Structure of Chloroplast: Chloroplasts are green plastids containing green pigment called chlorophyll. Chloroplasts are oval shaped organelles having a diameter of 2-10 micrometer and a thickness of 1-2 micrometer
(i) Envelope: Chloroplast envelope has outer and inner membranes which is seperated by intermembrane space.
(ii) Stroma: Matrix present inside to the membrane is called stroma. It contains DNA, 70 S ribosomes and other molecules required for protein synthesis
(iii) Thylakoids: It consists of thylakoid membrane that encloses thylakoid lumen.
(iv) Grana: Some of the thylakoids are arranged in the form of discs stacked one above the other. These stacks are termed as grana, they are interconnected to each other by membranous lamellae called Fret channels.

(i) A lens is an optically transparent medium bounded by two spherical refracting surfaces or one plane and one spherical surface.
(ii) Lens is basically classified into two types. They are: (i) Convex Lens (ii) Concave Lens.

(i) Mg
(ii) Si
(iii) CI

The name of the segment of DNA is gene.

Posterior lobe (Neurohypophysis) of pituitary gland.

The characters on the background of a Scratch window are known as Sprite. Usually a cat appears as a sprite when the Scratch window is opened. The software provides facilities to make alternations in sprite.

(i) Aquatic animals live more in cold regions.
(ii) More amount of dissolved oxygen is present in the water of cold regions.
(iii) This shows that the solubility of oxygen in water is more at low pressure.

(i) Resistance is inversely proportional to area of cross section. \(\mathbf{R} \alpha \frac{l}{A}\)
(ii) A thicker wire has Iarger area of cross section and hence the resistance decreases.

(i) Relative atomic mass of an element is the ratio between the average mass of its isotopes to \(\frac{1}{12^{\text {th }}}\) part of the mass of a carbon-12 atom.
(ii) It is denoted as A_r.
(iii) It is otherwise called "Standard atomic weight".

(i) Since, space station and astronauts have equal acceleration, they are under free fall condition.
(ii) Both the astronauts and the space station are in the state of weightlessness.
(iii) They are not actually floating but falling freely around the earth due to the huge orbital velocity.

Selection is one of the oldest methods of plant breeding in which individual plants or groups of plants are sorted out from a mixed population based on the morphological characters.
Methods of selection:
There are three methods of selection. They are
a) Mass selection
b) Pure line selection
c) Clonal selection
a) Mass selection:
(i) Seeds of best plants showing desired characters are collected from a mixed population.
(ii) The collected seeds are allowed to raise the second generation. This process is carried out for seven or eight generations.
(iii) At the end, they will be multiplied and distributed to the farmers for cultivation. Eg: Groundnut varieties like TMV-2 and AK-10.
b) Pureline selection :
(i) Pureline is "the progeny of a single individual obtained by self breeding":
(ii) This is also called as individual plant selection. In pureline selection large numbers of plants are selected from a self pollinated crop and harvested individually.
(iii) Individual plant progenies from them are evaluated separately. The best one is released as a pureline variety.
(iv) Progeny is similar both genotypically and phenotypically.
c) Clonal selection:
(i) A group of plants produced from a single plant through vegetative or asexual reproduction are called clones.
(ii) All the plants of a clone are similar both in genotype and phenotype.
(iii) Selection of desirable clones from the mixed population of vegetatively propagated crop is called clonal selection.

(i) Simple Microscope: Simple microscope is a convex lens of short focal length. It is held near the eye to get enlarged image of small objects.
(ii) Let an object (AB) of height 'h' is placed at a point within the principal focus (u < f) of the convex lens and the observer's eye placed just behind the lens.
(iii) As per this position the convex lens CJ produces an erect, virtual and enlarged image (A'B'). The image formed is in the same side of the object and the distance equal to the least distance of distinct vision (D) (D = 25 cm for normal human eye).

(iv) Magnification power (M): Ratio of the height of the image produced by the microscope h₂) to the original height of the object (h₁) is called magnification.
M=\(\frac { A'B' }{ AB } \)
Magnification M =\(\frac { height\ of\ the\ image }{ height\ of\ the\ object } \)
=\(\frac { { h }_{ 2 } }{ { h }_{ 1 } } \)
Magnification can also be defined as the ratio of distance of image (v) to the distance of object (u)
Magnification M =\(\frac { Distance\ of\ the\ image }{ Distance\ of\ the\ object } \)
=\(\frac { v }{ u } \)
Since, the image is virtual and erect; the Magnification (M) is taken as positive In simple Microscope,
when the image is formed at the near point
M=1+\(\frac { D }{ f } \)
when the image is formed at infinity
M=\(\frac { D }{ f } \).

a) To measure the speed of an automobile:
(i) An electromagnetic wave is emitted by a source attached to a police car. The wave is reflected by a moving vehicle, which acts as a moving source.
(ii) There is a shift in the frequency of the reflected wave.
(iii) From the frequency shift, the speed of the car can be determined. This helps to track the over speeding vehicles.
b) Tracking a satellite:
(i) The frequency of radio waves emitted by a satellite decreases as the satellite passes away from the earth.
(ii) By measuring the change in the frequency of radio waves, the location of satellites is studied.
c) RADAR (Radio Detection And Ranging):
(i) In RADAR, radio waves are sent, and the reflected waves are detected by the receiver of the RADAR station.
(ii) From the frequency change, the speed and location of aeroplanes and aircrafts are tracked.
(d) SONAR: In SONAR, by measuring the Q change in the frequency between the sent signal and received signal, the speed of marine animals and submarines can be determined.

Supply Voltage is same
Amount of heat produced is also same
Time taken to heat water t = 15 min.
\(H=\frac { { V }^{ 2 } }{ R } .t\quad [H={ I }^{ 2 }Rt\Rightarrow (V=IR)\therefore I=\frac { V }{ R } ]\)
\(\therefore \frac { { R }_{ 1 } }{ { R }_{ 2 } } =\frac { { t }_{ 1 } }{ { t }_{ 2 } } \quad ...(1)\)
\(R\alpha l\quad \quad \quad [R=\rho \frac { l }{ A } ]\)
\(\frac { { R }_{ 1 } }{ { R }_{ 2 } } =\frac { { l }_{ 1 } }{ { l }_{ 2 } } \quad \quad \quad ...(2)\)
From (i) and (2)
\(\frac { { l }_{ 1 } }{ { l }_{ 2 } } =\frac { { t }_{ 1 } }{ { t }_{ 2 } } =\frac { 15 }{ { t }_{ 2 } } \quad \quad ....(iii)\)
\({ l }_{ 2 }=\frac { 1 }{ 3 } { l }_{ 1 }\Rightarrow \frac { { l }_{ 1 } }{ { l }_{ 2 } } =3\)
Sub in (iii) \(3=\frac { 15 }{ { t }_{ 2 } } \)
\(\therefore { t }_{ 2 }=\frac { 15 }{ 3 } =5min\)
Time taken to heat \(\frac{1}{3}\) length of the wire = 5 min

1. Click events in script option

3. Click Looks in script option. Drag “say” to script area.

4. Type “Hello “ word in say tab.

5.Click sounds in script option. Drag play sound to script area. Choose the hello sound from the audio file.

6. From File menu choose the Save option.
7. Click the green flag at the top right corner of the stage window to run the program

Factors	Type I - Insulin-dependent diabetes mellitus(IDDM)	Type II - Non-insulin-dependent diabetes mellitus(NIDDM)
Prevalence	10 - 20%	80 - 90%
Age of Onset	Juvenile onset(<20 years)	Maturity onset(>30 years)
Body weight	Normal or Underweight	Obese
Defect	Insulin deficiency due to  destruction of \(\beta \) - cells	Target cells do not respond to insulin
Treatment	Insulin administration is necessary	Can be controlled by diet, exercise, and medicine.

Soddy and Fagan's displacement laws:
(i) When a parent element emits an alpha particle the new element is formed with a decrease in atomic number by two and mass number by four.
(ii) When a parent nucleus emits a beta particle the daughter nucleus is formed with an increase in atomic number by one and has same mass number of the parent nucleus.
(iii) In gamma decay, the energy level of nucleus only changes. No new elements are formed.

A) Since the resistors are connected in parallel, the potential difference across each resistor is same (i.e. V=10V)
Therefore, the current through R₁ is,
\({ I }_{ 1 }=\frac { V }{ { R }_{ 1 } } =\frac { 10 }{ 5 } =2A\)
Current through \({ R }_{ 2 }={ I }_{ 2 }=\frac { V }{ { R }_{ 2 } } =\frac { 10 }{ 10 } =1A\)
Current through \({ R }_{ 3 }={ I }_{ 3 }=\frac { V }{ { R }_{ 3 } } =\frac { 10 }{ 20 } =0.5A\)
B) Total current in the circuit, I = I₁ + I₂ + I₃
= 2 + 1 + 0.5 = 3.5 A
C) Total resistance in the circuit  \(\frac { 1 }{ { R }_{ P } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } +\frac { 1 }{ { R }_{ 3 } } \)
\(=\frac { 1 }{ 5 } +\frac { 1 }{ 10 } +\frac { 1 }{ 20 } \)
\(=\frac { 4+2+1 }{ 20 } \)
\(\frac { 1 }{ { R }_{ P } } =\frac { 7 }{ 20 } \)
Hence, \({ R }_{ P }=\frac { 20 }{ 7 } =2.857\Omega \) 

Area of copper rod,   A_o = 10 m²
Changes of Area of cross section,
Initial temperature \(\Delta \mathrm{A} =11 -10 =1 \mathrm{~m}^{2}  \)
\(\mathrm{~T}_{1} =90 \mathrm{~K}  \)
\(a_{\mathrm{A}} =0.0021 / \mathrm{K}  \)
\(\mathrm{T}_{2} =?  \)
\(\frac{\Delta A}{A_{0}} =a_{\mathrm{A}} \Delta \mathrm{T}  \)
\(\frac{1 }{10 } =0.0021\left[\mathrm{~T}_{2}-90\right]  \)
\(0.1 =0.0021\left[\mathrm{~T}_{2}-90\right]=\frac{0.1}{0.0021}+90=\mathrm{T}_{2}  \)
\(\mathrm{~T}_{2} =137.61 \mathrm{~K}\)
So the final temperature of a copper rod is 137.61 K

Given: The ratio of masses of two planets \(M_{1}: M_{2} \) is 2: 3 The ratio of their radii \(R_{1}: R_{2}\) is 4: 7
\(g_{1}: g_{2}= ? \)
\(\mathrm{g}=\frac{\mathrm{G \times M}}{\mathrm{R}^{2}} ; \quad \mathrm{g}_{1}=\frac{\mathrm{G\times M}_{1}}{\mathrm{R}_{1}^{2}} ; \quad \mathrm{g}_{2}=\frac{\mathrm{G \times M}_{2}}{\mathrm{R}_{2}^{2}} \)
\(\frac{g_{1}}{g_{2}}=\frac{\frac{G \times M_{1}}{R_{1}^{2}}}{\frac{G \times M_{2}}{R_{2}^{2}}} ; \quad \frac{g_{1}}{g_{2}}=\frac{M_{1}}{R_{1}^{2}} \times \frac{R_{2}^{2}}{M_{2}}=\frac{2 \times 7 \times 7}{4 \times 4 \times 3}=\frac{49}{24}\)
∴ \(g_{1}: g_{2}\) = 49: 24

(i) No, because Nickel is more reactive than Copper.
(ii) So Nickel easily reacts and displaces copper from copper sulphate solution.
(iii) \(\mathrm{Ni}_{(\mathrm{s})}+\mathrm{CuSO_4}_{(\mathrm{aq})} \rightarrow \mathrm{NiSO_4}_{(\mathrm{aq})}+\mathrm{Cu}_{(\mathrm{S})}\)

(i) 0°C ice is colder than water.
(ii) This can be explained in terms of the latent heat of fusion (336 kJ) which is required to convert ice at 0°C to water at 0°C.
(iii) In simpler terms water at 0°C has higher heat content than ice at the same temperature.
(iv) Thus we conclude ice at 0°C is colder, from either of your hands.
(v) Then, you would felt more chillness on those hand, you kept ice at 0°C.

DNA is a large molecule consisting of millions of nucleotides. Hence, it is also called a polynucleotide. Each nucleotide consists of three components.
(i) A sugar molecules - Deoxyribose sugar.
(ii) A nitrogenous base.
There are two types of nitrogenous bases in DNA.
They are
(a) Purines (Adenine and Guanine)
(b) Pyrimidines (Cytosine and Thymine)
(iii) A phosphate group
Nucleoside and Nucleotide:
Nucleoside = Nitrogen base + Sugar
Nucleotide = Nucleoside + Phosphate
The nucleotides are formed according to the purines and pyrimidines present in them.
Watson and Crick model of DNA:
(i) DNA molecule consists of two polynucleotide chains.
(ii) These chains form a double helix structure with two strands which run anti-parallel to one another.
(iii) Nitrogenous bases in the centre are linked to sugar-phosphate units which form the backbone of the DNA.
(iv) Pairing between the nitrogenous bases is very specific and is always between purine and pyrimidine linked by hydrogen bonds.
a) Adenine (A) links Thymine (T) with two hydrogen bonds (A = T)
b) Cytosine (C) links Guanine (G) with three hydrogen bonds( C ≡ G) This is called complementary base pairing.
(v) Hydrogen bonds between the nitrogenous bases make the DNA molecule stable.
(vi) Each turn of the double helix is 34 A^o (3.4 nm). There are ten base pairs in a complete turn.
(vii) The nucleotides in a helix are joined together by phosphodiester bonds.
Significance of DNA:
(i) DNA is responsible for the transmission of hereditary information from one generation to next generation.
(ii) It contains information required for the formation of proteins.
(iii) It controls the developmental process and life activities of an organism.

(i) Saturated solution:
a) A solution in which no more solute can be dissolved in a definite amount of the solvent at a given temperature is called saturated solution.
b) E.g. 36 g of sodium chloride in 100 g of water at 25° C forms saturated solution.
c) Further addition of sodium chloride, leaves it undissolved.
(ii) Unsaturated solution:
a) Unsaturated solution is one that contains less solute than that of the saturated solution at a given temperature.
b) E.g. 10 g or 20 g or 30 g of Sodium chloride in 100 g of water at 25°C forms an unsaturated solution.

Locomotion in leech takes place by
(i) Looping or crawling movement
(ii) Swimming movement.
(i) Looping or Crawling movement:
i). This type of movement is brought about by the contraction and relaxation of muscles.
ii).The two suckers serve for attachment during movement on a substratum.
(ii) Swimming movement:
i). Leeches swim very actively and perform undulating movements in water.

1 Bq = 2.703 x 10^-11 curie
∴ 4.3 x 10³ GBq = 4.3 x 103 x 10⁹ x 2.703 x10^-11
= 11.62 x 10¹
= 116.2 curie