1. Xtra Edge February 2012 - Career Point
1. Xtra Edge February 2012 - Career Point 1. Xtra Edge February 2012 - Career Point
(b) Inductive Reactance XL = ωL (3) Impendence = Z = (4) XC ν 2 R + ( X XL L – X 25. Force = F = Bil For current (i) 2 2 ⎛ emf ⎞ Bvl B vl i = ⎜ ⎟ = ⇒ F = ⎝ R ⎠ R R 26. Cyclotron work on the fact that a positively charged particle can be accelerated to a sufficiently high energy with the help of smaller values of oscillating electric field by making it to cross the same electric field time and again with use of strong magnetic field. North Magnet H.F oscillator Dees D2 27. ∫ r qin E. ds = ∈ E.2πrl = E = λ 2π ∈0 0 λ· l ∈ 0 r South B (Magnetic Field) + + + + + + + + + + + + + + λ r Dees D1 l C ) ν 2 target 28. Bending of light from sharp edges is called diffraction. Difference between interference and diffraction. Interference Diffraction (i) Two coherent sources (i) One source is are necessary Necessary (ii) All fringes are of (ii) CM has double same width width than all other fringes (iii) All bright fringes has (iii) as order of bright equal intensity fringes increases, intensity goes down (iv) For bright fringes (iv) For bright fringes, path difference = nλ Path difference λ = (2n – 1) . 2 (v) For dark fringes, path (v) For dark fringes, difference = (2n – path difference λ 1) 2 = nλ Angular width of CM = 2λ 2× 5000× 10 = −3 a 0. 5× 10 = 2 × 10 –3 Radian OR When white light incidents over prism then it divide it into component colours. This phenomenon is called diffraction and angle between end colours is called angular dispersion. Prism deviates violet colour maximum as its refractive index for violet colour is maximum. XtraEdge for IIT-JEE 90 FEBRUARY 2012 ω = µ v − µ µ −1 y 1. 5318 −1. 5140 ∴ ω = 1. 5170 −1 ∴ ω = .034 29. white Types of spectrum Spectrum is of 2 types. (i) Emission (ii) Absorption. R −10
(i) Emission : - If radiations emitted from a substance is obtained ones screen after passing through the prism then spectrum is called emission spectrum. It is line emission spectrum for substances at atomic level in gaseous state. It is continuous spectrum for substances in liquid or solid state. (ii) Absorption spectrum : - The substance of which absorption spectrum is to be find out is placed in a transparent tube and white light is passed through it now substance absorb radiation corresponding to some particular wavelength and black lines are obtained for these absorbed radiations. These black lines are called absorption spectrum. OR Energy of emitted photons from H-atom = 10.2 eV Work – function of metallic surface 12400 = eV = 3.1 eV 4000 Now According to Einstein's Law of Photo-electric effect ' E = w + K.E.max. 10.2 = 3.1 + K.E.max ∴ K.E.max = 7.1 eV Ans 30. Moving Coil Galvanometer – Principle : When a coil carrying current is placed in a magnetic field, it experiences a Torque τ = NiAB sin α. This Torque is proportional to the current passed through it. T1 Phosphor Bronze Copper coil T2 N Spring Torsion Head Mirror Frame S Core Working : When a current is passed through the coil, the two vertical limbs experience a force normal to it, equal in magnitude and parallel but opposite in sense. No forces act on the horizontal sides as they are parallel to the field .Since the field is radial, the forces acting on the vertical parts of coil remain always perpendicular to the plane of the coil in all its positions so that the perpendicular distance between the forces is always equal to breadth of the coil. Thus the coil is subjected to a torque whose magnitude is given by τ = NiAB … (i) Under the action of this deflecting torque, the coil rotates on its axis. Due to elastic forces, a restoring torque τr is produced in the suspension coil, which is perpendicular to the twist θ produced in the wire. τr = θ or τr = Cθ … (ii) where C is restoring torque per unit twist, and it is called the constant of twist for the suspension. As soon as restoring torque becomes equal to the deflecting torque, equilibrium is established. The coil does not rotate further. Let φ be the angle through which the coil rotates till it reaches the equilibrium position. τ = τr or NiAB = Cφ … (iii) C i = φ = K.φ NAB C K = is constant for the instrument, called NAB Galvanometer constant. ∴ i ∝ φ . Thus deflection of coil is proportional to the current passing through it. The deflection can be measured by using a lamp and scale arrangement. OR Consider two infinitely long thin conductors carrying currents in opposite directions. Magnetic field B1 due to I1 at P on conductor CD is given by XtraEdge for IIT-JEE 91 FEBRUARY 2012 B1 = µ 0I1 2πr F1B2 B I1 Q r D P B1F2 A C The magnetic field B1 is perpendicular to plane of paper and directed inward. This field will produce a force/length F2 on conductor given CD by I2
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(i) Emission : - If radiations emitted from a substance<br />
is obtained ones screen after passing through the prism<br />
then spectrum is called emission spectrum.<br />
It is line emission spectrum for substances at atomic<br />
level in gaseous state.<br />
It is continuous spectrum for substances in liquid or<br />
solid state.<br />
(ii) Absorption spectrum : - The substance of which<br />
absorption spectrum is to be find out is placed in a<br />
transparent tube and white light is passed through it<br />
now substance absorb radiation corresponding to some<br />
particular wavelength and black lines are obtained for<br />
these absorbed radiations. These black lines are called<br />
absorption spectrum.<br />
OR<br />
Energy of emitted photons from H-atom = 10.2 eV<br />
Work – function of metallic surface<br />
12400<br />
= eV = 3.1 eV<br />
4000<br />
Now According to Einstein's Law of Photo-electric<br />
effect '<br />
E = w + K.E.max.<br />
10.2 = 3.1 + K.E.max<br />
∴ K.E.max = 7.1 eV Ans<br />
30. Moving Coil Galvanometer – Principle : When a<br />
coil carrying current is placed in a magnetic field, it<br />
experiences a Torque τ = NiAB sin α. This Torque<br />
is proportional to the current passed through it.<br />
T1<br />
Phosphor<br />
Bronze<br />
Copper coil<br />
T2<br />
N<br />
Spring<br />
Torsion Head<br />
Mirror<br />
Frame<br />
S<br />
Core<br />
Working : When a current is passed through the<br />
coil, the two vertical limbs experience a force normal<br />
to it, equal in magnitude and parallel but opposite in<br />
sense. No forces act on the horizontal sides as they<br />
are parallel to the field .Since the field is radial, the<br />
forces acting on the vertical parts of coil remain<br />
always perpendicular to the plane of the coil in all its<br />
positions so that the perpendicular distance between<br />
the forces is always equal to breadth of the coil. Thus<br />
the coil is subjected to a torque whose magnitude is<br />
given by<br />
τ = NiAB … (i)<br />
Under the action of this deflecting torque, the coil<br />
rotates on its axis. Due to elastic forces, a restoring<br />
torque τr is produced in the suspension coil, which is<br />
perpendicular to the twist θ produced in the wire.<br />
τr = θ or τr = Cθ … (ii)<br />
where C is restoring torque per unit twist, and it is<br />
called the constant of twist for the suspension. As<br />
soon as restoring torque becomes equal to the<br />
deflecting torque, equilibrium is established. The coil<br />
does not rotate further. Let φ be the angle through<br />
which the coil rotates till it reaches the equilibrium<br />
position.<br />
τ = τr or NiAB = Cφ … (iii)<br />
C<br />
i = φ = K.φ<br />
NAB<br />
C<br />
K = is constant for the instrument, called<br />
NAB<br />
Galvanometer constant.<br />
∴ i ∝ φ .<br />
Thus deflection of coil is proportional to the current<br />
passing through it. The deflection can be measured<br />
by using a lamp and scale arrangement.<br />
OR<br />
Consider two infinitely long thin conductors<br />
carrying currents in opposite directions.<br />
Magnetic field B1 due to I1 at P on conductor CD is<br />
given by<br />
<strong>Xtra</strong><strong>Edge</strong> for IIT-JEE 91 FEBRUARY <strong>2012</strong><br />
B1 =<br />
µ 0I1<br />
2πr<br />
F1B2<br />
B I1<br />
Q<br />
r<br />
D<br />
P<br />
B1F2<br />
A<br />
C<br />
The magnetic field B1 is perpendicular to plane of<br />
paper and directed inward. This field will produce a<br />
force/length F2 on conductor given CD by<br />
I2
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