[3 Mark Questions]

Question 13 Marks

If in figure,$R _1=10 \Omega, R _2=40 \Omega, R _3=30 \Omega, R _4=20 \Omega, R _5=60 \Omega$, and a 12 V battery is connected to the arrangement. Calculate

a. the total resistance in the circuit, and
b. the total current flowing in the circuit.

Answer

Suppose we replace the parallel resistors $R _1$ and $R _2$ by an equivalent resistor of resistance, R′. Similarly we replace the parallel resistors $R _3, R _4$ and $R _5$ by an equivalent single resistor of resistance R". Then using eq. $\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$,we have Similarly, $\frac{1}{R^{\prime \prime}}=\frac{1}{30}+\frac{1}{20}+\frac{1}{60}=\frac{6}{60}$
that is $R^{\prime \prime}=10 \Omega$
Thus, the total resistance, $R = R ^{\prime}+ R ^{\prime \prime}=18 \Omega$
To calculate the current, we use Ohm’s law, and get
$I=\frac{V}{R}=\frac{12 V}{18 \Omega}=0.67 A$

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Question 23 Marks

$a.$ Two electric lamps rated $100 W,220 V$ and $60 W,220 V$ are connected in parallel to electric mains supply. Calculate the current drawn from the mains if the supply voltage is $220 V?$
$b.$ A lamp consumes $50 W$ and is lighted $2 h$ daily in month of April. How many units of electric energy is consumed ?

Answer

$(a)$ We know that
$P =\frac{V^2}{R}$
Therefore, $R =\frac{V^2}{P}$
Resistance of $1^{st}$ lamp,
$ R _1=\frac{V^2}{P}=\frac{220 \times 220}{100}$
$=484 \Omega$
Resistance of $2^{nd}$ lamp,
$R _2=\frac{220 \times 220}{60}$
$=\frac{2420}{3} \Omega$
Since, two lamps are connected in parallel,
so its equivalent resistance is given by
$\frac{1}{R}=\frac{1}{R_1+} \frac{1}{R_2}=\frac{1}{484}+\frac{3}{2420}=\frac{8}{2420}$
$R =\frac{2420}{8} \Omega$
By Ohm's Law,current drawn from the mains:
$I =\frac{V}{R}=\frac{220 \times 8}{2420}$
$=0.73 A$
$\therefore$ The current drawn from the mains is $0.73 A$
$(b)$ Energy consumed $=\frac{\text { Watt } \times \text { hour }}{1000}$
$=\frac{50 \times(2 \times 30)}{1000}$
$=\frac{3000}{1000}$
$=3$ unit or $3\ kWh$

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Question 33 Marks

A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil to be bent to the same extent, if instead of water we use liquids like, kerosene or turpentine? Support your answer with reasons.

Answer

We know that pencil appears to be bent at the interface of air and water because of refraction of light. The degree of refection depends on refractive index of a given liquid. Refraction indices of kerosene, water and other liquids would be different. Hence, degree of bend would be different in case of different liquids.

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Question 43 Marks

A pea plant with purple flowers were crossed with white flowers producing 40 plants with only purple flowers. On selfing, these plants produced 470 plants with purple flowers and 162 with white flowers. What genetic mechanism account for these results.

Answer

The ratio of purple flowers to white flowers in F2 generation was approximately 3 : 1. This ratio is termed Mendelian ratio or
Monohybrid ratio. It explains:
1. $F _1$hybrids always exhibited only one of the parental form of a trait and showed dominance / recessive mechanism.
2. Both parental forms of trait segregate and were expressed in $F _2$ (second filial) generation.
2. Both parental forms of trait segregate and were expressed in $F _2$ (second filial) generation.
3) The form of trait that appeared in $F _1$ offspring i.e. the dominant form was present in the $F _2$ generation about three times as frequently as its alternate form (470 : 162). It is approximately 3 : 1. It is due to mechanism of segregation at the time of gamete formation.

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Question 53 Marks

Describe the alimentary canal of man.

Answer

Mammalian (human) alimentary canal comprises mouth, buccal cavity, pharynx, oesophagus, stomach, small intestine
(duodenum, jejunum and ileum), large intestine (caecum, colon and rectum) and anus.
Digestive glands include salivary glands, gastric glands, intestinal glands, liver and pancreas.
Buccal cavity contains teeth and tongue, and receive saliva from 3 pairs of salivary glands (parotid, submaxillary, sublingual).
Teeth are meant for cutting and mastication of food. Teeth are the codent, heterodont and diphyodont.
Buccal cavity of man: The mouth is a slit-like aperture bounded by two lips. It leads to buccal cavity. Its roof is formed of hard
palate, continues as soft palate, and hangs as uvula. The sides of buccal cavity are formed by cheeks. At the floor is present a
musculo-sensory organ called tongue.
The oral cavity is supported by bones of jaws. The oral cavity contains the glands, tongue and teeth.

Pharynx: It is a vertical tube. It is a cross passage for food and air. It has uvula and epiglottis which closes the internal nares and glottis respectively during swallowing of food to ensure the passage of food into oesophagus (food pipe).
Oesophagus: It is 25 cm long narrow muscular straight tube. It runs downward through the chest behind heart and passes through the diaphragm into the abdomen. Here, it opens into stomach. Oesophagus propels the swallowed food into stomach.
Stomach of man: It is a sac-like structure situated in the upper part of abdominal cavity below the diaphragm. Large part of this
sac is situated left of the median line. The stomach is divisible into three parts namely cardiac, fundus or body and pylorus. At the junction of stomach and duodenum is a pyloric constriction having pyloric sphincter.
Small intestine: It is the longest part of alimentary canal. It is thin walled and highly coiled tubular structure. It is about 3-3.5
metres long and occupies most part of abdominal cavity . It is coiled upon itself. It is differentiated into three regions, viz.
duodenum, jejunum and ileum.
Duodenum is 25 cm long C-shaped. It receives the opening of bile-pancreatic duct. It also receives the opening of stomach.
Jejunum is 90 cm long. It lies above and left beyond duodenum.
Ileum is 180 cm long. Its inner lining is thrown into numerous villi.
Large intestine: The large intestine is about 1.5 metres long. It is divided into following parts, ie. the vermiform appendix, the
colon and the rectum. Caecum is blind tube and represented by vermiform appendix (5-8 cms) and is present below the junction of small and large intestine. Rectum is the last part and opens to outside by anus guarded by anal sphincter.

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Question 63 Marks

a. Name the following:
i. Metal that can be cut by knife
ii. Lustrous non-metal
iii. Metal that exists in liquid state at room temperature
iv. Most malleable and ductile metal
v. Metal that is best conductor of electricity
vi. Non-metal that can exist in different forms
b. How are alloys better than metals? Give composition of solder and amalgam.

Answer

a. i. Sodium (Na) / Potassium (K)
ii. Iodine (I)
iii. Mercury (Hg)
iv. Gold (Au)
v. Silver / Copper
vi. Carbon / Sulphur / Phosphorous
b. Alloys are better than metals as they are generally stronger / have high tensile strength / high electrical resistivity / resistant to
corrosion.
Composition of Solder : Lead + Tin / Pb + Sn
Composition of Amalgam - an alloy in which mercury is one of the constituent / Any metal + mercury

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Question 73 Marks

An alkali metal A gives a compound B (molecular mass = 40) on reacting with water. The compound B gives a soluble compound C on treatment with aluminium oxide. Identify A, B and C and give the reactions involved.

Answer

Let the atomic mass of alkali metal A be x. According to the question, the metal A reacts with water and forms a compound B of
molecular mass 40.
$2 A+2 H _2 O \rightarrow \underset{(B)}{2 AOH }+ H _2 \uparrow$
According to the question,
x + 16 + 1 = 40
$\therefore x=40-17=23$
and, 23 is the atomic mass of sodium(Na). Therefore, the alkali metal (A) is Na and the reaction is written below:
$\underset{A}{2 Na (s)}+2 H _2 O (l) \rightarrow \underset{(B)}{2 NaOH }(a q)+ H _2(g)$
So, compound B is sodium hydroxide (NaOH). Sodium hydroxide reacts with aluminium oxide $\left( Al _2 O _3\right)$ to give sodium aluminate $\left( NaAIO _2\right)$
$Al _2 O _3(s)+2 \underset{B}{2 NaOH (a q)} \rightarrow \underset{(C)}{2 NaAl _2(a q)}+ H _2 O (l)$
Thus, compound C is sodium aluminate $\left( NaAIO _2\right)$.

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Question 83 Marks

How can a layer of aluminium oxide on an aluminium object be made thicker? What is this process called?

Answer

Aluminium develops a thin oxide layer when exposed to air. This aluminium oxide coat makes it resistant to further corrosion.
The resistance can be improved further by making the oxide layer thicker. This process is called anodising.
During anodising, a clean aluminium article is made the anode and is electrolysed with dilute sulphuric acid. The oxygen gas
evolved at the anode reacts with aluminium to make a thicker protective oxide layer. This oxide layer can be dyed easily to give
aluminium articles an attractive finish.

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