1. Xtra Edge February 2012 - Career Point
1. Xtra Edge February 2012 - Career Point 1. Xtra Edge February 2012 - Career Point
28 Parallel light rays incidenting over a lens converges to a point (convex lens) or seems to diverge from the point (concave lens) after refraction from lens. Then this point is called principle focus of lens. When lens is dipped into a liquid of refractive index greater than lens material refractive index then nature of lens gets changed. Focal length of lens is maximum for red and minimum for violet colour. I – µ1 = 1 µ1 = 1 + µ2 II R1 = + 20; R2 → ∞ ; 1 ⎛ 1 1 ⎞ = (µ2 – 1) ⎜ − ⎟ f ⎝ + 20 ∞ ⎠ 1 ⎛ ⎞ = (2 –1) ⎜ − ⎟⎠ f ⎝ ∞ 1 1 20 ∴ f = + 20cm OR O µ1 α i u P M R By snell's law m1 sin i = µ2 sin r r β γ v C µ2 But for small aperture MP; µ1i = µ2r …(i) In ∆OCM; i = α + β and In ∆ICM; β = r + γ ⇒ r = β – γ. ∴ By (i) µ1 (α + β) = µ2 (β – γ) ⎡ MP MP ⎤ ⎡ MP MP ⎤ µ1 ⎢ + ⎥ = µ2 ⎣ OP PC ⎢ − ⎥ ⎦ ⎣ PC PI ⎦ ⎡ 1 1 ⎤ ⎡ 1 1 ⎤ µ1 ⎢ + ⎥ = µ2 ⎣− u + R ⎢ − ⎥ ⎦ ⎣ + R + v ⎦ – µ 1 µ 1 µ 2 µ 2 + = − u R R v µ 2 µ 1 µ 2 − µ 1 − = v u R I 29. Principal of van deGraff Generator : (i) Let a small charged conducting shell of radius r be located inside a larger charged conducting shell of radius R. If they are connected with a conductor, then charge q from the small shell will move to the outer surface of bigger shell irrespection of its own charge Q. Here potential difference XtraEdge for IIT-JEE 76 FEBRUARY 2012 R r q = V(r) – V(R) q = 4πε 0 ⎟ ⎛ 1 1 ⎞ ⎜ – ⎝ r R ⎠ In this way, the potential of the outer shell increases considerably. (ii) Sharp pointed surfaces have larger charge densities, so these can be used to set up discharging action. Conducting S Shell Grounded Steal Tank C1 , C2 metal Comb H vR C2 Q Target Insulating Column Working :Let spray comb C1 be charged to a high +ve potential which spray +ve charge to the belt which in turn becomes positively charged. Since belt is moving up, so it carries this positive charge upward. Opposite charge appears on the teeth of collecting comb C2 by induction from the belt. As a result of this, positive charge appears on the outer surface of shell S. As the belt is moving continuously, so the charge on the shell S increase continuously. Consequently, the potential of the shell (S) rises, to a very high value. Now the charged particles at the top of the tub (T) are very high potential with respect to the lower end of the tube which is earthed. Thus these particles get accelerated downward and hit the target emerging from the tube.
Use : It can be used to accelerate particles which are used in nuclear physics for collision experiments. Or (i) Intensity of the Electric field at a point on the Axis of a Diople : A l l B E2 P E1 –q O +q r The intensities E1 and E2 are along the same line in opposite directions. Therefore, the resultant intensity E at the point P will be equal to their difference and in the direction BP (since E1 > E2). That is, E = E1 – E2 1 q 1 q = = = = 4πε K 0 q 4πε K 0 2 ( r – l) ⎡ 1 ⎢ ⎣( r – l) 2 – – 4πε K ( r 1 0 + 2 l) q 4πε0K ⎥ ⎥ ⎡ 2 2 ( r + l) – ( r – l) ⎤ ⎢ 2 2 2 ⎢⎣ ( r – l ) ⎦ q 4πε K 0 ⎡ 4lr ⎤ ⎢ 2 2 2 ⎥ ⎣( r – l ) ⎦ ⎤ ⎥ ⎦ ( r + l) 1 ⎡ 2( 2q l) r ⎤ = ⎢ 2 2 2 ⎥ 4πε0K ⎣( r – l ) ⎦ But 2ql = p (electric dipole moment). 1 2p r ∴ E = 2 2 2 4πε0K ( r – l ) If l is very small compared to r (l
- Page 27 and 28: PHYSICS FUNDAMENTAL FOR IIT-JEE The
- Page 29 and 30: etween total mass and number of mol
- Page 31 and 32: KEY CONCEPT Organic Chemistry Funda
- Page 33 and 34: KEY CONCEPT Inorganic Chemistry Fun
- Page 35 and 36: 1. It is possible to supercool wate
- Page 37 and 38: z 1 K2 = = y( 0. 48M − z) 1. 26×
- Page 39 and 40: MATHEMATICAL CHALLENGES 1. as φ (a
- Page 41 and 42: 9. Point P (x, 1/2) under the given
- Page 43 and 44: 10. 11 ∴ n(E) = 10 + 9 + 8 + ...
- Page 45 and 46: dx (ii) ∫ 2 ax + bx + c This can
- Page 47 and 48: MATHS Functions with their Periods
- Page 49 and 50: a PHYSICS Questions 1 to 9 are mult
- Page 51 and 52: R P ⊗ B O ω0 14. Magnitude of cu
- Page 53 and 54: XtraEdge for IIT-JEE 51 FEBRUARY 20
- Page 55 and 56: XtraEdge for IIT-JEE 53 FEBRUARY 20
- Page 57 and 58: 7. NaOH can be prepared by two meth
- Page 59 and 60: (C) X = 100 ml of SO2 at 1 bar, 25
- Page 61 and 62: 21. Column-I Column-II (A) If y = 2
- Page 63 and 64: (A) potential energy (PE) increases
- Page 65 and 66: 2. In the reaction 3 Cu + 8 HNO3
- Page 67 and 68: (D) P A T B (S) Pressure is increas
- Page 69 and 70: XtraEdge for IIT-JEE 67 FEBRUARY 20
- Page 71 and 72: 14. Derive the expression for the c
- Page 73 and 74: (a) temperature and (b) Pressure ?
- Page 75 and 76: should be bought to meet the requir
- Page 77: This work done is stored inside the
- Page 81 and 82: 10. (i) (ii) Benzene H CH3 - C = O
- Page 83 and 84: NH3 NH3 NH3 Co +3 NH3 NH3 NH3 No un
- Page 85 and 86: 7. f(A) = A 2 - 5A + 7 I2 ⎡ 3 1
- Page 87 and 88: y 2 = - (2 - x) 2 + 9 This is the e
- Page 89 and 90: (0, 4) ( 2 3, 2) (4, 0) ⎡ 2 4 2
- Page 91 and 92: 15. X = ? Y = 20 Ω l1 = 40 cm l -
- Page 93 and 94: (i) Emission : - If radiations emit
- Page 95 and 96: OH (ii) + Br2 Phenol H2O 13. I st M
- Page 97 and 98: 28. (a) Acetylene is first oxidized
- Page 99 and 100: OR Let A → total of 8 in first th
- Page 101 and 102: = x 2 ⎡ ⎢ ⎢x ⎢ ⎣ 0 1 2 1
- Page 103 and 104: 28. x x y ∴ x → side of square
- Page 105 and 106: C 'XtraEdge for IIT-JEE IIT JEE bec
- Page 107 and 108: XtraEdge for IIT-JEE 105 FEBRUARY 2
Use : It can be used to accelerate particles which are<br />
used in nuclear physics for collision experiments.<br />
Or<br />
(i) Intensity of the Electric field at a point on the<br />
Axis of a Diople :<br />
A<br />
l l<br />
B E2 P E1<br />
–q O +q<br />
r<br />
The intensities E1 and E2 are along the same line in<br />
opposite directions. Therefore, the resultant intensity<br />
E at the point P will be equal to their difference and<br />
in the direction BP (since E1 > E2). That is,<br />
E = E1 – E2<br />
1 q 1 q<br />
=<br />
=<br />
=<br />
=<br />
4πε<br />
K<br />
0<br />
q<br />
4πε<br />
K<br />
0<br />
2<br />
( r – l)<br />
⎡ 1<br />
⎢<br />
⎣(<br />
r – l)<br />
2<br />
–<br />
–<br />
4πε<br />
K<br />
( r<br />
1<br />
0<br />
+ 2<br />
l)<br />
q<br />
4πε0K<br />
⎥ ⎥ ⎡ 2 2<br />
( r + l)<br />
– ( r – l)<br />
⎤<br />
⎢ 2 2 2<br />
⎢⎣<br />
( r – l ) ⎦<br />
q<br />
4πε<br />
K<br />
0<br />
⎡ 4lr<br />
⎤<br />
⎢ 2 2 2 ⎥<br />
⎣(<br />
r – l ) ⎦<br />
⎤<br />
⎥<br />
⎦<br />
( r<br />
+ l)<br />
1 ⎡ 2(<br />
2q<br />
l)<br />
r ⎤<br />
= ⎢ 2 2 2 ⎥<br />
4πε0K<br />
⎣(<br />
r – l ) ⎦<br />
But 2ql = p (electric dipole moment).<br />
1 2p<br />
r<br />
∴ E =<br />
2 2 2<br />
4πε0K<br />
( r – l )<br />
If l is very small compared to r (l
Change language