Answer 1 :Given that, S_n = 4n − n²
First term will be, a = S₁ = 4(1) − (1)² = 4 − 1 = 3
Sum of first two terms = S₂ = 4(2) − (2)² = 8 − 4 = 4
Second term, a₂ = S₂ − S₁ = 4 − 3 = 1
Common difference, d = a₂ − a = 1 − 3 = −2

Now we find the n^th term is using formula
a_n = a + (n − 1)d = 3 + (n − 1) (−2) = 3 − 2n + 2 = 5 − 2n

Therefore, a₃ = 5 − 2(3) = 5 − 6 = −1
a₁₀ = 5 − 2(10) = 5 − 20 = −15

Hence, the sum of first two terms is 4. The second term is 1.

The 3^rd, 10^th, and n^th terms are −1, −15, and 5 − 2n respectively.

Answer 2:

For the given, A.P. 11, 8, 5, 2, …, Here First term is a = 11, common difference is d = a₂ − a₁ = 8 − 11 = −3

Let −150 be the n^th term of this A.P. Here we know, for an A.P, a_n = a + (n − 1) d, substitute the value from previous steps, we get

–150 = 11 + (n – 1)(–3)
–150 = 11 – 3n + 3
–164 = –3n
n = 164/3

Clearly, n is not an integer but a fraction, therefore, –150 is not a term of this A.P.

In a carbon atom, total number of electrons 6. The distribution of electrons in carbon atom is given by: First orbit or K-shell = 2 electrons and Second orbit or L-shell = 4 electrons. So, we can write the distribution of electrons in a carbon atom as 2, 4.

In a sodium atom, total number of electrons is 11. The distribution of electrons in sodium atom is given by:First orbit or K-shell = 2 electrons, Second orbit or L-shell = 8 electrons and Third orbit or M-shell = 1 electron. So, we can write distribution of electrons in a sodium atom as 2, 8, 1.

A polyatomic ion is a group of atoms carrying a charge (positive or negative).

For example,
Nitrate : NO₃^-1, Chlorate : ClO_₃^⁻1, Hydroxide : OH^⁻1, Acetate : C₂H₃O₂^-1, Carbonate : CO_₃^⁻2, Sulfate : SO_₄^⁻2, Phosphate : PO_₄^⁻3, Ammonium : NH_₄^⁺1, Cynaide : CN^-1

a) i) Phenol is treated with concentrated HNO₃ to obtain 2,4,6-trinitrophenol picric acid.

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

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

b) Butan−1−ol and butan−2−ol can be distinguished using Lucas reagent (ZnCl₂ + HCL), where butan−2−ol  would react with Lucas reagent in arround 5 minutes to give a white precipitate of 2-chlorobutene,  whereas butan−1−ol would not give any reaction at room temperature.

c) Increasing order of acidity can be given as

Ethanol < Water < Phenol

a) The cell can be represented as Al | Al³⁺ (0.01 M) | Ni²⁺ (0.1) | Ni

The cell potential is given by the following equation in this case

E_cell =  E^o_cell – RT / 2F ln [ (Al³⁺)² / (Ni²⁺)³]

Given that concentration of Al³⁺ ions case 0.01 M and Ni²⁺ ions is 0.1 M, E^o_cell = 1.41, putting the values in the equation, we get

E_cell =  1.41 V – 0.059 / 2log [ (0.01)² / (0.1)³]

E_cell =  1.38 V log [ 1 × 10^-4 / 1 × 10^-3]

E_cell =  1.38 – (–1) = 1.38 + 1 = 2.38 V

So EMF of the cell is 2.38 volt. 

b) For strong electrolytes the molar conductivity is increased only slightly on dilution. A strong electrolyte is completely dissociated in solution and thus furnishes all irons for conductance. However, at higher concentrations, the dissociated ions are close to each other and thus, the interionic attractions are greater. These forces retard the motion of the iron and thus conductivity is low.

With a decrease in concentration dilution, the ions move away from each other thereby feeling less attractive forces from the counterions. This results in an increase in molar conductivity with dilation. The molar conductivity approaches a maximum limiting value at infinite dilution designated as ⋀⁰_m

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

So, limiting molar conductivity for weak electrolytes is obtained by using KOHLRAUSCH LAW from the limiting molar conductivities of individuals ions (λ°).

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

⋀⁰_m = λ°₊ + λ°

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

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

c) Deficiency of vitamin B12 causes pernicious anemia.

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

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

Nucleotides: They are the building blocks of DNA RNA. These consist of a pentose sugar moiety attached to a nitrogenous base at one position and a phosphoric acid molecule at five position

Example: 

a) Conductivity ⋀_m of the solution is given by the following equation ⋀_m =  k/c, where k is dissociation constant and c is the concentration of solution. Here, given

conductivity k = 4.95 × 10^–5 S cm^–1

Limiting molar conductivity ⋀⁰_m = 390.5 S cm² mol^–1 

Concentration c = 0.001 mol L^–1 = 1 × 10^–3 mol L^–1

Substituting  the given values in the above equation

Molar conductivity ⋀_m = (4.95 × 10^–5 S cm^–1 / 1 × 10^–3 mol L^–1) 10³ cm³ L^–1 

= 49.5 S cm² mol^–1

Now, degree of dissociation α is given by α = ⋀_m / ⋀⁰_m . Putting the values

α = 49.5 / 390.5 S cm² mol^–1 = 0.1267

Dissociation constant, k, for acetic acid, can be given as 

k = cα² / (1 – α)

= [1 × 10^–3 mol L^–1 × (0.1267)² ] / (1 – 0.1267)

= 0.016 × 10^–3 mol L^–1 / 0.8733

= 1.8 × 10^–5 mol L^–1

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

E_cell = E°_cell – RT / 2F ln( [Al³⁺]² / [Al²⁺]³ )

c) A secondary battery can be recharged after use, bypassing current through it in the opposite direction so that it can be used again. For example, the most important secondary cell is the lead storage cell. It consists of a lead anode and the grid of lead packed with lead dioxide as a cathode. A total 38 percent solution of sulfuric acid is used as an electrolyte.

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

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

b) The key difference between red and white phosphorus is that the red phosphorus appears as dark red colored crystals whereas the white phosphorus exists as a translucent waxy solid that quickly becomes yellow when exposed to light.

Phosphorus is a chemical element that occurs in several different allotropes. The most common allotropes are red and white forms, and these are solid compounds. Furthermore, when exposed to light, the white form converts into the red form. However, there are several differences between these two allotropes. Let us discuss more details on the difference between red and white phosphorus.

c) XeF₆ reacts with NaF as follows : XeF₆ + NaF → Na⁺ [XeF₆]^–

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

e) The increasing order of acidic character can be written as HF < HCL < HBr < HI

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

(iii) F₂ is more reactive than CIF₃ because of small size of fluorine atom F-F bond, bond dissociation enthalpy is low thus is reactive whereas CIF₃ is more reactive than Cl₂ because CIF₃ is an interhalogen halogen compound with weak CI-F bond means compared to cl-cl bond due to the difference in atomic sizes (hence in ineffective overlap of orbitals).

Structure of XeF₂ is a linear:

Structure of h₄P₂O₇

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

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

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

a) Bithionol is the odd one here, as it is an antiseptic whereas others are tranquilizers.

b) Liquid dishwashing detergent are non-ionic type.

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

Two poisonous gas prepared from chlorine is phosgene (COCl₂) and tear gas (CCl₃NO₂).

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

Cu²⁺ (aq) (Blue) + 4NH₃(aq) ⇋ [Cu(NH₃)₄]²⁺ (aq) (Deep Blue)

a) Cl₂ is passed through slaked lime to give bleaching powder Ca(OCl)₂.

2Ca(OH)₂ + 2Cl₂ → Ca(OCl)₂ + CaCl₂ + 2H₂0

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

2Fe³⁺ + SO₂ +2H₂O → 2Fe₂+ +SO₄^2– +4H⁺

Antibiotics: These are the compounds produced by a microorganism and synthetically which either inhabit the growth of bacteria or kill a bacteria. Example is Penicillin.

Antiseptics : These are the chemical used to kill or prevent the growth of microorganism when applied to the living tissues. Example is Soframicun. 

Anionic detergents : These are sodium salts of sulphonated long-chain alcohols or hydrocarbons. In these the anionic part of the molecule is involved in the cleansing action. Example sodium lauryl sulphate.

Solution a) Transaction elements have partly filled d-orbitals due to which they have variable oxidation states which enables them to bind with a variety of ligands and hence from complex compounds.

Solution b) Oxidation of Zn and Zn²⁺ leads to a completely filled d¹⁰ configuration in in Zn²⁺, making it more stable. Also Mn/Mn²⁺ conversion leads to a half-filled stable d⁵ configuration of Mn²⁺ ion. Hence, E^o value of Zn/Zn² and Zn/Zn²⁺ conversion have negative values.

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

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

Germanium : Zone referring is used for refining Germanium. This method is based on principle that the impurities are more soluble in the melt than in the solid state of the matter.
Titanium : Titanium is refined by van Arkel method. This method is used for removal of oxygen and nitrogen present as impurity. The crude metal is heated in as an evacuated vessel with iodine to obtain metal iodide, which volatilizes being covalent. Later this metal iodide is decomposed through electrical heating to obtain the pure metal.

Two solutions having same as osmotic pressure at a given temperature are called isotonic solution.

Now in the given problem, the KCl and urea solutions are given to be isotonic. Osmotic pressure π is given by the equation

π =  (N₂/V)RT, where, n₂ = molecules of solute, V = volume of a solution in litre.

Also n₂ = w₂ / M_2, where w₂ = grams of solute and M₂ = 74.5 g mol^–1 Weight of KCL , w₂ = 1.9 g, V = 100 mL.

So, for KCl

π = (w₂/M₂ × V)RT

πRT_KCL= 1.9/(74.5 × 100) = 2.55 × 10^–4

Now as the solutions are isotonic at the same temperature : π RT_KCL= π RT_UREA

Hence, substituting the values of urea : 2.55 × 10^–4 = 3/M₂ × 100 ⇒ M₂ = 117.6

So, the experimentally determine molecular weight of urea is found to be as 117.6, so the degree of this consecration can be given as : 

Osmotic pressure (π) = experimentally determined. Molecular weight/ Actual molecular weight = 117.6/60 = 1.96 ≈ 2.

α = i – 1/ n – 1 = 1.96 – 1 / 2 – 1 = 96 %. 

So, Urea diamerised in the given experimental solution.

Solution a) In dust the dispersed phase is solid particles and dispersion medium is air means gas.

Solution b) Physisorption occurs only because of physical attractive forces like van der Walls forces between molecules of absorbate and absorbent. Therefore, that can be reversed on application of bigger forces, but chemisorption occurs due to chemical reaction between molecules of odsorbate and absorbent, hence it can't be reversed.

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

The reaction is a A + 2B → C

Solution a) It can be seen that when concentration of A is doubled keeping B is a constant then the rate increases by a factor of 4 (4.2 × 10^–2 to 1.68 × 10^–1). This indicates that the rate depends on the square of the concentration of the reactant A. Also when concentration of reactant B is made four times, keeping the concentration of reactant A is constant, the reaction rate also becomes four times (2.40 × 10^–2 to 6.0 × 10^–3). This indicates that the rate depends on concentration of reactant B to the first power

Solution b) So, the rate equation will be : Rate = k [A]² [B]. Overall order of reaction will be 2 + 1 = 3.

Solution c) Rate constant can be calculated by putting the given values in formula, we get

4.2 × 10^–2 M min^–1 = k (0.2)² (0.3) M

k = 0.042/0.012 = 3.5 min^–1

Solution a)

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

Ansertr b)

Electromagnetic electronic configuration of d⁵ Δ₀ < P is given by t_2g³ e_g² and Electromagnetic electronic configuration of d⁵ Δ₀ > P is given by t_2g⁵ e_g².

Boiling point increase on increasing the pressure in case of liquids. Water used for cooking attends higher temperature than usual boiling temperature inside the pressure cooker due to the existing high pressure inside the pressure cooker vessel. This leads so faster flow of water inside the vegetables or grains etc. resulting in faster cooking of food in a pressure cooker then in the cooking pan. 

Red Blood Cells shrink when placed in salline water because of exosmosie that is water comes out from the cell of surrounding more concentrated to equate the concentration. Whereas, when placed his distilled water concentration within the cell becomes more than the surroundings, as water comes inside and endosmosis takes place to equate the concentrations.

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

Hydrolysis of DNA yields a pentose sugar (i.e. β−D−2deoxyribose), phosphoric acid and nitrogen containing heaterocyclic compounds called bases (Adenine, Guanine, Cytosine and Thymine).

Cyclohexyl chloride his more reactive towards nucelophilic substitution reaction, because the carbon bearing the chlorine atom is deficient in electron and seeks a  nucleophile. In chlorobenzene the carbon bearing the halogen is a part of aromatic ring and is electron rich due to the electron density in the ring.

Cyclohexyl chloride his more reactive towards nucelophilic substitution reaction, because the carbon bearing the chlorine atom is deficient in electron and seeks a  nucleophile. In chlorobenzene the carbon bearing the halogen is a part of aromatic ring and is electron rich due to the electron density in the ring.

Here C₁ = C, C₂ = C, V₁ = V, V₂ = 0. Now initially, energy stored in first capacitor as second capacitor is uncharged.

U₁ = 1/2 CV₁² = 1/2 CV² ....(i)

Now, when C₁ and C₂ are connected the two capacitor form a parallel combination. Equivalent capacitance,

C' = C₁ + C₂ = C + C = 2C.

Final potential = Total charge / Total capacitance 

V ' = −q/2C [Total charge will be of first capacitor, which is distrubuted]

V ' = CV/2C = V/2

Final energy is stored in a combination.

U₂ = 1/2 C' V'² = 1/2 × 2C × (V/2)² = CV²/4 ...(ii)

On dividing equation first with second, we get

U₂/U₁ = (1/4 CV²) / (1/2 CV²)  = 1/2

U₂ : U₁ = 1 : 2

When the plates of the parallel plate capacitor is connected to a battery. Then the first insulated metal plates gets the positive charge till its potential become maximum. Then the charge will be leak to surroundings. So the negative charge will be induced on the nearer face of the second plate and the positive charge will be induced on its further plate.

Consider a capacitor of capacitance C. Initial charge and potential difference be zero. Let, a chart Q be given in small steps. Let any instant when charge on capacitor be q, the potential difference between its plates V = q/C.

Now work done in giving an additional charge dq is, dW = V dq = q/C dq 

Total work done in giving charge from 0 to Q is, 

W = ∫₀^Q Vdq = ∫₀^Q q/C dq

= 1/C [q²/2]₀^Q = 1/C [q²/2 − 0/2] = Q²/2C

W =(CV)²/2C = 1/2 × CV² = 1/2 QV [Q = CV]

Threfore, Electrostatic potential energy U = Q²/2C = 1/2 CV² = 1/2 QV

Solution a)

Consider an electric dipole of −q and +q separated by a distance 2a and placed in a free space. Let P be a point on equitorial line of dipole at a distance r from the center of the dipole.

Let E_A and E_B be the electric field at point P due to charges −q and +q. Then resultant electric field at point P is E = E_A and E_B, Now,

| E_A | = 1/4πε₀ • q/AP² = 1/4πε₀ • q/(r² + a²)²  (Along PA) and 

| E_A | = 1/4πε₀ • q/BP² = 1/4πε₀ • q/(r² + a²)²  (Along BP)

THus, the resultant intensity is the vector sum of E_A and E_B. E_A and E_B can be resolved into two components. The y-components cancel out each other and x-component will add up to give the resultant field. Therefore,

E = E_A cos θ + E_B cos θ. Now the right triangle ORB

cos θ = OP/BP = a / √(r² + a²)

E = 2 × 1/4πε₀ × q / (r² + a²) × a / √(r² + a²)

E = 2qa / 4πε₀ × (r² + a²)^3/2

Threfore the required expression is E = 1/4πε₀ × p/(r² + a²)^3/2 [because 2qa = p]

Solution b)

Let the two charges of +q each placed at point A and B at a distance 2 m apart in air.

Suppose the third charge Q which is unknown magnitude and charge is placed at a point O, on the line joining the other two charges such as OA equals x and OB equals 2 − x. For a system to be in equilibrium, net force on each three charges must be equal to zero.

If we assume that charge Q placed at point O is positive, then the force on it at O may be zero. But the force on charge q at point A or B will not be zero. It is because, the forces on a charge q due to the other chrges will act in the same direction. 

If the charge Q is negative, then the force is on q due to other two charges will act in opposite direction. 

Hence, Q will be negative in nature. For charge −Q to be in equilibrium. Force at charge −q due to the charge +q at point A should be equal and opposite to charge +Q at B

1/4πε₀ × Qq/x² = 1/4πε₀ • Qq/(2 − x)²

or (2 − x)² = x²

⇒ x = 2 − x

⇒ 2x = 2

⇒ x = 1 m

Therefore for the system to be in equilibrium a charge −Q is placed at a midpoint between the two charges of +q each.

Condition for total internal reflection is

• The ray must travel from a denser medium into a rarer medium.
• The angle of incidence in the denser medium must be greater than the critical angle for the pair of media.

Relation between critical and angle and refractive index :  

When a ray of light travel from denser to rarer medium, the ray bends away from the normal. When the angle of incidence is equal to the critical angle then the refracted ray grazes the surface of separation represented by ray  and .

By Snell's law, sin i/sin r = 1/n, where i = C, r = 90 degree, and n = refractive index of a denser medium. Therefore, 

sin C / sin 90 = 1/n ⇒ n = 1/sin C

Solution b)

For first lens, u₁ = −30, f₁ = 10 cm. Therefore, using from lens formula, we get

1/f₁ = 1/v₁ − 1/u₁

⇒ 1/v₁ = 1/f₁ + 1/u₁ = 1/10 − 1/30

⇒ v₁ = 15 cm

This means that image formed by first lens is at a distance of 15 cm to the right of the first lens. This images serves as the vertical object for second lens. Therefore, for second lens we have

f₂ = −10 cm, u = 15 − 5 = 10 cm

Therefore, 1/v₂ = 1/f₂ + 1/u₂ = −1/10 + 1/10 = 0 

⇒ v₂ = ∞

This means that the real image is formed by second lens his infinite distance. This acts as an object for third lens. For third lens f₃ = 30 cm and u₃ = ∞

1/v₃ = 1/f₃ + 1/u₃ = 1/30 + 1/∞ = 1/30 

⇒ v₃ = 30 cm

Final image is formed at a distance of 30 cm to the right of third lens.

Interference

• It is the result of interaction of light coming from two different wavefronts originating from two coherent sources.
•  All the bright fringes are the same intensity.

Diffraction

• It is the result of interaction of light come from different parts of same wavefronts.
• The bright fringes are of varying intensity i.e. intensity of bright fringes decreases from central bright fringe on either sides).

A sources of monochromatic light illuminates two narrow slits S₁ and S₂. The two illuminated slits act as the two coherent sources. The two slits is very close to each other and at equal distance from the source. The wavefront S₁ and S₂ is spread in all the direction and superpose and produces dark and bright fringe on a screen. Let the displacement of waves from S₁ and S₂ at point P on screen at time t is

y₁ = a₁ sin ωt

y₂ = a₂ sin (ωt + φ)

The resultant displacement at point P is given by y = y₁ + y₂

= a₁ sin ωt + a₂ sin (ωt + φ)

= a₁ sin ωt + a₂ sin ωt + a₂ cos φ + a₂ cos ωt + a₂ sin φ

= (a₁ + a₂ cos φ) sin ωt + a₂ sin φ cos ωt .....(i)

Let 

a₁ + a₂ cos φ = A cos φ .......(ii)

a₂ sin φ = A sin φ .......(ii)

Therefore, equation (i) becomes

y = A cos θ sin ωt + A sin θ cos ωt = A cos (ωt + θ)

y = A cos (ωt + θ), this is the resultant displacement. 

Now, squaring and adding both the equation second and third, we get 

A² cos² θ + A² sin² θ = (a₁ + a₂ cos φ)² + a₂² sin² φ

⇒ A² (cos² θ + sin² θ) = a₁²+ a₂² cos² φ + 2 a₁ a₂ cos φ + a₂² sin² φ

⇒ A² = a₁²+ a₂² (cos² φ + sin² φ) + 2 a₁ a₂ cos φ

⇒ A² = a₁²+ a₂² + 2 a₁ a₂ cos φ

This is intensity of light is directly proportional to the square of the amplitude. i.e.

I = a₁²+ a₂² + 2 a₁ a₂ cos φ

This is the expression for intensity at a point of interference pattern.

I = I₁+ I₂ + 2√(I₁ I₂) cos φ

Solution b)

Here, λ = 620 nm = 620 × 10⁻⁹, a = 3 × 10⁻³ m, D = 1.5 m

Distance of first order minimum from the center.

y₁ = Dλ/a = 1.5 × 620 × 10⁻⁹ / 3 × 10⁻³ 

y₁ = 3.1 × 10⁻⁴ m

Distance of third order Maxima on the same side is given by

y₂ = 7Dλ/2a = 7× 1.5 × 620 × 10⁻⁹ / 2 × 3 × 10⁻³

y₂ = 10.85 × 10⁻⁴ m

Separation between them y = y₂ − y₁

= 10.85 × 10⁻⁴ − 3.1 × 10⁻⁴ = (10.85 − 3.1) × 10⁻⁴ 

= 7.75 × 10⁻⁴ m

For a full wave rectifier, we used two junctions diodes as shown in the figure

During first half cycle of input AC signal the terminal S₁ is positive relative to S and S₂ is negative. Then diode D₁ is in forward biased and diode D₂ is reverse biased. Therefore, current flow in D₁ not in D₂. In the next half cycle S₁ is negative and S₂ is positive relative to S. Then D₁ is in reverse biased and D₂ is in forward biased. Therefore, current flows in D₂ not in D₁. Thus for input AC signal the output current is a continuous series of unidirectional pulse. The input and output waveforms are shown in the figure below.

Solution a) Step Down Transformer 

It is a device  used for converting high alternating voltage at low current into low alternating voltage at high current and vice versa. The device works on the principle of mutual induction. That means if the current of magnetic flux linked with a coil changes than an EMF is induced in the other coil. In step-down transformer N_P >N_S and transformation ratio is less than 1.

Energy losses are :

Copper losses : 

Due to resistance of winding in primary and secondary coils,  some electrical energy is converted into heat energy. 

Flux losses : 

Some of the flux produced in primary coil is not linked up with the secondary coils.

Hysteresis losses :

When the iron core is subjected to a cycle of magnetisation the core gets heated up dure to hysteresis known as is known as hysteresis loss.

Iron Losses :

The varying magnetic flux produces eddy current in the iron core which leads to the wastage of energy in the form of heat.

Solution b)

Length of wire line = 20 × 2 = 40 km

Resistance of wire line r = 40 × 0.5 = 50 Ω

Power to be supplied = 1200 kW = 1200 × 10³ W 

Voltage at which power supplied = 4000 V

Since, P = VI ⇒ I = P/V

⇒ I = 1200 x 10³ / 4000 = 300 A

Therefore, line power loss = I² × R

⇒ (300)² × 200 = 36 × 10⁵ W = 360 kW

Therefore, Line power loss in the form of heat is 360 kW.

Solution a) Consider an alternating emf is connected in series with an inductor, resistance R and capacitance C. Let, E and I be the instantaneous values of e.m.f. and current in the LCR circuit V_L, V_C and V_R be the instantaneous values of voltage across inductor, capacitor and resistor respectively. Then,

V_L = I X_L , V_C = IX_C and V_R = IR Here, X_L = ωL and X_C = 1/ωC

In an AC circuit V_R and I are in same phase, V_L leads I by π/2 and V_C lags I by π/2.

From the graph,  in right angled ΔOEA.

OE = √(OA² + AE²) = √(OA² + OD²) 

OE = √[ V_R² + (V_L − V_C)² ]

Therefore,

E = √[ (IR)² + (IX_L − IX_C)² ]

E = I √ [ R² + (X_L − X_C)² ]

I = E / √ [ R² + (X_L − X_C)² ] 

Therefore, effective resistance or impedance of LCR circuit is

Z = √ [ R² + (X_L − X_C)² ]  = √ [ R² + (ωL − 1/ωC)² ] 

b) At resonance X_L = X_C i.e. i X_L = i X_C or V_L = V_C

The voltages across inductance and capacitance are equal and have a phase difference of 180 degree at resonance

Solution c) Since the reactance of an inductor is zero for DC circuit, but the inductor offers resistance to an AC circuit. Therefore, the current decreases for the same inductor then it is connected from the AC source. 

When inductor is connected in the AC circuit :  

V = 200 V, f = 50 Hz, i = 0.5 A

⇒ i = V/R = V/XL = V/ωL

⇒ 0.5 = 200 / (2π × 50) [ Because ω = 2πf ]

L = 200 / (100 × 0.5 × 3.14) = 3/3.14 = 1.27H

Solution a) Energy corresponding to the wavelength 600 nm is −

E = hc/λ

= (6 × 10⁻³⁴ × 3 × 10⁸) / 600 × 10⁻⁹ J = 3.3 × 10⁻¹⁹

= 3.3 × 10⁻¹⁹ / 1.6 × 10⁻¹⁹ eV = 2.06 eV

The photon energy E = 2.06 eV is greater than the band gap for diode D₂ only. Hence, diode D₁ and D₃ will not be able to detect the given wavelength. 

Solution b)

A photodiode is operated reverse bias because in reverse bias it is easier to observe change in current with change in light intensity.

1. Properties : State

Ferro-magnetic Materials : They are solid.

Para-magnetic Materials : They can be solid, liquid or gas.

Dia-magnetic Materials : They can be solid, liquid or gas.  

2. Properties :  Effect of Magnet

Ferro-magnetic Materials : Strongly attracted by a magnet  

Para-magnetic Materials : Weakly attracted by a magnet 

Dia-magnetic Materials : Weakly repelled by a magnet  

3. Properties : Effect of temperature

Ferro-magnetic Materials : Above curie point, it becomes a paramagnetic 

Para-magnetic Materials : With the rise of temperature, it becomes a diamagnetic.  

Dia-magnetic Materials : No effect.

4. Properties : Examples  

Ferro-magnetic Materials : Iron, Nickel, Cobalt 

Para-magnetic Materials : Lithium, Molybdenum, and Magnesium 

Dia-magnetic Materials :  Copper, Silver and Gold

Solution a) Gauss law of magnetism : 

If a closed surface is imagined in a magnetic field the number of lines of force emerging from the surface must be equal to the number entering it. That is, the net magnetic flux out of any closed surface is zero. Gauss law signifies that magnetic monopoles does not exist.

Solution b)

• In a bar magnet, each lines of force, starts from a north pole and reaches the south pole externally and then goes from south pole to our north pole internally. Thus magnetic line of force forms a closed loop.
• No two lines of force will never intersect each other. 
• In a uniform field, the lines are parallel and equidistant from each other. 
• The lines of force are crowded near the poles.

Decay constant of a radioactive sample is defined as the ratio of its instantaneous rate of disintegration to the number of atoms present at a time.

dN/dt = −λn ⇒ λ = −dN/dt/N

Let the initial number of nuclei be N₀ at t = 0, dN/dt = 10000 at t = 20 Hr and dN'/dt = 5000 at t = 30 Hr.

We have, dN/dt = λn

For first case 

⇒ 10000 = λN, 

⇒ N = 10000/λ and N = N₀ e^−λt

⇒ 10000/λ = N₀ e^−20λ ...(i)

For second case

⇒ 5000 = λN', 

⇒ N' = 5000/λ and N' = N₀ e^−λt

⇒ 5000/λ = N₀ e^−30λ ...(ii)

On dividing equation first by equation second, we get

⇒ e^−20λ/e^−30λ = 2 

Taking log both the sides 

⇒ −20λ + 30λ = log 2

⇒ 10λ = 2.302 × 0.3010

⇒ λ = 0.0693

Therefore, Half life is τ = 0.693/0.0693 = 10 Hr

Wave theory cannot explain the following laws of photoelectric effect.

• The instantaneous emission of photo electrons
• Existence of threshold frequency for metal surface
• Kinetic energy of emitted electrons is independent of intensity of light and depends on frequency.

The concept of photon explain that energy is not only emitted and absorbed a in depends energy quanta, but also it propagates through space in definite quanta with the speed of light. 

It can explain all the above photoelectric effect, which wave theory cannot explain.

Given i_p = 30°,

By Brewster's law, µ = tan i_p. Therefore

⇒ µ = tan 30° = 1/√3

Since, µ = C/v

v = velocity of light in medium = C/µ

= 3 × 10⁸ / 1/√3

= 3√3 × 10⁸ m/s

= 5.196 × 10⁸ m/s

a) Self-Inductance : Self inductance of a coil is numerically equal to the amount of magnetic flux linked with the coil when and current flows through the coil. The S.I. unit of self inductance is henery (H) or weber per ampere 1 H = 1wb/A.

b) The force on the side AB of a rectangle is attractive as the current is flowing in the same direction and on the side CD will be repulsive the current is flowing in the opposite direction with respect to the straight conductor. The resultant magnetic force on sides AD and BC is zero. The side AB is the straight wire. So the net force will be attractive and rectangular loop will move towards the straight wire.

Now, the force between AB and straigjht wire is 

F₂ = µ₀ / 2π × I₁ I₂ / (a + b)

Therefore,

F_net = F₁ − F₂ = µ₀/2π × I₁ I₂ / a − µ₀/2π × I₁ I₂ / (a + b)

F_net = µ₀/2π × I₁ I₂ [1/a − 1/(a + b)]

F_net = µ₀/2π × I₁ I₂ [b/a(a + b)]

F_net = µ₀I₁ I₂ b/2πa(a + b)

a) Half life period = 0.693/λ = 0.693 τ = 69.3% of average life.

b) We have, N = N₀e^−λt

For the time be t after which, N_A/N_B = 2/1 i.e.t_A = t_B = t. Therefore, 

N₀e^{−λA t} = N₀e^{−λB t}

⇒ (λ_B t − λ_A t) = log_e 2

⇒ (λ_B − λ_A) t = log_e 2

⇒ [log_e 2/T_B − log_e 2/T_A) t = log_e 2 [Because λ = log_e2 / T]

⇒ [1/30 − 1/60] t = 1

⇒ t = 30 x 60 / 30 = 60 years.

a)

b) Since, two charges are on Z-axis. Therefore, the electric potential on an arrival position at (0, 0, ±2) is given by

V₁ = 1 / 4πε₀ × 2lq / (r² − l²)

= 1 / 4πε₀ × 2aq / (r² − a²) [because l = a  and r = ±z ]

Now electric potential at the position (, y, 0) is given by

V = 1 / 4πε₀ × 2aq cos θ / r² but θ = 90 degree. Therefore, 

V₂ = 0 at the position of (x, y, 0) due to an electric dipole, placed on Z-axis.

Futures of photoelectric equation which can not be explained be wave theory: 

• Maximum kinetic energy of the emitted photoelectrons is independent of intensity of incident light.
• A the wave theory could not explain the instantaneous process of photoelectric effect.

a) When a current flows through the conductor coil, a torque acts on it due to the external radical magnetic field. Counter torque due to suspension balances coil after appropriate deflection due to current in the circuit. 

b) A galvanometer can be used as such to measure current due to the following two reasons.

• A galvanometer has a finite large resistance and is connected in series in the circuit, so it will increase the resistance of circuit and hence change the value of current in the circuit.
• A galvanometer is a very sensitive device. It gives a full scale deflection for the current of the order of micro ampere, hence if connected as such it will not measure current of the order of ampere.

c) Voltage sensitivity : It is defined as the deflection produced in the galvanometer when a unit voltage is applied across it.

Current sensitivity : The ratio of deflection produced by the coil φ to the current in the coil is called the current sensitivity. It is the deflection of the meter per unit current.

According to the Brewester's law

tan i_p = n, and n = 60°

Therefore, n = tan 60 ⇒ n = 1.732

Radius of circular path r = mv/bB ⇒ r = p/qB  [∵ p = mv]

Momentum of deuteron and alpha particle are same and in same magnetic field. Therefore, 

r_d = p/q_d B = p/eB   [∵ q_d = q_α = e]

r_α = p/q_α B = p/2eB   [∵ q_α = 2e ] 

Therefore, r_d : r_α = 2 : 1

When a charged particle enter in a magnetic field making an angle 30 degree then velocity component is resolved into 2 components one is v cos θ along the magnetic field and other one other is v sin θ normal to the magnetic field. 

As the charged particle moves along xy plane due to velocity component v sine θ, it also advances linearly due to the velocity component v cos θ . As a result, the charged particle will move in a helical path as shown in the figure.

With the raise in temperature the collision of electrons occurs more frequently, so relaxation time decreases and hence drift velocity increases. 

v_d = eE/m × τ ⇒ v_d ∝ τ

Let the wire A₃ placed at a distance x from the wire A₁ and the distance of A₃ from A₂ be (d − x). The current flowing through them is I, 1.5 I and 2 I in A₁, A₃ and A₂ respectively. Let  the current flowing in the wire A₃ be the same direction as A₁ and A₂.

Therefore, force between A₁ and A₃ is F₁ = µ₀/2π × (I × 1.5 I)/x .....(i)

Force between A₃ and A₂ is F₂ = µ₀/2π × (2 I × 1.5 I)/ (d − x) .....(ii)

Since, the net force on A₃ is zero, therefore F₁ = F₂

⇒ µ₀/2π × (I × 1.5 I)/x = µ₀/2π × (2 I × 1.5 I)/ (d − x)

⇒ 1.5/x = 3/(d − x)

⇒ 1.5(d − x) = 3x

⇒ 1.5d − 1.5 x = 3x

⇒ 4.5 x = 1.5d

⇒ x = d/3

The wire A₃ is placed at a distance of d/3 from A₁ and 2d/3 from A₂.

No, at a same distance, the force of the wire A₃ is independent of the direction of the current. As if current is in opposite direction, then F₁ and F₂ will be in opposite direction, but will be in equilibrium position.

In Rutherford's scattering experiment of an alpha particle, it was observed that the fast and heavy alpha particles could be deflected through 180 degree. But only very small number of particles that is 1 in about 8,000 alpha particles are deflected through 180 degree that too from center only.

So by this it was assumed that the size of central part. That is nucleus is about 1/10,000 th of the size of the atom and whole positive charge is concentrated in it. 

Consider an atom whose mass number is A and R be the radius of the nucleus. If we neglect the mass of orbital electrons, than mass of the nucleus of the atom of mass number A = A a.m.u.

Mass of nucleus, m = A × 1.66 × 10⁻²⁷ kg

Volume of nucleus V = 4/3 πR³ = 4/3 π(1.1 × 10⁻¹⁵)³ × A m³

Density of nucleus, ρ is equal to mass of nucleus divides by volume of nucleus. Therefore

ρ = A × 1.66 × 10⁻²⁷ × A / 4/3 π(1.1 × 10⁻¹⁵)³ × A

ρ = 2.97 × 10¹⁷ kg/m³

Therefore, density of nucleus is independent of mass number. 

Unpolarized light :

The light having way vibration of electric field vector in all possible directions are perpendicular to direction of wave propagation the light is known as unpolarized light.

Linearly polarized light :

The light having vibrations of electric field vector in only one direction of perpendicular to the direction of a propagation of light is as plane or linearly polarized light.

Threshold frequency : The minimum frequency of incident light which is just capable of ejecting electrons from a metal is called the threshold frequency. It is denoted by v₀.

Stopping potential : The minimum retarding potential applied to anode of a photoelectric tube which is just capable of stopping photoelectric current is called the stopping potential. It is donated by V₀ or V_s.

Principle of a cyclotron :  

It is based on a principle that a positive ion can acquire sufficiently charged energy from a comparatively smaller alternating potential difference by making it to cross the same electric field again and again by making use of a strong magnetic field. 

Working of a cyclotron :

The positive ions are produced from the source at the center are accelerated by a dee which is at negative potential at that moment. Due to the presence of perpendicular magnetic fields the ion will move in a circular path inside the dees. The magnetic field and the frequency of AC Source are so to that as the ions comes out of a dee. It changes it polarity and the ion is further accelerated and moves with higher velocity along a circular path of greater radius. This phenomena is continued till the ion reaches at the periphery of the dees where an deflecting plates deflects the accekerated ion on the target to be bombarded. 

Expression of cyclotron frequency:

Suppose a position ion with charge q moving with a velocity v, then qvb = mv²\r.

⇒ r = mv/qb

Therefore angular velocity ω is ω = v/r = qB/m

The time taken by iron in describing a semicircle is t = π/ω = πm/qB

This is a semi periodic time T/2 = t = πm/qB. i.e. T = 2πm/qB, time period of revolution. Therefore, frequency of revolution is f = 1/T = qB/2πm. This frequency is called the cyclotron frequency. 

The magnitudes of the current electric field due to the two charges +q and −q given by

E_+q = q/4πε₀ × 1/(r² + a²)

E_−q = q/4πε₀ × 1/(r² + a²)

⇒ E_+q = E_−q

The directions of E_+q and E_−q are shown in the given above figure. The components normal to the dipole axis cancel away. The components along the dipole axis add up. Therefore, Total electric field is 

E = −(E_+q + E_−q) cos θ

E = − 2qa / 4πε₀ × (r² + a²)^3/2 

In this expression negative sign shows that the field is opposite to . Ar large distance (r >>a), this reduces to 

E = − 2qa / 4πε₀ × r³

= q × 2a 

Therefore, E = −/ 4πε₀ (r >>a)

Given, λ = 590 nm = 590 × 10⁻⁹, KE = ? 

For some de-Broglie wavelength, (λ) = h/mv where h is a planck constant

h/m_e × 590 × 10⁻⁹ = v 

KE = 1/2 m_e v²

= 1/2 m_e h²/m_e² × (590 × 10⁻⁹)²

KE = 6.91 × 10⁻¹⁷ J

Optical fibers works on principle of total internal reflection. When angle of incidence is greater than critical angle then incident rays are totally reflected back in same media. When θ_i > θ_c, Total internal reflection occurs and if θ_i < θ_c, refraction occurs. Application Optical fibers are used for communication due to very high bandwidth of media.

Oscillating charges are responsible for generation of periodically varying electric field in the space. The oscillating charges generate varying electric current which in turn is responsible for the generation of periodical varying magnetic field. This way the electromagnetic waves are generated.

Photodiodes are reverse biased for working in photo conductive mode. These reduces the response time because the additional reverse bias increases the width of the depletion layer, which decreases the junction capacitance.

The reverse bias also increases the dark current without much change in the photo current.

Given, W = 2.5 − 2.28 eV

E = hc/λ

λ₁ = hc/E₁ 

= 6.63 × 10⁻³⁴ × 3 × 10⁸ / 2.5 × 1.6 × 10⁻¹⁹

= 3 × 6.63 × 10⁻⁷ / (−2.5) × 1.6

λ₁ = 497.25 nm

λ₂ = hc/E₂ 

= 6.63 × 10⁻³⁴ × 3 × 10⁸ / 2.8 × 1.6 × 10⁻¹⁹

λ₂ = 443.97 nm

(i) Condition of minimum deviation is A = 180° − 2α and µ = sin i / sin(90° − β) 

when r₁ = r₂ = r > critical angle

r₁ + r₂ = 180° − 2α

2r = 180° − 2α [Because r₁ = r₂]

r = 90° − α

β = 90° − r₁ = 90° − (90° − α) 

β = α

Condition when QR have total internal reflection

∠QRC ≥ critical angle for the prism

∠180° − β ≥ critical angle or ∠180° − α ≥ critical angle

Because 180° − α = ∠BAC. Therefore, ∠BCA ≥ critical angle

Given c = 100 µF, d = 4 × 10^-3 m, V = 250 V , k = 5 and

Q = CV = 200 × 100 × 10⁻⁶ = 2 × 10⁻² coulomb

As dielectric of 4 mm is inserted between the plates of capacitor and the spacing between the plates is doubled then it will both acts as following see in figure A and figure B.

Here, C' will be capacitance with dielectric of 4 mm and 8 mm separation between the plates.

C' = KC = 5 × 100 × 10⁻⁶ = 0.5 × 10⁻³ F

(i) Equivalent Capacitance is get by adding 1/C and 1/C' i.e.

1/C_eq = 1/C +1/C' = 1 / 100 × 10⁻³ + 1 / 0.5 × 10⁻³

= 10 × 10³ + 2 × 10³

= 12 × 10³

C_eq = 1/12 × 10⁻³ = 8.33 µF

(ii) Electric field inside dielectric will be

E' = E/K = 50 / (5 × 4 × 10⁻³) = 1000 V/m

and electric field inside capacitor but out of dielectric area will be E = 50×10³ V/m

(iii) Energy density of the capacitor is given by U = Q²/2C

⇒ U' + U = Q²/2C + Q²/2C'

= Q²/2 [1/C + 1/C']

= (2 × 10⁻² × 2 × 10⁻²)/2 × (12 × 10³)

= 2 × 10¹ × 12 = 2.4 J

Solution a)

From Bohr's model, an atom has a number of stable orbits in which an electron can reside without the emission of radiant energy. Each orbit correspondent to a certain energy level. Electron revolves in his circular orbit is given by

The motion of an electron in circular orbits is restricted in such a manner that its angular momentum is an integral multiple of h/2π.

Thus, L = mvr = nh/2π

E_n = −13.6/r² • z²eV

Z = 1 for H₂ atom

E_n = −13.6/r² • eV

From de-broglie hypothesis λ = h/p = h/mv and from Bohr model

⇒ nλ = 2πr

⇒  nh/2π = mvr = L

Solution b)

n_i = 1, n_f = 4 and Possible transitions are: ⇒ 4 → 3, 4 → 2, 4 → 1, ⇒ 3 → 2, 3 → 1 and ⇒ 2 → 1. Also six lines are possible.

• Has a smallest Wavelength : 4 → 1
• Paschen Series : 4 → 3
• Balmer Series : 4 → 2, and 3 → 2
• Lyman Series : 4 → 1, 3 → 1 and 2 → 1

From Ampere's Law, ∮ BdI = μ₀i(t). Let the case 1, where a point P is considered outside the capacitor charging. From the Ampere's Law magnetic field at Point P will be :

CASE 1

B•(2πr) – μ₀ i(t) ⇒ B = μ₀ i(t) / 2πr

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

CASE 2

Hence, there is a contradiction. Therefore, this Ampere's law was modified with addition of displacement current inside the capacitor.

φ_E = | E | A = 1/ • Q/A • A = Q/ε₀

dφ_E/dt = 1/ε₀ • dQ/dt

ε₀ dφ_E/dt = dQ/dt = i_d, where i_d is the displacement current.

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

i = i_c + i_d = ic + ε₀dφ_E/dt

outside the capacitor, i_d = 0, hence, i = i_c and inside the capacitor, i_c = 0, hence i = i_d. Thus, i_c = i_d as capacitor gets charged.

From diagram, it is clear that incidence angle at face KM is 60 degree. Therefore, 

sin C = 1/µ = 1/ 2/√3 = √3/2

Hence, critical angle is also 60 degree. Therefore incident light ray will not emerge from KM face due to total internal reflection at the face. So, it will move along face KM. Angle of emergence = 90 degree therefore, Angle of deviation = 30 degree.

Then K₂ is open, balance is obtained at l₁ = AN₁, ε = φl₁, when k₂ is closed V = φl_2.

(Circuit for determining Internal Resistance of a Cell) 

ε/V = l₁ / l₂

But ε = I (r  + R) and V = IR

ε/V = r + R / R = l₁ / l₂

r = R (l₁ / l₂ − 1)

where, r is the internal resistance of the cell. 

Solution b) 

Potentiometer is preferred over voltmeter for measurement of EMF of cell because a voltmeter draws some current from the cell where potentiometer draws no current. Therefore the potentiometer measures the actual EMF of the cell whereas voltmeter measures the terminal voltage. 

Solution c) 

For current through a AB

I = V/R = 5/500 = 10 mA

10 mA current passes through AB. Thus, voltage drop through AB = V_AB = 10 × 10⁻³ × 50 =  500 mV

Voltage drop per unit length = 500mV / 10 = 50 mV/m. 

Balancing point is at 300 mV. Hence l will be 6 m.

When 5V is replaced with 2V, then 

I_AB = 2/500 = 4mA

V_AB = 4 mA × 50 = 200 mV

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

Solution a):

As the armature coil is rotated in the magnetic field, angle θ between the field and normal to the coil changes continuously. Therefore, magnetic flux linked with the coil changes. An emf is induced in the coil. According to Fleming's right-hand rule current induced in AB is from A to B, and it is from C to D in CD in the external circuit current flows from B₂ to B₁.

To to calculate the magnitude of EMF induced :

Suppose A is area of each turn of the coil, N is the Number of turns in the coil, B is the trength of magnetic field and θ is the angle which normal to the coil makes.

Therefore, magnetic flux linked with the coil in this position.

φ = N (•) = NBA cos θ = NBA cos ωt ...(i) where ω is the angular velocity of the coil as the coil. 

As the coil rotates, angle θ changes. Therefore magnetic flux φ linked of with the coil changes and therefore the emf is induced in the coil. At this instant t, if e is the emf induced in the coil, then

e = − d/dt (θ) = − d/dt (NBA cos ωt)

= − NBA d/dt (cos ωt)

= − NBA (−sin ωt) ω

Therefore, e = NAB ω sin ωt 

e = ε₀ sin ωt (Here ε₀ = NAB ω)

Solution b) 

We have, 

Number of spokes = N = 100

Length of each spoke = L = 0.5 m

Magnetic field = B = 0.4 × 10⁻⁴ = 4 × 10⁻⁵ T

Frequency f = 120 rpm = 2 rps

Induced EMF between axle and rim his given by

e = N × B × l² × π × f

= 100 × 4 × 10⁻⁵ × (0.5)² × 3.14 × 2

e = 6.28 × 10⁻³ V 

Solution a: 

Meter bridge is the practical apparatus working which works on principle of Wheat-Stone bridge. It is used to measure unknown resistance experentally.

Hence, as per Wheat Stone Bridge balance condition

R₁/R₂ =  Resistance of wire AD / Resistance of wire DC

where, D is the point of balance. 

R₁/R₂ = ρl₁ /  ρl₂ = l₁ / l₂ = l₁/(100 − l₁)  (bacause l₁ + l₂ = 100 cm

ρ = Resistance per unit length of wire

R₁ = R₂ l₁ / (100 − l₁)

Solution b) 

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

(ii) Connection between registers are made of thick copper is strips so that it will have maximum resistance and location of point of balance D will be more accurate which results in correct measurements of unknown resistance. 

Solution c)

From KCl (kirchhoff's current law) at point D

Current flowing through 4 Ohm resistance is i = i₁ + i₂

kirchhoff's current law in loop DEFCD is ∑V = 0 i.e

−4(i₁ + i₂) − i₂ + 6 = 0

−4i₁ − 5i₂ + 6 = 0 (Voltage drop is negative and gain his positive) 

4i₁ + 5i₂ = 6 ...(i)

kirchhoff's current law in loop AE FBA is ∑V = 0 i.e.

−4(i₁ + i₂) − 2i₁ + 8 = 0

−6i₁ − 4i₂ + 8 = 0

6i₁ + 4i₂ = 8 ...(2)

After solving both the equations, we get i₁ = 8/7 and i₂ = 2/7

Solution a)

Phenomenon of interference can't be observed by iluminating two pin holes with two sodium lamps because these sources are not coherent sources. That means it is they are not in same phase. 

Solution b)

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

From the principle of superposition y = y₁ + y₂

=  a sin ωt + b sin (ωt + φ)

=  a sin ωt + b sin ωt cos φ + b cos ωt sin φ

=  (a + b cos φ) sin ωt + b cos ωt sin φ

let a + b cos φ = A cos ς

b sin φ = A sin ς

⇒ y = A sin ωt cos ς + A cos ωt sin ς

⇒ y = A sin (ωt + ς)

A = √(a² + b² + 2ab cos φ)

tan ς = b sin φ / (a + b cos φ)

1. Consecutive interference 

For I - Maxima, I ∝ A² and for A to be maximum cos φ = 1 i.e. 

cos φ = cos 2nπ, n = 1, 2, 3, 4, ...

φ = cos 2nπ

and the differene is Δx = nλ, A_max = a + b  and I → I_max = k(a + b)²

⇒ Destructive interference

For I - Minima cos φ = −1, Δφ = (2n + 1)π, 

Path difference is Δx = (2n + 1)λ/2, A_min = a − b, I → I_min = k(a − b)²

Solution c)

θ = (n + 1/2) λ/a, where a = 2 × 10⁻⁶

θ₁ = λ/2a = (590 × 10⁻⁹) / (4 × 10⁻⁶) = 147.5 × 10⁻³

θ'₁ = λ'/2a = (596 × 10⁻⁹) / (4 × 10⁻⁶) = 149 × 10⁻³

θ₂ − θ₁ = 1.5 × 10⁻³

λ₁ = 596 nm, λ₂ = 590 nm, a = 2 × 10⁻³, D = 1.5 m, y = 3λD/2a

y₁ − y₂ = 3D/2a (λ₁ − λ₂)

 = (3 × 1.5) / (2 × 2 × 10⁶)  ×  [596 − 590] × 10⁻⁹ m

 = 4.5 / 4 × 10⁶  × 6 × 10⁻⁹ m

 = 6.78 nm

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

Intensity of a light after passing through the polarized P₁(I),

⇒ I₁ = I₀/2

Intensity of light after passing through P₃,

⇒ I₂ = I₁ cos² 45° =  I₁ /2 = I/4

Intensity of light passing through P₂

⇒ I₃ = I₂ cos² 45° =  I/4 × I/2 = I/8

a) Some features of nucleus force are

• It is independent of the charges protons and neutrons i.e. charge independent.
• Nuclear forces are very strong binding forces i.e. attractive forces.
• It depends on the spins of the nucleones.

b) Plot showing variation of potential energy of a pair of nucleons as a function of separation mark attractive and repulsive region. In the following figure, the x-axis shows separations between pair of nucleons and the y-axis shows the variation of potential energy with respect to the separation in 10⁻¹⁵ m.

The symbols and V present anti-neutrino and neutrino respectively during β-decay both the neutral particles with very little or no mass. These particles are emitted from the nucleus along with the electron or positron during the decay process. Neutrons interact very weakly with matter. They can even penetrate the earth without being absorbed. It is for this reason that they are deduction is extremely difficult and their presence went unnoticed for a long.

Decay constant

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

Derivation of mean life

Let us consider N₀ the total number of radioactive atoms present initially. After some time t, total number of atoms present undecayed be N, Further dt time dN be the number of atoms distinguished. So the life of dN atoms ranges lies between t + dt and dt. Since dt is very small time, so the most appropriate life of dN atom is t. So the total life of N atom is equal to t • dN. 

Sum of edges of all atoms = ∫₀^N0 t dN .... (i)

N = N₀ e^−λt

dN = N₀ (−λt)e^−λt dt

Now, substituting the values of dN and changing the limits, we get the final result

= ∫_∞⁰ N₀ (−λt)e^−λt dt

= N₀ λ ∫_∞⁰ t e^−λt dt

= Sum of life of all atoms

Now, Mean Life = τ = N₀ λ ∫_∞⁰ t e^−λt dt / N₀

τ = λ ∫_∞⁰ t e^−λt dt

τ = λ × 1/λ²

τ =  1/λ

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

µ = − (e/2m) × L where, negative sign indicates µ direction is opposite to L. As Bohr's atomic model, L = mvr. Therefore, 

µ = − (e/2m) × mvr

µ = evr/2

But from Bohr's second postulate,

m_ev_r = nh/2π × h/2π (for the value of n = 1)

v_r = nh/2π m_e

Hence, the magnetic movement is

m = e/2 × h/2π m_e (Here n = 1)

Therefore, m = eh/4πm_e

i) Source frequency will be same as resonance frequency of LC circuit. 

f_R = 1 / 2π√LC

= 1 / 2π √(400 × 10⁻⁶)

= 1 / 2π × 2 × 10⁻²

= 100 / 4π = 7.957 Hz [Because ω = 2πf]

ω_r = 50 Hz

(ii) Impedance of a circuit 

z_i = RωL + 1/jωC

= 40 + 50j × 5 + 1/ ( j × 50 × 80 × 10⁻⁶)

= 40 + 250j + 10³/ 4j

= 40 + 250j − 250j (at resonance)

z_i = 40Ω

Now, Amplitude of a current is

I = 230/z_i = 230/40 = 5.75 A

iii)  As at the resonance frequency impedance of combination of L and C is 0. Hence, voltage drop across LC combination is zero at resonating frequency.

Given that V_C = 120, V_R = 90 V,  I = 3A and we find Z = ?

(i) From Kirchhoff's law,

V = V_C + V_R

= 230 V

I = V/Z ⇒ Z = V/I = 230/3 = 76.67 Ω

(ii)

cos φ = R/Z = 1

R = Z = R + j (ωL − 1/ωC)

Hence L = 1/C

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

m = angle β/ / angle α

m = h/f_e • f₀/h = f₀/f_e

The resolving power of microscope is the reciprocal of the maximum distance. Therefore, we have 

R.P = 1/d_min = 2n sin β / 1.22 λ

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

Given χ_mg at 300 K = 1.2 × 10⁵, χ'_mg at t temperature = 1.44 × 10⁵ , t = ?

From curies law, χ ∝ 1/T

χ'_mg / χ_mg = 300/T

⇒ 300/T = 1.44 × 10⁵ /1.2 × 10⁵

⇒ T = 300 ×1.2 /1.44

⇒ T = 250 K

Given, tan⁻¹ (x + 1) + tan⁻¹ (x − 1) = tan⁻¹(8/31)

We know that tan⁻¹ x + tan⁻¹ y = tan⁻¹ (x+y / 1 − xy).

Therefore, 

tan⁻¹ (x + 1) + tan⁻¹ (x − 1)  = tan⁻¹ [(x + 1) + (x − 1) / (1 − (x + 1)(x − 1))

= tan⁻¹ [2x / (1 − (x² − 1))

= tan⁻¹ [2x / (2 − x²))

Now, tan⁻¹ [2x / (2 − x²)) = tan⁻¹ (8/31)

⇒ 2x/(2 − x²) = 8/31

⇒ 62x = 8(2 − x²)

⇒ 31x = 8 − 4x²

⇒ 4x² + 31x − 8 = 0

⇒ 4x(x + 8) − 1(x + 8) = 0

⇒ (4x − 1) (x + 8) = 0

⇒ x = 1/4, −8

Given, (a + bx) e^y/x = x

⇒ e^y/x = x/(a + bx)

⇒ y/x = log [ x/ (a + bx)]

⇒ y/x = log [ x/ (a + bx)]

⇒ y/x = log x − log (a + bx)

Differertiating with respect to x, we get

⇒ (x dy/dx − y)/x² = 1/x − b/(a + bx)

⇒ x dy/dx − y = x² (1/x − b/(a + bx))

⇒ x dy/dx − y = x − bx²/(a + bx)

⇒ x dy/dx − y = ax/(a + bx) ...(i)

Again differentiating with respect to x, we get

⇒ d/dx (x dy/dx − y) = d/dx (ax/(a + bx))

⇒ x d²y/dx² + dy/dx − dy/dx = [(a + bx)a − ax × b] / (a+ bx)²

⇒ x d²y/dx² = [(a² + abx − abx] / (a+ bx)²

⇒ x d²y/dx² = a²/(a+ bx)²

On multiplying x² both the sides, we get

⇒ x³ d²y/dx² = a²x²/(a+ bx)²

⇒ x³ d²y/dx² = [ax/(a+ bx) ]²

y = e^√(3x)

On differentiating with respect to x, we get

dy/dx = d/dx (e^√(3x))

= e^√(3x) d/dx √(3x)

= e^√(3x) √3 d/dx √x

= √3//2 × e^√(3x)/√x

Given equation of circles are

x² + y² = 4 ...... (i)

(x − 2)² + y² = 4 ......(ii)

Equation x² + y² = 4 is a circle with center O at  the origin (0, 0) and radius is 2 units and equation (x − 2)² + y² = 4 is a ciecle with center is (2, 0) and radius is 2 units.

On Solving both the equations, we get 

(x − 2)² + y² = x² + y²

⇒ x² − 4x + 4 + y²= x² + y²

⇒ x = 1 which gives y = ±√3

Thus, the points of intersection of the given circle are A(1, √3) and A' (1, −√3).

Therefore, required area of the enclosed region in the figure is OACA'O between these two circles. 

= 2[Area of Region ODCAO]

= 2[Area of Region ODAO + Area of Region OCAD]

= 2[∫₀¹ ydx + ∫₁² ydx ]

= 2[∫₀¹ √(4 − (x − 2)² ) dx + ∫₁² √(4 − x²) dx ]

= 2[1/2 (x − 2) √ (4 − (x − 2)² ) + 1/2 × 4 sin⁻¹ x ( (x − 2 )/ 2) ]₀¹ + 2 [ 1/2 x √ (4 − x²) + 1/2 × 4 sin⁻¹ x/2]₁²

= [(x − 2) √(4 − (x − 2)² ) + 4 sin⁻¹ x ((x − 2 )/ 2) ]₀¹ + [ x √(4 − x²) + 4 sin⁻¹ x/2]₁²

= [−√3 + 4 sin⁻¹ (−1/2) − 4 sin⁻¹ (−1)] + [ 4 sin⁻¹ 1 − √3 −4 sin⁻¹ 1/2]

= [(−√3 + 4 × π/6) + 4 × π/2] + [ 4 × π/2 −√3 −4 × π/6]

= 8π/3 − 2√3

Area of the enclosed figure OACA'O = 8π/3 − 2√3

Let A(1, 0), B(2, 2) and C(3, 1) be the vertices of a triangle.

Area of Triangle ABC = Area of Triangle ABD + Area of Trapezium BDEC − Area of Triangle AEC

Now, first we find the equation of side AB using the formula y − y₁ = m (x − x₁) where m = (y₂ − y₁) / (x₂ − x₁)

m = (2 − 0) / (2 − 1) = 2/1 = 1

⇒ y − 0 = 2 (x − 1)

⇒ y = 2(x − 1)

Similarly, Equation of line BC

m = (1 − 2) / (3  − 2) = −1/1 = −1

⇒ y − 2 = −1 (x − 2)

⇒ y − 2 = −(x − 2)

⇒ y =  4 − x

Similarly, Equation of line AC

m = (1 − 0) / (3 − 1) = 1/2

⇒ y − 0 = 1/2 − (x − 1)

⇒ y =  1/2 (x − 1)

Hence, area of a Triangle ABC

= ∫₀² 2(x − 1) dx − ∫₁³ 1/2 (x − 1) dx + ∫₂³ (4 − x) dx

= 2[x²/2 − x]₀² − 1/2 [x²/2 − x]₁³ + [4x −  x²/2]₂³

= 2[ (2²/2 − 2) − (0²/2 − 2)] − 1/2 [(3²/2 − 3) − (1²/2 − 1)] + [(4 × 3 − 3²/2) − (4 × 2 − 2²/2)]

= 2[ 0− (− 2)] − 1/2 [(9/2 − 3) − (1/2 − 1)] + [(12 − 9/2) − (6)]

= 3/2 square units.

∫ (sin³ x + cos³ x)/ sin² x cos² x dx

= ∫ sin³ x / sin² x cos² x dx + ∫ cos³ x/sin² x cos² x dx

= ∫ sin x / cos² x dx + ∫ cos x/sin² x dx

= ∫ tan x sec x dx + ∫ cot x cosec x dx

= sec x − cosec x + C

[ ∫ tan x sec x dx = sec x and ∫ cot x cosec x dx = − cosec x ]

∫ (sin² x − cos² x)/sin x cos x dx

= − ∫ (cos² x − sin² x)/sin x cos x dx

= − ∫ cos 2x / sin 2x dx

= − ∫ cot 2x dx

= − log sin 2x + C

Let  y = tan⁻¹ [(1 + cos x)/sin x]

= tan⁻¹(2cos² x/2 / 2sinx/2 cos x2)

= tan⁻¹(cos x/2 / sinx/2)

= tan⁻¹(cot x/2)

= tan⁻¹(tan (π/2 − x/2))

y = π/2 − x/2

Now, differentiating with respect to x, we get

dy/dx = d/dx (π/2) − d/dx (x/2) = 0 − 1/2

dy/dx = −1/2

Let  =  − 2 + 3 and  = 3 − 2 +

We know that,  •  = |  | |  | cos θ

⇒ cos θ =  •  / |  | |  |

= ( − 2 + 3) • (3 − 2 + ) / |  − 2 + 3 | | 3 − 2 +  |

= 1(3) + (−2)(−2) + 3(1) /√(1+4+9) √(9+4+1)

= 3 + 4 + 3 / √(14) √(14)

= 10 /14 = 5/7

Now sin θ = √(1 − sin² θ)

= √ [1 − (5/7)²]

= √ [1 − 25/49]

= √ [24/49]

sin θ = 2√6 / 7

Given R { (a, b) : a, b ∈ A, |a − b| is divisible by 4}

⇒ Reflexivity : 

for any a ∈ A

|a − a| = 0, which is divisible by 4 

(a, a) ∈ R

So R is reflexive.

⇒ Symmertry : 

for any (a, b) ∈ R

⇒ |a − b| is divisible by 4

⇒ |b − a| is divisible by 4

Because, |a − b| = |b − a|

⇒ (b, a) ∈ R

So R is symmetric.

⇒ Transitive : 

Let (a, b) ∈ R and (b, c) ∈ R

⇒ |a − b| is divisible by 4

⇒ |a − b| = 4k

⇒ a − b = ±4k, where k ∈ Z

Also 

⇒ |b − c| is divisible by 4

⇒ |b − c| = 4m

⇒ b − c = ±4q, where q ∈ Z

Adding both the equations, we get

a − b + b − c = ±4 (k + q)

⇒ a − c = ±4(k + q) 

⇒ |a − c| is divisible by 4

⇒ (a, c) ∈ R

So R is transitive.

⇒ R is reflexive, symmetric and transitive. Therefore, R is an equivalence relation.

Now again, let x be an element of R such that (x, 1) ∈ R. Then |x − 1| is divisible by 4. i.e. 

x − 1 = 0, 4, 8, 16 ...

x = 1, 5, 9 (since x ≤ 12)

Therefore, set of all the elements of A which are related to 1 are {1,5, 9}.

Equivalance class of 2 i.e. 

[2] = {(a, 2) : a ∈ A. |a − 2| is divisible by 4}

⇒ |a − 2| = 4k (where k is a whole number, k ≤ 3)

⇒ a = 2, 6, 10

Therefore. equivalence class [2] is [2, 6, 10]

Equation of the line is = (2 −  + 2) + λ(3 + 4 + 2) ....(i)

Coordinates of any point on this line are (2 + 3λ) + (−1 + 4λ) + (2 + 2λ)

Given equation of the plane is ( −  + ) = 5 .....(ii)

Since the point on line lies on the plane ( −  + ) = 5, Therefore,

[(2 + 3λ) + (−1 + 4λ) + (2 + 2λ) ] ( −  + )  = 5

⇒ (2 + 3λ) − (−1 + 4λ) + (2 + 2λ) = 5

⇒ 2 + 1 + 2 + 3λ − 4λ + 2λ = 5

⇒ 5 + λ = 5 

⇒ λ = 0

So substitute the value λ = 0 in equation (i), we have

= (2 −  + 2) + 0 (3 + 4 + 2)

⇒ = (2 −  + 2) .... (iii)

Let point of intersection be (x, y, z). Therefore (x + y + z) ......(iv)

Comparing equations (iii) and (iv) we get x = 2, y = −1 and z = 2. Thererfore, point of intersection is (2, −1, 2) and (−1, −5, −10) is

= √ [(−1 −2)² + (−5 + 1)² + (−10 − 2)²]

= √ [(−3)² + (−4)² + (−12)²]

= √ [9 + 16 + 144]

= √169

= 13 units

×Given curve is x² + y² = 32 ⇒ x² + y² = (4√2)². It is a cricle with center is (0, 0) and radius is 4√2. Also given line is y = x. Solving both the given equations, we get

x² + y² = 32 ⇒ x² + x² = 32 ⇒ 2x² = 16 ⇒ x² = 16 ⇒ x = ± 4

Therefore, the Point of Intersection is (4, 4).

Now, Required Area = Area OMA

= Area OMP + Area MPA

= ∫₀⁴ x dx + ∫₀^4√2 √ [ (4√2)² − x²] dx

= [x²/2]₀⁴ + [ x/2 ( (4√2)² − x²) ) + (4√2)²/2 sin⁻¹ (x/4√2)]₀⁴

= 16/2 + [{4√2/2 ( (4√2)² − (4√2)²) + 32/2 sin⁻¹ 1 } − {4/2 ( (4√2)² − (4)²) + 32/2 sin⁻¹ 1/√2 }]

= 8 + (2√2 (0) + 16 × π/2) − (2 × 4 + 16 × π/4)

= 8 + 8π − 8 − 4π

= 4π square units

Given, y = sin(sin x). Differentiating with respect to x we get

dy/dx = cos (sin x) • cos x ...(i)

Again differentiating with respect to x we get

d²y/dx² = d/dx (cos (sin x) • cos x)

= cos (sin x) d/dx (cos x) + cos x d/dx (cos (sin x))

= cos (sin x) (− sinx) + cos x (− sin (sin x) • cos x)

= cos (sin x) (− sinx) + cos x (− sin (sin x) • cos x)

d²y/dx² = − sinx cos (sin x) − cos² x sin (sin x) ...(i)

Now, find LHS → d²y/dx² + tan x dy/x + y cos² x

= − sinx cos (sin x) − cos² x sin (sin x) + tan x cos (sin x) • cos x + y cos² x

= − sinx cos (sin x) − cos² x sin (sin x) + tan x cos (sin x) • cos x + sin(sin x) cos² x

= − sinx cos (sin x) − cos² x sin (sin x) + sin x cos (sin x) + cos² x sin(sin x)

= 0

Given that x = a (2θ − sin 2θ) and y = a (1 − cos 2θ), differentiating with respect to θ, we get 

x = a (2θ − sin 2θ)

dx/dθ = a(2 − 2 cos 2θ) = 2a(1 − cos 2θ)

dy/dθ = a (2 sin 2θ) = 2a sin 2θ

Now dy/dx = dy/dθ / dx/dθ

= 2a sin 2θ / 2a(1 − cos 2θ)

= sin 2θ / (1 − cos 2θ)

= 2 sin θ cos θ / 2 sin² θ

= cos θ / sin θ

dy/dx = cot θ

Now, [dy/dx]_θ=π/3 =  cot π/3 = 1/√3

Given (x² + y²)² = xy

Differentiating with respect to x, we get

d/dx (x² + y²)² = d/dx (xy)

⇒ 2(x² + y²) d/dx (x² + y²) = x d/dx (y)+ y d/dx (x)

⇒ 2(x² + y²) (2x + 2y dy/dx) = x dy/dx + y

⇒ 4(x² + y²)(x + y dy/dx) = x dy/dx + y

⇒ 4x(x² + y²) + 4y(x² + y²) dy/dx = x dy/dx + y

⇒ 4y(x² + y²) dy/dx − x dy/dx = y − 4x(x² + y²)

⇒ (4x²y + 4y³ − x) dy/dx = y − 4x³ + 4xy²

⇒ dy/dx = (y − 4x³ + 4xy²) / (4x²y + 4y³ − x)

dy/dx = −[(x + y cos x)  / (1 + sin x)]

dy/dx = −x/(1 + sin x) − y cosx /(1 + sin x)

dy/dx + y cosx /(1 + sin x) = −x/(1 + sin x)

Compare this equation with dy/dx + Py = Q. we have

P = cosx /(1 + sin x) and

Q = −x/(1 + sin x)

Now Intergrating factor is IF = e ^{∫ P dx} 

= e ^{∫ cosx /(1 + sin x) dx}

= e ^{log (1 + sin x)}

IF = (1 + sin x)

Then, 

y × (1 + sin x) = ∫ −1/(1 + sin x) × (1 + sin x) dx + C

⇒ y × (1 + sin x) = y (1 + sin x) = − ∫xdx + C

⇒ y × (1 + sin x) = y (1 + sin x) = −x²/2 + C

Let x be the diameter of the base of a cylinder and let h be the height of the cylinder. 

In Triangle ABC, we have BC² + AB² = AC²

⇒ h² + x² = (2R)²

⇒ x² = 4R² − h² ... (i)

Now, the formula to find the volume of a cylinder is given by V = πr²h i.e.

⇒ V = π (x/2)² × h

⇒ V = π × x²/4 × h

⇒ V = π × (4R² − h²)/4 × h

⇒ V = 4πR²h/4 − πh³/4

⇒ V = πR²h − πh³/4 ....(ii)

On differentiating equation (ii) with respect to h, we get

dV/dh = πR² d/dh h − π/4 d/dh (h³)

⇒ dV/dh = πR² − 3πh²/4 ....(iii)

⇒ 0 = πR² − 3πh²/4 [Because dV/dh = 0]

⇒ 3h²/4 = R²

⇒ h = 2R/√3

Again differentiating equation (iii) with respect to h, we get

d²V/dh² = d/dh (πR²) − 3π/4 d/dh(h²)

⇒ d²V/dh² = 0 − 3π/4 (2h)

⇒ d²V/dh² = −3πh/2

At h = 2R/√3, we have

⇒ d²V/dh² = −3π/2 × 2R/√3 

⇒ d²V/dh² = −√3πR < 0 

Hence, h = 2R/√3 is a point of maxima. So V is maximum when h = 2R/√3. Therefore, the height of the cylinder of maximum volume that can be inscribed in a sphere of radius R is 2R/√3.

Also from equation (i) we have

x² = 4R² − h²

⇒ x² = 4R² − (2R/√3)²

⇒ x² = 4R² − 4R²/3 = 8R²/3

Now, maximum volume of a cylinder is π × (x/2)² × h

V = π/4 × x² × h

V = π/4 8R²/3 × 2R/√3

V = 4πR³/3√3

First five positive integers are 1 2 3 4 and 5. We select two positive integers in 5 times 4 equal to 20 ways. Out of these two numbers are selected at random. Let X denote larger of the two selected numbers. Then X can have the values 2 3 4 or 5. 

P(X = 2) = P(Larger Number is 2) = {(1, 2), (2, 1)}

⇒ P(X = 2) = 2/20

Similarly, 

⇒ P(X = 3) = 4/20

⇒ P(X = 4) = 6/20

⇒ P(X = 5) = 8/20

Thus, the probability distribution of X is

Therefore,

Mean = E(X) = ∑_i=1ⁿ x_i P(x_i)

= 2 × 2/20 + 3 × 4/20 + 4 × 6/20 + 5 × 8/20

= 4/20 + 12/20 + 24/20 + 40/20

= (4 + 12 + 24 + 40)/20 = 80/20 = 4

E(X²) = ∑_i=1ⁿ x_i² P(x_i)

= 2² × 2/20 + 3² × 4/20 + 4² × 6/20 + 5² × 8/20

= 4 × 2/20 + 9 × 4/20 + 16 × 6/20 + 25 × 8/20

= 8/20 + 36/20 + 96/20 + 200/20

= (8 + 36 + 96 + 200)/20 = 340/20 = 34/2 = 17

Variance = E(X²) − [E(X)]² = 17 - (4)² = 17 - 16 = 1.

Therefore, mean and variance are 4 and 1 respectively.

Let E₁ be the event that girl gets 1 or 2 on the role and E₂ be the event that girl gets 3, 4, 5 or six on the role of a die. Therefore, P(E1) = 2/6 = 1/3 and P(E2) = 4/6 = 2/3.

If she tossed coin only once and exactly one tail shows, then P(A/E₁) = 3/8

P(E₁/A) = P(E₂) P(A/(E₂) / [P(E₁) P(A/E₁) + P(E₂) P(A/E₂)]

= 2/3 × 1/2 / [1/3 × 3/8 + 2/3 × 1/2]

= 1/3 / [1/8 + 1/3]

= 1/3 / 11/24

P(E₁/A) = 8/11

Given differential equation is dy/dx + 2y tan x = sin x. It is a linear differential equation of the form dy/dx + Py = Q.

When we compare both, we get P = 2 tan x and Q = sin x. Now we find Integrating factor, with the formula 

Integrating Factor (IF) = e ^{∫ Pdx} 

= e ^{∫ 2 tan x dx}

= e^{2 ∫ tan x dx}

= e^{2 log sec x dx}

Integrating Factor = sec² x

Now solution of the equation (i) is given by

y • IF = ∫ Q • IF dx + C

⇒ y sec² x = ∫ sin x • sec² x dx + C

⇒ y sec² x = ∫ sec x • tan x dx + C [ ∵ ∫ sec x • tan x dx  = sec x + C]

⇒ y sec² x = sec x + C

Putting y = 0 when x = π/3

0 × sec² π/3 = sec π/3 + C

0 = 2 + C

C = −2

Hence the pariicular solution is y sec²x = sec x − 2 ⇒ y = cos x − 2 cos² x

Given differential equation is e^x tan y dx + (2 − e^x) sec² y dy = 0

⇒ e^x tan y dx = − (2 − e^x) sec² y dy

⇒ e^x tan y dx = (e^x − 2) sec² y dy

⇒ e^x/(e^x − 2) dx = tan y sec² y dy

Integrating both the sides, we get 

⇒ ∫ e^x/(e^x − 2) dx = ∫ tan y sec² y dy

⇒ log |e^x − 2| = log tan y + log C

⇒ e^x − 2 = C tan y ... (i)

Put y = π/4, when x = 0

e⁰ − 2 = C tan π/4

1 − 2 = C

C = −1

Therefore, from equation (i), 

e^x − 2 = −1 tan y

tan y = 2 − e^x

y = tan⁻¹(2 − e^x) is the required solution of differential equation. 

Let the length breadth and height of the open tank be x, x and y units respectively. Then

Volume (V) = x²y ....(i)

Total Surface Area (S) = x² + 4xy .....(ii)

S = x² + 4xy .....(ii)

S = x² + 4x × V/x²    [Using Equation (i) y = V/x²]

S = x² + 4V/x

Differentiating w.r.to x, we get

dS/dx = d/dx [x² + 4V/x²]

dS/dx = d/dx (x²) + 4V d/dx (1/x²)

dS/dx = 2x − 4V/x²

For critical points, put dS/dx = 0

⇒ 2x − 4V/x² = 0

⇒ 2x³ − 4V = 0

⇒ 2x³ = 4V

⇒ 2x³ = 4 x²y

⇒ x = 2y .... (iii)

Now, Again differentiating w.r.to x, we get

d²S/dx² = d/dx (2x − 4V/x²)

= 2 − 4 × (−2V/x³)

= 2 + 8V/x³

= 2 + 8V/(2y)³  [Using equation (iii)]

= 2 + 8V/8y³

= 2 + V/y³ > 0

Area is minimum, therefore the cost is minimum when x = 2y that means depth of the tank is half of the width.

Let E₁ be the event of choosing the First bag and E₂ be the event of choosing the Second bag and A be the event of drawing a red ball. Then, P(E₁) = P(E₂) = 1/2.

Now,

P(A/E₁) = 5/9 × 4/8 = 20/72

P(A/E₂) = 3/9 × 2/8 = 6/72

Now the probability of drawing a ball from Secong Bag, if it is given that it is red is P(A/E₂). Use Baye's theorem of probability, we have

P(E₂/A) = P(E₂) • P(A/E₂) / [ P(E₁) × P(A/E₁) + P(E₂) × P(A/E₂) ]

= 1/2 × 6/72 / [1/2 × 20/72 + 1/2 × 6/72]

= 6/(20 + 6)

= 6/26

P(E₂/A) = 3/13

Given

x = ae^t (sin t + cos t)

y = ae^t (sin t − cos t)

On differentiating above equation with respect to t, we get

x = ae^t (sin t + cos t)

⇒ dx/dt = a d/dt [e^t (sin t + cos t)]

= a [e^t d/dt(sin t + cos t) + (sin t + cos t) d/dt e^t]

= a [e^t (cos t − sin t) + (sin t + cos t) e^t]

= a [e^t cos t − e^t sin t + e^t sin t + e^t cos t]

dx/dt = a [2e^t cos t] = 2ae^t cos t ...(i)

y = ae^t (sin t − cos t)

⇒ dy/dt = a d/dt [e^t (sin t − cos t)]

= a [e^t d/dt(sin t − cos t) + (sin t − cos t) d/dt e^t]

= a [e^t (cos t + sin t) + (sin t − cos t) e^t]

= a [e^t cos t + e^t sin t + e^t sin t − e^t cos t]

dy/dt = a [2e^t sin t] = 2ae^t sin t ...(ii)

On dividing equation (ii) by equation (i), we get

dy/dx = dy/dt / dx/dt

dy/dx = 2ae^t sin t / 2ae^t cos t

dy/dx = sin t /cos t

dy/dx = tan t

Also we solve (x + y)/(x − y) i.e.

x + y = ae^t (sin t + cos t) + ae^t (sin t − cos t)

= ae^t [sin t + cos t + sin t − cos t]

x + y = 2 ae^t sin t

x − y = ae^t (sin t + cos t) − ae^t (sin t − cos t)

= ae^t [sin t + cos t − sin t + cos t]

x + y = 2 ae^t cos t

Now (x + y)/(x − y) = 2 ae^t sin t/2 ae^t cos t

⇒ (x + y)/(x − y) = sin t/cos t

⇒ (x + y)/(x − y) = sin t/cos t = tant

Hence Proved

dy/dx = (x + y)/(x − y)

Given y = cos √(3x). On differentiating w.r.t x, we get

dy/dx = −sin √(3x) × √3 × 1/2√x

dy/dx = −√3 sin√(3x) / 2√x

Given that y = ae^2x + 5. On differentiating w.r.t x, we get

dy/dx = ae^2x • (2) [Using Chaing Rule]

⇒ dy/dx = 2ae^2x

Again differentiating w.r.t x, we get

d²y/dx² = 2[ae^2x • (2)]

⇒ d²y/dx² = 2[dy/dx]

⇒ d²y/dx² = 2 dy/dx

⇒ d²y/dx² − 2 dy/dx = 0

The given equations of the sides of triangle is y = 2x + 1 ...(i), y = 3x + 1 ...(ii), and x = 4 ...(iii). Preapre the table of this equations and draw the graph shown below.

Now, from the graph, equation y = 2x + 1 meets x and y axes at (−1/2, 0) and (0, 1). By joining these points we obtain the graph of x + 2y = 2. Similarly graph of other equations are drawn and points are found. On solving all the given three equations in pairs, we obtain intersection point. This method we obtain the coordinates of the vertices of Triange ABC are A(0, 1), B(4, 13) and C(4, 9).

Then area of the ∆ABC = Area (OLBAO) + Area (OLCAO)

= ∫₀⁴ (3x + 1)dx − ∫₀⁴ (2x + 1)dx

= ∫₀⁴ (3x + 1 − 2x − 1)dx

= ∫₀⁴ x dx

= [x²/2]₀⁴

=1/2 [4² − 0²]

=1/2 × 16

= 8 square units

Given function is f(x) = x⁴/4 − x³ −5x² + 24 x + 12.

Differentiate w.r.to x, we get

⇒ f '(x) = 4x³/4 − 3x² − 5(2x) + 24

⇒ f '(x) = x³ − 3x² − 10x + 24

For critical points, put f '(x) = 0. Therefore x³ − 3x² − 10x + 24 = 0

⇒ (x − 2) (x² − x − 12) = 0

⇒ (x − 2) (x − 4) (x + 3) = 0

⇒ x = 2, 4, −3

Therefore, we have the intervals (−∞, 3), (−3, 2), (2, 4) and (4, ∞).

Since, f '(x) > 0 in (−3, 2) U (4, ∞). Therefore f(x) in strictly increasing in interval (−3, 2) U (4, ∞) and f '(x) < 0 in (−∞, −3) U (2, 4). Therefore f(x) is strictly decreading in the interval (−∞, −3) U (2, 4).

Given curve is 16x² + 9y² = 145 .......(i)

Since the point (x₁, y₁) lies on the given curve therefore 16x₁² + 9y₁² = 145.

Here, given x₁ = 2,

⇒ 16(2)² + 9y₁² = 145

⇒ 64 + 9y₁² = 145

⇒ 9y₁² = 145 − 64 = 81

⇒ y₁² = 9

⇒ y₁ = 3 [∵ y1 > 0]

Therefore, Point of contact is (2, 3)

Differentiating equation (i) with respect to x, which will give the slope of the tangent i.e.

32x + 18y dy/dx = 0

⇒ dy/dx = −32x/18y

Now, slope of the tangent at the point (2, 3) is

m = (dy/dx)_{(2 ,3)} = −32(2)/18(3) = −64/54 = −32/27

Therefore, the equation of the tangent is y − 3 = m(x − 2)

⇒ y − 3 = −32/27 (x − 2)

⇒ 27(y − 3) = −32(x − 2)

⇒ 27y − 81 = −32x + 64

⇒ 32x + 27y = 64 + 81

⇒ 32x + 27y = 145

Now, the slope of the normal = −1/Slope of tangent = −1/(−32/27) = 27/32

Equation of the normal is

⇒ y − 3 = 27/32 (x − 2)

⇒ 32(y − 3) = 27(x − 2)

⇒ 32y − 96 = 27x − 54

⇒ 27x − 32y = 54 − 96

⇒ 27x − 32y = − 42

Equation of the tanget is 32x + 27y = 145 and normal is 27x − 32y = − 42

The sample space has 36 outcomes.

Let A be the event that the sum of observations is 8 :

A = {(2, 6), (3, 5), (5, 3), (4,4), (6, 2)}

∴ n(A) = 5, P(A) = 5/36

Let B be event that observation on red die is less than 4

B = {(1, 1), (2, 1), (3, 1), (4,1), (1, 1), , (6, 1), , (1, 2), , (2, 2), , (3, 2), , (4, 2), (5, 2), , (6, 2), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3)}

∴ n(B) = 18 ⇒ P(B) = 18/36 = 1/2

Clearly A ∩ B = {(5, 3), (6, 2)}

∴ n(A ∩ B) = 2 P(A ∩ B) = 2/36 = 1/18

Now P(A/B) = P(A ∩ B) /P(B) = (1/18)/(1/2) = 2/18 = 1/9

Given curve is y = ae^{bx + 5}

Differentiating equation with respect to x, we get

⇒ dy/dx = ae^{bx + 5} d/dx (bx + 5)

⇒ dy/dx = ae^{bx + 5} b

⇒ dy/dx = by ......(i)

Differentiating again with respect to x, we get

⇒ d²y/dx² = b dy/dx

⇒ d²y/dx² = dy/dx (dy/dx • 1/y) [∵ b = dy/dx • 1/y ... from (i)]

⇒ y d²y/dx² = (dy/dx)²

Cost function is given by C(x) = 0.005x³ − 0.02x² + 30x + 5000

Marginal Cost (MC) = d/dx [C(x)]

= d/dx (0.005x³ − 0.02x² + 30x + 5000)

= 0.005(3x²) − 0.02(2x) + 30

= 0.015x² − 0.04x + 30

When x = 3,

MC = 0.015 × 3² − 0.04 × 3 + 30

= 0.135 − 0.12 + 30

= 30.015

Let and are the two such vectors. Now • = || || cos θ .......(i)

It is given that || = ||, θ = 60° and • = 9/2

⇒ • = || || cos θ

⇒ 9/2 = ||² 60°

⇒ 9/2 = ||² 1/2

⇒ ||² = 9

⇒ || = 3

⇒ || = || = 3

Let the number of packets of screw a manufactured in a day be x and the screw B be y. Therefore,x ≥ 0, and  y ≥ 0.

Then the constants are

⇒ 4x + 6y ≤ 240 or 2x + 3y ≤ 120

⇒ 6x + 3y ≤ 240 or 2x + y ≤ 80

and total profit is Z = 0.7 x + y.

So the LPP (Linear Programming Problem) will be Max Z = 0.7 x + y

Subject to the constraints : 2x + 3y ≤ 120, 2x + y ≤ 80 and x ≥ 0, y ≥ 0

Now we solve the table for the equation 2x + 3y ≤ 120, 2x + y ≤ 80. i, e. 

For 2x + 3y = 120

For 2x + y ≤ 80

X + 3 y less than equal to 120 2x + Y less than is equal to 80 and x y greater Now 2 X + 3 Y less than is equal to 120 and 2 X + Y less than is equal to 180 draught a Floating the points on the graph we get feasible region oabc is shown in the shaded region

Given curve is y² = 4x and say point P(x, y) is on the curve which is nearest to the given point (2, −8). Because y² = 4x ⇒ x = y²/a. Therefore, a point P(x, y) will be P(y²/4, y).

Now, the distance between both the points A and P is given by this distance formula √ [(x₁ − x₂)² + (y₁ − y₂)²] i.e.

⇒ AP = √ [(x − 2)² + (y + 8)²]

⇒ AP = √ [(y²/4 − 2)² + (y + 8)²]

⇒ AP = √ [(y⁴/16 − 2 × y²/4 × 2 + 4) + (y² + 2 × y × 8 + 64)]

⇒ AP = √ [(y⁴/16 − y² + 4) + (y² + 16y + 64)]

⇒ AP = √ [y⁴/16 − y² + 4 + y² + 16y + 64)]

⇒ AP = √ [y⁴/16 + 16y + 68)]

Let Z = AP² = y⁴/16 + 16y + 68

Now differentiating Z with respect to y, we get

⇒ dZ/dy = 1/16 × 4y³ + 16 = y³/4 + 16

For maximum of minimum value of Z, we have dZ/dy = 0 i.e.

⇒ y³/4 + 16 = 0

⇒ y³ + 64 = 0

⇒ (y + 4) (y² − 4y + 16) = 0

(∵ y² − 4y + 16 = 0 gives the imaginary value of y)

⇒ y = −4

Now, again differentiating with respect to y, we get

⇒ d²Z/dy² = d/dy(y³/4 + 16)

⇒ d²Z/dy² = 3y²/4

At y = −4, d²Z/dy² = 3(−4)²/4 = 3 × 16/4 = 3 × 4 = 12 > 0. Thus, Z is maximum when y = −4. Substituting y = −4 in equation of given curve y² = 4x.

x = (−4)²/4 = 16/4 = 4

Hence, the point P(4, −4) on the curve y² = 4x is nearest to the point (12, −8).

let x be the length of side, V be the Volume and S be the surface area of cube. Then V = x³ and S = 6 x², where x is a function of time t. Now, dV/dt = 8cm³/s (it is given), therefore

8 = dV/dt = d/dt (x³) = 3x² dx/dt

⇒ 8 = 3x² dx/dt

⇒ 8/3x² = dx/dt .......(i)

Now, dS/dt = d/dt (6x²)

⇒ dS/dt = 12x dx/dt

⇒ dS/dt = 12x (8/3x²) .... (from equation i)

⇒ dS/dt = 32/x

Hence, when x = 12 cm, then

dS/dt = 32/12 = 8/3 cm²/s

We have

(a + bx) e^y/x = x

⇒ e^y/x = x/(a + bx)

⇒ y/x = log [ x/(a + bx) ]

⇒ y/x = log x − log (a + bx)

Now, differentiating with respect to x, we get

⇒ [x dy/dx − y] / x² = 1/x − 1/(a + bx) × b

⇒ [x dy/dx − y] (1/x²) = 1/x − b/(a + bx)

⇒ x dy/dx − y = x²[1/x − b/(a + bx)]

⇒ x dy/dx − y = ax/(a + bx) ..........(i)

Again. differentiating both the sides w.r.to x, we get

x d²y/dx² + dy/dx − dy/dx = [ (a + bx) a − ax × b ] / (a + bx)²

⇒ x d²y/dx² = [ a² + abx − abx ] / (a + bx)²

⇒ x d²y/dx² = a²/(a + bx)²

On multiplying both the sides by x², we get

⇒ x³ d²y/dx² = a²x²/(a + bx)²

⇒ x³ d²y/dx² = [ax / (a + bx) ]² ..........(ii)

From equation (i) and (ii), we get

⇒ x³ d²y/dx² = [ax/(a + bx)]²

Given dy/dx = e^{x + y}

⇒ dy/dx = e^x e^y

⇒ dy/e^y = e^x dx

On Integrating both the sides, we get

⇒ ∫dy/e^y = ∫e^x dx

⇒ ∫e^−ydy = ∫e^x dx

⇒ e^−y = e^x + c, which is the required solution.

Given that || = 2, || = 7 and × = 3 + 2 + 6.then the angle between and is given by sin θ = | × | / | | ||

| × | = √(3² + 2² + 6²) = √(9+4+36) = √49 = 7

sin θ = 7/2×7 = 7/14 = 1/2 = sin π/6 ⇒ θ = π/6

We know that if a, b and c are the edges of a cuboid, then volume of a coboid is given by • (×). Here

= −3 + 7 − 5

= −5 + 7 − 3

= 7 − 5 − 3

• (×) = −3 (−21 − 15) −7 (15 + 21) + 5 (25 − 49)

= −3 × (−36) − 7 × 36 + 5 × (−24)

= 108 − 252 − 120

= −264 cubit units

The line passing through the point with position (2, −1, 4) and has a direction ratio proportional to (1, 1, −2). Then cartesian equation of a line is

= (x − x₁)/ a = (y − y₁)/b = (z − z₁)/c

= (x − 2)/1 = (y + 1)/1 = (z − 4)/−2

There am two types of goods, A and B and let units of type A be x and units of type B be y.

Then, Total Profit of goods is P = 40x + 50y, x ≥ 0, y ≥ 0

Hence, the mathematical formulation of the problem is as follows :

Maximize P = 40x + 50y

Subject to the constraints

x + 2y ≤ 8,

3x + y ≤ 9

x ≥ 0, y ≥ 0

To solve this LPP, we draw the lines x + 2y = 8, 3x + y = 9, x = 0 and y = 0.

and

Plotting these points on the graph

The shaded region is the required feasible region.

Clearly, P is Maximum at Point B (2, 3) and the maximum profit is Rs. 230.

Given, there are three coins. Let, E₁ = Coin is two headed, E₂ = Biased coin, E₃ = Unbiased coin and A = Shows only head. Here P(E₁) = P(E₂) = P(E₃) = 1/3. Then

P(A/E₁) = 1

P(A/E₂) = 75/100 = 3/4 (This is Given value)

P(A/E₃) = 1/2

Now, the probability of two headed coin is

P(E₁/A) = P(E₁) P(A/E₁) / [P(E₁) × P(A/E₁) + P(E₂) × P(A/E₂) + P(E₃) × P(A/E₃)]

= (1/3 × 1) / [1/3 × 1 + 1/3 × 3/4 + 1/3 × 1/2]

= 1/[1/3 + 1/4 + 1/6]

= 1/(9/4)

= 4/9

The equation of the given lines are (x – 1)/–3 = (y – 2)/2λ = (z – 3)/2 and (x – 1)/3λ = (y – 1)/2 = (z – 6)/–5

if the lines are perpendicular then it satisft the condition a₁a₂ + b₁b₂ + c₁c₂ = 0

⇒ –3 × 3λ + 3λ × 2 + 2 × (–5) = 0

⇒ –9λ + 4λ –10 = 0

⇒ –5λ = 10

⇒ λ = –2

Now, the lines are (x – 1)/–3 = (y – 2)/–4 = (z – 3)/2 and (x – 1)/–6 = (y – 1)/2 = (z – 6)/–5. The coordinates of any point on first line are given by

⇒ (x – 1)/–3 = (y – 2)/–4 = (z – 3)/2 = α

⇒ (x – 1) = –3α ⇒ x = –3α + 1

⇒ (y – 2) = –4α ⇒ y = –4α + 2

⇒ (z – 3) = 2α ⇒ z = 2α + 3

Therefore, the coordinates of any point on first line are (–3α + 1, –4α + 2, 2α + 3).

The coordinates of any point on second line are given by

⇒ (x – 1)/–6 = (y – 1)/2 = (z – 6)/–5 = β

⇒ (x – 1) = –6β ⇒ x = –6β + 1

⇒ (y – 1) = 2β ⇒ y = 2β + 1

⇒ (z – 6) = –5β ⇒ z = –5β + 6

Therefore, the coordinates of any point on second line are (–6β + 1, 2β + 1, –5β + 6). If the linses intersect then they have a common point, So, for some value of α and β

–3α + 1 = –6β + 1

⇒ –3α = –6β

⇒ α = 2β

and

–4α + 2 = 2β + 1

⇒ –4α + 1 = 2β.

On solving we get α = 1/5 and β = 1/10

The values of α and β do not satisfy the third equation. Hence, lines do not intersect each other.

Given xdy/dx = y – x tan (y/x)

Let y = vx, (v = y/x) differentiate w.r.to x, we get

dy/dx = v + x dv/dx

Substitute dy/dx in the given differentiation, we get

⇒ x(v + xdv/dx) = vx – x tan (y/x)

⇒ xv + x² dv/dx = vx – x tan v

⇒ x² dv/dx = – x tan v

⇒ x dv/dx = –tan v

⇒ dv/tan v = –dx/x

⇒ cot v dv = –dx/x

⇒ ∫cot v dv = –∫dx/x

⇒ log sin v = –log x + log c

⇒ log sin v = log c/x

⇒ sin v = c/x

⇒ sin (y/x) = c/x

⇒ x sin (y/x) = c

Suppose the normal at point P(x₁, y₁) on the given equation of parabola x² = 4y passes through (–1, 4). Since the point (x₁, y₁) lies on the parabola x² = 4y

⇒ x₁² = 4y₁ ...............(i)

The equation of the curve is x² = 4y

Differentiating with respect to x, we get

2x = 4 dy/dx

dy/dx = 2x/4 = x/2

dy/dx at point P(x₁, y₁) = x₁/2

m (Slope) = dy/dx = x₁/2

Now, the equation (x₁, y₁) of normal at P(x₁, y₁) is given by y – y₁ = –1/m (x – x₁)

y – y₁ = –2/x₁ (x – x₁) ..........(ii)

This is passing through the given point (–1, 4). So putting the value as x = –1 and y = 4 in the above equation, we get the result

4 – y₁ = –2/x₁ (–1 – x₁)

x₁(4 – y₁) = 2(1 + x₁)

4x₁ – x₁y₁ = 2 + 2x₁

x₁y₁ = 4x₁ – 2x₁ – 2

x₁y₁ = 2x₁ – 2

y₁ = (2x₁ – 2)/x₁ ................(iii)

Now, in next step we elemination y₁ from equation (i) x₁² = 4 y₁, we get

x₁² = 4(2x₁ – 2)/x₁

x₁³ = 8x₁ – 8

⇒ x₁ = 2

Substitute x = 1 in equation (iii), we get y₁ = 1. Putting these values of x₁ and y₁ in equation (ii), we get

y – 1 = –1 (x – 2)

y – 1 = –x + 2

y – 1 + x – 2 = 0

x + y – 3 = 0

Which is the required equation of normal to the given curve.

Given (cos x)^y = (sin y)^x, taking log both the sides, we get

log (cos x)^y = log (sin y)^x ⇒ y log (cos x) = x log (sin y)

Differentiating with respect to x, we get

d/dx [ y log (cos x)] = d/dx [ x log (sin y)]

y d/dx log (cos x) + log (cos x) d/dx (y) = x d/dx log (sin y) + log (sin y) d/dx (x)

y • 1/cos x (– sinx) + log (cos x) dy/dx = x 1/sin y • cos y dy/dx + log (sin y)

– y tan x + log (cos x) dy/dx = x cot y dy/dx + log (sin y)

log (cos x) dy/dx – x cot y dy/dx = log (sin y) + y tan x

dy/dx [log (cos x) dy/dx – x cot y ] = log (sin y) + y tan x

dy/dx = [log (sin y) + y tan x] / [log (cos x) dy/dx – x cot y]

Given that x√(1 + y) + y√(1 + x) = 0

x√(1 + y) =–y√(1 + x)

After Squaring both the sides, we get

⇒ x²(1 + y) = y²(1 + x)

⇒ x² + x²y = y² + y²x = 0

⇒ x² + x²y – y² – y²x = 0

⇒ x² – y² + x²y – y²x = 0

⇒ (x + y)(x – y) + xy (x – y) = 0

⇒ (x – y)[x + y + xy] = 0

⇒ x + y + xy = 0 [∵ x ≠ y ]

⇒ y(1 + x) + x = 0

⇒ y(1 + x) = –x

⇒ y(1 + x) = – x/(1 + x)

On differentiating above equation with respect to x, we get

dy/dx = (1 + x)d/dx (–x) – (–x) d/dx (1 + x) / (1 + x)²

dy/dx = (1 + x)(–1) + x(0 + 1) / (1 + x)²

dy/dx = –1 – x + x / (1 + x)²

dy/dx = –1/(1 + x)²

sin (cos^–1 4/5 + tan^–1 2/3)

= sin (tan^–1 3/4 + tan^–1 2/3) [ ∵ cos^–1 4/5 = tan^–1 3/4) ]

We know that tan^–1 x + tan^–1 y = tan^–1 (x + y)/(1 – xy)

= sin(tan^–1 (3/4 + 2/3)/(1 – 3/4x2/3)

= sin(tan^–1 ( (9+8)/12 / (1 – 6/12) )

= sin(tan^–1 ( 17/12 / 6/12) )

= sin(tan^–1 (17/6) )

Now, tan^–1 17/6 = sin^–1 17/√325

= sin(sin^–1 (17/√325) )

= 17/√325

First, consider any one arbitrary element of Y. By the definition of y, y = 4x + 3, for some value of x in the domain N. This shows that

4x = y – 3 ⇒ x = (y – 3)/4

Now, define g : Y → N by g(y) =& (y – 3)/4

gof (x) = g ( f(x) ) = g(4x + 3) = (4x + 3 – 3)/4 = 4x/4 = x

fog (y) = f ( g(y) ) = g( (y – 3)/4) ) = 4( y – 3)/4 + 3 = y – 3 + 3 = y

This shows that gof = I_N and fog = I_Y which implies that f is invertible and g is the inverse of f. Therefore Inverse of f = g(y) = (y – 3)/4

Here R = {(a, b) : b = a + 1}, therefore R = {(a, a + 1) : a, a + 1 ∈ (1, 2, 3, 4, 5, 6) } ⇒ R = {(1, 2),(2, 3),(3, 4),(4, 5),(5, 6)}

• R is not reflexive as (a, a) ∉ R ∀ a
• R is not symmetric as (1, 2) ∈ R but (2, 1) ∉ R
• R is not transitive as (1, 2) ∈ R, (2, 3) ∈ R but (1, 3) ∈ R

Here, A coin is tossed 5 times, therefore n = 5, p = 1/2 and q = 1/2,

We know that p(x) = ⁿC_x p^x q^{n – x}, therefore p(x) = ⁵C_x (1/2)^x (1/2)^{5 – x}.

Now for 3 heads, x = 3, then the probability of getting 3 heads is

P(3) = ⁵C₃ (1/2)³ (1/2)^{5 – 3} = ⁵C₃ (1/2)³ (1/2)² = ⁵C₃ (1/2)⁵

ⁿC_r = n! / r! (n – r)! ⇒ ⁵C₃ = 5! / 2! (5 – 2)! = 5x4x3! / 2! x 3! = 10

Hence P(3) = 10 (1/2)⁵ = 5/16

Probability of getting at most 3 heads is P(x ≤ 3) = P(0) + P(1) + P(2) + P(3), then

P(0) = ⁵C₀ (1/2)⁰ (1/2)^{5 – 0} = ⁵C₅ (1/2)⁵ = 1/32

P(1) = ⁵C₁ (1/2)¹ (1/2)^{5 – 1} = ⁵C₁ (1/2)⁵ = 5/32

P(2) = ⁵C₂ (1/2)² (1/2)^{5 – 2} = ⁵C₂ (1/2)⁵ = 5/16

P(3) = ⁵C₃ (1/2)³ (1/2)^{5 – 3} = ⁵C₃ (1/2)⁵ = 5/16

Therefore, P(x ≤ 3) = 1/32 + 5/32 + 5/16 + 5/16 = 26/32 = 13/16

Given

P(not A) = 0.7 ⇒ P(A) = 1 – 0.7 = 0.3
P(B) = 0.7
P(B/A) = 0.5

We know that P(B/A) = P(A ∩ B) / P(A), therefore

0.5 = P(A ∩ B)/0.3

⇒ P(A ∩ B) = 0.5 x 0.3 = 0.15

Now, Also we know P(A/B) = P(A ∩ B) / P(B), therefore

P(A/B) = 0.15/0.7 = 15/7

Given differential equation is y = Ae^2x + Be^–2x. ___ (i)

Now differentiating this equation w.r.t. x, we get

dy/dx = 2Ae^2x – 2Be^–2x ____(ii)

Again differentiating equation w.r.t. x, we get

d²y/dx² = 4Ae^2x + 4Be^–2x

d²y/dx² = 4 (Ae^2x + Be^–2x) = 4y

Therefore, required differential equation is d²y/dx² – y = 0.

Let the direction cosines of the line make and angle α with each of the coordinate axex and direction cosines be l, m, and n. Therefore, l = cos α, m = cos α and n = cos α.

l² + m² + n² = 1 ⇒ cos² α + cos² α + cos² α = 1

⇒ 3 cos² α = 1

⇒ cos² α = 1/3

⇒ cos α = ±1/√3

Therefore, the direction cosine are (±1/√3, ±1/√3, ±1/√3)

⇒ Given that A'A = I

Then, | A' A | = | I |

⇒ | A' | | A | = | I |

⇒ | A | | A | = | I | [∵ | A' | = | A| ]

⇒ | A | ² = 1 [∵ | I | = 1 ]

⇒ | A | = ±1 [∵ | I | = 1 ]

(a) Magnetic field inside the solenoid

∮B • dl = µ₀ I₀= µ₀ naI ............................(i)

∮B • dl = ∮_∞ B•dl + ∮_OR B•dl + ∮₀ B•dl + ∮₀ B•dl

∮B • dl = B∮ dl cos 0 + ∮ B•dl cos 90 + 0 dl cos 0 + ∮ B•dl cos 90

B ∮ dl = B a .............................(ii)

From equation (i) and (ii),

B = µ₀ naI

µ₀ I₀ – µ₀naI (where, n = Number of turns per unit length | I = Current passing through)

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

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

Solenoid Coil

Toroidal Coil

From Ampere's Law, ∮ BdI = μ₀i(t). Let the case 1, where a point P is considered outside the capacitor charging. From the Ampere's Law magnetic field at Point P will be :

CASE 1

B•(2πr) – μ₀ i(t) ⇒ B = μ₀ i(t) / 2πr

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

CASE 2

Hence, there is a contradiction. Therefore, this Ampere's law was modified with addition of displacement current inside the capacitor.

φ_E = | E | A = 1/ • Q/A • A = Q/ε₀

dφ_E/dt = 1/ε₀ • dQ/dt

ε₀ dφ_E/dt = dQ/dt = i_d, where i_d is the displacement current.

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

i = i_c + i_d = ic + ε₀dφ_E/dt

outside the capacitor, i_d = 0, hence, i = i_c and inside the capacitor, i_c = 0, hence i = i_d. Thus, i_c = i_d as capacitor gets charged.

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

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

Construction : The construction of a simple cyclotron is shown in figure above.

It consist of two-semi cylindrical boxes D₁ and D₂, which are called Dees. They are enclosed in an evacuated chamber. Chamber is kept between the pules of a powerful magnet so that uniform magnetic field acts perpendicular to the plane of the dees. An alternating voltage is applied in the gap between the two dees by the help of a high quency oscillator. The electric field is zero inside the dees.

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

qvB =mv²/r

From the above equation we get,

r = mv/qvB

The period of revolution is given by,

T = 2πr/v = 2π/v • mv/qB

T = 2πm/qB

The frequency of revolution is given by,

f ≈ 1/T = qB/2πm

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

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

As we know 1/2 m_pv² = eV [during acceleration of proton, P.R will converted to kinetic energy]
v = √(2eV/m_p)

When V is doubled, V' = 2V
v' = √(2e2V/m_p) = √2v
qvB = m_pv/r
r = m_pv/qB
r' = m_p/qB v'
r' = m_p/qB √2v
r' = √2r

The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. This is a phenomenon due to combined effect of dispersion, refraction and reflection of Sunlight by spherical water droplets of rain.

The conditions for observing a rainbow are that the Sun should be shining in one part of the sky (say near western horizontal while it is raining in the opposite part of the sky (say eastern horizon).

Difference between Primary and Secondary Rainbow

1. Primary Rainbow : Three steps process (Reflection-Reflection and Refraction). | Secondary Rainbow : Four Step process (Refraction - Reflection and Refraction).
2. Primary Rainbow : Appearance intensity better than Secondary. | Secondary Rainbow : Appearance intensity lesser than Primary.
3. Primary Rainbow : Single reflection occurs. | Secondary Rainbow : Double reflection occurs.
4. Primary Rainbow : Two Degree range occurs. | Secondary Rainbow : Three Degree range.

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

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