MOCK TEST-2 (SOLUTION)
MOCK TEST– 2 PUBLISHED IN JANUARY ISSUE
6
P h −−
×
= ×
=λ = 1.32 × 10–27 kg-m/s 4. R = R0A1/3 ⇒ R ∝ A1/3
∴
2 1
R
R = 3
1
2 1
A A
⇒
R2
R =
3 / 1
27 8
=
3 2
∴ R2 = 2
R 3
5. It should be very high i.e. in KHz to GHz 6. =µc
v ; v2 < v3 <v1
7. Resistivity or specific resistance of a material is defined as the resistance offered by a wire made by that material having unit area of cross-section and unit length.
8. V = q
W = –7
5 –
10 2 . 3
10 6 . 1
×
× = 50 volt
9. Objective lens must have greater focal length than eye-piece because magnifying power of lens
=
e P f0
M f increases in this case.
10. Conditions of interference
(i) waves must travel in the same medium in the same direction.
(ii) Their frequencies and wavelengths must be same
(iii) Their plane of polarisation must be same
(iv) Their must be maintained a constant phase difference between the waves.
11. Ionisation energy:- Minimum energy required to remove electron from its ground state.
Excitation energy:- Energy required to transit an electron to one of its excited state from ground state KE =
r 2
KZe2 ; U = – r KZe2
∴ U=−2KE 12. Emax. = 12V; Emin. = 4V
Ma =
2 1 4 12
4 12 E
E E E
min max
min
max =
+
= − +
−
∴ ma in percentage = 2
1 × 100 = 50%
13. Modulating signal being of low frequency cannot travel to large distance so it is modulated with high frequency carrier. Frequency modulation is virtually free from noise where as amplitude modulation suffers from noise pollution.
14. Semiconductor atoms are tetravalent. The form covalent bonds by sharing their electrons with the neighboring atoms. When donor impurity of valency S is doped, the donor atoms takes the place of a semiconductor atom in the crystal lattice. Donor atoms forms covalent bond by sharing its four electrons whereas the fifth electron of donor atom is so loosely attached to the atom such that it behaves as a free electron at room temperature.
15. (i) width of depletion layer of zener diode becomes very small due to heavy doping of p and n regions.
(ii) The field across depletion layer becomes very high
=
d E V Q
16. Let at any instant charge on the capacitor is q, then p.d. V =
C q
Then work done to put additional charge dq is dW = (dq) V =
C q dq
∴ Total work done to give charge Q is W =
∫
Q 0
C dq
q =
C 2 Q2 =
2
1CV2 (Q Q = CV)
MOCK TEST-2 (SOLUTION)
MOCK TEST– 2 PUBLISHED IN JANUARY ISSUE
This work done is stored inside the capacitor as P.E.
∴ CV2
2 W 1
U= =
17. (i) A galvanometer can be converted into ammeter by connecting a low resistance called shunt parallel to the galvanometer coil.
(ii) A galvanometer can be converted into voltmeter by connecting a large resistance in series with the galvanometer coil.
18. (i) Electromagnetic waves are produced by accelerated charge.
(ii) The velocity of electromagnetic wave in free space is equal to the velocity of light in free space.
(iii) It is transverse in nature.
19. α = 2º β = ? fo = 20 cm fe = 4 cm MP =
α
β and MP =
e o
f f ∴
e o
f
=f α
β ⇒ β =
e o
f f α
∴ β =
4
20 × 2 = 10
20. 2mqV
= h λ for proton
meV 2
= h
λ …(i)
For α-particle λ = 2 4m 2eV'
h
×
× …(ii)
By (i) and (ii)
2meV = 2 × 4m × 2eV' 8
/ V ' V= 21.
ab c d e Fe 56
a → 2He4 b → 4Be8 c → 6C12 d → 8O16 e → 10Ne20 BE/A
A
22. 2 is NAND and 1 is OR gate
A B Y' Y
0 1 1 0
23. R = (AB × 10c ± 5)%
Yellow A = 4 Violet B = 7 Brown C = 1
∴ R = (47 × 10 ± 5)%
= (470 ± 5)%
24. (i) To make strong electromagnet (ii) To make high speed computers 25. tanθ =
H V
B B
∴ BV = BH tanθ = 0.4 × 10–4 tan60º = 0.69 × 10–4 T and BH = Be cos θ
⇒ Be =
º 60 cos
10 4 . 0 cos
BH × −4
θ= = 0.8 × 10–4 T 26. Biot–Savart′s law
This law is applicable to determine magnetic field due to small current element. Magnetic field due to small element d→l at point A is given as
→r A
→
d→ll d
i
3 0
r ) r d ( 4 dB i
→
→ →×
π
=µ l
or 0 2
r sin id
dB 4 θ
π
=µ l
where µ0 ⇒ Permeability of free space, θ ⇒ between d→l and →r .
27. (i) On increasing distance between the coils, flux linked decreases, hence mutual inductance decreases
(ii) Q M = l
A N N1 2 µ0
∴ on increasing no. of turns, M increases (iii) when iron sheet (µr) is inserted Then M =
l A N N1 2
r 0µ
µ
∴ M increased, because for iron µr > > 1
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.
µ2 µ1 = 1 µ1 = 1
– +
I II
R1 = + 20; R2 → ∞ ; f
1 = (µ2 – 1)
−∞ +
1 20 1
f
1 = (2 –1)
−∞1 20
1
∴ f =+20cm
OR
i M r
β γ
α
u P R v
µ2
I O C
µ1
By snell's law m1 sin i = µ2 sin r
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)
µ1
+ PC MP OP
MP = µ2
− PI MP PC
MP
µ1
++
− R
1 u
1 = µ2
−+
+ v
1 R
1
– u R R v
2 2 1
1+µ =µ −µ
µ
R u v
1 2 1
2 −µ =µ −µ
µ
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
r Q
q R
= V(r) – V(R)
= 4 0
q
πε
R – 1 r 1
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 Shell
H vR
Insulating Column
Target Grounded
Steal Tank
C1 , C2
metal Comb
C2
S
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 : 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,
But 2ql = p (electric dipole moment).
∴ E = be neglected in comparison to r2. Then the electric intensity at the point P due to the dipole is given by
E =
Intensity of the Electric Field at a point on the Equatorial Line of a Dipole.
A θ components parallel and perpendicular to AB, the components perpendicular to AB (E1 sin θ and E2 sin θ) cancel each other (because they are equal and opposite) while the components parallel to AB (E1 cos θ and E2 cos θ), being in the same direction ,add up. Hence the resultant intensity of electric field at the point P is can be neglected in comparison to r2. Then the electric intensity at the point P due to the dipole is E = The direction of the electric field E is parallel to the axis of the dipole from the positive charge towards the negative charge.
Comparing eq. (i) and (ii) we see that for a short dipole the intensity of the electric field on an axial point is twice the intensity at the same distance on the equatorial line.
(ii) W = Uf – Ui =PE (1 – cos θ) = PE (1 – cos 180°) = PE (1 + 1) = 2 PE
30.
A
t φ
E