1. Xtra Edge February 2012 - Career Point
1. Xtra Edge February 2012 - Career Point 1. Xtra Edge February 2012 - Career Point
MOCK TEST-3 (SOLUTION) PHYSICS 1. There are many atoms in an atomic hydrogen gas sample and electrons of different atoms can make different transitions and different wavelength radiations are emitted. 2. Frequency remains constant when light passes from one medium to other. 3. Those sources which have constant phase difference are called coherent sources. These are derived from a single sources. 4. Kinetic energy depends on frequency of incident radiations not on intensity so kinetic energy remains same. 5. Average power over a complete cycle of ac through an ideal inductor (Resistance = 0) is zero. Pav = (Irms) × (Vrms) cos φ R cos φ = = 0 Z Pav = 0 6. A 1A 2Ω 1Ω + – B VA – VB = 1 × 2 + 2 + 1 × 1 = 5 V 7. X–rays –→ 3 × 10 19 to 1 × 10 16 Hz Microwave –→ 3 × 10 11 to 1 × 10 9 Hz U-V rays –→ 5 × 10 17 to 8 × 10 14 Hz Radio Waves –→ 3 × 10 7 to 3 × 10 4 Hz 8. φ = → → q E . A , ∆ φ = ε φi = EA cos 180° φf = EA cos 0° φi = – EA φf = + EA φi = – 5 × 10 5 V– m φf = 4 × 10 5 V– m – 5 × 10 5 + 4 × 10 5 = q 0 ε 0 , q = – ε0 × 10 5 Cb 9. Intensity of radiations is 64 = 2 6 times the safe value. So after six half lines it would be safe so t = 6T = 6 × 2 = 12 hours. MOCK TEST– 3 PUBLISHED IN SAME ISSUE 10. On temperature rise more valance band electrons become free and more free electrons and hole pair will be produced and conductivity of semiconductor increases. 11. In optical fibres refractive index of cladding is less than core. So that electromagnetic wave traveling in core when incidents other cladding at an angle greater than ic. Total internal reflection will occur which is necessary for transmission of signals from one place to other. 12. Object should be placed at 2F of convex lens so as to form image of same size. It can't happen in case of concave lens because it always forms a diminished image. 13. Using d = 2 Rh T + 2Rh R = R ( h h ) 2 T R + = 20.8 5 km = 46.5 km 14. When a charge +Q is given to the insulated plate, then a charge –Q is induced on the nearer face of the plate. + Q – Q XtraEdge for IIT-JEE 88 FEBRUARY 2012 E Net electric field between the plate σ V E = also, E = ε0 d ∴ V = Ed = with dielectric. C' = σ d Q ε0A = ⇒ C = ε C d 0 ε0ε A r d
15. X = ? Y = 20 Ω l1 = 40 cm l – l1 = 60 cm X B Y 16. A l1 V1 D V1 ∝ l1 V2 ∝ (l – l1) X Y l ( l – l = 1 1 ) G l – l1 V2 X 20 X 1 = or = or 40 60 40 3 40 X = Ω 3 When X and Y are interchanged Y B X 2 A D G l2 l – l2 Y l = X 20 40 ⇒ = ( l – l ) l 3 ( 100 – l 2 1 2 = l 2 3( 100 – l 2) 300 – 3l2 = 2l2 or 5l2 = 300 l2 = 60 cm λ1λ 2 F = 2πε d 0 17. (i) Vcom = (ii) Ui = Uf = 2 1 λ1 2 C1V1 + C2V2 C + C 1 C 2 2 1V1 + 2 1 C 2 1 (C1 + C2) V d 2 V 2 com 2 2 λ2 2 C ) C E0 18. B0 = c 19. Work-function : Minimum energy needed by a free electron to remove it from metallic surface. Threshold frequency : Minimum frequency required for photo-electron emission. Threshold wavelength : Maximum wavelength of incident radiations required for photoelectric emission. Stopping potential : Minimum potential required across the photo cell to completely stop the photo current. 20. (En) = – 13. 6 eV, for ground state n = 1 2 n ∴ (E1) = – 13.6 eV For hydrogen like atom, (En) HLA = Z 2 (En) H XtraEdge for IIT-JEE 89 FEBRUARY 2012 2 × Z 13. 6 = eV 2 n For He + atom, Z = 2 and n = 2 ⎡ – 13. 6⎤ ∴ (E2)He = 4 ⎢ ⎥ = – 13.6 eV ⎣ 4 ⎦ 5Dλ −2 21. = 1. 5× 10 m d d = 0.56 × 10 –3 m D = 2.8 m 5× 2. 8× λ − ∴ = 1. 5× 10 −3 0. 56× 10 ∴ λ = 6 × 10 –7 m ∴ λ = 6000 Å c 22. Source frequency ν = λ 23. x Band width of system = ν = 100 2 × λ c x 100 Number of channels N which can be transmitted simultaneously can be found out by dividing band width by the system with band width of one channel. i.e., R R = R R 1 2 3 4 N = ⎛ xc ⎞ ⎜ ⎟ ⎝100× λ ⎠ xc = F 100λF 24. (1) Resistance = R (2) Reactance are of two types (a) Capacitive Reactance XC = C 1 ω
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15. X = ?<br />
Y = 20 Ω<br />
l1 = 40 cm<br />
l – l1 = 60 cm<br />
X<br />
B<br />
Y<br />
16.<br />
A<br />
l1<br />
V1<br />
D<br />
V1 ∝ l1<br />
V2 ∝ (l – l1)<br />
X Y<br />
l ( l – l<br />
=<br />
1<br />
1<br />
)<br />
G<br />
l – l1<br />
V2<br />
X 20 X 1<br />
= or = or<br />
40 60 40 3<br />
40<br />
X = Ω<br />
3<br />
When X and Y are interchanged<br />
Y B X<br />
2<br />
A<br />
D<br />
G<br />
l2 l – l2<br />
Y<br />
l =<br />
X 20 40<br />
⇒ =<br />
( l – l ) l 3 ( 100 – l<br />
2<br />
1 2<br />
=<br />
l 2 3(<br />
100 – l 2)<br />
300 – 3l2 = 2l2 or 5l2 = 300<br />
l2 = 60 cm<br />
λ1λ<br />
2 F =<br />
2πε<br />
d<br />
0<br />
17. (i) Vcom =<br />
(ii) Ui =<br />
Uf = 2<br />
1<br />
λ1<br />
2<br />
C1V1<br />
+ C2V2<br />
C + C<br />
1<br />
C<br />
2<br />
2<br />
1V1<br />
+<br />
2<br />
1<br />
C<br />
2<br />
1<br />
(C1 + C2) V<br />
d<br />
2<br />
V<br />
2<br />
com<br />
2<br />
2<br />
λ2<br />
2<br />
C<br />
)<br />
C<br />
E0 18. B0 =<br />
c<br />
19. Work-function : Minimum energy needed by a free<br />
electron to remove it from metallic surface.<br />
Threshold frequency : Minimum frequency<br />
required for photo-electron emission.<br />
Threshold wavelength : Maximum wavelength of<br />
incident radiations required for photoelectric<br />
emission.<br />
Stopping potential : Minimum potential required<br />
across the photo cell to completely stop the photo<br />
current.<br />
20. (En) =<br />
– 13.<br />
6<br />
eV, for ground state n = 1<br />
2<br />
n<br />
∴ (E1) = – 13.6 eV<br />
For hydrogen like atom, (En) HLA = Z 2 (En) H<br />
<strong>Xtra</strong><strong>Edge</strong> for IIT-JEE 89 FEBRUARY <strong>2012</strong><br />
2 ×<br />
Z 13.<br />
6<br />
= eV<br />
2<br />
n<br />
For He + atom, Z = 2 and n = 2<br />
⎡ – 13.<br />
6⎤<br />
∴ (E2)He = 4 ⎢ ⎥ = – 13.6 eV<br />
⎣ 4 ⎦<br />
5Dλ −2<br />
2<strong>1.</strong> = <strong>1.</strong><br />
5×<br />
10 m<br />
d<br />
d = 0.56 × 10 –3 m<br />
D = 2.8 m<br />
5× 2.<br />
8×<br />
λ<br />
−<br />
∴<br />
= <strong>1.</strong><br />
5×<br />
10<br />
−3<br />
0.<br />
56×<br />
10<br />
∴ λ = 6 × 10 –7 m<br />
∴ λ = 6000 Å<br />
c<br />
22. Source frequency ν =<br />
λ<br />
23.<br />
x<br />
Band width of system = ν =<br />
100<br />
2<br />
×<br />
λ<br />
c x<br />
100<br />
Number of channels N which can be transmitted<br />
simultaneously can be found out by dividing band<br />
width by the system with band width of one channel.<br />
i.e.,<br />
R R<br />
=<br />
R R<br />
1<br />
2<br />
3<br />
4<br />
N =<br />
⎛ xc ⎞<br />
⎜ ⎟<br />
⎝100×<br />
λ ⎠ xc<br />
=<br />
F 100λF<br />
24. (1) Resistance = R<br />
(2) Reactance are of two types<br />
(a) Capacitive Reactance<br />
XC = C<br />
1<br />
ω
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