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SEE Model Paper 4

Full marks: 75Time: 3 hours

All questions are compulsory.

Write your answers on paper first. Then open each answer and its marking scheme, and give yourself one mark for each point you got. In multiple choice, your first choice counts.

Section A

10 × 1 = 10

1. Choose the correct option for each of the following questions.

a.

An old telegraph code stood for each letter or sign with a pattern of 5 bits. How many different letters and signs could it stand for?

Pick an answer to see why.

b.

What are the two forms of a gene, such as T (tall) and t (dwarf), called?

Pick an answer to see why.

c.

As Nepal gets warmer, many plants and animals are now found higher up the mountains than before. Which of them are at the greatest risk of dying out?

Pick an answer to see why.

d.

The acceleration due to gravity is greater at the poles (9.83 m/s²) than at the equator (9.78 m/s²). What is the main reason?

Pick an answer to see why.

e.

A wooden boat of mass 300 kg carries two people of total mass 120 kg across the Phewa Lake. What mass of water does it displace?

Pick an answer to see why.

f.

When white light passes through a glass prism, which color is deviated least, and why?

Pick an answer to see why.

g.

A light-year is a unit of

Pick an answer to see why.

h.

Which of the following ores is roasted (heated strongly in a supply of air) during metallurgy?

Pick an answer to see why.

i.

In winter, people rub glycerine on dry, cracked skin. Why?

Pick an answer to see why.

j.

A statement and two arguments are given below.

Statement: The use, import and export of DDT are banned in Nepal.

Argument 1: DDT doesn’t break down easily, so it builds up in fat and becomes more concentrated at each step of a food chain.

Argument 2: DDT kills only harmful insects and doesn’t harm any other living things.

Pick an answer to see why.

Section B

9 × 1 = 9

2. Write very short answers to the following questions.

a.

Write the SI unit of luminous intensity.

Answer

The candela (cd).

Marking scheme
  • Candela (cd).
b.

What is a gene?

Answer

A gene is a segment of DNA in a chromosome that carries the code for a particular character of an organism, and passes it from parents to offspring.

Marking scheme
  • A segment of DNA (in a chromosome) that carries the code for a character.
c.

How does a submarine dive below the surface of the sea?

Answer

It lets seawater into its ballast tanks. Its weight increases and becomes more than the upthrust on it, so it goes down.

Marking scheme
  • It fills its ballast tanks with water, so its weight becomes more than the upthrust.
d.

How much heat is needed to raise the temperature of 3 kg of kerosene from 10 °C to 30 °C? (Specific heat capacity of kerosene = 2,010 J/kg °C)

Answer

Q=ms(T2−T1)=3×2010×(30−10)=120,600 J Q = ms(T_2 - T_1) = 3 \times 2010 \times (30 - 10) = 120{,}600\ \text{J}

Marking scheme
  • Q = ms(T₂ − T₁) = 120,600 J.
e.

Why does a well-cut diamond sparkle?

Answer

Diamond has a very high refractive index (2.42), so its critical angle is very small, only about 24°. Light that enters it meets its faces inside at more than 24° almost every time, so it is totally internally reflected again and again and comes back out of the top towards the eye.

Marking scheme
  • Its critical angle is very small (about 24°), so light is totally internally reflected many times inside it before coming out.
f.

The coil of a simple a.c. generator with two poles turns 25 times a second. What is the frequency of the a.c. it produces, and how many times a second does the current change direction?

Answer

Each turn of the coil gives one complete cycle, so the frequency is 25 Hz. The current changes direction twice in each cycle: 2 × 25 = 50 times a second.

Marking scheme
  • 25 Hz, and 50 changes of direction a second.
g.

What happens to a star like the Sun when the hydrogen in its core runs out?

Answer

It swells into a red giant, puffs off its outer layers as a planetary nebula, and its core is left as a white dwarf, which slowly cools and fades.

Marking scheme
  • It becomes a red giant, then a white dwarf (after shedding a planetary nebula).
h.

Lime water turns milky when carbon dioxide is bubbled through it, but it becomes clear again if the gas is bubbled on for a long time. Why?

Answer

The extra carbon dioxide and water turn the insoluble calcium carbonate (the milkiness) into calcium hydrogencarbonate, which dissolves:
CaCO3+H2O+CO2→Ca(HCO3)2 \text{CaCO}_3 + \text{H}_2\text{O} + \text{CO}_2 \rightarrow \text{Ca(HCO}_3)_2

Marking scheme
  • The insoluble calcium carbonate changes into soluble calcium hydrogencarbonate.
i.

A company plans to get aluminium by heating alumina (aluminium oxide from bauxite) with coke in a blast furnace, as iron is made. Will it work? Give a reason.

Answer

No. Aluminium is very reactive (high in the reactivity series), so its oxide is very stable and carbon can’t take the oxygen away from it. Aluminium oxide is reduced only by electrolysis.

Marking scheme
  • No: aluminium is very reactive, so its oxide can’t be reduced by carbon; electrolysis is needed.

Section C

14 × 2 = 28

Write short answers to the following questions.

3.
[2]

Nischal hangs loads of 1 N, 2 N, 3 N and 4 N, one after another, on the same spring, and measures its length each time with a meter scale.

  • i.Write the independent variable and the dependent variable of his experiment.
  • ii.Write any one controlled variable. Which variable should he plot on the x-axis of his graph?
Answer
  • i.Independent variable: the load (weight) he hangs on the spring, which he chooses. Dependent variable: the length (or extension) of the spring, which he measures.
  • ii.Any one controlled variable: the same spring; the same meter scale, read in the same way; the same way of hanging the loads. The load, the independent variable, goes on the x-axis (and the length on the y-axis).
Marking scheme
  • i. Independent: the load; dependent: the length (extension) of the spring.
  • ii. Any one controlled variable (same spring, same scale, same way of hanging), and the load on the x-axis.
4.
[2]

Write any two characteristics of the phylum Mollusca.

Answer

Any two:

  • a.They are soft-bodied animals.
  • b.The body has a head, a visceral mass, a muscular foot, and a mantle, which in most makes a hard shell of calcium carbonate.
  • c.The head carries tentacles and eyes.
  • d.The muscular foot is used for gliding or swimming.
  • e.They breathe through the body surface, gills or a lung-like pulmonary sac.
Marking scheme
  • One correct characteristic.
  • A second correct characteristic.
5.
[2]

After a walk through a wet forest in Ilam, Ritu found a leech on her leg, and under a stone she found a centipede. Both have long bodies divided into segments. The leech has a moist skin and no legs; the centipede has a hard outer covering and a pair of jointed legs on each segment.

  • i.Write the phylum of each animal.
  • ii.Write one feature from the description that separates the two phyla.
Answer
  • i.The leech belongs to Annelida; the centipede to Arthropoda.
  • ii.Any one: the centipede has jointed legs, the leech has no legs; the centipede has a hard exoskeleton (of chitin), the leech a soft, moist skin (through which it breathes).
Marking scheme
  • i. Leech: Annelida; centipede: Arthropoda.
  • ii. One separating feature: jointed legs or a hard exoskeleton (arthropod) against no legs and a moist skin (annelid).
6.
[2]

In a hive, the queen and the workers all come from fertilized eggs.

  • i.Why does one such larva grow into a queen while the others grow into workers?
  • ii.When the old queen leaves the hive with a swarm, how do the workers left behind get a new queen?
Answer
  • i.It depends on the food. All larvae are fed royal jelly for the first three days. A future queen keeps getting royal jelly for as long as she is a larva; a future worker is switched to bee bread (pollen mixed with honey).
  • ii.They choose a young larva from a fertilized egg, not more than about three days old, enlarge its cell, and feed it only on royal jelly, so it grows into a queen.
Marking scheme
  • i. The queen larva is fed royal jelly all the time; worker larvae are switched to bee bread after three days.
  • ii. They feed a young larva from a fertilized egg (in an enlarged cell) only on royal jelly.
7.
[2]

Write any two differences between mitosis and meiosis.

Answer

Any two:

Mitosis Meiosis
happens in body (somatic) cells happens only in the cells that make gametes, in the testes and ovaries
one cell gives two daughter cells one cell gives four daughter cells
the daughter cells have the same number of chromosomes as the mother cell (diploid) the daughter cells have half the number (haploid)
the daughter cells are identical the daughter cells are genetically different (crossing over)
it has one division (equational division) it has two divisions (reductional division)
for growth, development and repair for making gametes, and for variation
Marking scheme
  • One correct difference.
  • A second correct difference.
8.
[2]

One strand of a DNA molecule has the base sequence A T G G C A T C.

  • i.Write the base sequence of the other strand facing it.
  • ii.In a DNA molecule, 30% of the bases are adenine. What percentage of its bases are guanine?
Answer
  • i.A always pairs with T, and G with C, so the other strand is T A C C G T A G.
  • ii.Each A pairs with a T, so thymine is also 30%. A and T together make 60%, leaving 40% for G and C. Each G pairs with a C, so they are equal: guanine is 40 ÷ 2 = 20%.
Marking scheme
  • i. T A C C G T A G.
  • ii. 20% (A = T = 30%, so G = C = 20%).
9.
[2]

A goat farmer in Syangja has bred his goats only with their own close relatives for many years, to keep his herd “pure”. Now many of the kids are born weak, and some have defects.

  • i.Name the method of breeding he used, and write why it caused this problem.
  • ii.What should he do instead, and how will it help?
Answer
  • i.Inbreeding. Closely related animals carry many of the same genes, so mating them generation after generation makes it more likely that a kid gets two copies of a harmful gene: inherited defects become more common.
  • ii.He should crossbreed them: mate his does with a healthy, unrelated buck, for example of a good breed such as Boer or Jamunapari. The kids get different genes from the two parents and are often stronger than either (hybrid vigor).
Marking scheme
  • i. Inbreeding; mating close relatives makes inherited defects more likely.
  • ii. Crossbreeding with an unrelated buck, which gives hybrid vigor (stronger, healthier kids).
10.
[2]

At a health camp in Solukhumbu, the blood pressures of four adults were measured.

Person Dorje Kamala Bishnu Laxmi
Blood pressure (mmHg) 152/96 118/76 86/56 132/86
  • i.Whose blood pressure is low, whose is normal, whose is a little raised, and whose is high?
  • ii.Dorje says, “I have no headache and feel perfectly well, so my blood pressure can’t be harming me.” Is he right? Give a reason.
Answer
  • i.Low: Bishnu (below 90/60). Normal: Kamala. A little raised: Laxmi (between 120/80 and 139/89). High: Dorje (both numbers above 140/90), if it is the same when measured again.
  • ii.No. High blood pressure often has no symptoms for years (it is called the “silent killer”), but all the while it damages the arteries, heart, kidneys, eyes and brain, and can lead to a heart attack or a stroke. He should see a doctor and have it checked regularly.
Marking scheme
  • i. Bishnu low, Kamala normal, Laxmi a little raised, Dorje high.
  • ii. No: high blood pressure often has no symptoms but still damages the heart and arteries (the silent killer).
11.
[2]
  • i.How does a glacial lake form, and why can it burst?
  • ii.Write one way in which Nepal has made its dangerous glacial lakes safer.
Answer
  • i.As a glacier melts back, its meltwater collects in a lake behind the moraine, a ridge of loose rock and buried ice that the glacier pushed up long ago. This natural dam is weak: an avalanche falling into the lake can send a wave over it, or the buried ice can melt, and the dam suddenly gives way (a glacial lake outburst flood, GLOF).
  • ii.Any one: cutting channels to lower the water, as at Tsho Rolpa in Dolakha (3 m, in 2000) and Imja Tsho (3.4 m, in 2016); setting up early-warning sirens in the valleys below.
Marking scheme
  • i. Meltwater collects behind a weak moraine dam of loose rock and ice, which can give way (an avalanche wave or melting ice).
  • ii. One way: lowering the lake by cutting a channel (Tsho Rolpa, Imja), or early-warning sirens.
12.
[2]

Manish holds a concave lens of focal length 15 cm so that a candle stands 30 cm in front of it.

  • i.Draw a ray diagram to show how the image of the candle is formed.
  • ii.Write any two characteristics of the image.
Answer

i. A ray from the top of the candle parallel to the principal axis spreads out after the lens as if it came from the principal focus F₁ on the candle’s side; a ray through the optical center O goes straight on. The two rays spread apart and never meet, but traced backwards (dashed) they seem to meet between F₁ and O, where the image forms.

2F₁F₁F₂2F₂Oprincipal axisobjectvirtual image
Figure 1 A candle 30 cm from a concave lens of focal length 15 cm: the image is virtual, upright and diminished, between F₁ and O.

ii. Any two: it is virtual (it can’t be caught on a screen), upright (erect), diminished (smaller than the candle), and between F₁ and O on the same side as the candle (10 cm from the lens).

Marking scheme
  • i. Two correct rays (parallel, then spreading as if from F₁; straight through O) and the virtual image where their backward lines meet.
  • ii. Any two: virtual, upright, diminished, between F₁ and O.
13.
[2]

On a clear evening, Rojina sees a very bright “star” low in the western sky just after sunset. It shines steadily, while the stars around it twinkle.

  • i.Why do stars twinkle? What is the steady one most likely to be, and why doesn’t it twinkle?
  • ii.We see the Sun a little before it has actually risen above the horizon. Why?
Answer
  • i.The layers of warm and cool air keep moving, so the light from a star is bent now a little this way, now a little that way, and because a star is so far away that it is just a point of light, its brightness and position flicker. The steady one is most likely a planet (such as Venus). A planet is much closer and looks like a tiny disc; each point of the disc flickers on its own, but the flickers average out, so it shines steadily.
  • ii.Light from the Sun still below the horizon passes from rarer into denser air and bends towards the normal, down round the curve of the Earth (atmospheric refraction). So it reaches our eyes, and we see the Sun about 2 minutes before it actually rises.
Marking scheme
  • i. Moving layers of air keep bending a star’s point of light, so it flickers; the steady one is a planet, whose disc’s flickers average out.
  • ii. Atmospheric refraction: light from the Sun below the horizon bends round the Earth’s curve in the denser air.
14.
[2]

In the laboratory preparation of ammonia, a mixture of ammonium chloride and calcium hydroxide is heated in a hard-glass test tube.

  • i.Why is the test tube clamped sloping slightly downwards towards its mouth?
  • ii.Why is the ammonia dried with quicklime and not with concentrated sulphuric acid?
Answer
  • i.The reaction makes water. If the tube sloped upwards, drops of cool water would run back onto the hot glass at the bottom and crack it.
  • ii.Ammonia is a base, so it reacts with sulphuric acid, an acid (making ammonium sulphate), and would be lost. Quicklime (calcium oxide) is a base too, so it takes up only the water and leaves the ammonia alone.
Marking scheme
  • i. So that the water formed doesn’t run back onto the hot part and crack the tube.
  • ii. Ammonia (a base) reacts with sulphuric acid; quicklime, also a base, absorbs only the water.
15.
[2]
  • i.Write the IUPAC names of CH₃–CH₂–CH₂–CH₂–CH₂–CH₃ and CH₂=CH–CH₂–CH₂–CH₃.
  • ii.Write the molecular formula and the condensed structural formula of the alkyne with four carbon atoms, and name it.
Answer
  • i.Six carbons with only single bonds: hexane. Five carbons with a double bond: pentene (pent-1-ene).
  • ii.For an alkyne, CₙH₂ₙ₋₂ with n = 4 gives C₄H₆: CH≡C–CH₂–CH₃, butyne (but-1-yne).
Marking scheme
  • i. Hexane and pentene (pent-1-ene).
  • ii. C₄H₆, CH≡C–CH₂–CH₃, butyne.
16.
[2]

In the mango season, a food inspector in Birgunj finds a trader ripening green mangoes by keeping packets of calcium carbide among them.

  • i.What gas does calcium carbide give off with moisture? Write the equation, and give one reason why carbide-ripened fruit is harmful.
  • ii.How can a buyer recognize carbide-ripened mangoes? Suggest one safe way to ripen mangoes at home.
Answer

i. It gives off acetylene (ethyne):
CaC2+2H2O→C2H2+Ca(OH)2 \text{CaC}_2 + 2\text{H}_2\text{O} \rightarrow \text{C}_2\text{H}_2 + \text{Ca(OH)}_2
Industrial calcium carbide carries traces of poisonous arsenic and phosphorus compounds; it can also cause skin irritation, burns and eye irritation. (Nepal’s food rules forbid selling fruit ripened with it.)
ii. The skin is evenly bright yellow while the stalk end is still green; there may be a grayish powder or a garlic-like smell; the inside is hard and sour. A safe way: wrap the mangoes in a jute sack, or pack them in hay, husk or asuro leaves, or keep them in a closed bag with a ripe fruit, so that their own ethylene ripens them.

Marking scheme
  • i. Acetylene, CaC₂ + 2H₂O → C₂H₂ + Ca(OH)₂, and one harm (poisonous arsenic and phosphorus traces, skin or eye irritation).
  • ii. One sign (bright yellow skin with a green stalk end, gray powder, garlic smell, sour inside) and one safe method (jute sack, hay, asuro leaves, bag with a ripe fruit).

Section D

7 × 4 = 28

Write long answers to the following questions.

17.
[4]

Sita records a voice message on her phone in Dhangadhi and sends it over the internet to her brother in Kathmandu, who plays it on his phone.

  • i.Write, in order, the components of the digital communication system that carry her voice to her brother’s ear. (2)
  • ii.Write one difference between the work of the encoder and the work of the modulator. (1)
  • iii.On the way, her message travels hundreds of kilometers through an optical fiber together with thousands of other messages. Is this baseband or broadband transmission? Give a reason. (1)
Answer

i. Source (Sita’s voice) → input transducer (the microphone) → encoder → modulator → channel (radio waves and optical fiber) → demodulator → decoder → output transducer (the loudspeaker) → output (her voice, as sound, to her brother’s ear).

Sourceyour voiceInputtransducermicrophoneEncoderto bits, compressedModulatorbits onto a waveOutputthe voice againOutputtransducerloudspeakerDecoderbits to messageDemodulatorbits off the wavechannel(wire, fiber, air)noise gets in hereSENDING (transmitter)RECEIVING (receiver)
Figure 2 A digital communication system: the receiver undoes the sender’s steps in reverse order.
  • ii.The encoder decides which 0s and 1s to send: it turns the signal into bits, compresses them and can add check bits or a code. The modulator decides how to carry them: it puts the bits onto a carrier wave (a radio wave or light) to travel through the channel.
  • iii.Broadband. Each message is modulated onto a carrier of its own frequency, so many messages share the same fiber at once; broadband is used over long distances. (Baseband sends the bits as they are, one signal at a time, over short distances.)
Marking scheme
  • i. Sender’s side in order: source → input transducer (microphone) → encoder → modulator → channel.
  • i. Receiver’s side in order: demodulator → decoder → output transducer (loudspeaker) → output.
  • ii. Encoder codes (and compresses) the message into bits; modulator puts the bits on a carrier wave.
  • iii. Broadband, as many signals share one channel on different carrier frequencies over a long distance.
18.
[4]
  • i.What is diabetes? (1)
  • ii.Write any two symptoms of high blood glucose. (1)
  • iii.How is uric acid formed in the body, and which disease of the joints does too much of it cause? (1)
  • iv.Write any two ways to prevent a high level of uric acid. (1)
Answer
  • i.Diabetes (diabetes mellitus) is a condition in which the blood has more glucose than normal, because the pancreas makes too little insulin or the cells stop responding to it.
  • ii.Any two: great thirst and hunger; frequent urination; blurred vision; losing weight and feeling tired; frequent dizziness; muscle cramps or numbness in the hands and feet; wounds that heal slowly.
  • iii.Purines, from food and from the body’s own DNA and RNA, are broken down, mainly in the liver, into uric acid, which the kidneys pass out in the urine. Too much of it forms crystals in the joints and causes gout (and stones in the kidneys).
  • iv.Any two: exercise regularly; drink enough water, a little at a time, many times a day; avoid alcohol (especially beer) and smoking; eat less red meat, organ meat and seafood, and fewer purine-rich foods; avoid sugary drinks.
Marking scheme
  • i. More glucose in the blood than normal (too little insulin, or cells not responding to it).
  • ii. Any two correct symptoms.
  • iii. From the breakdown of purines (mainly in the liver); gout.
  • iv. Any two correct ways of prevention.
19.
[4]

Students of a school in Dhulikhel dropped a small stone from rest from windows at three heights, and timed each fall to the ground with a stopwatch. Their results are given below. Ignore air resistance.

Trial Height h (m) Time t (s)
1 5.0 1.00
2 10.0 1.44
3 15.0 1.75
  • i.Using h=ut+12at2 h = ut + \tfrac{1}{2}at^2 , calculate the acceleration of the stone in each trial. (1)
  • ii.Find the average acceleration, and the speed with which the stone hit the ground in trial 3. (1)
  • iii.The result of trial 1 is farthest from 9.8 m/s². Give a likely reason. (1)
  • iv.Sanjay says, “A stone twice as heavy, dropped from the same 15 m window, would reach the ground in about half the time.” Is he right? Give a reason. (1)
Answer

i. The stone starts from rest (u = 0), so h=12at2 h = \tfrac{1}{2}at^2 and a=2ht2 a = \dfrac{2h}{t^2} .
Trial 1: a1=2×5.01.002=10.0 m/s2 a_1 = \dfrac{2 \times 5.0}{1.00^2} = 10.0\ \text{m/s}^2
Trial 2: a2=2×10.01.442=202.0736=9.65 m/s2 a_2 = \dfrac{2 \times 10.0}{1.44^2} = \dfrac{20}{2.0736} = 9.65\ \text{m/s}^2
Trial 3: a3=2×15.01.752=303.0625=9.80 m/s2 a_3 = \dfrac{2 \times 15.0}{1.75^2} = \dfrac{30}{3.0625} = 9.80\ \text{m/s}^2
ii. Average: 10.0+9.65+9.803=9.82 m/s2 \dfrac{10.0 + 9.65 + 9.80}{3} = 9.82\ \text{m/s}^2 , about 9.8 m/s².
Speed in trial 3: v=u+at=0+9.80×1.75=17.15 m/s v = u + at = 0 + 9.80 \times 1.75 = 17.15\ \text{m/s} , about 17 m/s. (Or v=2ah=2×9.80×15=294=17.1 m/s v = \sqrt{2ah} = \sqrt{2 \times 9.80 \times 15} = \sqrt{294} = 17.1\ \text{m/s} .)

  • iii.The fall from 5 m is the shortest, only about 1 s. Pressing the stopwatch takes a reaction time of about 0.1–0.2 s, and a small error in timing a short fall makes a big error in a = 2h/t² (t is squared). A taller drop, and several trials averaged, give a better result.
  • iv.No. The acceleration due to gravity, g=GMR2 g = \dfrac{GM}{R^2} , doesn’t depend on the mass of the falling body: the heavier stone is pulled harder, but it is also harder to speed up by exactly the same factor. With no air resistance it falls with the same acceleration and takes the same time, about 1.75 s.
Marking scheme
  • i. a = 2h/t²: 10.0, 9.65 and 9.80 m/s².
  • ii. Average about 9.8 m/s², and v ≈ 17 m/s in trial 3.
  • iii. The short fall: a small timing error (reaction time) makes a big error in a.
  • iv. No: g does not depend on the mass of the falling body, so it takes the same time.
20.
[4]

At a teahouse in Namche Bazaar, Pasang heats water for tea in an aluminium kettle on a stove. The table gives his data. Assume that no heat is lost to the surroundings.

Mass Specific heat capacity Starting temperature Final temperature
Water 1.5 kg 4,200 J/kg °C 9 °C 89 °C (it boils)
Aluminium kettle 0.5 kg 884 J/kg °C 9 °C 89 °C
  • i.Calculate the heat taken in by the water. (1)
  • ii.Calculate the heat taken in by the kettle. (1)
  • iii.Why does the water boil at 89 °C at Namche Bazaar, and not at 100 °C? (1)
  • iv.Pasang pours 0.5 kg of this boiling water into a jug that holds 0.3 kg of water at 9 °C. Calculate the final temperature of the mixture, ignoring the heat taken by the jug. (1)
Answer
  • i.Q1=ms(T2−T1)=1.5×4200×(89−9)=1.5×4200×80=504,000 J Q_1 = ms(T_2 - T_1) = 1.5 \times 4200 \times (89 - 9) = 1.5 \times 4200 \times 80 = 504{,}000\ \text{J}
  • ii.Q2=0.5×884×80=35,360 J Q_2 = 0.5 \times 884 \times 80 = 35{,}360\ \text{J} . (So the stove supplies at least 504,000 + 35,360 = 539,360 J in all.)
  • iii.Namche Bazaar is high (about 3,440 m), where the air is thinner and its pressure is lower. A liquid boils when its vapor can push bubbles open against the pressure of the air above, so at a lower air pressure water boils at a lower temperature, about 89 °C.
  • iv.Heat lost by the hot water = heat gained by the cold water. Let the final temperature be T; s is the same and cancels:

0.5×(89−T)=0.3×(T−9) 0.5 \times (89 - T) = 0.3 \times (T - 9)
44.5−0.5T=0.3T−2.7⇒0.8T=47.2⇒T=59 °C 44.5 - 0.5T = 0.3T - 2.7 \quad\Rightarrow\quad 0.8T = 47.2 \quad\Rightarrow\quad T = 59\ °\text{C}
(It lies between 9 °C and 89 °C, nearer 89 °C, because there is more hot water.)

Marking scheme
  • i. Q = ms(T₂ − T₁) = 504,000 J.
  • ii. 35,360 J.
  • iii. Lower air pressure at that height, so water boils at a lower temperature.
  • iv. Heat lost = heat gained, T = 59 °C.
21.
[4]

The table gives some information about three transformers. Assume that each transformer is ideal.

Transformer Primary voltage Secondary voltage Primary turns Secondary turns
P, in a microwave oven 220 V 2,200 V 400 ?
Q, in a laptop adaptor 220 V ? 1,100 100
R, on an electric pole 11,000 V 220 V ? 60
  • i.Calculate the number of secondary turns of P. (1)
  • ii.Calculate the secondary voltage of Q and the number of primary turns of R. (1)
  • iii.Which of the transformers are step-up and which are step-down? Give a reason. (1)
  • iv.Bimal says, “Transformer P gives ten times the voltage, so it gives out ten times the power it takes in.” Is he right? Give a reason. (1)
Answer
  • i.NpNs=VpVs⇒Ns=Np×VsVp=400×2200220=4,000 \dfrac{N_p}{N_s} = \dfrac{V_p}{V_s} \Rightarrow N_s = \dfrac{N_p \times V_s}{V_p} = \dfrac{400 \times 2200}{220} = 4{,}000 turns.
  • ii.Q: Vs=Vp×NsNp=220×1001100=20 V V_s = \dfrac{V_p \times N_s}{N_p} = \dfrac{220 \times 100}{1100} = 20\ \text{V} .

R: Np=Ns×VpVs=60×11000220=3,000 N_p = \dfrac{N_s \times V_p}{V_s} = \dfrac{60 \times 11000}{220} = 3{,}000 turns.

  • iii.P is step-up: its secondary has more turns than its primary (4,000 > 400), so its output voltage is higher. Q and R are step-down: their secondaries have fewer turns (100 < 1,100; 60 < 3,000), so their output voltages are lower.
  • iv.No. A transformer doesn’t make energy. Power = voltage × current, and almost all the input power comes out, never more. With ten times the voltage, the output current is about ten times smaller: a transformer trades voltage for current.
Marking scheme
  • i. Nₛ = 4,000.
  • ii. Vₛ of Q = 20 V and Nₚ of R = 3,000.
  • iii. P step-up (more secondary turns); Q and R step-down (fewer secondary turns).
  • iv. No: power = voltage × current; the output current is about ten times smaller, and no energy is made.
22.
[4]

The table gives some information about five elements, A to E. The letters are not their real symbols.

Element Electrons in the shells What it is like
A 2, 8, 1 a soft metal that floats on water and fizzes as it reacts with it
B 2, 8, 8, 1 a soft metal that bursts into a lilac flame on water
C 2, 8, 7 a non-metal gas that combines readily with metals
D 2, 7 a gas that reacts even more readily than C
E 2, 8, 8 a gas that hardly ever reacts
  • i.In which group and period of the modern periodic table is C? (1)
  • ii.B reacts with water more violently than A. Explain why. (1)
  • iii.B is more reactive than A, which is above it in its group, but D is more reactive than C, which is below it in its group. Explain this difference. (1)
  • iv.Write the molecular formula of the compound formed by A and C, and give a reason why E forms no compound with A. (1)
Answer
  • i.C has 7 electrons in its outermost shell, the third: group VIIA (17), period 3. (It is chlorine.)
  • ii.A and B (sodium and potassium) are both in group IA with one valence electron, but B has one more shell, so its atom is bigger and its outer electron is farther from the nucleus and more loosely held. B loses it more easily, so it is more reactive.
  • iii.Metals react by losing electrons and non-metals by gaining them. Down a group the atoms get bigger: an outer electron is lost more easily (so the reactivity of metals increases), but a new electron is pulled in less strongly (so the reactivity of non-metals decreases). D (fluorine) is small, with a strong pull, so it gains an electron more readily than C.
  • iv.A has valency 1 (it gives one electron) and C has valency 1 (it takes one), so the formula is AC (NaCl, common salt). E has a full outer shell of eight electrons (an octet), so it has no need to lose, gain or share electrons: it is a noble gas.
Marking scheme
  • i. Group VIIA (17), period 3.
  • ii. B has one more shell, so its outer electron is farther from the nucleus and lost more easily.
  • iii. Metals lose electrons (easier down a group) but non-metals gain them (harder down a group, as atoms get bigger).
  • iv. AC (NaCl), and E has a full octet (noble gas).
23.
[4]

Pratik did four experiments in the laboratory and noted what he saw in the table below.

Test tube What he did What he saw
P added silver nitrate solution to common salt solution a white precipitate formed
Q dropped a piece of zinc into dilute sulphuric acid bubbles of a gas came off
R heated limestone powder strongly a gas came off that turned lime water milky
S added dilute hydrochloric acid, drop by drop, to sodium hydroxide solution colored pink with phenolphthalein the pink color disappeared and the tube became warm
  • i.Write the type of chemical reaction in P and in R. (1)
  • ii.Write the balanced chemical equation of the reaction in Q, and name the gas. (1)
  • iii.In S, why did the pink color disappear? Write the name of this type of reaction. (1)
  • iv.Would bubbles come off if he used a strip of copper instead of zinc in Q? Give a reason. (1)
Answer
  • i.P: double displacement (NaCl + AgNO₃ → AgCl↓ + NaNO₃: two compounds swap partners, and silver chloride is the white precipitate). R: decomposition (CaCO₃ → CaO + CO₂: one reactant breaks into two products, by heat).
  • ii.Zn+H2SO4→ZnSO4+H2 \text{Zn} + \text{H}_2\text{SO}_4 \rightarrow \text{ZnSO}_4 + \text{H}_2 . The gas is hydrogen. (It is a single displacement reaction.)
  • iii.The acid used up (neutralized) the base: HCl + NaOH → NaCl + H₂O. Once no base was left, phenolphthalein, which is pink only in a base, became colorless. This is an acid–base (neutralization) reaction, and it gives out heat.
  • iv.No. Copper is below hydrogen in the reactivity series (K > Na > Ca > Mg > Al > Zn > Fe > Pb > (H) > Cu > Hg > Ag > Au), so it is less reactive than hydrogen and can’t displace it from the acid. Zinc is above hydrogen, so it can.
Marking scheme
  • i. P double displacement, R decomposition.
  • ii. Zn + H₂SO₄ → ZnSO₄ + H₂; hydrogen.
  • iii. The acid neutralized the base, so the indicator lost its pink color; neutralization (acid–base reaction).
  • iv. No: copper is below hydrogen in the reactivity series, so it can’t displace hydrogen.

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