Question

A sounding body emitting a frequency of 150 Hz is dropped from a height. During its fall under gravity it crosses a balloon moving upwards with a constant velocity of 2m/s one second after it started to fall. The difference in the frequency observed by the man in balloon just before and just after crossing the body will be: (given that -velocity of sound = 300m/s; g = 10m/s2)

Answer 1

Title:
Pata ni kyu

Content:
Ajkal muje bhook todi jayada lagati hai
Par mein kam khane ka try karta hoo
Pata ni kyu
Shayad vo muje bhookad kehti h
Isliye

Hum frds baaten kar rahe the
Ki future me kya banna h
Me khi magan ( khoya hua) tha
Kisi yad mein
Tapak se bol diya
Pagak banna h mujr to
Pata ni kyu
Shayd vo muje pagal kehti h
Isliye

Ab khavab tode jayada aane lage h
Dost kehte h tu
Sote hue bhi hasta kyu h
Pata ni kyu
Shayd khavab me vo aati h
Isliye

Pata ni kyu
jab bhi room ka gate dekhta hoo
To ajeeb lagta hai
Face par ekdam smile aa jati hai
Ek alag ehsaas hota h
Pata ni kyu
Shayd dil ko lagata h
Koi hai us par khada
Bas usko godi me lene ka man karta hai

Kal ki hi baat hai
Mein aur mera dist baate kar rahe the
Last mein jab jane ko hue
To mera dost bola
" Chal Bhag"
Me hasne lag gaya
Laga tum hi ho samane
Jake ek tight vala hug kar loo
Fir khi aur nahi jaane doo
Par jab tum kehti ho
Ki ab jayada baate ni hogi(11th hostel)
To pata nhi ajib lagta hai
Aankhen bhar aati hai
Par dil man jata hai
Pata ni kyu
Shayd dil ko lagata h
Tumne kha , sahi hi kaha hoga

Note:
✈ Yeh mere khud ke vichar hai,
Jayada khush na hove
♨ Jalebi to banti hai is bat par

Answer 2

after 1 sec, falling point source (with frequency 150 Hz) will attain velocity of

Vs = 10 ms⁻¹ , and observer sitting on the balloon is coming upward with the velocity

V₀ = 2 ms⁻¹, say frequencies observed

by observer before and after crossing are n₁ and n₂

respectively, then

n₁ = ( 300 + 2 / 300 - 10 ) * 150 = 156.20 Hz

n₂ = ( 300 - 2 /  300 + 10 ) * 150 = 144.19 Hz

n₁ - n₂ = 12 Hz ( approx )