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We know that tangent at any point of a circle is perpendicular to radius through the point of contact.

OA ⊥ PA and OB ⊥ PB

∴ ∠OAP = ∠OBP = 90°

⇒ ∠OAP + ∠OBP = 90° + 90° = 180°

In quadrilateral PAOB

∠PAO + ∠AOB + ∠OBP + ∠BPA = 360°

⇒ ∠AOB + ∠BPA = 360° – 180°

⇒ ∠AOB + ∠BPA=180°

∵ Sum of opposite angle of quadrilateral is = 180°.

∴ Quadrilateral PAOB is a cyclic quadrilateral

(d) 70.4

√4096 + √40.96

= 64 + 6.4

= 70.4

Answer is (C) 2\(\sqrt { 3 }\) cm

Join OA. since AT is the tangent on the circle an OA is the radius of the circle. and OA is the radius of the circle.

So, AT ⊥ OA i.e., ∠OAT = 90°

From the right angled ΔOAT

∠OTA = 30° and OT = 4 cm

cos 30° = \(\frac { AT }{ OT }\)

⇒ \(\frac { \sqrt { 3 } }{ 2 }\) = \(\frac { AT }{ 4 }\)

⇒ AT = \(\frac { 4\sqrt { 3 } }{ 2 } \)

= 2\(\sqrt { 3 }\)

Let OP intersect chord AB at point C.

To prove : AC = BC.

Proof : In ΔPCA and ΔPCB

PA = PB (tangents at circle from external point P.)

PC = PC (common side)

∠APC = ∠BPC [∵ tangents PA and PB, make equal angle with OP]

by SAS congruence,

ΔPCA = ΔPCB

⇒ AC = BC

Hence, OP bisects line segment AB.

(b) 64

= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15

= 64

From figure,

OP ⊥ PT 

i.e. ∠OPT = 90°

∠OPQ = ∠OPT – ∠QPT

∠OPQ = 90° – 60°

= 30°

In ΔOPQ, (equal radii of circle)

∠OQP = ∠OPQ = 30°

(Opposite angles are equal to equal sides)

Now

∠OQP + ∠OPQ + ∠POQ = 180°

Sum of angles = 30° + 30° + ∠POQ = 180°

∠POQ = 180° – 60° = 120°

Exterior ∠POQ = 360° – 120° = 240°

We know that angle subtended by centre of circle is twice the angle subtended on the circumference of circle.

∠POQ = 2∠PRQ

⇒ 240° = 2∠PRQ

⇒ ∠PRQ = \(\frac { { 240 }^{ \circ } }{ 2 } \) = 120°

Hence ∠PRQ = 120°.

(a) VO₂⁺< Cr₂O²⁻ ₇ < MnO^–₄  

greater the oxidation state, higher is the oxidising power.

Given 2 tan^-1(cosx) = tan^-1 (2cosecx)

⇒ tan^-1(2cosx/(1 - cos²x)) = tan^-1(2cosecx) [∵ 2tan^-1x = tan^-1(2x/(1 - x²))]

⇒ tan^-1(2cosx/sin2x) = tan^-1(2/sinx)

⇒ 2cosx/sin²x = 2/sinx ⇒ cosx/sinx = 1

⇒ cotx = 1 ⇒ cotx = cotπ/4 ⇒ x = π/4

a) 10¹⁰ = 10000000000

b) 2¹⁰ = 1024

c) 10³ + 10¹ = 1000 + 10 = 1010

∴ a) 10¹⁰, c) 10³ + 10¹ are divisible by ’10’.

[∵ Their units digits are ‘0’.]

b) 1024 is not divisible by 10.

[∵ Its units digit is 4.]

a) 205, b) 4560, d) 105, e) 235785 are divisible by 5. 

[∵ The units digit of the above numbers are either 0 (or) 5.] 

c) 402 is not divisible by 5. [∵ Its units digit is 2.]

(i) + 32 – (+12) = 32 – 12 = 20 

(ii) + 7 – (+15) = 7 – 15 = – 8 

(iii) (-14) – (-20) = – 14 + 20 = 20 – 14 = 6 

(iv) (-30) – (-15) = – 30 + 15 = 15 – 30 = – 15 

(v) 23 – (-10) = 23 + 10 = 33 

(vi) (-27) – 22 = – 27 – 22 = – 49

(i) 11 + (- 2) = 11 – 2 = 9 

(ii) (- 4) + (- 6) = – 4 – 6 = – 10 

(iii) (- 250) + 150 = – 250 + 150 = – 100 

(iv) (- 380) + (- 270) = – 380 – 270 = – 650 

(v) (- 14) + 4 = – 14 + 4 = – 10 (vi) (- 180) + (- 80) = – 180 – 80 = – 260

Enzyme present in apple is Poly phenol oxidase.

Organic synthesis involves functional group inter conversions. A particular functional group can be converted into other functional group by reacting it with suitable reagents. 

For example, the carboxylic acid group (-COOH) present in organic acids can be transformed into -COCl by treating the acid with SOCl₂ reagent..

In the organic reactions, that series of simple steps which collectively represent the chemical change, from substrate to product, this is called as the mechanism of the reaction. The slowest step in the mechanism determines the overall rate of the reaction.

• Positive mesomeric effect occurs, when the electrons move away from substituent attached to the conjugated system. 
• It occurs, if the electron releasing substituents are attached to the conjugated system. 
• The attached group has a tendency to release electrons through resonance, these electron releasing groups are usually denoted as +R or +M groups 
• Examples: -OH, -SH, -OR, -SR, -NH₂ etc.

Two examples for free radical initiators

• Azobisisobutyronitnie (AIBN) 
• Benzoyl peroxide

Substrate is an organic molecule reacts with reagent, which may be an organic, inorganic or any agent like heat, photon etc, that brings about the chemical change to form a product, this is known as organic reactions.

• Apples contains an enzyme called polyphenol oxidase (PPO) also known as tyrosinase. 
• Cutting an apple exposes its cells to the atmospheric oxygen and oxidizes the phenolic compounds present in apples. This is called the “enzymatic browning” that turns a cut apple brown.
• In addition to apples, enzymatic browning is also evident in bananas, pears, avocados and even potatoes.

The cleavage of C-Br bond in tert-butyl bromide leads to formation of Carbocation.

• Free radicals can disrupt cell membranes. 
• Increase the risk of many forms of cancer. 
• Damage the interior lining of blood vessels. 
• leads to a high risk of heart disease and stroke.

Alcohol on refluxing with K₂Cr₂O₇ gives Carboxylic acid.

Ethane undergo thrombolytic cleavage to form Methyl free radical.

The cleavage of C-H bond in aldehydes Leads to formation of Carbanion.

Sources for human body produces free radicals,

• Human body is exposed to X-rays. 
• Cigarette smoke. 
• Industrial chemicals.
• Air pollutants.

Carbonyl compounds especially ketones undergo reduction to form Secondary alcohols.

• Use vitamins and minerals to counter the effects of free radicals. 
• Fruits contains antioxidants which decrease the effects of free radicals.

Electron displacement occurring in saturated compounds along a carbon chain is termed as Inductive effect.

The addition of H to alkene is an example of effect +E.

Hydrolysis of alkyl halide is an example for Nucleophilic substitution.

(a) Fertile – Supik

Acidity of phenol was explained by R-effect.

(b) T – Shaped

Option : (b) Square pyramidal

Iron objects are painted so that air and moisture can not come in contact with the iron objects and hence no rusting takes place.

Oils help to prevent corrosion by excluding moisture and oxygen from the metal surface. All metals will form a thin layer of oxide on the surface and oil etc. can penetrate into this, helping to stabilise it and prevent destructive corrosion

1. (ii) HCl gas

2. (i) Wet blue litmus paper

3. (i)TRUE

4. (ii) Acidic

5. (i) YES

(i) Increase in the strength of alkali. So the nature of solution will be basic.

(ii) Hydroxide ion or hydroxyl ion or OH^- ion.

(iii) A paper impregnated with the universal indicator is generally used to measure the pH of a solution.

1. (ii) 2 to 7

2. (i) 7 to 14

3. (ii) 6

4. (i) TRUE

5. (ii) FALSE

1. (ii) 6.2 - 7.6

2. (ii) TRUE

3. (i) Acids

4. (ii) Clean the mouth after eating

5. (ii) FALSE

The microbes used in fermentation of dhokla is Lactobacilli.

• HF is the most stable among all the hydrogen halides. 
• The thermal stability decreases from HF to HI. 
• This is due to a decrease in bond dissociation enthalpy and bond strength of H-X bond from F to I. 
• Down the group, from F to I, atomic size increases, bond length of H-X bond increases, bond polarity decreases and hence bond strength decreases. 
• F being of the lowest atomic size and the highest electronegativity, HF bond is stronger and highly thermally stable. 

• In the preparation of HNO₃, HCl, H₂PO₄, Na₂CO₃ sulphates, alums, alcohols, ethers, etc. 
• In the manufacture of dyes, fertilizers like ammonium sulphate, super phosphate, detergents. 
• As an electrolyte in lead storage battery. 
• As a dehydrating agent. 
• As an oxidising agent. 
• For refining petroleum. 
• As a pickling agent for removing layers of basic oxides from the metal surfaces like Fe, Cu, etc. before the metals are galvanized, electroplated, etc. 
• It is used as a laboratory reagent. 
• It is used in the manufacture of nitrocellulose products. 

Chlorine is used for :

• for sterilization of drinking water. 
• bleaching wood pulp required for the manufacture of paper and rayon, cotton and textiles are also bleached using chlorine.
• in the manufacture of organic compounds like CHCl₃, CCl₄, DDT, dyes and drugs.
• in the extraction of metals like gold and platinum.
• in the manufacture of refrigerant like Freon (i.e., CCl₂F₂).
• in the manufacture of several poisonous gases like mustard gas (Cl-C₂H₄ -S-C₂H₄ -Cl), phosgene (COCl₂) used in warfare.
• in the manufacture of tear gas (CCl₃NO₂).

Ozone decomposes to liberate nascent oxygen, hence it is a powerful oxidising agent.

O_3(g) → O_2(g) + O

For example :

(i) It oxidises lead sulphide (PbS) to lead sulphate (PbSO₄).

pbS_(s) + 4O_3(g) → pbSO_(s) + 4O_2(g)

(ii) Potassium iodide, KI is oxidised to iodine, l₂ in the solution.

2Kl_(aq) + H₂O₍₁₎ + O_3(g) → 2KOH_(aq) + I_2(s) + O_2(g)

• The H-F bond is stronger than H-Cl bond. 
• Hence, HF ionises less readily than HCl in H₂O, to give H⁺ions. 

Therefore,

HF is a weaker acid than HCl.

Correct answer is

D. HF

For HCl, pK_a = -7.0, hence its dissoClation constant is, K_a = 1 x 10^-7. 

For HI pK_a = – 10.0, hence its dissoClation constant is K_a = 1 x 10^-7. 

Hence HCl dissoClates more than HI. 

Therefore HCl is a stronger acid than HI.

• Bond length of HCl is less than HI. 
• Cl is more electronegative than I. 
• Hence bond strength of HCl is more than HI. 
• Therefore HCl dissociates less than HI, which makes HCl a weaker acid than HI.