Just as Rohan was about to give up on finding a downloadable PDF, he spotted a link that said "Download All Important Derivations of Physics Class 11 PDF". His heart racing, he clicked on the link and was redirected to a Google Drive page.
The minimum speed required for a body to permanently escape a planet's gravitational pull.Work done to move a mass slightly by from distance
y=usinθ(xucosθ)−12g(xucosθ)2y equals u sine theta open paren the fraction with numerator x and denominator u cosine theta end-fraction close paren minus one-half g of open paren the fraction with numerator x and denominator u cosine theta end-fraction close paren squared
T=2usinθgcap T equals the fraction with numerator 2 u sine theta and denominator g end-fraction Using all important derivations of physics class 11 pdf download
ΔU=Δm⋅g(h2−h1)cap delta cap U equals delta m center dot g of open paren h sub 2 minus h sub 1 close paren The change in kinetic energy ( ) of the fluid mass is:
τi=Fi⋅ri=(miriα)⋅ri=miri2αtau sub i equals cap F sub i center dot r sub i equals open paren m sub i r sub i alpha close paren center dot r sub i equals m sub i r sub i squared alpha
∫uvv⋅dv=a∫0sdsintegral from u to v of v center dot d v equals a integral from 0 to s of d s Just as Rohan was about to give up
dL⃗dt=r⃗×F⃗the fraction with numerator d modified cap L with right arrow above and denominator d t end-fraction equals modified r with right arrow above cross modified cap F with right arrow above
[v22]uv=a[s]0sopen bracket the fraction with numerator v squared and denominator 2 end-fraction close bracket sub u to the v-th power equals a open bracket s close bracket sub 0 to the s-th power
Proving that work done equals the change in kinetic energy (for both constant and variable forces). Potential Energy of a Spring: Deriving g′=g(1−dR)g prime equals g of open paren 1
∫ω0ωdω=∫0tα⋅dtintegral from omega sub 0 to omega of d omega equals integral from 0 to t of alpha center dot d t ω−ω0=αtomega minus omega sub 0 equals alpha t ω=ω0+αtomega equals omega sub 0 plus alpha t 2. Second Equation:
R=(ucosθ)⋅(2usinθg)cap R equals open paren u cosine theta close paren center dot open paren the fraction with numerator 2 u sine theta and denominator g end-fraction close paren
Derive the minimum velocity required at the lowest point to complete a vertical circle. Centripetal Force: Derive using vector method.
g′=g(1−dR)g prime equals g of open paren 1 minus the fraction with numerator d and denominator cap R end-fraction close paren 2. Escape Velocity (
This guide provides a complete, chapter-wise breakdown of every crucial derivation you need to know, along with details on where you can download these comprehensive notes for free.