[5 marks Questions]

Question 15 Marks

As shown in the figure above $A$ and $B$ are two lamps. Lamp $A$ is rated at $12\ V, 24\ W.$ Lamp $B$ is rated at $6.0 \ V.$ When lamp $B$ operates at its rated voltage, the current in it is $3.0 \ A$. The values of $R_1$ and $R_2$ are chosen so that both lamps operate at their rated voltages.
Based on the information given, answer the following.
$(i)$ Calculate the current in Lamp $A$.
$(ii)$ State and give reason for the reading of the Voltmeter.
$(iii)$ Calculate the resistance of $R_2$.
$(iv)$ Find the value of the resistance $R_1$.

Answer

$(i)$ Use of $P = IV (0.5)$
$I = P \div V =24\ W \div 12 V$
Current in lamp $A - 2 A (0.5)$
$(ii)$ Voltmeter reading $= 12\ V (0.5)$
Lamp $A$ and Lamp $B$ are in parallel.
Hence $p.d$. across the arm containing $A = p.d.$ across arm
containing $B$
$= 12\ V ($from a$)$
$(iii)\ p.d.$ across $R_2 = 12\ V - 6\ V = 6\ V (0.5)$
$p.d.$ across $B = 6\ V ($given$)$
Hence$ p.d$. across $R_2 = 12\ V - 6\ V = 6\ V$
Current through $R_2$ = Current through $B = 3A ($given$) (0.5)$
Use of $R = V/I$
$R_2=6V/3A = 2\Omega$
$(iv)$ Current through $R_1 =$ Total Current $= 3A+2A = 5A (0.5)$
$p.d.$ across $R_1 = 15V - 12 V = 3V$
$R_1 = 3V/5A = 0.6 \ \Omega\ (0.5)$

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Question 25 Marks

The above circuit is a part of an electrical device. Use the information given in the question to calculate the following.
$(i)$ Potential Difference across $R_2.$
$(ii)$ Value of the resistance $R_2.$
$(iii)$ Value of resistance $R_1$

Answer

$(i)$ p.d. across $4 \Omega$ resistor $=$ p.d. Across $R_2$ as both are in parallel. $(0.5)$
$1.5\ (A) \times 4(\Omega)=6\ V\ (0.5)$
$(ii)$ Total Current through $4 \Omega$ and $R_2=2.0 A($ given$).$
Current through $4 \Omega = 1.5 A ($given$)$
Hence current through $R _2=2-1.5=0.5 A$
Using Ohm's law for $R _2$ we get $(0.5)$
$6 V=0.5 A \times R _2$
Hence $R_2=6 / 0.5=12 \Omega\ (0.5)$
$(iii)$ p.d. across $R _1=$ Total p.d. (p.d. across $R_2$) $-$ (p.d. across $2.0 \Omega )\ (0.5)$
$p.d$. across $2.0 \Omega = 2x2 = 4 V$
p.d. across $R_2 = 6 V ($calculated before$)\ (0.5)$
Hence p.d. across $R _1 = 12-6-4=2V\ (0.5)$
Current through $R _1 = 2A (0.5)$
Using Ohm's Law$,$ we get
$R _1 = 2V/2A = 1\Omega\ (0.5)$
Alternative method
Total Resistance $= R1 +\frac{(4 \times 12)}{(4+12)}+2=12 V / 2 A=6 \Omega\ (0.5)$
$R1 = 6 - (3 + 2) = 1 \Omega\ (0.5)$

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Question 35 Marks

The image below shows a developing fetus in the mother's womb. The developing fetus is connected to the placenta by means of umbilical cord. The Umbilical vein and artery run inside the umbilical cord.

(i) Name two substance that moves through the blood vessels. (1)
(ii) If the placenta has less villi how will it affect the baby's growth? (1)
(iii) Name the region where the embryo develops inside the female body. Explain how this region is adapted for nourishing the baby. (1)
(iv) Some of the fetal cells fall off into the amniotic fluid and can be collected by careful procedure. The cells were screened and found to contain XY chromosome. (2)
(a) What is the sex of the foetus?
(b) How is this prenatal sex determination misused?

Answer

(i) Nutrients/glucose/oxygen/ waste. (any two) (1)
(ii) less surface area for nutrients (glucose/oxygen) to pass from mother to embryo slow growth. (1)
(iii) uterus; has thick lining with rich supply of blood to nourish the embryo. (1)
(iv) (a) male child
(b) misused as if the foetus is female, some people engage in aborting the child leading to female foeticide. (2)

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Question 45 Marks

The image below shows a banana plant which is growing with the help of suckers. These suckers are small plant stem outgrowths which can be separated from the main plant and planted separately and they will grow into a new plant subsequently.

(i) Give the name and type of reproduction that is shown in the image above. (1)
(ii) List two advantages the farmer will have on using the type of reproduction mentioned above. (2)
(iii) The above plant produces male flowers. Explain how this plant will be involved in the process of pollination. (1)
(iv) Why is the offspring of this banana plant not absolutely identical to its parent plant? (1)

Answer

(i) Vegetative propagation/ asexual reproduction. (0.5+0.5)
(ii) More crops in same time interval, genetically identical, flower fruit faster, no need to depend on pollinators. (1+1)
(iii) Cross pollination, the pollen from anther will be transferred the stigma of another banana plant using agents like wind, water, or insects etc. (0.5+0.5)
(iv) There would be minor changes/some variation during the process of copying of the DNA. (1)

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Question 55 Marks

$(i) '\ A\ '\ '\ B\ '$ are sodium salts of long$-$chain carboxylic acid and long chain Sulphonic acid respectively. Which one of $A$ or $B$ will you prefer as a cleansing agent while using underground water $($hand pump water$)$?
Give the reason for your answer.
$(ii)$ Elaborate on the process of cleansing action. Illustrate micelle with the help of labelled diagram.
$(iii)$ Write the chemical equation of the preparation of soap from an ester
$CH_3COOCH_3$. What is the name of this process?

Answer

$(i) '\ B\ '$ is preferred for cleansing action. The calcium and magnesium salts present in underground water are precipitated with carboxylic acids. The$ Ca^{++}$ and $Mg^{++}$ salts of sulphonic acid are soluble in water. $'\ B\ '$ is a more effective cleansing agent in presence of $Ca$ and $Mg$ salts. $(1)$
$(ii)$ Soaps are molecules in which the two ends have differing properties, one is hydrophilic, that is, it dissolves in water, while the other end is hydrophobic, that is, it dissolves in hydrocarbons.
The molecules of soap are sodium or potassium salts of long$-$chain carboxylic acids. The ionic end of soap dissolves in water while the carbon chain dissolves in oil. The soap molecules, thus form structures called micelles where one end of the molecules is towards the oil droplet while the ionic end faces outside. This forms an emulsion in water. The soap micelle thus helps in dissolving the dirt in water and we can wash our clothes clean

The process is saponification.

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Question 65 Marks

(i) "Keerthi thinks that Substitution reaction occurs in saturated Hydrocarbons, on the contrary Krishi thinks, it occurs in unsaturated Hydrocarbons." Justify with valid reasoning whose thinking is correct.
(ii) "Methane and Propane and their Isomers are used as fuels" Comment. Draw the electron dot structure of the immediate lower homologue of Propane. Give any two characteristics of homologues of a given homologous series.
(iii) A mixture of oxygen and ethyne is burnt for welding. Can you predict why a mixture of ethyne and air is not used?

Answer

(i) Keerthi’s thinking is correct as substitution reactions occur in
saturated hydrocarbons, hydrogen atoms are replaced with
heteroatoms in saturated hydrocarbons. Whereas in unsaturated
hydrocarbons an addition reaction occurs, simple molecules are
added across double and/or triple bonds. (1)
(ii) Methane and propane undergo combustion reaction in presence of
oxygen and produce large amount of energy. (1)
The lower homologue of propane is ethane has the following electron
dot structure:

ANY TWO CHARACTERISTICS
- Difference in -CH2-/14u molecular mass of any two adjacent homologues.
- Same general formula/ functional group
- Similar chemical properties
- Gradual change in physical properties (1)
(iii) The mixture of ethyne and oxygen in sufficient amounts undergoes complete combustion to fire a clean blue flame. In pressure of insufficient supply of oxygen or in presence of air, ethyne does not undergo complete combustion and produces sooty flame. (1)

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Science [5 marks Questions]

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