A point source is kept at the center of a spherically enclosed detector. If the volume of the detector increased by 8…

Q: A point source is kept at the center of a spherically enclosed detector. If the volume of the detector increased by 8 times, the intensity will

Since volume V propto R3, if volume is increased by 8 times: Since intensity I propto (1/R2): If R arrow 2R, the area A = 4pi R2 increases by 22 = 4 times (A arrow 4A). Therefore, intensity becomes I arrow (I0/4). Volume ratio cubed root dictates radius multiplier. The square of that radius multiplier is the area multiplier, which is the exact inverse of the intensity multiplier. Class 11 Physics: Waves

Year	2026	2025	2024	Total
Questions	7	10	5	22

Q 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,

y₁(x,t) = 4 (kx - ω t)

and

y₂(x,t) = 2 (kx - ω t + (2π)/(3))

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

ω

)

A. $[6, (2π)/(3)]$ $\left[6, \frac{2\pi}{3}\right]$
B. $[6, (π)/(3)]$ $\left[6, \frac{\pi}{3}\right]$
C. $[\sqrt(3), (π)/(6)]$ $\left[\sqrt{3}, \frac{\pi}{6}\right]$
D. $[2\sqrt(3), (π)/(6)]$ $\left[2\sqrt{3}, \frac{\pi}{6}\right]$

Solution

Related Formula

Aᵣₑₛ = \sqrt(A₁² + A₂² + 2 A₁ A₂ φ)

θ = (A₂ φ)/(A₁ + A₂ φ)

where, $A₁, A₂$ $A_1, A_2$ = amplitudes of individual harmonic waves $φ$ $\phi$ = phase difference between the waves $Aᵣₑₛ$ $A_{\text{res}}$ = resultant amplitude $θ$ $\theta$ = resultant phase angle relative to the first wave

Core Logic

Given parameters:

$A₁ = 4$ $A_1 = 4$
$A₂ = 2$ $A_2 = 2$
Phase difference, $φ = (2π)/(3) = 120^°$ $\phi = \frac{2\pi}{3} = 120^\circ$

Resultant phasor diagram for amplitude and phase

Aᵣₑₛ = \sqrt(4² + 2² + 2(4)(2) 120^°)

Aᵣₑₛ = \sqrt(16 + 4 + 16 (-0.5)) = \sqrt(20 - 8) = \sqrt(12) = 2\sqrt(3)

Calculate Resultant Phase (angle $θ$ $\theta$):

θ = (2 120^°)/(4 + 2 120^°) = 2 ( \sqrt(3)2)4 + 2 (-0.5) = \sqrt(3)3 = 1\sqrt(3)

θ = (π)/(6)

Step 1: Result Format

Represent the resultant amplitude and phase as a pair:

[2\sqrt(3), (π)/(6)]

Pattern Recognition

Sees: Vector-like addition of wave amplitudes. Shortcut: Solve it like a vector addition problem. Vector $A₁$ $A_1$ along horizontal ( $0$ $0$), and Vector $A₂$ $A_2$ at $120^°$ $120^\circ$. The magnitude is $\sqrt(4²+2²-2(4)(2)(0.5)) = 2\sqrt(3)$ and direction is $π/6$ $\pi/6$. ✓

Chapter Mix

Class 11 Physics: Waves

Q13 jee_main_2025_08_april_evening Wave Speed

Two strings with circular cross section and made of same material, are stretched to have same amount of tension. A transverse wave is then made to pass through both the strings. The velocity of the wave in the first string having the radius of cross section

R

is

v₁

, and that in the other string having radius of cross section

R/2

is

v₂

. Then

v₂v₁ =

A. $\sqrt(2)$ $\sqrt{2}$
B. $2$ $2$
C. $8$ $8$
D. $4$ $4$

Solution

Related Formula

v = \sqrt((T)/(μ))

μ = ρ A = ρ (π R²)

where, $v$ $v$ = velocity of transverse wave on a string $T$ $T$ = tension on string $μ$ $\mu$ = linear mass density $ρ$ $\rho$ = density of material $A$ $A$ = cross-sectional area

Core Logic

Since both strings are made of the same material, their density $ρ$ $\rho$ is the same. Also, they are stretched to have the same amount of tension $T$ $T$.

Substitute the formula for $μ$ $\mu$ into the velocity equation:

v = \sqrt((T)/(ρ π R²)) = (1)/(R) \sqrt((T)/(ρ π))

Thus, the wave velocity is inversely proportional to the radius of the cross section of the string:

v \propto (1)/(R)

(v₂)/(v₁) = (R₁)/(R₂)

Step 1: Ratio Computation

Given:

$R₁ = R$ $R_1 = R$
$R₂ = R/2$ $R_2 = R/2$

Substitute these values:

(v₂)/(v₁) = (R)/(R/2) = 2

Pattern Recognition

Sees: Circular cross-section strings + transverse wave speed relation. Shortcut: Wave velocity $v \propto 1\sqrt(μ) \propto (1)/(R)$ . If the radius is halved, the mass per unit length decreases by 4 times, which makes the speed increase by $\sqrt(4) = 2$ $\sqrt{4} = 2$ times. ✓

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

Solution

Related Formula

v = Bρ

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 $ρ$ $\rho$ is higher for solids, the corresponding elastic modulus parameter increases by several orders of magnitude, dominating the structural velocity index.

Chapter Mix

Class 11 Physics: Waves

Q18 jee_main_2025_04_april_evening Wave Parameters

Displacement of a wave is expressed as

x(t)=5 (628t+(π)/(2)) 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) = A (ω t + φ)

v = (ω)/(K)

K = (2π)/(λ)

Core Logic

From the given wave equation, angular frequency $ω = 628 rad/s$ $\omega = 628\text{ rad/s}$. Given wave velocity $v = 300 m/s$ $v = 300\text{ m/s}$. Using the relation $v = (ω)/(K)$ $v = \frac{\omega}{K}$:

300 = (628)/(K) K = (628)/(300)

Step 1: Compute Wavelength

Substitute $K = (2π)/(λ)$ $K = \frac{2\pi}{\lambda}$:

(2π)/(λ) = (628)/(300)

Since $2π \approx 2 \times 3.14 = 6.28$ $2\pi \approx 2 \times 3.14 = 6.28$, the expression simplifies neatly:

(6.28)/(λ) = (628)/(300) λ = 3 m

Pattern Recognition

Notice standard values like $ω = 628 = 200π$ $\omega = 628 = 200\pi$, which means the frequency is exactly $100 Hz$ $100\text{ Hz}$. Using $v = fλ 300 = 100λ λ = 3 m$ avoids setting up fractions.

Chapter Mix

Class 11 Physics: Waves

Q jee_main_2025_04_april_morning Speed of Sound in Gases

Consider the sound wave travelling in ideal gases of

He

,

CH₄

, and

CO₂

. All the gases have the same ratio

(P)/(ρ)

, where

P

is the pressure and

ρ

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

vHe : v_CH₄ : v_CO₂

is given by

A. $\sqrt((7)/(5)) : \sqrt((5)/(3)) : \sqrt((4)/(3))$
B. $\sqrt((5)/(3)) : \sqrt((4)/(3)) : \sqrt((7)/(5))$
C. $\sqrt((5)/(3)) : \sqrt((4)/(3)) : \sqrt((4)/(3))$
D. $\sqrt((4)/(3)) : \sqrt((5)/(3)) : \sqrt((7)/(5))$

Solution

Related Formula

Laplace correction equation for speed of sound:

v = \sqrt((γ P)/(ρ))

Given that $(P)/(ρ)$ $\frac{P}{\rho}$ is constant for all three gases:

v \propto \sqrt(γ)

where $γ = 1 + (2)/(f)$ $\gamma = 1 + \frac{2}{f}$ (adiabatic constant).

Core Logic

Determine the $γ$ $\gamma$ factor based on molecular atomic structures:

$He$ $\mathrm{He}$ (Monatomic) $f = 3 γHe = (5)/(3)$
$CH₄$ $\mathrm{CH}_{4}$ (Polyatomic/Non-linear) $γ_CH₄ \approx (4)/(3)$ based on experimental references.
$CO₂$ $\mathrm{CO}_{2}$ (Triatomic linear/vibrational modes) $γ_CO₂ \approx (4)/(3)$ as provided in textbook standard testing matrices.

Step 1: Construct the Ratio

Substitute these values into the proportionality:

vHe : v_CH₄ : v_CO₂ = \sqrt((5)/(3)) : \sqrt((4)/(3)) : \sqrt((4)/(3))

Pattern Recognition

When $(P)/(ρ)$ $\frac{P}{\rho}$ is locked down constant, sound speed depends strictly on internal degrees of freedom via $γ$ $\gamma$. Keep standard experimental values of complex gases like $CH₄$ $\mathrm{CH}_4$ and $CO₂$ $\mathrm{CO}_2$ memorized.

Chapter Mix

Class 11 Physics: Waves Class 11 Physics: Kinetic Theory

A point source is kept at the center of a spherically enclosed detector. If the volume of the detector increased by 8 times, the intensity will

Solution & Explanation

Related Formula

Core Logic

Step 1: Intensity Relation

Pattern Recognition

Chapter Mix

Reference Study Guides

More Waves Previous-Year Questions — Page 3

Solution

Related Formula

Core Logic

Step 1: Result Format

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Ratio Computation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Final Conclusion

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute Wavelength

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Construct the Ratio

Pattern Recognition

Chapter Mix

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