CBSE Exam 2021 : 10th & 12th Board Exam Dates Revised, Check Here

The Central Board of Secondary Education (CBSE) has released the revised date sheets for Class 10 and Class 12 students. Check The full-time table using the following link:

10th Exam Time Table

12th Exam Time Table

As per the new date sheet, the Class 10 Science exam will be held on May 21. Earlier, the Maths paper was scheduled for this date, which will be now conducted on June 2. Other Class 10 exams which have been rescheduled include French, German, Arabic, Sanskrit, Malayalam, Punjabi, Russian, and Urdu.

Apart from these changes, no changes have been made to the start and end dates of CBSE Class 10 board exams. Class 10 theory exams will begin on May 4 and end on June 7.

Class 12 Physics exam will now be held on June 8, instead of May 13. The Maths paper will be held on May 31. Earlier, the paper was scheduled to take place on June 1.

Class 12 Geography paper will be now held on June 3. Earlier, the paper was scheduled for June 2. There will be no exams on May 13, 14 for Class 12 students.

However, Class 12 papers will begin on May 4 and end on June 14. As per the old date sheets, the exams would have ended on June 11.

CBSE board exams this year will be conducted based on a reduced syllabus, a measure introduced by the board last year to make up for the academic loss caused by the COVID-19 pandemic.

 

5 March, 2021, 7:13 pm

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CBSE 12th Exam 2021 : Maths Case Study Problems

In Coming Exams, CBSE will ask two Case Study Questions in the CBSE class 12 maths questions paper. Each theme will have five questions and students will have a choice to attempt any four of them.

Here are some example questions Based On Case Study Problems:

Question-1

Read the case study carefully and answer any four out of the following questions:

Three friends Ganesh, Dinesh, and Ramesh went for playing a Tug of war game. Team A, B, and C belong to Ganesh, Dinesh, and Ramesh respectively.
Teams A, B, C have attached a rope to a metal ring and are trying to pull the ring into their own area (team areas shown below).
Team A pulls with F1 = 4i + 0j KN
Team B → F2 = -2i+ 4j  KN
Team C → F3 = -3i - 3j KN

  1. Which team will win the game?
    1. Team B
    2. Team A
    3. Team C
    4. No one
  2. What is the magnitude of the team's combine Force?
    1. 7 KN
    2. 1.4 KN
    3. 1.5 KN
    4. 2 KN
  3. In with direction approx the ring getting pulls:
    1. 2.0 radian
    2. 2.5 radian
    3. 2.4 radian
    4. 3 radian
  4. What is the magnitude of the force of Team B?
    1. 25–√25 KN
    2. 6 KN
    3. 2 KN
    4. 6–√6 KN
  5. How many KN Force is applied by Team A?
    1. 5 KN
    2. 4 KN
    3. 2 KN
    4. 16 KN

Answer Key:

  1. (a) Team B
  2. (b) 1.4 KN
  3. (c) 2.4 radian
  4. (a) 25–√25 KN
  5. (b) 4 KN

Question-2

Read the case study carefully and answer any four out of the following questions:

Three shopkeepers Ram Lal, Shyam Lal, and Ghansham are using polythene bags, handmade bags (prepared by prisoners), and newspaper's envelope as carry bags. It is found that the shopkeepers Ram Lal, Shyam Lal, and Ghansham are using (20,30,40), (30,40,20), and (40,20,30) polythene bags, handmade bags, and newspapers envelopes respectively. The shopkeeper's Ram Lal, Shyam Lal, and Ghansham spent ₹250, ₹270, and ₹200 on these carry bags respectively. 

  1.  What is the cost of one polythene bag?
    1. ₹ 1
    2. ₹ 2
    3. ₹ 3
    4. ₹ 5
  2. What is the cost of one handmade bag?
    1. ₹ 1
    2. ₹ 2
    3. ₹ 3
    4. ₹ 5
  3. What is the cost of one newspaper bag?
    1. ₹ 1
    2. ₹ 2
    3. ₹ 3
    4. ₹ 5
  4. Keeping in mind the social conditions, which shopkeeper is better?
    1. Ram Lal
    2. Shyam Lal
    3. Ghansham
    4. None of these
  5. Keeping in mind the environmental conditions, which shopkeeper is better?
    1. Ram Lal
    2. Shyam Lal
    3. Ghansham
    4. None of these

Answer Key:

  1. (a) ₹1
  2. (b) ₹ 2
  3. (d) ₹ 5
  4. (b) Shyam Lal
  5. (a) Ram Lal

Question-3

Read the case study carefully and answer any four out of the following questions:

Three friends Ravi, Raju, and Rohit were doing buying and selling stationery items in a market. The price per dozen of Pen, Notebook, and toys are Rupees x, y, and z respectively.
Ravi purchases 4 dozens of notebooks and sells 2 dozens of pens and 5 dozens of toys. Raju purchases 2 dozens of toys and sells 3 dozens of pens and 1 dozen of notebooks. Rohit purchases one dozen of pens and sells 3 dozens of notebooks and one dozen toys. In the process, Ravi, Raju, and Rohit earn ₹ 1500, ₹ 100, and ₹400 respectively.

Answer the following questions using the matrix method:

  1. What is the price of one dozen of pens?
    1. ₹ 100
    2. ₹ 200
    3. ₹ 300
    4. ₹ 400
  2. What is the total price of one dozen of pens and one dozen of notebooks?
    1. ₹ 100
    2. ₹ 200
    3. ₹ 300
    4. ₹ 400
  3. What is the sale amount of Ravi?
    1. ₹ 1000
    2. ₹ 1100
    3. ₹ 1300
    4. ₹ 1200
  4. What is the amount of purchases made by all three friends?
    1. ₹ 1200
    2. ₹ 1500
    3. ₹ 1300
    4. ₹ 1400
  5. What is the price of sales made by all three friends?
    1. ₹ 3000
    2. ₹ 2500
    3. ₹ 2700
    4. ₹ 2400

Answer Key:

  1. (a) ₹ 100
  2. (c) ₹ 300
  3. (d) ₹ 1200
  4. (b) ₹ 1500
  5. (c) ₹ 2700

Question-4

Read the case study carefully and answer any four out of the following questions:

Dinesh is having a jewelry shop at Green Park, Normally he does not sit in the shop as he remains busy in political meetings. The manager Lisa takes care of jewelry shop where she sells earrings and necklaces. She sells earrings for ₹ 30 and necklaces for 40. It takes 30 minutes to make a pair of earrings and 1 hour to make a necklace, and there are 10 hours a week to make jewelry. In addition, there are only enough materials to make 15 total jewelry items per week. The profit of 15 and 20 is made on each pair of earrings and a necklace respectively.

The graph of the equations formed is shown here:

Answer the following questions:

  1. What is the first inequality formed?
    1. 0.5x + y ≤ 10
    2. x +0.5 y ≤ 10
    3. 0.5x + y ≤ 15
    4. x + 0.5 y ≤ 20
  2. What is the second inequality formed?
    1. 0.5x + y ≤ 10
    2. x + y ≤ 10
    3. x + y ≤ 15
    4. x + 0.5 y ≤ 10
  3. In order to maximize the profits, per week how many pair of earrings and necklaces should be made?
    1. 10 and 5
    2. 5 and 10
    3. 15 and 5
    4. 5 and 15
  4. What would be the profit if 5 pairs of earrings and 5 necklaces are made?
    1. 150
    2. 300
    3. 175
    4. 250
  5. What would be the maximum possible profit?
    1. 150
    2. 300
    3. 175
    4. 250

Answer Key:

  1. 0.5x + y  ≤ 10
  2. x + y ≤ 15
  3. 10 and 5
  4. 175
  5. 250

Question-5

Read the case study carefully and answer any four out of the following questions:

A telephone company in a town has 500 subscribers on its list and collects fixed charges of 300 per subscriber per year. The company proposes to increases the annual subscription and it is believed that for every increase of 1 one subscriber will discontinue the service.

  1. If x be the annual subscription  then the total revenue of the company after increment will be:
    1. R(x) = -x2 + 200x + 150000
    2. R(x) = x2 - 200x - 140000
    3. R(x) = 200x2 + x + 150000
    4. R(x) = -x2 + 100 x + 100000
  2. To find maximum profit we put
    1. R'(x) = 0
    2. R'(x) > 0
    3. R'(x) < 0
    4. R''(x) = 0
  3. How much fee the company should increase to have maximum profit?
    1. Rs. 150
    2. Rs. 100
    3. Rs. 200
    4. Rs. 250
  4. Find the maximum profit that the company can make if the profit function is given by P(x) = 41 + 24x - 18x2.
    1. 25
    2. 44
    3. 45
    4. 49
  5. Find both the maximum and minimum value respectively of 3x4 - 8x3 + 48x + 1 on the interval [1, 4].
    1. -63, 257
    2. 258, -63
    3. 257, -63
    4. -63, -257

Answer Key:

  1. (a) -x+ 200x + 150000
  2. (a) R'(x) = 0
  3. (b) Rs.100
  4. (d) 49
  5. (c) 257, -63

 

4 March, 2021, 11:34 am

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CBSE 12th Exam 2021 : Chemistry Case Study Problems

In Coming Exams, CBSE will ask two Case Study Questions in the CBSE class 12 Chemistry questions paper. Each theme will have five questions and students will have a choice to attempt any four of them.

Here are some example questions Based On Case Study Problems:

Question-1

Read the passage given below and answer any four out of the following questions:
Ammonia is present in small quantities in air and soil where it is formed by the decay of nitrogenous organic matter e.g., urea. On a large scale, ammonia is manufactured by Haber’s process. In accordance with Le Chatelier’s principle, high pressure would favour the formation of ammonia. Ammonia is a colourless gas with a pungent odour. Its freezing and boiling points are 198.4 and 239.7 K respectively. In the solid and liquid states, it is associated with hydrogen bonds as in the case of water and that accounts for its higher melting and boiling points than expected on the basis of its molecular mass. Ammonia gas is highly soluble in water. Its aqueous solution is weakly basic due to the formation of OH ions. The presence of a lone pair of electrons on the nitrogen atom of the ammonia molecule makes it a Lewis base.

The following questions are multiple-choice questions. Choose the most appropriate choice 

  1. On a small scale, ammonia is obtained from ammonium salts which decompose when treated with
    1. caustic soda
    2. calcium chloride
    3. sodium hydroxide
    4. sodium chloride
  2. The optimum conditions for the production of ammonia are a pressure of
    1. ​200*105 Pa
    2. 400*105 Pa
    3. 100*105 Pa
    4. 300*105 Pa
  3. The catalyst which is used in the preparation of NH3 by Haber’s process
    1. Mg2O3 + K2O
    2. Al2O+ K2O
    3. NaO+ K2O
    4. None of these
  4. The ammonium molecule has:
    1. five bond pair and two lone pair
    2. four lone pair and one bond pair
    3. three bond pair and one lone pair
    4. three bond pair and two lone pair
  5. A compound reacts with ammonia to form deep colour solution, identify the compound
    1. Au2+
    2. Cu2+
    3. Al3+
    4. Mg2+

Questions-2

Read the passage and answer any four out of the following questions:
Colloidal particles always carry an electric charge. The nature of this charge is the same on all the particles in a given colloidal solution and may be either positive or negative. The charge on the sol particles is due to one or more reasons, viz., due to electron capture by sol particles during electrodispersion of metals. When two or more ions are present in the dispersion medium, preferential adsorption of the ion common to the colloidal particle usually takes place. When silver nitrate solution is added to the potassium iodide solution, the precipitated silver iodide adsorbs iodide ions from the dispersion medium, and negatively charged colloidal solution results. acquired a positive or a negative charge by selective adsorption on the surface of a colloidal particle The combination of the two layers of opposite charges around the colloidal particle is called Helmholtz electrical double layer. The presence of equal and similar charges on colloidal particles is largely responsible for providing stability to the colloidal solution.

In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. 

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion
  3. Assertion is correct statement and reason is wrong statement
  4. Assertion is wrong statement but reason is correct statement
  1. Assertion: The presence of equal and similar charges on colloidal particles is largely responsible in providing stability to the colloidal solution.
    Reason: The repulsive forces between charged particles having the same charge prevent them from aggregating and provide stability.
  2. Assertion: The first layer is mobile in Helmholtz electrical double layer.
    Reason: The potential difference between the fixed layer and the diffused layer of opposite charges is called zeta potential.
  3. Assertion: The sol particle in the colloid has a charge.
    Reason: The charge in sol is due to electron capture by sol particles during the electrodispersion of metals.
  4. Assertion: Methylene blue sol is a negatively charged sol.
    Reason: When KI solution is added to AgNO3 solution, positively charged sol formed.
  5. Assertion: If FeCl3 is added to an excess of hot water, a positively charged sol of hydrated ferric oxide is formed.
    Reason: When ferric chloride is added to NaOH a negatively charged sol is obtained with adsorption of OH- ions.

Question-3

Read the passage and answer the following questions:
The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionically arising purely from electrostatic interactions between the metal ion and the ligand. Ligands are treated as point charges in case of anions or point dipoles in case of neutral molecules. The five d orbitals in an isolated gaseous metal atom/ion have the same energy, i.e., they are degenerate. In an octahedral coordination entity with six ligands surrounding the metal atom/ion, there will be repulsion between the electrons in metal d orbitals and the electrons (or negative charges) of the ligands. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed crystal field splitting and the energy separation is denoted by Δ0. The colour in the coordination compounds can be readily explained in terms of the crystal field theory. 

In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for assertion.
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion.
  3. Assertion is correct statement but reason is wrong statement.
  4.  Assertion is wrong statement but reason is correct statement.
  1. Assertion: The dx2-y2 and dz2 orbitals which point towards the axes along the direction of the ligand will experience more repulsion. 
    Reason: The dxy, dyz and dxz orbitals which are directed between the axes will be lowered in energy.
  2. Assertion: The complex [Ti(H2O)6]3+, which is red in colour.
    Reason: The crystal field theory attributes the colour of the coordination compounds to d-d transition of the electron.
  3. Assertion: Ligands for which Δ0Δ0 < P are known as weak field ligands and form high spin complexes.
    Reason: If Δ0 > P, then the fourth electron enters one of the e​​​​​​g orbitals giving the configuration t2g3 eg1
  4. Assertion: In tetrahedral coordination entity formation, the d orbital splitting is inverted and is smaller as compared to the octahedral field splitting.
    Reason: Spectrochemical series is based on the absorption of light by complexes with different ligands. 
  5. Assertion: The crystal field model is successful in explaining the formation, structures, colour and magnetic properties of coordination compounds.
    Reason: The anionic ligands are found at the low end of the spectrochemical series.

Answer Key:

  1. (b) Assertion and reason both are correct statements and reason is not correct explanation for assertion.
  2. (c) Assertion is correct statement but reason is wrong statement.
  3. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
  4. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.
  5. (d) Assertion is wrong statement but reason is correct statement.

Question-4

Read the passage and answer any four out of the following question
Alfred Werner (1866-1919), a Swiss chemist was the first to formulate his ideas about the structures of coordination compounds. Werner proposed the concept of a primary valence and a secondary valence for a metal. The coordination entity constitutes a central metal atom or ion bonded to a fixed number of ions or molecules. In a coordination entity, the atom/ion to which a fixed number of ions/groups are bound in a definite geometrical arrangement around it is called the central atom or ion. The ions or molecules bound to the central atom/ion in the coordination entity are called ligands. Ligands may be simple ions such as Cl-, small molecules such as H2O or NH3, larger molecules such as H2NCH2CH2NH2 or N(CH2CH2NH2)3 or even macromolecules, such as protein. Ligands are unidentate, bidentate and polydentate. The coordination number (CN) of a metal ion in a complex is the number of ligand donor atoms to which the metal is directly bonded. 

In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

  1. Assertion and reason both are correct statements and reason is correct explanation for  assertion
  2. Assertion and reason both are correct statements but reason is not correct explanation for assertion
  3. Assertion is correct statement but reason is wrong statement
  4.  Assertion is wrong statement but reason is correct statement
  1. Assertion: Binary compounds such as CrCl3, have a primary valence of 3.
    Reason: Coordinate compound metals show only one type of linkage that is primary linkage.
  2. Assertion: CoCl3(NH3)3 is a coordination entity in which the cobalt ion is surrounded by three ammonia molecules and three chloride ion.
    Reason: The central atom/ion in the coordination entities: [NiCl2(H2O)4] is Ni2+.
  3. Assertion: H2NCH2CH2NH2 (ethane-1,2-diamine) ligand is said to be didentate.
    Reason: Didentate ligands are bind through two donor atoms.
  4. Assertion: The complex ions, [PtCl6]2- the coordination number of Pt is 4.
    Reason: Ligand which can ligate through two different atoms is called ambidentate ligand.
  5. Assertion: EDTA can bind through two nitrogen and four oxygen atoms to a central metal ion.
    Reason: The number of ligating groups attach to an atom is called the denticity of the ligand.

Answer Key:

  1. (c) Assertion is correct statement but reason is wrong statement
  2. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion
  3. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion
  4. (d) Assertion is wrong statement but reason is correct statement
  5. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion

Question-5

Read the passage given below and answer any four out of the following questions:
Nitrogen differs from the rest of the members of group 15 due to its smaller size, high electronegativity, high ionisation enthalpy, and non-availability of d orbitals. Nitrogen has a unique ability to form pπ-pπ multiple bonds with itself. Nitrogen exists as a diatomic molecule with a triple bond one s and two p between the two atoms. Phosphorus, arsenic and antimony from single bonds as P–P, As–As and Sb–Sb while bismuth forms metallic bonds in an elemental state. Dinitrogen is produced commercially by the liquefaction and fractional distillation of air. Liquid dinitrogen (b.p. 77.2 K) distils out first leaving behind liquid oxygen (b.p. 90 K). In the laboratory, dinitrogen is prepared by treating an aqueous solution of ammonium chloride with sodium nitrite. Dinitrogen is a colourless, odourless, tasteless and non-toxic gas. It has two stable isotopes 14N and 15N. It has very low solubility in water. The main use of dinitrogen is in the manufacture of ammonia and other industrial chemicals containing nitrogen.

The following questions are multiple-choice questions. choose the most appropriate answer

  1. N–N bond is weaker than the single P–P bond because 
    1. high interelectronic repulsion of the bonding electrons
    2. high interelectronic repulsion of the non-bonding electrons
    3. no repulsion between bonding electrons
    4. no repulsion between non-bonding electrons
  2. Very pure nitrogen can be obtained by the
    1. thermal decomposition of sodium
    2. thermal decomposition of barium azide
    3. thermal decomposition of ammonium dichromate
    4. both (a) and (b)
  3. Dinitrogen is rather inert at room temperature because of
    1. low bond enthalpy of N≡≡N bond
    2. high bond enthalpy of N≡≡N bond
    3. low freezing point
    4. low boiling point
  4. Dinitrogen combines with dioxygen only at very high temperature (at about 2000 K) to form
    1. nitric oxide
    2. nitrate
    3. nitrites
    4. nitric acid
  5.  Liquid dinitrogen is used as a refrigerant to
    1. preserve biological materials 
    2. preserve food items  
    3. in cryosurgery
    4. all of these

Answer Key:

  1. (b) high interelectronic repulsion of the non-bonding electrons
  2. (d) both (a) and (b)
  3. (b) high bond enthalpy of N≡≡N bond
  4. (a) nitric oxide
  5. (d) all of these

Question-6

Read the following passage and answer any four out of the following questions:
Transition metal oxides are generally formed by the reaction of metals with oxygen at high temperatures. The highest oxidation number in the oxides coincides with the group number. In vanadium, there is a gradual change from the basic V2O3 to less basic V2O4 and to amphoteric V2O5. V2O4 dissolves in acids to give VO​​​​​2+ salts. Potassium dichromate is a very important chemical used in the leather industry and as an oxidant for the preparation of many azo compounds. Dichromates are generally prepared from chromate. Sodium dichromate is more soluble than potassium dichromate. The latter is, therefore, prepared by treating the solution of sodium dichromate with potassium chloride. Sodium and potassium dichromates are strong oxidising agents; sodium salt has a greater solubility in water and is extensively used as an oxidising agent in organic chemistry. Potassium dichromate is used as a primary standard in volumetric analysis.

The following questions are multiple-choice questions. Choose the most appropriate answer.

  1. All transition metal reacts with oxygen to form MO oxide except
    1. scandium
    2. vanadium
    3. cupper
    4. zinc
  2. As the oxidation number of a metal increases, ionic character
    1. increases
    2. decreases
    3. remain the same
    4. none of these
  3. The shape of chromate ion is
    1. tetrahedral
    2. pyramidal
    3. square planer
    4. triangular
  4. Dichromates are generally prepared from chromate, which in turn are obtained by the fusion of 
    1. FeCr2O
    2. FeCr2O4
    3. Na2CrO4
    4. Na2Cr2O7
  5. The oxo cations stabilise V​​​​​IV
    1. VO
    2. VO​​​​​4+
    3. VO​​​​​2+
    4. all of these

Answer Key:

  1. (a) scandium
  2. (b) decreases
  3. (a) tetrahedral
  4. (b) FeCr2O4
  5. (c) VO​​​​​2+

 

3 March, 2021, 3:19 pm

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CBSE 12th Exam 2021 : Physics Case Study Problems

In Coming Exams, CBSE will ask two Case Study Questions in the CBSE class 12 Physics questions paper. Question numbers 15 and 16 are case-based questions where 5 MCQs will be asked based on a paragraph. Each theme will have five questions and students will have a choice to attempt any four of them.

Here are some example questions Based On Case Study Problems:

Question 1.

Read  the following source and answer any four out of the following questions:
Electric charge is the physical property of matter that causes it to experience a force when placed in an electromagnetic field. There are two types of charges positive and negative charges. Also, like charges repel each other whereas unlike charges attract each other.

  1. Charge on a body which carries 200 excess electrons is:
    1. -3.2 × 10-18 C
    2. 3.2 ×  10 18 C
    3. -3.2 × 10-17 C
    4. 3.2 × 10 -17  C
  2. Charge on a body  which carries 10 excess electrons is:
    1. -1.6 × 10-18 C
    2. 1.6 × 10 -18 C
    3. 2.6 × 10-18 C
    4. 1.6 × 10-21 C
  3. Mass of electron is:
    1. 9.1 × 10-31 kg
    2. 9.1 × 10-31 g
    3. 1.6 × 10-19 kg
    4. 1.6 × 10-19 g
  4. A body is positively charged, it implies that:
    1. there is only a positive charge in the body
    2. there is positive as well as negative charge in the body but the positive charge is more than negative charge
    3. there is equally positive and negative charge in the body but the positive charge lies in the outer regions
    4. the negative charge is displaced from its position
  5. On rubbing, when one body gets positively charged and other negatively charged, the electrons transferred from positively charged body to negatively charged body are:
    1. valence electrons only
    2. electrons of inner shells
    3. both valence electrons and electrons of the inner shell.
    4. none of the above

Question 2:

Read the following source and answer any four out of the following questions:
Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms. Also Resistivity, the electrical resistance of a conductor of unit cross-sectional area, and unit length. … A characteristic property of each material, resistivity is useful in comparing various materials on the basis of their ability to conduct electric currents.

  1. Resistivity is independent of:
    1. nature of material
    2. temperature
    3. dimensions of material
    4. none of the above
  2. As compare to short wires, long wires have _______ resistance.
    1. more
    2. less
    3. same
    4. zero
  3. As compare to thin wires, thick wires have _______ resistance.
    1. more
    2. less
    3. same
    4. zero
  4. The resistance of a wire depends upon:
    1. cross-sectional area
    2. length of wire
    3. wire’s nature
    4. all of the above
  5. A copper wire having the same size as steel wire have:
    1. more resistance
    2. less resistance
    3. same resistance
    4. none of the above

Question 3:

Read the source given below and answer any four out of the following questions:
The Bohr model of the atom was proposed by Neil Bohr in 1915. It came into existence with the modification of Rutherford’s model of an atom. Rutherford’s model introduced the nuclear model of an atom, in which he explained that a nucleus (positively charged) is surrounded by negatively charged electrons.

  1. Which of the following statements does not form a part of Bohr’s model of a hydrogen atom?
    1. The energy of the electrons in the orbit is quantized
    2. The electron in the orbit nearest the nucleus has the lowest energy
    3. Electrons revolve in different orbits around the nucleus
    4. The position and velocity of the electrons in the orbit cannot be determined simultaneously
  2. What is in the center of the Rutherford model?
    1. Single proton
    2. Multiple electrons
    3. A nucleus
    4. Neutrons
  3. When an electron jumps from its orbit to another orbit, energy is:
    1. emitted only
    2. absorbed only
    3. both (a) and (b)
    4. none of these
  4.  How were the limitations of the Rutherford model which could not explain the observed features of atomic spectra explained in Bohr’s model of a hydrogen atom?
    1.  It must emit a continuous spectrum
    2.  It loses its energy
    3. Gaining its energy
    4. A discrete spectrum
  5. When an electron remains between orbits its momentum is:
    1. quantized
    2. emitted
    3. dequantized
    4. none of the above

Question 4:

Read the source given below and answer any four out of the following questions:
The potentiometer consists of a long resistive wire(L) and a battery of known EMF, 'V' whose voltage is known as driver cell voltage. Assume a primary circuit arrangement by connecting the two ends of L to the battery terminals. One end of the primary circuit is connected to the cell whose EMF 'E' is to be measured and the other end is connected to galvanometer G. This circuit is assumed to be a secondary circuit.

 

 

  1. How can we increase the sensitivity of a potentiometer?
    1. Increasing the potential gradient
    2. Decreasing the potential gradient
    3. Decreasing the length of potentiometer wire
    4. Increasing resistance put in parallel
  2. If l1 and l2 are the balancing lengths of the potentiometer wire for the cells of EMFs ε1ε1 and ε2ε2, then
    1. ε1ε1+ ε2ε2 = l1 + l2
    2. ε1ε2=l1l2ε1ε2=l1l2
    3. ε1ε1ε2ε2 = l1l2
    4. None of these
  3. An example of a potentiometer is
    1. Mobile
    2. Modem
    3. Joystick
    4. All of these
  4. The emf of a cell is always greater than its terminal voltage. Why?
    1. Because there is some potential drop across the cell due to its small internal resistance
    2. Because there is some potential drop across the cell due to its large internal resistance
    3. Because there is some potential drop across the cell due to its low current
    4. Because there is some potential drop across the cell due to its high current
  5. Why is a ten-wire potentiometer more sensitive than a four-wire one?
    1. Small potential gradient
    2. Large potential gradient
    3. Large length
    4. None of these

Answer Key:

  1. (b) Decreasing the potential gradient
  2. (b) ε1ε2=l1l2ε1ε2=l1l2
  3. (c) Joystick
  4. (a) Because there is some potential drop across the cell due to its small internal resistance
  5. (a) Small potential gradient

Question 5:

Read the source given below and answer any four out of the following questions:
If two or more capacitors are connected in series, the overall effect is that of a single (equivalent) capacitor having the sum total of the plate spacing of the individual capacitors. If two or more capacitors are connected in parallel, the overall effect is that of a single equivalent capacitor having the sum total of the plate areas of the individual capacitors. ( figure (a) shows parallel combination and (b) shows series combination)

  1. Capacity can be increased by connecting capacitors in:
    1. parallel
    2. series
    3. both a  and b 
    4. none of these
  2. Three capacitors having a capacitance equal to 2F, 4F and 6F are connected in parallel. Calculate the effective parallel capacitance:
    1. 10 F
    2. 11 F
    3. 12 F
    4. 13 F
  3. When capacitors are connected in the series ________ remains the same.
    1. voltage
    2. capacitance
    3. charge
    4. resistance
  4. The plates of a parallel plate capacitor are 10 cm apart and have an area equal to 2m2. If the charge on each plate is 8.85×10−10C8.85×10−10C , the electric field at a point:

    1. between the plates will be zero
    2. outside the plates will be zero
    3. between the plates will change from point to point
    4. between the plates will be 25NC−125NC−1
  5. Four 10 F capacitors are connected in series, calculate the equivalent capacitance.
    1. 1.5 F
    2. 2.5 F
    3. 3.5 F
    4. 4.5 F

Answer Key:

  1. (a) parallel
  2. (c) 12 F
  3. (b) charge 
  4. (b) outside the plates will be zero
  5. (b) 2.5F

Question 6:

Read the source given below and answer any four out of the following questions:
A charge is a property associated with the matter due to which it experiences and produces an electric and magnetic field. Charges are scalar in nature and they add up like real numbers. Also, the total charge of an isolated system is always conserved. When the objects rub against each other charges acquired by them must be equal and opposite.

  1. The cause of charging is:
    1. the actual transfer of protons
    2. the actual transfer of electrons
    3. the actual transfer of neutrons
    4. none of the above
  2. Pick the correct statement.
    1. The glass rod gives protons to silk when they are rubbed against each other.
    2. The glass rod gives electrons to silk when they are rubbed against each other.
    3. The glass rod gains protons from silk when they are rubbed against each other.
    4. The glass rod gains electrons when they are rubbed against each other.
  3. If two electrons are each 1.5×10−10m1.5×10−10m from a proton, the magnitude of the net electric force they will exert on the proton is

    1. 1.97×10−8 N
    2. 2.73×10−8 N
    3. 3.83×10−8 N
    4. 4.63×10−8 N
  4. A charge is a property associated with the matter due to which it produces and experiences :
    1. electric effects only
    2. magnetic effects only
    3. both electric and magnetic effects
    4. none of these
  5. The cause of quantization of electric charges is:
    1. transfer of an integral number of neutrons
    2. transfer of  an integral number of protons
    3. transfer of an integral number of electrons
    4. none of the above

Answer Key:

  1. (b)  the actual transfer of electrons 
  2. (b) The glass rod gives electrons to silk when they are rubbed against each other.
  3. (a) 1.97×10−8N1.97×10−8N
  4. (c) both electric and magnetic effects
  5. (c) transfer of an integral number of electrons

 

2 March, 2021, 3:31 pm

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