Yes. As the skin of sheep secrete grease like oily substance, fleece is generally attracted with much dust and dirt.

Correct option is (c) common traits and personal disposition

Correct option is (a) a function of stimulation

a. Constant internal environment permits all biochemical reaction and physiological functions to proceed with maximal efficiency and thus enhance the overall fitness of the species. 

b. Organisms can overcome stressful external conditions with the following ways:

i. Regulation: Maintaining internal environment by maintaining constant body temperature or osmotic concentration. 

ii. Suspend (conform): By suspending metabolic activities through hibernation or aestivation diapause. 

iii. Migration: Organisms migrate temporarily to more hospitable areas.

Correct option is (b) surface and source trait

Correct option is (b) social theory

Correct option is (c) trait and type theory

Correct option is (C) 60

Let there are x number of boys and y number of girls in class.

According to given conditions, we have

8x + 4y = 400                    _____________(1)

And 5x + 10y = 400           _____________(2)

\(\Rightarrow\) 2x + y = 100                 _____________(3)    (By dividing both sides by 4 in equation (1))

And x + 2y \(=\frac{400}5=80\)      _____________(4)    (By dividing both sides by 5 in equation (2))

\(\Rightarrow\) 2 (80 - 2y) + y = 100   (From (3) & (4))

\(\Rightarrow\) 160 - 4y + y = 100

\(\Rightarrow\) 4y - y = 160 - 100

\(\Rightarrow\) 3y = 60

\(\Rightarrow y=\frac{60}3=20\)

\(\therefore\) x = 80 - 2y             (From (4))

= 80 - 40

= 40

\(\therefore\) x+y = 40+20 = 60

Total number of students in the class is 60.

Correct option is C) 60

Correct option is (C) 52

Let Ram, Shyam, Tarun and Varun had amount of Rs x, y, z and w respectively.

\(\because\) They had a total amount of Rs 240 together.

\(\therefore\) x+y+z+w = 240        ______________(1)

According to other given conditions, we have

\(x=\frac{y+z+w}2\)

\(\Rightarrow\) y+z+w = 2x        ______________(2)

\(y=\frac{x+z+w}3\)

\(\Rightarrow\) x+z+w = 3y        ______________(3)

and \(z=\frac{x+y+w}4\)

\(\Rightarrow\) x+y+w = 4z        ______________(4)

From (1) & (2), we obtain

x+2x = 240

\(\Rightarrow\) 3x = 240

\(\Rightarrow\) x = 80

From (1) & (3), we obtain

y+3y = 240

\(\Rightarrow\) 4y = 240

\(\Rightarrow\) y = 60

From (1) & (4), we obtain

z+4z = 240

\(\Rightarrow\) 5z = 240

\(\Rightarrow\) z = \(\frac{240}5\) = 48

Put x = 80, y = 60, z = 48 in (1), we get

80+60+48+w = 240

\(\Rightarrow\) w = 240 - 188

= 52

Hence, the amount which Varun has Rs 52.

Correct option is C) 52

Correct option is (C) 3

Let Ramu had x Rs 10, y Rs 50 & z Rs 100 notes.

\(\therefore\) x+y+z = 13         ______________(1)

\((\because\) Ramu had total 13 notes)

And 10x + 50y + 100z = 830

\(\Rightarrow\) x + 5y + 10z = 83     ______________(2)

\((\because\) Total value of notes with Ramu was Rs 830)

Ramu had more Rs 100 notes than Rs 50 notes.

i.e., z > y         ______________(3)

Since, Ramu has Rs 10, 50 & 100 notes with him and total notes are 13.

\(\therefore\) 0 < x < 13, 0 < y < 13 & 0 < z < 13    ______________(4)

Subtract equation (1) from (2), we get

4y + 9z = 83 - 13

\(\Rightarrow\) 4y + 9z = 70

\(\Rightarrow y=\frac{70-9z}4\)         ______________(5)

With the help of inequilities (3) & (4) and equation (5), we can conclude the value of y & z.

\(\because\) z > y > 0

\(\Rightarrow\) z > 1      \((\because y\geq1\) & z > 1)

(i) If z = 2 then \(y=\frac{52}4=13\)

but y < z

So, it is not possible.

(ii) If z = 3 then \(y=\frac{43}4\)      (Not possible)

(Because notes are never be in fraction)

(iii) If z = 4 then \(y=\frac{34}4=8.5\)   (Not possible)

(iv) If z = 5 then \(y=\frac{25}4=6.25\)   (Not possible)

(v) If z = 6 then \(y=\frac{16}4=4\)

Then x = 13 - y - z

= 13 - 6 - 4

= 13 - 10 = 3

Also, x + 5y + 10z = 3+20+60 = 83     (Satisfied)

\(\therefore\) Ramu has 3 Rs 10, 4 Rs 50 & 6 Rs 100 notes.

\(\therefore\) The number of Rs 10 notes with Ramu is 3.

Correct option is C) 3

1. 1. ungrateful

2. starve

3. right

4. Nothing

5. world

2. 1. unwilling

2. dead

3. found

4. disappointment

5. meeting

6. endure

6. free

(1) I had a narrow escape this morning when a speeding car unexpectedly swerved into my lane.

(2) Always try to remain calm and sensible when others are panicking.

(3) The apology of that poor man caught me into a tight situation to refuse.

(4) This has no relevance to anything that will happen today.

(5) I was terribly confused hearing her problem.

1. 1. for

2. the

3. then

4. of

5. across

6. if

2. 1. as though

2. and

3. of

4. from

5. the

6. under

1. 1. saying

2. would follow

3. had been carried

2. 1. to ache

2. was forced

3. followed

Correct option is (A) 3

Let Ram has x number of Rs 1 coins, y number of Rs 2 coins and z number of Rs 5 coins.

\(\therefore\) x+y+z = 18    ____________(1)    \((\because\) Ram has total 18 coins)

Their value is Rs 54

\(\therefore\) x + 2y + 5z = 54    ____________(2)

Given that number of Rs. 2 coins is more than of Rs. 5 coins.

\(\therefore\) y > z    ____________(3)

\(\because\) Total number of coins is 18.

\(\therefore0<x\leq18,0<y\leq18\;\&\;0<z\leq18\)    ____________(4)

Subtract equation (1) from (2), we obtain

y + 4z = 36    ____________(5)

By considering inequilities (3) & (4), we can conclude the possible values of y & z.

(i) If z = 1, then y = 32    (From (5))

which is contradictions of inequility (4).

(ii) If z = 2 then y = 28

which is not possible.

(iii) If z = 3 then y = 24 which is not possible.

(iv) If z = 4 then y = 20 which is not possible.

(v) If z = 5 then y = 16 then y+z = 16+5 = 21 but total coins are 18.

Hence, this case is not possible.

(vi) If z = 6 then y = 12 then y+z = 6+12 = 18

Then, x = 0 which is not possible because Ram has Rs 1 coin.

(vii) If z = 7 then y = 36 - 28 = 82    (From (5))

Then x = 18 - y - z

= 18 - 8 - 7

= 18 - 15 = 3

Also x + 2y + 5z = 3+16+35

= 54             (Satisfied)

Hence, Ram has 3 Rs 1 coin, 8 Rs 2 coin and 7 Rs 5 coin.

Hence, number of Rs 1 coin Ram has = 3.

Correct option is A) 3

Correct option is (D) x = 1

\(\sqrt{5x-1}+\sqrt {x-1} = 2\)       __________(1)

\(\Rightarrow\) \(\sqrt{5x-1}=2-\sqrt {x-1}\)

\(\Rightarrow\) \((\sqrt{5x-1})^2=(2-\sqrt {x-1})^2\)     (By squaring both sides)

\(\Rightarrow\) 5x - 1 = 4 + (x - 1) \(-4\sqrt{x-1}\)    \((\because(a-b)^2=a^2+b^2-2ab)\)

\(\Rightarrow\) 5x - 1 - 4 - x + 1 \(=-4\sqrt{x-1}\)

\(\Rightarrow\) 4x - 4 \(=-4\sqrt{x-1}\)

\(\Rightarrow\) \((4x-4)^2=16(x-1)\)

\(\Rightarrow\) \(16(x-1)^2=16(x-1)\)

\(\Rightarrow\) \((x-1)^2=x-1\)

\(\Rightarrow\) \((x-1)^2-(x-1)=0\)

\(\Rightarrow\) (x - 1) (x - 1 - 1) = 0

\(\Rightarrow\) (x - 1) (x - 2) = 0

\(\Rightarrow\) x - 1 = 0 or x - 2 = 0

\(\Rightarrow\) x = 1 or x = 2

Put x = 2 in equation (1), we obtain

\(\sqrt{5x-1}+\sqrt {x-1} = 2\)

\(\Rightarrow\) \(\sqrt{10-1}+\sqrt{2-1} = 2\)

\(\Rightarrow\) 3+1 = 2

\(\Rightarrow\) 4 = 2            (Not satisfied)

Hence, x = 2 is not a solution of given equation.

Put x = 1 in equation (1), we obtain

\(\sqrt{5x-1}+\sqrt {x-1}\) \(=\sqrt{5-1}+\sqrt{1-1}\)

= 2+0 = 2    (Satisfied)

Hence, x = 1 is a solution of given equation.

Correct option is D) x = 1

Solution :-

A. ‘Death’ has gone in search of another victim,

Correct option is (C) 12

Let the age of mother be x years and

age of her son be y years.

12 years ago, the mother age was (x - 12) years

14 years ago, the mother age was (x - 14) years

9 years hence, the mother's age will be (x+9) years

and her son's age will be (y+9) years

According to given conditions, we have

x+y = 2 (x - 12)

\(\Rightarrow\) x+y = 2x - 24

\(\Rightarrow\) y = x - 24          ______________(1)

And (x+9) + (y+9) = 3 (x - 14)

\(\Rightarrow\) x + y + 18 = 3x - 42

\(\Rightarrow\) x + x - 24 + 18 = 3x - 42

\(\Rightarrow\) 2x - 6 = 3x - 42

\(\Rightarrow\) 2x - 3x = 6 - 42

\(\Rightarrow\) -x = -36

\(\Rightarrow\) x = 36

\(\therefore\) y = 36 - 24 = 12      (From (1))

Hence, her son’s present age is 12 years.

Correct option is C) 12

Correct option is (A) no root

Given equation is

\(\sqrt{x+4}-\sqrt{x-3}+1=0\)       ______________(1)

\(\Rightarrow\) \(\sqrt{x+4}+1=\sqrt{x-3}\)

\(\Rightarrow\) \((\sqrt{x+4}+1)^2=x-3\)      (By squaring both sides)

\(\Rightarrow x+4+2\sqrt{x+4}+1=x-3\)

\(\Rightarrow2\sqrt{x+4}=x-3-x-5\)

\(\Rightarrow2\sqrt{x+4}=-8\)

\(\Rightarrow\) 4 (x+4) = 64        (By squaring both sides)

\(\Rightarrow\) x+4 = \(\frac{64}4\) = 16

\(\Rightarrow\) x = 16 - 4 = 12

Put x = 12 in equation (1), we get

\(\sqrt{12+4}-\sqrt{12-3}+1=0\)

\(\Rightarrow\) \(\sqrt{16}-\sqrt{9}+1=0\)

\(\Rightarrow\) 4 - 3 + 1 = 0

\(\Rightarrow\) 2 = 0        (Not satisfied)

Hence, x = 12 is not a solution of given equation.

Hence, the given equation has no root.

Correct option is A) no root

Correct option is (A) 3x – y = 1, 6x – 2y = 5

(A) \(\frac{a_1}{a_2}=\frac36=\frac12,\)

\(\frac{b_1}{b_2}=\frac{-1}{-2}=\frac12\)

and \(\frac{c_1}{c_2}=\frac{-1}{-5}=\frac15\)

\(\because\) \(\frac12\neq\frac15\)

\(\therefore\) \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\)

Therefore, this system has no solution.

Hence, this system of equations is inconsistent.

(B) \(\frac{a_1}{a_2}=\frac4{12}=\frac13,\)

\(\frac{b_1}{b_2}=\frac{6}{18}=\frac13\)

and \(\frac{c_1}{c_2}=\frac{-7}{-21}=\frac13\)

\(\therefore\) \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

\(\therefore\) This system has infinitely many solutions.

Hence, this system of equations is consistent.

(C) \(\frac{a_1}{a_2}=\frac49,\)

\(\frac{b_1}{b_2}=\frac98\)

\(\because\) \(\frac49\neq\frac98\)

\(\therefore\) \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\)

\(\therefore\) This system has unique solution.

Hence, this system of equations is consistent.

(D) \(\frac{a_1}{a_2}=\frac49,\)

\(\frac{b_1}{b_2}=\frac{12}{9}=\frac43\)

\(\because\) \(\frac49\neq\frac43\)

\(\therefore\) \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\)

\(\therefore\) This system has unique solution.

Hence, this system of equations is consistent.

Correct option is A) 3x – y = 1, 6x – 2y = 5 

A = {x ∈ Z : 0 ≤ x ≤ 12} = {0,1, 2,3,4,5,6,7,8,9,10,11,12}  and 

R = {(a, b): |a – b| is a multiple of 4} 

For any element a ∈ A, we have (a, a) ∈ R 

⇒ |a – a| = 0 is a multiple of 4. 

∴ R is reflexive. 

Now, let (a, b) ∈ R ⇒|a – b| is a multiple of 4. ⇒|–(a – b)| is a multiple of 4 ⇒|b – a| is a multiple of 4. ⇒ (b, a) ∈ R 

∴ R is symmetric. 

Now, let (a, b), (b, c) ∈ R. 

⇒|a – b| is a multiple of 4 and |b – c| is a multiple of 4. 

⇒(a – b) is a multiple of 4 and (b – c) is a multiple of 4. 

⇒(a – b + b – c) is a multiple of 4 

⇒(a – c) is a multiple of 4 

⇒|a – c| is a multiple of 4 

⇒ (a, c) ∈R 

∴ R is transitive. 

Hence, R is an equivalence relation. 

The set of elements related to 1 is {1, 5, 9} since 

|1 – 1| = 0 is a multiple of 4

|5 – 1| = 4 is a multiple of 4

|9 – 1| = 8 is a multiple of 4

Given that A = {1, 2, 3, 4, 5} and R = {(a, b) : |a – b| is even} 

It is clear that for any element a ∈A, we have (which is even). 

∴ R is reflexive. 

Let (a, b) ∈ R. 

⇒ |a – b| is even 

⇒(a – b) is even 

⇒– (a – b) is even

⇒(b – a) is even 

⇒ |b – a| is even 

⇒(b, a) ∈ R 

∴ R is symmetric. 

Now, let (a, b) ∈ R and (b, c) ∈ R. 

⇒ |a – b| is even and |b – c| is even 

⇒ (a – b) is even and (b – c) is even 

⇒ (a – c) = (a – b) + (b – c) is even (Since, sum of two even integers is even) 

⇒ |a – c| is even 

⇒ (a, c) ∈ R 

∴ R is transitive. 

Hence, R is an equivalence relation. 

Now, all elements of the set {1, 2, 3} are related to each other as all the elements of this subset are odd. Thus, the modulus of the difference between any two elements will be even. 

Similarly, all elements of the set {2, 4} are related to each other as all the elements of this subset are even. 

Also, no element of the subset {1, 3, 5} can be related to any element of {2, 4} as all elements of {1, 3, 5} are odd and all elements of {2, 4} are even. Thus, the modulus of the difference between the two elements (from each of these two subsets) will not be even.

\(\cos^{-1}\left(\frac{-1}{2}\right)\)\(=\pi-\cos^{-1}\left(\frac{1}{2}\right)\) [ Formula: cos ^-1(-x) = -cos ^-1(x) ]

\(=\pi-\frac{\pi}{3}=\frac{2\pi}{3}\)

tan(-1)= -tan(1) [ Formula: tan ^-1(-x)= -tan ^-1 (x) ]

[ We know that \(\tan\frac{\pi}{4}=1\), thus \(\tan^{-1}\frac{\pi}{4}=1\) ]

\(=-\frac{\pi}{4}\)

  \(\sec^{-1}(-2)=\pi-\sec^{-1}(2)\) [ Formula: sec ^-1(-x)= π – sec ^-1(x) ]

\(=\pi-\frac{\pi}{3}=\frac{2\pi}{3}\)

Yes, boron forms covalent compound.

i. False 

Covalent molecules have definite shape described with the help of bond lengths and bond angles. 

ii. False 

The reactivity of covalent inorganic compounds is better understood from their structures. i

ii. True

(a) Gold.

(b) Potassium.

(c) Mercury.

Sodium hydride,Hydrogen sulphide

 Mercury(Hg) exists in liquid state at room temperature.

(a) Carbon dioxide and sulphur dioxide.

(b) Sodium oxide and magnesium oxide.

Take a small strip of magnesium and burn it. It forms white ashes of magnesium oxide collect the ashes of magnesium and add some distilled water to it. Test the solution with red and blue litmus papers. Note the colour changes. The solution turns red litmus into blue. It indicates given solution is basic. So metallic oxides are basic in nature.

Take a small amount of powdered sulphur in a deflagrating spoon and heat it. As soon as sulphur starts burning, introduce the spoon into a gas jar or tumbler. Cover the tumbler with lid to ensure that the gas produced does not escape. The gas formed is sulphur dioxide. Remove the spoon after some time but try to keep the jar covered. Add a small quantity of water into the tumbler and quickly replace the lid. Shake the tumbler well. Check the solution with red and blue litmus paper. The solution turns blue litmus into red. It indicates the solution is acidic. So non-metallic oxides are acidic in nature.

The answer is (d) Hg

Electrovalent Bonding

When an atom donates one, two or three electrons from its valence shell to another atom, which has the ability to accept these electrons, it is known as electrovalency. As a result of electrovalency, both these atoms achieve the structure of an inert gas. When the chemical bond occurs by the transfer of electrons from the atom of an element to the atom or atoms of another it is called Ionic or Electrovalent bond.

Electronic Nature of Metals and Non-metals

The atoms of all elements, except noble gases, have an incomplete outermost shell. . Noble gases have their outermost shell complete and hence they are not reactive or "inert".

Most elements are reactive and try to achieve the stability of the noble or inert gases by electron transfer or by electron sharing. Elements that can donate electrons are called metals. They form positive ions by losing electrons.

The elements that accept electrons are called non-metals. They form negative ions by gaining electrons. Metals have 1 to 3 electrons in the outermost shell of their atom and non-metals have 4 to 8 electrons in the outermost shell.

There are two exceptions to this rule: Hydrogen and helium. Hydrogen is a non-metal having 1 electron in the valence shell and helium too is an inert gas having 2 electrons in the valence shell.

Bonding

The tendency of an atom to take part in chemical combination is determined by the number of valence electrons (electrons in the outermost shell of an atom). The atoms acquire the stable noble gas configuration of having eight electrons in the outermost shell (called octet rule) during chemical combination.

The combination of atoms occurs in two ways: either by electrovalent bonding or covalent bonding. In all chemical reactions, it is the electrons from the outermost shell of an atom that are involved in interacting with other atoms, either by their transfer or by sharing.

PCC (Pyridinium chlorochromate).

-Cl being electron withdrawing group stabilizes the CICH₂COO^- anion and increases the

Therefore, chloroacetic acid has lower pKa value than acetic acid.

Due to electron withdrawing nature of -NO₂ group which increases the acidic strength and decreases the pk_a value.

Detailed Answer:

As -NO₂ group is an electron withdrawing group, removal of electon is easier. This increases the acidic strength and decreases the pK_a value

CH₃COOH contains α-hydrogens and hence gives HVZ reaction but HCOOH doesn’t contain α-hydrogen and hence doesn’t give HVZ reaction.

Metal X is obtained simply by heating their oxides with carbon. e.g., mercury is obtained from cinnabar.

2HgS(s)+3O₂(g) +Heat  →2HgO(s)+2SO₂(g)

2HgO(s) +Heat →2Hg(l)+O₂

Metals in the reactivity series are obtained by electrolytic reduction of molten ores e.g., in NaCl

NaCl → Na⁺+Cl^-

Na+e^-  →Na(at cathode)

2Cl^-  →Cl₂+2e^-(at anode)

Metals in the middle of the activity series can be obtained by heating with carbon, e.g.,

ZnO(s)+C(s)+Heat →Zn(s)+CO(g)

The moderately reactive metals are extracted by the reduction of their oxides with carbon, aluminium, sodium or calcium.

Example: When zinc sulphide (zinc blende ore) is strongly heated in air (roasted), it forms zinc oxide and sulphur dioxide. This process is called roasting. Then, zinc oxide is heated with carbon to form zinc metal. This process is termed as reduction.

Metals high up in reactivitv series are extracted by electrolytic reduction. 

While those in the middle are extracted first by converting into oxide and then reducing by carbon. 

Same method cannot be used because metals have more affinity for oxygen than carbon.  

Molten sodium chloride is taken for electrolytic reduction. The metals are deposited at the cathode and chlorine is liberated at anode. 

At cathode : Na⁺ + e^-→ Na  

At anode : 2Cl^-→ Cl₂ + 2e^-

Different methods are used for extracting metals belonging to category of highly reactive metals, moderately reactive metals and less reactive metals. This is because the extraction of a metal from its concentrated ore is essentially a process of reduction of the metal compound present in the ore. For example: Manganese metal is obtained by the reduction of its oxide with aluminium powder and not carbon. This is because carbon is less reactive than manganese. Carbon, which is a non-metal, is more reactive than zinc and it can be placed just above Zn in the reactivity series. Hence, carbon can reduce the oxides of zinc and all other metals below zinc to form metals

NO₂CH₂COOH > FCH₂COOH > C₆H₅COOH

[Hint : electron withdrawing effect.]

Electrolytic reduction: Aluminium and Sodium;

Reduction with carbon : Zinc, Iron and Tin;

Reduction with aluminium: Manganese

Solution A - Basic solution

Because, [H⁺] is lesser than 1.0 x 10^-7.

Solution B - Acidic solution

Because [H⁺] is Greater than 1.0 x 10^-7 m