Que-Ans (Each of 5 Mark )

Question 15 Marks

$i.$ To make a saturated solution, $36 g$ of sodium chloride is dissolved in $100 g$ of water at $293K$. Find its concentration at this temperature.
$ii.$ Calculate the mass of glucose and mass of water required to make $200g$ of $25\%$ solution of glucose.

Answer

$\text { i. Concentration of sol }=\frac{\text { Mass of solute }}{\text { Mass of solution }} \times 100$
$=\frac{3 si }{136} \times 100$
$=26.4 \% \text { (hy mass) }$
$\text { ii. Given mass of solution(M) }-200 g$
$\text { Concentration of solution }=25 \%$
$\text { Since, Mass by Mass percentage of solution }-\frac{\text { Mass af eolutes }}{\text { Mass of of ulution }} \times 100$
$\Rightarrow 25- m \times \frac{100}{200} g$
$\Rightarrow m =25 \times \frac{200}{100}=50 g$
$\therefore \text { mass of solute }-50 g$
$\text { mass of solvent (water) }=M-m=200 g-50 g=150 g$

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Question 25 Marks

Differentiate between
i. Cell wall and cell membrane.
ii. Nuclear region of a bacterial cell and nuclear region of an animal cell.
iii. Prokaryotic cell & eukaryotic cell.

Answer

Cell wall	Cell membrane
It is present in bacteria, fungi, and plant cells. It is absent in animal cells and protozoans.	It is present in all cells.
There is no other name of the cell wall.	The cell membrane is also known as the plasma membrane or plasmalemma.
The cell wall is completely permeable.	The cell membrane is semi-permeable.
The cell wall is made up of cellulose.	The cell membrane is made up of lipids and proteins.

ii.

Nuclear region of bacterial cell	Nuclear region of an animal cell
Smaller in size.	Larger in size.
The nuclear membrane is absent, the nucleolus is absent. The nucleus is regarded as the nucleoid.	Nuclear membrane with nucleolus present.

iii.

Prokaryotic cell	Eukaryotic cell
The size of a cell is generally small.	The size of a cell is generally large.
The true nucleus is absent.	The true nucleus is present.
It contains a single chromosome.	Contains more than one chromosome.
Membrane-bound cell organelles absent.	Membrane-bound cell organelles present.

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Question 35 Marks

Write the main functions of atleast ten cell components.

Answer

The ten cell components are:
i. Plasma membrane: It acts as a semipermeable membrane and allows only selective substances to pass through it.
ii. Chromosomes: To carry hereditary characters of an organism from one generation to another.
iii. Lysosomes: Breakdown of unwanted macromolecules is the main function of these organelles.
iv. Ribosomes: These help in protein synthesis.
v. Nucleus: Control centre of the cell. It contains cellular DNA (genetic information) in the form of genes.
vi. Mitochondria: The main function of mitochondria in aerobic cells is the production of energy by the synthesis of ATP.
vii. Nucleolus: Biosynthesis of ribosomal RNA (rRNA) and acts as a platform for protein synthesis.
viii. Cell wall: It provides protection and rigidity to the plant cell.
ix. Chloroplasts: These are the sites of photosynthesis within plant cells.
x. Endoplasmic reticulum: Serves as channels for transport of materials.

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Question 45 Marks

i. Write the formula to find the magnitude of the gravitational force between the earth and an object on the earth's surface.
ii. Derive how does the value of gravitational force F between two objects change when
a. distance between them is reduced to half and
b. mass of an object is increased four times.

Answer

i. Formula to find the magnitude of gravitational force:
$
F=\frac{G M m}{k^2} .
$
where, $M =$ mass of the earth
$m =$ mass of the object
$R =$ distance between centres of the earth and an object.
and universal gravitational constant, $G =6.67 \times 10^{-11} N- m ^2 / kg ^2$
ii. a. Let gravitational force be F when the distance between them is R , $F =\frac{G M m}{R^3} . .$. (i)
Now, when the distance reduces to half,
$
F^{\prime}=\frac{G M m}{\left(\frac{\pi}{2}\right)^2}=\frac{4 G M m}{R^2}=4 F
$
i.e. the force of gravitation becomes 4 times the original value.
b. When the mass becomes 4 times,
$
F^{\prime}=\frac{G M(4 m)}{R^2}=4 F
$
i.e. the force of gravitation becomes 4 times the original value.

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Question 55 Marks

i. Suppose the mass of the earth somehow increases by 10% without any change in its size. What would happen to your weight?
ii. Suppose the radius of the earth becomes twice of its present radius without any change in its mass. What will happen to your weight?

Answer

i. We know that, Original weight, $W _{ o }= mg =\frac{G M m}{R^2}$, where M is the mass of the earth, $m = mass$ of body.Let the new mass of earth $= M ^{\prime}$
According to question, New mass, $M ^{\prime}= M +10 \%$ of $M = M +\frac{10}{100} M = M +\frac{ M }{10}=\frac{11 M }{10}=1.1 M$
$\therefore$ New weight, $W _{ n }=\frac{G M^{\prime} m}{R^2}=\frac{G \times 1.1 Mm }{R^2}$
Now, Ratio of new weight to original weight $=\frac{N_{\text {ew weight }}}{\text { Original weight }}=\frac{1.1 GMm / R^2}{G Mm / R^2}=1.1$
New weight becomes 1.1 times the original weight of body.
i.e., weight of body will increase by $10 \%$.
ii. Again, Original Weight, $W _{ o }=\frac{G M m}{I^2}$, where R is the radius of the earth.According to question, when $R$ changes to $2 R$, the new weight is given by,
New weight, $W _{ n }=\frac{G M m}{4 R^2}$
Now, Ratio of new weight to original weight $=\frac{\text { Newweight }}{\text { Oniginal weight }}=\frac{G M m / 4 R^2}{G M m / R^2}=\frac{1}{4}$
Therefore, New weight becomes $\frac{1}{4}$ times of original weight

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Science Que-Ans (Each of 5 Mark )

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