A very long pillar

A

B

C System containing

the players

Motor system

Moving wire

- 2 -

AL/2019/01/E-I,II

- 3 -

AL/2019/01/E-I,II

- 4 -

AL/2019/01/E-I,II

Part A- Structured essay Answer all of the questions on this paper itself

(Acceleration due to gravity 𝑔 = 10𝑁𝑘𝑔−1)

01.

(a) The stress vs strain graph of most material will have the following shape. Identify the point A, B in this graph

If both two are correct--------------------(1)

(b)

(i) When a mass m is hung freely on a string of length l and cross-sectional area A, the extension that took

place was e. Give an equation for the Young’s modulus of the material of the rod in terms of the quantities

given.

y =Fl

Ae=

mgl

Ae -------------------(1)

(ii) Give the condition range in which the equation that you wrote in (b) (i) is valid ?

Within elastic limit Or 𝐶𝐴------------(1)

(c) The experimental set-up that is normally used in an experiment to find the Young’s modulus of a cylindrical

wire is given below.

Additional weights (N) are added to the experimental wire and their corresponding additional extension (e) is

measured and a graph is drawn from it.

A: Elastic limit

B: Breaking point

Stress

Strain

A

B

C

Reference wire

Main scale

Stationary load

Vernier scale

Experimental wire

- 5 -

AL/2019/01/E-I,II

(i) Find the gradient of the graph

2 × 10−5𝑚𝑁−1 𝑜𝑟 2 × 10−5𝑠2𝑘𝑔−1 -------------------(1) (ii) What is the measurement that you need to take in order to obtain the cross-sectional area of the

experimental wire? What is the measuring instrument that you can use for this?

Measurement : Diameter of the wire (No marks if the answer is written as radius of the wire)

Measuring instrument : Micrometer screw gauge

If both two are correct--------------------(1)

(iii) The cross-sectional area of the experimental wire was 1 × 10−6𝑚2and the length of the wire is given as

2m. Find the young’s modulus of the wire.

𝑦 =2

2×10−5×1×10−6

= 1 × 1011𝑁𝑚−2 𝑜𝑟 1 × 1011𝑃𝑎

(iv) Give two reasons for using the two wires: reference wire and the experimental wire.

1. The error due to the lowering of stand is decreased.

2. The error due to the effect of temperature is decreased. If both two are correct ---------(1)

0 05 10 15 20 25 30 35

Additional weight (N)

Additional

extension(mm)

0.1

0.2

0.3

0.4

0.5

●

●

●

●

●

0.7

0.6 P

Q

- 6 -

AL/2019/01/E-I,II (v) How will you confirm that the experimental wire did not exceed the proportionality limit?

By observing nearly equal readings when loading and unloading weights------------(1)

(vi) Give reason for why the points P, Q differ from the other points in the graph.

Experimental wires may be slipped over from the stand above------------(1)

(vii) A system created for a game is shown in the

figure. Here the motor system is used to move

the system containing the players up and down

using the moving wire along the very long

pillar. It moves up from A to B with a small

constant velocity. Then moves from B to C

under gravity, finally it decelerates from C to A

and comes to rest at A. According to the given

motion, from A to B, B to C and from C to A,

where does the moving wire most likely to

break?

𝐶𝐴 ------------(1)

02. (a) You have been asked to find the latent heat of fusion of ice using the method of mixture. For this you have

been provided with warm water, appropriate calorimeter with stirrer, small ice cubes and blotting paper.

(i) Give another major thing that you would need to carry out this experiment successfully?

Electronic balance or triple beam balance or four beam balance------------(1)

(ii) Give the experimental procedure that needs to take place when preparing the ice cubes and adding to the

water

Preparing: Break the ice cubes at melting point into small pieces and blot them with blotting

paper ------------(1)

Adding to the water ;: By dissolving one by one such that no water is spilt out ------------(1)

(iii) What is the experimental procedure that needs to be carried out in order to minimize the heat exchange

with the surrounding?

By increasing the temperature of the water by some degrees (or 5) higher than the room

temperature at the starting of experiment and then adding ice cubes until the temperature is

decreased by that same amount of degrees lower than room temperature. ------------(1) (No marks for lagged calorimeter)

A very long pillar

A

B

C System containing

the players

Motor system

Moving wire

- 7 -

AL/2019/01/E-I,II

(iv) Assume that you have correctly carried out the experiment and obtained the following readings

Measurements Value

The mass of the calorie meter and the stirrer 150𝑔

The initial mass of the calorie meter, the stirrer and the water 250𝑔

The initial temperature of the system 350𝐶

The final minimum temperature of the system 250𝐶

The final mass of the system 261𝑔

You have been given that the heat capacity of the calorie meter is 40𝐽𝐾−1and the specific heat capacity of

the water is 4 × 103𝐽𝑘𝑔−1𝐾−1.

From the above data find the latent heat of fusion of ice.

The heat gained by ice cube = The heat lost by (Water; + Calorimeter) ------------(1)

11 × 10−3 × 𝐿 + 11 × 10−3 × 4 × 103 × 25 = [100 × 10−3 × 4 × 103 + 40](35 − 25)

𝐿 = 3 × 105𝐽𝑘𝑔−1------------(1)

(b) Instead of the method of mixture, the following set-up can be used to find the latent heat of fusion of ice. Small

ice cubes at melting point are placed in a large funnel. In the experimental process (beaker A), an electric heater

is used and in the controlled process (beaker B), electric heater is not used. Both of the set-ups are identical

Except the electricl heater part. They are placed in the same environment.There is a very small net at the

bottom of the wire. And the melted water is collected in the beakers.

The stop watch is started after confirming that the experimental process is supplied with electricity and the rate at

which water leaves each funnel remains constant separately. The reading of the electronic balance is obtained after

t seconds. The mass of water collected in the beakers after t seconds from the experimental process and from the

controlled process was m1 and m2 respectively. The power of the heater used was P watt.

m1kg

Connected to the

electricity

Ice cube

beaker𝐴

Melted water

Electonic balance

Electric heater

experimental process

Not Connected to the electricity

beaker𝐵

Ice cube

Electric heater

Melted water

Electronic balance

m2kg

controlled process

- 8 -

AL/2019/01/E-I,II

(i) What is the adjustment that you need to make on the electronic balance when starting the stopwatch?

By adjusting the stop watch to zero------------(1)

(ii) What is the purpose in carrying out this experiment with the experimental process and controlled process?

For reducing the errors due to the heat exchange with environment ------------(1)

(iii) Give the expression for the latent heat of fusion of ice cube (L) in terms of 𝑚1, 𝑚2, 𝑃, 𝑡

𝐿 =𝑃𝑡

(𝑚1−𝑚2)------------(1)

(iv) If different measurements of m1 and m2corresponding to different time measurements (t) were obtained and a

𝑚1 − 𝑚2 Vs (t) graph is drawn the gradient of the graph was found to be 2𝑔𝑠−1. Find the power of the heater.

(Use the final measurement that you calculated in question (a)(iv)

(𝑚1−𝑚2)

𝑡=

𝑃

𝐿

𝑃 = 2 × 10−3 × 3 × 105

𝑃 = 600𝑊 ------------(1)

03. You have been provided with an ideal spectrometer, equilateral glass prism and a sodium light source in order to

find the refractive index n of glass. The scale of the spectrometer increases in clockwise direction.

(a)

(i) In what order would you adjust the parts A, B, C and D of the spectrometer.

𝐶, 𝐴, 𝐷, 𝐵------------(1)

(ii) Draw how you would place an equilateral prism on B when adjusting it in the given diagram

For drawing a face of the prism perpendicularly to the dotted lines and for drawing such that

light from the collimator reaches the faces of prism and the vertex of the prism should be

placed near the center of the prism table ------------(1)

𝐷 Component 𝐷

- 9 -

AL/2019/01/E-I,II

(iii) In the experiment carried out to find the prism angle, light

ray is incident on the sides of an equilateral prism and

reflects in the paths P and Q. the corresponding reading are

taken correctly. The reading obtained at the position P is

40050’. What is the possible reading that is obtained at the

position Q?

𝑄−40050′

2= 600

𝑄 = 160050′ ------------(1) (iv) What are the two measurements that need to be taken to find the minimum deviation angle of the light ray

through the prism?

1. The reading when the prism is removed from the prism table and telescope is placed

directly in front of collimator.

2. If the two readings are correct in the minimum deviation stage.------------(1)

(v) If the two measurements that you obtained in (iv) are 143029’ and 180041’ then find the minimum deviation

angle. ( assume that when taking the measurements the scale doesn’t go through 3600)

180041′ − 143029′ = 37012′ ------------(1)

(vi) By taking the prism angle as 600 find the refractive index of the glass. (take sin 48036′ = 0.75)

𝑛 =𝑠𝑖𝑛(

𝐴+𝐷

2)

𝑠𝑖𝑛(𝐴

2)

=𝑠𝑖𝑛(

60+37012′

2)

𝑠𝑖𝑛(60

2)

=sin (48036′)

1

2

= 2 × 0.75 = 1.5 ---------For substitution ---(1)

(b) By using the spectrometer, you have been asked to find the deviation angle (d) of the prism for different

values of incident angle (i) and draw a graph that shows the variation of d with i. The light ray coming from

the part D of the spectrometer is incident on the surface of the prism at O. This ray gets refracted, travels

through the prism and emerges out from the prism. The ray that emerges is observed by using the part A of

the spectrometer and the reading Z is taken. The part of the light ray that gets reflected at O is also observed

by using the part A of the spectrometer and the reading X is taken. The prism is removed from the prism table

and the reading Y is taken

P Q

X

Y Z

- 10 -

AL/2019/01/E-I,II

(i) Give the expression for the incident angle (i) and the deviation angle (d) in terms of X, Y and Z.

𝑖 =180−(𝑌−𝑋)

2, 𝑑 = 𝑍 − 𝑌------------(2)

(ii) Draw the I Vs. d graph that you obtained in the following axis, Denote the minimum deviation angle D.

(ii) -------------(1)

(iii) ---------------(2)

Comparing with the above curve, draw the graph that you would obtain if you use a green colour light instead

of a sodium light source and label it as G

04. (a)The figure shows a potentiometer circuit. In the circuit E is cell with an electromotive force E and negligible

internal resistance. And a variable resistance R is connected to the circuit.

What is the use of a variable resistance R in a potentiometer circuit?

Changing the potential difference across the potentiometer wire -----------(1)

.(b) The following circuit is constructed to find the nature of a zener diode using a potentiometer circuit. The

voltmeter and all the mili-ammeter are ideal. The breaking voltage of zener diode used here is 5V

The touch key connected to the end X is placed such that it is in contact with the potentiometer circuit and is

allowed to slide from the end Q to the end P. QO =l

r=0

( 𝐸 𝑅

𝑆

P Q

)

r=0

( 𝐸 𝑅

𝑆

P Q

)

K

O

V

1k

V

mA 𝐴2

Y V

𝐴1 𝑉𝑍 = 5𝑉

mA

mA

2k

X V

i

d

D

G

- 11 -

AL/2019/01/E-I,II

(i) When the touch key is in contact with the potentiometer between PQ, is the zener diode at forward bias

or reverse bias?

Reverse bias -----------(1)

(ii) Give the readings of the mili-ammeter A1 and A2, when the reading of the voltmeter is found to be 6V.

𝐴1 = 0, 𝐴2 = 2𝑚𝐴 ------ If both two are correct-----(1)

(iii) Find the readings of the mili-ammeter A1 and A2, when the reading of the voltmeter is found to be 9V.

𝐴1 = 1.5𝑚𝐴, 𝐴2 = 2.5𝑚𝐴 ------If both two are correct-----(1)

(iv) Draw a graph showing the variation of the reading of the mili-ammeter( 𝐴1) with the distance QO(l)

(c) The following circuit is constructed in a way to find the work function of a material using a potentiometer circuit. A

light source S emits white light beams and a colour filter is used to send blue colour light beams to the cathode.

(i) The touch key connected to the end A of the photoelectric cell is kept so that it is in contact with the

potentiometer wire and is moved from the end Q to P step by step. What will happen to the reading of the

mili-ammeter during this?

No change in reading -----------(1)

(ii) The ends of the cell E are reversed and again the touch key is moved from the end Q to P step by step. What

will happen to the reading of the mili-ammeter during this?

Reading decreases -----------(1)

𝐴1

𝑙

A B

Colour filter

mA

V

K

r=0 (

𝐸 𝑅 𝑆

P Q

)

- 12 -

AL/2019/01/E-I,II

(iii) By using the knowledge that you obtained from (c)(i), (ii) draw a possible graph that sows the variation of

the reading of the ammeter(I) with the reading of the voltmeter(V). (Take the reading of the volt meter in the

question (c)(i) as positive and the reading of the voltmeter in the question (c)(ii) as negative)

(iv) If the brightness of the light source S is increased, draw the graph that you will obtain in (c)(i) in the above

axis comparing with the previous graph and label it as M.

(v) If the point at which the curve intersects the voltage axis is given as Vs, plank’s constant is given as h, the

frequency of the blue light ray from the filter is f and the charge of an electron is e, write an expression for the

work function 𝜙 for the material of the cathode.

𝑒𝑉𝑠 = ℎ𝑓 − 𝜑 -----------(1)

𝑉𝑠

𝐼

𝑀

- 13 -

AL/2019/01/E-I,II

5.

(a)

i) Speed of train compartment = 36 𝑘𝑚ℎ−1 = 10𝑚𝑠−1-------------------(1)

w = 𝑣

𝑟

w = 10

0.2 = 50𝑟𝑎𝑑𝑠−1-------------------(1)

ii) Normal reaction on a wheel given by the track = 1

10× 4 × 105

= 4 × 104 𝑁

∴ Reaction force on a wheel given by the track = 4 ×104

sin 60 -------------------(1)

= 4.62 × 104 𝑁-------------------(1)

(b)

i)

-------------------(1)

ii) length will increase / spring will extend-------------------(1)

(c)

i) Speed of A = 10 ×11

10= 11𝑚𝑠−1-------------------(1)

Speed of B = 10 ×9

10= 11𝑚𝑠−1-------------------(1)

ii) V = rω , wheels have different speed by changing effective radius of wheel without changing its

angular velocity about its axis. This is done by the sideward movement of wheel set.

-------------------(1)

iii) Reaction on wheel A increases. -------------------(1)

Reaction on wheel B decreases. -------------------(1)

iv) By using �⃗� = 𝑚𝑎 towards centre of curve,

Fc = 𝑚𝑣2

𝑟-------------------(1)

=4×104×102

10

= 4 × 105 𝑁

∴ Horizontal resultant force acts on 5 wheel set = 4 × 105𝑁-------------------(1)

∴Horizontal resultant force on one-wheel set. =4×105

5 = 8 × 104 𝑁

(d)

i) Speed of train should be decreased at the turnings. -------------------(1)

ii) Yes, If wheel designed as circular disc shape effective radius cannot be changed by sideward

motion.

Or If wheel designed as circular disk shape individual speeds of wheels cannot be changed.

-------------------(1)

B

A

- 14 -

AL/2019/01/E-I,II

6.

(a)

1. 𝑇𝐿 = 𝑊0 − 𝑊1

𝑇𝐿 = 120 − 110

𝑇𝐿 = 10𝑑𝐵-------------------(1)

2. 𝛽 = 10 log10 (𝐼0

𝐼′)

120 = 10 log10 (𝐼0

1×10−12)-------------------(1)

𝐼0 = 1𝑊𝑚−2

𝛼 =𝐼

𝐼0

0.3 =𝐼

1 -------------------(1)

I = 0.3𝑊𝑚−2-------------------(1)

3. Sound intensity of refracted sound beam = 𝐼0 − 𝐼 = 1 − 0.3

= 0.7𝑊𝑚−2-------------------(1)

4. 𝛽 = 10 log10 (𝐼1

𝐼′)

110 = 10 log10 (𝐼1

1×10−12)-------------------(1)

𝐼0 = 0.1𝑊𝑚−2

Sound intensity of absorbed sound beam= 𝐼 − 𝐼0

= 0.3 − 0.1

= 0.2𝑊𝑚−2-------------------(1)

5.

a. A-------------------(1)

b. B-------------------(1)

6. 𝑇𝐿 = 𝑊0 − 𝑊1

𝑇𝐿 = 10 log10 (𝐼0

𝐼′) − 10 log10 (

𝐼1

𝐼′′)

𝑇𝐿 = 10 log10 (𝐼0

𝐼′×

𝐼′

𝐼′)

𝑇𝐿 = 10 log10 (𝐼0

𝐼′) For correct proof-------------------(1)

- 15 -

AL/2019/01/E-I,II

(b)

1.

------------------(1)

2.

--------------(1)

------------------(1)

3.

------------------(1)

4.

------------------(1)

7.

a)

1. 𝑄 =𝜋𝑟4∆𝑃

8𝜂𝑙or

𝑣

𝑡=

𝜋𝑟4∆𝑃

8𝜂𝑙-------------------(1)

2. Right hand side dimension =𝑀

𝐿×𝑇×

𝐿3

𝑇= 𝑀𝐿2𝑇−2

Left hand side dimension; = 𝑀𝐿2𝑇−2

The equation is dimensionally correct because the dimension of right hand side and left hand side are

equal -------------------(1)

t

4a

x

t a

x

Wave sensed by the mic 2

t

a

x

Phase changed wave

t

5a

x

t

5a

x

- 16 -

AL/2019/01/E-I,II

3. 𝑄 =𝜋𝑟4∆𝑃

8𝜂𝑙-------------------(First equation)

𝑝 = ∆𝑃𝑄-------------------(Second equation)

By removing ∆𝑃 from two equations

𝑝 =8𝜂𝑙

𝜋𝑟4𝑄2

For a particular tube, 𝑙 𝑎𝑛𝑑 𝑟 are constants. So 𝑝 ∝ 𝑄2 -------------------For suitable proof(1)

b)

1. 𝑄 =𝜋𝑟4∆𝑃

8𝜂𝑙

1.5 =3×(0.2)4×∆𝑃

8×1×10−3×100×103---------------For substitution(1)

∆𝑃 = 2.5 × 105𝑃𝑎 ------------------(1)

2. ∆𝑃 =2.5×105

100× 20

𝑃𝐶 − 1.5 × 105 = 5 × 104

𝑃𝐶 = 1.9 × 105𝑃𝑎-------------------(1)

3. 𝑄 =𝜋𝑟4∆𝑃

8𝜂𝑙

0.3 =3×(0.2)4×∆𝑃

8×1×10−3×100×103--------------- For substitution (1)

∆𝑃 = 5 × 104𝑃𝑎 -------------------(1)

4. 𝑝 ∝ 𝑄2

𝑝1 ∝ (1.5)2-------------------(First relation)

𝑝2 ∝ (0.3)2-------------------(Second relation) 𝑝1

𝑝2= 25 -------------------(1)

5. 𝑄 =𝜋𝑟4∆𝑃1

8𝜂𝑙-------------------( First relation)

𝑄 =𝜋(0.9𝑟)4∆𝑃2

8𝜂𝑙-------------------(Second relation)

From two relations, ∆𝑃2

∆𝑃1= (

10

9)

4-------------------(1)

= 1.524 c)

1. 𝑉 = (6 × 60 × 60 × 1.5) + (18 × 60 × 60 × 0.3)-------------------(1)

𝑉 = 51 840𝑚3

2. 𝑃𝐶 = ℎ𝜌𝑔 = 1.9 × 105𝑃𝑎

ℎ × 1000 × 10 = 1.9 × 105--------------- For substitution (1)

ℎ = 19𝑚

3. 𝑄 ∝ ∆𝑃

Pressure difference now = 17 × 1000 × 10 = 1.7 × 105𝑃𝑎

Percentage decrease in rate of water flow =1.9×105−1.7×105

1.9×105 × 100

= 10.526%-------------------(1)

- 17 -

AL/2019/01/E-I,II

4. 𝑃 = ℎ𝜌𝑔

𝑃 = 2 × 103 × 10

𝑃 = 2 × 104𝑃𝑎-------------------(1)

5. Seen as high -------------------(1)

8. a)

i) Electric field strength = 𝑄

4𝜋 0𝑟2 ------------------(1)

Electric potential = 𝑄

4𝜋 0𝑟 ------------------(1)

ii) By Gauss’ law ,

E.A = 𝑄

0 ------------------(1)

E.A = 𝐴 . 𝜎

0

E = 𝛿

0

iii) By Gauss’ law,

E . 2𝜋𝑟𝑙 = 𝑙 𝜆

0 ------------------(1)

𝐸 =𝜆

2𝜋𝜀0𝑟

b)

i) The potential of the inside surface of the drum = (−)2.3

2𝜋 0 × 8.34 × 10−7 × log10 10 -----------(1)

= −4.13 × 1010 × 8.34 × 10−7 × 1

= −34.44 × 103V ------------------(1)

ii) The amount induced on outer surface = 8.34 × 10−7 × 0.4

= 3.34 × 10−7𝐶 ------------------(1)

iii)

A

++ + + + +++++++++++++

- - - - - - - - - - - - - - - - - - - -

Q

A A

l

------------------(1) •

- 18 -

AL/2019/01/E-I,II

The charge density on outer surface= 3.34 × 10−7

4𝜋 ×(5 × 10−2)2

= 1.06 × 10−5𝐶𝑚−2 ------------------(1)

( 1 × 10−5𝐶𝑚−2 𝑡𝑜 1.1 × 10−5𝐶𝑚−2 )

c)

i) Potential difference= 𝐸 . 𝑑

= 𝜎

0 . 𝑑 ------------------(1)

= 1.06×10−5×2×10−3×4.13×1010×2𝜋

2.3

= 2391 V ( 2391V − 2400V ) ------------------(1)

ii)

-------------(1)

iii) Positive charge ------------------(1) iv) Negative charge ------------------(1) v) To neutralize/remove the charges remaining on the drum after copying ------------------(1)

9.

A.

(a)

1) Hunter’s Organ, Main Organ -------------------(1)

2) To communicate, to know the location ------------If both two are correct(1) 3) 5000 × 10 = 50 000 cells-------------------(1)

4) 0.25Ω × 5000 = 1250Ω-------------------(1) 1250

10= 125Ω-------------------(1)

5) 0.15𝑉 × 5000 = 750𝑉-------------------(1)

6) 𝑉 = 𝐼𝑅-------------------(1)

𝐼 =750

125+500= 1.2𝐴-------------------(1)

7) 𝑉 = 𝐸 − 𝐼𝑅 -------------------(1)

𝑉 = 750 − 1.2 × 125

𝑉 = 600𝑉-------------------(1)

OR

V

t

V

t

- 19 -

AL/2019/01/E-I,II

OR

𝑉 = 𝐼𝑅 -------------------(1)

𝑉 = 1.2 × 500

𝑉 = 600𝑉-------------------(1)

8) 𝑃 = 𝐼𝑉

𝑃 = 1.2 × 600

𝑃 = 720𝑊-------------------(1)

𝐸 = 𝑃𝑡

𝐸 = 720 × 2 × 10−3

𝐸 = 1.44𝐽-------------------(1)

9)

10) The prey can be seen at the part where higher electric field strength is there because it bends more its

body than usual. Prey experiences higher electric shock. -------------------(1)

B.

a)

1. Inverting amplifier-------------------(1)

2. 𝑥 =𝑉0

−1

1 =𝑉0

−1

𝑉0 = −1 +1 − 0

20 × 103=

0 − (−1)

𝑅1

𝑅1 = 20𝑘Ω-------------------(1)

3. 𝑥 =𝑉0

−1

3 =𝑉0

−1

𝑉0 = −3

+1 − 0

20 × 103+

+1 − 0

𝑅2=

0 − (−3)

20 × 103

𝑅2 = 10𝑘Ω-------------------(1)

-------------------(2)

- 20 -

AL/2019/01/E-I,II

4.

(a) 1−0

5+

1−0

10=

0−𝑉0

20

2+1

10=

−𝑉0

20

𝑉0 = −6𝑉-------------------(1)

(b) 𝑥 =𝑉0

−1

𝑥 =−6𝑉

−1

𝑥 = 6𝑉 -------------------(1)

(a) 13=1101Two

One-------------------(1)

(b) 1−0

𝑅+

1−0

5+

1−0

20=

1−(−13)

20

𝑅 = 2.5𝑘Ω-------------------(1)

5. No -------------------(1)

𝑥 =𝑉0

−1

16 =𝑉0

−1

𝑉0 = −16𝑉

−16𝑉 can’t be obtained in V0 . Because −15𝑉 is the given potential difference-------------------(1)

(c)

1. 𝑖𝑐 = 𝛽𝑖𝐵-------------------(1)

𝑖𝑐 = 200 × 50 × 10−6

𝑖𝑐 = 10𝑚𝐴 𝑜𝑟 10 × 10−3𝐴-------------------(1)

2. 9𝑉 = 2𝑉 + 220 × 10 × 10−3 + 𝑉𝐶𝐸

𝑉𝐶𝐸 = 9 − 2 − 2.2

𝑉𝐶𝐸 = 4.8𝑉-------------------(1)

𝑉𝐶𝐸 = 𝑉𝐶𝐵 + 𝑉𝐵𝐸

4.8 = 𝑉𝐶𝐵 + 0.7

𝑉𝐶𝐵 = 4.1𝑉-------------------(1)

3. Advantage: We are able to know whether current goes through coated wires. OR The polar part is

not important for testing-------------------(1)

Disadvantage: Only alternate current can be tested. OR Direct current cannot be tested---------- (1)

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AL/2019/01/E-I,II

10.

A)

a)

i) P1𝑉1

𝑇1 =

P2𝑉2

𝑇2-------------------(1)

8×104×200

300=

𝑃×20

600

P = 1.6 × 106𝑃𝑎-------------------(1)

ii) PV = nRT

P ∝ nT

P1

𝑛1𝑇1 =

P2

𝑛2𝑇2

1.6×106

𝑛×600=

𝑃

1.25𝑛×1800

P = 6 × 106𝑃𝑎-------------------(1)

iii) P1𝑉1

𝑇1 =

P2𝑉2

𝑇2

6×106×20

1500=

𝑃×200

900

P = 3 × 105𝑃𝑎 -------------------(1)

iv)

i. W = P∆V -------------------(1)

W = 3.15 × 106 × (200 − 20)10−6

𝑊 = 567𝐽 -------------------(1)

ii. W = P∆V

W = 1.5 × 106 × (200 − 20)10−6

𝑊 = 270𝐽 -------------------(1)

iii. Work done by gas= 567 − 270

= 297J -------------------(1)

iv.

1. 297

𝑄× 100 = 40

𝑄 = 742.5𝐽 -------------------(1)

2. Mass of fuel = 742.5

4×104

= 1.856× 10−2g/ 1.856 x 10-5 kg-------------------(1)

v. The number of rotations in one hour= 3000

2× 60

= 90000 -------------------(1)

The required amount of fuel = 90 000 × 4.3562 × 10−2𝑔

= 1.67kg-------------------(1)

Density(d) =𝑚

𝑣

v = 1.67

1005 -------------------(1)

v = 1.66 × 10−3 m3

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AL/2019/01/E-I,II

vi. By Newton’s cooling law,

Temperature diffrence ∝ Surface Area

300 – 30 ∝ 𝐴--------------- 1st equation

250 - 30 ∝ 𝐴1--------------- 2nd equation

From first and second equations, -------------------(1)

𝐴1= 27

22𝐴

Percentage increase = 𝐴1−𝐴

𝐴× 100

=

27

22𝐴−𝐴

𝐴× 100

= 22.72% -------------------(1)

B)

a) 𝐸 = 𝜎𝑇4-------------------(1)

b)

i) By Wien’s law,

λ𝑚 𝑇 = 𝐶-------------------(1)

λm = 3 × 10−3

2 × 104

λm = 1.5 × 10−7 or 150nm-------------------(1)

ii) v = fλ

f = 3×108

1.5×10−7

f = 2 × 1015𝐻𝑧 -------------------(1)

iii) Area= The energy emitted from unit surface of the black body in one second

Area= 𝜎𝑇4

= 5.7 × 10−8 × (2 × 104)4 -------------------(1)

= 9.1 × 109

iv)

(1) The mass destroyed during a nuclear fusion reaction = 4 × 1.67 × 10−27 − 6.65 × 10−27

=3 × 10−29𝑘𝑔 -------------------(1)

(2) E = 𝑚𝑐2

𝐸 = 3 × 10−29 × (3 × 108)2

𝐸 = 2.7 × 10−12𝐽 -------------------(1)

(3) The energy emitted in one year = 2.7 × 10−12 × 2.1 × 1046

= 5.67 × 1034𝐽-------------------(1)

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AL/2019/01/E-I,II

The energy emitted in one second =5.67×1034

3×107 -------------------(1)

𝜎𝑇4 × 4𝜋𝑟2 = 1.89 × 1027

9.1 × 109 × 4 × 3 × 𝑟2 = 1.89 × 1027-------------------(1)

r = 1.314 × 108 m

𝜎𝑇4 × 4𝜋𝑟2 = 1.89 × 1027

(4)

5.7 × 10−8 × (6 × 103)4 × 4 × 3 × 𝑟2 = 1.89 × 1027-------------------(1)

r = 1.46 × 109𝑚

OR

𝑇4 ∝ 1

𝑟2

(2 × 104)4 ∝ 1

(1.314×108)2 --------------- 1st equation

(6 × 103)4 ∝ 1

r2 --------------- 2nd equation

From first and second equations,

r = 1.46 × 109𝑚-------------------(1)

c)

i) 2.3×106×6×1023

23×10−3 = 6 × 1031 cells-------------------(1)

ii) 𝑇1

2

= ln 2

𝜆

λ = 0.7

15×60×60-------------------(1)

λ = 1.296 × 10−5𝑠−1

iii) A = λN

A = 1.296 × 10−5 × 6 × 1031-------------------(1)

A = 7.776 × 1026𝑠−1

iv) 60 hours; = 4 Half-life time

∴ The activity after 60 hours = 𝐴

16

= 7.776×1026

16-------------------(1)

= 4.86 × 1025𝑠−1