Class 11 Physics vector Notes pdf Download

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class 11 physics vector notes

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  • Speed: Distance covers by a body per unit time is call speed

speed:distance/time

  • Velocity:Distance covers by a body per unit time in a definite direction is call velocity.

velocity:displacement/timedimension [V]= [LT(inverse of -1]

  • Displacement:Distance covers in a definite direction is call displacement .

Displacement =velocity ⨉ time Accelaration:[LT(inverse of -2)]

Class 11 Physics vector Notes pdf Download
Class 11 Physics vector Notes pdf Download


Establish the relation between Newton and Dyne . 1N = 1kgmS-2          =1 ⨉ 1000⨉100⨉ Sec-2       = 105  Dy Establish the relation between Jule and erg. Answer:1J = 1Kgm2sec-2

                        =1 × 1000 × (100)2                                                  =107gmcm2sec-2 


Parallelogram law of vectore  addition Class 11 Physics vector Notes pdf Download


If two arms of a parallelogram represent two vectore acting at a point then the hypothesis draw from the point of action represent the result and vectore .            
          Let ABCD is a parallelogram .AB arm represent P and AD arm represent Q.  ө is the angle between P and Q.
As shown in figure CE is the perpendicular draw from C therefore CEA is the right angle triangle .                   AC2 = AE2 + EC2

                           =(AB + BC)2 + CE2
                           =AB2 +BC2 2AB.BC + CE2                           =P2 + (BE2 +CE2) +2P.BE                           =P2 + BC2 + 2P.BE
                           =P2 + Q2 + 2P.BE ———————————(1)


from ∇CEB
cosθ =BE/BC
BE=BCcosθ
BE=Qcosθ
Therefore (1)

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AC2=P2 + Q2 + 2P.Qcosθ
R=√(P2 + Q2 + 2P.Qcosθ)



Direction of resultant vectore

parallelogram Class 11 Physics vector Notes pdf Download

Let us consider angle between resultant vectore and P is ∝ . From ∇CEA
tan∝ = CE/AE         =CE/AB+BE         =CE/P+BE ——————(1)



∇CEA
cosθ =BE/BC                                       
   BE=BCcosθ
     BE=Qcosθ Class 11 Physics vector Notes pdf Download  

sinθ=CE/BC
CE=Qsinθ

therefore (1)
tan∝ = Qsinθ/P + Q cosθ
∝ = 1/tan(Qsinθ/P + Q cosθ 







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