantheta = fracA_2 sinphiA_1 + A_2 cosphi

where,

A_1, A_2

= amplitudes of individual harmonic waves

phi

= phase difference between the waves

A_textres

= resultant amplitude

theta

= resultant phase angle relative to the first wave ### Core Logic Given parameters: -

A_1 = 4

A_2 = 2

- Phase difference,

phi = frac2pi3 = 120^circ

Calculate Resultant Amplitude:

A_textres = sqrt4^2 + 2^2 + 2(4)(2) cos 120^circ

A_textres = sqrt16 + 4 + 16 left(-0.5right) = sqrt20 - 8 = sqrt12 = 2sqrt3

Calculate Resultant Phase (angle

theta

tantheta = frac2 sin 120^circ4 + 2 cos 120^circ = frac2 left(fracsqrt32right)4 + 2 left(-0.5right) = fracsqrt33 = frac1sqrt3

theta = fracpi6

### Step 1: Result Format Represent the resultant amplitude and phase as a pair:

left[2sqrt3, fracpi6right]

### Pattern Recognition Sees: Vector-like addition of wave amplitudes. Shortcut: Solve it like a vector addition problem. Vector

A_1

along horizontal (

0

), and Vector

A_2

120^circ

. The magnitude is

sqrt4^2+2^2-2(4)(2)(0.5) = 2sqrt3

and direction is

pi/6

. ✓ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

Q8 jee_main_2025_28_jan_morning Speed of Sound in Medium

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: A sound wave has higher speed in solids than gases. Reason R: Gases have higher value of Bulk modulus than solids. In the light of the above statements, choose the correct answer from the options given below.

A. $textBoth A and R are true and R is the correct explanation of A$
B. $textA is false but R is true$ $\text{A is false but R is true}$
C. $textBoth A and R are true but R is NOT the correct explanation of A$
D. $textA is true but R is false$ $\text{A is true but R is false}$

Solution

### Related Formula

v = sqrtfracmathrmBrho

### Core Logic Assertion A: Sound velocity relies on structural elasticity bounds. Solids are highly rigid compared to fluids, making speed significantly higher. (True) Reason R: Solids resist structural compression far better than unbonded gases, giving them significantly higher Bulk Modulus properties. Thus, statement R is completely false. ### Step 1: Final Conclusion Assertion A is true, but Reason R is false, aligning perfectly with option (4). ### Pattern Recognition Even though density

rho

is higher for solids, the corresponding elastic modulus parameter increases by several orders of magnitude, dominating the structural velocity index. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

Q18 jee_main_2025_04_april_evening Wave Parameters

Displacement of a wave is expressed as

x(t)=5cosleft(628t+fracpi2right)text m

. The wavelength of the wave when its velocity is 300 m/s is:

A. 5 m
B. 3 m
C. 0.5 m
D. 0.33 m

Solution

### Related Formula

x(t) = Acos(omega t + phi)

v = fracomegaK

K = frac2pilambda

### Core Logic From the given wave equation, angular frequency

omega = 628text rad/s

. Given wave velocity

v = 300text m/s

. Using the relation

v = fracomegaK

300 = frac628K implies K = frac628300

### Step 1: Compute Wavelength Substitute

K = frac2pilambda

frac2pilambda = frac628300

Since

2pi approx 2 times 3.14 = 6.28

, the expression simplifies neatly:

frac6.28lambda = frac628300 implies lambda = 3text m

### Pattern Recognition Notice standard values like

omega = 628 = 200pi

, which means the frequency is exactly

100text Hz

. Using

v = flambda implies 300 = 100lambda implies lambda = 3text m

avoids setting up fractions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

Q7 jee_main_2025_04_april_morning Speed of Sound in Gases

Consider the sound wave travelling in ideal gases of

mathrmHe

mathrmCH_4

, and

mathrmCO_2

. All the gases have the same ratio

fracP ho

, where

P

is the pressure and

ho

is the density. The ratio of the speed of sound through the gases

v_mathrmHe : v_mathrmCH_4 : v_mathrmCO_2

is given by

A. $sqrtfrac75 : sqrtfrac53 : sqrtfrac43$
B. $sqrtfrac53 : sqrtfrac43 : sqrtfrac75$
C. $sqrtfrac53 : sqrtfrac43 : sqrtfrac43$
D. $sqrtfrac43 : sqrtfrac53 : sqrtfrac75$

Solution

### Related Formula Laplace correction equation for speed of sound:

v = sqrtfracgamma P ho

Given that

fracP ho

is constant for all three gases:

v propto sqrtgamma

where

gamma = 1 + frac2f

(adiabatic constant). ### Core Logic Determine the

gamma

factor based on molecular atomic structures: 1.

mathrmHe

(Monatomic)

implies f = 3 implies gamma_mathrmHe = frac53

mathrmCH_4

(Polyatomic/Non-linear)

implies gamma_mathrmCH_4 approx frac43

based on experimental references. 3.

mathrmCO_2

(Triatomic linear/vibrational modes)

implies gamma_mathrmCO_2 approx frac43

as provided in textbook standard testing matrices. ### Step 1: Construct the Ratio Substitute these values into the proportionality:

v_mathrmHe : v_mathrmCH*4 : v*mathrmCO2 = sqrtfrac53 : sqrtfrac43 : sqrtfrac43

### Pattern Recognition When

fracP ho

is locked down constant, sound speed depends strictly on internal degrees of freedom via

gamma

. Keep standard experimental values of complex gases like

mathrmCH_4

and

mathrmCO_2

memorized. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves Class 11 Physics: Kinetic Theory

Q16 jee_main_2025_04_april_morning Organ Pipes and Standing Waves

In an experiment with a closed organ pipe, it is filled with water by

left(frac15 ight)

th of its volume. The frequency of the fundamental note will change by

A. 25%
B. 20%
C. -20%
D. -25%

Solution

### Related Formula Fundamental frequency of a closed organ pipe:

f_1 = fracv4l

where

l

is the acoustic air column column length. ### Core Logic Initially, full air column length =

l

. Filling

frac15

of its space with fluid reduces the available vibrating air tract space down to:

l_2 = l - frac15l = frac45l

Resonating length air column profile initial state for Q16 - JEE Main 2025 Morning

### Step 1: Calculate New Frequency The modified acoustic frequency response is:

f_2 = fracv4l_2 = fracv4left(frac45l ight) = frac5v16l = frac54f_1

Resonating length air column profile initial state for Q16 - JEE Main 2025 Morning

### Step 2: Determine Percentage Shift

Delta f\% = fracf_2 - f_1f_1 times 100 = left( frac54 - 1 ight) times 100 = 25\%

### Pattern Recognition Shortening the resonance tube length raises pitch frequency proportionally. Shifting length down to

80\%

drives frequency up to

125\%

, yielding a positive

25\%

upward change jump. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Waves

YOUR FIRST PREP STEP STARTS HERE

We Map Every Repeating Question in Competitive Exams.

Say goodbye to generic mock test fatigue. RankBit uses smart analysis to group past exam questions into their foundational Repeating Question Types. Find chapter weightage, track repeating questions, and score higher with targeted practice.

Select Your Target Exam

Choose an exam track below to find formulas per chapter and patterns.

Syncing Exam Intelligence

Mapping formulas and patterns across all tracks…

PATH A — FULL LENGTH PRACTICE

Full Mock Test Hub

Simulate real NTA exam conditions with fully tracked mocks. Time yourself against past papers.

Now Live Open

PATH B — TARGETED PRACTICE

Topic-wise Practice Hub

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.

Loading Questions... Browse Topics

Exam Analysis Tools

Chapter weightage trends and topic-wise practice — all mapped from past shift papers.

Year-wise PYQ

Chapter Weightage & Trends

See high-yield chapters, year-wise counts and the frequency-based chance each topic repeats next exam.

Weightage Board Open

Targeted PYQs

Topic-wise Practice Hub

Every past-year question for a chapter pulled from all papers, with its key formulas alongside.

One chapter at a time Open

Latest from the Blog

View all →

Loading articles...

Review & Tools

Mistakes To Revise

No conceptual mistakes detected.

🔥 Practice Session Live

Showing All Questions

Exam Questions

Test your concepts live. Choose options or enter numerical values, then verify your answer to reveal the double-box solution matrix.

Solution & Explanation

Reference Study Guides

More Waves Previous-Year Questions — Page 2

Solution

Solution

Solution

Solution

Solution

More Waves Questions — jee_main_2026_21_jan_morning

We Map Every Repeating Question in Competitive Exams.

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Solution & Explanation

Reference Study Guides

More Waves Previous-Year Questions — Page 2

Solution

Solution

Solution

Solution

Solution

More Waves Questions — jee_main_2026_21_jan_morning

We Map Every Repeating Question in Competitive Exams.

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back