JEE Main 31 January 2023 Shift 2 question paper with solutions

JEE Main 31 January 2023 Shift 2: all 90 questions in paper order (Physics, Chemistry, Maths) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.

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Physics

Question 1

Physics · Motion in a Straight Line · Single correct

A body is moving with constant speed, in a circle of radius 10 m. The body completes one revolution in 4 s. At the end of 3rd second, the displacement of body (in m) from its starting point is:

  1. 30
  2. 15$\pi$
  3. 5$\pi$
  4. 10$\sqrt{2}$

Answer: (d)

Solution

No Solution

Question 2

Physics · Physical World, Units and Measurements · Single correct

Match List-I with List-II. Choose the correct answer from the options given below:

  1. A-I, B-IV, C-III, D-II
  2. A-III, B-I, C-IV, D-II
  3. A-II, B-III, C-IV, D-I
  4. A-IV, B-II, C-I, D-III

Answer: (b)

Solution

No Solution

Question 3

Physics · Laws of Motion · Single correct

A stone of mass 1 kg is tied to end of a massless string of length 1 m. If the breaking tension of the string is 400 N, then maximum linear velocity, the stone can have without breaking the string, while rotating in horizontal plane, is:

  1. 20 ms$^{-1}$
  2. 40 ms$^{-1}$
  3. 400 ms$^{-1}$
  4. 10 ms$^{-1}$

Answer: (a)

Solution

No Solution

Question 4

Physics · Motion in a Plane · Numerical

Two bodies are projected from ground with same speeds 40 $\mathrm{ms^{-1}}$ at two different angles with respect to horizontal. The bodies were found to have same range. If one of the body was projected at an angle of $60^\circ$, with horizontal then sum of the maximum heights, attained by the two projectiles, is ______ m. (Given $g=10\mathrm{ms^{-2}}$)

Answer: 80

Solution

No Solution

Question 5

Physics · Laws of Motion · Single correct

A body of mass 10 kg is moving with an initial speed of 20 m/s. The body stops after 5 s due to friction between body and the floor. The value of the coefficient of friction is: (Take acceleration due to gravity $g = 10 \, \mathrm{m/s^2}$)

  1. 0.2
  2. 0.3
  3. 0.5
  4. 0.4

Answer: (d)

Solution

No Solution

Question 6

Physics · System of Particles and Rotational Motion · Numerical

A ball is dropped from a height of 20 m. If the coefficient of restitution for the collision between ball and floor is 0.5, after hitting the floor, the ball rebounds to a height of _______ m.

Answer: 5

Solution

No Solution

Question 7

Physics · System of Particles and Rotational Motion · Numerical

Two discs of same mass and different radii are made of different materials such that their thicknesses are 1 cm and 0.5 cm respectively. The densities of materials are in the ratio 3:5. The moment of inertia of these discs respectively about their diameters will be in the ratio of $\frac{x}{6}$. The value of $x$ is.

Answer: 5

Solution

No Solution

Question 8

Physics · Gravitation · Single correct

A body weight $W$, is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be:

  1. $\frac{W}{91}$
  2. $\frac{W}{100}$
  3. $\frac{W}{9}$
  4. $\frac{W}{3}$

Answer: (b)

Solution

No Solution

Question 9

Physics · Mechanical Properties of Solids · Single correct

Under the same load, wire A having length 5.0 m and cross section $2.5 \times 10^{-5} \, \mathrm{m}^2$ stretches uniformly by the same amount as another wire B of length 6.0 m and a cross section of $3.0 \times 10^{-5} \, \mathrm{m}^2$ stretches. The ratio of the Young's modulus of wire A to that of wire B will be:

  1. 1 : 4
  2. 1 : 1
  3. 1 : 10
  4. 1 : 2

Answer: (b)

Solution

No Solution

Question 10

Physics · Thermal Properties of Matter · Numerical

A water heater of power $2000 \, \mathrm{W}$ is used to heat water. The specific heat capacity of water is $4200 \, \mathrm{J} \, \mathrm{kg}^{-1} \, \mathrm{K}^{-1}$. The efficiency of heater is $70\%$. Time required to heat $2 \, \mathrm{kg}$ of water from $10^{\circ}\mathrm{C}$ to $60^{\circ}\mathrm{C}$ is ________ $\mathrm{s}$. (Assume that the specific heat capacity of water remains constant over the temperature range of the water).

Answer: 300

Solution

No Solution

Question 11

Physics · Thermodynamics · Single correct

Heat energy of 735 J is given to a diatomic gas allowing the gas to expand at constant pressure. Each gas molecule rotates around an internal axis but do not oscillate. The increase in the internal energy of the gas will be:

  1. 525 J
  2. 441 J
  3. 572 J
  4. 735 J

Answer: (a)

Solution

No Solution

Question 12

Physics · Thermodynamics · Single correct

A hypothetical gas expands adiabatically such that its volume changes from 08 litres to 27 litres. If the ratio of final pressure of the gas to initial pressure of the gas is $\frac{16}{81}$. Then the ratio of $\frac{C_P}{C_V}$ will be.

  1. $\frac{4}{3}$
  2. $\frac{3}{1}$
  3. $\frac{1}{2}$
  4. $\frac{3}{2}$

Answer: (a)

Solution

No Solution

Question 13

Physics · Mechanical Properties of Solids · Single correct

For a solid rod, the Young's modulus of elasticity is $3.2 \times 10^{11} \, \mathrm{Nm^{-2}}$ and density is $8 \times 10^{3} \, \mathrm{kg \, m^{-3}}$. The velocity of longitudinal wave in the rod will be.

  1. $145.75 \times 10^{3} \, \mathrm{ms^{-1}}$
  2. $3.65 \times 10^{3} \, \mathrm{ms^{-1}}$
  3. $18.96 \times 10^{3} \, \mathrm{ms^{-1}}$
  4. $6.32 \times 10^{3} \, \mathrm{ms^{-1}}$

Answer: (d)

Solution

No Solution

Question 14

Physics · Waves · Numerical

The displacement equations of two interfering waves are given by $$y_1 = 10 \sin \left( \omega t + \frac{\pi}{3} \right) \, \mathrm{cm},$$ $$y_2 = 5 \left[ \sin (\omega t) + \sqrt{3} \cos \omega t \right] \, \mathrm{cm}$$ respectively. The amplitude of the resultant wave is _______ $\mathrm{cm}$.

Answer: 20

Solution

No Solution

Question 15

Physics · Electric Charges and Fields · Single correct

Considering a group of positive charges, which of the following statements is correct?

  1. Net potential of the system cannot be zero at a point but net electric field can be zero at that point.
  2. Net potential of the system at a point can be zero but net electric field can't be zero at that point.
  3. Both the net potential and the net field can be zero at a point.
  4. Both the net potential and the net electric field cannot be zero at a point.

Answer: (a)

Solution

No Solution

Question 16

Physics · Electrostatic Potential and Capacitance · Numerical

Two parallel plate capacitors $C_1$ and $C_2$ each having capacitance of $10 \, \mu \mathrm{F}$ are individually charged by a $100 \, \mathrm{V}$ D.C. source. Capacitor $C_1$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor $C_2$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_1$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be V. (Assuming Dielectric constant = 10)

Answer: 55

Solution

No Solution $$= 2.25 \, Joule$$

Question 17

Physics · Current Electricity · Single correct

The number of turns of the coil of a moving coil galvanometer is increased in order to increase current sensitivity by 50$\%$. The percentage change in voltage sensitivity of the galvanometer will be:

  1. 100$\%$
  2. 50$\%$
  3. 75$\%$
  4. 0$\%$

Answer: (d)

Solution

No Solution

Question 18

Physics · Current Electricity · Single correct

The $H$ amount of thermal energy is developed by a resistor in $10 \, \mathrm{s}$ when a current of $4 \, \mathrm{A}$ is passed through it. If the current is increased to $16 \, \mathrm{A}$, the thermal energy developed by the resistor in $10 \, \mathrm{s}$ will be:

  1. $H$
  2. $16H$
  3. $\frac{H}{4}$
  4. $4H$

Answer: (b)

Solution

No Solution

Question 19

Physics · Current Electricity · Numerical

For the given circuit, in the steady state, $|V_B - V_D|$ = $V$.

Answer: 1

Solution

No Solution

Question 20

Physics · Moving Charges and Magnetism · Single correct

A long conducting wire having a current I flowing through it, is bent into a circular coil of N turns. Then it is bent into a circular coil of n turns. The magnetic field is calculated at the centre of coils in both the cases. The ratio of the magnetic field in first case to that of second case is:

  1. N : n
  2. n^2 : N^2
  3. N^2 : n^2
  4. n : N

Answer: (c)

Solution

No Solution

Question 21

Physics · Alternating Current · Single correct

An alternating voltage source $V = 260 \sin (628t)$ is connected across a pure inductor of $5 \, \mathrm{mH}$. Inductive reactance in the circuit is:

  1. $3.14\Omega$
  2. $6.28\Omega$
  3. $0.5\Omega$
  4. $0.318\Omega$

Answer: (a)

Solution

No Solution

Question 22

Physics · Alternating Current · Numerical

A series LCR circuit consists of $R=80\,\Omega$, $X_L=100\,\Omega$, and $X_C=40\,\Omega$. The input voltage is $2500 \cos(100\, \pi t)$ V. The amplitude of current, in the circuit, is

Answer: 25

Solution

No Solution

Question 23

Physics · Electromagnetic Waves · Single correct

Match List-I with List-II. Choose the correct answer from the option given below:

  1. A-II, B-IV, C-III, D-I
  2. A-IV, B-I, C-II, D-III
  3. A-IV, B-III, C-I, D-II
  4. A-III, B-II, C-I, D-IV

Answer: (c)

Solution

No Solution

Question 24

Physics · Ray Optics and Optical Instruments · Single correct

A microscope is focused on an object at the bottom of a bucket. If liquid with refractive index $\frac{5}{3}$ is poured inside the bucket, then microscope have to be raised by 30 cm to focus the object again. The height of the liquid in the bucket is :

  1. 75 cm
  2. 50 cm
  3. 18 cm
  4. 12 cm

Answer: (a)

Solution

No Solution

Question 25

Physics · Wave Optics · Numerical

Two light waves of wavelengths 800 and 600 nm are used in Young's double slit experiment to obtain interference fringes on a screen placed 7 m away from plane of slits. If the two slits are separated by 0.35 mm, then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be

Answer: 48

Solution

No Solution

Question 26

Physics · Dual Nature of Radiation and Matter · Subjective

If the two metals A and B are exposed to radiation of wavelength 350 nm. The work functions of metals A and B are 4.8 eV and 2.2 eV. Then choose the correct option (1) Metal B will not emit photo-electrons (2) Both metals A and B will emit photo-electrons (3) Both metals A and B will not emit photo-electrons (4) Metal A will not emit photo-electrons

  1. Metal B will not emit photo-electrons
  2. Both metals A and B will emit photo-electrons
  3. Both metals A and B will not emit photo-electrons
  4. Metal A will not emit photo-electrons

Answer: (d)

Solution

No Solution

Question 27

Physics · Atoms · Single correct

The radius of electron's second stationary orbit in Bohr's atom is $R$. The radius of 3rd orbit will be

  1. $\frac{R}{3}$
  2. $2.25R$
  3. $3R$
  4. $9R$

Answer: (b)

Solution

No Solution

Question 28

Physics · Atoms · Numerical

If the binding energy of ground state electron in a hydrogen atom is 13.6 $\mathrm{eV}$, then, the energy required to remove the electron from the second excited state of $\mathrm{Li}^{2+}$ will be: $x \times 10^{-1}$ $\mathrm{eV}$. The value of $x$ is

Answer: 136

Solution

No Solution

Question 29

Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct

Given below are two statements: Statement I: In a typical transistor, all three regions emitter, base and collector have same doping level. Statement II: In a transistor, collector is the thickest and base is the thinnest segment. In the light of the above statements, choose the most appropriate answer from the options given below.

  1. Both Statement I and Statement II are correct
  2. Both Statement I and Statement II are incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Statement I is correct but Statement II is incorrect

Answer: (c)

Solution

No Solution

Question 30

Physics · Communication Systems · Single correct

Given below are two statements Statement I: For transmitting a signal, size of antenna $(l)$ should be comparable to wavelength of signal (at least $l = \frac{\lambda}{4}$ in dimension). Statement II: In amplitude modulation, amplitude of carrier wave remains constant (unchanged). In the light of the above statements, choose the most appropriate answer from the options given below.

  1. Both Statement I and Statement II are correct
  2. Both Statement I and Statement II are incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Statement I is correct but Statement II is incorrect

Answer: (d)

Solution

No Solution

Chemistry

Question 31

Chemistry · Some Basic Concepts of Chemistry · Single correct

When a hydrocarbon A undergoes complete combustion it requires 11 equivalents of oxygen and produces 4 equivalents of water. What is the molecular formula of A?

  1. C_9H_8
  2. C_{11}H_4
  3. C_5H_8
  4. C_{11}H_8

Answer: (a)

Solution

Given the reaction: $$\mathrm{C_xH_y} + \left( x + \frac{y}{4} \right) \mathrm{O_2} \rightarrow x\mathrm{CO_2} + \frac{y}{2} \mathrm{H_2O}$$ From the equation, $\($ $\frac{y}{2}$ = 4 $\)$, therefore $\($ y = 8 $\)$. Substituting $\($ y = 8 $\)$ into the equation: $$x + \frac{8}{4} = 11$$ Therefore, $\($ x = 9 $\)$. Thus, the hydrocarbon will be $\($ $\mathrm{C_9H_8}$ $\)$.

Question 32

Chemistry · Some Basic Concepts of Chemistry · Numerical

Assume carbon burns according to following equation : $$2\mathrm{C}_{(s)} + \mathrm{O}_2{(g)} \rightarrow 2\mathrm{CO}_{(g)}$$ When 12 g carbon is burnt in 48 g of oxygen, the volume of carbon monoxide produced is ______ $\times 10^{-1}$ L at STP [nearest integer] [Given : Assume CO as ideal gas, Mass of C is $12 \, \mathrm{g \, mol^{-1}}$, Mass of O is $16 \, \mathrm{g \, mol^{-1}}$ and molar volume of an ideal gas at STP is $22.7 \, \mathrm{L \, mol^{-1}}$]

Answer: 227

Solution

The reaction is given by $$2\mathrm{C(s)} + \mathrm{O_2(g)} \rightarrow 2\mathrm{CO(g)}$$ with 1 mol of carbon and 1.5 mol of oxygen. The limiting reagent is carbon. One mole of carbon produces one mole of CO. Hence, the volume at STP is $$227 \times 10^{-1} litre$$

Question 33

Chemistry · Structure of Atom · Single correct

Arrange the following orbitals in decreasing order of energy? $(A)$ $n = 3, l = 0, m = 0$ $(B)$ $n = 4, l = 0, m = 0$ $(C)$ $n = 3, l = 1, m = 0$ $(D)$ $n = 3, l = 2, m = 1$ The correct option for the order is:

  1. B > D > C > A
  2. D > B > C > A
  3. A > C > B > D
  4. D > B > A > C

Answer: (b)

Solution

$(A)$ $n = 3$; $l = 0$; $m = 0$; 3s orbital $(B)$ $n = 4$; $l = 0$; $m = 0$; 4s orbital $(C)$ $n = 3$; $l = 1$; $m = 0$; 3p orbital $(D)$ $n = 3$; $l = 2$; $m = 0$; 3d orbital As per Hund's rule energy is given by $(n+l)$ value. If value of $(n+l)$ remains same then energy is given by $n$ only.

Question 34

Chemistry · Classification of Elements and Periodicity in Properties · Single correct

Given below are two statements: one is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$ Assertion $(A)$: The first ionization enthalpy of 3d series elements is more than that of group 2 metals Reason $(R)$: In 3d series of elements successive filling of d-orbitals takes place. In the light of the above statements, choose the correct answer from the options given below:

  1. Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$
  2. Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A)$
  3. $(A)$ is false but $(R)$ is true
  4. $(A)$ is false but $(R)$ is true

Answer: (a)

Solution

From Sc to Mn ionization energy is less than that of Mg.

Question 35

Chemistry · Chemical Bonding and Molecular Structure · Numerical

Amongst the following, the number of species having the linear shape is ____. $\mathrm{XeF_2},\quad \mathrm{I_3^+},\quad \mathrm{C_3O_2},\quad \mathrm{I_3^-},\quad \mathrm{CO_2},\quad \mathrm{SO_2},\quad \mathrm{BeCl},\quad \text{and}\quad \mathrm{BCl_2^-}$

Answer: 5

Solution

Question 36

Chemistry · Thermodynamics · Numerical

Enthalpies of formation of $\mathrm{CCl_4(g)}$, $\mathrm{H_2O(g)}$, $\mathrm{CO_2(g)}$ and $\mathrm{HCl(g)}$ are $-105$, $-242$, $-394$ and $-92 \, \mathrm{kJ \, mol^{-1}}$ respectively. The magnitude of enthalpy of the reaction given below is ______ $\mathrm{kJ \, mol^{-1}}$ (nearest integer) $\mathrm{CCl_4(g) + 2H_2O(g) \rightarrow CO_2(g) + 4HCl(g)}$

Answer: 173

Solution

Given (Q8 (173)) $\Delta_r H = \sum H_P - \sum H_R$ $= [(-394 + 4 \times (-92)) - (-105 + 2 \times (-242))]$ $= -173\,\mathrm{kJ\,mol^{-1}}$

Question 37

Chemistry · Equilibrium · Single correct

Incorrect statement for the use of indicators in acid-base titration is :

  1. Methyl orange may be used for a weak acid vs weak base titration.
  2. Methyl orange is a suitable indicator for a strong acid vs weak base titration
  3. Phenolphthalein is a suitable indicator for a weak acid vs strong base titration
  4. Phenolphthalein may be used for a strong acid vs strong base titration.

Answer: (a)

Solution

Methyl orange may be used for a strong acid vs strong base and strong acid vs weak base titration. Phenolphthalein may be used for a strong acid vs strong base and weak acid vs strong base titration.

Question 38

Chemistry · Equilibrium · Numerical

At 298 K, the solubility of silver chloride in water is $1.434 \times 10^{-3} \, \mathrm{g \, L^{-1}}$. The value of $-\log K_{sp}$ for silver chloride is . (Given mass of Ag is $107.9\,\mathrm{g}\ \mathrm{mol^{-1}}$ and mass of Cl is $35.5\,\mathrm{g}\ \mathrm{mol^{-1}}$)

Answer: 10

Solution

The reaction is given by: $$\mathrm{AgCl(s) \rightarrow Ag^+(aq.) + Cl^-(aq.)}$$ The solubility product is: $$K_{sp} = S^2 = \left( \frac{1.43}{143.4} \times 10^{-3} \right)^2 = 10^{-10}$$ Taking the negative logarithm: $$-\log K_{sp} = 10$$

Question 39

Chemistry · Hydrogen · Single correct

Given below are two statements : Statement I : $\mathrm{H_2O_2}$ is used in the synthesis of Cephalosporin Statement II : $\mathrm{H_2O_2}$ is used for the restoration of aerobic conditions to sewage wastes. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. Both Statement I and Statement II are correct
  2. Statement I is incorrect but Statement II is correct
  3. Statement I is correct but Statement II is incorrect
  4. Both Statement I and Statement II are incorrect

Answer: (a)

Solution

It is used in the synthesis of hydroquinone, tartaric acid and certain food products and pharmaceuticals (cephalosporin) etc. Restoration of aerobic conditions to sewage wastes etc.

Question 40

Chemistry · The s-Block Elements · Single correct

The element playing significant role in neuromuscular function and interneuronal transmission is :

  1. Be
  2. Ca
  3. Li
  4. Mg

Answer: (b)

Solution

Calcium plays important role in neuromuscular function, interneuronal transmission, cell membrane etc.

Question 41

Chemistry · The s-Block Elements · Numerical

The number of alkali metal(s), from Li, K, Cs, Rb having ionization enthalpy greater than 400 \, $\mathrm{kJ \, mol^{-1}}$ and forming stable super oxide is .

Answer: 2

Solution

K, Rb and Cs form stable super oxides but Cs has ionisation enthalpy less than 400 $\mathrm{kJ}$.

Question 42

Chemistry · The p-Block Elements (Group-13 and 14) · Single correct

The Lewis acid character of boron tri halides follows the order:

  1. BBr_3 > BI_3 > BCl_3 > BF_3
  2. BCl_3 > BF_3 > BBr_3 > BI_3
  3. BF_3 > BCl_3 > BBr_3 > BI_3
  4. BI_3 > BBr_3 > BCl_3 > BF_3

Answer: (d)

Solution

Extent of back bonding reduces down the group leading to more Lewis acidic strength. $\mathrm{BF_3} > \mathrm{BCl_3} > \mathrm{BBr_3} > \mathrm{BI_3}$ (extent of back bonding) $(2p-2p) \ (2p-3p) \ (2p-4p) \ (2p-5p)$. $\mathrm{BF_3} < \mathrm{BCl_3} < \mathrm{BBr_3} < \mathrm{BI_3}$ (Lewis acidic nature).

Question 43

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

In Dumas method for the estimation of $N_2$, the sample is heated with copper oxide and the gas evolved is passed over:

  1. Ni
  2. Copper gauze
  3. Pd
  4. Copper oxide

Answer: (b)

Solution

Duma's method. The nitrogen containing organic compound, when heated with $\mathrm{CuO}$ in an atmosphere of $\mathrm{CO_2}$, yields free $\mathrm{N_2}$ in addition to $\mathrm{CO_2}$ and $\mathrm{H_2O}$. $$\mathrm{C_xH_yN_z} + \left(2x + \frac{y}{2}\right) \mathrm{CuO} \rightarrow x \mathrm{CO_2} + \frac{y}{2} \mathrm{H_2O} + \frac{z}{2} \mathrm{N_2} + \left(2x + \frac{y}{2}\right) \mathrm{Cu}$$ Traces of nitrogen oxides formed, if any, are reduced to nitrogen by passing the gaseous mixture over heated copper gauze.

Question 44

Chemistry · Hydrocarbons · Single correct

A hydrocarbon 'X' with formula $C_6H_8$ uses two moles of $H_2$ on catalytic hydrogenation of its one mole. On ozonolysis, 'X' yields two moles of methane dicarbaldehyde. The hydrocarbon 'X' is:

  1. hexa-1, 3, 5-triene
  2. 1-methylcyclopenta-1, 4-diene
  3. cyclohexa-1, 3-diene
  4. cyclohexa-1, 4-diene

Answer: (d)

Solution

The ozonolysis of cyclohexa-1,4-diene ($C_6H_8$) in the presence of $\mathrm{Zn/H_2O}$ results in the formation of 2 moles of $\mathrm{H{-}C{-}CH_2{-}C{-}H}$ with carbonyl groups.

Question 45

Chemistry · Environmental Chemistry · Single correct

The normal rain water is slightly acidic and its pH value is 5.6 because of which one of the following?

  1. $\mathrm{CO_2 + H_2O \rightarrow H_2CO_3}$
  2. $\mathrm{4NO_2 + O_2 + 2H_2O \rightarrow 4HNO_3}$
  3. $\mathrm{2SO_2 + O_2 + 2H_2O \rightarrow 2H_2SO_4}$
  4. $\mathrm{N_2O_5 + H_2O \rightarrow 2HNO_3}$

Answer: (a)

Solution

We are aware that normally rain water has a pH of 5.6 due to the presence of $\mathrm{H^+}$ ions formed by the reactions of rain water with carbon dioxide present in the atmosphere. $$\mathrm{H_2O(l) + CO_2(g) \rightleftharpoons H_2CO_3(aq)}$$ $$\mathrm{H_2CO_3(aq) \rightleftharpoons H^+(aq) + HCO_3^-(aq)}$$

Question 46

Chemistry · Redox Reactions · Numerical

A sample of a metal oxide has formula $\mathrm{M}_{0.83}\mathrm{O}_{1.00}$. The metal $\mathrm{M}$ can exist in two oxidation states $+2$ and $+3$. In the sample of $\mathrm{M}_{0.83}\mathrm{O}_{1.00}$, the percentage of metal ions existing in $+2$ oxidation state is $\%$ (nearest integer)

Answer: 59

Solution

Given the reaction scheme: $M \rightarrow M^{2+} \rightarrow x$ and $M \rightarrow M^{3+} \rightarrow (0.83 - x)$ The equation is $2x + 3(0.83 - x) = 2$ Solving for $x$, we get $x = 0.49$ The percentage of $M^{2+}$ is calculated as $\frac{0.49}{0.83} \times 100$ Therefore, the percentage is $59\%$.

Question 47

Chemistry · Solutions · Single correct

Evaluate the following statements for their correctness. (A) The elevation in boiling point temperature of water will be same for 0.1 M NaCl and 0.1 M urea. (B) Azeotropic mixtures boil without change in their composition (C) Osmosis always takes place from hypertonic to hypotonic solution (D) The density of 32$\%$ $\mathrm{H_2SO_4}$ solution having molarity 4.09 $\mathrm{M}$ is approximately 1.26 $\mathrm{g \, mL^{-1}}$ (E) A negatively charged sol is obtained when $\mathrm{KI}$ solution is added to silver nitrate solution. Choose the correct answer from the options given below :

  1. B, D, and E only
  2. A, B, and D only
  3. A and C only
  4. B and D only

Answer: (d)

Solution

(A) $\Delta T_b \propto i \times c$ (B) Azeotropic mixtures have same composition in both liquid and vapour phase. (C) Osmosis always takes place from hypotonic to hypertonic solution. (D) $$M = \frac{30 \times 10 \times 1.26}{98} \approx 4.09 \, \mathrm{M}$$ (E) When KI solution is added to $\mathrm{AgNO_3}$ solution, positively charged solution results due to adsorption of $\mathrm{Ag^+}$ ions from dispersion medium. $$\frac{\mathrm{AgI}}{\mathrm{Ag^+}}$$ Positively charged

Question 48

Chemistry · Electrochemistry · Numerical

The resistivity of a 0.8 M solution of an electrolyte is $5 \times 10^{-3} \, \Omega \mathrm{cm}$. Its molar conductivity is _______ $\times 10^{4} \, \Omega^{-1} \mathrm{cm}^{2} \mathrm{mol}^{-1}$. (Nearest integer)

Answer: 25

Solution

Given $$\Lambda_m = \frac{\kappa \times 1000}{M}$$ $$\Lambda_m = \frac{1}{\rho} \times \frac{1000}{M}$$ $$\frac{1}{5 \times 10^{-3}} \times \frac{1000}{0.8}$$ Ans. $25 \times 10^4 \, \Omega^{-1} \, \mathrm{cm^{-2} \, mol^{-1}}$

Question 49

Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical

The rate constant for a first order reaction is $20 \, \mathrm{min}^{-1}$. The time required for the initial concentration of the reactant to reduce to its $\frac{1}{32}$ level is ______ $\times 10^{-2} \, \mathrm{min}$. (Nearest integer) (Given: $\ln 10 = 2.303$ $\log 2 = 0.3010$)

Answer: 17

Solution

Given $C = \frac{C_o}{2^n} = \frac{C_o}{32}$. $n = 5$. $t = 5 t_{1/2}$. $$= \frac{5 \times 0.693}{20} = \frac{0.693}{4}$$ $$= 0.17325 \, min = 17.325 \times 10^{-2} \, min.$$

Question 50

Chemistry · Surface Chemistry · Single correct

Match List-I with List-II Choose the correct answer from the options given below:

  1. A – II, B – III, C – I, D – IV
  2. A – III, B – IV, C – I, D - II
  3. A – IV, B – II, C – III, D - I
  4. A – II, B – I, C – IV, D – III

Answer: (d)

Solution

$(A)$ Physisorption $= 20 - 40 \, \mathrm{kJ/mol}$ and Chemisorption $= 80 - 240 \, \mathrm{kJ/mol}$ $(B)$ Physisorption is multi-layered and chemisorption is unimolecular layered. $(C)$ In heterogeneous catalysis, medium and catalyst are in different phases. $(D)$ Chromatography uses adsorption to purify/separate mixtures.

Question 51

Chemistry · General Principles and Processes of Isolation of Elements · Single correct

Which one of the following statements is incorrect?

  1. Boron and Indium can be purified by zone refining method.
  2. Van-Arkel method is used to purify tungsten.
  3. Cast iron is obtained by melting pig iron with scrap iron and coke using hot air blast.
  4. The malleable iron is prepared from cast iron by oxidising impurities in a reverberatory furnace.

Answer: (b)

Solution

Van-Arkel process is used for purification of Ti, Zr, Hf and B.

Question 52

Chemistry · Classification of Elements and Periodicity in Properties · Single correct

Which of the following elements have half-filled f-orbitals in their ground state ? (Given : atomic number Sm = 62; Eu = 63; Tb = 65; Gd = 64, Pm = 61 )

  1. B and D only
  2. A and E only
  3. A and B only
  4. C and D only

Answer: (a)

Solution

1. $^{62}\mathrm{Sm} : 4f^66s^2$ 2. $^{64}\mathrm{Gd} : 4f^75d^16s^2$ 3. $^{63}\mathrm{Eu} : 4f^76s^2$ 4. $^{65}\mathrm{Tb} : 4f^96s^2$ 5. $^{61}\mathrm{Pm} : 4f^56s^2$

Question 53

Chemistry · Co-ordination Compounds · Numerical

If the CFSE of $[\mathrm{Ti}(\mathrm{H}_2\mathrm{O})_6]^{3+}$ is $-96.0 \, \mathrm{kJ/mol}$, this complex will absorb maximum at wavelength ___ nm. (nearest integer) Assume Planck's constant $(h) = 6.4 \times 10^{-34} \, \mathrm{Js}$, Speed of light $(c) = 3.0 \times 10^8 \, \mathrm{m/s}$ and Avogadro's constant $(N_A) = 6 \times 10^{23} \, / \mathrm{mol}$.

Answer: 480

Solution

Given $[\mathrm{Ti(H_2O)_6}]^{3+}$ $\mathrm{Ti}^{3+}: 3d^1$ $\mathrm{CFSE}=-0.4\Delta_0$ $=-\dfrac{96\times10^3}{N_0}\,\mathrm{J}$ $\Delta_0=\dfrac{96\times10^3}{0.4\times6\times10^{23}}$ $\Rightarrow \dfrac{hc}{\lambda}=\dfrac{96\times10^3}{0.4\times6\times10^{23}}$ $\lambda=\dfrac{0.4\times6\times10^{23}\times6.4\times10^{-34}\times3\times10^8}{96\times10^3}$ $=0.48\times10^{-6}\,\mathrm{m}$ $=480\times10^{-9}\,\mathrm{m}$ $=480\,\mathrm{nm}$

Question 54

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

In the following halogenated organic compounds the one with maximum number of chlorine atoms in its structure is :

  1. Chloral
  2. Gammaxene
  3. Chloropicrin
  4. Freon -12

Answer: (b)

Solution

The question provides four chemical compounds: (1) Chloral, (2) Gammaxene, (3) Chloropicrin, and (4) Freon-12. Each compound is represented by its chemical structure.

Question 55

Chemistry · Amines · Single correct

Cyclohexylamine when treated with nitrous acid yields (P). On treating (P) with PCC results in (Q). When (Q) is heated with dil. NaOH we get ($R$) The final product ($R$) is :

Answer: (b)

Solution

The reaction starts with cyclohexylamine, which is treated with $\mathrm{HNO_2}$ to form cyclohexanol. This intermediate is then oxidized using PCC to form cyclohexanone. Finally, cyclohexanone undergoes an aldol condensation with itself in the presence of concentrated $\mathrm{NaOH}$ and heat to form the final product, which is a bicyclic compound with a ketone group.

Question 56

Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical

The number of molecules which gives haloform test among the following molecules is $\underline{\hspace{1cm}}$.

Answer: 3

Solution

Molecules having gives positive haloform test

Question 57

Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct

An organic compound $[A](C_4H_{11}N)$, shows optical activity and gives $N_2$ gas on treatment with $HNO_2$. The compound $[A]$ reacts with $PhSO_2Cl$ producing a compound which is soluble in $KOH$. The structure of $A$ is:

Answer: (d)

Solution

$\mathrm{C_4H_{11}N}$ releases $\mathrm{N_2}$ with $\mathrm{HNO_2}$ i.e. it is primary amine. After reacting with Hinsberg reagent it forms a compound which is soluble in $\mathrm{KOH}$, Hence, the amine is primary.

Question 58

Chemistry · Chemistry in Everyday Life · Single correct

Which of the following compounds are not used as disinfectants? $(A)$ Chloroxylenol $(B)$ Bithional $(C)$ Veronal $(D)$ Prontosil $(E)$ Terpineol Choose the correct answer from the options given below:

  1. A, B, E
  2. A, B
  3. B, D, E
  4. C, D

Answer: (d)

Solution

Veronal is neurological medicine, Prontosil is antibiotic, rest all are disinfectants.

Question 59

Chemistry · Analytical Chemistry · Single correct

Given below are two statements : Statement I : Upon heating a borax bead dipped in cupric sulphate in a luminous flame, the colour of the bead becomes green. Statement II : The green colour observed is due to the formation of copper(I) metaborate. In the light of the above statements, choose the most appropriate answer from the options given below :

  1. Both Statement I and Statement II are true
  2. Statement I is true but Statement II is false
  3. Both Statement I and Statement II are false
  4. Statement I is false but Statement II is true

Answer: (c)

Solution

(Borax Bead Test) On treatment with metal salt, boric anhydride forms metaborate of the metal which gives different colours in oxidising and reducing flame. For example, in the case of copper sulphate, following reactions occur. $CuSO_4$ + $B_2O_3$ $\overset{\text{Luminous\\flame}}{\longrightarrow}$ $Cu(BO_2)_2$ + $SO_3$ Cupric metaborate blue-green Two reactions may take place in reducing flame (Luminous flame) (i) The blue-green $\mathrm{Cu(BO_2)_2}$ is reduced to colourless cuprous metaborate as: 2$Cu(BO_2)_2$ + 2$NaBO_2$ + C $\overset{\text{Luminous\\flame}}{\longrightarrow}$ 2$CuBO_2$ + $Na_2B_4O_7$ + CO (ii) Cupric metaborate may be reduced to metallic copper and bead appears red opaque. $2Cu(BO_2)_2$ + $4NaBO_2$ + 2C $\overset{\text{Luminous\\flame}}{\longrightarrow}$ 2Cu + 2$Na_2B_4O_7$ + 2CO

Question 60

Chemistry · Biomolecules · Single correct

Compound A, $C_5H_{10}O_5$, given a tetraacetate with $Ac_2O$ and oxidation of A with $Br_2-H_2O$ gives an acid, $C_5H_{10}O_6$. Reduction of A with HI gives isopentane. The possible structure of A is :

Answer: (a)

Solution

Formation of tetraacetate with $\mathrm{Ac_2O}$ means compound A has four $-OH$ linkage. Reduction of A with HI gives Isopentane i.e. molecule contains five carbon atom.

Maths

Question 61

Maths · Complex Numbers and Quadratic Equations · Single correct

The equation $$e^{4x} + 8e^{3x} + 13e^{2x} - 8e^x + 1 = 0, x \in \mathbb{R}$$ has:

  1. two solutions and both are negative
  2. no solution
  3. four solutions two of which are negative
  4. two solutions and only one of them is negative

Answer: (a)

Solution

Given $e^{4x} + 8e^{3x} + 13e^{2x} - 8e^x + 1 = 0$. Let $e^x = t$. Now, $t^4 + 8t^3 + 13t^2 - 8t + 1 = 0$. Dividing equation by $t^2$, $$t^2 + 8t + 13 - \frac{8}{t} + \frac{1}{t^2} = 0$$ $$t^2 + \frac{1}{t^2} + 8 \left(t - \frac{1}{t}\right) + 13 = 0$$ $$\left(t - \frac{1}{t}\right)^2 + 2 + 8 \left(t - \frac{1}{t}\right) + 13 = 0$$ Let $t - \frac{1}{t} = z$. $$z^2 + 8z + 15 = 0$$ $$(z + 3)(z + 5) = 0$$ $z = -3$ or $z = -5$. So, $t - \frac{1}{t} = -3$ or $t - \frac{1}{t} = -5$. $$t^2 + 3t - 1 = 0$$ or $$t^2 + 5t - 1 = 0$$ $$t = \frac{-3 \pm \sqrt{13}}{2}$$ or $$t = \frac{-5 \pm \sqrt{29}}{2}$$ As $t = e^x$ so $t$ must be positive, $x=\ln\left(\frac{\sqrt{13}-3}{2}\right)$ or $x=\ln\left(\frac{\sqrt{29}-5}{2}\right)$.

Question 62

Maths · Complex Numbers and Quadratic Equations · Single correct

The complex number $z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}$ is equal to:

  1. $\sqrt{2} \left( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right)$
  2. $\cos \frac{\pi}{12} - i \sin \frac{\pi}{12}$
  3. $\sqrt{2} \left( \cos \frac{\pi}{12} + i \sin \frac{\pi}{12} \right)$
  4. $\sqrt{2} i \left( \cos \frac{5\pi}{12} - i \sin \frac{5\pi}{12} \right)$

Answer: (a)

Solution

Given $( Z = \frac{i-1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}} = \frac{i-1}{\frac{1}{2} + \frac{\sqrt{3}}{2} i} )$. $$= \frac{i-1}{\frac{1}{2} + \frac{\sqrt{3}}{2} i} \times \frac{\frac{1}{2} - \frac{\sqrt{3}}{2} i}{\frac{1}{2} - \frac{\sqrt{3}}{2} i} = \frac{\sqrt{3} - 1}{2} + \frac{\sqrt{3} + 1}{2} i$$ Apply polar form, $$r \cos \theta = \frac{\sqrt{3} - 1}{2}$$ $$r \sin \theta = \frac{\sqrt{3} + 1}{2}$$ Now, $( \tan \theta = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} )$ $$\theta = \frac{5\pi}{12}$$

Question 63

Maths · Sequences and Series · Single correct

Let $a_1, a_2, a_3, \ldots$ be an A.P. If $a_7 = 3$, the product $a_1 a_4$ is minimum and the sum of its first $n$ terms is zero, then $n! - 4a_{n+2}$ is equal to:

  1. 24
  2. $\frac{33}{4}$
  3. $\frac{381}{4}$
  4. 9

Answer: (a)

Solution

Given $a + 6d = 3$, $$Z = a(a + 3d) = (3 - 6d)(3 - 3d) = 18d^2 - 27d + 9$$ Differentiating with respect to $d$: $$36d - 27 = 0$$ $$\Rightarrow d = \frac{3}{4}$$, from (1) $a = \frac{-3}{2}$, (Z = minimum) Now, $$S_n = \frac{n}{2} \left( -3 + (n - 1) \frac{3}{4} \right) = 0$$ $$\Rightarrow n = 5$$ Now, $$n! - 4a_{n(n+2)} = 120 - 4(a_{35})$$ $$= 120 - 4 \left( a + (35 - 1)d \right)$$ $$= 120 - 4 \left( \frac{-3}{2} + 34 \cdot \left( \frac{3}{4} \right) \right)$$ $$= 120 - 4 \left( \frac{-6 + 102}{4} \right)$$ $$= 120 - 96 = 24$$

Question 64

Maths · Sequences and Series · Fill in the blank

The sum $1^2 - 2 \cdot 3^2 + 3 \cdot 5^2 - 4 \cdot 7^2 + 5 \cdot 9^2 - \ldots + 15 \cdot 29^2$ is___________.

Answer: 6952

Solution

Separating odd placed and even placed terms we get $$S = (1 \cdot 1^2 + 3 \cdot 5^2 + \ldots 15 \cdot (29)^2) - (2 \cdot 3^2 + 4 \cdot 7^2 + \ldots + 14 \cdot (27)^2)$$ $$S = \sum_{n=1}^{8} (2n-1)(4n-3)^2 - \sum_{n=1}^{7} (2n)(4n-1)^2$$ Applying summation formula we get $$= 29856 - 22904 = 6952$$

Question 65

Maths · Binomial Theorem · Numerical

The Coefficient of $x^{-6}$, in the expansion of $$\left(\frac{4x}{5} + \frac{5}{2x^2}\right)^9$$, is

Answer: 5040

Solution

$\left(\frac{4x}{5}-\frac{5}{2x^{2}}\right)^{9}$ Now $T_{r+1}=$ ${}^{9}C_{r} \left(\frac{4x}{5}\right)^{9-r} \left(\frac{5}{2x^{2}}\right)^{r}$ $= {}^{9}C_{r} \left(\frac{4}{5}\right)^{9-r} \left(\frac{5}{2}\right)^{r} x^{\,9-3r}$ Coefficient of$x^{-6}$ $\Rightarrow 9-3r=-6$ $\Rightarrow r=5$ So,Coefficient of$x^{-6}$ $= {}^{9}C_{5} \left(\frac{4}{5}\right)^{4} \left(\frac{5}{2}\right)^{5}$ $=5040$

Question 66

Maths · Binomial Theorem · Numerical

If the constant term in the binomial expansion of $$\left( \frac{x^{\frac{5}{2}}}{2} - \frac{4}{x^\ell} \right)^9$$ is $-84$ and the Coefficient of $x^{-3\ell}$ is $2^\alpha \beta$, where $\beta < 0$ is an odd number, Then $|\alpha \ell - \beta|$ is equal to

Answer: 98

Solution

In, $$\left( \frac{x^{\frac{5}{2}}}{2} - \frac{4}{x} \right)^9$$ $$T_{r+1} = \binom{9}{r} \left( x^{5/2} \right)^{9-r} \left( \frac{-4}{x} \right)^r$$ $$= (-1)^r \binom{9}{r} \frac{x^4}{2^{9-r}} \times 4^r \times x^{-r}$$ $$= 45 - 5r - 21r = 0$$ $$r = \frac{45}{5 + 21} ------ (1)$$ Now, according to the question, $$(-1)^r \binom{9}{r} \frac{4^r}{2^{9-r}} = -84$$ $$= (-1)^r \binom{9}{r} 2^{3r-9} = 21 \times 4$$ Only natural value of $r$ possible if $3r - 9 = 0$ $$r = 3 and \binom{9}{3} = 84$$ $$\therefore \; 1 = 5 \; from equation (1)$$ Now, coefficient of $x^{-3r} = x^{2 - \frac{5r - r}{2}}$ at $1 = 5$, gives $$r = 5$$ $$\therefore \; \binom{9}{5} (-1)^r \frac{45}{2^4} = 2^\alpha \times \beta$$ $$= -63 \times 2^7$$ $$\Rightarrow \; \alpha = 7, \; \beta = -63$$ $$\therefore \; value of |\alpha - \beta| = 98$$

Question 67

Maths · Permutations and Combinations · Numerical

If $$\frac{{^{2n+1}P_{n-1}}}{{^{2n-1}P_n}} = 11 : 21$$, then $n^2 + n + 15$ is equal to:

Answer: 45

Solution

Given $\($ $\frac{(2n+1)!(n-1)!}{(n+2)!(2n-1)!}$ = $\frac{11}{21}$ $\)$. $\[$ $\Rightarrow$ $\frac{(2n+1)(2n)}{(n+2)(n+1)n}$ = $\frac{11}{21}$ $\]$ $\[$ $\Rightarrow$ $\frac{2n+1}{(n+1)(n+2)}$ = $\frac{11}{42}$ $\]$ $\[$ $\Rightarrow$ n = 5 $\]$ $\[$ $\Rightarrow$ n^2 + n + 15 = 25 + 5 + 15 = 45 $\]$

Question 68

Maths · Conic Sections · Single correct

The set of all values of $a$ for which the line $x + y = 0$ bisects two distinct chords drawn from a point $P\left(\frac{1+a}{2}, \frac{1-a}{2}\right)$ on the circle $2x^2 + 2y^2 - (1+a)x - (1-a)y = 0$ is equal to:

  1. $(8, \infty)$
  2. $(4, \infty)$
  3. $(0, 4]$
  4. $(2, 12]$

Answer: (a)

Solution

Given the equation $x^2 + y^2 - \frac{(1+a)x}{2} - \frac{(1-a)y}{2} = 0$. The centre is $\left( \frac{1+a}{4}, \frac{1-a}{4} \right) \Rightarrow (h,k)$. Point $P\left( \frac{1+a}{2}, \frac{1-a}{2} \right) \Rightarrow (2h,2k)$. The equation of the chord $\Rightarrow T = S_1$ is $$(x-y)\lambda - \frac{2h(x+\lambda)}{2} - \frac{(2k)(y-\lambda)}{2} = 2\lambda^2 - 2h\lambda.$$ Now, $\lambda \ 2h$ satisfies the chord. Therefore, $$(2h-2k)\lambda - h(x+\lambda) - k(y-\lambda) \Rightarrow 2\lambda^2 + 4k\lambda - 4h\lambda + h\lambda - k\lambda + hx + ky = 0 \Rightarrow 2\lambda^2 + \lambda(3k-3h) + ky + hx = 0.$$ For $D > 0$, $$7k^2 + 7h^2 + 18kh 32 \Rightarrow t > 8 \Rightarrow a^2 > 8.$$

Question 69

Maths · Conic Sections · Numerical

Let S be the set of all $a \in \mathbb{N}$ such that the area of the triangle formed by the tangent at the point $P \,(b, c)$, $b, c \in \mathbb{N}$, on the parabola $y^2 = 2ax$ and the lines $x = b$, $y = 0$ is $16 \, unit^2$, then $\sum_{a \in S} a$ is equal to _____.

Answer: 146

Solution

As $P(b, c)$ lies on the parabola, $c^2 = 2ab$ --- (1). Now the equation of the tangent to the parabola $y^2 = 2ax$ in point form is $yy_1 = 2a \frac{(x + x_1)}{2}$, $(x_1, y_1) = (b, c)$. This implies $yc = a(x + b)$. For point $B$, put $y = 0$, now $x = -b$. So, the area of $\triangle PBA$, $\frac{1}{2} \times AB \times AP = 16$. $$\frac{1}{2} \times 2b \times c = 16$$ This implies $bc = 16$. As $b$ and $c$ are natural numbers, the possible values of $(b, c)$ are $(1, 16)$, $(2, 8)$, $(4, 4)$, $(8, 2)$, and $(16, 1)$. Now from equation (1), $a = \frac{c^2}{2b}$ and $a \in \mathbb{N}$, so values of $(b, c)$ are $(1, 16)$, $(2, 8)$, and $(4, 4)$. Now values of $a$ are $128$, $16$, and $2$. Hence the sum of values of $a$ is $146$.

Question 70

Maths · Conic Sections · Single correct

Let H be the hyperbola, whose foci are $(1 \pm \sqrt{2}, 0)$ and eccentricity is $\sqrt{2}$. Then the length of its latus rectum is ______________.

  1. 2
  2. 3
  3. $\frac{5}{2}$
  4. $\frac{3}{2}$

Answer: (a)

Solution

Given $2ae = \left| (1 + \sqrt{2}) - (1 + \sqrt{2}) \right| = 2\sqrt{2}$. Therefore, $ae = \sqrt{2}$ and $a = 1$. This implies $b = 1$ and $e = \sqrt{2}$, so the hyperbola is rectangular. Thus, $L.R = \frac{2b^2}{a} = 2$.

Question 71

Maths · Limits and Derivatives · Single correct

$$\lim_{x \to \infty} \frac{\left( \sqrt{3x+1} + \sqrt{3x-1} \right)^6 + \left( \sqrt{3x+1} - \sqrt{3x-1} \right)^6}{\left( x + \sqrt{x^2-1} \right)^6 + \left( x - \sqrt{x^2-1} \right)^6} x^3$$

  1. is equal to 9
  2. is equal to 27
  3. does not exist
  4. is equal to $\frac{27}{2}$

Answer: (b)

Solution

Given the limit expression: $$\lim_{x \to \infty} \frac{\left( \sqrt{3x+1} + \sqrt{3x-1} \right)^6 + \left( \sqrt{3x+1} - \sqrt{3x-1} \right)^6 x^3}{\left( x + \sqrt{x^2-1} \right)^6 + \left( x - \sqrt{x^2-1} \right)}$$ We simplify it as follows: $$\lim_{x \to \infty} x^3 \times \left\{ \left( \sqrt{3 + \frac{1}{x}} + \sqrt{3 - \frac{1}{x}} \right)^6 + \left( \sqrt{3 + \frac{1}{x}} - \sqrt{3 - \frac{1}{x}} \right)^6 \right\}$$ $$\times \left\{ x^6 \left( 1 + \sqrt{1 - \frac{1}{x^2}} \right)^6 + \left( 1 - \sqrt{1 - \frac{1}{x^2}} \right) \right\}$$ Evaluating the limit, we get: $$= \frac{\left( 2\sqrt{3} \right)^6 + 0}{2^6 + 0} = 3^3 = (27)$$

Question 72

Maths · Mathematical Reasoning · Single correct

The number of values of $r \in \{p, q, \sim p, \sim q\}$ for which $((p \land q) \Rightarrow (r \lor q)) \land ((p \land r) \Rightarrow q)$ is a tautology, is:

  1. 3
  2. 2
  3. 1
  4. 4

Answer: (b)

Solution

We know, $p \Rightarrow q$ is equivalent to $$\sim p \lor q$$ $$(\sim (p \land q) \lor (r \lor q)) \land (\sim (p \land r) \lor q)$$ $$\Rightarrow (\sim p \lor \sim q \lor r \lor q) \land (\sim p \lor \sim r \lor q)$$ $$\Rightarrow (\sim p \lor r \lor t) \land (\sim p \lor \sim r \lor q)$$ $$\Rightarrow (t) \land (\sim p \lor \sim r \lor q)$$ For this to be tautology, $(\sim p \lor \sim r \lor q)$ must be always true which follows for $r = \sim p$ or $r = q$.

Question 73

Maths · Statistics · Single correct

Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and $\alpha (> 0)$, and the mean and standard deviation of marks of class B of $n$ students be respectively 55 and $30 - \alpha$. If the mean and variance of the marks of the combined class of $100 + n$ students are respectively 50 and 350, then the sum of variances of classes A and B is:

  1. 500
  2. 650
  3. 450
  4. 900

Answer: (a)

Solution

Given: For group A: $x_1 = 40$, $\sigma_1 = \alpha$, $n_1 = 100$ For group B: $x_2 = 55$, $\sigma_2 = 30 - \alpha$, $n_2 = n$ For group A+B: $\bar{x} = 50$, $\sigma^2 = 350$, $n = 100 + n$ The combined mean is given by: $$\bar{x} = \frac{100 \times 40 + 55n}{100 + n}$$ Solving for $n$: $$5000 + 55n = 4000 + 50n$$ $$1000 = 5n$$ $$n = 200$$ Calculating $\sigma_1^2$: $$\sigma_1^2 = \frac{\sum x_i^2}{100} - 40^2$$ Calculating $\sigma_2^2$: $$\sigma_2^2 = \frac{\sum x_j^2}{100} - 55^2$$ The combined variance is: $$350 = \sigma^2 = \frac{\sum x_i^2 + \sum x_j^2}{300} - (\bar{x})^2$$ Substituting the values: $$350 = \frac{(1600 + \alpha^2) \times 100 + [(30 - \alpha)^2 + 3025] \times 200}{300} - (50)^2$$ Simplifying: $$2850 \times 3 = \alpha^2 + 2(30 - \alpha)^2 + 1600 + 6050$$ $$8550 = \alpha^2 + 2(30 - \alpha)^2 + 7650$$ Solving for $\alpha$: $$\alpha^2 + 2(30 - \alpha)^2 = 900$$ $$\alpha^2 - 40\alpha + 300 = 0$$ The solutions are: $$\alpha = 10, 30$$ Finally: $$\sigma_1^2 + \sigma_2^2 = 10^2 + 20^2 = 500$$

Question 74

Maths · Relations and Functions · Single correct

Among the relations $$S = \{(a, b) : a, b \in \mathbb{R} - \{0\}, 2 + \frac{a}{b} > 0\}$$ And $$T = \{(a, b) : a, b \in \mathbb{R}, a^2 - b^2 \in \mathbb{Z}\},$$

  1. S is transitive but T is not
  2. T is symmetric but S is not
  3. Neither S nor T is transitive
  4. Both S and T are symmetric

Answer: (b)

Solution

For relation $T = a^2 - b^2 = -I$. Then, $(b, a)$ on relation $R$. $$b^2 - a^2 = -I$$ Therefore, $T$ is symmetric. $$S = \left\{ (a, b) : a, b \in \mathbb{R} - \{0\}, 2 + \frac{a}{b} > 0 \right\}$$ $$2 + \frac{a}{b} > 0 \implies \frac{a}{b} > -2 , \implies \frac{b}{a} < -\frac{1}{2}$$ If $(b, a) \in S$ then $$2 + \frac{b}{a}$$ not necessarily positive. Therefore, $S$ is not symmetric.

Question 75

Maths · Matrices · Numerical

Let $A = [a_{ij}]$, $a_{ij} \in \mathbb{Z} \cap [0,4]$, $1 \leq i, j \leq 2$. The number of matrices $A$ such that the sum of all entries is a prime number $p \in (2,13)$ is

Answer: 196

Solution

As given $a + b + c + d = 3$ or $5$ or $7$ or $11$ If sum = 3 $$(1 + x + x^2 + \ldots + x^4)^4 \rightarrow x^3$$ $$(1 - x^5)^4 (1 - x)^{-4} \rightarrow x^3$$ $$\therefore \binom{4+3-1}{3} = \binom{6}{3} = 20$$ If sum = 5 $$(1 - 4x^5)(1 - x)^{-4} \rightarrow x^5$$ $$\Rightarrow \binom{4+5-1}{5} - 4x \binom{4.4+0-1}{0} = \binom{8}{5} - 4 = 52$$ If sum = 7 $$(1 - 4x^5)(1 - x)^{-4} \rightarrow x^7$$ $$\Rightarrow \binom{4+5-1}{4} - \binom{4.4+0-1}{0} = \binom{8}{5} - 4 = 52$$ If sum = 11 $$(1 - 4x^5 + 6x^{10})(1 - x)^{-4} \rightarrow x^{11}$$ $$\Rightarrow \binom{4+11-1}{11} - 4 \cdot \binom{4+6-4}{6} + 6 \cdot \binom{4+1-1}{1}$$ $$= \binom{14}{11} - 4 \cdot \binom{9}{6} + 6 \cdot 4 = 364 - 336 + 24 = 52$$ $$\therefore Total matrices = 20 + 52 + 80 + 52 = 204$$

Question 76

Maths · Matrices · Numerical

Let A be a $n \times n$ matrix such that $|A|=2$. If the determinant of the matrix $Adj \left( 2 \cdot Adj(2A^{-1}) \right)$ is $2^{84}$, then $n$ is equal to .

Answer: 5

Solution

Given $\left| Adj(2 Adj(2A^{-1})) \right|$. $$= \left| 2 Adj \left( Adj(2A^{-1}) \right) \right|^{n-1}$$ $$= 2^{n(n-1)} \left| Adj(2A^{-1}) \right|^{n-1}$$ $$= 2^{n(n-1)} \left| (2A^{-1}) \right|^{(n-1)(n-1)}$$ $$= 2^{n(n-1)} 2^{n(n-1)(n-1)} \left| A^{-1} \right|^{(n-1)(n-1)}$$ $$= 2^{n(n-1) + n(n-1)(n-1)} \frac{1}{\left| A \right|^{(n-1)^2}}$$ $$= \frac{2^{n(n-1) + n(n-1)(n-1)}}{2^{(n-1)^2}}$$ $$= 2^{n(n-1) + n(n+1)^2 - (n-1)^2}$$ $$= 2^{(n-1)(n^2-n+1)}$$ Now, $2^{(n-1)(n^2-n+1)} = 2^{84}$ So, $n = 5$

Question 77

Maths · Determinants · Single correct

If a point $P(\alpha, \beta, \gamma)$ satisfying $(\alpha$ $\beta$ $\gamma$)$$\begin{vmatrix} 2 & 10 & 8 \\ 9 & 3 & 8 \\ 8 & 4 & 8 \end{vmatrix} = (0\ 0\ 0)$$ lies on the plane $2x + 4y + 3z = 5$, then $6\alpha + 9\beta + 7\gamma$ is equal to:

  1. 1
  2. $\frac{11}{5}$
  3. $\frac{5}{4}$
  4. 11

Answer: (d)

Solution

Given the equations: $$2\alpha + 4\beta + 3\gamma = 5 (1)$$ $$2\alpha + 9\beta + 8\gamma = 0 (2)$$ $$10\alpha + 3\beta + 4\gamma = 0 (3)$$ $$8\alpha + 8\beta + 8\gamma = 0 (4)$$ Subtract (4) from (2): $$-6\alpha + \beta = 0$$ $$\beta = 6\alpha (5)$$ From equation (4): $$8\alpha + 48\alpha + 8\gamma = 0$$ $$\gamma = -7\alpha (6)$$ From equation (1): $$2\alpha + 24\alpha - 21\alpha = 5$$ $$5\alpha = 5$$ $$\alpha = 1$$ $$\beta = +6, \gamma = -7$$ Therefore: $$6\alpha + 9\beta + 7\gamma$$ $$= 6 + 54 - 49$$ $$= 11$$

Question 78

Maths · Inverse Trigonometric Functions · Single correct

Let (a, b) $\subset$ (0, 2$\pi$) be the largest interval for which $\sin^{-1}(\sin \theta) - \cos^{-1}(\sin \theta) > 0$, $\theta \in (0, 2\pi)$, holds. If $\alpha x^2 + \beta x + \sin^{-1}(x^2 - 6x + 10) + \cos^{-1}(x^2 - 6x + 10) = 0$ and $\alpha - \beta = b - a$, then $\alpha$ is equal to :

  1. $\frac{\pi}{48}$
  2. $\frac{\pi}{16}$
  3. $\frac{\pi}{8}$
  4. $\frac{\pi}{12}$

Answer: (d)

Solution

Given $\sin^{-1} \sin \theta - \left( \frac{\pi}{2} - \sin^{-1} \sin \theta \right) > 0$. This implies $\sin^{-1} \sin \theta > \frac{\pi}{4}$. Therefore, $\sin \theta > \frac{1}{\sqrt{2}}$. So, $\theta \in \left( \frac{\pi}{4}, \frac{3\pi}{4} \right)$. Thus, $\theta \in \left( \frac{\pi}{4}, \frac{3\pi}{4} \right) = (a, b)$. We have $b - a = \frac{\pi}{2} = \alpha - \beta$. This implies $\beta = \alpha - \frac{\pi}{2}$. Therefore, $\alpha x^2 + \beta x + \sin^{-1} \left[ (x-3)^2 + 1 \right] + \cos^{-1} \left[ (x-3)^2 + 1 \right] = 0$. For $x = 3$, $9\alpha + 3\beta + \frac{\pi}{2} + 0 = 0$. Hence, $9\alpha + 3 \left( \alpha - \frac{\pi}{2} \right) + \frac{\pi}{2} = 0$. This simplifies to $12\alpha - \pi = 0$. Therefore, $\alpha = \frac{\pi}{12}$.

Question 79

Maths · Relations and Functions · Single correct

Let $f : \mathbb{R} - \{2,6\} \rightarrow \mathbb{R}$ be real valued function defined as $f(x) = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$. Then range of $f$ is

  1. $\left(-\infty, -\frac{21}{4}\right] \cup [0, \infty)$
  2. $\left(-\infty, -\frac{21}{4}\right) \cup (0, \infty)$
  3. $\left(-\infty, -\frac{21}{4}\right] \cup \left[\frac{21}{4}, \infty\right)$
  4. $\left(-\infty, -\frac{21}{4}\right] \cup [1, \infty)$

Answer: (a)

Solution

Let $y = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}$. By cross multiplying, $yx^2 - 8xy + 12y - x^2 - 2x - 1 = 0$. $x^2(y-1) - x(8y+2) + (12y-1) = 0$. Case 1, $y \neq 1$. $D \geq 0$. This implies $(8y+2)^2 - 4(y-1)(12y-1) \geq 0$. This further implies $y(4y+21) \geq 0$. $y \in \left(-\infty, \frac{-21}{4}\right] \cup [0, \infty) - \{1\}$. Case 2, $y = 1$. $x^2 + 2x + 1 = x^2 - 8x + 12$. $10x = 11$. $x = \frac{11}{10}$. So, $y$ can be 1.

Question 80

Maths · Relations and Functions · Single correct

The absolute minimum value, of the function $f(x) = |x^2 - x + 1| + \left\lfloor x^2 - x + 1 \right\rfloor$, where $[t]$ denotes the greatest integer function, in the interval $[-1, 2]$, is :

  1. $\frac{3}{4}$
  2. $\frac{3}{2}$
  3. $\frac{1}{4}$
  4. $\frac{5}{4}$

Answer: (a)

Solution

Given $f(x) = |x^2 - x + 1| + |x^2 - x + 1|$; $x \in [-1, 2]$. Let $g(x) = x^2 - x + 1$. $$= \left(x - \frac{1}{2}\right)^2 + \frac{3}{4}$$ Thus, $|x^2 - x + 1|$ and $[x^2 - x + 2]$ both have minimum value at $x = \frac{1}{2}$. Therefore, minimum $f(x) = \frac{3}{4} + 0$. $$= \frac{3}{4}$$

Question 81

Maths · Integrals · Single correct

Let $\alpha > 0$. If $\int_{0}^{\alpha} \frac{x}{\sqrt{x+\alpha} - \sqrt{x}} \, dx = \frac{16 + 20\sqrt{2}}{15}$, then $\alpha$ is equal to:

  1. 2
  2. 4
  3. $\sqrt{2}$
  4. 2$\sqrt{2}$

Answer: (a)

Solution

After rationalising $$\int_0^\alpha \frac{x}{\alpha} \left( \sqrt{x + \alpha} + \sqrt{x} \right)$$ $$\int_0^\alpha \frac{1}{\alpha} \left[ (x + \alpha)^{3/2} - \alpha (x + \alpha)^{1/2} + x^{3/2} \right]$$ $$\frac{1}{\alpha} \left[ \frac{2}{5} (x + \alpha)^{5/2} - \alpha \frac{2}{3} (x + \alpha)^{3/2} + \frac{2}{5} x^{5/2} \right]_0^\alpha$$ $$= \frac{1}{\alpha} \left( \frac{5}{2} (2\alpha)^{5/2} - \frac{2\alpha}{3} (2\alpha)^{3/2} + \frac{2}{5} \alpha^{5/2} - \frac{2}{5} \alpha^{5/2} + \frac{2}{3} \alpha^{5/2} \right)$$ $$= \frac{1}{\alpha} \left( \frac{2^{7/2} \alpha^{5/2}}{5} - \frac{2^{5/2} \alpha^{5/2}}{3} + \frac{2}{3} \alpha^{5/2} \right)$$ $$= \alpha^{3/2} \left( \frac{2^{7/2}}{5} - \frac{2^{5/2}}{3} + \frac{2}{3} \right)$$ $$= \frac{\alpha^{3/2}}{15} \left( 24\sqrt{2} - 20\sqrt{2} + 10 \right) = \frac{\alpha^{3/2}}{15} \left( 4\sqrt{2} + 10 \right)$$ Now, $$\frac{\alpha^{3/2}}{15} \left( 4\sqrt{2} + 10 \right) = \frac{16 + 20\sqrt{2}}{15}$$ $$\Rightarrow \alpha = 2$$

Question 82

Maths · Differential Equations · Single correct

If $\phi(x) = -\frac{1}{\sqrt{x}} \int_{\frac{\pi}{4}}^{x} (4\sqrt{2} \sin t - 3\phi'(t)) \, dt, \quad x > 0,$ then $\phi'\left(\frac{\pi}{4}\right)$ is equal to :

  1. \frac{8}{\sqrt{\pi}}
  2. \frac{4}{6 + \sqrt{\pi}}
  3. \frac{8}{6 + \sqrt{\pi}}
  4. \frac{4}{6 - \sqrt{\pi}}

Answer: (c)

Solution

Given $$\phi'(x) = \frac{1}{\sqrt{x}} \left[ \left( 4\sqrt{2} \sin x - 3\phi'(x) \right) \cdot 1 - 0 \right] - \frac{1}{2} x^{-3/2}$$ we have $$\int_{\frac{\pi}{4}}^{x} \left( 4\sqrt{2} \sin t - 3\phi'(t) \right) \, dt,$$ $$\phi'\left( \frac{\pi}{4} \right) = \frac{2}{\sqrt{\pi}} \left[ 4 - 3\phi'\left( \frac{\pi}{4} \right) \right] + 0$$ $$\left( 1 + \frac{6}{\sqrt{\pi}} \right) \phi'\left( \frac{\pi}{4} \right) = \frac{8}{\sqrt{\pi}}$$ $$\phi'\left( \frac{\pi}{4} \right) = \frac{8}{\sqrt{\pi} + 6}$$

Question 83

Maths · Applications of Integrals · Numerical

Let the area of the region $\{(x, y) : |2x - 1| \leq y \leq x^2 - x, 0 \leq x \leq 1\}$ be $A$. Then $(6A + 11)^2$ is equal to ______.

Answer: 125

Solution

Given $y \geq |2x - 1|$, $y \leq |x^2 - x|$. Both curves are symmetric about $x = \frac{1}{2}$. Hence, $$A = 2 \int_{\frac{3 - \sqrt{5}}{2}}^{\frac{1}{2}} \left( (x - x^2) - (1 - 2x) \right) \, dx$$ $$A = 2 \int_{\frac{3 - \sqrt{5}}{2}}^{\frac{1}{2}} (-x^2 + 3x - 1) \, dx = 2 \left( \frac{-x^3}{3} + \frac{3}{2}x^2 - x \right)_{\frac{3 - \sqrt{5}}{2}}^{\frac{1}{2}}$$ On solving $6A + 11 = 5 \sqrt{5}$

Question 84

Maths · Differential Equations · Single correct

Let $y = y(x)$ be the solution of the differential equation $$(3y^2 - 5x^2) y \, dx + 2x(x^2 - y^2) \, dy = 0$$ such that $y(1) = 1$. then $$\left| (y(2))^3 - 12y(2) \right|$$ is equal to:

  1. 32$\sqrt{2}$
  2. 64
  3. 16$\sqrt{2}$
  4. 32

Answer: (a)

Solution

Given $(3y^2 - 5x^2)y \, dx + 2x(x^2 - y^2)dy = 0$. \[ \Rightarrow \frac{dy}{dx} = \frac{y \left(5x^2 - 3y^2\right)}{2x \left(x^2 - y^2\right)} \] Put $y = mx$. \[ \Rightarrow m + x \cdot \frac{dm}{dx} = \frac{m \left(5 - 3m^2\right)}{2 \left(1 - m^2\right)} \] \[ \Rightarrow x \cdot \frac{dm}{dx} = \frac{\left(5 - 3m^2\right)m - 2m \left(1 - m^2\right)}{2 \left(1 - m^2\right)} \] \[ \Rightarrow \frac{dx}{x} = \frac{2 \left(m^2 - 1\right)}{m \left(m^2 - 3\right)} dm \] \[ \Rightarrow \frac{dx}{x} = \left( \frac{2}{m} - \frac{4}{3} + \frac{4m}{m^2 - 3} \right) dm \] \[ \Rightarrow \int \frac{dx}{x} = \int \left( \frac{2}{3m} \right) + \int \frac{2}{3} \left( \frac{2m}{m^2 - 3} \right) dm \] \[ \Rightarrow \ln |x| = \frac{2}{3} \ln |m| + \frac{2}{3} \ln |m^2 - 3| + C \] Or, \[ \ln |x| = \frac{2}{3} \ln \frac{y}{x} + \frac{2}{3} \ln \left( \frac{y}{x} \right)^2 - 3 + C \] Put $(x = 1, y = 1)$: we get $c = -\frac{2}{3} \ln (2)$. \[ \Rightarrow \ln |x| = \frac{2}{3} \ln \frac{y}{x} + \frac{2}{3} \ln \left( \frac{y}{x} \right)^2 - 3 - \frac{2}{3} \ln (2) \]

Question 85

Maths · Vector Algebra · Single correct

Let : $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{c} = 5\hat{i} - 3\hat{j} + 3\hat{k}$ be there vectors. If $\vec{r}$ is a vector such that, $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ and $\vec{r} \cdot \vec{a} = 0$. Then $25|\vec{r}|^2$ is equal to

  1. 449
  2. 336
  3. 339
  4. 560

Answer: (c)

Solution

Given $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$, $\vec{c} = 5\hat{i} - 3\hat{j} + 3\hat{k}$. $\left( \vec{r} - \vec{c} \right) \times \vec{b} = 0$, $\vec{r} \cdot \vec{a} = 0$. This implies $\vec{r} - \vec{c} = \lambda \vec{b}$. Also, $\left( \vec{c} + \lambda \vec{b} \right) \cdot \vec{a} = 0$. This implies $\vec{a} \cdot \vec{c} + \lambda \left( \vec{a} \cdot \vec{b} \right) = 0$. Therefore, $\lambda = -\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}} = -\frac{8}{5}$. $\vec{r} = \frac{1}{5} \left[ 5 \left( 5\hat{i} - 3\hat{j} + 3\hat{k} \right) - 8 \left( \hat{i} - \hat{j} + 2\hat{k} \right) \right]$. $\vec{r} = \frac{1}{5} \left( 17\hat{i} - 7\hat{j} + \hat{k} \right)$. $|\vec{r}|^2 = \frac{1}{25} (289 + 50)$. $25 |\vec{r}|^2 = 339$.

Question 86

Maths · Vector Algebra · Numerical

Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $\left| \vec{a} \right| = \sqrt{31}$, $4 \left| \vec{b} \right| = \left| \vec{c} \right| = 2$ and $2 (\vec{a} \times \vec{b}) = 3 (\vec{c} \times \vec{a})$. If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2\pi}{3}$, then $\left( \frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}} \right)^2$ is equal to ________.

Answer: 3

Solution

Given $\mathbf{b} : 3 \mathbf{c} \times \mathbf{a}$. $\mathbf{a} \times (2\mathbf{b} + 3\mathbf{c}) = 0$. $\mathbf{a} = \lambda (2\mathbf{b} + 3\mathbf{c})$. $|\mathbf{a}|^2 = \lambda^2 |2\mathbf{b} + 3\mathbf{c}|^2$. $|\mathbf{a}|^2 = \lambda^2 \left(4|\mathbf{b}|^2 + 9|\mathbf{c}|^2 + 12\mathbf{b} \cdot \mathbf{c}\right)$. $31 = 31\lambda^2 \Rightarrow \lambda = \pm 1$. $\mathbf{a} = \pm (2\mathbf{b} + 3\mathbf{c})$. $\frac{|\mathbf{a} \times \mathbf{c}|}{|\mathbf{a} \cdot \mathbf{b}|} = \frac{2|\mathbf{b} \times \mathbf{c}|}{2\mathbf{b} \cdot \mathbf{b} + 3\mathbf{c} \cdot \mathbf{b}}$. $|\mathbf{b} \times \mathbf{c}|^2 = |\mathbf{b}|^2 |\mathbf{c}|^2 - (\mathbf{b} \cdot \mathbf{c})^2 = \frac{3}{4}$. $\frac{|\mathbf{a} \times \mathbf{c}|}{|\mathbf{a}|^2} = \frac{2 \times \frac{\sqrt{3}}{2}}{31}$.

Question 87

Maths · Three Dimensional Geometry · Single correct

Let the plane $P: 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ be parallel to the line $L: \frac{x+2}{2} = \frac{y-3}{3} = \frac{z+4}{5}$. If the intercept of $P$ on the y-axis is 1, then the distance between $P$ and $L$ is:

  1. $\sqrt{14}$
  2. $\frac{6}{\sqrt{14}}$
  3. $\sqrt{\frac{2}{7}}$
  4. $\sqrt{\frac{7}{2}}$

Answer: (a)

Solution

P: $8x + \alpha_1 y + \alpha_2 z + 12 = 0$ L: $\frac{x+2}{2} = \frac{y-3}{3} = \frac{z+4}{5}$ Therefore, P is parallel to L. $$\Rightarrow 8(2) + \alpha_1 (3) + 5(\alpha_2) = 0$$ $$\Rightarrow 3\alpha_1 + 5(\alpha_2) = -16$$ Also, the y-intercept of plane P is 1. $$\Rightarrow \alpha_1 = -12$$ And $\alpha_2 = 4$. $$\Rightarrow Equation of plane P is 2x - 3y + z + 3 = 0$$ $$\Rightarrow Distance of line L from Plane P is$$ $$= \frac{|0 - 3(6) + 1 + 3|}{\sqrt{4 + 9 + 1}}$$ $$= \sqrt{14}$$

Question 88

Maths · Three Dimensional Geometry · Single correct

Let P be the plane, passing through the point $(1, -1, -5)$ and perpendicular to the line joining the points $(4, 1, -3)$ and $(2, 4, 3)$. Then the distance of P from the point $(3, -2, 2)$ is

  1. 6
  2. 4
  3. 5
  4. 7

Answer: (c)

Solution

Equation of Plane: $$2(x-1) - 3(y+1) - 6(z+5) = 0$$ Or $$2x - 3y - 6z = 35$$ Therefore, the required distance is: $$\frac{|2(3) - 3(-2) - 6(2) - 35|}{\sqrt{4 + 9 + 36}}$$ $$= 5$$

Question 89

Maths · Vector Algebra · Single correct

The foot of perpendicular from the origin $O$ to a plane $P$ which meets the co-ordinate axes at the points $A$, $B$, $C$ is $(2, a, 4)$, $a \in \mathbb{N}$. If the volume of the tetrahedron $OABC$ is $144 \, unit^3$, then which of the following points is NOT on $P$?

  1. (2, 2, 4)
  2. (0, 4, 4)
  3. (3, 0, 4)
  4. (0, 6, 3)

Answer: (c)

Solution

Equation of Plane: $$\left(2\mathbf{i} + a\mathbf{j} + 4\mathbf{k}\right) \cdot \left[ (x-2)\mathbf{i} + (y-a)\mathbf{j} + (z-4)\mathbf{k} \right] = 0$$ $$\Rightarrow 2x + ay + 4z = 20 + a^2$$ $$\Rightarrow \mathbf{A} \equiv \left( \frac{20 + a^2}{2}, 0, 0 \right)$$ $$\mathbf{B} \equiv \left( 0, \frac{20 + a^2}{a}, 0 \right)$$ $$\mathbf{C} \equiv \left( 0, 0, \frac{20 + a^2}{4} \right)$$ $$\Rightarrow Volume of tetrahedron = \frac{1}{6} \left[ \mathbf{a} \; \mathbf{b} \; \mathbf{c} \right]$$ $$= \frac{1}{6} \; \mathbf{a} \cdot \left( \mathbf{b} \times \mathbf{c} \right)$$ $$\Rightarrow \frac{1}{6} \left( \frac{20 + a^2}{2} \right) \cdot \left( \frac{20 + a^2}{a} \right) \cdot \left( \frac{20 + a^2}{4} \right) = 144$$ $$\Rightarrow \left( 20 + a^2 \right)^3 = 144 \times 48 \times a$$ $$\Rightarrow a = 2$$ $$\Rightarrow Equation of plane is 2x + 2y + 4z = 24$$ Or $$x + y + 2z = 12$$ $$\Rightarrow (3, 0, 4) Not lies on the Plane$$ $$x + y + 2z = 12$$

Question 90

Maths · Probability · Numerical

Let A be the event that the absolute difference between two randomly chosen real numbers in the sample space $[0, 60]$ is less than or equal to $a$. If $P(A) = \frac{11}{36}$, then $a$ is equal to ________.

Answer: 10

Solution

Given $|x-y| -a$. $$P(A) = \frac{ar(OACDEG)}{ar(OBDF)}$$ $$= \frac{ar(OBDF) - ar(ABC) - ar(EFG)}{ar(OBDF)}$$ $$\Rightarrow \frac{11}{36} = \frac{(60)^2 - \frac{1}{2}(60-a)^2 - \frac{1}{2}(60-a)^2}{3600}$$ $$\Rightarrow 1100 = 3600 - (60-a)^2$$ $$\Rightarrow (60-a)^2 = 2500 \Rightarrow 60-a = 50$$ $$\Rightarrow a = 10$$