
The velocity component vPS(t) along the line joining the source and the observer is given by:
vPS(t) = v(t) cosΟ
where Ο is the angle between the velocity vector and the line joining S and P.
From the geometry of the problem, the distance between the source and the observer is:
r(t) = β(x(t)Β² + LΒ²) = β(aΒ² sinΒ²(Οt) + LΒ²)
The angle Ο satisfies:
cosΟ = L / r(t)
Substituting v(t) = v0 cos(Οt) and cosΟ = L / r(t), we have:
vPS(t) = v0 cos(Οt) Γ (L / β(aΒ² sinΒ²(Οt) + LΒ²))
The component along the line of sight becomes:
vPS(t) = (1/2) Γ (a v0 / β(aΒ² sinΒ²(Οt) + LΒ²)) sin(2Οt)
Thus, statement (A) is correct.
The observed frequency is given by the Doppler effect, which depends on the relative velocity component vPS(t).
Hence, statement (B) is correct.
The correct statements are (A) and (B).
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hookeβs law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Time period of a simple pendulum is longer at the top of a mountain than that at the base of the mountain.
Reason (R): Time period of a simple pendulum decreases with increasing value of acceleration due to gravity and vice-versa. In the light of the above statements.
choose the most appropriate answer from the options given below:

At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/mΒ³, which is dimensionally incorrect for a spectral density. The correct unit J/(mΒ³Β·Hz) or JΒ·s/mΒ³ is used here for the solution.)