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FIITJEE MOCK TEST PAPER

for XIIth Board CBSE (Set-I)

PHYSICS

SOLUTIONS

1. 10 x 101 ± 5% Ω.

2

3.

4. zero as φ = π/2 5. No Effect.

6. 1 1P 0f

7. When man stands beyond focus is i.e. between focus and centre of curvature, his real and

inverted image is formed beyond C (beyond him) and thus he cannot see the image. But when he stands beyond C, image is formed between focus and center of curvature is in front of him and thus he is able to see his image.

8. (i) Due to very large size of antenna. (ii) To avoid mixing of signal with other signals, modulation is required.

9. As E = hcn nhv

MOCK TEST PAPER (2015)-CBSE-XII-(Sol.)-(Set-I)-PHYSICS-1


10. Side base band .1MHz +15 KHz and 1MHz - 15 KHz i.e. 1.015 MHz and 0.985 MHz. 11. Δm = [ M(U) - M(Th) - M(He) ] = (238.05079 – 234.04363 – 4.00560) = 0.00456 U Q = 0.00456 x 931 mev

12. Lyman series equation

For shortest wavelength

13. Due to increase in Intensity no effect on kinetic energy of photo electrons. As due to

increase in Intensity , there is only an increase in the number of photons per unit area , and not the incident energy.

14. The impedance of L-C-R circuit is

Z = 2 2 22L CX X R 50 50 10 10

Now the r.m.s. current flowing in the circuit is

i.e. i = amp2010200

ZE

The potential difference across the inductance is VL = iXL = 2050 = 1000V, VR = iR = 2010 = 200V The potential difference across L-C is (50 50)20 = 0 15.

A

B

Y

Y B A 0 1 0 1

0 0 1 1

1 1 1 0



Here, output y of NOR gate is used as the input of NOT gate (made from NOR gate by joining two inputs). The Boolean expression is

y = BA , and y = BA = A+B

A B Y Y

0 0 1 0

1 0 0 1

0 1 0 1

1 1 0 1 16. Density

m = 1.67 × 10–27 kg, Ro = 1.1 × 10–15 m. Order of density is 1017 kg/m3

17. n 213.6E eVn

Energy of the photon emitted during a transition of the electron from the first excited state to its ground state

2 1E E E = -3.40 + 13.6 = 10.2 eV. It belongs to ultraviolet region. 18. on1 + 3Li6 → 1H3 + 2He4 + Q

Δm = (6.01512126 + 1.0086654) – (3.010000 + 4.0026044) = 0.01118226 x 931 mev

= 10.41 mev

19.

+

OUTPUT

VCC

vo

+

RL

vS vCE

~ vBE

VBB

INPUT OUTPUT

INPUT Ib

Ic

Ie

Figure shows the circuit diagram of a n-p-n common emitter transistor amplifier. Amplified output voltage is obtained across the emitter collector i.e. VCE

, VCE = VCC - ICRL. During input +ve half, input junction is more forward biased due to which, ib increases which in turn increases ic, due to which ic RL increase thus VCE decreases in positive i.e. it becomes more negative with respect initial value. During – ve half, input junction is less forward biased ib decreases ic decreases ic RL decreases



Vce becomes more positive, i.e. in common emitter amplifier, input signal and output amplified signals are out of phase i.e. at a phase difference of 180o. voltage gain It is defined as the ratio of the output voltage across the load resistance to input voltage i.e.

Voltage gain = o L c

i i b

V R iV R i

voltage gain = current gain L

i

RR

20. Given θ = 30o , = 0.016Nm. B = 800 x 10-4 T (i) = MB sinθ M = / B sinθ = 0.016/800 x 10-4 x 1/2 = 16/40 =0.4Am2 (ii) Wθ=0→θ to180 = MB(cosθ1 – cosθ2) = 2MB = 0.032J 21. (i) N – type (ii) P type (iii) P type 22. = 6000 Å = 6000 × 10–10 m 1 = 30°, m = 1 (1) For first maximum

sinm =

1m2

a

; sin1 = 3

2a

or a = 7

o1

3 3 6 102sin 2 sin 30

= 1.8 × 10–6 m

(2) For first minimum

sinm = ma ; or sin1 =

a

a = 1sin

= 7

o6 10sin 30

a = 1.2 × 10–6 m 23. Sensitivity – sensitivity of a potentiometer is related to potential gradient. Smaller the

potential gradient, more sensitive be the potentiometer.

(i) E2 = kl2 1.5 = k x 60 x 10-2

VAB = k × lAB = 2.5



(ii) No, emf of driver cell is less then balancing cell.

OR

Resistance offered by a conductor of unit length and unit area of cross section is known as resistivity. Derivation - I = neAvd

...(1)

...(2)

24. For equilibrium of charge at A

Which indicate nature of charge at M must be +ve

25. Given 2L = 2cm, θ = 60o, E = 105N/C and = 8 (i) = PE sin θ = q(2L) E sin θ 8 = q x ( 2 x 10-2) x (105) sin 60 8 = q x 2 x 103 x /2 8 x 10-3 = q = 8mc



(ii) U = –PE cos θ = –(8 x 10-3) (2 x 10-2) cos 60 = –8 x 10–5 joule

26. Polaroids: A Polaroid is a material which polarizes light. Tourmaline is a natural polarizing material (I) light waves are transverse in nature. The electric field associated with a propagating light

wave is always at right angles to the direction of propagation of the wave. If a light wave is incident on the Polaroid, the electric vectors along the direction of the aligned molecules get absorbed

(ii) If an unpolarised light wave is incident on such a Polaroid then the light wave get linearly polarized with the electric vector oscillating along a direction perpendicular to the aligned molecules

27. (a)

A capacitor of capacitance is charged with an external source. The graphical Variation of charge with applied potential is shown. The work done to

charge upto Maximum value QO is given by Area of OABi.e W = 1/2 OB x AB = ½ VoQo = ½VoCVo W = ½CVo

2 = U

As ⇒ u = ½ V02 Put V0 = E0 x d

U = ½ ; Energy stored per unit volume =

(b) C1 = 10μF C2 = 50μF V1 = 30V V2 = 0 VCom = ? ΔU = ? As VCommon = = 5 Volt

ΔU = Loss in energy = ½ = 3.75 × 10–3 J 28. Interference: Redistribution of energy carried by two or more than two waves having same

frequency, constant phase difference and moving in same direction is known as Interference



When one sleet is open.

Also β = Dd

Fringe Separation will Increases as β D. 29. (a) α – decay , 92U238 → 90Th234 + 2He4 + Q

(b) Activity – Activity is the total decay rate of sample of one or more radionuclide. Unit -

Curie OR

4 –3I1 – 2I3 = 0 …(i) 12 – 4I2 – 2I3 = 0 …(ii) Solving (i) and (ii), I1 = 0A, I2 = I3 = 2A


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