CBSE Class 10 - Science

By chance students may forget to think about valence electrons to solve this questions.

Properties of elements depends on the number of valence electrons. Elements with same number of valence electrons have similar properties.

a.Since all the elements in a group has the same number of valence electrons they exhibit similar properties to one another.

b.Similarly elements in the same period have different valence electrons and hence they exhibit different properties.

Assigning position to an element in periodic table is difficult.

While finding the valency of an element there is a chance to commit a mistake.

From the electronic configuration a student can write the valence electrons.

The number of valence electrons is equal to its group number.The valence shell or outer most shell number is equal to its period number.

Valency = 8 - its group number

a.Group 16 or (VI)

b.3rd period

c.6 valence electrons

d.Two possibilities - +6 or -2

Only vegetative propagation should be discussed and not all the asexual modes.

Listing the advantages of vegetative propagation with examples.

Advantages of vegetative propagation:

1)To produce exact copies of the parent plant with desired characters.

2)To propagate the plants which do not produce viable seeds.

3)For proper transferring of characteristics from parent plant to the progeny.

Sugarcane, Potato, Ginger are few examples of the plants which are vegetatively propagated.

The role of the special structure should be specified and not the tissues involved.

Functions of placenta, the structure established between embryo and the mother are discussed.

The placenta is a connective tissue established between foetus and the mother. It provides a large surface area for the nutrients and oxygen to pass from mother to the embryo.

It also plays a critical role in transferring the waste generated in the embryo to the mother’s blood that undergoes purification.

Use scale to draw light rays and mark the arrows properly.

Recall and list the two rays required to construct a ray diagram for image formation by concave mirrors. State their path after reflection. Draw the ray diagram for the image formed by a concave mirror when an object is placed between the centre of curvature and the focus.

The two rays that can be used to locate an image are:

(i) a ray parallel to the prinicipal axis of the concave mirror after reflection will pass through the principal focus;

(ii) a ray passing through the principal focus will emerge parallel to the principal axis after reflection.

If the object is placed between the center of curvature and the focus of a concave mirror, the ray diagram will look like this.

It is scattering and not refraction which causes the blue colour of the sky.

Recall that the colour of the sky is blue during daytime and the reason for the blue colour of the sky is the blue component of the sunlight getting scattered more than the other colours.

The sky appears blue during daytime. The scattering of blue component of the white sun light by air molecules present in the atmosphere causes the blue colour of sky. The air molecules and other fine particles in the atmosphere have size smaller than the wavelength of visible light. When sunlight passes through the atmosphere, the fine particles in air scatter the blue colour (shorter wavelengths) more strongly than red. The scattered blue light enters our eyes. That is why sky appears blue.

The rays emerging from the prism should bend towards the base of the prism.

Violet should emerge near the base of the prism.

Draw a triangle to represent a prism. Draw an incident ray and show that each colour bends differently inside the prism. Draw the spectrum on the other side.

When a ray of light falls on a prism, refraction occurs and different components of white light travel with different speeds through the glass prism. Hence a set of seven colours appear as a patch on the screen kept on the other side of the prism.

Only the gases evolved by the combustion of fossil fuels should be listed.

Gases released by combustion of fossil fuels and their effects on environment are discussed.

Fossil fuels with constituents like carbon, hydrogen, nitrogen and sulphur, on combustion, release water, carbon dioxide, oxides of nitrogen and sulphur.

These in turn form greenhouse gases in the atmosphere.

Increased levels of carbon dioxide and other greenhouse gases in atmosphere are the major causes of global warming.

Global warming causes melting of polar ice caps.

General topic of dams should not be discussed. Advantages of dams should be avoided.

Problems arising due to construction of dams are to be emphasised. Solutions should be given.

There are mainly three types of problems that arise due to construction of a big dam. They are Social problems, Environmental problems and Economic problems.

Social problems - People lose their habitat to live in. People lose their livelihood.

Environmental problems - Destruction of forests an excess loss of wildlife. Enormous damage is caused to aquatic life.

Economic problems - Lot amount of investment is made readily, which fetches economic gains after a long time.

The above problems can be overcome by relocating the people who lose their habitat, providing employment for them and afforestation.

Students commit mistakes while writing the general formulae of Saturated and unsaturated hydrocarbons.

Student commit mistakes while drawing the structures of hydrocarbons.

Generally addition reactions occur in unsaturated hydrocarbons. Both alkenes and alkynes are comes under unsaturated hydrocarbons. Sometimes student write the addition reaction on alkenes only. But they should write addition reaction on alkynes also.

First write an alkane with two carbon atoms.Try to write the general formula of of that alkane.

Then reduce two hydrogens from that general formula you will get the general formula for alkenes.

From this alkenes remove two hydrogen atoms you will get the general formula for alkynes.

In addition reaction unsaturated hydrocarbons are converted into saturated hydrocarbons.

To get an alkane from alkene add 1 mole of hydrogen (two hydrogen atoms) in the presence of a catalyst.

To get an alkane from alkyne add 2 moles of hydrogen (four hydrogen atoms) in the presence of a catalyst.

Compounds that are formed by hydrogen and carbon are called hydrocarbons.

Saturated hydrocarbon: Compounds of carbon and hydrogen that contain only carbon-carbon single bonds are known as saturated hydrocarbons.

General formula of alkane is C_nH_2n+2

Example: Ethane - C₂H₆

(ii) Unsaturated Hydrocarbons - Compounds of carbon and hydrogen that contain atleast one double covalent bond or a triple covalent bond between carbon atoms are called unsaturated hydrocarbons.

Alkenes and Alkynes are comes under unsaturated hydrocarbons

General formula of alkene - C_nH_2n

Example:

Ethene C₂H₄

General formula of alkyne - C_nH_{2n - 2}

Example:

Ethyne C₂H₂ whose structural formula is H-C ≡ C-H

An unsaturated hydrocarbon is made to saturated hydrocarbon by doing addition reaction with hydrogen in the presence of a catalyst.

For example Ethene And Ethyne are converted into Ethane by reacting with hydrogen in the presence of Ni or Platinum catalyst.

Detergent chemical formula is different from soap chemical formula by chance they will write the soap formula instead of detergent formula.

Detergents works effectively in hard water because the charged end of the compounds do not form insoluble precipitates with the calcium and magnesium ions that are present in the hard water.

Detergents are ammonium or sulphonate salts of long chain carboxylic acids.

The main advantages of Detergents are

(a)Detergents works better in hard water

(b)Detergents can clean better than soap

Demerits of detergents

(i)Detergents are not bio degradable.

(ii)As the detergents are non biodegradable they cause environmental pollution like water and soil pollution.

One major advantage of detergents is that it is an alternative to soap when washing has to be done with hard water. As far as detergents are concerned the charged end of the compounds do not form insoluble precipitates with the calcium and magnesium ions that are present in the hard water. Hence, they are useful when washing with hard water.

Heredity and inheritance are related. But acquired and adaptation are not completely the same.

Concept of heredity and adaptation should be clearly understood.

Inherited characters are those which are transmitted from one generation to another. i.e Parents transmit certain characters to their offspring.

Acquired characters are the changes produced in an individual as a response to the environment. These acquired characters are not purposefully inherited.

Acquired characters are not exactly passed on to next generation.

Example of inherited character - Shape of nose. shape of the nose can be inherited by the offspring from one of its parents.

Example of acquired characteristic - Malfunctioning of kidney affected by disease. This character present in the parent need not be passed on to the child.

Fossilisation should not be compared with sedimentation.

The process of fossil formation is to be discussed. Methods for determining the age of fossils are mentioned.

Formation of fossils is called as fossilisation.Fossils are in one way a glimpse to the history of the earth in terms of what animals and plants existed many years ago.

The quick burial of plant and animal remains in moist sediments that are acted over by particular pressure and temperature results in the formation of fossils.

The age of fossils can be identified either by determining the layer in which the fossil is found or by the technique of radiocarbon dating.

Alleles of the cross should not be confused with characters.

Monohybrid cross should be represented. Results are obtained in the form of ratio.

In this condition, tall pea plant is crossed with dwarf pea plant.

(a) In monohybrid cross, the F₁ generation were all tall plants. F₂ had tall plants and dwarf plants.

(b) When the plants in the F₁ generation were crossed again, there were short plants in the F₂ generation in the ratio 3:1. The genes consisting of hereditary information were being passed on from one generation to the other.

(c)There were no dwarf plants in the F₁ generation while there were dwarf plants in the F₂ generation.

Object distance is to be taken as negative.

Use the lens formula and the given data to find the position, size and nature of the image formed.

Given: Object height h = 5 cm

Focal length of convex lens f = 12 cm

Object distance u = −8 cm

Lens formula is given by

$\frac{1}{v}$ − $\frac{1}{u}=\frac{1}{f}$

$\frac{1}{v}-\frac{1}{-8}=\frac{1}{12}$

⇒ v = −24 cm

The image is formed on the same side of the convex lens as the object, at a distance of 24 cm.

Magnification m = h'/h = v/u

h'/5 = −24/(−8) = 3

⇒h' = 15 cm

Magnification is positive. The image formed is erect and virtual.

The image is enlarged as its height is 3 times that of the object height.

Remember that a convex mirror has a wide field of view but can form only dimished images but a concave mirror can form magnified images and can produce a parallel beam of light.

A concave mirror is used as a reflector in the headlights of vehicles because it produces a strong parallel beam of light when the bulb is placed at its principal focus. In case of a rear view mirror, the primary need is to give as wide a view to the driver as possible. Hence a convex mirror is used for that. It also gives an image that is erect.

A concave mirror is used as a reflector in the headlights of vehicles because it produces a strong parallel beam of light when the bulb is placed at its principal focus. In case of a rear view mirror, the primary need is to give as wide a view to the driver as possible. Hence a convex mirror is used for that. It also gives an image that is erect.

A person with hypermetropia or long sight can also see only distant objects clearly, not the nearby objects. But if it is due to old age it is presbiopia. This is because the image is formed behind the retina.

Recall the name of the defect which makes a person to see only distant objects clearly. Write how it can be corrected. Draw the ray diagrams for hypermetropia and its correction.

The old man is suffering from presbiopia. It is the defect of vision due to which an old person cannot see nearby objects clearly due to loss of power of accommodation of the eye. This defect can be corrected by wearing eye-glasses with converging (convex) lenses of appropriate focussing power. The ray diagrams for the (i) defect of vision and (ii) for its correction, respectively, are given below.

By chance student commit a mistake while assigning the nature of alcohol.

They even try to write the reaction of alcohol with sodium hydrogen carbonate which is not possible.

Acids turns blue litmus to red.whereas alcohol do not show any effect on litmus paper.

There is no reaction of alcohol with sodium hydrogen carbonate.But acids reacts with sodium hydrogen carbonate

We need to use an oxidising agent like alkaline KMnO₄ (potassium permanganate) or K₂Cr₂O₇(acidified potassium dichromate) for oxidising acids from alchohols.

To differentiate between ethanol and ethanoic acid.

(i)Litmus test - Ethanol will have no effect on the litmus paper. Ethanoic acid will change blue litmus paper into red.

(ii)Reaction with sodium hydrogen carbonate - Ethanol has no reaction with sodium hydrogen carbonate (NaHCO₃) but in the case of Ethanoic acid it reacts with NaHCO₃ which gives CH₃COONa, CO₂ and H₂O. The reaction is evident as the carbon dioxide will turn the lime water milky.

Contraception should not be confused with conception.

Methods of contraception are to be named and explained.

(a) Oral contraceptives - Fertilisation is prevented by preventing eggs from being released. This is achieved by using oral drugs or tablets.

(b) Barrier method - Fertilisation is prevented by using barriers to avoid contact between the penis of males and vagina of females . Condoms act as barriers to prevent fertilisation.

(c) Implants and surgical methods - Planting contraceptive devices like copper-t and loop in uterus is also used as a contraception method. Surgical methods like vasectomy and tubectomy also prevent fertilisation. Vasectomy is done to block sperms from being transferred by blocking vas deferens. Tubectomy is when fallopian tubes are blocked preventing release of eggs to uterus.

Students may think Atomic number increases from left to right in a period so atomic size also increases. But this is not correct.

They may think reactivity is directly related to number of valence electrons.This is not correct.

Na,Mg,Al belongs to same period. In period while moving from left to right there is no change in the number of shells. But electrons are added to the same shell. Hence nuclear attractions on the valence electrons increases. Relate this sentence to atomic size and reactivity.

All these elements belong to the third period and among them Na has the largest atomic radius.

As we move from left to right in a period the nuclear charge increases and it pulls the electrons closer to the nucleus. This increased attraction will cause the atomic size to reduce.

Since Al is smaller in size it has higher attractive force and hence tends to hold the electron together with a stronger force.

Remember: Object distance is considered to be negative. Virtual, diminished images are possible only with convex mirrors. Use scale to draw the ray diagrams and mark the arrows properly.

List the sign conventions for spherical mirrors. Using the given data draw the ray diagram following sign conventions and find the focal length and hence the nature of the mirror.

(OR)

Define refraction and state Snell's law. Express it using symbols. Write the formula connecting the refractive index and speed of light in two media. Using the given data and the formula, find the speed of light in air and in water.

The Cartesian sign convensions are as follows.

(i) The object is always placed to the left of the mirror, i.e. the light from the object falls on the mirror from the left-hand side.

(ii) All distances parallel to the principal axis are measured from the pole of the mirror. (The principal axis is taken along the x-axis with pole as the origin.)

(iii) All the distances measured to the right of the pole are taken as positive while those measured to the left of the pole are taken as negative.

(iv) Distances measured perpendicular to and above the principal axis are taken as positive.

(v) Distances measured perpendicular to and below the principal axis are taken as negative.

Object distance u = −18 cm

Magnification =$\frac{1}{3}$ = $\frac{-v}{u}$

⇒ v = 6 cm (image distance)

v is positive and hence the image is formed on the other side of the mirror.

So we have a diminished erect virual image at the back of the mirror. So the mirror is a convex mirror.

Now let us use the mirror formula to calculate the focal length.

Mirror formula:

$\frac{1}{v}$+ $\frac{1}{u}$ = $\frac{1}{f}$

$\frac{1}{6}$ + $\frac{1}{-18}$ = $\frac{1}{f}$

⇒ f = 9 cm.

(The positive focal length confirms that the mirror is convex.)

OR

The bending of a light ray when it travels from one transparent medium to another is called refraction. Refraction is due to change in the speed of light that occurs when it enters from one transparent medium to another. The speed of light in a medium is dependent on the refractive index of the medium. The speed of light is higher in a rarer medium (medium with lower refractive index) than in a denser medium (medium with greater refractive index). Thus, a ray of light travelling from a rarer medium to a denser medium slows down and bends towards the normal. When it travels from a denser medium to a rarer medium, it speeds up and bends away from the normal.

Snell’s law states that the ratio of the sine of the angle of incidence to the sine of angle of refraction is a constant.

Mathematically it can be expressed as follows:

$\frac{\sin i}{\sin r}=\mathrm{cons}\tan t=$^n_ab