100525 + Interview Questions in GENERAL QA IN GENERAL AWARENESS

1.	Which of the following is not an example of mechanical energy?(a) Kinetic energy(b) Potential energy(c) Internal energy(d) Work
Answer» Correct answer is (c) Internal energy Easiest explanation: Internal energy is not a mechanical energy.

Discussion

2.	In humans, there is a 50% probability of the birth of a boy and 50 % probability that a girl will be born. Justify the statement on the basis of the mechanism of sex-determination in human beings.
Answer» In human beings, the genes inherited from our parents decide whether we will be boys or girls. Women have a perfect pair of sex chromosomes (XX). But, men have a mismatched pair (XY). All children will inherit an X chromosome from their mother regardless of whether they are boys or girls. Thus, the sex of the children will be determined by what they inherit from their father. A child who inherits an X chromosome from her father will be a girl, and one who inherits a Y chromosome from him will be a boy.

2.

In humans, there is a 50% probability of the birth of a boy and 50 % probability that a girl will be born. Justify the statement on the basis of the mechanism of sex-determination in human beings.

Answer»

In human beings, the genes inherited from our parents decide whether we will be boys or girls. Women have a perfect pair of sex chromosomes (XX). But, men have a mismatched pair (XY).

All children will inherit an X chromosome from their mother regardless of whether they are boys or girls. Thus, the sex of the children will be determined by what they inherit from their father. A child who inherits an X chromosome from her father will be a girl, and one who inherits a Y chromosome from him will be a boy.

Discussion

3.	After self-pollination in pea plants with round, yellow seeds, following types of seeds were obtained by Mendel:SeedsNumberRound, yellow630Round, green216Wrinkled, yellow202Wrinkled, green64Analyse the result and describe the mechanism of inheritance which explains these results.
Answer» The ratio obtained is 9:3:3:1 in which parental as well as new combinations are observed. This indicates that progeny plants have not inherited a single whole gene set from each parent. Every germ cell takes one chromosome from the pair of maternal and paternal chromosomes. When two germ cells combine, segregation of one pair of characters is independent of other pair of characters.

3.

After self-pollination in pea plants with round, yellow seeds, following types of seeds were obtained by Mendel:SeedsNumberRound, yellow630Round, green216Wrinkled, yellow202Wrinkled, green64Analyse the result and describe the mechanism of inheritance which explains these results.

Answer»

The ratio obtained is 9:3:3:1 in which parental as well as new combinations are observed. This indicates that progeny plants have not inherited a single whole gene set from each parent.

Every germ cell takes one chromosome from the pair of maternal and paternal chromosomes. When two germ cells combine, segregation of one pair of characters is independent of other pair of characters.

Discussion

4.	I haven’t a chair ____. A) to sit B) for to sit on C) to sit on D) for sitting
Answer» Correct option is C) to sit on

Discussion

5.	Which process allows complete conversion of mechanical energy?(a) Irreversible(b) Reversible(c) Reversible & Irreversible(d) None of the mentioned
Answer» The correct answer is (b) Reversible For explanation: Reversible process so it allows complete conversion of mechanical energy.

Discussion

6.	Read the following and answer any four questions.Sumati wanted to see the stars of the night sky. She knows that she needs a telescope to see those distant stars. She finds out that the telescopes, which are made of lenses, are called refracting telescopes and the ones which are made of mirrors are called reflecting telescopes.So she decided to make a refracting telescope. She bought two lenses, L1 and L2. out of which L1 was bigger and L2 was smaller. The larger lens gathers and bends the light, while the smaller lens magnifies the image. Big, thick lenses are more powerful. So to see far away, she needed a big powerful lens. Unfortunately, she realized that a big lens is very heavy.Heavy lenses are hard to make and difficult to hold in the right place. Also since the light is passing through the lens, the surface of the lens has to be extremely smooth. Any flaws in the lens will change the image. It would be like looking through a dirty window.(i) Based on the diagram shown, what kind of lenses would Sumati need to make the telescope? a) Concave lenses b) Convex lenses c) Bifocal lenses d) Flat lenses(ii) If the powers of the lenses L1 and L2 are in the ratio of 4:1, what would be the ratio of the focal length of L1 and L2? a) 4:1 b) 1:4 c) 2:1 d) 1:1(iii) What is the formula for magnification obtained with a lens? a) Ratio of height of image to height of object b) Double the focal length. c) Inverse of the radius of curvature. d) Inverse of the object distance.(iv) Sumati did some preliminary experiment with the lenses and found out that the magnification of the eyepiece (L2) is 3. If in her experiment with L2 she found an image at 24 cm from the lens, at what distance did she put the object? a) 72 cm b) 12 cm c) 8 cm d) 6 cm(v) Sumati bought not-so-thick lenses for the telescope and polished them. What advantages, if any, would she have with her choice of lenses? a) She will not have any advantage as even thicker lenses would give clearer images. b) Thicker lenses would have made the telescope easier to handle. c) Not-so-thick lenses would not make the telescope very heavy and also allow considerable amount of light to pass. d) Not-so-thick lenses will give her more magnification.
Answer» (i) b) Convex (ii) P=\(\frac{1}{f}\) P₁= \(\frac{1}{f_1}\) and P₂= \(\frac{1}{f_2}\) \(\frac{P_1}{P_2}=\frac{4}{1}\), hence \(\frac{\frac{1}{f_1}}{\frac{1}{f_2}}\) = \(\frac{4}{1}\) Hence \(\frac{f_1}{f_2}=\frac{1}{4}\) b) 1:4 (iii) a) Ratio of height of image to height of object (iv) m=\(\frac{v}{u}\) 3=\(\frac{24}{u}\) Hence u = 8cm c) 8 cm (v) c) Not-so-thick lenses would not make the telescope very heavy and they will also allow considerable amount of light to pass through them.

6.

Read the following and answer any four questions.Sumati wanted to see the stars of the night sky. She knows that she needs a telescope to see those distant stars. She finds out that the telescopes, which are made of lenses, are called refracting telescopes and the ones which are made of mirrors are called reflecting telescopes.So she decided to make a refracting telescope. She bought two lenses, L1 and L2. out of which L1 was bigger and L2 was smaller. The larger lens gathers and bends the light, while the smaller lens magnifies the image. Big, thick lenses are more powerful. So to see far away, she needed a big powerful lens. Unfortunately, she realized that a big lens is very heavy.Heavy lenses are hard to make and difficult to hold in the right place. Also since the light is passing through the lens, the surface of the lens has to be extremely smooth. Any flaws in the lens will change the image. It would be like looking through a dirty window.(i) Based on the diagram shown, what kind of lenses would Sumati need to make the telescope? a) Concave lenses b) Convex lenses c) Bifocal lenses d) Flat lenses(ii) If the powers of the lenses L1 and L2 are in the ratio of 4:1, what would be the ratio of the focal length of L1 and L2? a) 4:1 b) 1:4 c) 2:1 d) 1:1(iii) What is the formula for magnification obtained with a lens? a) Ratio of height of image to height of object b) Double the focal length. c) Inverse of the radius of curvature. d) Inverse of the object distance.(iv) Sumati did some preliminary experiment with the lenses and found out that the magnification of the eyepiece (L2) is 3. If in her experiment with L2 she found an image at 24 cm from the lens, at what distance did she put the object? a) 72 cm b) 12 cm c) 8 cm d) 6 cm(v) Sumati bought not-so-thick lenses for the telescope and polished them. What advantages, if any, would she have with her choice of lenses? a) She will not have any advantage as even thicker lenses would give clearer images. b) Thicker lenses would have made the telescope easier to handle. c) Not-so-thick lenses would not make the telescope very heavy and also allow considerable amount of light to pass. d) Not-so-thick lenses will give her more magnification.

Answer»

(i) b) Convex

(ii) P=\(\frac{1}{f}\)

P₁= \(\frac{1}{f_1}\) and P₂= \(\frac{1}{f_2}\)

\(\frac{P_1}{P_2}=\frac{4}{1}\), hence \(\frac{\frac{1}{f_1}}{\frac{1}{f_2}}\) = \(\frac{4}{1}\)

Hence \(\frac{f_1}{f_2}=\frac{1}{4}\)

b) 1:4

(iii) a) Ratio of height of image to height of object

(iv) m=\(\frac{v}{u}\)

3=\(\frac{24}{u}\)

Hence u = 8cm

c) 8 cm

(v) c) Not-so-thick lenses would not make the telescope very heavy and they will also allow considerable amount of light to pass through them.

Discussion

7.	Name two chemicals obtained from common salt. Write their preparations from common salt. Also, write two uses of each chemical obtained from common salt.
Answer» The two chemicals obtained from common salt are: Sodium hydroxide and bleaching powder. Preparation of sodium hydroxide: When electricity is passed through an aqueous solution of sodium chloride (called brine), it decomposes to form sodium hydroxide. The process is called the chlor-alkali process because of the products formed, chlor for chlorine and alkali for sodium hydroxide. 2Nacl (aq.) + 2H₂O (l) ⇾ 2NaOH (aq.) + Cl₂ (g) + H₂ (g) Uses: i. For de- greasing metals ii. For making soaps and detergents, paper iii. For making artificial fibres. Preparation of bleaching powder: Thus, we see that chlorine is produced during the electrolysis of aqueous sodium chloride (brine). This chlorine gas is used for the manufacture of bleaching powder. Bleaching powder (CaOCl₂) is produced by the action of chlorine on dry slaked lime[Ca(OH)₂] Ca(OH)₂ + Cl₂ ⇾ CaOCl₂ + H₂O Uses: i. For bleaching cotton and linen in the textile industry, for bleaching wood pulp in paper factories and for bleaching washed clothes in laundry, ii. As an oxidising agent in many chemical industries, iii. To make drinking water free from germs.

7.

Name two chemicals obtained from common salt. Write their preparations from common salt. Also, write two uses of each chemical obtained from common salt.

Answer»

The two chemicals obtained from common salt are:

Sodium hydroxide and bleaching powder.

Preparation of sodium hydroxide:

When electricity is passed through an aqueous solution of sodium chloride (called brine), it decomposes to form sodium hydroxide. The process is called the chlor-alkali process because of the products formed, chlor for chlorine and alkali for sodium hydroxide.

2Nacl (aq.) + 2H₂O (l) ⇾ 2NaOH (aq.) + Cl₂ (g) + H₂ (g)

Uses:

i. For de- greasing metals

ii. For making soaps and detergents, paper iii. For making artificial fibres.

Preparation of bleaching powder:

Thus, we see that chlorine is produced during the electrolysis of aqueous sodium chloride (brine). This chlorine gas is used for the manufacture of bleaching powder. Bleaching powder (CaOCl₂) is produced by the action of chlorine on dry slaked lime[Ca(OH)₂]

Ca(OH)₂ + Cl₂ ⇾ CaOCl₂ + H₂O

Uses:

i. For bleaching cotton and linen in the textile industry, for bleaching wood pulp in paper factories and for bleaching washed clothes in laundry,

ii. As an oxidising agent in many chemical industries,

iii. To make drinking water free from germs.

Discussion

8.	Read the following and answer any 4 questions. A solenoid is a long helical coil of wire through which a current is run in order to create a magnetic field. The magnetic field of the solenoid is the superposition of the fields due to the current through each coil. It is nearly uniform inside the solenoid and close to zero outside and is similar to the field of a bar magnet having a north pole at one end and a south pole at the other depending upon the direction of current flow. The magnetic field produced in the solenoid is dependent on a few factors such as, the current in the coil, number of turns per unit length etc. The following graph is obtained by a researcher while doing an experiment to see the variation of the magnetic field with respect to the current in the solenoid. The unit of magnetic field as given in the graph attached is in milli-Tesla (mT) and the current is given in Ampere.(i) What type of energy conversion is observed in a linear solenoid? a. Mechanical to Magnetic b. Electrical to Magnetic c. Electrical to Mechanical d. Magnetic to Mechanical(ii) What will happen if a soft iron bar is placed inside the solenoid? a. The bar will be electrocuted resulting in short-circuit. b. The bar will be magnetised as long as there is current in the circuit. c. The bar will be magnetised permanently. d. The bar will not be affected by any means.(iii) The magnetic field lines produced inside the solenoid are similar to that of … a. a bar magnet b. a straight current carrying conductor c. a circular current carrying loop d. electromagnet of any shape(iv) After analysing the graph a student writes the following statements. I. The magnetic field produced by the solenoid is inversely proportional to the current. II. The magnetic field produced by the solenoid is directly proportional to the current. III. The magnetic field produced by the solenoid is directly proportional to square of the current. IV. The magnetic field produced by the solenoid is independent of the current. Choose from the following which of the following would be the correct statement(s). a. Only IV b. I and III and IV c. I and II d. Only II(v) From the graph deduce which of the following statements is correct. a. For a current of 0.8A the magnetic field is 13 mT b. For larger currents, the magnetic field increases non-linearly. c. For a current of 0.8A the magnetic field is 1.3 mT d. There is not enough information to find the magnetic field corresponding to 0.8A current.
Answer» (i) c. Electrical to Mechanical (ii) b. The bar will be magnetised as long as there is current in the circuit (iii) a. A bar magnet (iv) d. Only II (v) a. For a current of 0.8A the magnetic field is 13 mT

8.

Read the following and answer any 4 questions. A solenoid is a long helical coil of wire through which a current is run in order to create a magnetic field. The magnetic field of the solenoid is the superposition of the fields due to the current through each coil. It is nearly uniform inside the solenoid and close to zero outside and is similar to the field of a bar magnet having a north pole at one end and a south pole at the other depending upon the direction of current flow. The magnetic field produced in the solenoid is dependent on a few factors such as, the current in the coil, number of turns per unit length etc. The following graph is obtained by a researcher while doing an experiment to see the variation of the magnetic field with respect to the current in the solenoid. The unit of magnetic field as given in the graph attached is in milli-Tesla (mT) and the current is given in Ampere.(i) What type of energy conversion is observed in a linear solenoid? a. Mechanical to Magnetic b. Electrical to Magnetic c. Electrical to Mechanical d. Magnetic to Mechanical(ii) What will happen if a soft iron bar is placed inside the solenoid? a. The bar will be electrocuted resulting in short-circuit. b. The bar will be magnetised as long as there is current in the circuit. c. The bar will be magnetised permanently. d. The bar will not be affected by any means.(iii) The magnetic field lines produced inside the solenoid are similar to that of … a. a bar magnet b. a straight current carrying conductor c. a circular current carrying loop d. electromagnet of any shape(iv) After analysing the graph a student writes the following statements. I. The magnetic field produced by the solenoid is inversely proportional to the current. II. The magnetic field produced by the solenoid is directly proportional to the current. III. The magnetic field produced by the solenoid is directly proportional to square of the current. IV. The magnetic field produced by the solenoid is independent of the current. Choose from the following which of the following would be the correct statement(s). a. Only IV b. I and III and IV c. I and II d. Only II(v) From the graph deduce which of the following statements is correct. a. For a current of 0.8A the magnetic field is 13 mT b. For larger currents, the magnetic field increases non-linearly. c. For a current of 0.8A the magnetic field is 1.3 mT d. There is not enough information to find the magnetic field corresponding to 0.8A current.

Answer»

(i) c. Electrical to Mechanical

(ii) b. The bar will be magnetised as long as there is current in the circuit

(iii) a. A bar magnet

(iv) d. Only II

(v) a. For a current of 0.8A the magnetic field is 13 mT

Discussion

9.	A student has two resistors- 2 Ω and 3 Ω. She has to put one of them in place of R2 as shown in the circuit. The current that she needs in the entire circuit is exactly 9A. Show by calculation which of the two resistors she should choose.
Answer» The overall current needed = 9A. The voltage is 12V Hence by Ohm’s Law V=IR, The resistance for the entire circuit = \(\frac{12}{9}\) = \(\frac{4}{3}\Omega\) = R R₁and R₂ are in parallel. Hence, R=\(\frac{(R_1R_2)}{(R_1+R_2)}\)= \(\frac{4R_2}{(4+R_2)}\) \(\because\, R\) = \(\frac{4}{3}\) \(\frac{4}{3}=\)\(\frac{4R_2}{(4+R_2)}\) \(16+4R_2=12R_2\\8R_2=16\) R₂ = 2\(\Omega\)

9.

A student has two resistors- 2 Ω and 3 Ω. She has to put one of them in place of R2 as shown in the circuit. The current that she needs in the entire circuit is exactly 9A. Show by calculation which of the two resistors she should choose.

Answer»

The overall current needed = 9A. The voltage is 12V

Hence by Ohm’s Law V=IR,

The resistance for the entire circuit = \(\frac{12}{9}\) = \(\frac{4}{3}\Omega\) = R

R₁and R₂ are in parallel.

Hence, R=\(\frac{(R_1R_2)}{(R_1+R_2)}\)= \(\frac{4R_2}{(4+R_2)}\)

\(\because\, R\) = \(\frac{4}{3}\)

\(\frac{4}{3}=\)\(\frac{4R_2}{(4+R_2)}\)

\(16+4R_2=12R_2\\8R_2=16\)

R₂ = 2\(\Omega\)

Discussion

10.	(I) what is a neutralisation reaction? Give an example. (II) What colour change do you observe when CuO is added to dilute HCl, while stirring it? Write balanced chemical equation for the above reaction? (III) Why metallic oxides are said to be basic oxides?
Answer» (i) A neutralization reaction is when an acid and a base react to form water and salt and involves the combination of hydrogen ions and hydroxyl ions to generate water. The neutralization of a strong acid and strong base has a P^H equal to 7. Example 1: When sodium hydroxide is added to hydrochloric acid, sodium chloride and water are formed. NaOH + HCl ⇾ NaCl + H₂O Example 2: Milk of magnesia, which is a base, is given as antacid in the case of indigestion, to neutralize the more acid produced in the stomach. Mg(OH)₂ + 2HCl ⇾ MgCl₂ + 2H₂O. (ii) When dilute HCl is added to copper oxide powder, blue- green colour is imparted due to the formation of copper chloride. The reaction is: 2HCl + CuO ⇾ CuCl₂ + H₂O This is a double displacement reaction. (iii) When metallic oxide reacts with water, it gives base. E.g. MgO + H₂O ⇾ Mg(OH)₂ That is why metallic oxides are said to be basic oxides. It turns red litmus paper blue. Secondly, metal oxides reacts with acids to form salt and water, which is a primary property of bases.

10.

(I) what is a neutralisation reaction? Give an example. (II) What colour change do you observe when CuO is added to dilute HCl, while stirring it? Write balanced chemical equation for the above reaction? (III) Why metallic oxides are said to be basic oxides?

Answer»

(i) A neutralization reaction is when an acid and a base react to form water and salt and involves the combination of hydrogen ions and hydroxyl ions to generate water. The neutralization of a strong acid and strong base has a P^H equal to 7.

Example 1: When sodium hydroxide is added to hydrochloric acid, sodium chloride and water are formed.

NaOH + HCl ⇾ NaCl + H₂O

Example 2: Milk of magnesia, which is a base, is given as antacid in the case of indigestion, to neutralize the more acid produced in the stomach.

Mg(OH)₂ + 2HCl ⇾ MgCl₂ + 2H₂O.

(ii) When dilute HCl is added to copper oxide powder, blue- green colour is imparted due to the formation of copper chloride.

The reaction is:

2HCl + CuO ⇾ CuCl₂ + H₂O

This is a double displacement reaction.

(iii) When metallic oxide reacts with water, it gives base.

E.g. MgO + H₂O ⇾ Mg(OH)₂

That is why metallic oxides are said to be basic oxides. It turns red litmus paper blue. Secondly, metal oxides reacts with acids to form salt and water, which is a primary property of bases.

Discussion

11.	A student observes the above phenomenon in the lab as a white light passes through a prism. Among many other colours, he observed the position of the two colours Red and Violet.What is the phenomenon called? What is the reason for the violet light to bend more than the red light?
Answer» The phenomenon is called dispersion. Speed of Violet Light inside the prism is slowest and that of Red is highest. Hence, deviation of Violet Light is maximum and that of Red is minimum.

11.

A student observes the above phenomenon in the lab as a white light passes through a prism. Among many other colours, he observed the position of the two colours Red and Violet.What is the phenomenon called? What is the reason for the violet light to bend more than the red light?

Answer»

The phenomenon is called dispersion.

Speed of Violet Light inside the prism is slowest and that of Red is highest. Hence, deviation of Violet Light is maximum and that of Red is minimum.

Discussion

12.	Bile juice does not have any digestive enzyme but still plays a significant role in the process of digestion. Justify the statement.
Answer» Bile juice makes the acidic food coming from the stomach alkaline for the action of pancreatic enzymes. Bile salts break the large globules of fat in the intestine to smaller globules increasing the efficiency of enzyme action. This is similar to the emulsifying action of soaps on dirt.

Discussion

13.	(a) Which gas is produced when dilute hydrochloric acid is added to a reactive metal? Write the chemical reaction when iron reacts with dilute H2SO4. (b) What would you observe when Zinc is added to a solution of iron (II) sulphate? Write the chemical reaction that takes place.
Answer» (a) When dilute HCl react with reactive metals like alkali metal or alkaline metal, hydrogen gas is liberated. For example: Mg + 2HCl ⇾ MgCl₂ + H₂ Reaction when iron reacts with dilute H₂SO₄ Fe(s) + H₂SO₄ (aq.) ⇾ FeSO₄(aq.) + H₂(g). (b) When zinc is added to solution of iron (II) sulphate, the colour of iron sulphate solution changes from green to colourless.It is because zinc is more reactive than iron, therefore, it displaces iron from its solution of iron sulphate and a grey precipitate of iron and a colourless solution of zinc sulphate is formed. Zn(s) + FeSO₄ (aq.) ⇾ ZnSO₄ (aq.) +Fe(s).

13.

(a) Which gas is produced when dilute hydrochloric acid is added to a reactive metal? Write the chemical reaction when iron reacts with dilute H2SO4. (b) What would you observe when Zinc is added to a solution of iron (II) sulphate? Write the chemical reaction that takes place.

Answer»

(a) When dilute HCl react with reactive metals like alkali metal or alkaline metal, hydrogen gas is liberated.

For example: Mg + 2HCl ⇾ MgCl₂ + H₂

Reaction when iron reacts with dilute H₂SO₄

Fe(s) + H₂SO₄ (aq.) ⇾ FeSO₄(aq.) + H₂(g).

(b) When zinc is added to solution of iron

(II) sulphate, the colour of iron sulphate solution changes from green to colourless.It is because zinc is more reactive than iron, therefore, it displaces iron from its solution of iron sulphate and a grey precipitate of iron and a colourless solution of zinc sulphate is formed.

Zn(s) + FeSO₄ (aq.) ⇾ ZnSO₄ (aq.) +Fe(s).

Discussion

14.	In birds and mammals the left and right side of the heart are separated. Give reasons.
Answer» The separation keeps oxygenated and deoxygenated blood from mixing allowing a highly efficient supply of oxygen to the body. This is useful in animals that have high energy needs (birds and mammals) which constantly use energy to maintain their body temperature.

Discussion

15.	Explain the meanings of malleable and ductile. Give example of each, a malleable element and a ductile element.
Answer» Ductility: It is the property of metals of being drawn into wire. It is a desirable property in machine components that are subjected to unanticipated overloads or impact loads. The ductility of metals decreases as the temperature increases because metals become weak at increasing temperature. Examples: copper, aluminium, steel, gold. Malleability: It is the ability of the material to deform to a greater extent before signs of cracks appear when it is subjected to a compressive force. In general, malleability increases with an increase in temperature. Processes like forging or rolling gives shapes to hot ingots or slabs. Examples: gold, silver, iron, aluminium, copper, lead.

15.

Explain the meanings of malleable and ductile. Give example of each, a malleable element and a ductile element.

Answer»

Ductility:

It is the property of metals of being drawn into wire. It is a desirable property in machine components that are subjected to unanticipated overloads or impact loads. The ductility of metals decreases as the temperature increases because metals become weak at increasing temperature.

Examples: copper, aluminium, steel, gold.

Malleability:

It is the ability of the material to deform to a greater extent before signs of cracks appear when it is subjected to a compressive force. In general, malleability increases with an increase in temperature. Processes like forging or rolling gives shapes to hot ingots or slabs.

Examples: gold, silver, iron, aluminium, copper, lead.

Discussion

16.	The following observations were made by a student on treating four metals P, Q, R and S with the given salt solutions:SampleMgSO4(aq)Zn(NO3)2(aq)CaSO4(aq)Na2SO4(aq)PNo reactionReaction occursReaction occursNo ReactionQReaction occursReaction occursReaction occursReaction occursRNo ReactionReaction occursNo ReactionNo ReactionSNo ReactionNo ReactionNo ReactionNo ReactionBased on the above observations: (a) Arrange the given samples in the increasing order of reactivity (b) Write the chemical formulae of products formed when Q reacts with CuSO4 solution.
Answer» i. S>R>P>Q ii. Cu and QSO₄

16.

The following observations were made by a student on treating four metals P, Q, R and S with the given salt solutions:SampleMgSO4(aq)Zn(NO3)2(aq)CaSO4(aq)Na2SO4(aq)PNo reactionReaction occursReaction occursNo ReactionQReaction occursReaction occursReaction occursReaction occursRNo ReactionReaction occursNo ReactionNo ReactionSNo ReactionNo ReactionNo ReactionNo ReactionBased on the above observations: (a) Arrange the given samples in the increasing order of reactivity (b) Write the chemical formulae of products formed when Q reacts with CuSO4 solution.

Answer»

i. S>R>P>Q

ii. Cu and QSO₄

Discussion

17.	State the events occurring during the process of photosynthesis. Is it essential that these steps take place one after the other immediately?
Answer» 6CO₂ (Carbon dioxide) + 6H₂O (water) \( \underset{chlorophyll}{\stackrel{Sunlight}{\rightarrow}} \) C₆H₁₂O₆(Carbohydarate) + 6O₂ (oxygen) Absorption of light energy by chlorophyll. Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen. Reduction of carbon dioxide to carbohydrates. These steps need not take place one after the other immediately. For example, desert plants take up carbon dioxide at night and prepare an intermediate which is acted upon by the energy absorbed by the chlorophyll during the day

17.

State the events occurring during the process of photosynthesis. Is it essential that these steps take place one after the other immediately?

Answer»

6CO₂ (Carbon dioxide) + 6H₂O (water) \( \underset{chlorophyll}{\stackrel{Sunlight}{\rightarrow}} \) C₆H₁₂O₆(Carbohydarate) + 6O₂ (oxygen)

Absorption of light energy by chlorophyll.

Conversion of light energy to chemical energy and splitting of water molecules into hydrogen and oxygen.
Reduction of carbon dioxide to carbohydrates.

These steps need not take place one after the other immediately. For example, desert plants take up carbon dioxide at night and prepare an intermediate which is acted upon by the energy absorbed by the chlorophyll during the day

Discussion

18.	The number of carbon compounds is more than those formed by all other elements put together. Justify the statement by giving two reasons.
Answer» Due to self linking ability of carbon/catenation Since carbon has a valency of four it can form bonds with four other atoms of carbon or atoms of some other mono-valent element. Due to small size of carbon it forms very strong and (or) stable bonds with other elements

Discussion

19.	(I) Why the taste or smell of fat change when left for a long time? What can you do to prevent it?(II) Name the gas used to prevent the chips from getting oxidised?
Answer» (i) When food materials containing fats and oils are left for a long time, they become rancid (i.e) fat and oil present in them get oxidised. The methods which are used to slow down oxidation are: a) Keeping food in air tight container b) Flushing the bags containing food with nitrogen gas c) Refrigerating food also helps to slow down rancidification. (ii) Nitrogen is filled inside of chips packet to make atmosphere inside it as inert to prevent oxidation of chips.

19.

(I) Why the taste or smell of fat change when left for a long time? What can you do to prevent it?(II) Name the gas used to prevent the chips from getting oxidised?

Answer»

(i) When food materials containing fats and oils are left for a long time, they become rancid (i.e) fat and oil present in them get oxidised.

The methods which are used to slow down oxidation are:

a) Keeping food in air tight container

b) Flushing the bags containing food with nitrogen gas

c) Refrigerating food also helps to slow down rancidification.

(ii) Nitrogen is filled inside of chips packet to make atmosphere inside it as inert to prevent oxidation of chips.

Discussion

20.	Samples of four metals A, B, C, and D were taken and added to the following solutions one by one. The results obtained have been tabulated on follows: Metals Iron (II) sulphate Copper sulphate (II) Zinc sulphateANo reactionDisplacementBDisplacementNo reactionCNo reactionNo reactionNo reactionCNo reactionNo reactionNo reactionI. What would you observe if B is added to a solution of copper (II) sulphate? II. Arrange the metals A, B, C, D in the order of decreasing reactivity.
Answer» (i) If B is added to CuSO₄ solution, then it will displace Cu because Fe is more reactive than Cu, so the metal which can displace Fe, can also displace Cu. (ii) B > A > C > D

20.

Samples of four metals A, B, C, and D were taken and added to the following solutions one by one. The results obtained have been tabulated on follows: Metals Iron (II) sulphate Copper sulphate (II) Zinc sulphateANo reactionDisplacementBDisplacementNo reactionCNo reactionNo reactionNo reactionCNo reactionNo reactionNo reactionI. What would you observe if B is added to a solution of copper (II) sulphate? II. Arrange the metals A, B, C, D in the order of decreasing reactivity.

Answer»

(i) If B is added to CuSO₄ solution, then it will displace Cu because Fe is more reactive than Cu, so the metal which can displace Fe, can also displace Cu.

(ii) B > A > C > D

Discussion

21.	Give a test that can be used to confirm the presence of carbon in a compound. With a valency of 4, how is carbon able to attain noble gas configuration in its compounds?
Answer» - Burn compound in air/ oxygen; Gas evolved turns lime water milky - By sharing its four valence electrons with other elements.

21.

Give a test that can be used to confirm the presence of carbon in a compound. With a valency of 4, how is carbon able to attain noble gas configuration in its compounds?

Answer»

- Burn compound in air/ oxygen; Gas evolved turns lime water milky

- By sharing its four valence electrons with other elements.

Discussion

22.	Give a test that can be used to confirm the presence of carbon in a compound. Write its three allotropes.
Answer» (i) Presence of carbon can be detected by heating the compound with cupric oxide (Cuo) strongly, when carbon is oxidised to carbon dioxide (CO₂), which can be tested by lime water test. C + 2CuO ⇾ CO₂ + 2Cu (In the compound) CO₂+ Ca(OH)₂(Milky)⇾ CaCO₃+H₂O (ii) Also, as an another test, when saturated hydrocarbons are heated, they give a clean flame, while unsaturated hydrocarbons will give a yellow flame with lots of black smoke. This results in a sooty deposit on the metal plate. Three allotropes of carbon: (i) Graphite (ii) Diamond (iii) Fullerenes

22.

Give a test that can be used to confirm the presence of carbon in a compound. Write its three allotropes.

Answer»

(i) Presence of carbon can be detected by heating the compound with cupric oxide (Cuo) strongly, when carbon is oxidised to carbon dioxide (CO₂), which can be tested by lime water test.

C + 2CuO ⇾ CO₂ + 2Cu (In the compound)

CO₂+ Ca(OH)₂(Milky)⇾ CaCO₃+H₂O

(ii) Also, as an another test, when saturated hydrocarbons are heated, they give a clean flame, while unsaturated hydrocarbons will give a yellow flame with lots of black smoke. This results in a sooty deposit on the metal plate.

Three allotropes of carbon:

(i) Graphite

(ii) Diamond

(iii) Fullerenes

Discussion

23.	Explain inductive effect and resonance effect
Answer» Inductive effect is used for describing the transmission of electrical charges between the atoms whereas resonance effect is used for describing the transmission of electron pairs between the atoms.

Discussion

24.	Which of the following respresent the variation of conducatnes of solution if weak base `NH_(4)OH` is titrated with dilute HCl?A. B. C. D.
Answer» Correct Answer - 2 `NH_(4)OH` has almost constant conductance but after neturallization conductance`uarr`.

Discussion

25.	Find the `DeltapH(` initial `pH -` final `pH)` when `100 ml 0.01M HCl` is added in a solution containing `0.1m` moles of `NaHCO_(3)` solution of negligible volume `(k_(a_(1))=10^(-7),k_(a_(2))=10^(-11)` for `H_(2)CO_(3)) :`A. 6+2 log3B. 6-log3C. 6+2log2D. 6-2log3
Answer» Correct Answer - A pH of `NaHCO_3` solution =9 (initial) Now `H^+ + HCO_3^(-) to H_2CO_3` `:.` no . of mmole of HCl remaining =1-0.1=0.9 mmole `:.` pH=-log `(9xx10^(-3))=-2 log 3+3` (Final)

Discussion

26.	In the connection shown in the figure the switch K is open and the capacitor is uncharged. Then we close the switch and let the capacitor charge up to the maximum and open the switch again. Then (a) the current through `R_(1)` be `I_(1)` immediately after closing the switch, (b) the current through `R_(2)` be `I_(3)` immediately after reopening the switch, Find `I_(1)/(I_(2) I_(3))` (in `ampere^(-1)`) (Use the following data: `V_(0)=30VR_(1)=10 K Omega R_(2) = 5 k Omega)`
Answer» Correct Answer - `0750` `(Cq_(1))/k=(Q_(0)-q_(1)) Rightarrow q_(1)=Q_(0)/((1+C/K)) Rightarrow (Cq_(2))/K=K(q_(1)-q_(2))` `q_(2)=q_(1)/((1+C/K)) Rightarrow q_(10)= Q_(0)/((1+C/K)^(10)) Rightarrow Q_(0)/2 = Q_(0)/((1+C/K)^(10))` `(1+C/K)=10 sqrt(2) Rightarrow Q_(0)/3=Q_(0)/((1+C/K)^(10+n))` `10sqrt(2)=10+nsqrt(3)` `(l n2)/10=(l n3)/(l n2)`

Discussion

27.	The indicator constant for an acidic indicator, `HIn` is `5xx10^(-6)M`. This indicator appears only in the colour of acidic form when `([In])/([HIn])le(1)/(20)` and it appears only in the colour of basic form when `([HIn])/([In])le(1)/(40)` . The pH range of indicator is `[` Given `: log 5=0.7]`A. 3.7-6.9B. 4.0-6.6C. 4.0-6.9D. 3.7-6.6
Answer» Correct Answer - C `pH=pKln+log_(10)[(In^(-))/(HIn)]` `pH=[6-log_10 5]+log_10[1/20]=4` `pH=pKln+log_(10)[(In^(-))/(HIn)]` `=pKIn+log_10[40/1]=6-log_10 5+log_10 40 =6-0.7+1.6=6.9`

Discussion

28.	2 conducting objects one with charge of +Q and another with -Q are kept on x-axis at `x=-3` and `x -+4` respectively. The electric field on the x - axis is given by `3Q(x^(2)+4/3)` What is the capacitance C of this configuration of objects. Fill `1/C` (in `F^(-1))` in OMR sheet.
Answer» Correct Answer - `119` Just after closing switch no current flows through `R_(2)` so `I_(1) = 3` mA Long time after closing switch no current flows through C so `I_(2)=2` mA Directly after re-opening the switch no current flows through `R_(1)` and the capacitor will discharge through `R_(2)` so `I_(3)=2` mA

Discussion

29.	When 100 mL of 0.1 M NaCN solution is titrated with 0.1 M HCl solution the variation of pH of solution with volume of HCl added will be :A. B. C. D.
Answer» Correct Answer - C Initially pH will decrease fast, then slowly due to buffer formation and then will decrease fast as buffer action diminishes.

Discussion

30.	To a 100 mL of 0.1 M weak acid HA solution 22.5 mL of 0.2 M solution of NaOH are added.Now, what volume of 0.1 M NaOH solution be added into above solution, so that pH of resulting solution be 4.7: [Given :`(K_b(A^(-))=5xx10^(-10)]`A. 5 mLB. 20 mLC. 10 mLD. 15 mL
Answer» Correct Answer - A `HA+NaOHtoNaA+H_2O` `t=0 " " 10 " " 4.5 " " 0 " " 0` t=50% `" " 5 " " 0 " " 5` 5 mL NaOH is required

Discussion

31.	`0.01` moles of an ideal diatomic gas is enclosed in an adiabatic cylinder of cross-sectional area `A=10^(-4)m^(2)` In the arrangement shown, a block of mass `M= 0.8` kg is placed on a horizontal support, and piston of mass `m=1` kg is suspended from a spring of stiffness constant `k=16 N//m` Initially, the springs is relaxed and the volume of the gas is `V= 1.4 xx10^(-4)m^(3)` When the gas in the cylinder is heated up the piston starts moving up and the spring gets compressed so that the block M is just lifted up. Determine the heat supplied (in Joule) Take atmospheric pressure `P_(0)=10^(5) Nm^(-2) , g=10 m//s^(2)`
Answer» Correct Answer - `0075` J `DeltaV=f/2nRDeltaT=5/2(P_(2)V_(2)-P_(1)V_(1))=63` J `mgx+1/2kx^(2)+P_(0)Ax=omega_(gas)` `Omega_(gas)=12 j Rightarrow DeltaQ=75J`

Discussion

32.	(a)When 100 ml of 0.1M NaCN solution is titrated with 0.1 M HCl solution the variation of pH of solution with volume of HCl added will be : (b)Variation of degree of dissociation `alpha` with concentration for a weak electrolyte at a particular temperature is best represented by : ( c)0.1 M acetic acid solution is titrated against 0.1 M NaOH solution. The difference in pH between 1/4 and 3/4 stages of neutralization of acid will be 2 log 3.A. T,F,TB. F,F,FC. T,T,TD. F,T,F
Answer» Correct Answer - C (a)Initially pH will decrease fast, then slowly due to buffer formation and then will decrease fast as buffer action diminishes. (b) For a weak electrolyte `K_a=(Calpha^2)/((1-alpha))` when `alphaltlt 1` then `alpha=sqrt(k_a/C)` as C increases `implies` `alpha` decreases as C is tending to zero `implies` `alpha` will be unity ( c) At `1//4^(th)` neutralisation `CH_3COOH+NaOHtoCH_3COONa+H_2O` `(0.1xx3/4) " " (0.1xx1/4)` `pH=pK_a+"log" ([CH_3COO^(-)])/([CH_3COOH])=pK_a+ "log" (1/3)` At `3//4^(th)` neutralisation `pH=pK_a+"log" 3` so difference in pH=`Delta(pH)="log"3-"log"1/3=2"log"3`

Discussion

33.	In a linear particle accelerator two large plane uniformly charged are placed parallel to each other as shown in the figure. figure also shows variation of potential between planes A charged particle carrying a charge `q=1.6xx10^(-19)` C and having mass m `=16xx10^(-31)` kg is released from rest at `x=-10` cm determine time taken (in ns ) by particle to strike the other plane Neglect relativistic considerations.
Answer» Correct Answer - `40` Electric field on particle `E=E=(-DeltaV)/(Deltax)=E=(-[250-(-250)])/(20xx10^(-2))=-2500 V//m` Acceleration of charge particle `=(qE)/m=(1.6xx10^(-19)xx2500)/(16xx10^-31))=2500xx10^(11)m//s^(2)` thus time taken `= sqrt((2xx5)/a) {S=1/2 at^(2))` `=sqrt((2xx20xx10^(-2))/(2500xx10^(11))=4xx10^(-8) sec = 40 nS`

Discussion

34.	The visible portion of a lightning strike is preceded by an invisible stage in which a column of electrons extends downward from a cloud to the ground Assume the linear charge density along the column is `1.00xx10^(-3)C//m` Treat the column of charge as if it were straight and infinitely long. At what distance (in m) from the column of electrons does the electric field have a magnitude of `3.00xx10^(6) V//m` the dielectric strength for air? This is an estimate of the radius of a visible lightning bolt. (Round off to nearest integer) (value of `in_(0)is 8.85xx10^(-12) C^(2)//Nm^(2))`
Answer» Correct Answer - `6 m` `lambda/(2pi in_(0)r)=E_(break)` `r=lambda/(2piin_(0)E_(break))` `=10^(-3)/(2xx3.14xx8.85xx10^(-12)xx3xx10^(6))` `=1/(2xx3.14xx8.85)xx10^(3)` `=5.99m approx 6m`

Discussion

35.	A particle is unchanged and is thrown vertically upward from ground level with a speed of `5sqrt(5)` m//s in a region of space having uniform electric field As a result, it attains a maximum height h. The particle is then given a positive charge +q and reaches the same maximum height h when thrown vertically upward with a speed of `13 m//s` Finally the particle is given a negative charge -q Ignoring air resistance determine the speed (in m//s) with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the same maximum height h.
Answer» Correct Answer - `0009` In uniform electric in vertical direction if (+ve) charge feels extra acceleration in downward direction then (-ve) charge will feel acceleration in upward direction. `v_(uncharged)=5sqrt(5)m//sec` `v=0, h=height` `v^(2)-u^(2)=-2(g) h` `-(5sqrt(5))=-2gh` `u_(q+)=13m//sec` `v=0 h = h` `v^(2)-u^(2)=2(g+F_(E)/m)h` `0-(13)^(2)=-2(g+(F_(E))/m)h` Let `u_(q-)=u(say)` `v=0 h=ht` `v^(2)-u^(2)=-2(g-(F^(E))/m)h` `-u^(2)=-2(g-(F_(E))/m)h,u=9m//sec`

Discussion

36.	In single slit diffraction pattern :A. central fringe has negligible width than othersB. all fringes are of some widthC. central fringe does not existD. none of the above
Answer» Correct Answer - D In a single slit diffraction, the central fringe has maximum intensity and has width double than other fringe.

Discussion

37.	verify Euler's theorem for the following functions, u=ax2+2hxy+by2
Answer» According to Euler theorem, x du/dx + y du/dy = nu; where n is degree of the homogeneous equation. Now, u = ax²+ 2hxy + by² du/dx = 2ax + 2hy du/dy = 2hx + 2by L.H.S = x du/dx + y du/dy x(2ax + 2hy) + y(2hx + 2by) = 2ax² + 2hxy + 2hxy + 2by² = 2(ax²+ 2hxy + by²) = 2u Thus Euler theorem verified.

37.

verify Euler's theorem for the following functions, u=ax2+2hxy+by2

Answer»

According to Euler theorem, x du/dx + y du/dy = nu; where n is degree of the homogeneous equation.

Now,
u = ax²+ 2hxy + by²
du/dx = 2ax + 2hy
du/dy = 2hx + 2by
L.H.S = x du/dx + y du/dy
x(2ax + 2hy) + y(2hx + 2by)
= 2ax² + 2hxy + 2hxy + 2by²
= 2(ax²+ 2hxy + by²) = 2u

Thus Euler theorem verified.

Discussion

38.	An electric field is given by `vec(E)=4hat(i)+3(y^(2)+2)hat(j)` pierces gaussian cube of side 1 m placed at origin such that one of its corners is at origin & rest of sides are along positive side of coordinate axis. If the magnitude of net charge enclosed is n `varepsilon_(0)` then n (in SI units) will be equal to
Answer» Correct Answer - `0003` Along z axis `vec(E) . D vec(A) =0` Along x axis `vec(E) =const` `:.phi_(x)=0` for y=0 `int vec(E) . Dvec(A)= int 3(0+2)hat(j)dA (-hat(j))=6intdA =-6` y=1 `intvec(E) . Dvec(A) = int3(1+2)hat(j).dA(hat(j))=9intdA=9` `:.phi_(net)=+3in_(0)`

Discussion

39.	A liquid is contained in a vertical tube of semi-circular cross-section. The angle of contact is zero fig. What is the ratio of the force of surface tension on the curved part and the flat part of the tube? A. `2 : pi`B. `1 : pi`C. `3 : pi`D. `2.7 : pi`
Answer» Correct Answer - A Effective length of flat part = 2r Effective length of curved part is `pi r` So, required ratio is `2 : pi`

Discussion

40.	Solve dx/dt +2y =5e^t ; dy/dt-2x = 5e^t, given that x = -1, y=3 when t=0
Answer» \(\frac{d\mathrm x}{dt}+2y = 5e^t\) \(\frac{dy}{dt}-2\mathrm x=5e^t\) By taking \(\frac{d}{dt}\) = D, we get Dx + 2y = 5et ........(1) Dy – 2x = 5et .........(2) Now, operate D on both sides of equation (1), we get D²x + 2Dy = 5De^t = 5e^t .........(3) \((\because De^t = \frac{d}{dt}e^t = e^t)\) Now, multiply equation (2) by 2, we get 2Dy – 4x = 10e^t ............(4) By subtracting equation (4) from equation (3), we get D²x + 4x = –5e^t. ⇒ (D² + 4)x = –5e^t. ........(5) Its auxiliary equation is m² + 4 = 0 ⇒ m = \(\pm\)2i Therefore, complementary function is x = C₁cos2t + C₂sin2t Particular integral is x \(=\frac{1}{D^2+4}(-5e^t)\) \(\Rightarrow \mathrm x=\frac{-5e^t}{1^2+4}\) \(=\frac{-5e^t}{1+4}\) \(=\frac{-5e^t}{5}\) \(=-e^t\) Hence, complete solution of equation (5) is x = C.F + P.I = C₁cos2t + C₂sin2t – \(e^t\) \(\Rightarrow \frac{d\mathrm x}{dt}\) = –2C₁sin2t + 2C₂cos2t – \(e^t\) \(\Rightarrow\) 2y = 5e^t – \(\frac{d\mathrm x}{dt}\) (From equation (1)) = 5e^t –(–2C₁sin2t + 2C₂cos2t – e^t) = 6e^t + 2C₁sin2t – 2C₂cos2t ⇒ y = 3e^t + C₁sin2t – C₂cos2t Hence, the solution of given system of equation is x = C₁cos2t + C₂sin2t – e^t .........(6) And y = C₁sin2t – C₂cos2t + 3e^t .......(7) Given that when t = 0, then x = –1 and y = 3 Then from equation (6) and (7), we get –1 = C₁ – 1 ⇒ C₁ = –1 + 1 = 0 And 3 = –C₂ + 3 ⇒ C₂ = 3 – 3 = 0 Hence, the solution of given system of equations is x = –e^t And y = 3e^t (By putting C₁ = 0 & C₂ = 0 in equations (6) and (7))

40.

Solve dx/dt +2y =5e^t ; dy/dt-2x = 5e^t, given that x = -1, y=3 when t=0

Answer»

\(\frac{d\mathrm x}{dt}+2y = 5e^t\)

\(\frac{dy}{dt}-2\mathrm x=5e^t\)

By taking \(\frac{d}{dt}\) = D, we get

Dx + 2y = 5et ........(1)

Dy – 2x = 5et .........(2)

Now, operate D on both sides of equation (1), we get

D²x + 2Dy = 5De^t = 5e^t .........(3) \((\because De^t = \frac{d}{dt}e^t = e^t)\)

Now, multiply equation (2) by 2, we get

2Dy – 4x = 10e^t ............(4)

By subtracting equation (4) from equation (3), we get

D²x + 4x = –5e^t.

⇒ (D² + 4)x = –5e^t. ........(5)

Its auxiliary equation is m² + 4 = 0

⇒ m = \(\pm\)2i

Therefore, complementary function is

x = C₁cos2t + C₂sin2t

Particular integral is x \(=\frac{1}{D^2+4}(-5e^t)\)

\(\Rightarrow \mathrm x=\frac{-5e^t}{1^2+4}\) \(=\frac{-5e^t}{1+4}\) \(=\frac{-5e^t}{5}\) \(=-e^t\)

Hence, complete solution of equation (5) is

x = C.F + P.I = C₁cos2t + C₂sin2t – \(e^t\)

\(\Rightarrow \frac{d\mathrm x}{dt}\) = –2C₁sin2t + 2C₂cos2t – \(e^t\)

\(\Rightarrow\) 2y = 5e^t – \(\frac{d\mathrm x}{dt}\) (From equation (1))

= 5e^t –(–2C₁sin2t + 2C₂cos2t – e^t)

= 6e^t + 2C₁sin2t – 2C₂cos2t

⇒ y = 3e^t + C₁sin2t – C₂cos2t

Hence, the solution of given system of equation is

x = C₁cos2t + C₂sin2t – e^t .........(6)

And y = C₁sin2t – C₂cos2t + 3e^t .......(7)

Given that when t = 0, then x = –1 and y = 3

Then from equation (6) and (7), we get

–1 = C₁ – 1 ⇒ C₁ = –1 + 1 = 0

And 3 = –C₂ + 3 ⇒ C₂ = 3 – 3 = 0

Hence, the solution of given system of equations is

x = –e^t

And y = 3e^t (By putting C₁ = 0 & C₂ = 0 in equations (6) and (7))

Discussion

41.	5. What do you understand by Gaussian surface? Is always closed surface?
Answer» A gaussian surface is a 3 dimensional closed surface which is used to calculate flux. A gaussian surface is always a closed surface.

Discussion

42.	Assertion:- At point P object is speeding down. ltbegt Reason:- If acceleration is positive, object must speed up.A. If both Assertion `&` Reason are True `&` the Reason is a correct explanation of the Assertion.B. If both Assertion `&` Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion `&` Reason are False.
Answer» Correct Answer - C

Discussion

43.	Assertion:- Two stones are simultaneously projected from level ground from same point with same speeds but different angles with horizontal Both stones move in same vertical plane. Then the two stones may collide in mid air. Reason:- `For two stones projected simultaneously from same point with same speed at different angles with horizontal, their trajectories must intersect at some point except projection point.A. If both Assertion `&` Reason are True `&` the Reason is a correct explanation of the Assertion.B. If both Assertion `&` Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion `&` Reason are False.
Answer» Correct Answer - D

Discussion

44.	Figure shows sequence of large number of photograph of on object moving vertically under gravity. A motion picture of this photograph is run backward. `overset(circ)overset(circ)overset(circ)overset(circ)overset(circ)overset(circ)(circ)` Statement-1: In time reversal sequence the gravitational acceleration will appear to be upward. Statement-2: A time reversal operation changes every `vec(v)to -vec(v)`.A. Statement -1 is true, statement-2 is true and statement-2 is correct explanation for statement-1B. Statement-1 is true, statement-2 is true and statement -2 is NOT the correct explanation for statement-1C. Statement-1 is true, statement-2 is false.D. Statement-1 is false, statement -2 is true.
Answer» Correct Answer - D The velocity will be upwards but decreasing, so acceleration will g downwards.

Discussion

45.	I-V characteristics of a copper wire of length L and area of cross-section A is shown in figure. The slope of the curve becomes(1) More if the experiment is performed at higher temperature (2) More if a wire of steel of same dimension is used(3) More if the length of the wire is increased (4) Less if the length of the wire is increased
Answer» Correct option (4) Less if the length of the wire is increased Explanation: Slope of V-i curve =R(=ρl/A). But in given curve axis of i and V are interchanged. So slope of given curve =1/R(=A/ρl) i.e. with the increase in length of the wire. Slope of the curve will decrease.

45.

I-V characteristics of a copper wire of length L and area of cross-section A is shown in figure. The slope of the curve becomes(1) More if the experiment is performed at higher temperature (2) More if a wire of steel of same dimension is used(3) More if the length of the wire is increased (4) Less if the length of the wire is increased

Answer»

Correct option (4) Less if the length of the wire is increased

Explanation:

Slope of V-i curve =R(=ρl/A). But in given curve axis of i and V are interchanged. So slope of given curve =1/R(=A/ρl) i.e. with the increase in length of the wire. Slope of the curve will decrease.

Discussion

46.	When temperature increases, the frequency of a tuning forkA. increasesB. decreasesC. remains sameD. Increases or decreases depending on the material.
Answer» Correct Answer - B The length of the arms of the tuning fork will decrease

Discussion

47.	Calculate relative rate of effusion of `O_(2)` to `CH_(4)` through a container containing `O_(2)` and `CH_(4)` in 3:2 mass ratio:A. `(3sqrt(2))/(4)`B. `(3sqrt(2))/(8)`C. `(3)/(2sqrt(2))`D. `(3sqrt(2))/(2)`
Answer» `(r_(o_(2)))/(r_(CH_(4)))=(n_(o_(2)))/(n_(CH_(4)))=sqrt((M_(CH_(4)))/(M_(O_(2))))` `=(3)/((2)/(2))=sqrt((16)/(32))` `=(3)/(4sqrt(2))`

Discussion

48.	The standard EMF for the given cell reaction `Zn + Cu^(2+) rarr Cu + Zn^(2+)` is `1.10 V` at `25^(@)C`. The EMF for the cell reaction, when `0.1 M Cu^(2+)` and `0.1 M Zn^(2+)` solutions are used, at `25^(@)C` is -A. `1.10 V`B. `0.110 V`C. `-1.10 V`D. `-0.110 V`
Answer» Correct Answer - A `E_("cell") = E_("cell")^(@)-(0.059)/(2) "log" ([Zn^(+2)])/([Cu^(+2)])` `= 1.10 - (0.059)/(2)"log"(0.1)/(0.1)` `= 1.10 V`

Discussion

49.	Which type of defect is product when `NaCl` is dopped with `AlCl_(3)`?
Answer» Impurity defect. Two cation vacancies are created. A sustitutional solid solution is formed because `3 Na^(o+)` ions are replaced by on `Al^(3+)` ion in the lattice site.

Discussion

50.	The relative rates of effusion of `O_(2)` to `CH_(4)` through a containe containing `O_(2)` and `CH_(4)` in `3 : 2` mass ratio will be-A. `(3sqrt(2))/(4)`B. `(3)/(4sqrt(2))`C. `(3)/(2sqrt(2))`D. None of these
Answer» Correct Answer - B `{:(pi prop (n)/(sqrt(Mw)) , n = "mole"),(,M_(w)=mol.mass):}` `(r_(O_(2)))/(r_(Ch_(4)))= ((3)/(32sqrt(32)))/((2)/(16sqrt(16)))=3/(4sqrt(2))`

Discussion

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(a) Which gas is produced when dilute hydrochloric acid is added to a reactive metal? Write the chemical reaction when iron reacts with dilute H2SO4. (b) What would you observe when Zinc is added to a solution of iron (II) sulphate? Write the chemical reaction that takes place.

In birds and mammals the left and right side of the heart are separated. Give reasons.

Explain the meanings of malleable and ductile. Give example of each, a malleable element and a ductile element.

State the events occurring during the process of photosynthesis. Is it essential that these steps take place one after the other immediately?

The number of carbon compounds is more than those formed by all other elements put together. Justify the statement by giving two reasons.

(I) Why the taste or smell of fat change when left for a long time? What can you do to prevent it?(II) Name the gas used to prevent the chips from getting oxidised?

Give a test that can be used to confirm the presence of carbon in a compound. With a valency of 4, how is carbon able to attain noble gas configuration in its compounds?

Give a test that can be used to confirm the presence of carbon in a compound. Write its three allotropes.

Explain inductive effect and resonance effect

Which of the following respresent the variation of conducatnes of solution if weak base `NH_(4)OH` is titrated with dilute HCl?A. B. C. D.

Find the `DeltapH(` initial `pH -` final `pH)` when `100 ml 0.01M HCl` is added in a solution containing `0.1m` moles of `NaHCO_(3)` solution of negligible volume `(k_(a_(1))=10^(-7),k_(a_(2))=10^(-11)` for `H_(2)CO_(3)) :`A. 6+2 log3B. 6-log3C. 6+2log2D. 6-2log3

2 conducting objects one with charge of +Q and another with -Q are kept on x-axis at `x=-3` and `x -+4` respectively. The electric field on the x - axis is given by `3Q(x^(2)+4/3)` What is the capacitance C of this configuration of objects. Fill `1/C` (in `F^(-1))` in OMR sheet.

When 100 mL of 0.1 M NaCN solution is titrated with 0.1 M HCl solution the variation of pH of solution with volume of HCl added will be :A. B. C. D.

To a 100 mL of 0.1 M weak acid HA solution 22.5 mL of 0.2 M solution of NaOH are added.Now, what volume of 0.1 M NaOH solution be added into above solution, so that pH of resulting solution be 4.7: [Given :`(K_b(A^(-))=5xx10^(-10)]`A. 5 mLB. 20 mLC. 10 mLD. 15 mL

In single slit diffraction pattern :A. central fringe has negligible width than othersB. all fringes are of some widthC. central fringe does not existD. none of the above

verify Euler's theorem for the following functions, u=ax2+2hxy+by2

A liquid is contained in a vertical tube of semi-circular cross-section. The angle of contact is zero fig. What is the ratio of the force of surface tension on the curved part and the flat part of the tube? A. `2 : pi`B. `1 : pi`C. `3 : pi`D. `2.7 : pi`

Solve dx/dt +2y =5e^t ; dy/dt-2x = 5e^t, given that x = -1, y=3 when t=0

5. What do you understand by Gaussian surface? Is always closed surface?

When temperature increases, the frequency of a tuning forkA. increasesB. decreasesC. remains sameD. Increases or decreases depending on the material.

Calculate relative rate of effusion of `O_(2)` to `CH_(4)` through a container containing `O_(2)` and `CH_(4)` in 3:2 mass ratio:A. `(3sqrt(2))/(4)`B. `(3sqrt(2))/(8)`C. `(3)/(2sqrt(2))`D. `(3sqrt(2))/(2)`

The standard EMF for the given cell reaction `Zn + Cu^(2+) rarr Cu + Zn^(2+)` is `1.10 V` at `25^(@)C`. The EMF for the cell reaction, when `0.1 M Cu^(2+)` and `0.1 M Zn^(2+)` solutions are used, at `25^(@)C` is -A. `1.10 V`B. `0.110 V`C. `-1.10 V`D. `-0.110 V`

Which type of defect is product when `NaCl` is dopped with `AlCl_(3)`?

The relative rates of effusion of `O_(2)` to `CH_(4)` through a containe containing `O_(2)` and `CH_(4)` in `3 : 2` mass ratio will be-A. `(3sqrt(2))/(4)`B. `(3)/(4sqrt(2))`C. `(3)/(2sqrt(2))`D. None of these