KCET 2020 April Question Paper with Answer Key and Solutions

KCET 2020 Question Paper All Questions

Karnataka Common Entrance Test - 2020

Chemistry

With regard to photoelectric effect, identify the CORRECT statement among the following:

a)	Number of e^- ejected increases with the increase in the frequency of incident light.	b)	Number of e^- ejected increases with the increase in work function.
c)	Number of e^- ejected increases with the increase in the intensity of incident light.	d)	Energy of e^- ejected increases with the increase in the intensity of incident light.

Chemistry - Structure of Atom

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Correct Option: c

Solution :

The number of ejected electrons can only be increased by increasing the intensity of the light. Any change in wavelength or frequency can only change kinetic energy of ejected electrons.

Therefore, Option(3) is correct.

0.4g of dihydrogen is made to react with 7.1g of dichlorine to form hydrogen chloride. The volume of hydrogen chloride formed at 273K and 1 bar pressure is

a)	4.54L	b)	90.8L
c)	45.4L	d)	9.08L

Chemistry - States of Matter : Gases and liquids

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Correct Option: a

Solution :

The reaction occurs as follows:

$\mathrm{H_{2}\: +\: Cl_{2}\: \rightarrow \: 2HCl}$

Now, clearly 2g of hydrogen reacts with 71g of chlorine.

Thus, 0.4g of hydrogen will react with 14.2g of chlorine.

Thus, no reactant is the limiting reagent here. Both the reactants will consume completely.

Thus, 2g of hydrogen will produce 73g of HCl.

Thus, 0.4g of hydrogen will produce 14.6g of HCl.

Now, moles of HCl produced = 7.3/73 = 0.2 moles

Now, according to the ideal gas equation, we have:

PV = nRT

Thus, V = (nRT)/P

V = (0.2 x 0.0821 x 273)/1

V = 4.54L

Therefore, Option(1) is correct.

The conjugate base of NH₃ is:

a)	NH₄OH	b)	NH₂OH
c)	NH₂^-	d)	NH₄⁺

Chemistry - Equilibrium

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Correct Option: c

Solution :

Ammonia is amphoteric, meaning that it can behave as either an acid or a base. It can accept a proton and therefore acts as a base. It can donate a proton and act as an acid. NH₂^- is therefore the conjugate base of ammonia.

Therefore, Option(3) is correct.

A gas mixture contains 25% He and 75% CH₄ by volume at a given temperature and pressure. The percentage by mass of methane in the mixture is approximately _________

a)	25%	b)	92%
c)	8%	d)	75%

Chemistry -

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Correct Option: b

Solution :

Let total moles of He and CH₄ be x.

Thus, moles of He = 0.25x

And moles of CH₄ = 0.75x

Now, mass of He = (0.25x) x (4) = x

Further, mass of CH₄ = (0.75x) x 16 = 12x

Thus, mass percentage of methane = [(12x)/(13x)] x 100 = 92%

Therefore, Option(b) is correct

The last element of the p-block in 6^th period is represented by the outermost electronic configuration:

a)	$5 f^{14} 6 d^{10} 7 s^{2} 7 p^{6}$	b)	$4 f^{14} 5 d^{10} 6 s^{2} 6 p^{4}$
c)	$4 f^{14} 5 d^{10} 6 s^{2} 6 p^{6}$	d)	$7 s^{2} 7 p^{6}$

Chemistry - Classification of elements and periodicity in properties

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Correct Option: c

Solution :

6th period starts with the filling of 6s-orbital and ends when 6p-orbitals are completely filled. In between 4f and 5d-orbitals are filled in accordance with aufbau principle. Thus, the outmost electronic configuration of the last element of the p-block in the 6th period is 4f¹⁴5d¹⁰6s²6p⁶.

Therefore, Option(3) is correct.

The formal charge on central oxygen atom in ozone is

a)	0	b)	+2
c)	+1	d)	-1

Chemistry - Chemical Bonding and Molecular Structure

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Correct Option: c

Solution :

The figure is given below:

$\\\text{Formal charge =valence - non bonding - bonding electrons} \\\\\mathrm{For 1:\: \: \: \: \: \: 6-4-\frac{4}{2}=0}\\\\ \mathrm{For 2:\: \: \: \: \: \: 6-2-\frac{6}{2}=6-2-3=1}\\\\ \mathrm{For 3:\: \: \: \: \: \: 6-6-\frac{2}{3}=-1}$

Therefore, Option(3) is correct.

When the same quantity of heat is absorbed by a system at two different temperatures T₁ and T₂, such that T₁> T₂, change in entropies are $$\Delta$ S₁ and $$\Delta$ S₂ respectively. Then :

a)	$\Delta \mathrm{S}_{1}=\Delta \mathrm{S}_{2}$	b)	$\mathrm{S}_{2}>\mathrm{S}_{1}$
c)	$\Delta S_{2}<\Delta S_{1}$	d)	$\Delta \mathrm{S}_{1}<\Delta \mathrm{S}_{2}$

Chemistry - Thermodynamics

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Correct Option:

Solution :

The formula for entropy change is given as:

$\mathrm{\Delta S=\frac{q}{T}}$

Now, since q is the same for both cases. Thus, more is the T, less will be entropy change.

Thus, for T₁ > T₂, we have:

$\Delta \mathrm{S}_{1}<\Delta \mathrm{S}_{2}$

Therefore, Option(4) is correct.

The percentage of s-character in the hybrid orbitals of nitrogen in NO₂⁺, NO₃^- and NH₄⁺ respectively are:

a)	$33.3 \%, 25 \%, 50 \%$	b)	$50 \%, 33.3 \%, 25 \%$
c)	$25 \%, 50 \%, 33.3 \%$	d)	$33.3 \%, 50 \%, 25 \%$

Chemistry - Chemical Bonding and Molecular Structure

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Correct Option: b

Solution :

$\\$\mathrm{NO}_{2}^{+}$\\ \text{Number of electron pairs =2}\\\text{Number of bond pairs =2}\\\text{Number of lone pair =0}\\\text{So, the species is linear with sp hybridisation.}\\\\ $\mathrm{NO}_{3}^{-}$\\ \text{Number of electron pairs =3}\\ \text{Number of bond pairs =3} \\\text{Number of lone pair =0}\\\text{So, the species is trigonal planar with sp2 hybridisation}\\\\ $\mathrm{NH}_{4}^{+}$\\\\ \text{Number of electron pairs = 4} \\\text{Number of bond pairs = 4}\\\text{Number of lone pair = 0}\\\text{So, the species is tetrahedral with sp3 hybridisation.}$

Thus percentage character of s in the hybrid orbitals of nitrogen in NO₂⁺, NO₃^- and NH₄⁺ respectively are:

50%, 33.3% and 25%.

Therefore, Option(2) is correct.

A Lewis acid 'X' reacts with LiAlH₄ in ether medium to give a highly toxic gas, This gas when heated with NH₃ gives a compound commonly known as inorganic benzene. The gas is

a)	$\mathrm{B}_{2} \mathrm{H}_{6}$	b)	$\mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6}$
c)	$\mathrm{B F_{3}}$	d)	$\mathrm{B}_{2} \mathrm{O}_{3}$

Chemistry - P-block elements

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Correct Option: c

Solution :

First the toxic gas that reacts with ammonia and produces borazine, then it according to the reaction given below, it must diborane.

$3 \mathrm{B}_{2} \mathrm{H}_{6}+6 \mathrm{NH}_{3} \stackrel{\text { High temp. }}{\longrightarrow} 2 \mathrm{B}_{3} \mathrm{N}_{3} \mathrm{H}_{6}+12 \mathrm{H}_{2}$

Now, formation of diborane using LiAlH₄ is only possible if the reactant involved in the reaction is BF₃.

$4 \mathrm{BF}_{3}+3 \mathrm{LiAlH}_{4} \longrightarrow 2 \mathrm{B}_{2} \mathrm{H}_{6}$

It is prepared by treating boron Trifluoride with LiAlH₄ in diethyl ether.

Therefore, Option(3) is correct.

The oxide of potassium that does not exist is

a)	$\mathrm{KO}_{2}$	b)	$\mathrm{K}_{2} \mathrm{O}_{2}$
c)	$\mathrm{K}_{2} \mathrm{O}_{3}$	d)	$\mathrm{K}_{2} \mathrm{O}$

Chemistry - S - Block Elements

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Correct Option:

Solution :

Potassium oxide (K₂O) is an ionic compound of potassium and oxygen. This pale yellow solid, the simplest oxide of potassium, is a rarely encountered, highly reactive compound.

Therefore, Option(4) is correct.

The oxidation number of nitrogen atoms in NH₄NO₃ are

a)	-3, +5	b)	+3, -5
c)	-3, -3	d)	+5, +5

Chemistry - P-block elements

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Correct Option: a

Solution :

Since ammonium (NH4) has a charge of +1 then N + 4 H = +1, hydrogen has a +1 Oxidation state so then N + 4 = +1 which means that the nitrogen must be -3.

Since nitrate (NO3) has a charge of -1 then N + 3 O = -1, oxygen has a -2 Oxidation so then N + (-6) = -1 or N - 6 = -1 which means that this nitrogen must be +5.

Therefore, Option(1) is correct.

Which of the following is NOT a pair of functional isomers?

a)	$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} \text { and } \mathrm{CH}_{3} \mathrm{OCH}_{3}$	b)	$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{NO}_{2} \text { and } \mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{COOH}$
c)	$\mathrm{CH}_{3} \mathrm{COOH} \text { and } \mathrm{HCOOCH}_{3}$	d)	$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OC}_{2} \mathrm{H}_{5} \text { and } \mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OCH}_{3}$

Chemistry - Organic Chemistry - some Basic principles and techniques

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Correct Option: d

Solution :

Two or more compounds having the same molecular formula but different functional groups are called functional isomers.

In C₂H₅OC₂H₅ and C₃H₇OCH₃, the functional group is same, i.e, ether. Only position of oxygen atom is changing. Thus, C₂H₅OC₂H₅ and C₃H₇OCH₃, are not functional isomers.

Therefore, Option(4) is correct.

Identify 'X' in the following reaction:

a)		b)
c)		d)

Chemistry - Haloalkanes and Haloarenes

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Correct Option: b

Solution :

When benzene is treated with excess Cl₂ in the presence of sunlight, addition reaction takes place to form hexachlorocyclohexane. This is a free radical reaction, in presence of sunlight, Cl₂ undergoes homolytic fission to give 2Cl $.$ radicals which add to the benzene ring.

Therefore, Option(2) is correct.

The metal that produces H₂ with both dil HCl and NaOH(aq) is

a)	Mg	b)	Ca
c)	Fe	d)	Zn

Chemistry - d - and f - BLOCK ELEMENTS

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Correct Option: d

Solution :

Zinc is a transition metal which shows moderate reactivity. Zinc reacts with NaOH to form sodium zincate with the evolution of hydrogen gas. The reaction occurs as follows:

$\mathrm{Zn\: +\: 2NaOH \rightarrow Na_{2}ZnO_{2}\: +\: H_{2} \uparrow}$

The reaction between zinc and hydrochloric acid is a single replacement reaction where zinc metal displaces the hydrogen to form hydrogen gas and zinc chloride.

$\mathrm{Zn}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{ZnCl}_{2(\mathrm{aq})}+\mathrm{H}_{2(\mathrm{g})}$

Therefore, Option(4) is correct.

A metal exists as an oxide with formula M_0.96O. Metal M can exist as M⁺² and M⁺³ in its oxide M_0.96O. The percentage of M⁺³ in the oxide is nearly

a)	4.6%	b)	5%
c)	9.6%	d)	8.3%

Chemistry - d - and f - BLOCK ELEMENTS

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Correct Option: d

Solution :

M_0.98O is non-stoichiometric compound and is a mixture of M²⁺ and M³⁺ ions. Let x atoms of M³⁺ are present in the compound. This means that x M²⁺ have been replaced by M³⁺ ions.

$\therefore$ Number of M²⁺ ions = 0.96-x
For electrical neutrality, positive charge on compound = negative charge on compound.

$\begin{array}{l} \therefore 2(0.96-x)+3 x=2 \\ 1.92-2 x+3 x=2 \\ \text {or } x=2-1.92=0.08 \end{array}$

$\therefore \quad \% \text { of } \mathrm{M}^{3+} \text { ions }=\frac{0.08}{0.96} \times 100=8.33 \%$

Therefore, Option(4) is correct.

A metal crystallises in face centred cubic structure with metallic radius $\sqrt{2} \mathrm{A}^{\circ} .$ The volume of the unit cell in m³ is

a)	$6.4 \times 10^{-29}$	b)	$4 \times 10^{-9}$
c)	$6.4 \times 10^{-30}$	d)	$4 \times 10^{-10}$

Chemistry - Solid state

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Correct Option: a

Solution :

We know that the in FCC, the unit cell edge length is given as:

$\\\mathrm{a\: =\: 2\sqrt{2}r}\\\mathrm{a\: =\: 2\sqrt{2}\: x\:\sqrt{2}}\\\\\mathrm{Thus,\: a\: =\: 4A^{o}\: or\: 4\: x\: 10^{-10}m}$

Now, volume of unit cell = a³

Thus, volume = (4 x 10^-10) x (4 x 10^-10) x (4 x 10^-10) = 64 x 10^-30

Thus, volume = 6.4 x 10^-29m³

Therefore, Optioin(1) is correct.

Which of the following is NOT a green house gas?

a)	CO₂	b)	O₂
c)	NO₂	d)	CFC

Chemistry - Environmental Chemistry

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Correct Option: b

Solution :

Naturally occurring greenhouse gases include water vapor, carbon dioxide, ozone, methane and nitrous oxide, and together create a natural greenhouse effect. The greenhouse gases have three or more atoms in their molecules. CO₂, H₂O, CFC all have three atoms.

Oxygen is not greenhouse gases, because it is transparent to infrared light. These molecules are invisible because when you stretch one, it doesn't change the electric field. These are symmetric molecules, made of two identical atoms whose electric fields just cancel each other out.

Therefore, Option(2) is correct.

The pair of electrolytes that possess same value for the constant (A) in the Debye - Huckel - Onsagar equation, $\lambda_{m}=\lambda_{m}^{\circ}-A \sqrt{C}$ is

a)	$\mathrm{NH}_{4} \mathrm{Cl}, \mathrm{NaBr}$	b)	$\mathrm{NaBr}, \mathrm{MgSO}_{4}$
c)	$\mathrm{NaCl}, \mathrm{CaCl}_{2}$	d)	$\mathrm{MgSO}_{4}, \mathrm{Na}_{2} \mathrm{SO}_{4}$

Chemistry -

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Correct Option: a

Solution :

The pair of electrolytes that possess same value for the constant (A) is NH₄Cl and NaBr.

Therefore, Option(1) is correct.

Which of the following pair of solutions is isotonic?

a)	$\begin{array}{l} 0.001 \mathrm{M}\: \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} \text { and } 0.01 \mathrm{M}\: \mathrm{BaCl}_{2} \end{array}$	b)	$\begin{aligned} \\0.001 \mathrm{M} \: \mathrm{CaCl}_{2} \text { and } 0.001 \mathrm{M}\: \mathrm{Al_{2}}\left(\mathrm{SO}_{4}\right)_{3} \end{aligned}$
c)	$\mathrm{0.01M\: B a C l_{2} \: \text { and } \: 0.001 & M\: CaCl_{2}}$	d)	$0.01 \mathrm{M}\: \mathrm{BaCl}_{2} \text { and } 0.015 \mathrm{M}\: \mathrm{NaCl}$

Chemistry - Solutions

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Correct Option: d

Solution :

Isotonic solutions are those solutions which have the same osmotic pressure. If osmotic pressures are equal at the same temperature, concentrations must also be equal.

$\mathrm{\pi=iCRT}$

For BaCl₂, we have:

i = 3

Thus, C of ions = 3 x 0.01 = 0.03M

Now, for NaCl, we have:

i = 2

Thus, C of ions = 2 x 0.015 = 0.03M

Thus as the concentrations are same, the solutions are isotonic.

Therefore, Option(4) is correct.

Silicon doped with gallium forms

a)	both n and p type semiconductor	b)	an intrinsic semiconductor
c)	p - type semiconductor	d)	n - type semiconductor

Chemistry - P-block elements

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Correct Option: c

Solution :

Gallium has only three outer electrons. When mixed into the silicon lattice, they form "holes" in the lattice where a silicon electron has nothing to bond to.

Therefore, Option(3) is correct.

Given and The equilibrium constant for the reaction taking place in galvanic cell consisting of above two electrodes is

a)	1x10⁹	b)	3x10⁸
c)	5x10¹²	d)	1x10⁷

Chemistry - Electrochemistry Electrochemistry

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(Asked in: AEEE - 1970)

Correct Option: d

Solution :

GIven,

$E^0 _{Fe^{+3}/Fe^{+2}}= 0.76 ($ This is Cathode$)\\\\ E^0 _{I_{2}/I^{-1}}= 0.55 ($ This is Anode$a)\\\\ E^0 _{cell}=E^0 _{C} - E^0 _{A} \\\\ = 0.76 - .55 = .21 \\\\ $ The reaction involved is $\\\\ E^0 _{cell} = \frac{0.059}{n} \log K_c \\\\ 0.21 = \frac{0.059}{2} \log K_c\\\\ \log K_c = 7 \\\\ $ Therefore $ , K_c = 10^7$

Hence, Option D is correct

If an aqueous solution of NaF is electrolyzed between inert electrodes, the product obtained at anode is

a)	H₂	b)	Na
c)	O₂	d)	F₂

Chemistry - Electrochemistry Electrochemistry

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(Asked in: AEEE - 1970)

Correct Option: c

Solution :

When an aqueous solution of NaF is electrolyzed between inert electrodes, the product obtained at anode is Oxygen gas (O₂) .

Hence Option C is correct

Solute ‘X’ dimerises in water to the extent of 80%. 2.5g of ‘X’ in 100g of water increases the boiling point by 0.3 0C. The molar mass of ‘X’ is [Kb=0.52K kg mol-1]

a)	52	b)	65
c)	26	d)	13

Chemistry - Solutions

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Correct Option: c

Solution :

Given,

‘X’ dimerises in water to the extent of 80%.

2.5g of ‘X’ in 100g of water increases the boiling point by 0.3 ⁰C.

K_b=0.52K kg mol-1 $i = 1 + \alpha (\frac{1}{n}-1)\\\\ i = 1 + 0.8 (\frac{1}{2}-1)\\\\ i = 1 - 0.4 = 0.6\\\\ $ According to depression in boiling point equation, we have $ \\\\ \Delta T_b = K_b \times \frac{W}{m} \times \frac{100}{W} \times i \\\\ 0.3 = 0.52 \times \frac{2.5}{m} \times \frac{1000}{100} \times 0.6 \\\\ $ therefore molar mass will be = $\\\\ \frac{0.52 \times 2.5 \times 10 \times 0.6}{0.3} \\\\ = 26$

Therefore, option C is correct

The time required for 60% completion of a first order reaction is 50 min. The time required for 93.6% completion of the same reaction will be

a)	83.8 min	b)	50 min
c)	150 min	d)	100 min

Chemistry - Chemical kinetics

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Correct Option: c

Solution :

For 60% completionof reaction of the reaction,

$K = \frac{2.303}{t} \log\frac{R_0}{R}\\\\ K = \frac{2.303}{50} \log\frac{100}{40}\\\\ K = \frac{2.303}{50} \times 0.397\\\\ $ 93.6 \% completation $ \\\\ K = \frac{2.303}{t} \log\frac{R_0}{R} \\\\ \frac{2.303}{t} \log\frac{100}{6.4} = \frac{2.303}{50} \times 0.397 \\\\ $ t = 150 minutes$

Therefore the time required for the completion of 93.6% is 150 minutes

Therefor option C is correct

For an elementary reaction 2A+3B ------> 4C+D the rate of appearance of C at time ‘t’ is 2.8x10^-3 mol L-1S-1. Rate of disappearance of B at ‘t’ t will be

a)		b)
c)		d)

Chemistry - Chemical kinetics

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Correct Option: a

Solution :

Given equation,

Rate of appearance of C at time ‘t’ is 2.8x10-3 mol L^-1S^-1

Rate of disappearance of B at ‘t’ t will be-

According to rate law,

$\frac{-1}{3}\frac{dB}{dt} = \frac{1}{4}\frac{dC}{dt}\\\\\\ \frac{-dB}{dt} = \frac{3}{4}\frac{dC}{dt} \\\\ = \frac{3}{4}(2.8 \times 10^{-3}) \text{mol }L^{-1}s^{-1}$

Hence, Option A is correct

In which of the following cases a chemical reaction is possible ?

a)	AgNO3 solution is stirred with a copper spoon	b)	Conc. HNO3 is stored in a platinum vessel
c)	gold ornaments are washed with dil HCl	d)	ZnSO4(aq) is placed in a copper vessel

Chemistry - Some basic concepts in chemistry

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Correct Option: a

Solution :

When AgNO₃ solution is stirred with a copper spoon, displacement reaction will take place which will form a new compound called as copper nitrate.

In all other cases no chemical reaction takes place between the components taken.

Hence only option in which chemical reaction is possible is when AgNO₃ solution is stirred with a copper spoon

Therefore option A is correct

A sol of AgI is prepared by mixing equal volumes of 0.1M AgNO3 and 0.2M KI, which of the following statement is correct ?

a)	Sol obtained is a positive sol with Ag+ adsorbed on AgI	b)	Sol obtained is a positive sol with K+ adsorbed on AgI
c)	Sol obtained is a negative sol with I- adsorbed on AgI	d)	Sol obtained is a negative sol with NO3^- adsorbed on AgI

Chemistry - Surface Chemistry

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Correct Option: c

Solution :

Since KI is in excess (higher concentratopn), Hence due to I^- ions, sol obtained is a negative sol with I^-adsorbed on AgI.

Therefore, Option C is correct

During Adsorption of a gas on a solid

a)		b)
c)		d)

Chemistry - Thermodynamics

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Correct Option: d

Solution :

When a gas is adsorbed on a solid surface, its movement is restricted leading to a decrease in the entropy of the gas i.e., ΔS is negative. Now for a process to be spontaneous, ΔG should be negative. Since ΔS is negative, ΔH has to be negative to make ΔG negative.

Hence

Option D is correct

The rate constant of a reaction is given by k=P Ze-Ea/RT under standard notation. In order to speed up the reaction, which of the following factors has to be decreased ?

a)	Both Z and T	b)	Ea
c)	T	d)	Z

Chemistry - Chemical kinetics

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Correct Option: b

Solution :

k=P Ze^-Ea/RT

From the above reaction it is clear that,

Speed of the reaction is directly proportional to temperature(T), Z and inversly proportional to E_a,

Hence In order to speed up the reaction, the only factor which needs to be decreased is E_a

Therefore the correct option is B

Function of potassium ethyl xanthate in froth floatation process is to make the ore

a)	a) Lighter	b)	b) Hydrophobic
c)	c) Hydrophilic	d)	d) Heavier

Chemistry - General Principle and process of Isolation of metals

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Correct Option: b

Solution :

Function of potassium ethyl xanthate in froth floatation process is to make the ore Hydrophobic as the polar part of xanthate molecule attaches to the ore particles with the non-polar hydrocarbon part sticking out and forming a hydrophobic layer.

Sulphide ore on roasting gives a gas X. X reacts with Cl2 in the presence of activated charcoal to give Y. Y is:

a)	SCl₂	b)	SCl₆
c)	SOCl₂	d)	SO₂Cl₂

Chemistry - General Principle and process of Isolation of metals

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Correct Option: d

Solution :

Sulphide ore on Burning gives SO₂ gas, which on reacting with Cl₂ in the presence of activated charcoal gives SO₂Cl₂

Below reaction describes the process-
SO₂+Cl₂→SO₂Cl₂

Hence, Correct Option is D

Copper is extracted from copper pyrites by

a)	Reduction by coke	b)	Electrometallurgy
c)	Auto reduction	d)	Thermal decomposition

Chemistry - General Principle and process of Isolation of metals

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Correct Option: c

Solution :

The reaction occurs as follows:

$\mathrm{Cu}_{2} \mathrm{O}+\frac{1}{2} \mathrm{Cu}_{2} \mathrm{S} \rightarrow 3 \mathrm{Cu}+\frac{1}{2} \mathrm{SO}_{2}$

This reaction includes reduction of copper (I) oxide by copper (I) sulphide. In this process, copper is reduced by itself hence this process is known as autoreduction and the solidified copper. So, obtained is known as blister copper.

Therefore, Option(3) is correct.

Bond angle in PH₄⁺ is more than that of PH₃. This is because

a)	PH₄⁺ has square planar structure	b)	PH₃ has planar trigonal structure
c)	hybridisation of P changes when PH₃ is converted to PH₄	d)	lonepair - bond pair repulsion exists in PH₃

Chemistry - Chemical Bonding and Molecular Structure

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Correct Option: d

Solution :

P in both PH₃ and PH₄⁺ is sp³ hybridized. Due to the absence of lone pair-bond pair repulsion and presence of four identical bond pair-bond pair interactions, PH4+ assumes tetrahedral geometry with a bond angle to 109° 28'. But PH₃ has three bond pairs and one lone pair around P. Due to greater lone pair-bond pair repulsion than bond pair-bond pair repulsion, the tetrahedral angle decreases from 109° 28' to 93.6°. As a result, PH₃ is pyramidal.

Therefore, Option(4) is correct.

Incorrectly matched pair is :

a)	XeF₄ - tetrahedral	b)	XeF₆ - distorted octahedral
c)	XeOF₄- square pyramidal	d)	XeO₃- pyramidal

Chemistry - Chemical Bonding and Molecular Structure

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Correct Option: a

Solution :

The lone pairs of Xenon lie in the perpendicular plane in an octahedral arrangement. Therefore, XeF4 molecular geometry is square planar. The bond angles are 90^o or 180°. The lone pairs lie on the opposite sides of the molecule basically at 180° from each other.

Therefore, Option(1) is correct.

Aqueous solution of a salt (A) forms a dense white precipitate with BaCl₂ solution. The precipitate dissolves in ditute HCl to produce a gas (B) which decolourises acidified KMnO₄ solution.

A and B respectively are:

a)	$\mathrm{BaSO}_{4}, \mathrm{H}_{2} \mathrm{S}$	b)	$\mathrm{BaSO}_{3}, \mathrm{H}_{2} \mathrm{S}$
c)	$\mathrm{BaSO}_{4}, \mathrm{SO}_{2}$	d)	$\mathrm{BaSO}_{3}, \mathrm{SO}_{2}$

Chemistry - S - Block Elements

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Correct Option: d

Solution :

The reactions occur as follows:

$\mathrm{BaSO} _{3}+\mathrm{HCl}\: \rightarrow \: \mathrm{BaCl} _{2}+\mathrm{H} _{2} \mathrm{O}+\mathrm{SO} _{2}$

Now SO₂ is the gas which can decolourise acidified KMnO₄ solution.

Therefore, Option(4) is correct.

Identify the set of paramagnetic ions among the following

a)	$\mathrm{Ni}^{2+} \cdot \mathrm{Cu}^{2+}, \mathrm{Zn}^{2+}$	b)	$\mathrm{T i^{3+}, C u^{2+}, M n^{3+}}$
c)	$\mathrm{Sc}^{3+}, \mathrm{Ti}^{3+}, \mathrm{V}^{3+}$	d)	$\mathrm{V}^{2+}, \mathrm{Co}^{2+}, \mathrm{Ti}^{4+}$

Chemistry - Co-ordination Compounds

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Correct Option: b

Solution :

The electronic configurations are:

Ti³⁺ = [Ar] 3d¹

It has one unpaired electron and thus paramagnetic.

Cu²⁺ = [Ar] 3d⁹

It has also one unpaired electron and thus paramagnetic.

Mn³⁺ = [Ar] 3d⁴

It has 4 unpaired electrons and thus paramagnetic.

Therefore, Option(2) is correct.

How many moles of acidified K₂Cr₂O₇ is required to liberate 6 moles of I₂from an aqueous solution of I^-?

a)	1	b)	0.25
c)	0.5	d)	2

Chemistry - d - and f - BLOCK ELEMENTS

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Correct Option: d

Solution :

The reaction occurs as follows:

$\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{I}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{I}_{2}$

Thus, for producing 6 moles of I₂, 2 moles of acidified K₂Cr₂O₇ is required.

Therefore, Option(4) is correct.

Phosphorus pentachloride

a)	on hydrolysis gives an oxo acid of phosphorus which is a good reducing agent.	b)	has all the five equivalent bonds.
c)	exists as an ionic solid in which cation has octahedral structure and anion has tetrahedral structure.	d)	on hydrolysis gives an oxo acid of phosphorus which is tribasic.

Chemistry - P-block elements

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Correct Option: d

Solution :

The reaction of water with phosphorus pentachloride (PCl5) is called a hydrolysis reaction. The products of the reaction are phosphoric acid (H3PO4) and hydrochloric acid (HCl).

According to the structure of phosphoric acid (H3PO4) as given below, it is tribasic in nature.

Therefore, Option(4) is correct.

The co-ordination number of Fe and Co in the complex ions, [Fe(C₂O₄)₃]^3- and [Co(SCN)₄]^2- are respectively:

a)	6 and 8	b)	4 and 6
c)	6 and 4	d)	3 and 4

Chemistry - Co-ordination Compounds

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Correct Option: c

Solution :

Coordination number is the number of atoms or ions immediately surrounding a central atom in a complex or crystal.

Now, (C₂O₄)^2- is bidentate ligand and (SCN)^- is monodentate ligand. Thus, the coordination numbers of [Fe(C₂O₄)₃]^3- and [Co(SCN)₄]^2- are 6 and 4 respectively.

Therefore, Option(3) is correct.

Number of stereoisomers exhibited by [Co(en)₂Cl₂] is

a)	2	b)	5
c)	3	d)	4

Chemistry - Co-ordination Compounds

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Correct Option: c

Solution :

Number of possible isomers for the complex [Co(en)2Cl2]⁺ will be (en=ethylenediamine) 3. There are two geometrical isomers cis and trans. The trans isomer is optically inactive whereas the cis form exists in d and l forms.

Therefore, Option(3) is correct.

Give the IUPAC name of $\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{4}\right]\left[\mathrm{PtCl}_{4}\right]$ is

a)	tetra ammine platinate (II) tetra chlorido platinum (II)	b)	tetra ammine platinate (o) tetra chlorido platinum (IV)
c)	tetra ammine platinum (II) tetra chlorido platinate (II)	d)	tetra ammine platinum (o) tetra chlorido platinum (IV)

Chemistry - Co-ordination Compounds

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Correct Option: c

Solution :

tetra ammine platinum (II) tetra chlorido platinate (II)

Is correct.

Cu₂Cl₂ and CuCl₂ in aqueous medium

a)	CuCl₂ is more stable than Cu₂Cl₂	b)	Stability of Cu₂Cl₂ is equal to stability CuCl₂
c)	Both are unstable	d)	Cu₂Cl₂ is more stable than CuCl₂

Chemistry - d - and f - BLOCK ELEMENTS

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Correct Option: d

Solution :

Cu₂Cl₂ is more stable than CuCl₂ is correct.

Which of the following halide shows the highest reactivity towards S_N1 reaction?

a)	$\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Cl}$	b)	$\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{CH}_{2} \mathrm{I}$
c)	$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}$	d)	$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Cl}$

Chemistry - Organic Chemistry - some Basic principles and techniques

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Correct Option: d

Solution :

Option 4 is correct.

Rate of SN1 reaction is directly proportional to stability of carbocation or Reactivity of SN1 reaction is influenced by stability of carbonation.

$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2}^+$ is most stable among them.

In the reaction

The number of possible isomers for the organic compound X is

a)	5	b)	3
c)	2	d)	4

Chemistry - Organic Compounds Containing Nitrogen

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Correct Option: c

Solution :

2-methylpropane has 2 isomers.

Another is butane.

Prolonged exposure of chloroform in humans may cause damage to liver. It is due to the formation of the following compound

a)	COCl₂	b)	CH₂Cl₂
c)	Cl₂	d)	CCl₄

Chemistry - Haloalkanes and Haloarenes

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Correct Option: a

Solution :

$\mathrm{CHCl}_{3} \longrightarrow \mathrm{COCl}_{2}+\mathrm{HCl}$

. The steps involved in the conversion of propan -2-ol to propan -1-ol are in the order

a)	Heating with PCl5, heating with alc. KOH, acid catalysed addition of water	b)	Heating with PCl5, heating with alc. KOH, hydroboration oxidation
c)	Dehydration, addition of HBr, heating with aq. KOH	d)	Dehydration, the addition of HBr in the presence of peroxide, heating with alc. KOH

Chemistry - Alcohols, Phenols and Ethers

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Correct Option: c

Solution :

Option 2 is correct.

. Which of the following on heating gives an ether as major products?

$\\$\mathrm{P}: \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}_{2} \mathrm{Br}+\mathrm{CH}_{3} \mathrm{ONa}$\\ $\mathrm{Q}: \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{ONa}+\mathrm{CH}_{3} \mathrm{Br}$\\ $\mathrm{R}:\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Cl}+\mathrm{CH}_{3} \mathrm{ONa}$\\ $\mathrm{S}: \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CH}=\mathrm{CHCl}+\mathrm{CH}_{3} \mathrm{ONa}$$

a)	Both P and R	b)	Both Q and S
c)	Both P and Q	d)	Both R and S

Chemistry - Alcohols, Phenols and Ethers

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Correct Option: d

Solution :

Primary alkyl halides/benzyl halides react with alkoxide/phenoxide through the SN2 mechanism gives ethers. Vinyl and aryl halides least reactive towards SN1

a)		b)
c)		d)

Chemistry - Aldehydes, Ketones and Carboxylic Acids

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Correct Option:

Solution :

is the answer.

Which of the following has the lowest boiling point?

a)	$\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$	b)	$\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{3}$
c)	HCOOH	d)	$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}$

Chemistry - Organic Chemistry - some Basic principles and techniques

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Correct Option: b

Solution :

$\mathrm{CH}_{3}-\mathrm{O}-\mathrm{CH}_{3}$ has not hydrogen polar bond.

So, due to the absence of H-bonding it has lowest boiling point.

Which of the following is the strongest base?

a)	Cl^-	b)	OH^-
c)	CH₃O^_-	d)	CH₃COO^_-

Chemistry - Organic Chemistry - some Basic principles and techniques

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Correct Option: c

Solution :

CH₃O^- strongest base.

a)		b)
c)		d)

Chemistry - Organic Compounds Containing Nitrogen

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Correct Option: a

Solution :

Option no. 1 correct.

Hinsberg's reagent is

a)	$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl}$	b)	$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{NH}_{2}$
c)	$\mathrm{CH}_{3} \mathrm{COCl} / \text { pyridine }$	d)	$\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O} / \text { pyridine }$

Chemistry - Organic Compounds Containing Nitrogen

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Correct Option: a

Solution :

$\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{2} \mathrm{Cl}$ is Hinsberg's reagent.

The carbonyl compound that does not undergo aldol condensation is

a)	Di chloro acetaldehyde	b)	Tri chloro acetaldehyde
c)	Acetaldehyde	d)	Acetone

Chemistry - Aldehydes, Ketones and Carboxylic Acids

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Correct Option: b

Solution :

For aldol condensation need alpha hydrogen.

Cl3C-CHO has not alpha Hydrogen.

Hypothyroidism is caused by the deficiency of

a)	Adrenalin	b)	Thyroxine
c)	Glucocorticoid	d)	Vitamin B-12

Chemistry - Biomolecules

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Correct Option: b

Solution :

Thyroxine produced in the thyroid gland is an iodinated derivative of amino acid tyrosine. Abnormally low level of thyroxine leads to hypothyroidism which is characterised by lethargyness and obesity.

Increased level of thyroxine causes hyperthyroidism.

Low level of iodine in the diet may lead to hypothyroidism and enlargement of the thyroid gland. This condition is largely being controlled by adding sodium iodide to commercial table salt (“Iodised” salt).

C₁-C₄ glycosidic bond is NOT found in

a)	Sucrose	b)	Lactose
c)	Starch	d)	Maltose

Chemistry - Biomolecules

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Correct Option: a

Solution :

Sucrose has 1-2 linkage.

Which of the following polymer has the strongest intermolecular forces of attraction?

a)	Terylene	b)	Polythene
c)	Polystyrene	d)	Neoprene

Chemistry - Polymers

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Correct Option: a

Solution :

Increasing of Molecular Forces Polymers:-

Fibres > Thermoplastic polymers > Elastomers

Terylene (Fibres) is the answer.

Which one of the following vitamins is not stored in adipose tissue?

a)	B₆	b)	D
c)	E	d)	A

Chemistry - Biomolecules

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Correct Option: a

Solution :

Fat-soluble vitamins: Vitamins that are soluble in fat and oils but insoluble in water are kept in this group. These are vitamins A, D, E, and K. They are stored in liver and adipose (fat-storing) tissues.

Water-soluble vitamins: B group vitamins and vitamin C are soluble in water so they are grouped together. Water-soluble vitamins must be supplied regularly in the diet because they are readily excreted in urine and cannot be stored (except vitamin B12) in our body.

So, Vitamin B6 cannot be stored in adipose tissue.

Therefore, the correct option is (1).

A food additive that acts as an antioxidant is

a)	Saccharin	b)	Sugar syrup
c)	Salt	d)	BHA

Chemistry - Chemistry in Everyday Life

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Correct Option: d

Solution :

The anti -Oxidants are important and necessary food additives. These help in food preservation by retarding the action of oxygen on food. These are more

reactive towards oxygen than the food material which they are protecting. The two most familiar antioxidants are butylated hydroxytoluene (BHT) and butylated hydroxyanisole (BHA).

Butylated hydroxyanisole (BHA) is an antioxidant.

Therefore, the correct option is (4).

Which of the following is not related to drug enzyme interaction?

a)	Antagonist	b)	Co-enzymes
c)	Enzyme inhibitor	d)	Allosteric site

Chemistry - Chemistry in Everyday Life

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Correct Option: a

Solution :

Co-enzymes, enzyme inhibitor, allosteric site, all are related to Drug-enzyme interaction.

But antagonist is not related to Drug-enzyme interaction.

It is part of receptors.

Drugs that bind to the receptor site and inhibit its natural function are called antagonists.

Therefore, the correct option is (1).

Which of the following monomers can undergo condensation polymerization?

a)	Glycine	b)	Isoprene
c)	Propene	d)	Styrene

Chemistry - Polymers

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Correct Option: a

Solution :

Condensation polymers:

The condensation polymers are formed by repeated condensation reaction between two different bi-functional or tri-functional monomeric units. In these polymerization reactions, the elimination of small molecules such as water, alcohol, hydrogen chloride, etc. take place.

Glycine is bifunctional and rest are mono functional.

Maths

If $a_1, a_2, a_3 \ldots \ldots a_9$ are in A.P. then the value of $\left|\begin{array}{ccc}a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9\end{array}\right|_{\text {is }}$

a)	$a_1+a_9$	b)	$\log _e\left(\log _c e\right)$
c)	1	d)	$\frac{9}{2}\left(a_1+a_9\right)$

Maths - Algebra

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Correct Option: b

Solution :

Let a be first term and d be common difference

$
\begin{aligned}
& \left|\begin{array}{ccc}
a_1 & a_2 & a_3 \\
a_4 & a_5 & a_6 \\
a_7 & a_8 & a_9
\end{array}\right|=\left|\begin{array}{ccc}
a & a+d & a+2 d \\
a+3 d & a+4 d & a+5 d \\
a+6 d & a+7 d & a+8 d
\end{array}\right| \\
& \text { by applying } C_2 \rightarrow C_2-C_1 \\
& C_3 \rightarrow C_3-C_2 \\
& =\left|\begin{array}{ccc}
a & d & d \\
a+3 d & d & d \\
a+6 d & d & d
\end{array}\right|=0
\end{aligned}
$

Now by looking into the option, only $\log \left(\log _e e\right)$ equals 0.

$
\log \left(\log _e e\right) \quad[\because \log 1=0]
$
Hence, the answer is the option 2.

If $A$ is a square matrix of order 3 and $|A|=5$, then $\mid A$ adj. $A \mid$ is

a)	5	b)	25
c)	125	d)	625

Maths - Algebra

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Correct Option: c

Solution :

Given $|A|=5$ and the order of $A$ is $3(n=3)$ :

1. Calculate $|\operatorname{adj}(A)|$ :

$
|\operatorname{adj}(A)|=|A|^{n-1}=5^{3-1}=5^2=25 .
$

2. Compute $|A \cdot \operatorname{adj}(A)|$ : Using the property of determinants:

$
|A \cdot \operatorname{adj}(A)|=|A|^n=5^3=125
$

Hence, the answer is the option 3.

If $A$ and $B$ are square matrices of same order and $B$ is a skew symmetric matrix, then $A^{\prime} B A$ is

a)	Symmetric matrix	b)	Null matrix
c)	Diagonal matrix	d)	Skew symmetric matrix

Maths - Algebra

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Correct Option: d

Solution :

Given, $B=-B^{\prime}$

$\operatorname{Now}\left(A^{\prime} B A\right)^{\prime}=(B A)^{\prime}\left(A^{\prime}\right)^{\prime}\left[\because(A B)^{\prime}=B^{\prime} A^{\prime}\right]$
$=A^{\prime} B^{\prime} A$
$=-A^{\prime} B A$

Hence, the answer is the option 4.

If $f(x)=\left|\begin{array}{ccc}x^3-x & a+x & b+x \\ x-a & x^2-x & c+x \\ x-b & x-c & 0\end{array}\right|$, then

a)	$f(1)=0$	b)	$f(2)=0$
c)	$f(0)=0$	d)	$f(-1)=0$

Maths - Algebra

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Correct Option: c

Solution :

$f(0)=\left|\begin{array}{ccc}
0 & a & b \\
-a & 0 & c \\
-b & -c & 0
\end{array}\right|$

The determinant of a skew-symmetric matrix of odd order (here, $3 \times 3$ ) is always zero. This is a fundamental result in linear algebra.

$f(0) =0$

Hence, the answer is option 3.

If $\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] A=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$, then the matrix A is

a)	$\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]$	b)	$\left[\begin{array}{cc}-2 & 1 \\ 3 & -2\end{array}\right]$
c)	$\left[\begin{array}{cc}2 & -1 \\ 3 & 2\end{array}\right]$	d)	$\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right]$

Maths - Algebra

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Correct Option: a

Solution :

We know that $A^{-1} A=A A^{-1}=I$ or, $B A=A B=I$
Let $B=\left[\begin{array}{ll}2 & 1 \\ 3 & 2\end{array}\right] \Rightarrow|B|=4-3=1$
$\operatorname{adj} B=\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]$
$B^{-1}=\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]$
$\therefore A=B^{-1}=\left[\begin{array}{cc}2 & -1 \\ -3 & 2\end{array}\right]$

Hence, the answer is the option 1.

If A ={ a, b, c }, then the number of binary operations on A is

a)	$3^{1}$	b)	$3^{3}$
c)	$3^{6}$	d)	$3^{9}$

Maths - Algebra

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Correct Option: d

Solution :

$A =\{ a , b , c \}$

The number of binary operations are $n^{n^{2}}=3^{3^{2}}=3^{9}$

$A=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right] {\text {then } \mathrm{A}^4 \text { is equal to }}$

a)	I	b)	A
c)	2A	d)	4A

Maths - Algebra

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Correct Option: a

Solution :

$
\begin{aligned}
& A^2=A A=\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]=I \\
& A^3=A^2 \times A=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]=I
\end{aligned}
$

$
A^4=A^3 A=I \times A=I
$

Hence, the answer is the option 1.

$f(x)=\left\{\begin{array}{cl}\frac{1-\cos K x}{x \sin x}, & \text { If } x \neq 0 \\ \frac{1}{2}, & \text { If } x=0 \text { is continuous at } \mathrm{x}=0, \text { then the value of } \mathrm{K} \text { is }\end{array}\right.$

$\pm \frac{1}{2}$

$\pm 1$

$\pm 2$

$0$

Maths - Calculus

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Correct Option: b

Solution :

$\begin{aligned} & \because \mathrm{f} \text { is continuous at } \mathrm{x}=0 . \\ & \text { Then } \mathrm{f}(0)=\lim _{x \rightarrow 0} \mathrm{f}(\mathrm{x}) \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{1-\cos k x}{x \sin x}=1 / 2 \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{k \sin k x}{x \cos x+\sin x}=1 / 2 \text { (By L'Hopital Rule) } \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{k^2 x \frac{\sin k x}{k x}}{x\left(\cos x+\frac{\sin x}{x}\right)}=1 / 2 \\ & \Rightarrow \lim _{x \rightarrow 0} \frac{k^2 \times \frac{x}{x} \times \frac{\sin k x}{k x}}{\cos x+\frac{\sin x}{x}}=1 / 2 \\ & \Rightarrow \mathrm{k}^2 \frac{\lim _{k x \rightarrow 0} \frac{\sin k x}{k x}}{\lim _{x \rightarrow 0} \cos x+\lim _{x \rightarrow 0} \frac{\sin x}{x}}=1 / 2 \\ & \Rightarrow \frac{k^2 \times 1}{\cos 0+1}=1 / 2\left(\because \lim _{x \rightarrow 0} \sin \mathrm{x} / \mathrm{x}=1\right) \\ & \Rightarrow \mathrm{k}^2 / 2=1 / 2 \\ & \Rightarrow \mathrm{k}^2=1 \\ & \Rightarrow \mathrm{k}= \pm 1\end{aligned}$

Hence, the answer is the option 2.

The value of $\cos \left(\sin ^{-1} \frac{\pi}{3}+\cos ^{-1} \frac{\pi}{3}\right)$ is

a)	-1	b)	0
c)	1	d)	Does not exist

Maths - Sets, Relations and Functions

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Correct Option: b

Solution :

$\cos \left(\sin ^{-1} \frac{\pi}{3}+\cos ^{-1} \frac{\pi}{3}\right)=\cos \frac{\pi}{2}=0$

Hence, the correct answer is the option 2.

The domain of the function defined by $f(x)=\cos ^{-1} \sqrt{x-1}$ is

$[1,2]$

$[0,2]$

$[-1,1]$

$[0,1]$

Maths - Sets, Relations and Functions

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Correct Option: a

Solution :

The domain of the inverse cosine function is $[-1,1]$

$-1 \leq \cos ^{-1}(\sqrt{x-1}) \leq 1$

The square root is always non-negative $(\sqrt{x-1} \geq 0)$

$\begin{aligned} & 0 \leq \sqrt{x-1} \leq 1 \\ & 0 \leq(x-1) \leq 1 \\ & 1 \leq x \leq 1+1 \Rightarrow[1,2]\end{aligned}$

Hence, the correct answer is the option 1.

If $A, B, C$ are three mutually exclusive and exhaustive events of an experiment such that $P(A)=2 P(B)=$ $3 P(C)$, then $P(B)$ is equal to

a)	$\frac{1}{11}$	b)	$\frac{2}{11}$
c)	$\frac{3}{11}$	d)	$\frac{4}{11}$

Maths - Statistics and Probability

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Correct Option: c

Solution :

$\begin{aligned} & \frac{P(A)}{6}=\frac{2 P(B)}{6}=\frac{3 P(C)}{6} \\ & \frac{P(A)}{6}=\frac{P(B)}{3}=\frac{P(C)}{2}=K \\ & P(A)=6 K, P(B)=3 K, P(C)=2(K) \\ & P(A)+P(B)+P(C)=1 \Rightarrow 11 K=1, K=\frac{1}{11} \\ & P(B)=\frac{3}{11}\end{aligned}$

Hence, the correct answer is the option 3.

Let $f:[2, \infty) \rightarrow R {\text {be the function defined }} f(x)=x^2-4 x+5$, then the range of f is

a)	$(-\infty, \infty)$	b)	$[1, \infty)$
c)	$(1, \infty)$	d)	$[5, \infty)$

Maths - Sets, Relations and Functions

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Correct Option: b

Solution :

$\begin{aligned} & y=f(x)=x^2-4 x+5 \\ & =(x-2)^2+1 \\ & 1+(x-2)^2 \geq 0+1 \\ & y \geq 1 \Rightarrow y \in[1, \infty)\end{aligned}$

Hence, the correct answer is the option 2.

If a relation R on the set {1, 2, 3} be defined by R={(1, 1)}, then R is

a)	Reflexive and symmetric	b)	Reflexive and transitive
c)	Symmetric and transitive	d)	Only symmetric

Maths - Sets, Relations and Functions

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Correct Option: c

Solution :

$
R=\{(1,1)\} \text { on a set }\{1,2,3\}
$

$R$ is symmetric as $(a, b)$ is related to $(b, a) \Rightarrow(1,1)$ is related to $(1,1)$

R is only symmetric.

Hence, the answer is option 4.

$\lim _{x \rightarrow 0}\left(\frac{\tan x}{\sqrt{2 x+4}-2}\right)$ is equal to

a)	6	b)	4
c)	3	d)	2

Maths - Calculus

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Correct Option: d

Solution :

$\begin{aligned} & \lim _{x \rightarrow 0}\left(\frac{\tan x}{\sqrt{2 x+4}-2}\right) \\ & \text { Rationalize } \\ & \lim _{x \rightarrow 0} \frac{\tan x(\sqrt{2 x+4}+2)}{2 x+4-4} \\ & \lim _{x \rightarrow 0} \frac{\tan x}{x} \times \frac{\lim }{x \rightarrow 0} \frac{(\sqrt{2 x+4}+2)}{2} \\ & \Rightarrow1 \times \frac{2+2}{2}=2\end{aligned}$

Hence, the correct answer is the option 4.

Events $E_1$ and $E_2$ from a partition of the sample space $S$. $A$ is any event such that $P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}, P\left(E_2 / A\right)=\frac{1}{2}$ and $P\left(A / E_2\right)=\frac{2}{3}$, then $P\left(E_1 / A\right)$ is

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{2}{3}$

$1$

Maths - Statistics and Probability

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Correct Option: b

Solution :

$\begin{aligned} & \frac{P\left(E_2 \cap A\right)}{P(A)}=\frac{1}{2}, \frac{P\left(A \cap E_2\right)}{P\left(E_2\right)}=\frac{2}{3} \\ & P(A)=\frac{2}{3}, \quad P\left(A \cap E_2\right)=\frac{1}{3} \\ & A \cap\left(E, \cup E_2\right)=A \\ & P\left(A \cap E_1\right)+P\left(A \cap E_2\right)=P(A) \\ & P\left(A \cap E_1\right)=\frac{1}{3} \\ & P\left(E_1 / A\right)=\frac{P\left(E_1 \cap A\right)}{P(A)}=\frac{1 / 3}{2 / 3}=1 / 2\end{aligned}$

Hence, the correct answer is the option 2.

The standard deviation of the data $6,7,8,9,10$ is

a)	$\sqrt{2}$	b)	$2$
c)	$\sqrt{10}$	d)	$10$

Maths - Statistics and Probability

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Correct Option: a

Solution :

The given data is 6,7,8,9,10.

From the given data, $n=5$

Mean of the given data, $\bar{x} = \frac{6+7+8+9+10}{5}$ = 8

$\begin{gathered}\text { S.D }=\sqrt{\frac{1}{n}\left(x_1^2\right)-x^{-2}}=\sqrt{66-64}=\sqrt{2}\end{gathered}$

Hence, the correct answer is the option 1.

The negation of the statement "For all real numbers $x$ and $y, x+y=y+x$ " is

For all real numbers x and $\mathrm{y}, x+y \neq y+x$

For some real numbers x and $\mathrm{y}, x+y=y+x$

For some real numbers x and $\mathrm{y}, x+y \neq y+x$

For some real numbers x and $\mathrm{y}, x-y=y-x$

Maths - Mathematical reasoning

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Correct Option: c

Solution :

Negation : For some real numbers $x$ and $y, x+y \neq y+x$

Hence, the correct answer is the option 3.

If the sum of $n$ terms of an A.P is given by $S_n=n^2+n$ then the common difference of the A.P is

a)	1	b)	2
c)	4	d)	6

Maths - Algebra

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Correct Option: b

Solution :

$\begin{aligned} & S_n=n^2+n \\ & S_1=1+1=2=T_1 \\ & S_2=2^2+2=6=T_1+T_2 \\ & T_2=S_2-S_1=6-2=4 \\ & d=T_2-T_1=4-2=2\end{aligned}$

Hence, the answer is the option 2.

If the parabola $x^2=4$ ay passes through the point $(2,1)$, then the length of the latus rectum is

a)	1	b)	2
c)	4	d)	8

Maths - Co-ordinate geometry

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Correct Option: c

Solution :

Substitute the value of $x,y$ from the point $(2,1)$.

$\begin{aligned} & x^2=4 a y \\ \Rightarrow & 2^2=4 a \times 1 \\ \Rightarrow & a=1\end{aligned}$

Length of latus rectum,

$4 a =4 \times 1=4$

Hence, the answer is the option 3.

The two lines $l^{\prime} x+m^{\prime} y=n^{\prime}$ and $l x+m y=n$ are perpendicular if

a)	$l l^{\prime}+m m^{\prime}=0$	b)	$l m^{\prime}=m l^{\prime}$
c)	$l m+l^{\prime} m^{\prime}=0$	d)	$l m^{\prime}+m l^{\prime}=0$

Maths - Co-ordinate geometry

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Correct Option: a

Solution :

Slope of the line $l^{\prime} x+m^{\prime} y=n^{\prime}$ is $m_1 = -\frac{l^{\prime}}{m^{\prime}}$

Slope of the line $l x+m y=n$ is $m_2 = -\frac{l}{m}$

For two lines to be perpendicular, $m_1m_2 = -1$

$\begin{aligned} & m_1=-\frac{l^{\prime}}{m^{\prime}} \quad m_2= -\frac{l}{m}\\ & m_1 \times m_2=-1 \\ & \left(\frac{-l}{m}\right)\left(\frac{-l^{\prime}}{m^{\prime}}\right)=-1 \\ & l l^{\prime}+m m^{\prime}=0\end{aligned}$

$ll^\prime=-mm^\prime$

Hence, the answer is the option 1.

If $P(n): 2^n<n!$. Then the smallest positive integer for which $\mathrm{P}(\mathrm{n})$ is true if

a)	5	b)	4
c)	3	d)	2

Maths - Algebra

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Correct Option: b

Solution :

For $n=1$, $2^1 < 1!$ is not true.

For $n=2$, $2^2 < 2! \Rightarrow 4<2$ is not true

For $n=3$, $2^3 < 3! \Rightarrow 4<6$ is not possible.

For $n=4$, $2^n<4!\Rightarrow 16<24$ is true.

The smallest number satisfying the condition $2^n < n!$ is 4.

Hence, the answer is the option 2.

The number of terms in the expansion of $(x+y+z)^{10}$ is

a)	11	b)	66
c)	110	d)	142

Maths - Algebra

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Correct Option: b

Solution :

The number of distinct terms:

$
\begin{aligned}
& { }^{(n+r-1)} C_{r-1} \\
& n=10, r=3 \\
& { }^{(10+3-1)} C_{31}={ }^{12} C_2=\frac{12 \times 11}{2}=66
\end{aligned}
$

Hence, the answer is the option 2.

The value of ${ }^{16} C_9+{ }^{16} C_{10}-{ }^{16} C_6-{ }^{16} C_7$ is

a)	${ }^{17} C _{10}$	b)	${ }^{17} C _3$
c)	1	d)	0

Maths - Algebra

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Correct Option: d

Solution :

$\begin{aligned} & { }^n C_r+{ }^n C_{r-1}={ }^{(n+1)} C_r \\ & { }^{16} C_9+{ }^{16} C_{10}-\left({ }^{16} C_6+{ }^{16} C_7\right) \\ & { }^{17} C_{10}-{ }^{17} C_7 \\ & { }^{17} C_{17-10}-{ }^{17} C_7= { }^{17} C_{7}-{ }^{17} C_7 = 0\left(\because{ }^n C_r={ }^n C_{n-r}\right)\end{aligned}$

Hence, the answer is the option 4.

If $z=x+i y$, then the equation $|z+1|=|z-1| {\text {represents }}$

a)	a circle	b)	a parabola
c)	x-axis	d)	y-axis

Maths - Algebra

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Correct Option: d

Solution :

$\begin{aligned} & |z+1|=|(x+1)+i y|=\sqrt{(x+1)^2+y^2} \\ & |z-1|=|(x-1)+i y|=\sqrt{(x-1)^2+y^2}\end{aligned}$

Substitute in the given equation,

$\sqrt{(x+1)^2+y^2}=\sqrt{(x-1)^2+y^2}$

$\begin{gathered}(x+1)^2+y^2=(x-1)^2+y^2\end{gathered}$

$4 x=0$

$x=0$

Since $x=0$, the equation represents the y-axis.

Hence, the answer is option 4.

If $A = {1,2,3,4,5,6} $ then the number of subsets of A which contain at least two elements is

a)	57	b)	58
c)	63	d)	64

Maths - Sets, Relations and Functions

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Correct Option: a

Solution :

Subsets of A are $P(A)=2^6=64$

Subsets with fewer than two elements = {empty set, 6 singleton sets} = 7

Subsets with at least two elements $=$ Total subsets - Subsets with fewer than two elements

Subsets of A which contain at least two elements $P(A)-7=64-7=57$

Hence, the correct answer is the option 1.

If $\tan A+\cot A=2$ then the value of $\tan ^4 A+\cot ^4 A=$ ?

a)	1	b)	2
c)	4	d)	5

Maths - Sets, Relations and Functions

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Correct Option: b

Solution :

$\begin{aligned} & \tan A+\frac{1}{\tan A}=2 \\ & (\tan A-1)^2=0 \\ & \tan A=1 \quad \Rightarrow A=45^{\circ} \\ & \tan ^4 45^{\circ}+\cot ^4 45= 1^4 + 1^4 = 1+1 = 2\end{aligned}$

Hence, the answer is the option 2.

The value of $\sin ^2 51^{\circ}+\sin ^2 39^{\circ}$ is

a)	$\sin 12^{\circ}$	b)	$\cos 12^{\circ}$
c)	0	d)	1

Maths - Sets, Relations and Functions

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(Asked in: JEE Main - 1970)

Correct Option: d

Solution :

$\sin ^2 51^{\circ}+\sin ^2 39^{\circ}=\cos ^2 39^{\circ}+\sin ^2 39^{\circ}=1$

Hence, the answer is the option(4).

If n(A) = 2 and total number of possible relations from Set A to set B is 1024, then n(B) is

a)	5	b)	10
c)	20	d)	512

Maths - Sets, Relations and Functions

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(Asked in: JEE Main - 1970)

Correct Option: a

Solution :

$\begin{aligned} & n(A)=2 \\ & 2^{m n}=1024 \\ & \left(2^2\right)^n=2^{10} \\ & 2^{2 \times 5}=2^{10} \\ & n(B)=5\end{aligned}$

Hence, the answer is the option(1).

The probability of solving a problem by three persons A, B and C independently is $\frac{1}{2}, \frac{1}{4} \text { and } \frac{1}{3}$ respectively. Then the probability of the problem is solved by any two of them is

a)	$\frac{1}{12}$	b)	$\frac{1}{24}$
c)	$\frac{1}4$	d)	$\frac{1}8$

Maths - Statistics and Probability

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Correct Option: c

Solution :

$\\ \quad P ( A )=\frac{1}{2}, P ( B )=\frac{1}{4}, P ( C )=\frac{1}{3}\\\\ \text{Probability of the problem solved by any two } \\\\= P ( A ) P ( B ) P (\overline{ C })+ P ( A ) P (\overline{ B }) P ( C )+ P (\overline{ A }) P ( B ) P ( C ) =\frac{1}{4}$

If A and B are two events such that $P(A)=\frac{1}{3}, P(B)=\frac{1}{2}$ and $P(A \cap B)=\frac{1}{6}_{\text {, then }} P\left(A^{\prime} / B\right)_{\text {is }}$

a)	$\frac{2}{3}$	b)	$\frac{1}{3}$
c)	$\frac{1}{2}$	d)	$\frac{1}{12}$

Maths - Statistics and Probability

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(Asked in: JEE Main - 1970)

Correct Option: a

Solution :

$\begin{aligned} & P\left(A^{\prime} / B\right)=\frac{P\left(A^{\prime} \cap B\right)}{P(B)} \\ & =\frac{P(B)-P(A \cap B)}{P(B)} \\ & =\left(\frac{1}{2}-\frac{1}{6}\right) 2 \\ & =2\left[\frac{3-1}{6}\right]=\frac{2}{3}\end{aligned}$

Hence, the answer is the option(1).

A die is thrown 10 times, the probability that an odd number will come up atleast one time is

a)	$\frac{1}{1024}$	b)	$\frac{1023}{1024}$
c)	$\frac{11}{1024}$	d)	$\frac{1013}{1024}$

Maths - Statistics and Probability

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Correct Option: b

Solution :

$\text { Given } n=10 \quad p =\frac{1}{2}, \quad q =\frac{1}{2} \\$

$\begin{aligned} \text { Required probability } &=1- P ( X =0) \\ &=1-{ }^{10} C _{0}\left(\frac{1}{2}\right)^{10-0}\left(\frac{1}{2}\right)^{0} \\ &=1-\frac{1}{2^{10}}=1-\frac{1}{1024} \\ &=\frac{1023}{1024} \end{aligned}$

The feasible region of an LPP is shown in the figure. If $Z=11 x+7 y$, then the maximum value of Z occurs at

a)	( 0, 5)	b)	( 3, 3)
c)	( 5, 0)	d)	( 3, 2)

Maths - Linear Programming

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(Asked in: MHT-CET - 1970)

Correct Option: d

Solution :

Corner points $(0,3),(0,5),(3,2)$
$\therefore$ maximum value at $(3,2)$

Hence, the answer is the option(4).

Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1) and (3, 0). Let z = px + qy, where p, q>0. Condition on p and q so that the minimum of z occurs at (3, 0) and (1, 1) is

a)	$p=2 q$	b)	$p=\frac{q}{2}$
c)	$p=3 q$	d)	$p=q$

Maths - Linear Programming

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(Asked in: MHT-CET - 1970)

Correct Option: b

Solution :

Given corner points are $(0,3),(1,1),(3,0)$

$
z=p x+q y
$

At $(3,0) \quad z=3 p$
At $(1,1) \quad z=p+q$

$
\begin{aligned}
& \Rightarrow 3 p=p+q \\
& 2 p=q \Rightarrow p=\frac{q}{2}
\end{aligned}
$
Hence, the answer is the option(2).

If a line makes an angle of $\frac{\pi}{3}$ with each of x and y-axis, then the acute angle made by z-axis is

a)	$\frac{\pi}{4}$	b)	$\frac{\pi}{6}$
c)	$\frac{\pi}{3}$	d)	$\frac{\pi}{2}$

Maths - Vectors and three-dimentional geometry

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(Asked in: JEE Main - 1970)

Correct Option: b

Solution :

$\begin{aligned} & \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\ & \frac{1}{4}+\frac{1}{4}+\cos ^2 \gamma=1 \\ & \cos ^2 \gamma=1-1 / 2 \\ & =1 / 2 \\ & \cos \gamma=\frac{1}{\sqrt{2}} \\ & \gamma=\pi / 4\end{aligned}$

Hence, the answer is the option(2).

The sine of the angle between the straight line $\frac{x-2}{3}=\frac{3-y}{-4}=\frac{z-4}{5}$ and the plane $2 x-2 y+z=5$ is

a)	$\frac{3}{\sqrt{50}}$	b)	$\frac{3}{50}$
c)	$\frac{4}{5 \sqrt{2}}$	d)	$\frac{\sqrt{2}}{10}$

Maths - Vectors and three-dimentional geometry

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(Asked in: JEE Main - 1970)

Correct Option: d

Solution :

Given line is $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ and plane is $2 x-2 y+z=5$ w.k.t. $\sin \theta=\left|\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\right|$

$
\begin{aligned}
& \sin \theta=\frac{(3 \hat{i}+4 \hat{j}+5 \hat{k}) \cdot(2 \hat{i}-2 \hat{j}+\hat{k})}{\sqrt{9+16+25} \sqrt{4+4+1}} \\
& =\frac{6-8+5}{\sqrt{50} \sqrt{9}}=\frac{3}{5 \sqrt{2} .3}=\frac{1}{5 \sqrt{2}}=\frac{1}{5 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{10}
\end{aligned}
$

Hence, the answer is the option(4).

The distance of the point $(1,2,-4)$ from the line $\frac{x-3}{2}=\frac{y-3}{3}=\frac{z+5}{6}$ is

a)	$\frac{293}{7}$	b)	$\frac{\sqrt{293}}{7}$
c)	$\frac{293}{49}$	d)	$\frac{\sqrt{293}}{49}$

Maths - Vectors and three-dimentional geometry

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(Asked in: JEE Main - 1970)

Correct Option: b

Solution :

$\begin{aligned} & \cos \theta=\frac{4+3-6}{\sqrt{6} \cdot 7}=\frac{1}{7 \sqrt{6}} \\ & \sin \theta=\frac{h}{\sqrt{6}} \\ & h=\sqrt{6} \sqrt{1-\cos ^2 \theta} \\ & =\frac{\sqrt{293}}{7}\end{aligned}$

Hence, the answer is the option(2).

If the vectors $2 i-3 j+4 k, 2 i+j-k$ and $\lambda i-j+2 k$ are coplanar, then the value of $\lambda$ is

a)	5	b)	-5
c)	6	d)	-6

Maths - Vectors and three-dimentional geometry

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(Asked in: JEE Main - 1970)

Correct Option: c

Solution :

$\begin{aligned} & \left|\begin{array}{ccc}2 & -3 & 4 \\ 2 & 1 & -1 \\ \lambda & -1 & 2\end{array}\right|=0 \\ & 2(2-1)+3(4+\lambda)+4(-2-\lambda)=0 \\ & \lambda=6\end{aligned}$

Hence, the answer is the option(3).

The point (1,-3, 4) lies in the octant

a)	Second	b)	Third
c)	Fourth	d)	Eighth

Maths - Vectors and three-dimentional geometry

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(Asked in: JEE Main - 1970)

Correct Option: c

Solution :

Fourth quadrant

Hence, the answer is the option(3).

${ }_{\text {If }}|\vec{a} \times \vec{b}|^2+|\vec{a} \cdot \vec{b}|^2=144$ and $|\vec{a}|=6$, then $|\vec{b}|_{\text {is equal to: }}$

a)	2	b)	3
c)	4	d)	6

Maths - Vectors and three-dimentional geometry

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(Asked in: JEE Main - 1970)

Correct Option: a

Solution :

$\begin{aligned} & |\vec{a}|^2|b|^2=144 \\ & \left.\vec{b}\right|^2=4 \\ & \left.\vec{b}\right|^2=2\end{aligned}$

Hence, the answer is the option(1).

The curve passing through the point (1, 2) given that the slope of the tangent at any point (x, y) is $\frac{3 x}{y}$represents

a)	Circle	b)	Hyperbola
c)	Ellipse	d)	Parabola

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: b

Solution :

$
\begin{array}{r}
\frac{d y}{d x}=\frac{2 x}{y} \\
y d y=2 x d x \\
\frac{y^2}{2}=x^2+c
\end{array}
$

passes through $(1,2)$

$
\begin{aligned}
& \Rightarrow c=1 \\
& \frac{y^2}{2}=x^2+1 \\
& \frac{x^2}{1}-\frac{y^2}{2}=-1
\end{aligned}
$

Represents hyperbola

Hence, the answer is the option(2).

If $\vec{a}$ and $\vec{b}$ are unit vectors and $\theta$ is the angle between $\vec{a}$ and $\vec{b}$, then $\sin \frac{\theta}{2}$ is

a)	$\|\vec{a}+\vec{b}\|$	b)	$\frac{\|\vec{a}+\vec{b}\|}{2}$
c)	$\frac{\|\vec{a}-\vec{b}\|}{2}$	d)	$\|\vec{a}-\vec{b}\|$

Maths - Vectors and three-dimentional geometry

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(Asked in: JEE Main - 1970)

Correct Option: c

Solution :

$\begin{aligned} & |\vec{a}-\vec{b}|=|\vec{a}|^2+|\vec{b}|^2-2 \vec{a} \cdot \vec{b} \\ & =1+1-2 \cos \theta \\ & =2(1-\cos \theta) \\ & |\vec{a}-\vec{b}|^2=2 \cdot 2 \sin ^2(\theta / 2) \\ & |\vec{a}-\vec{b}|=2 \sin (\theta / 2)\end{aligned}$

Hence, the answer is the option(3).

The two vectors $\square$ and $\hat{i}+3 \hat{j}+5 \hat{k}$ represent the two sides $\overrightarrow{A B}$ and $\overrightarrow{A C}$ respectively of a $\qquad$ $\triangle A B C$. The length of the median through $A$ is

a)	$\frac{\sqrt{14}}{2}$	b)	14
c)	7	d)	$\sqrt{14}$

Maths - Vectors and three-dimentional geometry

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(Asked in: JEE Main - 1970)

Correct Option: d

Solution :

Consider A as the origin

$\begin{gathered}\overrightarrow{A D}=\frac{\overrightarrow{A B}+\overrightarrow{A C}}{2}=\hat{1}+2 \hat{\jmath}+3 \hat{k} \\ \text { Length of } A D=\sqrt{1+4+9}=\sqrt{14}\end{gathered}$

Hence, the answer is the option(4).

The area of the region bounded by the line y=2x+1, x-axis and the ordinates x=-1 and x=1 is

a)	2.25	b)	2.5
c)	2	d)	5

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: b

Solution :

$\begin{aligned} & \text { Area }=\frac{1}{2} \times\left(\frac{1}{2} \times 1\right)+\frac{1}{2} \times\left(\frac{3}{2} \times 3\right) \\ & \text { Area }=\frac{1}{4}+\frac{9}{4}=\frac{5}{2}\end{aligned}$

Hence, the answer is the option(2).

The general solution of the differential equation $x^2 d y-2 x y d x=x^4 \cos x d x$ is

a)	$y=x^2 \sin x+c x^2$	b)	$y=x^2 \sin x+c$
c)	$y=\sin x+c x^2$	d)	$y=\cos x+c x^2$

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: a

Solution :

$\begin{aligned} & \frac{d y}{d x}-\frac{2}{x} y=x^2 \cos x \\ & I \cdot F=e^{\int \frac{-2}{x} d x}=\frac{1}{x^2} \\ & \text { G.S } \quad y \cdot \frac{1}{x^2}=\int \frac{1}{x^2}\left(x^2 \cos x\right) d x \\ & \frac{y}{x^2}=\sin x+c \\ & y=x^2 \sin x+c x^2\end{aligned}$

Hence, the answer is the option(1).

The order of the differential equation obtained by eliminating arbitrary constants in the family of curves $c_1 y=\left(c_2+c_3\right) e^{x+c_4}$ is

a)	1	b)	2
c)	3	d)	4

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: a

Solution :

$\begin{aligned} & c_1 y=k_1 e^x \cdot e^{c_4} \\ & y=\frac{k_1 e^{c_4}}{c_1} e^x \\ & y=K e^x \\ & \text { order }=1\end{aligned}$

Hence, the answer is the option(1).

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos x}{1+e^x} d x$

a)	2	b)	0
c)	-2	d)	1

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: d

Solution :

$
\begin{aligned}
I= & \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x \ldots(1) \\
I= & \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^{-x}} d x \\
I= & \int_{-\pi / 2}^{\pi / 2} \frac{e^x(\cos x)}{1+e^x} d x \ldots(2) \\
& \ldots(1)+(2) \ldots \\
2 I= & \int_{-\pi / 2}^{\pi / 2} \cos x d x=\left.\sin x\right|_{-\pi / 2} ^{\pi / 2} \\
2 I= & 2 \\
I= & 1
\end{aligned}
$

Hence, the answer is the option(4).

The area of the region bounded by the curve y²=8x and the line y=2x is

a)	$\frac{16}{3}$ sq.units	b)	$\frac{4}{3}$ sq.units
c)	$\frac{3}{4}$ sq.units	d)	$\frac{8}{3}$ sq.units

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: b

Solution :

$\begin{aligned} & A=\int_0^2(2 \sqrt{2} \sqrt{x}-2 x) d x \\ & =\frac{4 \sqrt{2}}{3} x^{3 / 2}-\left.x^2\right|_0 ^2 \\ & =\left(\frac{42^{1 / 2}}{3} 2^{3 / 2}-4\right) \\ & =\frac{4}{3}\end{aligned}$

Hence, the answer is the option(2).

The value of $\int_0^1 \frac{\log (1+x)}{1+x^2} d x$ is

a)	$\frac{\pi}{2} \log 2$	b)	$\frac{\pi}{4} \log 2$
c)	$\frac{1}{2}$	d)	$\frac{\pi}{8} \log 2$

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: d

Solution :

Let $x=\tan \theta$

$
\begin{aligned}
& d x=\sec ^2 \theta d \theta \\
& I=\int_0^{\frac{\pi}{4}} \frac{\log (1+\tan \theta)}{\sec ^2 \theta} \times \sec ^2 \theta d \theta \\
& I=\int_0^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta \ldots(1) \\
& I=\int_0^{\frac{\pi}{4}} \log \left(1+\tan \left(\frac{\pi}{4}-\theta\right)\right) d \theta \\
& I=\int_0^{\frac{\pi}{4}} \log \left(\frac{2}{1+\tan \theta}\right) d \theta \\
& I=\log 2 \int_0^{\frac{\pi}{4}} 1 d \theta-I \\
& 2 I=\log 2 \cdot \frac{\pi}{4} \\
& I=\frac{\pi}{8} \log 2
\end{aligned}
$

Hence, the answer is the option(4).

$\int_{\text {If }} \int \frac{3 x+1}{(x-1)(x-2)(x-3)} d x=A \log |x-1|+B \log |x-2|+C \log |x-3|+C$ then the values of $\mathrm{A}, \mathrm{B}$ and C are respectively.

a)	$5,-7,-5$	b)	$2,-7,-5$
c)	$5,-7,5$	d)	$2,-7,5$

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: d

Solution :

$
\begin{aligned}
& \frac{3 x+1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \\
& 3 x+1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)
\end{aligned}
$

Taking $x=1,2,3$

$
A=2, B=-7, C=5
$

Hence, the answer is the option(4).

The value of $\int_{-\frac{1}{2}}^{\frac{1}{2}} \cos ^{-1} x d x$ is

a)	$\pi$	b)	$\frac{\pi}{2}$
c)	1	d)	$\frac{\pi^2}{2}$

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: b

Solution :

$
\begin{aligned}
& I=\int_{-1 / 2}^{1 / 2} \cos ^{-1}(x) d x \ldots(1 \\
& I=\int_{-1 / 2}^{1 / 2} \cos ^{-1}(-x) d x \\
& I=\int_{-1 / 2}^{1 / 2} \pi-\cos ^{-1} x d x \ldots \\
& \ldots(1)+(2) \ldots \\
& 2 I=\int_{-1 / 2}^{1 / 2} \pi d x \\
& 2 I=\pi(1) \\
& I=\pi / 2
\end{aligned}
$

Hence, the answer is the option(2).

The value of $\int e^{\sin x} \sin 2 x d x$ is

a)	$2 e^{\sin x}(\sin x-1)+C$	b)	$2 e^{\sin x}(\sin x+1)+C$
c)	$2 e^{\sin x}(\cos x+1)+C$	d)	$2 e^{\sin x}(\cos x-1)+C$

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: a

Solution :

Let $\sin x=t$

$
\begin{aligned}
& \cos x d x=d t \\
& 2 \int e^t t d t=2 e^t[t-1]+c \\
& =2 e^{\sin x}[\sin x-1]+c
\end{aligned}
$

Hence, the answer is the option(1).

The maximum value of $\frac{\log _e x}{x}$, if $x>0$ is

a)	$\frac{1}{e}$	b)	$-\frac{1}{e}$
c)	1	d)	e

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: a

Solution :

$\begin{aligned} & f^{\prime}(x)=\frac{1-\log x}{x^2} \\ & f^{\prime}(x)=0 \\ & x=e \\ & f^{\prime \prime}(e)<0 \\ & \therefore \text { maximum value }=\frac{1}{e}\end{aligned}$

Hence, the answer is the option(1).

The value of $\int \frac{1+x^4}{1+x^6} d x$ is

a)	$\tan ^{-1} x+\tan ^{-1} x^3+C$	b)	$\tan ^{-1} x+\frac{1}{3} \tan ^{-1} x^3+C$
c)	$\tan ^{-1} x-\frac{1}{3} \tan ^{-1} x^3+C$	d)	$\tan ^{-1} x+\frac{1}{3} \tan ^{-1} x^2+C$

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: b

Solution :

$
\begin{aligned}
& \int \frac{\left(x^4-x^2+1\right)+x^2}{\left(x^2+1\right)\left(x^4-x^2+1\right)} d x \\
= & \int \frac{1}{x^2+1} d x+\frac{1}{3} \int \frac{3 x^2}{x^6+1} d x
\end{aligned}
$

Let $x^3=t, 3 x^2 d x=d t$

$
=\tan ^{-1} x+\frac{1}{3} \tan ^{-1}\left(x^3\right)+c
$

Hence, the answer is the option(2).

If $(x e)^y=e^y$, then $\frac{d y}{d x}$ is

a)	$\frac{\log x}{(1+\log x)^2}$	b)	$\frac{1}{(1+\log x)^2}$
c)	$\frac{\log x}{(1+\log x)}$	d)	$\frac{e^x}{x(y-1)}$

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: a

Solution :

$
(x e)^y=e^x
$

Takelogonbothsides

$
\begin{aligned}
& y(1+\log x)=x \\
& y=\frac{x}{1+\log x} \\
& \frac{d y}{d x}=\frac{\log x}{(1+\log x)^2}
\end{aligned}
$

Hence, the answer is the option(1).

If the side of a cube is increased by 5%, then the surface area of a cube is increased by

a)	10%	b)	60%
c)	6%	d)	20%

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: a

Solution :

$\begin{aligned} & \frac{\delta x}{x} \times 100=5 \% \\ & \frac{\delta S}{S} \times 100=\frac{12 x}{6 x^2} \times \delta x \times 100 \\ & =2 \times \frac{\delta x}{x} \times 100 \\ & =2 \times 5=10 \%\end{aligned}$

Hence, the answer is the option(4).

If the curves 2x=y²and 2xy=K intersect perpendicularly, then the value of K² is

a)	2	b)	$2 \sqrt{2}$
c)	4	d)	8

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: d

Solution :

$\begin{aligned} & 2 x=y^2, \quad 2 x y=k \\ & \frac{d y}{d x}=\frac{1}{y}, \frac{d y}{d x}=\frac{-y}{x} \\ & m_1 m_2=-1 \\ & \frac{1}{y} \times \frac{y}{n}=-1 \\ & X=1 \\ & \therefore y^2=2 \\ & 4 x^2 y^2=k^2 \\ & k^2=4(1)(2) \\ & k^2=8\end{aligned}$

Hence, the answer is the option(4).

If $y=2 x^{n+1}+\frac{3}{x^n}$, then $x^2 \frac{d^2 y}{d x^2}$ is

a)	$x \frac{d y}{d x}+y$	b)	$6 n(n+1) y$
c)	$n(n+1) y$	d)	$y$

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: c

Solution :

$\begin{aligned} & \frac{d y}{d x}=2(n+1) x^n-\frac{3 n}{x^{n+1}} \\ & \frac{d^2 y}{d x^2}=2 n(n+1) x^{n-1}+\frac{3 n(n+1)}{x^{n+2}} \\ & \frac{x^2 d^2 y}{d x^2}=n(n+1)\left[2 x^{n+1}+\frac{3}{x^n}\right] \\ & \frac{x^2 d^2 y}{d x^2}=n(n+1) y\end{aligned}$

Hence, the answer is the option(3).

The right-hand and left-hand limit of the function

$f(x)=\left\{\begin{array}{cl}\frac{e^{1 / x}-1}{e^{1 / x}+1}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0 \text { are respectively. }\end{array}\right.$

a)	1 and 1	b)	1 and -1
c)	-1 and -1	d)	-1 and 1

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: b

Solution :

$R H L \quad \operatorname{Lt}_{x \rightarrow 0+} f(x)=\operatorname{Lt}_{h \rightarrow 0} \frac{e^{1 / h}-1}{e^{1 / h}+1}=\operatorname{Lt}_{h \rightarrow 0} \frac{1-e^{-1 / h}}{1+e^{-1 / h}}=1$
$L H L \quad \operatorname{Lt}_{h \rightarrow 0^{-}} f(x)=\operatorname{Lt}_{h \rightarrow 0} \frac{e^{-1 / h}-1}{e^{-1 / h}+1}=\frac{0-1}{0+1}=-1$

Hence, the answer is the option(2).

If $f(x)=\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)$, then $f^{\prime}(\sqrt{3})_{\text {is }}$

a)	$-\frac{1}{2}$	b)	$\frac{1}{2}$
c)	$\frac{1}{\sqrt{3}}$	d)	$-\frac{1}{\sqrt{3}}$

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: b

Solution :

$\begin{aligned} & f(x)=2 \tan ^{-1} x \\ & f^{\prime}(x)=\frac{2}{1+x^2} \\ & f^{\prime}(\sqrt{3})=\frac{2}{4}=\frac{1}{2}\end{aligned}$

Hence, the answer is the option(2).

If $2^{x+2}=2^{x+y}$, then $\frac{d y}{d x}$ is

a)	$2^{y-x}$	b)	$-2^{y-x}$
c)	$2^{x-y}$	d)	$\frac{2^y-1}{2^x-1}$

Maths - Calculus

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(Asked in: JEE Main - 1970)

Correct Option: b

Solution :

$\begin{aligned} & \frac{d y}{d x}=\frac{2^x\left(1-2^y\right)}{2^y\left(2^x-1\right)} \\ & =\frac{2^x-\left(2^x-2^y\right)}{2^x+2^y-2^y} \\ & =-2^{y-x}\end{aligned}$

Hence, the answer is the option(2).

Physics

$\\ \text{Rain is falling vertically with a speed of} \ 12 ms ^{-1} .$ A woman rides a bicycles with a speed of $12 ms ^{-1}$ in east to west direction. What is the direction in which she should hold her umbrella?$

a)	$30^{\circ}$ towards West$	b)	$45^{\circ}$ townrds West$
c)	$30^{\circ}$ towards East$	d)	$45^{\circ}$ townrds East$

Physics - Motion in a Plane

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Correct Option: b

Solution :

In Fig. v_r represents the velocity of rain and v_b, the velocity of the bicycle, the woman is riding. Both these velocities are with respect to the ground.Since the woman is riding a bicycle, the velocity of rain as experienced by here is the velocity of rain relative to the velocity of the bicycle she is riding. That is v_rb=v_r−v_b This relative velocity vector as shown in Fig. makes an angle θ with the vertical. It is given by

$\tan \theta=\frac{v_{b}}{v_{r}} = \frac{12}{12} = 1 \\ \\ \theta = 45^o$

Therefore, the woman should hold her umbrella at an angle of about 45^o with the vertical towards the west

$\\ \text{A cylindrical wire has a mass} \ (0.3 \pm 0.003) g$. radius $(0.5 \pm 0.005)$ $mm$ and length $(6 \pm 0.06) cm$. The maximum percentage error in the measurement of its density is$

a)	3	b)	4
c)	1	d)	2

Physics - Units and Measurement

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Correct Option: b

Solution :

$\begin{array}{l} \text { Density, } \rho=\frac{M}{V}=\frac{M}{\pi r^{3} L} \\ \\ \Rightarrow \frac{\triangle \rho}{\rho}=\frac{\triangle M}{M}+2 \frac{\triangle r}{r}+\frac{\triangle L}{L} \\ \\ =\frac{0.003}{0.3}+2 \times \frac{0.005}{0.5}+\frac{0.06}{6} \\ \\ =0.01+0.02+0.01=0.04 \\ \\ \text { percentage error }=\frac{\triangle \rho}{\rho} \times 100=0.04 \times 100=4 \% \end{array}$

At a metre station, a girl walks up a stationary escalator in 20 sec. If she remains stationary on the escalator, then the escalater take her to walk up in 30 sec. The time taken by her to walk up on the moving escalator will be

a)	12 sec	b)	10 sec
c)	25 sec	d)	60 sec

Physics -

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Correct Option: a

Solution :

$\\ \text{We have to find net velocity with respect to the Earth that will be equal to velocity of the girl plus velocity of escalator. } \\ \text{Let dispalcement is} \ L$, then Velocity of girl $v_{g}=\frac{L}{t_{1}}$ \\ Velocity of escalator $v_{e}=\frac{L}{t_{2}}$ \\ \\ Net velocity of the girl $=v_{g}+v_{e}=\frac{L}{t_{1}}+\frac{L}{t_{2}}$ \\ If $t$ is total time taken is covering distance $L,$ then $\frac{L}{t}=\frac{L}{t_{1}}+\frac{L}{t_{2}} \\ \\ \Rightarrow t=\frac{t_{1} t_{2}}{t_{1}+t_{2}}$$ Putting the value -

$\Rightarrow t=\frac{20 \times 30}{20+30}$ = 12 \ sec$

$\\ \text{A thin uniform rectangular plate of mass} \ 2 kg$ is placed in $X - Y$ plane as shown in figure. The moment of inertia about $x$ -axis is $I_{x}=0.2 kg m ^{2}$ and the moment of inertia about $y$ -axis is $I _{y}=0.3 kg m ^{2}$. The radius of gyration of the plate about the axis passing through $O$ and perpendicular to the plane of the plate is$

a)	38.7 cm	b)	31.6 cm
c)	50 cm	d)	5 cm

Physics - System of particles and rigid body

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Correct Option: c

Solution :

Here, I_xx = 0.2 kg-m² and I_yy = 0.3 kg-m²

As we know -

I_zz = I_xx+I_yy = 0.2 kg-m² + 0.3 kg-m² = 0.5 kg-m²

Now,

mk² = I_zz

where k = Radius of gyration

So,

k² = 0.5/2 = 0.25

k = 0.5 m = 50 cm

$\\ \text{One end of a string of length} \ ^{\prime} l^{ \prime}$ is connected to a particle of mass ^{\prime} $m ^{\prime}$ and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed " $v$ ", the net force on the particle (directed towards the centre) is : (T is the tension in the string)$

a)	$T+\frac{m v^{2}}{l}$$	b)	$0$
c)	$T$$	d)	$\quad T-\frac{m v^{2}}{I}$$

Physics - Work Energy and Power

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Correct Option: c

Solution :

When a particle connected to a string revolves in a circular path around a center, the centripetal force is provided by the tension produced in the string. Hence, in the given case, the net force on the particle is the tension T, i.e.,

$F=T=\frac{m v^{2}}{l}$

So we can say -

Net force on the particle in uniform circualar motion is centripetal force, which is-provided by the tension in string.

Option B

A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time 't' is proportional to

a)	$t^{3 / 2}$	b)	$t^{2}$$
c)	$t^{1 / 2}$	d)	$t$

Physics - Work Energy and Power

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Correct Option: d

Solution :

We know that,

$\\ Power \ delivered, \ $P=F v$$

Now,

$\\ \text{From} \ v=u+a t$ \\ \\ $v=0+a t=a t$ \\ \\ As power, $P=F \times v \quad \\ \\ \therefore P=(m a) \times a t=m a^{2} t$ \\ \\ As $m$ and a are constants, therefor, $P \propto t$$

A wheel starting from rest gains an angular velocity of 10 rad/s after uniformly accelerated for 5 sec. The total angle through which it has turned is

a)	$25 \pi \ rad$	b)	$50 \ \pi\ rad \ about\ a\ vertical \ axis$
c)	$25 rad$	d)	$100 rad$

Physics - System of particles and rigid body

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Correct Option: c

Solution :

$\\ \omega_f = \omega_i + \alpha.t \\ \\ 10 = 0 + \alpha \times 5 \\ \alpha = 2 \ rad/s^2$

Now,

$\\ \theta = \omega_i . t + \frac{1}{2}\alpha.t^2 \\ \\ \theta = 0 + \frac{1}{2} .2 \times 5^2 \\ \theta = 25 \ rad$

$\\ \text {Iceberg floats in water with part of it submerged. What is the fraction of the volume of iceberg} \\ \text {submerged if the density of ice is} \ \rho_{i}=0.917 g cm ^{-3} ?$$

a)	$0.458$	b)	$0$
c)	$0.917$	d)	$1$

Physics - Mechanical properties of fluids

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Correct Option: c

Solution :

Let the volume of the iceberg be V. The weight of the iceberg is ρ_iVg. If x is the fraction submerged, then the volume of water displaced is xV. The buoyant force is ρ_wxVg where ρ_w is the density of water.

$\begin{array}{l} \rho_{i} V{g}=\rho_{w} x V g \\ x=\rho_{i} / \rho_{w}=0.917 \end{array}$

The value of acceleration due to gravity at a height of 10 km from the surface of earth is x. At what depth inside the earth is the value of the acceleration due to gravity has the same value x ?

a)	10 km	b)	15 km
c)	5 km	d)	20 km

Physics - Gravitation

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Correct Option: d

Solution :

According to the question -

$g_{d}=g_o\left(1-\frac{ d }{ R }\right) = g_h = g_{0}\left(1-\frac{2 h}{R}\right)$

So,

$g_o\left(1-\frac{ d }{ R }\right) = g_{0}\left(1-\frac{2 h}{R}\right)$

So,

$d = 2h$

h = 10 km (given)

so,

d = 20 km

Young"s modulus of a perfect rigid body is

a)	infinity	b)	between zero and unity
c)	rero	d)	unity

Physics - Mechanical properties of solids

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Correct Option: a

Solution :

$Y = \frac{Stress}{Strain}$

For perfectly rigid body, strain is zero. SO the Young's modulus (Y) tends to infinity.

A certain amount of heat energy is supplied to a monoatomic ideal gas which expands at constant pressure. What fraction of the heat energy is converted into work?

a)	$\frac{2}{5}$$	b)	$\frac{5}{7}$$
c)	$1$	d)	$\frac{2}{3}$$

Physics -

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Correct Option: a

Solution :

$\\ \text{At constant pressure,}\\ $W =P \Delta V=n R \Delta T$$

Here, n =1 and we know that C_p - C_v = R

So,

$\therefore \Delta W=\left(C_{p}-C_{v}\right) \Delta T$ \\ Fraction of heat converted into work $\frac{\Delta W}{\Delta Q}=\frac{\left(C_{p}-C_{v}\right) \Delta T}{C_{p} \Delta T}\ \\ \\ =1-\frac{1}{\gamma}=\left(1-\frac{3}{5}\right)=\frac{2}{5}$$

A sphere, a cube and a thin circular plate all of same material and same mass initially heated to same high temperature are allowed to cool down under similar conditions. Then the

a)	plate will cool the fastest and sphere the slowest.	b)	cube will cool the fastest and plate the slowest.
c)	plate will cool the fastest and cube the slowest.	d)	sphere will cool the fastest and cube the slowest.

Physics - Mechanical properties of solids

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Correct Option: a

Solution :

$\text { Rate of cooling } = \frac{\Delta \theta}{t}=\frac{A \varepsilon \sigma\left(T^{4}-T_{0}^{4}\right)}{m c} \Rightarrow \frac{\Delta \theta}{t} \propto A$

The circular plate will cool fastest and the sphere will cool the slowest. The rate of heat transfer is directly proportional to the surface area. And the circular plate has a greater surface area. Hence, circular plate will cool faster and sphere slowest(Minimum surface area)

In an adiabatic expansion of an ideal gas the product of pressure and volume

a)	Remains constant	b)	At first increases and then decreases
c)	Decreases	d)	Increases

Physics - Thermodynamics

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Correct Option: c

Solution :

In an adiabatic expansion, internal decreases and hence temperature decreases. Becasue in adiabatic expansion, no heat is loss and since it is expanding means it is expanding on the expense of Internal energy.

For an ideal gas -

$\\ \text{from equation of state of ideal gas} \ PV = nRT$ \\ $\Rightarrow$ the product of $P$ and $V$ decreases$

A tray of mass 12 kg is supported by two identical springs as shown in figure. When the tray is pressed down slightly and then released, it executes SHM with a time period of 1.5 s. The spring constant of each spring is

a)	$105 Nm ^{-1}$	b)	$\infty$
c)	$50 Nm ^{-1}$	d)	$0$

Physics - Oscillations

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Correct Option: a

Solution :

Spring constant of both the springs be 'k' (Identical spring)

Let the combine spring constant be K.

K = k + k = 2k

$\begin{array}{l} T=2 \pi \sqrt{\frac{M}{K}} \\ T=2 \pi \sqrt{\frac{M}{2 k}} \\ M=12 kg \\ 1.5=2 \pi \sqrt{\frac{12}{2 k}} \end{array}$

On solving we get, k = 105.26 N/m

So, Option (1) is correct

A train whistling at constant frequency 'n' is moving towards a station at a constant speed V. The train goes past a stationary observer on the station. The frequency 'n' of the sound as heard by the observer is plotted as a function of time 't'. Identify the correct curve

a)		b)
c)		d)

Physics - Waves

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Correct Option: b

Solution :

Whistling train is the source of sound, v_s= V

Before crossing a stationary observer on station, frequency heard is

$Apparent \ frequency =\frac{v n}{\left(v-v_{s}\right)}=\frac{v n}{v-V}$

The apparent frequency is not a function of time and constant w.r.t. time, also apparent frequency > n.

Here , $v$ is velocity of sound in air and n is actual frequency of whistle.

After crossing the stationary observer, frequency heard is -

$Apparent \ frequency =\frac{v n}{\left(v-(-v_{s}\right))}=\frac{v n}{v+V}$

The apparent frequency is also not a function of time and constant w.r.t. time, also apparent frequency < n.

Therefore, the expected curve is (B).

$\\ \text{An infinitely long thin straight wire has uniform charge density of} \ \frac{1}{4} \times 10^{-2} cm ^{-1}$ What is the magnitude of electric field at a distance $20 cm$ from the axis of the wire?$

a)	$2.25 \times 10^{8} NC ^{-1} \\$	b)	$9 \times 10^{8} NC ^{-1} \\$
c)	$1.12 \times 10^{8} NC ^{-1} \\$	d)	$4.5 \times 10^{8} NC ^{-1}$

Physics - Electric charges and fields

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Correct Option: a

Solution :

We have to find magnitude of electric frld at point P

For an infinitely long wire, electyric field at a distance 'r' away from its axis -

$E = \frac{2K \lambda}{r}$

$E = \frac{2 \times 9 \times 10^9 \times \frac{1}{4} \times 10^{-2}}{0.2} = 2.25 \times 10^8 \ N C^{-1}$

A point charge "q" is placed at the comer of a cube of side 'a' as shown in the figure. What is the electric flux through the face ABCD?

a)	$\frac{q}{6 \varepsilon_{0}}$$	b)	$\frac{q}{72 \varepsilon_{0}}$$
c)	$0$	d)	$\frac{q}{24 \varepsilon_{0}}$$

Physics - Electric charges and fields

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Correct Option: c

Solution :

We can see that the side ABCD is parallel to the electric field lines produced by the charge 'q'. Also we know that the area vector of ABCD is perpendicular to it. SO the angle between electric field and the area vector is 90^o. So the electric flux is given by -

$\phi = \oint \vec{E}. \vec{dA} = \oint E.dA.Cos 90^o = 0$

The electric field lines on the left have twice the separation on those on the right as shown in figure. If the magnitude of the field at A is $40 Vm ^{-1}$ what is the force on $20 \mu C$ charge kept at B ?

a)	$16 \times 10^{-4} Vm ^{-1} \\$	b)	$1 \times 10^{-4} Vm ^{-1} \\$
c)	$4 \times 10^{-4} Vm ^{-1} \\$	d)	$8 \times 10^{-4} Vm ^{-1}$

Physics - Electric charges and fields

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Correct Option: c

Solution :

Since separation between electric field lines at point B is twice as that at the point A, electric field at point B,

$E_{B}=40 \times \frac{1}{2}=20 NC ^{-1}$

Now force on charge at B =

F = q.E = 20 x 10^-6 x 20 = $4 \times 10^{-4} Vm ^{-1} \\$

Figure shows three points A, B and C in a region of uniform electric ficld $\overrightarrow{ E }$ . The line AB is perpendicular and BC is parallel to the field lines. Then which of the following holds good ? $\left( V _{ A }, V _{ B }\right. and \ $V_{C}$ represent the electric potential at points A, B and C respectively)

$V_{A}=V_{B}<V_{C} \\$

$V_{A}>V_{B}=V_{C} \\$

$V_{A}=V_{B}=V_{C} \\$

$V_{A}=V_{B}>V_{C}$

Physics - Electrostatic potential and capacitance

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Correct Option: d

Solution :

Electric lines of force flow from higher potential to lower potential so,

$V_{A}=V_{B}>V_{C}$

Here, A and B is at same position so they have same potential and C is ahead of B in the direction of electric field. So, $V_{A}=V_{B}>V_{C}$

A dipole of dipole moment "P" and moment of inertia I is placed in a uniform electric field $\overrightarrow{ E }$ . If it is displaced slightly from its stable equilibrium position, the period of oscillation of dipole is

a)	$\frac{1}{2 \pi} \sqrt{\frac{P E}{I}}$$	b)	$\pi \sqrt{\frac{I}{P E}}$$
c)	$\sqrt{\frac{P E}{I}}$$	d)	$2 \pi \sqrt{\frac{I}{ PE }}$$

Physics - Electric charges and fields

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Correct Option: d

Solution :

$\begin{aligned} \tau &=\vec{P} \times \vec{E} \\\ &= PE \sin \theta \end{aligned}$\\ \\ For small angle $\theta$ \\ $\tau= PE \theta$ \\ \\$

$\\ Also, \\ \tau = I\alpha \\ \\ \alpha=\left(\frac{PE}{I}\right) \theta$$

Comparing with angular SHM

$w^{2}=\frac{P E}{I}$

So,

Time period for the dipole oscillations is -

$T=\frac{2 \pi}{\sqrt{\frac{p E}{I}}}$

$T =$ $2 \pi \sqrt{\frac{I}{ PE }}$$

The difference between equivalent capacitances of two identical capacitors connected in parallel to that in series is $6 \mu \mathrm{F}$ . The value of capacitance of each capacitor is

a)	$$4 \mu \mathrm{F}$^{}$	b)	$6 \mu \mathrm{F}$
c)	$2 \mu \mathrm{F}$	d)	$3\mu \mathrm{F}$

Physics - Electrostatic potential and capacitance

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Correct Option: a

Solution :

Let the value of capacitance of each capacitor=C

and equivalent capacitances of two identical capacitors connected in parallel =2C

and equivalent capacitances of two identical capacitors connected in series= $\frac{C}{2}$

According to the question

$2C-\frac{C}{2}=6 \mu \ F\\ \frac{3C}{2}=6 \mu \ F\\ \\ C=4 \mu \ F\\$

A metal rod of length $10 \mathrm{cm}$ and a rectangular cross-section of $1 \mathrm{cm} \times \frac{1}{2} \mathrm{cm}$$ is connected to a battery across opposite faces. The resistance will be

a)	maximum when the battery is connected across $10 \mathrm{cm} \times 1 \mathrm{cm}$ faces.	b)	same irrespective of the three faces.
c)	maximum when the battery is connected across $1 \mathrm{cm} \times \frac{1}{2} \mathrm{cm}$$ faces.	d)	maximum when the battery is connected across $10 \mathrm{cm} \times \frac{1}{2} \mathrm{cm}$$ faces.

Physics - Current Electricity

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Correct Option:

Solution :

The resistance is given as $R=\rho \times \frac{l}{A}$$

i.e $R \ \ \alpha \ \ \frac{l}{A}$

For maximum resistance A should be minimum and $l$ should be maximum

So for option C

length $l=10 \mathrm{cm}$$ is maximum
a rectangular cross-section $A=1 \mathrm{cm} \times \frac{1}{2} \mathrm{cm}$ is minimum

So R will be maximum in this case.

A car has a fresh storage battery of e.m.f. $12 \mathrm{V}$ and internal resistance $2 \times 10^{-2} \Omega .$ If the starter motor draws a current of 80 A. Then the terminal voltage when the starter is on is

a)	10.4 V	b)	9.3 V
c)	12 V	d)	8.4 V

Physics - Current Electricity

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Correct Option: a

Solution :

Let the terminal voltage of the battery when the starter is on is V

then $V=\varepsilon-I r=12-80 \times\left(2.0 \times 10^{-2}\right)=10. 4 \mathrm{V}$

When a soap bubble is charged

a)	The radius remains the same	b)	Its radius may increase or decrease.
c)	Its radius increases	d)	Its radius decreases

Physics - Mechanical properties of fluids

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Correct Option: c

Solution :

When a soap bubble is charged then Charge will be distributed uniformly on the surface of a bubble. And these charges will try to repel each other due to the electrostatic force.

So the bubble will expand.

I.e Its radius increases

(This will happen to both positive and negatively charged bubbles.)

A hot filament liberates an electron with zero initial velocity. The anode potential is $1200 \mathrm{V}$ . The speed of the electron when it strikes the anode is

a)	$2.1 \times 10^{7} \mathrm{ms}^{-1}$	b)	$2.5 \times 10^{8} \mathrm{ms}^{-1}$
c)	$1.5 \times 10^{5} \mathrm{ms}^{-1}$	d)	$2.5 \times 10^{6} \mathrm{ms}^{-1}$

Physics - Atoms

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Correct Option: a

Solution :

Let the speed of the electron when it strikes the anode is V

The gain in KE= loss in PE

$\Delta K=-\Delta U\\ \frac{1}{2}m_e(V^2-0)=Vq_e\\ V_e=\sqrt {\frac{2Vq_e}{m_e}}=\sqrt {\frac{2*1200*1.6*10^{-19}}{9*10^{-31}}}=2.1*10^{7} \ m/s$

Each resistance in the given cubical network has a resistance of $1 \ ohm$ ? and equivalent resistance between A and B is

a)	$\frac{5}{12} \Omega$	b)	$\frac{12}{5} \Omega$
c)	$\frac{5}{6} \Omega$	d)	$\frac{6}{5} \Omega$

Physics - Current Electricity

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Correct Option: c

Solution :

Equivalent resistance across the vertices of body diagonal of a cube is $\frac{5R}{6}$

For R= $1 \Omega$

we get

$R_{AB}=\frac{5}{6} \Omega$

A potentiometer has a uniform wire of length $$5 \mathrm{m}$ .A battery of $\mathrm{emf} 10 \mathrm{V}$ and negligible internal resistance is connected between its ends. A secondary cell connected to the circuit gives balancing length at $200 \mathrm{cm}$ . The emf of the secondary cell is

a)	2 V	b)	8 V
c)	4 V	d)	6 V

Physics - Current Electricity

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Correct Option: c

Solution :

In the potentiometer, the potential of wire AB (V) is proportional to the length AB(L).

i.e $V \ \alpha \ L$

As given A battery of $\mathrm{emf} 10 \mathrm{V}$ and negligible internal resistance is connected between its ends of a potentiometer.

So the potential of the whole wire of length 500 cm is 10 V.

Let $V_1=10 \ V$

Now A secondary cell connected to the circuit gives balancing length at $200 \mathrm{cm}$ .

So let The emf of the secondary cell is= $V_2$

As $\frac{V_1}{V_2}=\frac{500}{200}=\frac{5}{2}$

So $V_2=\frac{V_1*2}{5}=\frac{10*2}{5}=4 \ V$

The colour code for a carbon resistor
of resistance $$0.28 \mathrm{k} \Omega \pm 10 \%$ \ is$

a)	Red, Grey, Silver, Silver	b)	Red, Green, Silver
c)	Red, Grey, Brown, Silver	d)	Red, Green, Brown, Silver

Physics - Current Electricity

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Correct Option:

Solution :

Colour coding of Resistance -

Colour coding of Resistance

The carbon resistance has normally four coloured rings bands $A,B,C$ and $D$ .

Colour bands A and B Indicates the decimal figure.

Colour band C Indicates the decimal multiplier.

Colour band D Indicates the tolerance in per cent about the indicated value. Tolerance represents the percentage accuracy of the indicated value.

To remember the sequence of colour code

B B ROY Great Britain Very Good Wife

Tolerance of Gold is %

Tolerance of Silver is $\pm 10$ %

Tolerance if no colour $\pm20$ %

R= $$0.28 \mathrm{k} \Omega \pm 10 \%$=280\Omega \pm 10 \%$

Here A=2 So colour is Red

and B=8 so the colour is Grey

and C= $10^1$ ,so the colour is Brown

and D=10 so the colour is siver

as Tolerance of Silver is $\pm 10$ %

So the

The colour code for a carbon resistor of given resistance is Red, Grey, Brown, Silver

The magnetic field at the origin due to a current element $i\vec{dl}$ placed at a point With vector position $\vec{r}$$ is

a)	$\left(\frac{\mu_{0} i}{4 \pi}\right)\left(\frac{d \vec{l} \times \vec{r}}{r^{2}}\right)$	b)	$\left(\frac{\mu_{0} i}{4 \pi}\right)\left(\frac{ \vec{r} \times d \vec{l}}{r^{2}}\right)$
c)	$\left(\frac{\mu_{0} i}{4 \pi}\right)\left(\frac{d \vec{l} \times \vec{r}}{r^{3}}\right)$	d)	$\left(\frac{\mu_{0} i}{4 \pi}\right)\left(\frac{ \vec{r} \times d \vec{l}}{r^{3}}\right)$

Physics - Moving charges and magnetism

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Correct Option: d

Solution :

According to the Biot-Savart Law -

The magnetic field at a point P(at the position $\vec{r}$$ ), due to a current element placed at the origin , is given by:

$d \vec{B}=\frac{\mu_{0}}{4 \pi} i \frac{d \vec{l} \times \vec{r}}{r^{3}}$

So the magnetic field at the origin due to a current element $i\vec{dl}$ placed at a point With vector position $\vec{r}$$ is

$d \vec{B}=\frac{\mu_{0}}{4 \pi} i \frac{\vec{r} \times d \vec{l} }{r^{3}}$

I-V characteristic of a copper wire of length L and area of cross-section A is shown in the figure. The slope of the curve becomes

a)	Less if the area of the wire is increased.	b)	Less if the length of the wire is increased.
c)	More if the experiment is performed at a higher temperature.	d)	More if a wire of steel of the same dimension is used.

Physics - Semiconductor Electronics

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Correct Option: b

Solution :

From ohms law we have $V=IR$

and $R=\rho\frac{l}{A}$

Here Slope of graph is $\frac{I}{V}$

Slope of graph = $\frac{I}{V}=\frac{1}{R}=\frac{A}{\rho*l}$

So with the increase in the length of the wire. The slope of the curve will decrease.
So The slope of the curve becomes less if the length of the wire is increased.

In the given figure, the magnetic field at "O".

$\frac{3}{8} \frac{\mu_{0} I}{r}+\frac{\mu_{0} I}{4 \pi r}$

$\frac{3}{8} \frac{\mu_{0} I}{r}-\frac{\mu_{0} I}{4 \pi r}$

Physics - Moving charges and magnetism

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Correct Option:

Solution :

Magnetic field due to three quarter Semicircular Current-Carrying arc at the center is

$B_0=\frac{\mu_{o}}{4\pi}\:\frac{(2\pi-\frac{\pi}{2})i}{r}=\frac{3\mu_{o}i}{8r}$

And

Magnetic field due to a current-carrying wire at a point P which lies at a perpendicular distance r from the wire, as shown, is given as:

$B=\frac{\mu_{0}}{4 \pi} \cdot \frac{i}{r}\left(\sin \phi_{1}+\sin \phi_{2}\right)$

Considering the special case

When the linear Current-Carrying wire is of infinite length and the point P lies near one of its ends as shown in below figure

then

$B=\frac{\mu_{0}}{4 \pi} \frac{i}{r}\left[\sin 90^{\circ}+\sin 0^{\circ}\right]=\frac{\mu_{0}}{4 \pi} \frac{i}{r}$

So for the below figure

$B_0=\frac{3}{8} \frac{\mu_{0} I}{r}+\frac{\mu_{0} I}{4 \pi r}$

A paramagnetic sample shows a net magnetization of $8 \mathrm{Am}^{-1}$ when placed in an external magnetic field of $0.6 \mathrm{T}$$
at a temperature of $4 \mathrm{K}$ . When the same sample is placed in an external magnetic field of $0.2 \mathrm{T}$ at a temperature of $16 \mathrm{K}$ . The magnetization will be

a)	$6 \ Am^{-1}$	b)	$2.4 \ Am^{-1}$
c)	$\frac{32}{3} \ Am^{-1}$	d)	$\frac{2}{3} \ Am^{-1}$

Physics -

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Correct Option: d

Solution :

From Curie law of magnetization we have

$I$ or $M=C \frac{B}{T}$$

where

$i.e., I(\text { magnetization }) \propto \frac{B(\text { magnetic field induction })}{t(\text { temperature in kelvin })} \Rightarrow \frac{I_{2}}{I_{1}}=\frac{B_{2}}{B_{1}} \times \frac{t_{1}}{t_{2}}$$

using

$\begin{array}{l} I_{1}=8 \mathrm{Am}^{-1} \\ B_{1}=0.6 \mathrm{T}, \quad t_{1}=4 \mathrm{K} \\ B_{2}=0.2 \mathrm{T}, \quad t_{2}=16 \mathrm{K} \end{array}$

We get

$\frac{0.2}{0.6} \times \frac{4}{16}=\frac{I_{2}}{8}\\ \Rightarrow I_{2}=8 \times \frac{1}{12}=\frac{2}{3} \mathrm{Am}^{-1}$$

A long cylindrical wire of radius R carries a uniform current I flowing through it. The variation of the magnetic field with distance 'r' from the axis of the wire is shown by

a)		b)
c)		d)

Physics - Moving charges and magnetism

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Correct Option: a

Solution :

Inside the solid cylinder :

Current enclosed by loop (i') is lesser then the total current (i) -

Since the current density will remain same. So,

$J=J' \Rightarrow \quad i'=i \times \frac{A^{\prime}}{A}=i\left(\frac{r^{2}}{R^{2}}\right)$

$\text { Hence at inside point } \oint \overrightarrow{B_{\mathrm{in}}} \cdot d \vec{l}=\mu_{0} i'\Rightarrow B=\frac{\mu_{0}}{2 \pi} \cdot \frac{i r}{R^{2}}$

Outside the cylinder -

Then from the top it can be seen as -

with the help of Ampere's circuital law

$\oint \overrightarrow{B_}} d \vec{l}=\mu_{0} i$

$\\ B.2 \pi r = \mu_o i \\ \\ B = \frac{\mu_o i}{2 \pi r}$

On the surface (r=R)

$B_s = \frac{\mu_o i}{2 \pi r}$

From the above equations, we can plot a graph between B and different position 'r' as follows

So the correct option is A

A cyclotron is used to accelerate protons $_{1}^{1}\textrm{H}$ , Deuterons $_{1}^{2}\textrm{H}$ and
a-particles $_{2}^{4}\textrm{He}$ . While exiting under similar conditions, the minimum K.E. is gained by

Physics - Moving charges and magnetism

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Correct Option:

Solution :

In a cyclotron

The radius of the path traveled by the particle can be given as :

$r=\frac{mV}{qB}$

where V is the velocity, q is the charge and B is the magnitude of the magnetic field applied.

Maximum energy gained by the charged particle in Cyclotron is given as

$E_{max}=\left(\frac{q^2B^2}{2m}\right)r_{0}^2$

where $r_0$ =maximum radius of the circular path followed by the positive ion.

As B, $r_0$ , are same for all

So for Minimum KE the term $\frac{q^2}{m}$ should be minimum

For

protons $_{1}^{1}\textrm{H}$ , $\frac{q^2}{m}=\frac{e^2}{m_p}$

Deuterons $_{1}^{2}\textrm{H}$ , $\frac{q^2}{m}=\frac{e^2}{2m_p}$

and
a-particles $_{2}^{4}\textrm{He}$ , $\frac{q^2}{m}=\frac{(2e)^2}{4m_p}=\frac{(e)^2}{m_p}$

So for Deuterons KE gained is minimum

A rod of length 2 m slides with a speed of $5 \mathrm{ms}^{-1}$ on a rectangular conducting frame as shown in the figure. There exists
a uniform magnetic field of 0.04 T perpendicular to the plane of the figure. If the resistance of the rod is $3 \Omega$ .The current through the rod is

Physics - Moving charges and magnetism

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Correct Option:

Solution :

As we learn for the above figure Motional EMF is given by

$\varepsilon = Blv$
where

$B\rightarrow$ magnetic field

$l\rightarrow$ length of conducting

$v\rightarrow$ the velocity of rod perpendicular to a uniform magnetic field.

Now

Induced Current in the conducting rod is given as $I= \frac{\varepsilon }{r}= \frac{Blv}{r}$

where r is the resistance of the rod

So $I= \frac{\varepsilon }{r}= \frac{Blv}{r}=\frac{0.04*2*5}{3}=133 \ mA$

The ratio of the magnetic field at the center of a current-carrying circular coil to its magnetic moment is ‘x’ if the current and the radius both are doubled. The new ratio will become

a)	$\frac{x}{4}$	b)	$\frac{x}{8}$
c)	2x	d)	4x

Physics -

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Correct Option: b

Solution :

The magnetic field at the center of the current-carrying coil is,

$B=\frac{\mu_{o} i}{2 r}$$

The magnetic moment is given by:
$M=i \pi r^{2}$
Given ratio is:
$x=\frac{B}{M}\\ \Rightarrow x=\frac{\mu_{o}}{2 \pi r^{3}}$$

On doubling the current and radius,

$x^{\prime}=\frac{\mu_{0}}{2 \pi \times(2 r)^{3}}$$

$\therefore x^{\prime}=\frac{x}{8}$$

In a permanent magnet at room temperature

a)	Domains are partially aligned	b)	Domains are all perfectly aligned
c)	The magnetic moment of each molecule is zero	d)	The individual molecules have non-zero magnetic moment which are all perfectly aligned

Physics - Magnetism and matter

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Correct Option: b

Solution :

At room temperature, the permanent magnet retains ferromagnetic property for a long period of time.
The individual atoms in a ferromagnetic material possess a dipole moment as in a paramagnetic material.
However, they interact with one another in such a way that they spontaneously align themselves in a common direction over a macroscopic volume called domain. Thus, we can say that in a permanent magnet at room temperature, domains are all perfectly aligned.

The power factor of R-L circuit is $\frac{1}{\sqrt{3}}$$ If the inductive reactance is $2 \Omega .$ The value of resistance is

a)	$0.5 \Omega$	b)	$\frac{1}{\sqrt{2}}\Omega$
c)	$2 \Omega$	d)	$\sqrt{2} \Omega$

Physics - Alternating current

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Correct Option: d

Solution :

Power factor -

$\cos \phi = \frac{R}{Z}$

$R\rightarrow$ resistance

$Z\rightarrow$ impedance

And For RL circuit

$Z=\sqrt{R^2+(X_L)^2}$

where $X_L=inductive \ \ reactance$

$Power \ factor=\cos \phi = \frac{R}{Z}\\ \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{R}{\sqrt{R^2+2^2}}$

on Solving we get

$R=\sqrt{2} \Omega$

The current in a coil of inductance $0.2 \mathrm{H}$ changes from $5 \mathrm{A}$ to $2 \mathrm{A}$ in 0.5 sec . The magnitude of the average induced emf in the coil is

a)	30 V	b)	0.3 V
c)	0.6 V	d)	1.2 V

Physics - Electromagnetic Induction

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Correct Option: d

Solution :

As we know that,

$e = L \frac{di}{dt}$

$e =0.2 \left( \frac{5-2}{0.5} \right ) = \frac{6}{5}= 1.2 V$

Hence, option (4) is correct.

In the given circuit, the resonant frequency is

a)	1592 Hz	b)	15910 Hz
c)	15.92 Hz	d)	159.2 Hz

Physics - Alternating current

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Correct Option: a

Solution :

Resonant frequency is given as

$f = \frac{1}{2 \pi \sqrt{LC}}$

Here L = 0.5 mH = 0.5*10^-3

$C = 20 \mu F = 20 \times 10^{-6} F$

Therefore

$f = \frac{1}{2 \pi \sqrt{20 \times 10^{-6}\times 0.5 \times 10^{-3}}} \approx 1592$

In the given circuit the peak voltages across C, L and R are 30 V, 110 V and 60 V respectively. The rms value of the applied voltage is

a)	70.7 V	b)	141 V
c)	100 V	d)	200 V

Physics - Alternating current

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Correct Option: a

Solution :

Given, V_R=60V, V_L=110 V, V_C=30V

$V_o=\sqrt{V_R^2+(V_L-V_C)^2}=\sqrt{60^2+(110-30)^2}=\sqrt{60^2+80^2}=100V$

$V_{rms}=\frac{V_o}{\sqrt2}=\frac{100}{\sqrt2}=70.7V$

The refracting angle of a prism is A and refractive index of material of prism is $cot \frac{A}{2}$$ . The angle of minimum
deviation is

a)	90^o-A	b)	180^o-2A
c)	180^o-3A	d)	180^o+2A

Physics - Ray optics and optical instruments

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Correct Option: b

Solution :

$\begin{array}{l} \mu=\frac{\sin \left(\frac{A+D_{m}}{2}\right)}{\sin \frac{A}{2}} \\ \Rightarrow \cot \frac{A}{2}=\frac{\sin \left(\frac{A+D_{m}}{2}\right)}{\sin \frac{A}{2}} \\ \frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}=\frac{\sin \left(\frac{A+D_{m}}{2}\right)}{\sin \frac{A}{2}} \\ \sin \left(\frac{\pi}{2}-\frac{A}{2}\right)=\sin \left(\frac{A+D_{m}}{2}\right) \\ \Rightarrow \frac{\pi}{2}-\frac{A}{2}=\frac{A}{2}+\frac{D_{m}}{2} \\ D_{m}=\pi-2 A \\ D_{m}=180^{\circ}-2 A \end{array}$

A light beam of intensity 20 W/cm² is incident normally on a perfectly reflecting surface of sides 25 cm x 15 cm. The momentum imparted to the surface by the light per second is

a)	$5 \times 10^{-5} kgms^{-1}$	b)	$1.2 \times 10^{-5} kgms^{-1}$
c)	$2 \times 10^{-5} kgms^{-1}$	d)	$1 \times 10^{-5} kgms^{-1}^{}$

Physics - Dual Nature of Radiation and Matter

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Correct Option: a

Solution :

$\\ \text{Momentum carried by a photon of energy} \mathrm{E}$ is $\frac{E}{c}$\\ When a photon hits a perfectly reflecting surface, it reflects back in the opposite direction and goes with same energy and hence same momentum. Hence the change in momentum of the photo $=\frac{E}{c}-\left(-\frac{E}{c}\right)=\frac{2 E}{c}$$

$\\ \text{Now the momentum imparted per second}=\frac{\Delta P}{\Delta t}=\frac{2E}{ct}=\frac{2EA}{ctA}=\frac{2IA}{c}\\ \left [ \because I=\frac{E}{At} \right ]$

$\\ \text{So, momentum imparted per second= }\\ \frac{2\times 20 \times 10^4\times 25 \times 15 \times 10^{-4}}{3\times 10^8}=5 \times 10^{-5} kgms^{-1}$

An object approaches a convergent lens from the left of the lens with a uniform speed 5 m/s and stops at the focus, the image

a)	moves away from the lens with a non-uniform acceleration.	b)	moves towards the lens with a non-uniform acceleration.
c)	moves away from the lens with an uniform speed 5 m/s	d)	moves away from the lens with an uniform acceleration.

Physics - Ray optics and optical instruments

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Correct Option: a

Solution :

When an object approaches a convergent lens from the left of the lens with a uniform speed of 5m/s, the image moves away from the lens with a non uniform acceleration.

For example, if f=20m and u=−50m,−45m,−40m and −35m , we get corresponding values of v=33.3m, 36m, 40m and 46.7m using the lens formula. Clearly, image moves away from the lens with a non uniform acceleration.

Two poles are separated by a distance of 3.14 m. The resolving power of human eye is 1 minute of an arc. The maximum distance from which he can identify the two poles distinctly is

a)	188 m	b)	376 m
c)	10.8 km	d)	5.4 km

Physics - Ray optics and optical instruments

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Correct Option: c

Solution :

$\\ \theta=\frac{l}{x},$ where $\theta=1$ minute\\ $\theta=\frac{1}{60^{\circ}}=\left(\frac{\pi}{180} \times \frac{1}{60}\right) \mathrm{rad}$ and $l=3.14 \mathrm{m}$\\ $x=\frac{l}{\theta}=\frac{3.14 m}{\frac{\pi}{180} \times \frac{1}{60}}=10794m\approx 10.8 \mathrm{km}$$

The following figure shows a beam of light converging at point P. When a a concave lens of focal length 16 cm is introduced in the path of the beam at a place shown by dotted line such that OP becomes the axis of the lens, the beam converges at a distance x from the lens. The value of x will be equal to

a)	36 cm	b)	48 cm
c)	12 cm	d)	24 cm

Physics - Ray optics and optical instruments

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Correct Option: b

Solution :

Here , the point P on the right of the lens acts as a virtual object.

$\\ \text{Focal length of concave lens,} f=-16 \mathrm{cm}$ \\ \text{Object distance,} $u=12 \mathrm{cm}$ \\ $\begin{aligned} \frac{1}{v} &=\frac{1}{f}+\frac{1}{u} \\ &=-\frac{1}{16}+\frac{1}{12} \\ &=\frac{-3+4}{48} \\ &=\frac{1}{48} \\ \Rightarrow \quad v &=48 \mathrm{cm} \end{aligned}$$

Three polaroid sheets P₁, P₂ and P₃ are kept parallel to each other such that the angle between pass axes of P₁ and P₂ is 45^o and that between P₂ and P₃ is 45^o. If unpolarised beam of light of intensity 128 Wm^-2 is incident on P_1.What is the intensity of light coming out of P₃ ?

a)	16 Wm^-2	b)	64 Wm^-2
c)	128 Wm^-2	d)	0

Physics - Wave optics

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Correct Option: a

Solution :

$\\ \text{The ray of light passing through polaroid}\ P_{1}$ will have intensity reduced by half.\\ $I_{1}=\frac{I_{0}}{2}$\\ Now, the polaroid $\mathrm{P}_{2}$ is oriented at an angle $45^{\circ}$ with respect to $\mathrm{P}_ 1$.\\ Therefore the intensity is $I_{2}=I_{1} \cos ^{2} 45^o=\frac{I_{0}}{2} \times \frac{1}{2}=\frac{I_{0}}{4}$\\ Now, the polaroid $\mathrm{P}_{3}$ is oriented at an angle $45^{\circ} .$ \\ Therefore, the intensity is $I_{3}=I_{2} \cos ^{2} 45^o=\frac{I_{0}}{4} \cos ^{2} 45^o=\frac{I_{0}}{4} \times \frac{1}{2}=\frac{I_o}{8} \\ \text{And given,}\ I_{o}=128 Wm^{-2}\\ So, I_3=\frac{I_o}{8}=\frac{128}{8}=16 Wm^{-2}$

The following graph represents the variation of photo current with anode potential for a metal surface. Here I₁, I₂ and I₃ represents intensities and $\gamma_{1}, \gamma_{2}$, $\gamma_{3}$ represent frequency for curves 1,2 and 3 respectively, then

a)	$\gamma_{1}=\gamma_{2} \text { and } I_{1}=I_{2}$	b)	$\gamma_{2}=\gamma_{3} \text { and } I_{1}=I_{3}$
c)	$\gamma_{1}=\gamma_{2} \text { and } I_{1} \neq I_{2}$	d)	$\gamma_{1}=\gamma_{3} \text { and } I_{1}=I_{3}$

Physics - Dual Nature of Radiation and Matter

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Correct Option: c

Solution :

From the graph, we can say that the saturation current is same for curves 2 and 3 but different for curve 1. Therefore, intensities of 2 and 3 will be equal as $I_{sat}\propto \ Intensity$ but different from that of 1 i.e. $I_1\neq I_2$ .

As stopping potential is same for curves 1 and 2, so $\gamma_1=\gamma_2$ as $V_{stopping}\propto frequency$

In Young's Double Slit Experiment, the distance between the slits and the screen is 1.2m and the distance between the two slits is 2.4 m. If a thin transparent mica sheet of thickness1 µm and R.I. 1.5 is introduced between one of the interfering beams, the shift in the position of central bright fringe is

a)	0.125 mm	b)	0.25 mm
c)	2 mm	d)	0.5 mm

Physics - Wave optics

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Correct Option: b

Solution :

$\begin{array}{l} (\mu-1) t=\frac{x d}{D} \text { or } \mu=1+\frac{x d}{D t} \\ \\ \text { Or } x=\frac{(\mu-1)tD}{d}=\frac{(1.5-1)\times 1\times 10^{-6}\times 1.2}{2.4 \times 10^{-3}}=0.25 mm \end{array}$

The de-Broglie wavelength associated with electron of hydrogen atom in this ground state is

a)	6.26 A^o	b)	10 A^o
c)	0.3 A^o	d)	3.3 A^o

Physics - Dual Nature of Radiation and Matter

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Correct Option: d

Solution :

$\\ \text{Radius of ground state of hydrogen atom} $=0.53 \mathrm{A}^{\circ}=0.53 \times 10^{-10} \mathrm{m}$\\ \text{According to de Broglie relation}\ $2 \pi r=n \lambda$ \\ \text{For ground state},\ $n=1$ \\ $2 \times 3.14 \times 0.53 \times 10^{-10}=1 \times \lambda$\\ \therefore $\lambda=3.32 \times 10^{-10} \mathrm{m}$ $=3.32 \mathrm{A}^{o}$$

A beam of fast moving alpha particles were directed towards a thin film of gold. The parts A. B and C of the transmitted and reflected beams corresponding to the incident parts A, B and C of the beam are shown in the adjoing diagram. The number of alpba particles in

a)	A' will be minimum and in B' maximum	b)	`C' will be minimum and in B' maximum
c)	B' will be minimum and in C' maximum	d)	A^' will be maximum and in C' minimum

Physics - Atoms

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Correct Option: d

Solution :

Alpha particles are not attracted by nucleus as both are having positive charge.

The beam A is passing undeviated that means the number of alpha particles in A' will be maximum.

And beam is tracing its original path after reflection, so alpha particles in B' will be minimum.

And beam C is also getting deviated that means the alpha particles in C' will be less than that in A'

So according to options, option 4 is correct

The period of revolution of an electron revolving in n^th orbit of H-atom is proportional to

a)	n³	b)	independent of n
c)	n²	d)	$\frac{1}{n}$

Physics - Atoms

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Correct Option: a

Solution :

Time period of revolution of electron in n^th orbit

$\mathrm{T} \propto \frac{\mathrm{n}^{3}}{\mathrm{Z}^{2}}$

Angular momentum of an electron in hydrogen atom is $\frac{3 h}{2 \pi}$ (h is the Planck's constant). The K.E. of the electron is

a)	3.4 eV	b)	6.8 eV
c)	4.35 eV	d)	1.51 eV

Physics - Atoms

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Correct Option: d

Solution :

$\\ \begin{aligned} & \mathrm{L}=\frac{\mathrm{nh}}{2 \pi}=\frac{3 \mathrm{h}}{2 \pi} \\ \therefore \mathrm{n} &=3 \end{aligned}$ \\ In the electronlc third orblt, the energy of electron \\ $\mathrm{E}=13.6 \frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}}$ \\Or, $\mathrm{E}=13.6 \times \frac{1}{9}$ \\ Or, $\mathrm{E}=\frac{13.6}{9}=1.51 \mathrm{eV}$$

A radio-active element has half-life of 15 yeare. What is the fraction that will decay in 30 years ?

a)	0.75	b)	0.85
c)	0.25	d)	0.5

Physics - Nuclei

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Correct Option: a

Solution :

$\\ \text{fraction undecayed} =\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{\frac{t}{T}}$. \\ Here $\mathrm{t}=30$ yrs and $\mathrm{T}=15$ yrs. \\ $\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{\frac{30}{15}}=\left(\frac{1}{2}\right)^{2}=\frac{1}{4}$ \\ Fraction decayed $=1-\frac{1}{4}=\frac{3}{4}=0.75$$

Twe protons are kept at a separation of 10 nm. Let F_n and F_e be the nuclear force and the electromagnetic force between them

a)	$$F_{e}<<F_{n}$	b)	F_e and F_n differ only slightly
c)	$F_{e}=F_{n}$	d)	$F_{e}>>F_{n}$

Physics - Nuclei

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Correct Option: d

Solution :

Two protons exert strong attractive nuclear force and repulsive electrostatic force on each other. But the nuclear forces are short range forces existing in the range of a few fermimeters.

Therefore, at a separation of 10 nm which is much more tha fermimeter, so in this case, the electromagnetic force will be greater than the nuclear force, i.e. $F_{e}>>F_{n}$

During a $\beta^{-}$ decay

a)	A neutron in the nucleus decays emitting an electron.	b)	A proton in the nucleus decays emitting an electron.
c)	an atomic electron is ejected.	d)	an electron which is already present within the nucleus is ejected.

Physics - Nuclei

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Correct Option: a

Solution :

In beta-minus decay, a neutron breaks down to a proton and an electron, and the electron is emitted from the nucleus.

A positive hole in a semiconductor is

a)	absence of free electrons.	b)	an artificially created particle.
c)	an anti-particle of electron.	d)	a vacancy created when an electron leaves a covalent bond.

Physics - Semiconductor Electronics

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Correct Option: d

Solution :

In semiconductors, electrons often gets displaced from one site to another by breaking the covalent bond. When this happens, the initial site of that electron, theoretically, is assumed to acquire a positive charge. This site is called a hole.

A 220 V A.C. supply is connected between points A and B as shown in figure what will be the potential difference V across the capacitor?

a)	0V	b)	$220\sqrt2V$
c)	220 V	d)	110 V

Physics - Semiconductor Electronics

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Correct Option: b

Solution :

$\\ \text{Capacitor will be charged to maximum potential difference as the diode is forward biased}\\ $\therefore V=V_{\max }=V_{r m s} \times \sqrt{2}$ $=220 \sqrt{2}$ volts$

In the following circuit what are P and Q:

a)	P=0, Q=0	b)	P=1, Q=1
c)	P=1, Q=0	d)	P=0, Q=1

Physics - Semiconductor Electronics

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Correct Option: d

Solution :

Two long straight parallel wires are a distance 2d apart. They carry steady equal currents flowing out of the plane of the paper. The variation of magnetic field B along the line xx' is given by

a)		b)
c)		d)

Physics - Moving charges and magnetism

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Correct Option: d

Solution :

$\\ \text{We know, the magnetic field due to a long current carrying wire is given as}\\ B=\frac{\mu_{0} I}{2 \pi r}$ $\Rightarrow B \propto \frac{1}{r}$\\ Magnetic field in between the two conductors will be in opposite direction. Using right hand thumb rule, the magnetic field due to left wire will be in the $\hat{j}$ direction while due to the right wire in $(-\hat{j})$ direction. \\ Magnetic field at a distance $x$ from the left wire is given as: \\ $B_{\text {in between}}=\frac{\mu_{0} I}{2 \pi x} \hat{j}+\frac{\mu_{0} I}{2 \pi(2 d-x)}(-\hat{j})=\frac{\mu_{0} I}{2 \pi}\left[\frac{1}{x}-\frac{1}{(2 d-x)}\right] \hat{j}$$

$\\ At\ x=d, B=0 \\ At \ x> B= along( -\hat{j}) \\ At\ x<d, B= along\ \hat{j} \\ \text{On the left side of the left conductor, magnetic field due to the both}\\ \text{ the conductors will add up and will be along}\ (-\hat{j}) \ \ direction.\\ \text{On the right side of the right conductor, magnetic field due to the both}\\ \text{ the conductors will add up and will be along}\ \hat{j} \ direction.$

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