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Physics 12th Previous Year Question Paper 2019 SET-I (CBSE)

Physics

SET-I

Section-A

Q.1. Draw equipotential surfaces for an electric dipole.

Q.2. A proton is accelerated through a potential difference V, subjected to a uniform magnetic field acting normal to the velocity of the proton. If the potential difference is doubled, how will the radius of the circular path described by the proton in the magnetic field change ?

Answer: Given, proton accelerated through a potential difference V, the direction of magnetic field is normal to velocity of protons.

As we know

Q.3. The magnetic susceptibility X of magnesium at 300 K is 1-2 × 10⁵ . At what temperature will its magnetic susceptibility become 1-44 × 10⁵ ?

Q.3. The magnetic susceptibility X of a given material is – 0.5. Identify the magnetic material.

Answer:

Q.4. Identify the semiconductor diode whose V-I characteristics are as shown.

Answer: Photo diode.

Q.5. Which part of the electromagnetic spectrum is used in RADAR? Give its frequency range.

Q.5. How are electromagnetic waves produced by accelerating charges ?

Answer : Microwaves [1GHz to 100 GHz].

Answer: An oscillating electric field in space, produces an oscillating magnetic field, which in turn, is a source of oscillating electric field, and so on. The oscillating electric and magnetic fields thus regenerate each other.

Section – B

Q.6. A capacitor made of two parallel plates, each of area ‘A’ and separation W is charged by an external d.c.-source. Show that during charging, the displacement current inside the capacitor is the same as the current charging the capacitor.

Answer:

From Ampere’s law,

Let the case-1, where a point P is considered outside the capacitor charging.

From Ampere’s law magnetic field at point P will be :

Now, take case-2 where shape of surface under consideration covers capacitor’s plate as we consider there is no current through capacitor then this value of B will be zero.

Case-2

Hence, there is a contradiction.

Therefore, this Ampere’s law was modified with addition of displacement current inside capacitor.

Where, id is displacement current.

During charging of capacitor, outside the capacitor, i_c (conduction current) flows and inside i_c (displacement current) flows.

Q.7. A photon and a proton have the same deBroglie wavelength X. Prove that the energy of the photon is (2mλc/h) times the kinetic energy of the proton.

Answer:

Q.8. A photon emitted during the de-excitation of electrons from a state in the first excited state in a hydrogen atom, irradiates a metallic cathode of work function 2 eV, in a photocell, with a stopping potential of 0.55V. Obtain the value of the quantum number of the state n.

Q.8. A hydrogen atom in the ground state is excited by an electron beam 12-5 eV energy. Find out the maximum number of lines emitted by atom from its excited state.

Answer:

Q.9. Draw the ray diagram of an astronomical telescope showing image formation in the normal adjustment position. Write the expression for its magnifying power.

Q.9. Draw a labelled ray diagram to show the image formation by a compound microscope and write the expression for its resolving power.

Answer:

The magnifying power m is the ratio of the angle P subtended at the eye by the final image to the angle a which the object subtends at the lens or the eye. Hence,

Q.10. Write the relation between the height of a TV antenna and the maximum range up to which signals transmitted by the antenna can be received. How is this expression modified in the case of line of sight communication by space waves? In which range of frequencies, is this mode of communication used?

Answer : The curvature of the earth limits the distance up to which a signal can be transmitted by a tower.

If the height of the transmitting antenna is ‘h’ and the radius of the earth is ‘R’, then the optimum distance ‘d’between the receiving and the transmitting antenna is given by:

d= √2Rh

Q.11. Under which conditions can a rainbow be observed? Distinguish between a primary and a secondary rainbow.

Answer : The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. This is a phenomenon due to combined effect of dispersion, refraction and reflection of Sunlight by spherical water droplets of rain. The conditions for observing a rainbow are that the Sun should be shining in one part of the sky (say near western horizontal while it is raining in the opposite part of the sky (say eastern horizon).

Difference between Primary and Secondary Rainbow:

Q.12. Explain the following :

(a) Sky appears blue.

(b) The Sun appears reddish at (i) sunset, (ii) sunrise

Answer: (a) Light from the sun reaches the atmosphere that is comprised of the tiny particles of the atmosphere. These act as a prism and cause the different components to scatter. As blue light travels in shorter and smaller waves in comparison to the other colours of the spectrum. It is scattered the most , causing the sky to appear bluish.

(b) The molecules of the atmosphere and other particles that are smaller than the longest wavelength of visible light are more effective in scattering light of shorter wavelengths than light of longer wavelengths. The amount of scattering is inversely proportional to the fourth power of the wavelength. (Rayleigh Effect) Light from the Sun near the horizon passes through a greater distance in the Earth’s atmosphere than does the light received when the Sun is overhead. The correspondingly greater scattering of short wavelengths accounts for the reddish appearance of the Sun at rising and setting.

Section – C

Q.13. A capacitor (C) and resistor (R) are connected in series with an ac source of voltage of frequency 50 Hz. The potential difference across C and R are respectively 120 V, 90 V, and the current in the circuit is 3 A. Calculate

(i) the impedance of the circuit

(ii) the value of the inductance, which when connected in series with C and R will make the power factor of the circuit unity.

Q.13. The figure shows a series LCR circuit connected to a variable frequency 230 V source.

(a) Determine the source frequency which drives the circuit in resonance.

(b) Calculate the impedance of the circuit and amplitude of current at resonance.

Answer:

Answer: (a) Source frequency will be same as resonance frequency of LC circuit,

Hence, the voltage drop across LC combination is zero at resonating frequency.

Q.14. Give reason to explain why n and p regions of a Zener diode are heavily doped. Find the current through the Zener diode in the circuit given below :

(Zener breakdown voltage is 15 V)

Answer : By heavily doping both p and n sides of the junction, depletion region formed is very thin, i.e. < 10⁶ m. Hence, the electric field across the junction is very high (~5 × 10⁶V/m) even for a small reverse bias voltage. This can lead to a break down during reverse biasing.

Q.15. Draw a labelled diagram of cyclotron. Explain its working principle. Show that cyclotron frequency is independent of the speed and radius of the orbit.

Q.15. (a) Derive, with the help of a diagram, the expression for the magnetic field inside a very long solenoid having n turns per unit length carrying a current I.

(b) How is a toroid different from a solenoid?

Answer: Cyclotron : Cyclotron is a device by which the positively charged particles like protons, deuterons, etc. can be accelerated.

Principle : Cyclotron works on the principle that a positively charged particle can be accelerated by making it to cross the same electric field repeatedly with the help of a magnetic field.

Construction : The construction of a simple cyclotron is shown in figure above, it consists of two-semi cylindrical boxes D₁ and D₂, which are called Dees They are enclosed in an evacuated chamber.

Chamber is kept between the poles of a powerful magnet so that uniform magnetic field acts perpendicular to the plane of the dees. An alternating voltage is applied in the gap between the two dees by the help of a high frequency oscillator. The electric field is zero inside the dees.

Working and theory : At a certain instant, let D₁ be positive and D₂ be negative. A proton from an ion source will be accelerated towards D₂ , it describes a semi-circular path with a constant speed and is acted upon only by the magnetic field. The radius of the circular path is given by.

From the above equation it follows that the frequency f is independent of both v and r and is called cyclotron frequency. Also if we make the frequency of applied a.c. equal to f, then every time the proton reaches the gap between the dees, the direction of electric field is reversed and proton receives a push and finally it gains very high kinetic energy. The proton follows a spiral path and finally gets directed towards the target and comes out from it.

Answer: (a) Magnetic field inside the solenoid

(b) Toroid is a form in which a conductor is wound around a circular body. In this case we get magnetic field inside the core but poles are absent because circular body don’t have ends. Toroid is used in toroidal inductor, toroidal transformer.

Solenoid is a form in which conductor is wound around a cylindrical body with limbs. In this case magnetic field creates two poles N and S. Solenoids have some flux leakage. This is used in relay, motors, electro-magnetes.

Q.16. Prove that the magnetic moment of the electron revolving around a nucleus in an orbit of radius r with orbital speed v is equal to evr/2. Hence using Bohr’s postulate of quantization of angular momentum, deduce the expression for the magnetic moment of hydrogen atom in the ground state.

Answer:

Q.17. Two large charged plane sheets of charge densities and -2σ C/m² are arranged vertically with a separation of d between them. Deduce expressions for the electric field at points

(i) to the left of the first sheet,

(ii) to the right of the second sheet, and

(iii) between the two sheets.

Q.17. A spherical conducting shell of inner radius r₁ and outer radius r₂ has a charge Questions.

(a) A charge q is placed at the center of the shell. Find out the surface charge density on the inner and outer surfaces of the shell.

(b) Is the electric field inside a cavity (with no charge) zero; independent of the fact whether the shell is spherical or not? Explain.

Answer:

Q.18. A signal of low frequency f_m is to be transmitted using a carrier wave of frequency f_c . Derive the expression for the amplitude modulated wave and deduce expressions for the lower and upper side bands produced. Hence, obtain the expression for modulation index.

Q.19. Draw a plot of a-particle scattering by a thin foil of gold to show the variation of the number of the scattered particles with scattering angle. Describe briefly how the large angle scattering explains the existence of the nucleus inside the atom, Explain with the help of impact parameter picture, how Rutherford scattering serves a powerful way to determine and upper limit on the size of the nucleus.

Answer:

From the plot it is clear that Most of the a-particles passed through the foil,, only 0.14% of the incident particles scatter by more than 1% and about 1 in 8000 deflect by more than 90° a-particles deflected backward due to strong repulsive force. This force will come from positive charge concentrated at the center as most of the particles get deflected by small angles.

The α-particle’s trajectory depends on collision’s impact parameter (b) for a given beam of a-particles, distribution of impact parameters as beam gets scattered in different directions with different probabilities.

fig.2 shows a-particle close to nucleus suffers large scattering. Impact parameter is minimum for head on collision a-particles rebound by 180°. Impact parameter is high, for undeviated a-particles. With deflection angle = 0°.

As these of nucleus was 10^-14 m to 10^-15 m w.r.t. 10^-10 m size of an atom which is 10,000 to 100,000 times larger hence most of the space is empty, only a small % of the incident particles rebound back indicates that the number of α-particle goes head on collision. Hence most of the mass of the atom is concentrated in a small volume. Thus, Rutherford scattering is a strong tool to determine upper limit to the size of the nucleus.

Q.20. A 200 µF parallel plate capacitor having plate separation of 5 mm is charged by a 100 V dc source. It remains connected to the source. Using an insulated handle, the distance between the plates is doubled and a dielectric slab of thickness 5 mm and dielectric constant 10 is introduced between the plates. Explain with reason, how the

(i) capacitance,

(ii) electric field between the plates,

(iii) energy density of the capacitor will change ?

As dielectric of 5 mm is inserted with spacing between the dielectric doubled then it will act as following-Fig.A and Fig-B.

Q.21. Why is it difficult to detect the presence of an anti-neutrino during β -decay ? Define the term decay constant of a radioactive nucleus and derive the expression for its mean life in terms of the decay constant.

Q.21. (a) State two distinguishing features of nuclear force.

(b) Draw a plot showing the variation of potential energy of a pair of nucleons as a function of their separation. Mark the regions on the graph where the force is

(i) attractive, and (ii) repulsive.

Answer : The symbols v and v present antineutrino and neutrino respectively during j₃ decay both are neutral particles. With very little or no mass. These particles are emitted from the nucleus along with the electron or positron during the decay process. Neutrinos interact very weakly with matter, they can even penetrate the earth without being absorbed. It is for this reason that their detection is extremely difficult and their presence went unnoticed for long.

Decay constant: Decay constant of a radioactive element is the reciprocal of time during which the number of atoms left in the sample reduces to 1/r times the number of atoms in the original sample.

Derivation of mean life : Let us consider, No be the total number of radioactive atoms present initially. After time t, total no. of atoms present (undecayed) be N. In further dt time dN be the no. of atoms disintegrated. So, the life of dN atoms ranges lies between t + dt and dt. Since, dt is very small time, the most appropriate life of aN atom is t. So the total life of N atom = f.dN

Now, substituting the value of dN and changing the limit in equation (i) from (ii) we get

This expression gives the relation between mean life and decay constant. Hence, mean life is reciprocal of decay constant.

Answer: (a) Distinguish features of nuclear force are :

(i) Nuclear forces are very strong binding forces (attractive force.)

(ii) It is independent of the charges protons and neutrons (charge independent.)

(iii) It depends on the spins of the nucleons.

(b) Plot showing variation of potential energy of a pair of nucleons as a function of separation mark attractive and repulsive region.

X-axis shows separation between pair of nucleons and Y-axis shows variation of potential energy w.r.t. Separation.

(in x 10^-15 m).

Q.22. A triangular prism of refracting angle 60° is made of a transparent material of refractive index 2/√3. A ray of light is incident normally on the face KL as shown in the figure. Trace the path of the ray as it passes through the prism and calculate the angle of emergence and angle of deviation.

Answer: From the diagram it is clear that incidence angle at face KM is 60°.

Hence, critical angle is also 60°.

Therefore, incident light ray will not emerge from KM face due to total internal reflection at this face. Hence, it will move along face KM Angle of emergence = 90°.

Hence angle of deviation = 30° (from fig.)

Q.23. Prove that in a common-emitter amplifier, the output and input differ in phase by 180°. In a transistor, the change of base current by 30 µA produces change of 0 02 V in the base-emitter voltage and a change of 4 mA in the collector current. Calculate the current amplification factor and the load resistance used, if the voltage gain of the amplifier is 400.

Q.24. Show, on a plot, variation of resistivity of

(i) a conductor, and

(ii) a typical semiconductor as a function of temperature.

Using the expression for the resistivity in . terms of number density and relaxation time between collisions, explain how resistivity in the case of a conductor increases while it decreases in a semiconductor, with the rise of temperature.

Answer:

n → number of free electrons

t → Average time between collisions.

In metals n is not dependent on temperature to any appreciable extent and thus the decrease in the value of x with rise in temperature causes p to increases.

for semiconductors, n increases with temperature. This increases more than compensates any decrease in t, so that for such materials, p decreases with temperature.

Section – D

Q.25.(a) Derive an expression for the induced

EMF developed when a coil of N turns, and area of cross-section A, is rotated at a constant angular speed o in a uniform magnetic field B.

(b) A wheel with 100 metallic spokes each 0-5 m long is rotated with a speed of 120 rev/min in a plane normal to the horizontal component of the Earth’s magnetic field. If the resultant magnetic field at that place is 4 × 10^-4 and the angle of dip at the place is 30°, find the emf induced between the axle and the rim of the wheel. 

Q.25. (a) Derive the expression for the magnetic energy stored in an inductor when a current I develops in it. Hence, obtain an expression for the magnetic energy density.

(b) A square loop of sides 5 cm carries a current of 0-2 A in the clockwise direction is placed at a distance of 10 cm from an infinitely long wire carrying a current of 1 A as shown. Calculate

(i) the resultant magnetic force, and

(ii) the torque, if any, active on the loop.

Answer: (a) As the armature coil is rotated in the magnetic field, angle 0 between the field and normal to the coil changes continuously. Therefore, magnetic flux linked with the coil changes. An e.m.f. is induced in the coil. According to Fleming’s right hand rule, current induced in AB is from A to B and it is from C to D in CD in the external circuit current flows from B₂ to B₁.

To calculate the magnitude of e.m.f. Induced:

Suppose,

A → Area of each turn of the coil

N → Number of turns in the coil

B → Strength of magnetic field

θ → Angle which normal to the coil makes

Magnetic flux linked with the coil in this position.

Answer: (a) Energy stored in an inductor : When a current flows through an inductor, a back e.m.f. is set up which opposes the growth of current. So, work needs to be done against back e.m.f. (e) in building up the current. This work done is stored as magnetic potential energy.

Let I be the current through the inductor L at any instant t.

The forces acting on all sides of the square due to current of infinite length wire are lying in the plane of coil. Thus, there is no net torque. Thus torque is zero.

Q.26. Explain, with the help of a diagram, how plane polarized light can be produced by scattering of light from the Sun.

Two polaroids P₁ and P₂ are placed with their pass axes perpendicular to each other. Unpolarised light of intensity I is incident on P₁. a third polaroid P₃ is kept between P₁ and P₂such that its pass axis makes an angle of 450 with that of P₁. calculate the intensity of light transmitted P₁, P₂ and P₃.

Q.26. (a) Why cannot the phenomenon of interference be observed by illuminating two pin holes with two sodium lamps?

(b) Two monochromatic waves having displacements y₁= a cos ωf and y₂= a cos (ωt + Φ ) from two coherent sources interfere to produce an interference pattern. Derive the expression for the resultant in¬tensity and obtain the conditions for constructive and destructive interference.

(c) Two wavelengths of sodium light of 590 nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture 2 x 10⁶.

Answer: Molecules behave like dipole radiators and scatter no energy along the dipole axis by this way plane polarized light can be produced during scattering of light.

Answer: (a) Phenomenon of interference can’t be observed by illuminating two pin holes with two sodium lamps because these sources are not coherent source (it means they are not in the same phase).

(b) Consider two monochromatic coherent sources A and B with waves y₁ = a cos ωt and y₂ = a cos (ωt + Φ ) respectively.

Q.27. (a) Describe briefly, with the help of a circuit diagram, the method of measuring the internal resistance of a cell.

(b) Give reasons why a potentiometer is preferred over a voltmeter for the measurement of emf of a cell.

(c) In the potentiometer circuit given below, calculate the balancing length l. Give reason, whether the circuit will work, if the driver cell of emf 5 V is replaced with a cell of 2 V, keeping all other factors constant.

(a) State the working principle of a meter bridge used to measure an unknown resistance.

(b)Give reasons.

(i) why the connections between the resistors in a meter bridge are made of thick copper strips.

(ii) why is it generally preferred to obtain the balance length near the midpoint of the bridge wire.

Answer:

Where, r is the internal resistance of cell.

(b) Potentiometer is preferred over voltmeter for measurement of e.m.f. of cells because a voltmeter draws some current from the cell while potentiometer draws no current. Therefore, the potentiometer measures the actual e.m.f. of cell whereas voltmeter measures the terminal voltage.

Hence, balancing will not possible as it needs to cater to 300 mV.

Answer: (a) Meter bridge is the practical apparatus which works on principle of Wheat-Stone bridge. It is used to measure an unknown resistance experimentally.

(b) (i) Connection between resistors are made of thick copper strips so that it will have maximum resistance and location of point of balance (D) will be more accurate which results in correct measurement of unknown resistance.

(ii) It is preferred to obtain the balance length near the midpoint of the bridge wire because it increases the sensitivity of meter bridges.

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Chemistry 12th Previous Year Question Paper 2018 (CBSE)

Chemistry

Section-A

Q.1. The analysis shows that FeO has a non-stoichiometric composition with formula Fe_0.95O. Give reason.

Answer: Shows metal defeciency defect / It is a maxture of Fe²⁺ and Fe³⁺ / Some Fe²⁺ ions are replaced by Fe³⁺ / Some of the ferrousion get oxidised to ferric ions.

Q.2. CO(g) and H₂(g) react to give different products in the presence of different catalysts. Which ability of the catalyst is shown by these reactions?

Answer: CO(g) and H₂(g) react in the presence of different catalysts to give different products, this shows that the action of a catalyst is highly selective in nature.

Q.3. Write the coordination number and oxidation state of Platinum in the complex [Pt(en)₂Cl₂].

Answer: Coordination number: 6;

Oxidation state: +2

Q.4. Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why?

Answer: Benzyl chloride would be easily hydrolysed compared to chlorobenzene. In the given reaction conditions, hydrolysis proceeds by nucleophilic substitution mechanism and the benzyl carbonium ion formed after losing the leaving group (-Cl) is better stabilized (through resonating structures) hence reacts easily.

Q.5. Write the IUPAC name of the following:

Answer: The IUPAC name would be 3, 3- Methyl-pentan-2-ol.

Section-B

Q.6. Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol^-1) in 250 g of water. (K_f of water = 1.86 K kg mol^-1)

Answer: Molality (m) of a given solution of Glucose:

m = [(60/180) g mol^-1/250 g] × 1000 = 1.33 mol kg^-1

Now, depression in freezing point is given by, ΔT_f = K_fm

Putting the given values,

ΔT_f = K_fm = 1.86 × 1.33 = 2.5

So, the freezing point of the solution would be = 273.15 K – 2.5 K = 270.65 K.

Q.7. For the reaction

2N₂O₅>(g) → 4NO₂(g) + O₂(g) the rate of formation of NO₂(g) is 2.8 × 10^-3 Ms^-1.

Calculate the rate of disappearance of N₂O₅(g).

Answer: Rate of reaction for the given reaction can be given as,

Rate = 1/2 {-Δ[N₂O₅]/Δt} or {-Δ[N₂O₅]/Δt} = 1/2 {[NO₂/ Δt]}

So, the rate of disappearance of N₂O₅ would be half of the rate of production of NO₂ (given 2.8 × 10^-3Ms^-1).

So, the rate of disappearance of N₂O₅ is 1.4 × 10^-3 Ms^-1.

Q.8. Among the hydrides of Group-15 elements, which have the

(a) lowest boiling point?

(b) maximum basic character?

(d) maximum reducing character?

Answer : The Hydrides of the group 15 elements are NH₃, PH₃, AsH₃, SbH₃, BiH₃.

(a) The lowest boiling point is of PH₃

(b) Maximum basic character is shown by NH₃

(d) BiH₃ has the maximum reducing character.

Q.9. How do you convert the following?

(a) Ethanal to Propanone

(b) Toluene to Benzoic acid

Q.9. Account for the following:

(a) Aromatic carboxylic acids do not undergo Friedel- Crafts reaction.

(b) pK_a value of 4-nitrobenzoic acid is lower than that of benzoic acid.

Answer: (a) Conversion of ethanol to Propanone:

Answer: (a) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction because the carboxyl group is deactivating for electrophilic substitution reaction, secondarily, the catalyst aluminium chloride gets bonded to the carboxyl group.

(b) pK_a value of 4-Nitrobenzoic acid is lower than benzoic acid, which means 4-Nitrobenzoic acid is more acidic than benzoic acid. Being an electron-withdrawing group, the -NO₂ group withdraws electrons towards itself resulting in ease of carboxylic proton release, hence increasing the acidity.

Q.10. Complete and balance the following chemical equations:

(a) Fe²⁺ + MnO^–₄ + H⁺ →

(b) MnO^–₄ + H₂O + I^– →

Answer: (a) 5Fe²⁺ + MnO^–₄ + 8H⁺ → Mn²⁺ + 4H₂O + 5Fe³⁺

(b) 2MnO^–₄ + H₂O + I^– → 2MnO₂ + 2OH^– + IO^–₃.

Section-C

Q.11. Give reasons for the following:

(a) Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers.

(b) Aquatic animals are more comfortable in cold water than in warm water.

Answer: (a) Molar masses of macromolecules like polymers and proteins are measured through osmotic pressure method. The osmotic pressure method uses ‘molarity’ of solution (instead of molality) which has a large magnitude even for dilute solutions, given that polymers have poor solubility, osmotic pressure measurement is used for determination of their molar masses. Macromolecules such as proteins are not stable at high temperatures and because measurement of osmotic pressure is done at around room temperature, it is useful for determination of molar masses of proteins.

(b) The solubility of gases in liquids decreases on increasing the temperature. Hence, the availability of dissolved oxygen in water is more at lower temperatures hence, the aquatic animals feel more comfortable at lower temperatures than at higher temperatures.

(c) Elevation of boiling point is a colligative property and hence depends on the number of solute particles in the solution. Now, 1 M KCl would have twice the number of solute particles, as KCl dissociates into K⁺ and C1^– , compared to sugar solution (as sugar does not undergo any dissociation). So, the elevation of boiling point is nearly double for 1M KCl solution compared to 1M sugar solution.

Q.12. An element ‘X’ (At. mass = 40 g mol^-1) having f.c.c. the structure has a unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’. (NA = 6.022 × 10²³ mol^-1).

Answer :

Given:

Atomic mass of element(m)=40 gmol^-1 length of unit cell(a)=400 pm=4× 10^-8 cm; Z=4(Fcc).

Now density is given by the formula

Where Z is the number of an atom in the unit cell

M is the molecular mass

N_A is the Avogadro number

V is the volume of the unit cell.

V=a³=(4× 10^-8)³ cm =64× 10^-24 cm³

Putting the values

Density, D=160/38.5=4.1 g cm^-3

Q.13. A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK^-1 mol^-1)

Answer: Rate constant for a first-order reaction is given by,

Q.14. What happens when

(a) a freshly prepared precipitate of Fe(OH)³ is shaken with a small amount of FeCl³ solution?

(b) persistent dialysis of a colloidal solution is carried out?

Answer: (a) When FeCl³ is added to a freshly prepared precipitate of Fe(OH)₃, a positively charged sol of hydrated ferric oxide is formed due to adsorption of Fe³⁺ ions.

(b) When persistent dialysis of the colloidal solution is carried out, traces of electrolytes present in the sol are removed almost completely leaving the colloids unstable and finally, coagulation takes place.

Q.15. Write the chemical reactions involved in the process of extraction of Gold. Explain the role of dilute NaCN and Zn in this process.

Answer: Extraction of gold involves leaching the metal with a dilute solution of NaCN or KCN in the presence of air (for O₂) from which the metal is obtained later by replacement method (using Zinc).

The reactions involved are:

4Au(s) + 8CN^–(aq) + 2H₂O(aq) + O₂(g) → 4[AU(CN)₂]^– (aq) + 4OH^–(aq)

2 [AU(CN)₂]^– (aq) + Zn(s) → 2Au (s) + [Zn(CN)₄]^2- (aq)

Q.16. Give reasons:

(a) E⁰ value for Mn³⁺/Mn²⁺ couple is much more positive than that for Fe³⁺/Fe²⁺.

(b) Iron has a higher enthalpy of atomization than that of copper. (c) Sc³⁺ is colourless in aqueous solution whereas Ti³⁺ is coloured.

Answer: (a) Mn²⁺ has a d⁵ configuration, and the extra stability of half-filled d-orbitals is compromised when another electron is taken out to give Mn³⁺, On the contrary, Fe³⁺ attains a half-filled orbital configuration when Fe²⁺ is oxidized to Fe³⁺. Hence, the E⁰ value for Mn³⁺/ Mn²⁺ couple has more positive E⁰ value.

(b) Fe has a 3d⁶4s² outer electronic configuration whereas Cu has 3d¹⁰4s¹ configuration. Now, the more the number of impaired electrons in d-orbital, more favourable are interatomic attractions and thus higher atomization enthalpies. Hence, Fe having 4 unpaired d-electrons has more enthalpy of atomization than copper having no unpaired d-electron.

(c) Sc³⁺ has a 3d⁰ configuration whereas Ti³⁺ has a 3d¹ configuration. As there are no electrons in the d orbital for Sc³⁺ ion, there is no transition of electrons by absorption of energy and hence no emission in the visible range imparting colour to the Sc³⁺ ion.

Q.17. (a) Identify the chiral molecule in the following pair:

(b) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.

Answer: (a) The molecule (i) is a chiral molecule.

(b) Chlorobenzene reacts with methyl chloride in the presence of sodium metal and dry ether to give toluene. This reaction is known as Wurtz-Fittig reaction.

(c) In the 1-Bromo-1-methylcyclohexane, all β-hydrogen atoms are equivalent. Thus dehydrohalogenation takes place, in the reaction of this compound with KOH.

Q.18. (A), (B) and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C₄H₈O. Isomers (A) and (C) give positive Tollens test whereas isomer (B) does not give Tollen’s test but gives positive Iodoform test. Isomers (A) and (B) on reduction with Zn (Hg)/conc. HCl, give the same product (D).

(a) Write the structures of (A), (B), (C) and (D).

(b) Out of (A), (B) and (C) isomers, which one is least reactive towards the addition of HCN?

Answer: (a) Compound A and C give positive Tollens test which indicates that they are aldehydes. Compound C gives Iodoform test which means it contains a carbonyl group with a methyl group attached to the carbonyl carbon so, with formula C₄H₈O the structure of compound would be CH₃COCH₂CH₃ (Butanone).

Now upon reduction with Zn(Hg)/conc. HCl, the corresponding alkanes are obtained, so a reduction of B gives Butane (D), so the isomer A has to be a linear chain aldehyde (Butanal), giving Butane (compound D) on reduction. So, the last isomer possible is compound C, 2-Methyl propionaldehyde. The reactions involved are shown below with the structures of compounds:

(b) Out of the three isomers A, B and C, compound B’ (Butanone) would be least reactive towards the addition of HCl as the carbonyl carbon is sterically hindered and most reactive would be compound A (Butanal) towards the addition of HCN.

Q.19. Write the structures of the main products in the following reactions:

Answer: (i) Sodium borohydride doesn’t reduce esters, so the product would be,

Q.20. Answer the given questions:

(A)Why is bithional added to soap?

(B)What is the tincture of iodine? Write its one use.

(C)Among the following, which one acts as a food preservative?

Aspartame, Aspirin, Sodium Benzoate, Paracetamol

Answer: 1. Bithional is added to soaps to impart antiseptic properties to soap.

2. Tincture of iodine is 2-3 per cent mixture of iodine in the alcohol-water mixture. It is used as an antiseptic.

3. Sodium benzoate is used as a food preservative.

Q.21. Define the following with an example of each:

(a) Polysaccharides

(b) Denatured protein

Q.21. (a) Write the product when D-glucose reacts with conc. HNO₃.

(b) Amino acids show amphoteric behaviour. Why?

Answer: (a) Polysaccharides: Polysaccharides are food storage materials and most commonly found carbohydrates in nature. These are the compound which is formed of a large number of monosaccharide units joined together by glycosidic linkages. Example. Starch, main storage polysaccharide of plants.

(b) Denatured protein: Proteins have a unique three-dimensional structure in their native form. If the native form of protein is subjected to any physical change (such as temperature change) or any chemical change (such as a change in pH), the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled due to which protein loses its biological activity. This is called denaturation of the protein. During denaturation 2° and 3° structures of proteins are destroyed but 1° structure remains intact. Coagulation of egg white is an example of denaturation of the protein.

(c) Essential amino acids: amino acids which are not synthesized in our body and have to be obtained through diet are known as essential amino acids. Example: Tryptophan

Answer: (a) D-Glucose gets oxidized to give saccharic acid, a dicarboxylic acid on reacting with nitric acid.

(b) Amino acids show amphoteric behaviour due to the presence of both acidic (carboxylic group) and basic (amino group) in the same molecule. So, in the basic medium, the carboxyl group can lose a proton and in acidic medium, the amino group can accept a proton.

(c) In α-helix structure the polypeptide chain forms all possible hydrogen bonds by twisting into a right-handed screw (helix) with the -NH group of each amino acid residue gets hydrogen-bonded to the -C = O of an adjacent turn of the helix (Intramolecular bonding), whereas in β-structure all peptide chains are stretched out to nearly maximum extension and then laid side by side which are held together by intermolecular hydrogen bonds (intermolecular bonding).

Q.22. (a) Write the formula of the following coordination compound: Iron (III) hexacyanoferrate (II)

(b) What type of isomerism is exhibited by the complex [Co(NH₃)₅ Cl]SO₄?

(Atomic number of Co = 27)

Answer: (a) The molecular formula of Iron(III) α-cyanoferrate(II) is Fe₄[Fe(CN)₆]₃

(b) [CO(NH₃)₅Cl]SO₄ will show Ionisation isomerism and the possible isomers are [CO(NH₃)₅Cl]SO₄ and [Co (NH₃)₅SO₄]Cl

Electronic configuration of sp³d² hybridized (as F is a weak field ligand) orbitals of Co³⁺ , with six pairs of electrons from six F ions.

There are 4 impaired electrons in [CoF₆]₃.

Section-D

Q.23. Shyam went to a grocery shop to purchase some food items. The shopkeeper packed all the items in polythene bags and gave them to Shyam. But Shyam refused to accept the polythene bags and asked the shopkeeper to pack the items in paper bags. He informed the shopkeeper about the heavy penalty imposed by the government for using polythene bags. The shopkeeper promised that he would use paper bags in future in place of polythene bags.

Answer the following:

(a) Write the values (at least two) shown by Shyam.

(b) Write one structural difference between low-density polythene and high-density polythene.

(d) What is a biodegradable polymer? Give an example.

Answer: (b) Low-density polythene has a branched-chain structure, whereas the high-density polythene has a linear chain structure.

(d) Biodegradable polymers contain functional groups similar to functional groups present in biopolymers, so they get degraded in the environment by certain microorganisms and thus are environment-friendly.

For example Poly β -hydroxybutyrate-co-β-hydroxy valerate (PHBV).

Section-E

Q.24. (a) Give reasons:

(i) H₃PO₃ undergoes disproportionation reaction but H₃PO₄ does not.

(ii) When Cl₂ reacts with an excess of F₂, ClF₃ is formed and not FCl₃.

(iii) Dioxygen is a gas while Sulphur is a solid at room temperature.

(b) Draw the structures of the following:

(i) XeF₄

(ii) HClO₃

Q.24. (a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into a colourless solid (B).

(i) Identify (A) and (B).

(ii) Write the structures of (A) and (B).

(iii) Why does gas (A) change to solid on cooling?

(b) Arrange the following in decreasing order of their reducing character: HF, HCl, HBr, HI

XeF₄ + SbF₅ →

Answer: (a) (i) In H₃PO₃ (orthophosphoric acid) oxidation state of phosphorus is +3 and it contains one P-H bond in addition to P = O and P-OH bonds. These types of oxoacids tend to undergo disproportionation to give orthophosphoric acid (P has +5 state) and phosphine (P has +3 state). Whereas in H₃PO₄ (orthophosphoric acid), Phosphorus is in +5 state hence no disproportionation takes place in H₃PO₄.

(ii) When Cl₂ reacts with an excess of F₂, ClF₃ is formed-and not FCl₃ because Fluorine can’t expand its valency and can show only -1 oxidation state, whereas Cl can expand its valency due to the availability of d-orbitals.

(iii) Dioxygen is a gas while sulphur is a solid at room temperature this is because sulphur have S₈ molecules and these are packed to give different crystal structure, whereas dioxygen is a diatomic molecule (O₂) and it does not have enough intermolecular attraction and thus exists in gaseous form.

(a) (i) The brown gas A is NO₂ or nitrogen dioxide. On cooling, it dimerises to N₂O₄ and solidifies as a colourless solid.

(iii) Compound A, that is, NO₂ contains an odd number of valence electrons. It behaves as a typical odd molecule. On dimerization, it is converted to stable N₂O₄ molecule with an even number of electrons (thus colourless) and have better intermolecular forces to get solidified. Thus, it changes to solid on cooling.

(b) Decreasing order of reducing character: HI > HBr > HCl > HF

Q.25. (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K:

Sn(s) | Sn²⁺ (0.004 M) || H⁺> (0.020 M) | H²(g) (1 bar) | Pt(s)

(Given: E° Sn²⁺/Sn = – 0.14V)

(b) Give reasons:

(i) On the basis of E° values, O₂ gas should be liberated at anode but it is Cl₂ gas which is liberated in the electrolysis of aqueous NaCl.

(ii) The conductivity of CH₃COOH decreases on dilution.

Q.25. (a) For the reaction

2AgCl(s) + H₂(g) (1 atm) → 2Ag(s) + 2H⁺ (0.1M) + 2Cl^– (0.1M), ΔG° = -43600 J at 25°C.

Calculate the e.m.f. of the cell. [log 10^-n> = -n]

(b) Define fuel cell and write its two advantages.

Answer: (a) The half cell reactions can be written as;

The reaction at the anode with a lower value of Ecell is preferred and therefore, water should get oxidized to give O₂ but on account of overpotential of oxygen, Cl^– gets oxidized preferably, liberating Cl₂ gas.

(ii) The conductivity of CH₃COOH decreases on dilution because the number of ions per unit volume that carry the current in a solution decreases on dilution.

Answer:

(b) Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol etc. directly into electrical energy are called fuel cells.

Advantages of fuel cells are:

Fuel cells produce electricity with an efficiency of about 70% compared to thermal plants whose efficiency is about 40%.
Fuel cells are pollution-free.

Q.26. (a) Write the reactions involved in the following:

(i) Hofmann bromamide degradation reaction

(ii) Diazotisation

(iii) Gabriel phthalimide synthesis

(b) Give reasons:

(i) (CH₃)₂NH is more basic than (CH₃)₃N in an aqueous solution.

(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts.

Q.26. (a) Write the structures of the main products of the following reactions:

(b) Give a simple chemical test to distinguish between Aniline and N, N-dimethylaniline.

C₆H₅NH₂, C₂H₅NH₂, C₆H₅NHCH₃

Answer: (a) (i) Hofmann bromamide degradation reaction: Acetamide can be considered for example. In this reaction, Acetamide (CH₃CONH₂) undergoes Hofmann degradation in the presence of Bromine and NaOH to give Methanamine.

CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O.

(ii) Diazotisation: The conversion of primary aromatic amines into diazonium salts is known as diazotization.

(iii) Gabriel phthalimide synthesis: This reaction is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine.

(b) (i) (CH₃)₂ NH is more basic than (CH₃)₃N in aqueous solutions because in (CH₃)₃N the lone pair of electrons on the nitrogen atom is responsible for its basicity are quite hindered by the three methyl groups, hence are less available. Due to which it is less basic as compared to (CH₃)₂NH.

(ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts because the positive charge on the nitrogen atom is stabilized by the resonance with an attached phenyl group.

Answer:

(b) Aniline can be distinguished from N, N-dimethyl aniline by diazo coupling reaction. Aniline would react with benzene diazonium chloride to give a yellow dye, whereas N, N-dimethyl aniline won’t undergo this reaction.

Chemistry 12th Previous Year Question Paper 2019 (CBSE)

Chemistry

Set-I

Section – A

Q.1.Out of KCl and AgCl, which one shows Schottky defect and why?

Q.1.Why does ZnO appear yellow on heating?

Answer: KCl shows schottky defect because cation & anions are of similar size.

Answer: ZnO on heating loses oxygen leaving behind their electrons at that position due to electrons it appears yellow in colour.

Q.2. Arrange the following in decreasing order of basic character:

C₆H₅NH₂, (CH₃)₃N, C₂H₅NH₂

Answer: Decreasing order of basic character:

CH₃CH₂NH₂ > (CH₃)₃N > C₂H₅NH₂

Q.3. What type of colloid is formed when a solid is dispersed in a liquid ? Give an example.

Answer: Sols are formed when a solid is dispersed in a liquid.

Example – Paints.

Q.4. Out of Chlorobenzene and Cyclohexyl chloride, which one is more reactive towards nucleophilic substitution reaction and why?

Answer: Cyclohexyl chloride is more reactive towards nucleophilic substitution reaction because the carbon bearing the chlorine atom is deficient in electron and seeks a nucleophile. In Chlorobenzene the carbon bearing the halogen is a part of an aromatic ring and is electron-rich due to the electron density in the ring.

Q.5. What is the basic structural difference between starch and cellulose?

Q.5. Write the products obtained after hydrolysis of DNA.

Answer: Starch consists of two components- amylose and amylopectin. Amylose is a long linear chain of α-D -(+)-glucose units joined by C₁, C₄glycosidic linkage (α-link). Amylopectin is a branched-chain polymer of α-D-glucose units, in which the chain is formed by C₁-C₄ glycosidic linkage and the branching occurs by C₁-C₆ glycosidic linkage. On the other hand, cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C₁-C₄ glycosidic linkage (β-link).

Answer: Hydrolysis of DNA yields a pentose sugar (β-D-2deoxyribose), phosphoric acid and nitrogen-containing heterocyclic compounds called bases (Adenine, Guanine, Cytosine and Thymine).

Section – B

Q.6. Write balanced chemical equations for the following processes:

(a) Cl₂ is passed through slaked lime.

(b) SO₂ gas is passed through an aqueous solution of Fe (III) salt.

Q.6. (a) Write two poisonous gases prepared from chlorine gas. (b) Why does the Cu₂₊ solution give a blue colour on reaction with ammonia?

Answer: (a) Cl₂ is passed through slaked lime to give bleaching powder [Ca(OCl)₂]

2Ca(OH)₂ + 2Cl₂ → Ca(OCl)₂ + CaCl₂ + 2H₂O

(b) When SO₂ gas is passed through a Fe(III) aqueous solution, Fe(III) is reduced to Fe(II) ion:

2Fe₃₊ + SO₂ + 2H₂O → 2Fe²⁺ + SO₂^2- + 4H⁺

Answer: (a) Two poisonous gases prepared from chlorine – Phosgene (COCl₂) and tear gas (CCl₃NO₂).

(b) Nitrogen in ammonia has a lone pair of electrons, which makes it a Lewis base. It donates the electron pair and forms linkage with metal ions-

Q.7. Give reasons:

(a) Cooking is faster in a pressure cooker than in cooking pan.

(b) Red Blood Cells (RBC) shrink when placed in saline water but swell in distilled water.

Answer: (a) Boiling points increase in increasing the pressure in case of liquids. Water used for cooking attains a higher temperature than the usual boiling temperature inside the pressure cooker due to the existing high pressure inside the pressure cooker vessel. This leads to a faster flow of water inside the vegetables or grains etc. resulting in faster cooking of food in a pressure cooker than in the cooking pan.

(b) Red blood cells shrink when placed in saline water because of exosmosis, i.e., water comes out from the cell to surrounding (more concentrated) to equate the concentration. Whereas, when placed in distilled water concentration within the cell becomes more than the surrounding, hence water comes inside and endosmosis takes place to equate the concentrations.

Q.8. Define the order of the reaction. Predict the order of reaction in the given graphs:

where [R]₀ is the initial concentration of reactant and t_1/2is a half-life.

Answer: It is defined as the sum of powers to which the concentration terms are raised in the rate law equation.

(a) In this graph, as t_1/2 is independent of initial reactant concentration, it is a first-order reaction.

(b) In this graph, as tin is directly proportional to the initial concentration of reactant hence, it is a zero-order reaction.

Q.9. When FeCr₂O₄ is fused with Na₂CO₃ in the presence of air it gives a yellow solution of compound (A). Compound (A) on acidification gives compound (B). Compound (B) on reaction with KCl forms an orange coloured compound (C). An acidified solution of compound (C) oxidises Na₂O₃ to (D). Identify (A), (B), (C) and (D).

Answer:

Q.10. Write IUPAC name of the complex [Co(en)₂(NO₂)Cl]⁺. What type of structural isomerism is shown by this complex?

Q.10. Using IUPAC norms, write the formula for the following complexes:

(a) Hexaaquachromium (III) chloride

(b) Sodium trioxalatoferrate (III)

Answer: IUPAC name of [Co(en)₂(NO₂)Cl]⁺ is Chlorobis(ethane-1, 2-diamine)nitro cobalt(III).

This compound shows geometrical isomerism.

Answer: (a) Hexaaquachromium(III) chloride- [Cr(H₂O)₆]Cl₃

(b) Sodium trioxalatoferrate(III)- Na₃[Fe(C₂O₄)₃]

Q.11. (a) Although both [NiCl₄]_2- and [Ni(CO)₄] have sp³ hybridisation yet [NiCl₄]⁴ is paramagnetic and [Ni(CO)₄] is diamagnetic. Give reason. (Atomic no. of Ni = 28).

(b) Write the electronic configuration of d⁵ on the basis of crystal field theory when

(i) ∆₀ < P and

(ii) ∆₀ > P [2]

Answer: (a) [NiCl₄]^2- is a high spin complex and there are two impaired electrons with 3d8 electronic configuration of a central metal atom, hence it is paramagnetic. Whereas in [Ni(CO)₄] Ni is in zero oxidation state and contains no unpaired electrons, hence it is diamagnetic in nature.

(b) (i) Electronic configuration of d₅ when ∆o < P is given as t_2g³ e_g²

(ii) Electronic configuration of d⁵ when ∆o > P is given as t_2g⁵ e_g⁰.

Q.12. Write structures of main compounds A and B in each of the following reactions:

Answer:

Section – C

Q.13. The following data were obtained for the reaction:

A + 2B → C

(a) Find the order of reaction with respect to A and B.

(b) Write the rate law and overall order of the reaction.

(a) Find the order of reaction with respect to A and B.

(b) Write the rate law and overall order of the reaction.

Answer: The reaction is A + 2B → C

(a) It can be seen that when the concentration of A is doubled keeping B constant, then the rate increases by a factor of 4 (from 4.2 × 10^2- to 1.68 × 10^2-). This indicates that the rate depends on the square of the concentration of the reactant A. Also when the concentration of reactant B is made four times, keeping the concentration of reactant A constant, the reaction rate also becomes 4 times (2.4 × 10^2- to 6.0 × 10^3-). This indicates that the rate depends on the concentration of reactant B to the first power.

(b) So, the rate equation will be:

Rate = k[A]²[B]

Overall order of reaction will be 2 + 1 = 3.

4.2 × 10^2- M min^-1= k (0.2)²(0.3) M

k = 0.0420.0123.5 min^-1

Q.14. (a) Write the dispersed phase and dispersion medium of dust.

(b) Why is physisorption reversible whereas chemisorption is irreversible?

What is the charge of Agl colloidal particles formed in the test tube?

How is this sol represented?

Answer: (a) In dust, the dispersed phase is solid particles and the dispersion medium is air (gas).

(b) Physisorption occurs only because of physical attractive forces such as van der Waals forces between molecules of adsorbate and adsorbent, hence that can be reversed on the application of bigger forces but chemisorption occurs due to the chemical reaction between molecules of adsorbate and adsorbent, and hence can’t be reversed.

(c) When KI solution is added to AgNO³ a positively charged sol results due to absorption of Ag⁺ ions from dispersion medium-AgI/ Ag⁺(positively charged).

Q.15. An element X with an atomic mass of 81 u has density 10.2 g cm^3-. If the volume of the unit cell is 27 × 10^-23 cm³, identify the type of cubic unit cell. (Given: NA = 6.022 × 10 mol^-1.

Answer:

Q.16. A solution containing 19 g per 1oo mL of K (M = 74.5 g mol^-1) is isotonic with a solution containing 3 g per 100 mL of urea (M = 60 g mol^-1). Calculate the degree of dissociation of KCl solution. Assume that both the solutions have the same temperature.

Answer: Two solutions having the same osmotic pressure at a given temperature are called isotonic solution. Now in the given problem, the KCl and urea solutions are given to be isotonic.

Osmotic pressure π is given by the equation

π = (n²/V)RT, where n₂ = moles of solute,

V = volume of solution in litres.

Also, n₂ = w₂/M₂,

where W₂ = grams of solute and M₂ = molar mass of solute.

The other given information is

The molar mass of KCl = 74.5 g mol^-1

Weight of KCl, W₂= 1.9 g, V = 100 mL

So, for KCl

n = (w₂/M₂ × V)RT

nRTKCl = 1.9/(74.5 × 100) = 2.55 × 10^-4

Now as the solutions are isotonic at the same temperature:

πRTKCl = πRTurea

Hence, substituting the values for urea:

2.55 × 10^-4 = 3/M₂ × 100

M₂ = 117.6

So, the experimentally determined molecular weight of urea is found to be as 117.6, so the degree of dissociation can be given as:

Osmotic pressure (TT) = Experimentally determined.

So, Urea dimerized in the given experimental solution.

Q.17. Write the name and principle of the method used for refining of (a) Zinc, (b) Germanium, (c) Titanium.

Answer: (a) Distillation is used for refining zinc. As zinc is a low boiling metal, the impure metal is evaporated and the pure metal is obtained as a distillate.

(b) Zone refining is used for refining Germanium. This method is based on the principle that the impurities are more soluble in the melt than in the solid-state of the metal.

(c) Titanium is refined by van Arkel method. This method is used for the removal of oxygen and nitrogen present as an impurity. The crude metal is heated in an evacuated vessel with iodine to obtain metal iodide, which volatilizes being covalent. Later this metal iodide is decomposed through electrical heating to obtain the pure metal.

Q.18. Give reasons for the following:

(a) Transition metals form complex compounds.

(b) E⁰ values of (Zn²⁺/Zn) and (Mn²⁺/Mn) are more negative than expected.

Answer: (a) Transition elements have partly filled d-orbitals due to which they have variable oxidation states which enables them to bind with a variety of ligands and hence form complex compounds.
(b) Oxidation of Zn to Zn²⁺ leads to a completely filled d10 configuration in Zn²⁺, making it more stable. Also, Mn/Mn²⁺ conversion leads to a half-filled stable d⁵ configuration of Mn²⁺ ion. Hence, E⁰ value for Zn/Zn²⁺and Mn/ Mn²⁺conversion has negative values.

(c) Actinoids show a wide range of oxidation states due to their partially filled f-orbitals and they have comparable energies as well.

Q.19. Write the structures of monomers used for getting the following polymers:

(a) Nylon-6

(b) Terylene

Q.19. (a) Is [CH²-CH(C⁶H⁵)]ⁿ homopolymer or copolymer? Give reason.

(b) Write the monomers of the following polymers:

Answer: Structures of monomers

(a) Caprolactam is monomer of Nylon-6

Q.20. (a) Pick out the odd one from the following on the basis of their medicinal properties:

Equanil, Seconal, Bithional, Luminal

(b) What types of detergents are used in dishwashing liquids?

Q.20. Define the following terms with a suitable example of each:

(a) Antibiotics

(b) Antiseptics

Answer: (a) ‘Bithionol’ is the odd one here, as it is an antiseptic whereas others are tranquilisers.

(b) Liquid dishwashing detergents are non-ionic type.

(c) Aspartame is an artificial sweetener which is unstable at cooking temperature hence its use is limited to cold foods.

Answer: (a) Antibiotics: These are the compounds (produced by microorganisms or synthetically) which either inhibit the growth of bacteria or kill bacteria. Example: Penicillin.

(b) Antiseptics: These are the chemicals used to kill or prevent the growth of microorganisms when applied to the living tissues.

Example: Soframycin.

(c) Anionic detergents: These are sodium salts of sulfonated long-chain alcohols or hydrocarbons. In these, the anionic part of the molecule is involved in the cleansing action.

Example: Sodium lauryl sulphate.

Q.21. Among all the isomers of molecular formula C₄H₈Br, identify:

(a) the one isomer which is optically active.

(b) the one isomer which is highly reactive towards SN₂.

Answer:

(a) 2- Bromobutane is optically active as C-2 is a chiral carbon here having all the four different groups attached to it.

(b) 1-Bromobutane being primary alkyl halide is highly reactive towards SN₂ reaction.

Q.22. Complete the following reactions:

Q.22. How do you convert the following:

(a) N-phenylethylamine to p-bromoaniline

(b) Benzene diazonium chloride to nitro-benzene

Answer:

Answer:

Q.23. (a) Give reasons:

(i) Benzoic acid is a stronger acid than acetic acid.

(ii) Methanal is more reactive towards nucleophilic addition reaction than ethanal.

(b) Give a simple chemical test to distinguish between propanal and propanone.

Answer (a) (i) Benzoic acid is a stronger acid than acetic acid because the benzoate anion (the conjugate base of benzoic acid) formed after loss of H⁺ is stabilized by resonance, whereas acetate ion (CH³COO^–) has no such extra stability. Hence, Benzoic acid has more tendency of losing proton compared to acetic acid hence more acidic.

(ii) Methanal is more reactive towards nucleophilic addition reaction than ethanal because in ethanal there is a methyl group attached to the carbonyl carbon (centre for nucleophilic attack) and +1 effect of the methyl group decreases the nucleophilicity of carbonyl carbon by increasing the electron density at carbonyl carbon.

(b) Propanal and propanone can be distinguished using Tollen’s reagent by silver mirror test. Propanal being an aldehyde reacts with Tollen’s reagent to give silver deposition whereas propanone being a ketone does not give the reaction.

Q.24. (a) What is the product of hydrolysis of maltose?

(b) What type of bonding provides stability to the α-helix structure of a protein?

Q.24. Define the following terms:

(a) Invert sugar

(b) Native protein

Answer: (a) On hydrolysis maltose gives two molecules of α-D-glucose.

(b) α-Helix structure of proteins is stabilized by hydrogen bonds between -NH group of each amino acid and -COOH group of amino acid at adjacent turn.

Answer: (a) Invert sugar: It is a mixture of glucose and fructose obtained after hydrolysis of sucrose. Sucrose is dextrorotatory, but after hydrolysis gives a mixture of dextrorotatory glucose and levorotatory fructose which outweighs in magnitude and hence the whole mixture becomes levorotatory hence the mixture obtained is called invert sugar.

(b) Native protein: a Protein found in a biological system with a unique three-dimensional structure and biological activity is called a native protein.

(c) Nucleotide: They are building blocks of DNA/RNA. These consist of a pentose sugar moiety attached to a nitrogenous base at V position and a phosphoric acid molecule at 5′ position.

Example:

Section – D

Q.25. (a) The conductivity of 0.001 mol L^-1 acetic acid is 4.95 × 10^-5 S cm^-1. Calculate the dissociation constant if ∧⁰_m for acetic acid is 390.5 S cm² mol^-1.
(b) Write Nest equation for the reaction at 25°C:

2Al(s) + 3Cu²⁺ (aq) → 2 Al³⁺ (aq) + 3Cu (s) (c)

What are secondary batteries? Give an example.

Q.25. (a) Represent the cell in which the following reaction takes place:

2Al (s) + 3 Ni²⁺ (0.1M) → 2Al³⁺ (0.01M) + 3 Ni (s)

Calculate its emf if E⁰_cell = 1.41 V.

(b) How does molar conductivity vary with increase in concentration for strong electrolytes and weak electrolytes? How can you obtain limiting molar conductivity (∧m0) for weak electrolyte?

Answer: (a) Conductivity ∧_m of a solution is given by the following equation:

∧_m = K/C

where k is the dissociation constant and c is the concentration of the solution. Here, given.

Conductivity, k = 4.95 × 10^-5 S cm^-1 Limiting molar conductivity, ∧⁰_m = 390.5 S cm² mol^-1

Concentration,

c = 0.001 mol L^-1 = 1 × 10^-3mol L^-1

Substituting the given values in above equation

Molar conductivity,

(b) Nernst equation for the given reaction can be written as

(c) A secondary battery can be recharged after use, bypassing current through it in the opposite direction so that it can be used again.

Example: The most important secondary cell is lead storage cell. It consists of a lead anode and a grid of lead packed with lead dioxide as a cathode. A 38% solution of sulphuric acid is used as an electrolyte.

Answer: (a) The cell can be represented as

(b) For strong electrolytes, the molar conductivity is increased only slightly on dilution. A strong electrolyte is completely dissociated in solution and thus, furnishes all ions for conductance. However, at higher concentrations, the dissociated ions are close to each other and thus, the interionic attractions are greater. These forces retard the motion of the ions and thus, conductivity is low. With a decrease in concentration (dilution), the ions move away from each other thereby feeling less attractive forces from the counterions. This results in an increase in molar conductivity with dilution. The molar conductivity approaches a maximum limiting value at infinite dilution designated as ∧m0.

In the case of weak electrolytes as the solution of a weak electrolyte is diluted, its ionization is increased. This results in more number of ions in solution and thus, there is an increase in molar conductivity, also there is a large increase in the value of molar conductivity with dilution, especially near-infinite dilution. However, the conductance of a weak electrolyte never approaches a limiting value. Or in other words, it is not possible to find conductance at infinite dilution (zero concentration).

So, limiting molar conductivity for weak electrolytes are obtained by using Kohlrausch law, from the limiting molar conductivities of individual ions (λ0).

Kohlrausch law of independent migration of ions states that limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and the cation of the electrolyte.

∧_m⁰ = λ⁰⁺ + λ⁰

Q.26. (a) Give the equation of the following reactions:

(i) Phenol is treated with cone. HNO₃.

(ii) Propene is treated with B₂H₆ followed by H₂O₂/OH-.

(iii) Sodium t-butoxide is treated with CH₃Cl.

(b) How will you distinguish between butan-l-ol and butan-2-ol?

Q.26. (a) How can you obtain Phenol from (i) Cumene, (ii) Benzene sulphonic acid, (iii) Benzene diazonium chloride?

(b) Write the structure of the major product obtained from the denitration of 3-methyl phenol.

Answer: (a) (i) Phenol is treated with conc. HNO3 to obtain 2,4,6-trinitrophenol picric acid.

(ii) Propene undergoes hydroboration-oxidation when treated with B₂H₆ followed by hydrogen peroxide in basic medium to give propan-1-ol.

(iii) Methyl tert-butyl ether is produced when sodium tert-butoxide is treated with methyl chloride.

(b) Butan-l-ol and Butan-2-ol can be distinguished using Lucas reagent (ZnCl²⁺HCl), where butan-2-ol would react with Lucas reagent in around 5 minutes to give a white precipitate of 2-chlorobutane, whereas butan-l-ol won’t give any reaction at room temperature.

Answer: (a) (i) Phenol from cumene

(ii) Phenol from benzene sulphonic acid

(iii) Phenol from benzene diazonium chloride

(b) The combined influence of -OH and -CH₃groups determine the position of the entering groups, also the sterically hindered positions are not substituted.

(c) In Kolbe’s reaction phenol is reacted with CO₂ in the presence of sodium hydroxide, followed by acidification, to give a carboxylic acid group on 2-position of phenol-

Q.27. (a) Account for the following:

(i) The tendency to show – 3 oxidation state decreases from N to Bi in group 15.

(ii) Acidic character increases from H₂O to H₂Te.

(iii) F₂ is more reactive than CIF₃, whereas ClF₃ is more reactive than Cl₂.

(b) Draw the structure of (i) XeF₂ (ii) H₄P₂O₇.

Q.27. (a) Give one example to show the anomalous reaction of fluorine.

(b) What is the structural difference between white phosphorous and red phosphorous?

(d) Why is H₂S a better reducing agent than H₂O?

(e) Arrange the following acids in the increasing order of their acidic character: HF, HCl, HBr and HI

Answer: (a) (ii) Acidic character increases from H₂O to H₂Te due to decrease in E—H bond dissociation enthalpy down the group. Thus it becomes easy to lose proton going down the group.

(iii) F₂ is more reactive than ClF₃ because of the small size of fluorine atom F—F bond, bond dissociation enthalpy is low (thus is reactive).

Whereas ClF₃ is more reactive than Cl₂ because ClF₃ is an interhalogen compound with weak Cl—F bond (compared to Cl—Cl bond) due to the difference in atomic sizes (hence ineffective overlap of orbitals).

(b) (i) Structure of XeF₂ is linear.

(ii) Structure of H₄P₂O₇.

Answer: (a) Fluorine reacts with cold sodium hydroxide solution to give OF₂.

2F₂ (g) + 2NaOH (aq) → 2NaF (aq) + OF₂ (g) + H₂O(l)

XeF₆ + NaF → Na⁺[XeF₇]^–

(d) Ability to reduce is judged by the ease with which an atom can donate its electrons to the species which is getting reduced. Now, the size of oxygen atom in H₂O is smaller than that of Sulphur atom in H₂S, due to which the lone pair of electrons on oxygen are more attracted by the oxygen nucleus, making it difficult to donate electrons (by oxygen compared to Sulphur, while in H₂S the influence of the nucleus is less on lone pair of electrons of sulphur and hence, it can give away its electrons, easily compared to oxygen, and thus acts as a better reducing agent.

(e) The increasing order of acidic character can be written as

HF < HCl < HBr < HI

Chemistry

Set-II

Section – A

Q.2. Arrange the following in increasing order of pKb values:

C₆H₅CH₂NH₂, C₆H₅NHCH₃, C₆H₆NH₂

Answer: These can be arranged in increasing order of pKb values as follows:

C₆H₅CH₂NH₂ < C₆H₅NHCH₃ < C₆H₅NH₂

Q.3. What type of colloid is formed when a liquid is dispersed in a solid? Give an example.

Answer: When a liquid is dispersed in a solid, a ‘gel’ is formed.

Example: Butter.

Q.4. Out of chlorobenzene and p-nitrochloro-benzene, which one is more reactive towards nucleophilic substitution reaction and why?

Answer: p-Nitro chlorobenzene would be more reactive towards nucleophilic substitution reaction compared to chlorobenzene. In chlorobenzene the carbon bearing the halogen is a part of the aromatic ring and is electron-rich due to the electron density in the ring so it does not attract the nucleophile. The -NO₂ substitution lessens the electron density on the benzene ring due to its electron-withdrawing nature, making the electron density on ringless compared to chlorobenzene, hence p-nitro chlorobenzene attracts nucleophiles better.

Section -B

Q.7. Give reasons:

(a) A decrease in temperature is observed in mixing ethanol and acetone.

(b) Potassium chloride solution freezes at a lower temperature than water.

Answer: (a) Upon mixing molecules of ethanol and acetone have strong intermolecular attractions due to which heat is evolved from the reaction system and hence cooling of mixture is observed.

(b) Potassium chloride solution is a solution of non-volatile solute KCl and water solution. We know that, at the freezing point of a substance, the solid phase (here ice) is in dynamic equilibrium with the liquid phase. A solution freezes when its vapour pressure equals the vapour pressure of the pure solid solvent. Now, according to Raoult’s law when a non-volatile solid is added to the solvent (in this case it is KCl), its vapour pressure decreases and now it would become equal to that of solid solvent at a lower temperature. Thus, the freezing point of the solvent decreases.

Q.10. Define the following terms with a suitable example of each:

(a) Chelate complex

(b) Ambidentate ligand

Q.10. Using IUPAC norms, write the formula for the following complexes:

(a) Tetraamminedichlorocobalt (III) chloride

(b) Dibromobis (ethane-1,2-diamine) platinum (IV) nitrate

Answer: (a) Chelate complex: Chelate complexes are coordination or complex compound consisting of a central metal atom attached to a large molecule, called a ligand, in a cyclic or ring structure. The ligands are bi or polydentate i.e., they can attach to metal atom through two or more than two binding sites. An example of a chelate ring occurs in the ethylenediamine-nickel complex.

(b) Ambidentate ligand: Ligands which can ligate (attach to the metal atom) through two different atoms is called ambidentate ligand. One example of such ligand is NO^–₂ , this can bind through both the atoms, nitrogen and oxygen.

Answer: IUPAC names

Tetraamine Diaqua Cobalt(III)chloride- [Co(NH₃)₄(H₂O)₂]Cl₃

(b) Dibromobis(ethane-1,2-diamine) platinum(IV) nitrate -[PtBr₂(en)₂](NO₃)₂.

Section – C

Q.13. (a) Write the dispersed phase and dispersion medium of milk.

(b) Why is adsorption exothermic in nature?

Answer: (a) Dispersed phase of milk is liquid and the dispersion medium of milk is liquid.

(b) During the process of adsorption molecules of adsorbate and adsorbent come closer to form physical or chemical bonds hence getting stabilized, in this process heat is evolved leading the overall process to be exothermic.

It is an empirical relationship between the quantity of gas adsorbed by a unit mass of solid adsorbent and pressure at a particular temperature: x/m = kp_1/2(n > 1)

Where x is the mass of the gas adsorbed on mass m of the adsorbent at pressure p, k and n are constants which depend on the nature of the adsorbent and the gas at a particular temperature.

The relationship is generally represented in the form of a curve x/m is plotted against the pressure. This curve always approaches saturation towards high pressure, thus indicating that at adsorption through increases with increase in pressure till a limit and at high pressures, no further adsorption is observed.

Q.15. Write the name and principle of the method used for the refining of

(a) Tin,

(b) Copper,

Answer: (a) Tin: It is refined through liquidation. In this method, a low melting metal like tin is made to flow on a sloping surface, where the higher melting impurities are left behind and the lower melting metal is collected at the sloping end.

(b) Copper: It is refined through electrolytic refining. Anode is made of impure copper and pure copper stripes are taken as the cathode. They are dipped in acidified solution of copper sulphate, as an electrolyte. The net result of electrolysis is the transfer of copper in pure form from the anode to the cathode and the impurities get deposited as anode mud.

(c) Nickel: It is refined through Mond’s process. In this process, Nickel is heated in a stream of carbon monoxide forming a volatile complex, nickel tetracarbonyl.

The carbonyl is subjected to a higher temperature so that it is decomposed giving the pure metal.

Q.16. Give a reason for the following:

(a) Transition metals show variable oxidation states.

(b) E⁰ value of (Zn²⁺/Zn) is negative while that of (Cu²⁺/Cu) is positive.

Answer: (a) Transition metals show variable oxidation states because their d-orbitals are incompletely filled and different arrangements of electrons are possible according to the chemical environment of metal ions. hence, the ions can occupy variable oxidation states.

(b) E° value for Zn/Zn²⁺ is negative because the conversion of Zn to Zn²⁺ gives it a completely filled d⁵ configuration and extra stability gained by Zn²⁺. Whereas conversion of Cu to Cu²⁺ does not give any extra stability, hence it has a positive E⁰ value.

(c) Mn has the highest oxidation state of +4 with fluorine but with oxygen, it is +7. This is due to the ability of oxygen to form multiple bonds with the metal ion, whereas fluorine being of small size and devoid of d-orbitals can’t form multiple bonds.

Q.19. Write the structures of monomers used for getting the following polymers:

(a) Nylon-6, 6

(b) Bakelite

Q.19. (a) Write one example each of:

(i) Thermoplastic polymer

(ii) Elastomers

(b) Arrange the following polymers in the increasing order of their intermolecular forces:

Polythene, Nylon-6, 6, Buna-S

Answer: (a) Monomers of Nylon-6,6 are adipic acid and hexamethylenediamine.

(b) Monomers of bakelite are phenol and formaldehyde:

Answer: (a) (i) Example of thermoplastic polymer – polythene, polystyrene.

(ii) Example of elastomer – Neoprene.

(b) In increasing order of their intermolecular force, they can be arranged as:

Buna-S < Polythene < Nylon-6,6

(c) Strong intermolecular forces between the polymer molecules, such as hydrogen bonding lead to closed packed structure, thus imparting crystalline nature to the polymers.

Chemistry

Set-III

Section – A

Q.1. Out of Chlorobenzene and Cyclohexyl chloride, which one is more reactive towards nucleophilic substitution reaction and why?

Q.2. Arrange the following in decreasing order of solubility in water:

(C₂H₅)₂NH, C₂H₅NH₂, C₆H₅NH₂

Answer: Decreasing order of solubility in water is:

C₂H₅NH₂ > (C₂H₅)₂NH₂ > C₆H₅NH₂.

Q.3. What type of colloid is formed when a solid is dispersed in a gas? Give an example.

Answer: An aerosol is the type of colloid formed when solid is dispersed in gas.

Example: smoke and dust.

Q.5. What is the difference between amylose and amylopectin?

Q.5. Write the products obtained after hydrolysis of lactose.

Answer: Amylose is a long linear chain of α-D-(+)-glucose units joined by C₁-C₄ glycosidic linkage (α-link), whereas Amylopectin is a branched-chain polymer of α-D-glucose units, in which the chain is formed by C₁-C₄ glycosidic linkage and the branching occurs by C₁-C₆ glycosidic linkage.

Answer: Galactose and Glucose are the products obtained after hydrolysis of lactose.

Section – B

Q.7. Give reasons:

(a) An increase in temperature is observed on mixing chloroform and acetone.

(b) Aquatic animals are more comfortable in cold water than in warm water.

Answer: (a) A mixture of chloroform and acetone forms a solution with a negative deviation from Raoult’s law. This is because chloroform molecule is able to form a hydrogen bond with acetone molecule as shown by the following figure:

This decreases the escaping tendency of molecules for each component, and consequently, the vapour pressure decreases and the temperature of the solution is increased because of stability attained by the molecule by associating and releasing energy.

(b) The solubility of gases in liquids increases on decreasing temperature, hence cold water has more dissolved oxygen because of which aquatic species find themselves more comfortable in cold water as compared to hot water.

Q.10. Define the following terms with a suitable example of each:

(a) Polydentate ligand

(b) Homoleptic complex

Q.10. Using IUPAC norms, write the formula for the following complexes:

(a) Potassium tri (oxalato) chromate (III)

(b) Hexaaquamanganese (II) sulphate.

Answer: (a) Polydentate ligands: Ligands with several donor atoms are called polydentate ligands. These can bond with the metal ion in a complex with the different donor atoms present in them.

Example: N(CH₂CH₂NH₂)₃.

(b) Homoleptic complex: Complexes in which a metal atom is bound to only one kind of donor groups, e.g., [Co(NH₃)₆]²⁺ are known as homoleptic complex.

Answer: (a) K₃[Cr(C₂O₄)₃]

(b) [Mn (H₂O)₆] SO₄.

Q.12. Write structures of main compounds A and B in each of the following reactions:

Answer:

Section -C

Q.14. (a) Write the dispersed phase and dispersion medium of butter.

(b) Why does physisorption decrease with increase in temperature?

(c) A colloidal sol is prepared by the method given in the figure. What is the charge on AgI colloidal particles formed in the test tube? How is this sol represented?

Answer: (a) Butter is an example of ‘Gel’ type of colloid. Here the dispersed phase is liquid and dispersion medium is solid.

(b) Physisorption occurs because of physical attractive forces, like Vander Waals forces between molecules of adsorbate and adsorbent, hence that can be reversed on the application of bigger forces. Hence, when the temperature is increased, the movement of adsorbed molecules increases, resulting in disturbed attractive forces, detachment of adsorbed molecules from the adsorbent surface hence physisorption decreases.

(c) When the AgNO₃ solution is added to KI, silver iodide, AgI, is precipitated. The precipitated silver iodide adsorbs iodide ions from dispersion medium and negatively charged colloidal sol results. It can be shown as AgI/I_– (negatively charged).

Q.17. Write the principle of the following:

(a) Hydraulic washing

(b) Chromatography

Answer: (a) Hydraulic washing: This method of concentration of ores is based on the differences in gravity of the ore and gangue particles. It is a type of gravity separation. An upward stream of running water is used to wash the powdered ore. The lighter gangue particles are washed away and the heavier ores are left behind.

(b) Chromatography: Chromatography is a physical method of separation of a mixture in which the components to be separated are distributed between two phases, stationary and mobile phase. The stationary phase may be a solid or a liquid supported on a solid or a gel. The mobile phase may be either a liquid or a gas.

(c) Froth floatation process: Froth floatation is a physicochemical method of concentrating fine minerals. This process utilizes the difference in surface properties of valuable minerals and gangue (impurity) particles. For example, removal of gangue from sulphide ores.

Q.18. Give reasons for the following:

(a) Transition metals have high enthalpies of atomization.

(b) Manganese has a lower melting point even though it has a higher number of unpaired electrons for bonding.

Answer: (a) Transition element has high effective nuclear charge and a large number of valence electrons ((n-1) d electrons). So, as a result of the greater number of electrons participating, very strong metallic bonds are formed. As a result of the strong inter-atomic metallic bonding, the transition metals have high enthalpies of atomization.

(b) Manganese has a lower melting point even though it has a higher number of impaired electrons for bonding. Melting point depends on the intermolecular or interatomic forces. Stronger the forces, the higher the melting point. In Mn there is half-filled 3d subshell (3d⁵ configuration) which makes it stable and hence, it does not make additional covalent bonds with nearby atoms hence, it has less melting point.

(c) Ce⁴⁺ is a strong oxidising agent because Ce⁴⁺ oxidizes others and itself gets reduced to the common and preferred 3+ oxidation state of lanthanide elements.

Q.19. Write the structures of monomers used for getting the following polymers:

(a) Novolac

(b) Neoprene

Q.19. (a) Write on example each of

(i) Cross-linked polymer

(ii) Natural polymer

(b) Arrange the following in the increasing order of their intermolecular forces: Terylene, Buna-N, Polystyrene

Answer: (a) Novolac is the polymer of 2-hydroxymethyl phenol which is obtained by reaction of phenol and formaldehyde.

Answer:

(b) Increasing order of their molecular forces:

Buna-N < Polystyrene < Terylene

(c) Biodegradable polymer: These are synthetic polymers designed so as to contain functional groups similar to ones present in biopolymers. These are thus easily degraded by environmental degradation process hence, known as Biodegradable polymers.

Example: Poly (β-hydroxybutyrate-co-β-hydroxy valerate (PHBV).

Q.23. (a) Write the product when D-glucose reacts with Br₂aq.

Answer:

Biology 12th Previous Year Question Paper 2017 (CBSE)

Biology

SECTION – A

(Q. Nos. 1 – 5 are of one marks each)

Q.1. Name the type of cross that would help to find the genotype of a pea plant bearing violet flowers.

Answer: Test cross

Q.2. State two postulates of Oparin and Haldane with reference to the origin of life.

Answer: (i) First form of life could have come from pre-existing non-living organic molecules / RNA & Protein

(ii) Formation of life was preceded by chemical evolution / formation of diverse organic molecules from inorganic constituents.

Q.2. Bt -toxins are released as inactive crystals in the bacterial body. What happens to it in

the cotton boll worm body that it kills the boll worm.(SET-III)

Ans: It is converted into an active protein (due to alkaline pH of the gut of the boll worm) , the toxin binds to midgut cells / create pores / causes cell swelling and lysis that kills the bollworm.

Q.3. A herd of cattle is showing reduced fertility and productivity. Provide one reason and one suggestion to overcome this problem.

Answer: Reason: Inbreeding depression / continuous inbreeding.

Suggestion: Should be mated with unrelated superior cattle of the same breed / out – breeding / out – crossing.

Q.3. Name the specific type of gene that is incorporated in a cotton plant to protect the plant against cotton bollworm infestation. (SET-II)

Answer: cry I Ac / cry II Ab

Q.4. What are Cry genes ? In which organisms are they present ?

Ans. The genes which code for Bt toxin / Cry proteins / toxic proteins , Bacillus thuringiensis.

Q.5. An electrostatic precipitator in a thermal power plant is not able to generate high voltage of several thousands. Write the ecological implication because of it.

Answer: Air Pollution // particulate matter / dust particles released in the air.

SECTION – B

(Q Nos. 6-10 are of two marks each)

Q.6. A pollen grain in angiosperm at the time of dehiscence from an anther could be 2-celled or 3-celled. Explain. How are the cells placed within the pollen grain when shed at a 2-celled stage ?

Answer: • In 2-celled stage the mature pollen grain contains a generative and vegetative cell, whereas in 3- celled stage one vegetative cell and two male gametes are present.

• The generative cell floats in the cytoplasm of vegetative cell

Q.7. Differentiate between the genetic codes given below :

(a) Unambiguous and Universal

(b) Degenerate and Initiator

Answer:

(a) Unambiguous: One codon codes for only one amino acid.	Universal: Genetic code/codons are (nearly) the same for all organisms from bacteria to humans.
(b)Degenerate: More than one codon coding for the same amino acid.	Initiator: Start codon / AUG.

Q.8. Mention one application for each of the following :

(a) Passive immunization

(b) Antihistamine

(d) Cytokinin-barrier

Answer: (a) Provide preformed antibodies / antitoxins for quick response in case of infection by deadly microbes(tetanus) or snake bite.

(b) Reduces symptoms of allergy

(d) Protection of non-infected cells from further viral infection.

Q.9. Name the microbes that help production of the following products commercially:

(a) Statin

(b) Citric acid

(d) Butyric acid

Answer: (a) Monascus purpureus

(b) Aspergillus niger

(d) Clostridium butylicum

Q.10. List four benefits to human life by eliminating the use of CFCs.

Answer: (i) Delay in aging of skin

(ii) Prevent damage to skin cells

(iii) Prevent skin cancer

(iv) Prevent snow blindness / inflammation of cornea

(v) Prevent cataract (a) Unambiguous:

(vi) Prevents ozone depletion

(vii) Prevents global warming

(viii) Reduces greenhouse effect

(ix) Reduces odd climatic changes or El Nino effect

Q.10. Suggest two practices giving one example of each, that help protect rare or threatened species.

Ans: (1) In situ conservation , biodiversity hotspot / biosphere reserve / national parks /sanctuaries / Ramsar sites / sacred groves (Any one).

(2) Ex situ conservation , Zoological parks / botanical garden / wildlife safari parks /

cryopreservation techniques / Tissue culture / seed bank / pollen banks.

Q.6. Name the type of immunity the colostrum provides to a newborn baby. Write giving an example where this type of immunity should be provided to a person.

Answer: Passive Immunity.

In case of infection by deadly microbes(tetanus) / snake bite where quick immune response is required =1

Q.8. Write the binomials of two fungi and mention the products/bioactive molecules they help to produce.

Answer: Trichoderma polysporum , cyclosporin A

Aspergillus niger,citric acid

Monascus purpureus , statin

Saccharomyces cerevisiae, ethanol / alcohol

Penicillium notatum , Penicillin

Q.7. Give the binomials of two types of yeast and the commercial bioactive products they help to produce.

Ans: Saccharomyces cerevisiae- ethanol / alcohol Monascus purpureus- statin

Q.9. How many cells are present in the pollen grains at the time of their release from another ? Name the cells.

Ans. Pollen grain may be released at

2-celled stage , one vegetative and one generative cell ,

3-celled stage , one vegetative cell and two male gametes

Q.10. Name the group of cells the HIV enters after getting into the human body. What happens in these cells and what are these cells subsequently referred to as ? Name the next group of cells the HIV attacks from here.

Ans. Macrophages , Reverse transcription , HIV Factory , helper T-lymphocytes (TH)

SECTION – C

(Q Nos. 11-22 are of three marks each)

Q.11. (a) Can a plant flowering in Mumbai be pollinated by pollen grains of the same species growing in New Delhi ? Provide explanations to your answer.

(b) Draw the diagram of a pistil where pollination has successfully occurred. Label the parts involved in reaching the male gametes to its desired destination.

Answer: (a) Yes, By artificial means ( any relevant explanation).

(b) Diagram with following labellings Stigma , Pollen tube , Synergid / Filiform Apparatus , Micropyle.

Q.12. Both Haemophilia and Thalassemia are blood related disorders in humans. Write their causes and the difference between the two. Name the category of genetic disorder they both come under.

Answer:

Haemophilia	Thalassemia
Single protein involved in the clotting of blood is affected	Defects in the synthesis of globin leading to formation of abnormal haemoglobin.
Sex linked reccessive disorder.	Autosomal recessive disorder.
Blood does not clot.	Results in anaemia.

Q.13. (a) List the two methodologies which were involved in human genome project. Mention how they were used.

(b) Expand ‘YAC’ and mention what was it used for.

Answer: (a) Expressed Sequence Tags , Identifying all the genes that are expressed as RNA Sequence Annotation , sequencing the whole set of genome coding or non coding sequences and later assigning different region with functions.

(b) Yeast Artificial Chromosome , used as cloning vectors (cloning / amplification ).

Q.14. Write the characteristics of Ramapithecus , Dryopithecus and Neanderthal man.

Answer: Ramapithecus: hairy/ walked like gorillas and chimpanzees , more man like.

Dryopithecus: hairy/ walked like gorillas and chimpanzees , more ape- like.

Neanderthal man: brain size is 1400cc , used hides to protect their body / buried their dead.

Q.15. Name a human disease, its causal organism, symptoms (any three) and vector, spread by intake of water and food contaminated by human faecal- matter.

Answer: Amoebiasis (Amoebic dysentery) , Entamoeba histolytica , constipation / abdominal pain / cramps / stools with excess mucus / blood clots (Any three symptoms) , Housefly.

Ascariasis, Ascaris , internal bleeding / muscular pain / fever / anaemia / blockage of intestinal passage (Any three symptoms), Housefly.

Typhoid, Salmonella typhi, high fever / weakness / stomach pain / constipation / headache / loss of appetite (Any three symptoms), Housefly.

Q.15. (a) Why is there a fear amongst the guardians that their adolescent wards may get

trapped in drug/alcohol abuse ?

(b) Explain ‘addiction’ and ‘dependence’ in respect of drug/alcohol abuse in youth.

Defects in the synthesis of globin leading to formation of abnormal haemeoglobin Sex linked recessive disorder.

Autosomal recessive disorder Blood does not clot Results in anaemia

Answer: (a) Adolescents are easily affected by ( vulnerable to) peer pressure /adventure /curiosity / excitement / experimentation / media .

(b) Addiction -Psychological attachment to certain effects such as Euphoria / temporary feeling of well-being. Dependence-Tendency of the body to show withdrawal syndrome / symptoms if regular doses of drug / alcohol is abruptly discontinued.

Q.16. (a) Write the desirable characters a farmer looks for in his sugarcane crop.

(b) How did plant breeding techniques help north Indian farmers to develop cane with

desired characters ?

Answer: (a) High yield , thick stem,high sugar content , ability to grow in their areas.

(b) By crossing Saccharum officinarum / south Indian variety having desired characteristics with Saccharum barberi / north Indian low yield variety.

Q.17. Secondary treatment of the sewage is also called Biological treatment. Justify this statement and explain the process.

Answer: Involves biological organism such as aerobic and anaerobic microbes / bacteria and fungi to digest / consume organic waste.

Primary effluent is passed into aeration tank where vigorous growth of aerobic microbes (flocs) take place, BOD reduced (microbes consume major part of organic matter), effluent is passed to settling tank where flocs sediment to produce activated sludge , sludge is pumped to anaerobic sludge digester to digest bacteria and fungi.

Q.18. (a) Explain the significance of ‘palindromic nucleotide sequence’ in the formation of recombinant DNA.

(b) Write the use of restriction endonuclease in the above process.

Answer: (a) Palindromic nucleotide sequence is the recognition (specific) sequence present both on the vector and on a desired / alien DNA for the action of the same(specific) restriction endonuclease to act upon.

(b) Same restriction endonuclease binds to both the vector and the foreign DNA , cut each of the two strands of the double helix at specific points in their sugar phosphate backbone of recognition sequence for restriction endonucleases / palindromic sequence of vector and foreign DNA , to cut strand a little away from the centre of the palindrome sites, creates overhanging stretches /sticky ends .

(b) If depict diagrammatically showing the above mentioned value points it can be accepted.

Q.19. Describe the roles of heat, primers and the bacterium Thermus aquaticus in the process of PCR.

Answer: Heat – Denaturation / separation of DNA into two strands.

Primer- Enzyme DNA Polymerase extend the primers using the nucleotides provided in the reaction and the genomic DNA as template.

Thermus aquaticus – source of thermostable DNA polymerase / Taq polymerase.

Q.20. Explain the various steps involved in the production of artificial insulin.

Answer: Two DNA sequences corresponding to A and B polypeptide chains of human insulin were prepared , these were introduced into E.coli to produce A and B chains separately , these chains were extracted and combined by creating disulphide bonds.

Q.21. (a) “Organisms may be conformers or regulators.” Explain this statement and give one example of each.

(b) Why are there more conformers than regulators in the animal world ?

Answer: (a) Conformers- organisms which cannot maintain a constant internal environment under varying external environmental conditions / change body temperature and osmotic concentration with change in external environment eg. all plants / fishes / amphibians / reptiles.

Regulators – organisms which can maintain homeostasis (by physiological means or behavioural means ) // maintain constant body temperature and osmotic concentration eg. birds /mammals.

(b) Thermoregulation is energetically expensive for animals.

Q.22. Describe the inter-relationship between productivity, gross primary productivity and net productivity.

Answer: Productivity is the rate of biomass production per unit area over a period of time ,

Gross primary productivity is the rate of production of organic matter during photosynthesis in an ecosystem , Net productivity is the gross primary productivity minus respiration losses ®.

Q.13. Explain the process of pollination in Vallisneria. How is it different in water-lily, which is also an aquatic plant ?

Answe: In Vallisneria pollination takes place through water , the female flower reach the surface of water by long stalk , male flowers / pollen grain released on to the surface of water , carried passively by water current reaching the female flowers / stigma.

In Water lily pollination takes place through wind or insect , female flower emerges above the surface of water and gets pollinated

Q.15. What is disturbance in Hardy-Weinberg genetic equilibrium indicative of ? Explain how it is caused.

Answer: Disturbance in Hardy-Weinberg equilibrium is an indicator of change of frequency of alleles in a population , resulting in evolution.

It is caused by genetic drift / gene flow or gene migration / mutation / genetic recombination /natural selection.

Q.18. Different animals respond to changes in their surroundings in different ways. Taking one example each, explain “some animals undergo aestivation while some others hiberna

tion”. How do fungi respond to adverse climatic conditions ?

Answer: Some animals go into aestivation to avoid summer related problems ( heat and desiccation) , eg. snails / fish ( any other suitable eg.)

Some animals go into hibernation to avoid winter related problem ( extreme cold) eg. bear ( any other suitable eg.).

Fungi form thick walled spores and suspend their activities to respond to adverse climatic condition.

Q.11. Rearrange Ramapithecus, Australopithecus and Homo habilis in the order of thein evolution on the Earth. Comment on their evolutionary characteristics.

Answer: Ramapithecus Australopithecus Homo habilis.

Ramapithecus – hairy / walked like gorilla and chimpanzees / more man like.

Australopithecus – Hunted with stone weapons / ate fruit.

Homo habilis -Brain capacity 650- 800 cc / probably did not eat meat.

Q.16. (a) Trace the development of an endosperm after fertilisation with reference to coconut. Mention the importance of endosperm development.

(b) Write the importance of ‘pollen bank’.

Answer: (a) In coconut Primary Endosperm Nucleus (PEN-3n) undergoes successive nuclear divisions , give rise to free- nuclear endosperm known as coconut water , white kernel is the cellular endosperm , provides nourishment to the growing embryo.

(b) Storage / cryopreservation ( storage in liquid nitrogen at – 196 o C) , to use in crop breeding programmes.

Q.20. Describe the inter-relationship , between productivity, gross primary productivity and net productivity.

Answer: Productivity is the rate of biomass production per unit area over a period of time ,

Gross primary productivity is the rate of production of organic matter during photosynthesis in an ecosystem ,Net productivity is the gross primary productivity minus respiration losses ®.

Q.22. How do kangaroo rats and desert plants adapt themselves to survive in their extreme habitat ? Explain.

Answer: Kangaroo rats- internal fat oxidation where water is a byproduct , excretes concentrated urine.

Desert Plants -thick cuticle / sunken stomata / leaves reduced to spines / deep roots /

Special photosynthetic pathway / CAM.

SECTION – D

(Q No. 23 is of four mark)

Q.23 . It is commonly observed that parents feel embarrassed to discuss freely with their adoles- cent children about sexuality and reproduction. The result of this parental inhibition is that the children go astray sometimes.

(a) Explain the reasons that you feel are behind such embarrassment amongst some

parents to freely discuss such issues with their growing children.

(b) By taking one example of a local plant and animal, how would you help these parents

to overcome such inhibitions about reproduction and sexuality ?

Ans: (a) Illiteracy / conservative attitude / misconceptions / social myths / any other relevant point (Any two).

(b) If a student gives the clarity of the concept of reproduction and sexuality by taking any example of a plant and an animal with respect to reproductive organs, gamete formation, fertilization, sexual behaviour etc.

SECTION – E

(Q Nos. 24-26 are of five marks each)

Q.24. (a) When a seed of an orange is squeezed, many embryos, instead of one are observed.

Explain how it is possible.

(b) Are these embryos genetically similar or different ? Comment.

Ans: (a) Polyembryony , nucellar cells surrounding embryo sac start dividing , protrude into the embryo

sac and develop into many embryos.

(b) These embryos are genetically similar, as produced from nucellar cells by mitotic division / formed without fertilisation (but different from the embryo formed by fertilization).

Q.24. (a) Explain the following phases in the menstrual cycle of a human female:

(i) Menstrual phase

(ii) Follicular phase

(iii) Luteal phase

(b) A proper understanding of menstrual cycle can help immensely in family planning.

Do you agree with the statement ? Provide reasons for your answer.

Ans: (a) (i) Menstrual phase – first 3-5 days of the cycle where menstrual flow occurs due to break down of endometrial lining of the uterus, if the released ovum is not fertilised.

(ii) Follicular phase – from 5^thto 14^thday of the cycle where the primary follicles grow to become a fully mature Graafian follicle , endometrium of uterus regenerates , Graafian follicle ruptures to release ova (ovulation on 14^thday).

(iii) Luteal Phase – During 15^thto 28^thday remaining parts of graafian follicle transforms into the corpus luteum , secretion of progesterone (essential for maintenance of endometrium).

All these phases are under the influence of varying concentrations of pituitary and ovarian Hormone.

(b) Yes , can take appropriate precautions between 10^thto 17^thday of the menstrual cycle when the chances of fertilisation are high.

Q.25. (a) Compare, giving reasons, the J-shaped and S-shaped models of population growth of a species.

(b) Explain “fitness of a species” as mentioned by Darwin.

Answer:

J shaped – growth curve	S shaped- growth curve
Resources are unlimited	Resources are limited
Growth is exponential	Logistic Growth
As resources are unlimited all individuals survive and reproduce	Fittest individual will survive and reproduce
Growth Equation dN/dt=Rn (If explained)	Growth Equation dN/dt=rN (k-N/K) (If explained)

Note – Marks to be awarded only if the corresponding difference is written.

(b) When resources are limited , Competition occurs between individuals , the fittest will survive,

who reproduce to leave more progeny.

Q.25. (a) What is an ecological pyramid ? Compare the pyramids of energy, biomass and numbers.

(b) Write any two limitations of ecological pyramids.

Ans: (a) Graphical representation of the relationships among organisms at different trophic levels.

(b) It does not accommodate the food web / does not take into account the same species belonging to two or more trophic levels , Saprophytes are not given any place.

Q.26. (a) Describe the structure and function of a t-RNA molecule. Why is it referred to as an adapter molecule?

(b) Explain the process of splicing of hn-RNA in a eukaryotic cell.

Ans: (a) Clover-leaf shaped / inverted L shaped molecules has an anticodon loop with bases complementary to specific codon , has an amino acid acceptor end = 1+1

As it reads the code on one hand and binds with the specific amino acid on the other hand.

(b) Introns are removed and exons are joined in a definite order.

Process of splicing shown diagramatically.

Q.26. Growth Equation dN/dt=rN (k-N/K) (If explained) Pyramid of Energy Pyramid of BioMass Pyramid of Numbers Pyramid of Numbers Shows transfer of Energy from shows numbers of one trophic level to another

organisms at each trophic level. Always upright Mostly upright but can be inverted Mostly upright can be

inverted Shows transfer of amount of food/ biomass from one trophic level to another. Write the different components of a lac-operon in E.coli. Explain its expression while in an ‘open’ state.

Ans: It consists of one regulatory gene(i) , promoter gene , operator gene , and three structural genes(z,y,a).

Lactose/ inducer binds to the repressor protein , makes it inactive so it cannot bind with operator, allows RNA Polymerase access to the promoter and transcription proceeds ,β -galactosidase , permease , transacetylase formed (by translation process for Lactose metabolism).

Q.25. (a) Explain Polygenic inheritance and Multiple allelism with the help of suitable examples.

(b) “Phenylketonuria is a good example that explains Pleiotropy.” Justify.

Answer: (a) Traits that are generally controlled by three or more genes , the phenotype reflects the contribution of each allele / effect of each allele is additive.

eg. Human skin colour , controlled by three genes (A , B, C).

In multiple allelism more than two alleles , govern the same character / phenotype.

eg . Human blood group (ABO system) , controlled by three different alleles (I^A, I^B, i).

(b) In pleiotropy a single gene can exhibit multiple phenotypic expressions , in phenylketonuria single mutated gene express mental retardation and reduction in hair and skin pigmentation

Q.25. (a) What is an operon ?

(b) Explain how a polycistronic structural gene is regulated by a common promoter and a

combination of regulatory genes in a lac-operon.

Answer: (a) An operon is a polycistronic structural gene which is regulated by a common promoter and regulator

gene / transcriptionally regulated system in which polycistronic structural gene is controlled by a common promoter and regulator gene.

(b)

Lac operon consist of one regulatory gene i which codes for the repressor protein , promoter (P) and operator (o) are adjacent to gene i.
Structural genes z, y, a code for enzymes (â-galactosidase , permease and transacetylase respectively).
The regulator gene i synthesizes the repressor protein (all the time) , in the absence of inducer , the repressor protein binds to the operator region of the operon , prevents transcription (by RNA polymerase).
The repressor is inactivated in the presence of an inducer (lactose) that binds with it , this allows RNA polymerase access to promoter and transcription proceeds.

Q.24. (a) A pea plant bearing axial flowers is crossed with a pea plant bearing terminal flowers. The cross is carried out to find the genotype of the pea plant bearing axial flowers. Work out the cross to show the conclusions you arrive at.

(b) State the Mendel’s law of inheritance that is universally acceptable.

Answer: (i) If the plants is homozygous for the dominant trait

(ii) If the plants is heterozygous for the dominant trait A A a a (All plants with Axial Flower) A a a (50 % plants are with Axial ?ower and 50% plants with terminal ?ower) a

Conclusion : If all progeny show axial flowers ( dominant) the plant is homozygous (AA) ,

If 50 % of Progeny show Axial flower ( Dominant) and 50% Terminal flower ( Recessive) the plant is heterozygous.

(b) Law of Segregation , allelic pair segregate (separates) during gamete formation ( do not loose their identity ).

Q.24. (a) Absence of lactose in the culture medium affects the expression of a Lac-operon in E. coli. Why and how ? Explain.

(b) Write any two ways in which the gene expression is regulated in eukaryotes.

Answer: (a) • Lactose acts as inducer thus absence of lactose switches off the operon.

• Repressor protein produced by regulatory gene ( i-gene ) is free ( in the absence of inducer ) ,

• Repressor protein binds with the operator gene ( o-gene ) ,

• Preventing RNA polymerase to transcribe the structural gene and operon is switched off.

If the above mentioned points are properly represented with help of schematic diagram.

(b) • Transcriptional level ( formation of primary transcripts )

• Processing level ( regulation of splicing )

• Transport of messengar RNA from nucleus to the cytoplasm

• Translational level

Q.25. (a) When a seed of an orange is squeezed, many embryos, instead of one are observed. Explain how it is possible.

(b) Are these embryos genetically similar or different ? Comment.

Ans: (a) Polyembryony , nucellar cells surrounding embryosac start dividing , protrude into the embryosac and develop into many embryos = 1+ 1+ 1

(b) These embryos are genetically similar, as produced from nucellar cells by mitotic division / formed without fertilisation (but different from the embryo formed by fertilization)

Q.25. (a) Explain the following phases in the menstrual cycle of” a human female:

(i) Menstrual phase

(ii) Follicular phase

(iii) Luteal phase

(b) A proper understanding of menstrual cycle can help immensely in family planning.

Do you agree with the statement ? Provide reasons for your answer.

Ans: (a) (i) Menstrual phase – first 3-5 days of the cycle where menstrual flow occurs due to break down of endometrial lining of uterus, if the released ovum is not fertilised.

(ii) Follicular phase – from 5th to 14th day of the cycle where the primary follicles grow to become a fully mature Graafian follicle , endometrium of uterus regenerates , Graafian follicle ruptures to release ova (ovulation on 14th day)

(iii) Luteal Phase – During 15th to 28th day remaining parts of graafian follicle transform into corpus luteum , secretion of progesterone (essential for maintenance of endometrium)

All these phases are under the influence of varying concentrations of pituitary and ovarian hormone.

(b) Yes , can take appropriate precautions between 10th to 17th day of the menstrual cycle when the chances of fertilisation are high.

Biology 12th Previous Year Question Paper 2018 (CBSE)

Biology

SECTION-A

(Q. Nos. 1 – 5 are of one mark each)

Q.1. Write the dual purpose served by Deoxyribonucleoside triphosphates in polymerisation.

Ans. Acts as a substrate , provide energy (from the terminal two phosphates).

Q.2. Name two diseases whose spread can be controlled by the eradication of Aedes

mosquitoes.

Ans. Dengue , Chikungunya // Yellow Fever / Eastern Equine Encephalitis / West Nile Fever / Zika / Zika Virus Disease (Any two)..

Q.3. How do cytokine barriers provide innate immunity in humans ?

Ans. Interferon (proteins) , secreted by virus infected cells (protect non – infected cells from further viral infection).

Q.4. Write the names of the following :

(a) A 15 mya primate that was ape-like

(b) A 2 mya primate that lived in East African grasslands

Ans. (a) Dryopithecus.

(b) Australopithecines / Australopithecus / Homo habilis.

Q.5. Mention the chemical change that proinsulin undergoes, to be able to act as mature

insulin.

Ans. Removal of C – peptide (from proinsulin)

SECTION-B

(Q. Nos. 2 – 10 are of two marks each)

Q.6. Your advice is sought to improve the nitrogen content of the soil to be used for the cultivation of a non-leguminous terrestrial crop.

(a) Recommend two microbes that can enrich the soil with nitrogen.

(b) Why do leguminous crops not require such enrichment of the soil ?

Ans. (a) Azospirillum / Azotobacter / Anabaena / Nostoc / Oscillatoria / Frankia (Any

two correct names of microbes)..

(If cyanobacteria mentioned. , but if along with cyanobacteria Anabaena / Nostoc / Oscillatoria mentioned then No mark on cyanobacteria)

(b) They can fix atmospheric nitrogen , due to the presence of Rhizobium / N₂fixing bacteria

in their root nodules..

Q.7. With the help of an algebraic equation, how did Hardy-Weinberg explain that in a given population the frequency of occurrence of alleles of a gene is supposed to remain the same through generations ?

Ans. In a population of diploid organisms

If the frequency of allele A = p and frequency of allele a = q.

Expected genotype frequency under random mating are

AA = p2 (for the AA homozygotes)

aa = q2 (for the aa homozygotes)

Aa = 2pq (for the Aa heterozygotes).

(In absence of selection , mutation , genetic drift or other forces allelic frequency p and q are constant through generations)

Therefore p²+ 2pq +q²= 1 = 1

Q.7. Although a prokaryotic cell has no defined nucleus , yet DNA is not scattered throughout the cell. Explain.

Ans. DNA is negatively charged , positively charged proteins , hold it in places , in large loops (in a region termed as nucleoid). × 4

Q.8. How did a citizen group called Friends of the Arcata Marsh, Arcata, California, USA, help to improve water quality of the marshland using Integrated Waste Water Treatment ? Explain in four steps.

Ans.- Water is treated by conventional method / sedimentation / filtration / chlorination

– Water flows to six connected marshes

– The water in marshes is seeded with appropriate plants / algae / fungi /

bacteria

– Which helps to neutralise the pollutants / assimilate the pollutants / absorb pollutants / Remove heavy metals. × 4

Q.9. You have obtained a high yielding variety of tomatoes. Name and explain the procedure that ensures retention of the desired characteristics repeatedly in large populations of future generations of the tomato crop.

Ans. – Tissue culture / micropropagation / somaclonal propagation / apomixis.

– Explant / any part of plant taken out and grown (in a test tube / vessel) ,

– under sterile conditions ,

– in special nutrient medium (containing carbon source / sucrose , inorganic salt

vitamins / amino acids and growth regulator). × 3

Q.10. (a) Name the source plant of heroin drug. How is it obtained from the plant ?

(b) Write the effects of heroin on the human body.

Ans. (a) – Papaver somniferum / Poppy plant.

– Extracted from latex of the plant / acetylation of morphine (obtained from the

latex of plant).

(b) Depressant , slows down body function..

SECTION-C

(Q. Nos. 11 – 22 are of three marks each)

Q.11. Draw a diagram of a mature human sperm. Label any three parts and write their

functions.

(Any three labelling).

Plasma membrane – Envelope of the sperm

Acrosome – Filled with enzyme that help fertilization of ovum

Mitochondria – Energy source for swimming

Middle Piece – Possess mitochondria which is the energy source for swimming

Tail – For movement of sperm

Nucleus – Containing chromosomal material

(Functions of the parts labelled ). × 3

Q.12. (a) Expand VNTR and describe its role in DNA fingerprinting.

(b) List any two applications of DNA fingerprinting technique.

Ans. (a) VNTR – Variable Number of Tandem Repeat(s).

– used as a probe (because of its high degree of polymorphism).

(b) Forensic science / criminal investigation (any point related to forensic science) / determine population and genetic diversities / paternity testing / maternity testing / study of evolutionary biology (Any two).

Q.13. Differentiate between Parthenocarpy and Parthenogenesis. Give one example of each.

Ans.

Parthenocarpy	Parthenogenesis
– Formation of fruit without fertilization	– New organism develops without fertilization
– e.g. banana / grapes / any other correct example.	– e.g. Drones /male honey bee / turkey / rotifers / some lizards / any other correct example.

Q.14. Medically it is advised to all young mothers that breastfeeding is the best for their

newborn babies. Do you agree ? Give reasons in support of your answer.

Ans.Yes,

provides nutrition (calcium , fats , lactose ) / provides (passive) immunity / provides antibodies / Ig A.

Q.15. Explain the mechanism of ‘sex determination’ in birds. How does it differ from that of

human beings ?

Ans. In birds ;

Birds : female heterogamety / female produces (Z) type and (W) type of gametes.

Humans : male heterogamety / male produces (X) and (Y) type of gametes.

Q.16. (a) How has the development of bioreactor helped in biotechnology ?

(b) Name the most commonly used bioreactor and describe its working.

Ans. (a) Larger biomass / large volume of culture can be processed leading to higher yields of desired specific products (protein / enzymes) , under controlled conditions..

(b) Stirring type.

– Mixing of reactor contents evenly (with agitator system or a stirrer).

– Facilitates oxygen availability.

– Temperature / pH / foam control // under optimum conditions.

Q.17. Explain the roles of the following with the help of an example each in recombinant

DNA technology:

(a) Restriction Enzymes

(b) Plasmids

Ans. (a) It recognises a specific sequence of base pairs / palindromes, and cuts the DNA

strand at a specific site..

eg. EcoRI / Hindiii or any other correct example.

(b) Act as vectors / cloning of desired alien gene / foreign gene = 1

eg. pBR322 / plasmid of Salmonella / plasmid of Agrobacterium / Ti Plasmid / Tumour inducing Plasmid.

Q.18. Explain out-breeding, out-crossing and crossbreeding practices in animal husbandry.

Out breeding – Breeding of unrelated animals (which may be between individual of

same breed or between individuals of different species) = 1

Outcrossing – (a kind of out breeding) Mating of animals within the same breed but having no common ancestors on either side of their pedigree upto 4 – 6 generations = 1

Cross breeding – (another type of out breeding) Superior males of one breed are mated

with superior females of another breed = 1

Q.19. (a) Organic farmers prefer biological control of diseases and pests to the use of

chemicals for the same purpose. Justify.

(b) Give an example of a bacterium, a fungus and an insect that are used as

biocontrol agents.

Ans. (a) – Reduces dependence on toxic chemicals

– Protects our ecosystem or environment

– Protects and conserves non-target organisms / they are species – specific

– These chemicals being non-biodegradable may pollute the environment

permanently

– These chemicals being non-biodegradable may cause biomagnification

(b) Bacteria – Bacillus thuringiensis.

Fungus – Trichoderma.

Insect – Ladybird / Dragonfly / Moth or any other correct example.

Q.20. (a) Differentiate between analogous and homologous structures.

(b) Select and write analogous structures from the list given below :

(i) Wings of butterflies and birds

(ii) Vertebrate hearts

(iii) Tendrils of bougainvillea and cucurbita

(iv) Tubers of sweet potato and potato

Ans. (a) Analogous – Anatomically not similar though perform similar functions / are a

result of convergent evolution = 1

Homologous – Anatomically similar (but perform different functions) / are a result

of divergent evolution = 1

(b) Option (i) Wings of butterflies and birds / (iv) Tubers of sweet potato and potato

Q.21. (a) “India has greater ecosystem diversity than Norway.” Do you agree with the

statement ? Give reasons in support of your answer.

(b) Write the difference between genetic biodiversity and species biodiversity

that exists at all levels of biological organisation.

Ans. (a) Yes.

India / tropical region Norway / temperate region

– are less seasonal – more seasonal /

/ more constant / more predictable / less constant / less predictable

– promote niche specialisation – do not promote niche specialisation

leading to greater biodiversity leading to low biodiversity

– Species diversity increases as we – Species diversity decreases as we

move towards equator move away from equator

– More number of species exist – Less number of species exist

(b) Genetic diversity – Diversity / variation within a species over its distributional range / same explained with the help of a correct example = 1

Species diversity – Diversity / variation at a species level / same explained with

the help of a correct example = 1

Q.21. Explain the effect on the characteristics of a river when urban sewage is discharged into it.

Ans. – Rise in organic matter , leads to increased microbial activity / growth of microbes.

– It results in a decrease in dissolved oxygen / rise in BOD / rise in Biochemical Oxygen

Demand = 1

– Leads to fish mortality / algal bloom / colour change / foul odour / increase in

toxicity.

Q.22. How has the use of Agrobacterium as vectors helped in controlling Meloidogyne

incognitia infestation in tobacco plants ? Explain in correct sequence.

Ans. – Using Agrobacterium vector nematode specific genes introduced into host plant

– Sense and antisense strands of mRNA are produced

– ds RNA is formed

– ds RNA initiates RNAi

– Prevents translation of mRNA / silencing of mRNA of parasite / nematode

– Parasite will not survive

SECTION-D

(Q. Nos. 23 is of four marks)

Q.23. Looking at the deteriorating air quality because of air pollution in many cities of the country, the citizens are very much worried and concerned about their health. The doctors have declared health emergency in the cities where the air quality is very severely poor.

(a) Mention any two major causes of air pollution.

(b) Write any two harmful effects of air pollution to plants and humans.

(c) As a captain of your school Eco-club, suggest any two programmes you would plan to organise in the school so as to bring awareness among the students on how to check air pollution in and around the school.

Ans. (a) Vehicular discharge / smoke from industries / burning of agricultural wastes / smoke from incinerator / dust / smoke from thermal plants or any other correct cause

(b) Reduces growth of plants / reduces yields of crops / premature death of plants / respiratory problems / acid rain / any other relevant point (Any two – one from plant and one from human)

/ rallies / debates or any other

SECTION-E

(Q. Nos. 24 – 26 are of five marks each)

Q.24. (a) Describe any two devices in a flowering plant which prevent both autogamy and

geitonogamy.

(b) Explain the events upto double fertilisation after the pollen tube enters one of

the synergids in an ovule of an angiosperm.

Ans. (a) – Dioecy / production of unisexual flowers (in different plants)

– Self incompatibility.

(b) – Pollen tube releases 2 male gametes in the cytoplasm of synergid

– One male gamete fuses with egg cell / syngamy , resulting in diploid zygote

– Other male gamete fuses with polar nuclei / triple fusion , to form triploid PEN (Primary Endosperm Nucleus) / PEC (Primary Endosperm Cell).

Q.24. (a) Explain menstrual cycle in human females.

(b) How can the scientific understanding of the menstrual cycle of human females

help as a contraceptive measure ?

Ans. (a) – Menstrual Phase – Menstrual flow occurs / due to breakdown of endometrial

lining of uterus , when fertilization does not occur

– Follicular Phase – Primary follicles grow into mature graafian follicles and endometrium regenerates through proliferation , changes induced by pituitary and ovarian hormones

– Ovulatory Phase – LH surge , induces rupture of graafian follicle and release

of secondary oocyte / ovum during middle of cycle (i.e. 14th day)

– Luteal phase – Ruptured graafian follicle transforms into the corpus luteum which secrete large amounts of progesterone , essential for maintaining endometrium.

(b) Because ovulation occurs during mid cycle chances of fertilisation are very high

so , couples should abstain from coitus between day 10 – 17.

Q.25. (a) Write the scientific name of the organism Thomas Hunt Morgan and his colleagues worked with for their experiments. Explain the correlation between linkage and recombination with respect to genes as studied by them.

(b) How did Sturtevant explain gene mapping while working with Morgan ?

Ans. (a) Drosophila melanogaster.

They observed that two genes (located closely on a chromosome) did not segregate independently of each other (F₂ratio deviated significantly from 9 : 3 : 3 : 1).

Tightly linked genes tend to show fewer (less) recombinant frequency of parental traits / show higher (more) frequency of parental type.

Loosely linked genes show higher percentage (more) of recombinant frequency of parental traits / lower frequency percentage of parental type.

Genes present on the same chromosome are said to be linked and the recombinant frequency depends on their relative distance on the chromosome.

(b) He used the frequency of recombination between gene pairs on the same chromosome , as a measure of the distance between genes and mapped their position on the chromosome

Q.25. (a) State the ‘Central dogma’ as proposed by Francis Crick. Are there any

exceptions to it ? Support your answer with a reason and an example.

(b) Explain how the biochemical characterisation (nature) of ‘Transforming Principle’ was determined, which was not defined from Griffith’s experiments.

Ans. (a)

Yes, in some viruses flow of information is in reverse direction/reverse transcription.

(b) Protein and DNA and RNA were purified from heat killed S strain / smooth

Streptococcus / Diplococcus pneumoniae.

Protein + Protease → transformation occurred (R cell to S type).

RNA + RNA base → transformation occurred (R cell to S type).

DNA + DNAse → transformation inhibited.

Hence DNA alone is the transforming material.

Q.26. (a) Following are the responses of different animals to various abiotic factors.

Describe each one with the help of an example.

(i) Regulate

(ii) Conform

(iii) Migrate

(iv) Suspend

(b) If 8 individuals in a population of 80 butterflies die in a week, calculate the death rate of population of butterflies during that period.

Ans.

(a) (i) Regulate –	Maintain constant internal temperature / osmotic concentration /homeostasis. e.g. birds / mammals.
(ii) Conform –	Do not maintain constant internal temperature / osmotic concentration / No homeostasis. e.g. any one example of animals other than birds and mammals.
(iii) Migrate –	Temporary movement of organisms from the stressful of habitats to hospitable areas and return when stressful period is over. e.g. birds from Siberia / or any other correct example.
(iv) Suspend –	Reducing / minimising the metabolic activities during unfavourable conditions. e.g. Polar bear / amphibian / snails / fish / any other examples of animals.

(b) Death rate = 8/80 = 0.1, individuals per butterfly per week.

Q.26. (a) What is a trophic level in an ecosystem ? What is ‘standing crop’ with reference to it ?

(b) Explain the role of the ‘first trophic level’ in an ecosystem.

natural ecosystem ?

Ans. (a) Specific place of an organism in a food chain , mass of living material (biomass) at each trophic level at a particular time.

(b) First trophic level has producers / autotrophs , which trap solar energy / to produce

food (photosynthesis).

(c) Organisms of the Detritus food chain (DFC) are the prey to the Grazing food chain (GFC) organism , the dead remains of GFC are decomposed into simple inorganic materials which are absorbed by DFC organisms.

Maths 10th Previous Year Question Paper 2018 (CBSE)

Maths

Section – A

Q.1. If x = 3 is one root of the quadratic equation x² – 2kx – 6 = 0, then find the value of k.

Solution:

Given quadratic equation is, x² – 2kx – 6 = 0

x = 3 is a root of above equation, then

(3)² – 2k (3) – 6 = 0

⇒ 9 – 6k – 6 = 0

⇒ 3 – 6k = 0

⇒ 3 = 6k

⇒ k = ³/₆

⇒ k =½

Q.2. What is the HCF of the smallest prime number and the smallest composite number?

Solution:

Smallest prime number = 2

Smallest composite number = 4

Prime factorisation of 2 is 1 × 2

Prime factorisation of 4 is 1 × 2²

HCF (2, 4) = 2

Q.3.Find the distance of a point P(x, y) from the origin.

Solution:

The given point is P (x, y).

The origin is O (0, 0)

The distance of point P from the origin,

Q.4. In an AP if the common difference (d) = -4 and the seventh term (a⁷) is 4, then find the first term.

Solution:

Given,

d = -4, a⁷ = 4

a + 6d = 4

⇒ a + 6(-4) = 4

⇒ a – 24 = 4

⇒ a = 4 + 24

⇒ a = 28

Q.5. What is the value of (cos² 67° – sin² 23°) ?

Solution:

We have, cos² 67° – sin² 23°

= cos² 67° – cos² (90° – 23°) [∵ sin (90° – θ) = cos θ]

= cos² 67° – cos² 67°

= 0

Q.6. Given ΔABC ~ ΔPQR, if ^AB/_PQ=¹/₃ then find ^arΔABC/_arΔPQR

Solution:

Section – B

Q.7.Given that √2 is irrational, prove that (5 + 3√2) is an irrational number.

Solution:

Given, √2 is an irrational number.

Let √2 = m

Suppose, 5 + 3√2 is a rational number.

But ^a-5b/_3b is rational number, so m is rational number which contradicts the fact that m = √2 is irrational number.

So, our supposition is wrong.

Hence, 5 + 3√2 is also irrational.

Hence Proved.

Q.8.In fig. 1, ABCD is a rectangle. Find the values of x and y.

Solution:

Given, ABCD is a rectangle.

AB = CD

⇒ 30 = x + y

⇒ x + y = 30 …(i)

Similarly, AD = BC

⇒ 14 = x – y

⇒ x – y = 14 …(ii)

On adding eq. (i) and (ii), we get

2x = 44

⇒ x = 22

Putting the value of x in eq. (i), we get

22 + y = 30

⇒ y = 30 – 22

⇒ y = 8

So, x = 22, y = 8.

Q.9.Find the sum of the first 8 multiples of 3.

Solution:First 8 multiples of 3 are 3, 6, 9,….. up to 8 terms

We can observe that the above series is an AP with

a = 3, d = 6 – 3 = 3, n = 8

Sum of n terms of an A.P is given by,

Q.10.Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3). Hence find m.

Solution:Let P divides line segment AB in the ratio k : 1

Coordinates of P

Q.11. Two different dice are tossed together. Find the probability:

(i) of getting a doublet.

(ii) of getting a sum 10, of the numbers on the two dice.

Solution:

Total outcomes on tossing two different dice = 36

(i) A: getting a doublet

A = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

Number of favourable outcomes of A = 6

(ii) B: getting a sum 10.

B = {(4, 6), (5, 5), (6, 4)}

Number of favourable outcomes of B = 3

Q.12. An integer is chosen at random between 1 and 100. Find the probability that it is:

(i) divisible by 8.

(ii) not divisible by 8.

Solution: The total number are 2, 3, 4, …….. 99

(i) Let E be the event of getting a number divisible by 8.

E = {8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96} = 12

(ii) Let E’ be the event of getting a number not divisible by 8.

Then, P(E’) = 1 – P(E) = 1 – 0.1224 = 0.8756

Section – C

Q.13. Find HCF and LCM of 404 and 96 and verify that HCF × LCM = Product of the two given numbers.

Solution:

Prime factorization of 404 = 2 × 2 × 101

Prime factorization of 96 = 2 × 2 × 2 × 2 × 2 × 3

HCF = 2 × 2 = 4

And LCM = 2 × 2 × 2 × 2 × 2 × 3 × 101 = 9696

HCF = 4, LCM = 9696

Verification:

HCF × LCM = Product of the two given numbers

4 × 9696 = 404 × 96

38784 = 38784

Hence Verified.

Q.14. Find all zeroes of the polynomial (2x⁴ – 9x³ + 5x² + 3x – 1) if two of its zeroes are (2 + √3) and (2 – √3).

Solution:

Here, p(x) = 2x⁴ – 9x³ + 5x² + 3x – 1

And two of its zeroes are (2 + √3) and (2 – √3).

Quadratic polynomial with zeroes is given by,

{x – (2 + √3)}. {x – (2 – √3)}

⇒ (x – 2 – √3) (x – 2 + √3)

⇒ (x – 2)² – (√3)²

⇒ x² – 4x + 4 – 3

⇒ x² – 4x + 1 = g(x) (say)

Now, g(x) will be a factor of p(x) so g(x) will be divisible by p(x)

For other zeroes,

2x² – x – 1 = 0

2x² – 2x + x – 1 = 0

2x (x – 1) + 1 (x – 1) = 0

(x – 1) (2a + 1) = 0

x – 1 = 0 and 2x + 1 = 0

x = 1, x = -½

Zeroes of p(x) are

1, -½, 2 + √3 and 2 – √3.

Q.15. If A(-2, 1) and B(a, 0), C(4, b) and D( 1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides.

Q.15. If A(-5, 7), B(-4, -5), C(-1, -6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

Solution:

Given ABCD is a parallelogram.

Q.16. A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.

Solution: Let the usual speed of plane be x km/h.

Increased speed = (x + 100) km/h.

Distance to cover = 1500 km.

Time taken by plane with usual speed = ¹⁵⁰⁰/x hr

Time taken by plane with increased speed = ¹⁵⁰⁰/_{100 + x}

According to the question,

x² + 100x = 300000

x² + 100x – 300000 = 0

x² + 600x – 500x – 300000 = 0

x(x + 600) – 500(x + 600) = 0

(x + 600) (x – 500) = 0

Either x + 600 = 0 ⇒ x = -600 (Rejected)

x – 500 = 0 ⇒ x = 500

Usual speed of plane = 500 km/hr.

Q.17. Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.

Q.17. If the area of two similar triangles is equal, prove that they are congruent.

Solution: Let ABCD be a square with side ‘a’.

Q.18. Prove that the lengths of tangents drawn from an external point of a circle are equal.

Solution:

Given: A circle with centre O on which two tangents PM and PN are drawn from an external point P.

To Prove: PM = PN

Construction: Join OM, ON and OP

Proof: Since tangent and radius are perpendicular at point of contact,

∠OMP = ∠ONP = 90°

In ΔPOM and ΔPON,

OM = ON (Radii)

∠OMP = ∠ONP

PO = OP (Common)

ΔOMP = ΔONP (RHS cong.)

PM = PN (C.P.C.T)

Hence Proved.

Q.19. If 4 tanθ = 3, evaluate [(4sinθ – cosθ + 1)/ (4sinθ + cosθ – 1)]

Q.19. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Solution:

Given, 4 tan θ = 3

⇒ tan θ = ¾ (=P/B)

Solution: Given, tan 2A = cot (A – 18°)

⇒ cot (90° – 2A) = cot (A – 18°)

[∵ tan θ = cot (90° – θ)]

⇒ 90° – 2A = A – 18°

⇒ 90° + 18° = A + 2A

⇒ 108° = 3A

⇒ A = 36°

Q.20. Find the area of the shaded region in Fig. 2, where arcs are drawn with centres A, B, C and D intersect in pairs at mid-points P, Q, R and S of the sides AB, BC, CD and DA respectively of a square ABCD of side 12 cm. [Use π = 3.14]

Solution:

Given, ABCD is a square of side 12 cm.

P, Q, R and S are the midpoints of sides AB, BC, CD and AD respectively.

Area of shaded region = Area of square – 4 × Area of quadrant

= a² – 4 × ¼πr²

= (12)² – 3.14 × (6)²

= 144 – 3.14 × 36

= 144 – 113.04

= 30.96 cm²

Q.21. A wooden article was made by scooping out a hemisphere form each end of a solid cylinder, as shown in Fig. 3. If the height of the cylinder is 10 cm and its base is of radius 3.5 cm. Find the total surface area of the article.

Q.21. A heap of rice is in the form of a cone of base diameter 24 m and height 3.5 m. Find the volume of the rice. How much canvas cloth is required to just cover the heap?

Solution:

Given, Radius (r) of cylinder = Radius of hemisphere = 3.5 cm.

Total SA of article = CSA of cylinder + 2 × CSA of hemisphere

Height of cylinder, h = 10 cm

TSA = 2πrh + 2 × 2πr²

= 2πrh + 4πr²

= 2πrh (h + 2r)

= 2 × ²²⁄7× 3.5 (10 + 2 × 3.5)

= 2 × 22 × 0.5 × (10 + 7)

= 2 × 11 × 17

= 374 cm²

Solution: Base diameter of cone = 24 m.

Radius r = 12 m

Height of cone, h = 3.5 m

Volume of rice in conical heap = ⅓πr²h

= ⅓ × ²²⁄7 × 12 × 12 × 3.5 = 528 cm³

Q.22. The table below shows the salaries of 280 persons:

Calculate the median salary of the data.

Solution:

^N∕₂= ²⁸⁰∕₂ = 140

The cumulative frequency just greater than 140 is 182.

Median class is 10 -15.

l = 10, h = 5, N = 280, c.f. = 49 and f = 133

Section – D

Q.23. A motorboat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream.

Q.23. A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete the total journey, what is the original average speed?

Solution:

Given, speed of motorboat instil

water = 18 km/hr.

Let speed of stream = x km/hr.

Speed of boat downstream = (18 + x) km/hr.

And speed of boat upstream = (18 – x) km/hr.

Time of the upstream journey = ²⁴∕_18-x

Time of the downstream journey = ²⁴∕_18+x

According to the question,

⇒ x² + 48x – 324 = 0

⇒ x² + 54x – 6x – 324 = 0

⇒ x(x + 54) – 6(x + 54) = 0

⇒ (x + 54)(x – 6) = 0

Either x + 54 = 0 ⇒ x = -54

Rejected, as speed cannot be negative

x – 6 = 0 ⇒ x = 6

Thus, the speed of the stream is 6 km/hr.

Solution: Let the original average speed of train be x km/hr.

Increased speed of train = (x + 6) km/hr.

Time taken to cover 63 km with average speed = ⁶⁴∕_x hr.

Time taken to cover 72 km with increased speed = ⁷²∕_x+6 hr.

According to the question,

⇒ 135x + 378 = 3(x² + 6x)

⇒ 135x + 378 = 3x² + 18x

⇒ 3x² + 18x – 135x – 378 = 0

⇒ 3x² – 117x – 378 = 0

⇒ 3(x² – 39x – 126) = 0

⇒ x² – 39x – 126 = 0

⇒ x² – 42x + 3x – 126 – 0

⇒ x(x – 42) + 3(x – 42) = 0

⇒ (x – 42) (x + 3) = 0

Either x – 42 = 0 ⇒ x = 42

x + 3 = 0 ⇒ x = -3

Rejected (as speed cannot be negative)

Thus, average speed of train is 42 km/hr.

Q.24. The sum of four consecutive numbers in an AP is 32 and the ratio of the product of the first and the last term to the product of two middle terms is 7:15. Find the numbers.

Solution: Let the first term of AP be a and d be a common difference.

Let your consecutive term of an AP be a – 3d, a – d, a + d and a + 3d

According to the question,

a – 3d + a – d + a + d + a + 3d = 32

⇒ 4a = 32

⇒ a = 8 …(i)

Also,

(a – 3d) (a + 3d) : (a – d) (a + d) = 7 : 15

For d = 2, four terms of AP are,

a – 3d = 8 – 3 (2) = 2

a – d = 8 – 2 = 6

a + d = 8 + 2 = 10

a + 3d = 8 + 3(2) = 14

For d = -2, four term are

a – 3d = 8 – 3(-2) = 14

a – d = 8 – (-2) = 10

a + d = 8 + (-2) = 6

a + 3d = 8 + 3 (-2) = 2

Thus, the four terms of AP series are 2, 6, 10, 14 or 14, 10, 6, 2.

Q.25. In an equilateral ∆ABC, D is a point on side BC such that BD = ⅓ BC. Prove that 9(AD)² = 7(AB)².

Q.25. Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Solution:

Given, ABC is an equilateral triangle and D is a point on BC such that BD = ⅓ BC.

To prove: 9AD² = 7AB²

Construction : Draw AE ⊥ BC

Proof: BD = ⅓ BC …(i) (Given)

AE ⊥ BC

We know that perpendicular from a vertex of equilateral triangle to the base divides base in two equal parts.

BE = EC = ½ BC …(ii)

In ∆AEB,

AD² = AE² + DE² (Pythagoras theorem)

AE² = AD² – DE² …(iii)

Similarly, In ∆AEB,

AB² = AE² + BE²

Solution: Given: ∆ABC is a right angle triangle, right-angled at A.

To prove : BC² = AB² + AC²

Construction : Draw AD ⊥ BC.

Proof: In ∆ADB and ∆BAC,

∠B = ∠B (Common)

∠ADB = ∠BAC (Each 90°)

∆ADB ~ ∆BAC (By AA similarity axiom)

^AB∕_BC= ^BD∕_AB (CPCT)

AB² = BC × BD

Similarly,

∆ADC ~ ∆CAB

^AC∕_BC= ^DC∕_AC

AC² = BC × DC …(ii)

On adding equation (i) and (ii)

AB² + AC² = BC × BD + BC × CD = BC (BD + CD) = BC × BC

AB² + AC² = BC²

BC² = AB² + AC²

Hence Proved.

Q.26. Draw a triangle ABC with BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are ¾ of the corresponding sides of the ∆ABC.

Solution:

Draw a line segment BC = 6 cm.

Construct ∠XBC = 60°.

With B as centre and radius equal to 5 cm, draw an arc intersecting XB at A.

Join AC. Thus, ∆ABC is obtained.

Draw an acute angle ∠CBY below of B.

Mark 4-equal parts on BY as B₁, B₂, B₃ and B₄

Join B₄ to C.

From By draw a line parallel to B₄C intersecting BC at C’.

Draw another line parallel to CA from C’, intersecting AB at A’.

∆A’BC’ is required triangle which is similar to ∆ABC such that BC’ = ¾ BC.

Q.27.

Solution:

Q.28. The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm, find:

(i) The area of the metal sheet used to make the bucket.

(ii) Why we should avoid the bucket made by ordinary plastic? [Use π = 3.14]

Solution: Given, Height of frustum, h = 24 cm.

Diameter of lower end = 10 cm

Radius of lower end, r = 5 cm.

Diameter of upper end = 30 cm

Radius of upper end, R = 15 cm.

(i) Area of metal sheet used to make the bucket = CSA of frustum + Area of base

= πl(R + r) + πr²

= π[26 (15 + 5) + (5)²]

= 3.14 (26 × 20 + 25)

= 3.14 (520 + 25)

= 3.14 × 545

= 1711.3 cm²

(ii) We should avoid the bucket made by ordinary plastic because plastic is harmful to the environment and to protect the environment its use should be avoided.

Q.29. As observed from the top of a 100 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. [Use √3 = 1.732]

Solution: Let AB be the lighthouse and two ships are at C and D.