Verified Solution • 2025–26 Session moderate

QJEEAdv19Ph9bae: JEE Advanced 2019 Physics - Class 12 JEEAdv19Ph9bae Solution: A perfectly reflecting mirror of mass M mounted on a spring…

A perfectly reflecting mirror of mass M mounted on a spring constitutes a spring-mass system of angular frequency Ω\Omega such that 4πMΩh=1024m−2{{4\pi M\Omega } \over h} = {10^{24}}{m^{ - 2}} with h as Planck's constant. N photons of wavelength λ\lambda = 8 π\pi ×\times 10 −- 6 m strike the mirror simultaneously at normal incidence such that the mirror gets displaced by 1 μ\mu m. If the value of N is x ×\times 1012, then the value of x is ................ [Consider the spring as massless]

JEE Advanced 2019 Paper 2 Offline Physics - Dual Nature of Radiation Question 20 English

Step-by-Step Verified Solution

Momentum transferred on mirror = 2Nhλ{{2Nh} \over \lambda }

2Nhλ=MV(mean position){{2Nh} \over \lambda } = M{V_{(mean\,position)}}

V(mean position)=ΩA{V_{(mean\,position)}} = \Omega A (where, A = 1 μ\mu m)

2Nhλ{{2Nh} \over \lambda } =MΩA= M\Omega A

(where λ\lambda = 8 π\pi ×\times 10 −- 6)

N=MΩ(10−6)λ2hN = {{M\Omega ({{10}^{ - 6}})\lambda } \over {2h}}

=MΩ8π×10−6×10−62h= {{M\Omega 8\pi \times {{10}^{ - 6}} \times {{10}^{ - 6}}} \over {2h}}

N=4πMΩh×10−12N = {{4\pi M\Omega } \over h} \times {10^{ - 12}}

=10−24×10−12= {10^{ - 24}} \times {10^{ - 12}}

N=1×10−12⇒x=1N = 1 \times {10^{ - 12}} \Rightarrow x = 1

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring