PHYSICS QUESTION BANK FOR PLUS TWO STUDENTS

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PHYSICS
QUESTION BANK FOR PLUS TWO STUDENTS
CO:1
To get an idea about frictional electricity charges and their properties through simple experiments problem solving discussion and I.T.
Question Text:
Q. No:1
When a rubber sheet is rubbed with woolen carpet, the carpet is found to acquire a positive charge of 8 × 10-7C
a) In the above process charging is by
(i) conduction
(ii) friction
(iii) induction (iv) polarization
b) Among Rubber shoe and woolen carpet which one acquire both mass and charge during rubbing?
Explain.
c) “The charge acquired by the carpet is quantized”. Justify this statement.
Score (1+2+2 = 5),
Time : 10 mints
Scoring key
a) friction 1 Score (MP-1, 2)
b) Rubber shoe, Rubber shoe gains electrons, so it gains mass and charge 2 Score (MP - 1,2,7)
c) charge is an integral multiple of e (1.6 × 10-19)
q = ± ne ,
n=
q 8 × 10-7
,
= 5×1012 2 Score
e 1.6×10 -19
(MP 8, 7, 5)
Question Text :
Q. No: 2
Two point charges q1 and q2 are separated by a distance ‘r’ in space
A q1
(a) what happens to the force between the charges;
(i) When the magnitude of the point charges increases
(ii) When the distance of separation decreased
q3
q2
Hence establish the law governing it.
C
B
(b) Three point charges +2 µ C each are placed at the corners of an equilateral triangle of side 1m. Find the
magnitude of the force between
charge q1 and q2 [F12]
and q1 and q3 [F13]
r
r
(c) Draw F12 and F13 at A and find the total force acting on q1. Hence establish super position principle
(21/2)
(d) “The law governing force between two point charges can be established using charged pith balls”
Comment on the statement. (1/2)
[Total Score : 6]
Time : 8 mts
Scoring Key
Score
(1/2)
(1/2)
(1/2+ 1/2)
a) (i) increases
(ii) increases
Coulomb’s law statement
9 q1q 2
b) F = 9×10 r
2
(MP - 1,2)
(1/2)
(1/2)
(1/2)
Result
c) Figure
r r r
F= F12 +F13
(MP - 1,2,5)
(1/2)
F= F122 +F132 +2F12 F13Cos60
(1/2)
Answer
(1/2)
Super position
(1/2)
Principle
(1/2)
d) No
(1/2)
Q. Text
Q. No. 3
A metal sphere of radius R carrying q (+ve) charges is shown in the figure.
(MP - 1,2,5,7)
(MP = 1,2,5,7,10)
r
R
a) Draw the electric lines of force related to this metal sphere.
b) Derive the strength of electric field at a distance ‘r’ from the centre of sphere.
c) “A sphere of radius 1cm can hold a charge of 1 coulomb” Comment on the statement.
[Hint:- The dielectric strength of air is 3 mv/m]
Scoring key
Total Score 4
a)
(MP - 1,2)
b)
r
R
E.ds=
q
εo
E × 4π r 2 =
q
εo
q
E= 1 2
4πε o r
(1/2)
(MP - 1,2)
c) No, the charge will leak through the air
(MP - 1, 5, 7, 10)
Q. Text
Q. No. 4
An electric dipole is placed in a uniform electric field of Intensity E as shown in the figure.
E
+q
θ
-q
a)
b)
c)
d)
What are the forces on +q and -q?
What is the net force on the system?
Copy the diagram, mark the forces and derive an expression for torque.
If the dipole is placed in a nonuniform electric field what nature of motion does it show?
Time : 7 min
Score : 4
Scoring Key
Score
a) +qE, -qE
(1/2)
(MP - 1,2)
1
b) Zero
( /2)
(MP - 1,2)
c) fig, Proof (τ = P×E)
d) Both translation and rotational motion
Text :
No. 5
A charge of +5µC is placed in free space
2
(MP - 1,2,6)
(1/2)
(MP = 1, 2, 6, 7)
3mm
P
5µC
5mm
Q
a) The workdone to bring a + IC charge from infinity to a point ‘P’ is called
(i) capacitance (ii) dielectric constant (iii) potential energy (iv) electric potential
(b) Calculate the workdone in above process
(c) Calculate the workdone to move a +IC charge from ‘P’ to ‘Q’
Score (1+1 1/2+1 1/2 = 4)
Time 8 mts
Scoring Key
a) Electric potential
(1)
(MP -1,2)
(11/2)
(MP - 1,2,5,6)
1 q
b) Workdone = potential = 4πε
o r
−6
9×109 × 5×10 −3 =15×10 6 J
3×10
1
c) Workdone p.d = 4πε
o


q 1 − 1 
 r1 r2 
9
−6 
1 − 1 
= 9×10 ×5×10 
 3×10−3 5×10−3 
6  5−3 
= 9×5×10  15 


Question Text:
Q.No: 6
r2
O
6 2 
= 9×5×10 15 
 
P
r
r12
r
r3
= 6×10 6 J
(11/2)
Q
q1
r1
The above fig shows a charge q1 placed at a distance r1 from the origin.
a. What is the work done to bring a charge q2 at P from infinity
b) Derive equation for work done to bring another charge q3 on the point Q (q1 and q2 are present) from
infinity.
c) What is the total work done.
To arrange the three changes in the system and name the work done
Score 3
Time 5 min
Scoring Key
1 q1
a) Potential of P due to q1 = 4πε r
o 12
1 q1q 2
Work done to bring q2 from infinity to P = 4πε r
o
12
1 q1
1 q2
b. Potential at Q due to q1 and q2 = 4πε r + 4πε r
o 13
o 23
Workdone to bring q3 from infinity to Q
1
= 4π
εo
(1/2)
(MP - 1,2)
(1/2)
(MP - 1,2,5)
(1/2)
 q1 q 2 
 +  q3
 r13 r23 
(1/2)
(MP - 1,2,5)
1  q1q2 q1q 3 q 2 q 3 
c. Total workdone to make the system = 4πε  r + r + r 
o  12
13
23 
Total workdone is equal to potential energy
(1/2) (MP - 1,2,5,6,7)
Question Text
No: 7
A spherical surface enclosing a charged spherical shell of radius R and charge density σ is shown
r
R
a) which law establishes the total flux related to this surface and total charge on the spherical shell? State
the law
(11/2)
b) Using this law find the field at any point on the spherical surface.
(2)
c) If the entire charge on the shell is brought to its centre, does the field on the spherical surface change?
Justify.
(11/2)
Time : 9 min
Score : 5
Scoring Key
(1
a) Gauss’s Theorem
/2 )
(MP - 1)
Statement
(1)
(MP - 1,2)
z
r r
b. φ = E . ds =
4πR 2σ
(1)
ε0
and result
c. Does not change
Proof
(MP - 1,2,5)
(1)
(1/2)
(1 )
(MP- 1,2,5,7)
Question Text :
Q.No: 8
Classify the following molecules in to polar and nonpolar and justify your answer with proper explanation.
O
H
H
H
CO 2
H2
O
H
O
Cl
H
HCl
H
H2O
Time 5
Score 2
Scoring Key
Polar HCl, H2O
Non - Polar - H2CO2
Explanation based on dipole moment
Score
(1/2)
(1/2)
(1/2+1/2)
(MP - 1, 3)
(MP - 1, 3, 4, 7)
Question Text :
Q.No: 9
The figure below shows a non polar dielectric slab placed in between the plates of an uncharged parallel
plate capacitor.
Area of each plate = A
Distance of separation between the plates = d
a) If the dielectric slab is absent and the capacitor is charged to a surface charge density σ, the electric field
in between the plates is
σ2
(i) E = σ ε o
(ii) E = ε
o
σ
(iii) E = ε
(iv) E = σ ε o 2
o
b) Redraw the given figure, which shows the alignment of the nonpolar molecules, when the capacitor is
charged
c) Derive an expression for capacitance of the above capacitor with the dielectric slab in between the
plates
d) “Dielectric constant of a conductor is infinity” - Justify
Time 7 mts
Score 5
Scoring Key
Score
σ
(a) E ε
o
(1/2)
(MP - 1)
(b) Figure showing polarisation
(1)
(MP - 1,2,5)
 σ -σ p 
(c) E =  ε 
 o 
(1/2)
V = Ed
Q=σA
(1/2)
(1/2)
σ -σ p ) = 1 σ
K
(1/2)
ε o KA
(1/2)
(MP - 1,2,5,7)
d
d) Explanation based on ability to allow electric field lines to pass through
(1)
(MP 1,2,5,7)
Result C =
Question text
Q.No 10
a
b
c
C1
d
e
C2
f
g
C3
Three capacitors C1, C2 and C3 are connected to a cell of emf V as shown in the figure.
a) The arrangement of these three capacitors are called.................
(i) parallel combination
(ii) series combination
(iii) LCR combination (iv) c-c combination
b) Find the effective capacitance of the above combination.
c)
4V
2V
1V
g
a b c d e f
The above graph shows the variation of potential in going from a to g. From the graph the relation among
C1, C2 and C3 is
(i) C1 = C2 = C3
ii) 2C1 = 2C2 = C3
(iii) C1 =C2 = 2C3
iv) C1≠ C2 ≠ C3
Total score 4
Time 9min
(a) Series Combination
(1)
(MP - 1,2)
(b) V - V1 + V2 + V3
(1)
Q Q Q Q
= + +
C C1 C 2 C3
(1/2)
1= 1 + 1 + 1
C C1 C 2 C3
(1/2)
(MP- 1, 2, 3, 5)
c) Q is same, V1 = 1V, V2 = 1V, V3 = 2V
C1 : C2 : C3
1:1: 1
= 2C1 = 2C2 = C3
11 2
(Score 1)
(MP - 1, 2, 3, 5, 6, 7, 10)
Time : 10 min
Score : 5
Scoring Key
a) Capacitor
b) Electric field
1
CV2 (Proof)
2
d) Since V is constant while decreasing C
Energy also decreases (Capacitance decreases
with increase in distance)
c) E =
Q. Text
Q. No: 11
Score
(1)
(1)
(MP - 1)
(MP - 1,2)
(2)
(MP - 1,2,4,5)
(1)
(MP - 1, 2, 4, 5,7)
Y
X
d
Two metal plates X and Y of the area ‘A’ are separated by a distance ‘d’, charged + and - respectively.
a) This arrangement is called ...............................
b) The arrangement stored energy in the ..............................
(Magnetic field, Electric field, Electromagnetic field, Gravitational field)
c) Derive an expression for the energy stored the arrangement
d) When we increase separation between two plates by keeping V constant, what happens to total
energy stored in the system
Time : 10 min
Score : 5
Scoring Key
Score
a) Capacitor
(1)
(MP - 1)
b) Electric field
(1)
(MP - 1,2)
1
CV2 (Proof)
2
d) Since V is constant while decreasing C
Energy also decreases (Capacitance decreases
with increase in distance)
c) E =
Question ext:
Q No: 12
(2)
(MP - 1,2,4,5)
(1)
(MP - 1, 2, 4, 5,7)
Metal sphere
Motor
(iii) Photovoltaic cell
a) Identify the device
(i) Moving coil Galvanometer
(ii) Cyclotron
(iv) Van-de-Graff electrostatic generator
b) Explain the construction and working of the above device
c) What happens if the upper metal sphere is replaced by a cubical shaped metal? Explain.
Score - (1 + 2 + 2 = 5)
Time 9 mts
Scoring Key
a) (iv) Van-de-graff electrostatic generator
b) Construction
Working
c) The device will not work
The charge will get discharged
(1)
(1)
(1)
(1)
(1)
(MP - 1)
(M.P -1,2)
(MP 1, 2, 5)
(MP-1, 2, 5, 7)
Question Text
Q: No: 13
A rectangular conductor of length l and area of cross section A and electron density ‘n; is shown below
Y
X
a. When the face Y is given positive potential and X negative potential what will happen to the electrons
inside the block
b. What is meant by drift velocity? How is it related to the field inside the metal?
c. Deduce an expression connecting intensity of electric field and drift velocity.
d. Under the application of an electric field do all the electrons move in a same direction? Explain
Scoring Indicator
1
a. Drifts towards +ve and -ve plate
/2Score
(MP -1)
b. Defenition
1 Score
(MP- 1)
1
Relation
/2 Score
c. Derivation
2 Score
(MP - 1,2,5)
1
d. No
/2 Score
(MP -1,2,5)
1
Explanation
/2 Score
(MP - 1,2,5,7)
Time: 10 mts
Score : 5
Q: No: 14
To study the relation between potential difference and current in an electrical circuit, a student is
provided with a resistance wire, a cell and a key.
a) Draw a circuit which allows current flow through the resistance wire
b) Modify the circuit by introducing an ammeter, Voltmeter and a rheostat for varying the potential difference across the resistance and to measure that potential difference and the corresponding current.
c) Let in the above experiment the student obtained the following data
Voltage (V)
0
2
4
6
8
10
Current (I) A
0
0.1
0.2
0.3
0.4
0.5
Draw a graph connecting V and I using above data. Then establish the relation between V and I as a
law.
d) Instead of the resistance wire if the student uses a p-n junction diode in the forward biased condition
how the relation between V and I changes? Justify.
Score and time
(1 + 1 + 2 + 2 = 6)
Time 10 mts
Scoring Key
(a)
R
Key
E
(1)
(MP - 1,2)
(b)
V
A
R
K
Rh
(1)
(MP - 1,2)
E
(c) A graph connecting V and I , which is a straight line statement of ohms law. (Score 2) Mp (8, 9)
(d) Forward characteristics of a diode the graph is not a straight line V and I are not directly proportional,
up to knee voltage
(Score 2)
(MP - 1,2,6,8,9)
Question Text:
Q.No.15
Table given below shows the resistivity and temperature coefficient of resistivity of certain materials.
Material
Resistivity
Ω m at O C
o
Temperature coefficient
of resistance / (OoC)
Copper
1.7 × 10-8
0.0068
Carbon
3.5 × 10-5
-0.0005
Platinum
11 × 10-8
0.0039
Glass
1010 × 10-14
-
............
1014
-
Silicon
2300
0.07
a) Classify the materials into conductors, semiconductors and insulators and justify your answer
b) Temperature coefficient of resistance of silicon and carbon are negative. What does it mean? Justify
c) Carbon can not be used to make a pn junction diode. Why?
Time 6 mts
Score: 4
Scoring key
(a) Classification
(b) Resistivity decreases with increasing
temperature
(c) Four valance electrons of carbon is in the 2nd orbit,
where as in Silicon they are in the 3rd orbit.
Hence difficult to get free electrons in the case of Carbon.
Score
(1/2+ 1/2 + 1/2)
(1/2)
(MP - 1)
(MP-1,2)
(1)
(MP - 1,2,5,7)
Question Text :
Q. No: 16
A, B, C and D are four rings on a carbon resistor.
ABC D
A = yellow, B = violet, C = Yellow, D, Silver
(a) What is the value of resistance of above resistor?
(b)
Brown Brown Yellow Black
(i) 120 Ω
Green
Red
(ii) 45 Ω (iii) 165 Ω (iv) 35 Ω
The combined resistance of the above two resistor is
Time : 4
Score : 2
Scoring key
(a) 470 x 103Ω ± 10%
(Score 1)
(MP- 1, 2)
(b) 165 Ω
(Score 1)
(MP -1, 2, 5)
Question Text:
Q: No: 17
A cell and two resistors R1 and R2 are provided to you.
a) Draw different combinations of resistors using R1, R2 and the cell.
b) Derive an expression for the effective resistance of the circuit in which current is the same in both
resistors
c) If R1 = 4 Ω and R2 = 6 Ω ., in which combination effective resistance is minimum? find its value?
Scoring key
Score
Time : 7 min
a) parallel and series combination of resistors
(1/2+ 1/2)
(MP - 1)
Score: 4
b) Series combination, R = R1+ R2 (Proof)
2
(MP - 1,2)
c) Parallel Combination R =
4×6
4+6
1
(MP - 1,2,3,5)
Resistance
Qn: Text Q. No: 18
a) State whether the following statement is true or false
“The value of resistance of a metal increase with the rise of temperature”
b) Explain the reason
C
c)
B
A
Temperature
With the help of the above graph, match the following
I
A
B
C
II
Manganin
Iron
Wood
Carbon
d) Alloys like manganin, eureka, constantan etc are used in making standard resistance coils. Why?
Scoring key:a) True
(1)
(MP - 2)
b) As temperature increase the frequency of collision of free electrons increases. This reduces relaxation
time.
(11/2)
(MP - 2,5)
c)
II
I
A
B
C
Carbon
Manganin
Iron
(11/2) (MP - 2, 5, 6)
d) Due to high resistivity and low temperature coefficient of resistance. (1)
Question Text:
Q No: 19
(MP - 2, 5, 6, 9)
K
r
E
R
a) Name the pd between terminals of the cell when (i) key K is open (ii) K is closed
b) What is the reason for the difference in potential in the above two cases?
c) What happen to terminal potential if current is increased (i) For an ideal cell (ii) for ordinary cell
Scoring key
a. (i) emf
(ii) Terminal potential difference
Score 1/2+ 1/2 ( MP - 1,2)
b. Internal resistance
Score 1/2
(MP (detects similarities)
(MP - 1,2,3)
c. For ideal cell r = O, V = E, for ordinary cell I increases , V decreases
Score 2
MP (Cause - effect relations)
(MP -1,2,3,5,6)
Question text :
Q. No: 20
a. Potentiometer is better than voltmeter for measuring emf because
i) It is cheap
ii) Easy to handle
iii) Its measurement uses null method
b. Give the basic principle of potentiometer,
c.
3Ω K
2
G
E
J
Rh
K1
If K2 is open balancing length is 600cm, if K2 is closed 350cm is balancing length. Calculate the
internal resistance
Score : 4
Time : 7
a) iii
b) Principle
c) r =
R (l1 −l 2 )
l2
Score (1)
Score (1)
MP -1
(MP- 1, 2 )
Score 1/2
MP (1, 2, 5)
Substitution and answer
1 1/
Question No:21
A resistor network connected to a source of emf is shown
R1
R3
I1
I
R5
R2
R4
V
Applying Rule
a. What is the current through the resistance R2 shown in the figure
R1 R 3
b. If the current through the resistance R5 is zero show that R = R
2
4
c) In the above circuit if R1 is doubled and R2 is halved what change should be made in R4. So that the
current through the resistance R5 remains zero.
Scoring Key
a. I- I1,
b. Voltage across R1
Voltage across R2
Kirchoff’s II Law
R1 R 3
=
R2 R4
c. Substitution
Becomes one fourth
1 score
1
/2 Score
1
/2 Score
1
/2 Score
(MP - 1,2)
1
(MP - 1,2,5,6)
/2 Score
Time : 7
Score : 4
1
/2 Score
/2 Score
1
(MP - 1,2,5,6,7,10)
Q.No:22
“When electric current is passed through a resistance wire, it get heated up”
a) Name the law associated with this phenomenon.
b) What happens to the heat energy developed, if the current through the wire is doubled.
c) What happens to the heat energy developed in the wire, if the wire is stretched to double its length?
Explain
d) “Length of a fuse wire is immaterial”, comment on this statement.
Score ( 1/2 + 1 + 1 1/2 + 2 = 5)
Time 10 mts
Scoring Key
a) Joules law
(MP - 1,2)
b) H = I2Rt, when I = 2I, H = 4H , Heat becomes four times (Score -1) (MP - 1,2,5)
c) H = I2Rt, when L = 2L, R = 4R and H = 4H Heat becomes four times (Score -1/2) (MP - 1, 2, 5)
d) The statement is true
I 2ρ
Heat produced per unit surface area = 2 3
2π r
ρ -density, r -radius
Thus heat produced per unit area is independent of length of conductor
Question text :
Question No: 23
Data given below shows the measured thermo emf with temperature of the hot junction in the case of
Ni-Cu thermocouple.
Temperature
of hot junction OC
Thermo e.m.f
µV
0
100
200
300
400
500
600
700
800
900
1000
0
1525
2840
3945
4940
5525
6000
6265
6320
6135
5800
a) Draw a graph connecting temperature of hot junction and thermo emf of cold junction temperature is
found to be 0OC. (graph paper is suppled)
b) Extrapolate the graph to get temperature of inversion and hence note the values of neutral temperature
and temperature of inversion. What is the relation between temperature of cold junction, neutral temperature and temperature of inversion?
c) Observing the shape of the graph, give a relation between thermo emf and temperature difference
between hot and cold junction.
d) If the temperature of cold junction is increased, what happens to neutral temperature.
Time 9 mts
Score : 5
Scoring Key
a) Graph
b) Values of each relation
1 2
c) Parabola, E = αθ+ βθ
2
d) Neutral temperature remaining constant
Score
(11/2)
(1/2+1/2 + 1/2))
(MP - 1,2,3)
(MP - 1,2,3,5)
(1/2 + 1/2)
(MP - 1,2,5,6)
(1)
(MP - 1,2,5,6,10)
Q. Text
Q. No. 24
I
.
P
dl
Consider a conductor carrying current ‘i’, P is a point at a distance ‘r’ away from the conductor.
a) What is the direction of magnetic field at P?
b) What are the factors affecting magnetic field at P due the element dl carrying current i?
c) Derive an expression for magnetic field at P, if the current carrying conductor has infinite length?
d) Draw a graph connecting Intensity of magnetic field and distance.
Time : 11
Score : 5
Scoring key
MP
Score
a) Inward to the plane of paper ( ⊗ )
1
1
b) idl, r and sin θ
2
1
µo I
c) Proof -
2π r
21 / 2
6
7
d.
/2
1
B
r
Question No: 25
A current flows through a circular loop of radius r is shown in the figure
o
r
a) What is the direction of magnetic field at ‘o’ ?
b) Derive an equation for magnetic field at ‘o’ due to the circular loop carrying current i?
c) If the loop splits into two equal halves as shown in figure.
What will be the magnetic field at the centre ‘o’.
r
i
i
O
i
i
Time 11
Score : 41/2
Scoring key
a) Outward to the plane of the paper
b) Proof - B =
µ o Ir 2
4π (r 2 + x 2 ) 3/2
c) Zero
MP
1
Score
1
/2
4
3
5
1
Qn. Text :Question No: 26
The figure shows a long straight conductor carrying a current I. A magnetic field is produced around the
conductor.
I
P
x
l
dl
a) What is the magnitude of the magnetic induction at a point ‘P’ distant ‘x’ from the conductor?
b) What is the shape of the magnetic line of force?
c) If a small element of length ‘dl’ of this line of force is considered, what is the angle between B and dl?
d) Deduce the law relating current ‘I’ and magnetic induction ‘B’, by integrating the line integral of magnetic induction of the element
e) “The above law is nothing but Biot-Savart law” Justify.
Scoring Time
Score
Scoring key
a) B =
µoI
2π r
b) Concentric Circle
(1)
I
(MP -1)
(MP - 1)
B
c) Zero
d) Proof
(1/2)
(2)
(MP - 1, 2, 7)
(MP - 1, 2, 5, 9)
e) Biot-Savart law is the differential form of magnetic induction B or magnetising form H and amperes
r
circuital law is the integral form of B .
Question Text .
Question No: 27
When a changed particle entering normal to a uniform magnetic field, it take a circular path.
a. Name the particle accelerator using this principle.
b) Explain the working of that particle accelerator with relevant theory.
c) The neutrons can’t be accelerated using this principle accelerator. Why? Score 3
Scoring key
Time 5 min
a) Cyclotron
Score 1
(MP - 1)
b) Explanation
Score 1
Cyclotron frequency Score 1
maximum velocity of particle
Score 1
(MP - 1,3)
c) No charge on neutron
Score 1
(MP - 1,3,5)
Question text :
Question No: 28
Path of a chargeded particle in a particle accelerator is shown in the above figure
a) Name the particle accelerator
b) Write the principle behind the working of this particle accelerator
c) A particle of mass 6.65 × 10-27 kg having positive charge equal to two times of electron, moves with
a speed of 6 × 105 m/s in a direction perpendicular to that of a given magnetic field of flux density 0.4
Score 6
weber/m2. Find the acceleration of the particle
Time 8 min
Scoring key
a. Cyclotron
Score 1
MP 1, 2
b. Principle of cylcotron
Score 2
MP 1, 2
c. Formula
F = BqV
ma = BqV
Score 1
substitution with SI unit
Score 1
MP 1, 2, 5, 6
Answer with SI unit
Score 1
Question Text .
Question No: 29
A rectangular loop ABCD made of copper suspended in a magnetic field B is shown in this figure.
Here the conductors AD and BC are perpendicular to the field. A current I is passed through the loop.
B
A
B( magnetic field)
D
C
a. Write down the expression for the force acting on a current carrying conductor placed in a magnetic
field.
b. Using the above expression find the force loop and torque acting on this loop.
c. What is the difference in torque acting on a rectangular loop when it is rotated in a radial and parallel
magnetic field.
Time : 8 min, Score : 4
a. Expression
1 Score
(MP 1)
b. Forces on four sides
1 Score
(MP 1, 2, 5)
Torque
1 Score
1
c. Constant torque
/2 Score (MP - 1, 2, 5, 7 )
1
/ Score
varying torque
Question text :
Question No: 30
“Moving coil Galvanometer is a device used for detecting very feeble current”
a) What is the working principle of a moving coil galvanometer?
b) Describe the construction and working of a moving coil galvanometer
c) When a high current is passed through a moving coil galvanometer , it will get destroyed. How?
Score and Time
Scoring key
a) Working principle
(Score 1 )
(MP - 1)
b) Construction and working (Score 4)
(MP 1, 4, 5, 7)
c) High torque may distroy the suspension fibre
High current may burn out the coil
(Score 1)
(MP - 1, 4, 5, 7,9)
Question No: 31
The magnetic field along the axis of a circular coil is found to be B =
µ o Ia 2
2(r 2 +a 2 )3/2
a) What is the magnetic field along the axis if r>>a
b) Compare the above magnetic field with the electric field along the axis of an electric dipole and explain
what is magnetic dipole moment?
2
c) If the above circular coil is placed in a radial magnetic field, a torque will act. If the radial field is
replaced by a nonuniform magnetic field which is not radial, what will happen?
Time 6 mts
Score 4
Scoring key
Score
a) B =
µ o Ia 2
2r 3
b) Comparison
Definition
c) Translation motion
1
(MP - 1)
1
1
1
(MP - 1,2,5)
(MP - 1,2,5,10)
Q Text.
Q. No 32
A freely suspended bar magnet aligned in north-south direction, i.e, north pole of magnet indicates northern side of earth.
a) What is the direction of earth’s magnetic field?
b) Explain three magnetic elements of earth.
c) Write any one of the causes of earth’s magnetism?
d) What in the alignment of a freely suspended magnet at geographic north pole of earth?
Time : 7 Min
Score : 4
Scoring Key
Score
1
/2
(MP - 1)
a) Magnetic North to Magnetic South or
almost in Geographic South to geographic north
b) dip, declination, Bh-Explanation
c) Any logical statement
d) Vertical with north pole of magnet downward
11/2
1
1
(MP - 1,3)
(MP - 1,2,5)
(MP - 1,2,3,5)
Qn Text:
Question No: 33
A magnetic material contained in a curved glass plate, when placed in a nonuniform magnetic field, exhibits
a property as shown.
N
S
a) Which type of magnetic material is this? Explain the property
b) Write two examples for such a magnetic material. Explain how the property relates with temperature?
c) “The susceptibility of a material is small” what do you mean by this statement.
Scoring key
a) Paramagnetic
- 1/2
Explanation
- 1/2
(MP 2, 3, 4)
b. Aluminium, platinum
-1
As temperature increases paramagnetism decreases
-1
(MP 2, 5, 6, 7)
c) The material can’t be easily magnetized
-1
(MP 2, 9, 10)
Question Text :
Q No: 34
S
N
A
G
B
When a magnet is moved inside the coil, the galvanometer shows a deflection which is shown in figure.
a. Name the phenomena involved in this process
b) Write the laws governing this phenomena
c) Explain how conservation of energy is satisfied in the phenomena
Scoring Key
a. Electro magnetic induction
Score 1
(MP - 1)
b. Laws of electromangetic induction
Score 2
(MP - 1,2)
(MP - 1, 2, 5)
c. Explanation of conservation of energy
Score 1
Question No: 35
Metalic disc
Electro magnet
When AC is switched on the thin metallic disc is found to thrown up in air.
a. Which makes the disc to thrown
b. How will you explain the mechanism behind the movement of disc
c. Write the working principle of induction heater.
Scoring Key
a. Eddy current
Score 1
(MP - 1, 2)
b. Explanation on the basis of eddy current
Score 2
(MP - 1, 2, 5, 6)
c. Principle
Score 1
(MP - 1, 2, 5, 6, 9)
Question text:
Q. No: 36
A solenoid is a insulated copper wire wound in the form of cylinder.
a) When current increases flux linked with the solenoid is ..............................
b) Derive an expression for inductance of a solenoid
c) Calculate the inductance and energy stored in the magnetic field of a air cored solenoid 50cm in
diameter and 50m in length wound with 1000 turns, if it is carrying a current of 9A.
Scoring Key
a. Increases
Score 1
(MP - 1,2)
b. Derivation
Score 2
(MP - 1, 2, 3, 5, 6)
1
c. Formula
Score /2
Substitution
Score 1/2
Answer with unit
Score 1
(MP - 1, 2, 3, 5, 6, 9)
Question Text
Q.No 37.
A graph connecting the voltage generated by an a.c. source and time is shown.
V
200
100
O
a.
b.
c.
t
What is the maximum voltage generated by the source ?
Write the relation connecting voltage and time
This a.c. source when connected to a resistor
produces 40J of heat per second. Find the equivalent d.c.
voltage which will produce the same heat in this resistor.
Time : 7 min.
Scoring Indicators
a.
b.
c.
200 V
1 score
(MP -1)
Relation
1 score
(MP -1)
Score : 4
RMS voltage
1 score
Result
1 score
(MP -1,5)
Question Text.
Q.No.38.
In a tuner circuit of radio receivers an electrical circuit is familiar to you is used.
a.
Identify the circuit.
b.
Deduce an expression for impedance of that circuit.
c.
Explain the phenomenon which enables above circuit to select a particular frequency from a
number of frequency.
d.
Low cost radio sometime receives more than one station at a time. What may be the
problem associated with the tuner circuit of the radio ?
Time : (1+2+2+1 = 6)
Score : 12 min.
Scoring key :
a.
LCR series circuit (score 1) MP (1)
b.
2
Derivation of the equation Z = R + ( Lω −
1 2
)
Cω
(Score 2)
MP (1, 4, 5)
c.
Explanation of the phenomenon resonance
(Score 2)
MP (1,2,5,7)
d.
The Q factor of LCR circuit is small ie; the circuit is flat, not sharp (Score 1) MP(1,2,5,7)
Questions Text:
Q.No.39.
An alternating voltage V = V0sin ωt is applied to an LCR series circuit.
a.
What is the equation of current in the circuit ?
b.
Starting from the equation of instantaneous power, obtain an expression for average power
in the circuit. [Take the average value of cos function as zero]
c.
In an LCR series circuit, the voltage and current are found to be in phase. What does it
mean ?
Time : 7 min.
Score : 4
Scoring Key
a. I = I m sin ωt
Score [1]
(MP-1)
b. P(t) = V(t) I( t)
(½)
Substitution
(1)
(MP-1,2,5,7)
Result
(½)
c. Resonance
(1) [MP -1,2,5,6,10]
Question Text :
Q.No 40.
A.C. adaptor converts household ac into low voltage dc. A stepdown transformer is a essential
part of ac adapter.
a.
What is the use of step down transformer ?
b.
What is the principle of a transformer. Explain.
c.
N
V
1
1
Derive the relation N = V
2
2
d.
Can we call the current related to the primary of transformer as wattles current ? Justify
your answer [take the resistance of primary as zero]
Time : 13 min.
Score : 6
Scoring Key
a.
To decrease the input voltage - (1) MP - 1, 2
b.
Mutual induction - (½) MP - 1, 2,
c.
dφ 1
dφ 2
α N1 ,
α N 2 ..... (1).
dt
dt
Explanation - (1½)
∴
V1
N
= 1 ....... (2)
V2
N2
MP - 1, 2, 5, 6
T
d.
dφ 1
dt = N 1
dφ 2
N2
dt
VIdt
Average power Pav = ∫
=
T
0
T
∫
0
V0 Sin ω t x I 0 sin (ω t − π / 2)
T
[½]
[1½]
MP - 1, 2, 3, 5, 6, 10
Q.No41.
When a bar magnet is brought towards a coil, the compass needle placed near to the coil shows a
deflection
(a) The compass needle deflects due to
(i) magnetic effect of current
(ii) Electromagnetic induction.
(b) How can you explain lenzs law using this experiment
Time : 5 min.
Score : 2
Scoring Key
Score
a. Electromagnetic induction
(1
(MP-1)
b.
Explanation on
Lenz’s’s law
(1)
(MP-1.2,5,7)
Q.No42.
Fill in the blanks :
..................... predicted theoretically the existence of electromagnetic waves. Production of electro
magnetic waves was first demonstrated by ................. The first Indian Scientist who became successful in producing and detecting electromagnetic waves was .................. The first transmission of
electromagnetic waves over distance of many kilometer was done by ..................
(H. Hertz, E.C.G.. Sudarsan, G. Marconi, T.A. Edison, J.C. Maxwell, J.C. Bose)
Time : 4 min.
Score : 2
Scoring Key
J.C. Maxwell, H.Hertz, J.C. Bose, G.. Marconi.
2 score
MP-2
Q.No43
Match
. the following :
A
Radiowave
Infra red waves
X-rays
γ - ray
B
High energy electrons
Radioactive nuclei
accelerated motion of
charges in conducting wires
Hot bodies and molecules
Special vacuum tubes
C
destroy cancer cells
Radar systems
Green house effect
cellular phones
diagnostic purpose
Time : 5 min.
Score : ½ + ½ + ½ + ½ = 2
Scoring Key
A
Radiowave
Infrared waves
X-ray
γ - ray
B
accelerated motion of charges
in conducting wires
hot bodies and molecules
high energy electrons
radio active nuclei
C
Cellular phones
Green house effect
diagnostic purpose
destroy cancer cells
4x½=2
Question text :
Q.No 44.
Figure shows the path of the light rays through
a glass slab.
n1
i
n2
r
r
a.
b.
c.
Name the phenomena involved here.
Relate the values of n1, n2, i and r on the basis of one figure.
Copy the figure of glass and draw the path of ray when n2 < n1.
Scoring indicater
a.
b.
Refraction
n1 < n2
sin i
= 1n2
sin r
c. Figure
score -1
MP - 1,2
score -1
MP -1,2,5,6
score -1
MP -1,2,5,6,9
i
Time ; 7min
Score; 3
Q.No 45.
High precision optical instruments uses prisms instead of mirror to reflect light.
a.
Name the phenomena used for reflecting light using prism.
b.
What is the advantage of using prism instead of mirror for reflecting light ?
c.
The polarizing angle of water is 52O calculate the critical angle of water.
Time & Score
Time : 8 min.
Score : 5
Scoring Key
a.
Total internal refection of light
score - 1
MP - 1, 2
b.
100% light intensity is reflected by prism
where as mirror reflect a maximum of 95%.
score - 2
MP- 1, 2, 5
c.
Formula
score - ½ MP - 1,2,5,6,7
Application
score - ½
Answer with unit
score - 1
Question text :
Q.No 46
Refraction of a ray of light at a spherical surface separating two media having refractive indices n1 and n2 is
shown in the figure
n2
n1
O
C
A
Medium I
I
Medium II
a) Which of the two media is more denser ?
n 1 n 2 n 2 − n1
+
=
OA AI
AC
b) Using the above relation arrive at the thin lens formula
d) An object is placed on the principal axis of a convex lens at a distance 8 cm from it. Find the magnification
b) In the figure, show that
of the image if the focal length of the lens is 4 cm
a. Medium II
1 Score
MP -1,2
b. Expression for i
1/2 score
MP -1,2,5
Expression for r
1/2 score
Snells Law
1/2 score
Equation
1/2 score
c.First curved face equation
Second curved face equation
Case object at infinity
Equation
1/2 score
1/2 score
1/2 score
1/2 score
MP -1,2,5
d. Determination of V
Magnification
1/2 score
1/2 score
MP -1,2,5
Question Text: The image formed by a thin lens is shown in the figure
Q.No: 47
l
100cm l
Time: 12mts
Score: 6
a.
b.
c.
d.
What is the nature of the image
Find out the power of the image
Draw the ray diagram showing above lens forming a magnified erect, virtual image
If a convex lens of focal length 20cm is kept in contact with above lens. What is the focal length
and power of the combination
Score :( 1+ 1+1½ +1 ½ = 5)
Time : 10 Mnts .
Scoring key
a. Magnified, inverted and real
(score 1)
M.P 8
b. Power = 1/f = 1/10
(Score 1)
M.P (1,2, 5)
c. Figure
(Score1½)
M.P (1,2,5,6,9)
1 1 1
(Score1½)
M.P (1,2,5,6,9)
d. f = f + f and p =1/f
1
2
Question Text:
Q No: 48
In the figure given below, PQ represents an incident ray falling in the side AB of a prism, when monochromatic
light is used
A
Q
P
a.
b.
c.
d.
C
B
Draw the refracted ray, emergent ray and mark the angle of deviation
Derive an equation for refractive index of the material of the prism in terms of angle of minimum
deviation
Suggest the experimental procedure that can be used to measure the angle of minimum deviation
using spectrometer
Draw the incident ray and refracted ray, at the angle of minimum deviation
Total score 6
Time : 12mts
Scoring indicater
a. Diagram
b. A= r1 +r2
i1 +i2= A+d
1-d curve
angle of min deviation
result
d. Procedure (8)
e. Diagram (10)
(½ + ½ )
½
½
½
½
½
1½
1
Question Text:
Q. No: 49
Match the following
A
1. Colour of sky
2. Rainbow
3. different colours
seen in soap bubbles
B
1.
2.
3.
4.
5.
6.
7.
Interference
scattering
Dispersion
Diffraction
coherence
Looming
mirage
MP 1,2
MP 1,2,5
MP 1,2,5,6,
MP 1,2,5,6,10
Time : 2 mnt
Score: 1½
Scoring indicator
Colour of sky- Scattering
Rainbow – Dispersion
Different colours in soap bubbles – interference
Score(½ + ½ + ½ = 1½)
MP- 1,2,3
Question Text:
Q. No: 50
When white light is passed through a prism it splits into seven colours
a. Name the optical device which is used to observe pure spectra and to determine the refractive
index of the prism
b. Write down the main parts of the above device
c. Explain how the refractive index of the material of a prism can be determined using the above
device
Score: ( 1+ 2+ 4 = 7)
Scoring key
Time : 12 Mts
a. Spectrometer
Score 1
b. Collimator, Telescope, Prism Table, Vernier Table
Score 1
c. Initial arrangements
Score 1
Angle of the prism
Score 1
Angle of minimum deviation
Score 1
Refractive index
Score 1
Question Text:
Q. No: 51
Match the following
1.
2.
3.
4.
Longitudinal wave
Spherical wave front
A part of spherical wave front
Linear source
1.cylindrical wave front
2.plane wave front
3.Hygiene’s wave theory
4.Point source
Score :2
Time : 3min
Scoring Key
1.Huygen’s wave front
2.Point source
3.Plane wave front
4.Cylindrical wave front
MP 1,2,3
Question Text:
Q.No: 52
screen
A double slit placed in front of a screen is shown
a.When the double slit is illuminated by a monochromatic light, alternate bright and dark fringes is
observed in the screen. Name the phenomenon
b. Explain why certain part of the screen appears bright while certain other parts dark
c. Deduce the relation connecting the wave length of light and the fringe width
d. Two coherent sources have intensities in the ratio 25: 16. Find the ratio of intensities of maxima
to minima after the superposition of waves from the two source
Time: 12mts
Score : 6
Scoring indicator
a. Interference
Score 1
b. Explanation
Score 1
c. Derivation
Score 1
d. Ratio of ampltudes 5:4
MP 1,2
MP 1,2,5
MP 1,2,5
b g
b g
a +a
I max
= 1 1
I min
a1 − a1
2
2
81: 1
Score 1
MP 1,2,5,7
Question Text:
Q. No: 53
In the classroom while studying optics you might have seen the pattern formed on a screen, when the
narrow gap between two razor blades is illuminated by a monochromatic light. The pattern consists of a
central maximum and a few secondary maxima and minima on either sides.
a. Name the phenomenon due to which the pattern is formed
b. Derive an equation for the width of central maximum of above pattern
c. What happens to the width of central maximum if the frequency of the light used is increased
d. What happens to the width of central maximum when the distance between the blade and the
screen is increased. Justify your answer
Score ( 1+ 2+ 1+ 1= 5)
Ttime10 mnts
Scoring Key
a. Diffraction
Score 1
MP 1
b. Derivation of width
Score 1
MP 1,4
c. When frequency increases, λ decreases so width decreases Score 1
MP 1,5,6
d. When x is increased width incrreases
Score 1
MP 1,5,6,7.
Question Text:
Q. No: 54
Figures (a) and (b) given below show the pattern obtained by passing laser beam on a pair of closely
spaced slits and on a single slit
Fig (b) Single slit patterm
Fig (a) Double slit patterm
Make scientific not based on the patterns given above and the flowing hints
1. Interference and diffraction
2. Variation of intensity and band width
3. Graphical representation of the above two pattern
( Total Score :4 )
( Total time :8mints )
Scoring Key
Interference- diffraction
Variation of intensity and band width
Graphs
comparison
Score
Score
Score
Score
1
1
1
1
MP 1
MP 2,3
MP 2,3
MP 2,3,5
s
Question Text:
Q.No: 55
You might have seen “reading lens” (Simple microscope)
a. Which type of lens is used as reading lens?
b. Derive an expression for magnification of this reading lens
c. One student argues that blue colour of light give more resolving power than red colour in a simple
microscope. Do you agree with this argument? Explain
Time : 10
Score : 5
Scoring Key
Convex lens
Score 1
MP 1,2
Derivation of magnification
Score 3
MP 1,2,5,6
No resolving power indepentent of light used to illuminate the object Score 1
MP 1,2,5,6,10
Question Text:
Q.No 56
The figure represents a light ray after reflection is polarized. To get maximum polarization the angle made
57.50
θ
a. This angle is called ………………( Critical Angle, angle & deviation, Polarizing angle, angle of
refraction)
b. Calculate tan (57.50) and compare the result with refractive index of glass.
c. State the law connecting refractive index and angle of incidence at maximum polarization
d. At the stage of maximum polarization reflected and refracted and rays are mutually perpendicular.
Prove it
Tome : 10 mts
Scoring Key
Score : 5
Polarizing angle
Score(½)
MP1
Tan 57.5= 1.5
n= tanθ
Score(1)
MP1,2
Brewster’s law
Score(1½)
MP1,2
Proof
Score(2)
MP1,2,5,6
Photo electric
currentcurrent
Question Text:
Q. No: 57
The graph shows photoelectric current with anode potential
O
A
B
Anode potential
C
Photoelectric
current
a. The potential at ‘O’ is called
(i) accelerating potential (ii) retarding potential (iii) stopping potential (iv) saturation potential
b. Why current becomes constant in the region BC
c.
I1 , ν1
I ν2
2 ,
Intensity I1 > Intensity I2
ν2 > ν1
Is the above graph possible? Justify your answer
Anode potential
Score (1+ 2+ 2= 5)
Time10mts
Inten-
Scoring Key
a. Stopping potential
Score(1)
MP1,3
b. At a particular anode potential all the electrons from the cathode can reach at anode and the current
becomes maximum (saturation) hence the photocurrent will not increase further with increase in anode
potential
Score(2)
MP1,3,5,7
c.Yes. ν2>ν1.Hence stopping potentials are different. The stopping potential of ν2.must be greater than
stopping potential of ν1.This graph agrees with above condition
In this I1>I2. Hence photo current corresponding to intensity I1 must be greater than intensity I2
This condition also satisfies the above Graph
Score(1)
MP1,2,3,4,5,6,7,8,9,10
Question Text:
Q.No. 58
You are familiar with the graph obtained in an experiment which is shown below
500
θ
a. Name the scientist who conducted this experiment at first
b. Derive an equation for matter wavelength associated with a moving electron
c. Find out the wave length at the corresponds wave represented by the graph above
Score 4
time 8min
Scoring indicater
a. Davison and Germer
b. Derivation of wave length of moving electron
c. λ =
122
.
V
A0
Score(2)
MP1,3,5,6
Score(1)
MP1,3,5,6,7,8,9,10
Q.No: 59
“Moving particles of matter shows wave like properties under suitable conditions”
a. Who putforword this hypothesis ?
b. Explain the experiment that provide this hypothesis
c. A proton and an electron have been accelerated through same potential. Which one have higher
matter wave length. Write the reason
Scoring key
De brogle Score(1) MP1,3
Description of Davison germer experiment Score(1) MP1,3,5
Electron , Score(1) MP1,3,5
l=h
p , and p = 2mVe or explanation Score(1) MP1,3,5,6,7,9,10
Q.No: 60
a. Similar to fast moving electron
b. It is an electromagnetic wave
c. Similar to helium nucleus
d. Travel with 1/10th the velocity of light
e. Travel with 99/100th the velocity of light
f. Travel with the velocity of light
g. Positively charged
h. Negatively charged
i. Chargeless
Scoring Key
α → c,d,g
β → a,c,h
γ → b, j, i
Score (3)
(MP 1,4)
Question Text :
Q. No. 61
Suppose you are a health physicist and you are being consulted about a spill occured in a radiochemistry lab. The isotope spilled out in the lab was 500 micro cure of Ba 131 which has half life of 12-days.
a) what is the decay constant of Ba 131
b) What mass of Ba 131was spilled
c) Your recommendation is to clear the lab untill the radiation level is down to 1 micro curie.
How length will the lab have to be closed.
Score and Time - (1+1 1/2 + 1 1/2 = 4) 8 Mts
Scoring Keys
Score
MP
a) λ = T1 /
2
(1)
(1, 4)
b) Activity = λ N, So find N & mass of N atoms of B a 131
(11/2)
(1, 4, 7)
(11/2)
(1, 5)
0.693
c) N = N 0e − λ t Find t
N = 1µC ,
NO = 500 mC
Question No: 62
β - Particles does not exist inside a nucleus. But it is emitted from the nucleus !
a) What is β Particle (1)
b) What happens to nucleus of a atom, when a a particle is emitted ? Explain (11/2)
c) Why β particle is emitted from the nucleas ? (1)
Kinetic energy
d)
Number of particles
The distribution of the kinetic energies of β particles emitted from nucleus is given in the above graph. Is
this graph possible ?
Explain ( 11/2)
Scoring indication
a) Electron
b) z X A → z +1Y A + β
Score
(1)
Time : 9 mnts
Score : 5
MP
(1)
(1)
(1, 3, 6)
c) Atomic number increases but mass number remains constant (1/2)
To acquire to acquire stability
(1)
1
d) No. Distribution of energy is continous
( /2 + 1)
Question text :
Q. No : 63
Atomic mass of 8 O16 is found to be 16.0000u
a) what is the mass of 8 O16 nucleus
(1, 3, 5, 6, 7)
(Hint : mass of an electron = 0.00055u)
b) determine the total mass of the constituents particles of the 168 O nucleus and find the mass defect
(Hint : Mass of neutron = 1.00864 u
Mass of proton = 1.007274 u)
(11/2)
c) Give a general expression for mass defect and explain what is binding energy ?
d) Binding energy per nucleon is lower for both very light nuclei ( Z ≤ 10)
and very heavy nuclei ( Z ≥ 70) Justify a nucler fission and fusion (1)
(Total score : 5)
Time : 9 mnts
Score : 5
Scoring Key
Score
MP
1
a) (16.0000-8x0.00055)
( /2)
(1)
(1/2)
b) 8x1.00864 + 8 x 1.00727
(1)
(2, 5)
1
( /2)
c) ∆m = [(Z mp + (A-Z) mn) -- M]
(1/2)
Binding Energy
(1)
(5,8)
c) Explanation
based on
(1)
(10)
BE / Nucleon
Question no : 64
Classify the following statement in to nuclear fission and nuclear fusion
1) Products are radio active
2) Can be controlled
3) Reaction is spontaneous
4) Can’t proceed as a chain reaction
Time : 4, Score : 2
Scoring indication
1)
Nuclear fission
2)
Nuclear fission
3)
Nuclear fission
4)
Nuclear fussion
Question No. 65
(1/2)
(1/2)
(1/2)
(1/2)
MP - 1, 2
The fission of one nucleus of 92 U 239 release 200 Mev of energy
a) What is meant by fission
b) Express 200 Mev energy in joule
c) How many fission of
92
U 239 should occur per send for producing a power of 1 Mev
Time : 6, Score : 3
Scoring Indicators
a) Explanation of fission
b) 200 × 1.6 × 10-16 × 106J
T
c)
∫
VIdt
MP
(1, 2, 5, 6)
= 3.125 × 1016
T
0
Score
(1)
(1)
(1, 2, 5, 6, 10)
Question No. 66
CB
Eg
CB
CB
VB
VB
Eg
VB
Eg ≈ 5-9eV
Eg ≈ 1eV
Energy band diagrams of three materials are given
a)
According to energy gap, Classify them as metal, Insulator and semiconductor.
b)
From which of the above material we can eject electrons with minimum effort Explain ‘
c)
Photo current
Stopping potential
retarding potential
In Photo electric effect, while we are measuring photo current by varrying retrading potential the variations is as shown in graph. Explain the nature of graph on the basis of band theory.
Scoring Indicator
a) Insulator, Semiconductor, Metals
b) Metals, Explanation
c) Since Electrons are in different energy levels KE of ejected
electrons are different. Hence Voltage required to stop
electrons also different
Score
1
1
Score : 3
Time : 5
MP
1
3
1
7
Question No 67
+
_
A P N junction diode is conected to a cell as a shown in fig.
a. Name the type of biasing used here
b. Design a circuit diagram to draw the characteristics of the diode in above biasing.
c. Trace the characteristics curve if the polarity of battery is reversed
Score 4
Time 6 Min
a) Forward biasing
Score 1
MP (Retrives and retells)
b) Circuit diagram of forward biasing Score 2
MP (Applies reasoning and communicate)
c) Characteristics curve of reverse biasing
Score 1
MP (Applies reasoning and draw inference and communicate)
Question No: 68
The circuit diagram of a full wave rectifier is shown
D1
D2
a. Explain how its works ? Also draw the output wave form
(2)
b. If another diode is connected in series with D2, as shown below what will happen to the out
put wave form ?
D1
D2
c. If the frequency of a.c. at the input is 50Hz what will be the output frequency ?
Time & Score
Time : 7 Mins Score : 4
Scoring Indications
a. Working
Output wave form
b. Out Put wave form
c. 100 Hz
Score
(1½
(½ )
(1)
(1)
MP
(1, 2)
(7)
(7)
Question No. 69
A diode can be properly doped at the time of its manufacture, so that it have a sharp break down
voltage
a) To above diode is called
(i) Zener diode (ii) Photo diode (iii) Light emitting diode (iv) Solar cell
b) Compare V-I Characteristics of above diode with that of an ordinary diode
c) Explain how the above diode can be used as an voltage regulator
Score and Time - (1+2+2 = 5) 10 Mints.
Scoring Key
a) Zener diode (Score 1) MP 1
b) Draw two characteristics in forward biased condition both are similar in reverse biased
condition, the break down voltage of zener diode is small and sharpe (Score 2)MP ( 1,9,3)
c) Working of voltage regulator using zener diode the diagram (Score 2)
MP (1,9,7)
Question No : 70
A boy designs a circuit to study the input and output characteristics of an npn transistor
VBB
VCC
a) Identify the transistor configuration, input current and output current (1)
b) To measure the input voltage and input current, modify the circuit diagram by connecting a
voltmeter and a micrometer in the input region
(1)
c) By keeping the output voltage constant, the boy measures the input current by varying the
input voltage. If a graph is drawn, what is the nature of the input characteristic ? Justify your answer (2)
(Total Score : 4)
Time : 7 Mins
Score : 4
Scoring Key
Score
MP
a) CE Configuration
(½)
(1)
IB, Ic
(½)
(1)
b) Diagram
(1)
(3)
c) Forward biased
characteristics
(½)
(7)
Behaves as a pn junction diode
(½)
Question No : 71
A transistor in the common emitter mode can be used as an amplifier
a) Design a circuit to amplify as ac signal given in the input region
[Hint: Give forward biasing to input region, reverse biasing to output region and take output
across a resistor]
(1 ½ )
b) Derive expressions for voltage gain, current gain and power gain in the above transistor
configuration
(2 ½ )
c) Design a practical amplifier circuit which operates with a single voltage source
[Hint : Use potential divider arrangement with two resistors]
(1)
(Total score : 5)
Time : 8 Mins
Score : 5
Scoring Key
a)
Circuit
b)
Voltage gain
Current gain
Power gain
c)
Circuit
Score
(1½)
(1)
(1)
(½)
(1)
CO 65
To develop clear idea about junction transistor, transistor actions, characteristics of
transistor, transistor as an amplifier and oscillator through discussions, experiments and IT
Question Text Question No : 72
Figure below shows three different transistor configurations
(a)
(b)
(c)
Identify the transistor configurations.Biase the configurations properly using separate input and output
battery and define current gain in each configuration
(Total Score 5)
Time (8 mins)
Score : 5
Scoring
Identification
Biasing
Current gain equation
α, β, γ
Score
(1½)
(1½)
(½)
MP
(1,3)
(5)
(7)
(1½)
CO 65 To develop clear idea about junction transistor, transister actions,
characteristics of transistor, transistor as an amplifier and oscillator through
discussions, experiments and IT
Question No : 73
A girl constructed an LC Oscillator and observed the output wave form using a CRO as shown in the
block diagram [CRO - Cathode Ray Oscilloscope can be used to display electrical wave forms]
a) The angular frequency of oscillator is found to be
1
1
iii) ω=
iv) ω= LC
i) ω= LC ii) ω=
LC
LC
b) Since the oscillation is damped, the girl decides to construct an oscillatory circuit, which gives
undamped oscillations using a transistor. Help the girl to design a circuit and explain how it works.
[Hint: use npn transistor in CE configuration]
(2½)
c) “An oscillator is nothing but an amplifier” comment on the statement
(1)
(Total Score : 4)
Time: 7 min
Score : 4
Scoring Key
Score
MP
1
a) ω=
LC
Explanation
b) Circuit
c) Explanation based on the
concept of feed back
Question No. 74
A
B
(1/2)
(1)
(1/2)
(1)
(5, 8)
(1)
(9)
Switch
Switch
Bulb
Battery
An electric circuit containing a battery a bulb and two switches is give above
a) Identify the gate analogues to the above electric circuit
(1)
b) Explain the working of above gate using an equivalent circuit consisting of diodes, battery and LED
etc. (3)
Time : 8 Min
Score : 4
Score
MP
a) OR gate
(1)
(1,2)
b) Figure
(1 )
Explanation of working
( 2)
(1,2,3,5,6)
c)
Out put signal
(1)
(1,2,3,5,6,8)
Question NO: 75
An expert said that it is necessary to modulate low frequency signals to sent long distances.
a) What is meant by modulation
b) What are the different type of modulation
c) Take any two cases of modulation and compare their merits and demerits
Score : 4
Time : 8 min
Scoring Indicator
Score
MP
a) Modulator
1
1
b) Different type - Atleast
1
4
c) Comparison - 2 merits
2 demerits
Question No. 76
2
4
The senario of communication is so developed that the world shrinks into a small village
a) Name the device that connect one computer to another across ordinary telephone lines
b) Write a note on electronic transmission and reproduction of a document at a distant place (4)
Score (1+1+2=4)
Scoring Keys
Score
MP
Time : 7 Min
a) Modem
(1)
(1,2)
b) Fax
(1)
Description
(2)
(5,7,8)
Question No : 77
Match the following
1
2
3
4
Ground wave
Sky Wave
Space Wave
Satellite
Using microwave
Diffraction
Infra Red Wave
Total Interal reflection
Using antena
Score 2
Time : 2 min
Scoring Keys
1) Diffraction
2) Total Internal Reflection
3) Using Antena
4) Using Microwave
Satellite
Score ½ each
MP [Organise information appropriate)
Question No. 78
Communication through an artificial satellite is shown
Earth
a) What is the minimum number of satelites required to cover the entire globe ?
(1)
b) Write any of the frequency or frequency bands used for this type of communication (1)
c) Ionospheric reflection and ground wave transmission is not possible at this frequency why ? (2)
Time 7 min Score 4
MP
(1)
(1, 2)
Scoring Indication
Score
a. 3
(1)
b. Any frequency or band 1 score
(1)
c. Ionosphric reflection - not possible - reason
Ground wave transmission - not possible - reason
(1+1)
(5)
Question No : 79
Which of the following is essential for laser action to occur between two energy levels of an atom
a) There are more atoms in the upper level than in the lower level
b) The upper level is meta stable
c) The lower level is metastable
d) The lower level is the ground state of an action
e) The lasing medium is a gas
Scoring and Time = 2 2 mints
Scoring Keys
The essential conditions are a and b. (Score 2) MP ( 1,2,4)

PHYSICS QUESTION BANK FOR PLUS TWO STUDENTS

Transcription

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