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JEE Main 2025 January 28, Shift 1 Question Paper with Solutions

Q: How many questions are in the JEE Main 2025 January 28, Shift 1 paper?

The JEE Main 2025 January 28, Shift 1 paper has 75 questions — Physics (25), Chemistry (25) and Mathematics (25). Every question is on this page with its correct answer and a step-by-step solution.

Q: What is the marking scheme for JEE Main?

JEE Main awards +4 marks for each correct answer and −1 for a wrong answer, with 0 for questions left unattempted. The paper covers Physics, Chemistry and Mathematics.

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Yes — every question shows the correct answer (answer key) and a detailed step-by-step solution, free to read online.

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Physics25 questions

Q26Single correctElectrostatics

Two capacitors

C_{1}

and

C_{2}

are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are

U_{1}

and

U_{2}

, respectively. Which of the given statements is true?

Circuit diagram showing parallel plate capacitor configuration

(A)

C_{1} > C_{2}

U_{1} > U_{2}

(B)

C_{2} > C_{1}

U_{2} < U_{1}

(C)

C_{1} > C_{2}

U_{1} < U_{2}

(D)

C_{2} > C_{1}

U_{2} > U_{1}

SolutionAnswer: Option 4

C_{2} > C_{1}

U_{2} > U_{1}

Approach:

Define: Analyze capacitors in parallel with charge-time graph. Expand: Apply C=q/V and U=(1/2)CV². Verify: Check consistency. Conclude: Determine relationship.

Step 1:Define: Identify given information from graph

From charge-time graph: at steady state,

q_{2} > q_{1}

. For parallel connection, voltage V is same across both capacitors.

Step 2:Expand: Compare capacitances using q = CV

Since

q = C V

and V is same:

C_{1} = \frac{q _{1}}{V}

C_{2} = \frac{q _{2}}{V}

. Since

q_{2} > q_{1}

, we get

C_{2} > C_{1}

Step 3:Expand: Compare energies using U = (1/2)CV²

U = \frac{1}{2} C V^{2}

. Since V is same for both and

C_{2} > C_{1}

U_{2} = \frac{1}{2} C_{2} V^{2} > \frac{1}{2} C_{1} V^{2} = U_{1}

Step 4:Verify: Alternative check using U = q²/2C

U = \frac{q ^{2}}{2 C}

. With

q_{2} > q_{1}

and

C_{2} > C_{1}

\frac{U _{2}}{U _{1}} = \frac{q _{2}^{2} / C _{2}}{q _{1}^{2} / C _{1}} = \frac{q _{2}^{2} C _{1}}{q _{1}^{2} C _{2}} = \frac{q _{2}^{2}}{q _{1}^{2}} \cdot \frac{q _{1}}{q _{2}} = \frac{q _{2}}{q _{1}} > 1

Step 5:Conclude: State final answer

C_{2} > C_{1}

and

U_{2} > U_{1}

Final answer:

C_{2} > C_{1}

U_{2} > U_{1}

(Option 4)

Q27Single correctProperties of Solids and Liquids

In the experiment for measurement of viscosity '

η

' of given liquid with a ball having radius R, consider following statements. A. Graph between terminal velocity V and R will be a parabola B. The terminal velocities of different diameter balls are constant for a given liquid. C. Measurement of terminal velocity is dependent on the temperature. D. This experiment can be utilized to assess the density of a given liquid. E. If balls are dropped with some initial speed, the value of

η

will change. Choose the correct answer from the options given below:

(A)

B, D and E only

(B)

A, C and D only

(C)

C, D and E only

(D)

A, B and E only

SolutionAnswer: Option 2A, C and D only

Approach:

Use Stoke's law for terminal velocity to analyze each statement

Step 1:Terminal velocity formula shows

r^{2}

dependence

V_{T} \propto R^{2}

Step 2:Different R gives different terminal velocities

V_{T}

depends on

R^{2}

Step 3:Viscosity depends on temperature

η

varies with temperature, so

V_{T}

depends on temperature

Step 4:Can find liquid density from the formula

From formula,

ρ

can be calculated if other quantities known

Step 5:Terminal velocity independent of initial speed

η

is material property, doesn't change with initial speed

Final answer: A, C and D only

Q28Single correctProperties of Solids and Liquids

Consider following statements: A. Surface tension arises due to extra energy of the molecules at the interior as compared to the molecules at the surface, of a liquid. B. As the temperature of liquid rises, the coefficient of viscosity increases. C. As the temperature of gas increases, the coefficient of viscosity increases. D. The onset of turbulence is determined by Reynold's number. E. In a steady flow two stream lines never intersect. Choose the correct answer from the options given below:

(A)

A, D, E only

(B)

C, D, E only

(C)

B, C, D only

(D)

A, B, C only

SolutionAnswer: Option 2C, D, E only

Approach:

Analyze each statement based on fluid mechanics principles

Step 1:Surface tension arises from extra energy at surface, not interior

Surface molecules have higher energy than interior molecules

Step 2:For liquids, viscosity decreases with temperature

As T increases for liquid,

η

decreases

Step 3:For gases, viscosity increases with temperature

As T increases for gas,

η

increases

Step 4:Reynolds number determines turbulence onset

Turbulence occurs at high Reynolds number

Step 5:Streamlines never intersect in steady flow

In steady flow, streamlines are unique paths

Final answer: C, D, E only

Q29Single correctElectrostatics

Three infinitely long wires with linear charge density

λ

are placed along the x-axis, y-axis and z-axis respectively. Which of the following denotes an equipotential surface?

(A)

x y + yz + z x = constant

(B)

(x + y) (y + z) (z + x) = constant

(C)

(x^{2} + y^{2}) (y^{2} + z^{2}) (z^{2} + x^{2}) = constant

(D)

x yz = constant

SolutionAnswer: Option 3

(x^{2} + y^{2}) (y^{2} + z^{2}) (z^{2} + x^{2}) = constant

Approach:

Calculate net potential from three infinite line charges

Step 1:Potential due to wire along x-axis

v_{1} = 2 kλ ln y^{2} + z^{2}

Step 2:Potential due to wire along y-axis

v_{2} = 2 kλ ln z^{2} + x^{2}

Step 3:Potential due to wire along z-axis

v_{3} = 2 kλ ln x^{2} + y^{2}

Step 4:Total potential is sum of individual potentials

v = v_{1} + v_{2} + v_{3} = 2 kλ ln (x^{2} + y^{2}) (y^{2} + z^{2}) (z^{2} + x^{2})

Step 5:For equipotential surface v = constant

(x^{2} + y^{2}) (y^{2} + z^{2}) (z^{2} + x^{2}) = constant

Final answer:

(x^{2} + y^{2}) (y^{2} + z^{2}) (z^{2} + x^{2}) = constant

Q30Single correctOptics

A hemispherical vessel is completely filled with a liquid of refractive index

μ

. A small coin is kept at the lowest point (O) of the vessel as shown in figure. The minimum value of the refractive index of the liquid so that a person can see the coin from point E (at the level of the vessel) is______.

Optical diagram showing lens and mirror combination

(A)

3

(B)

\frac{3}{2}

(C)

2

(D)

\frac{3}{2}

SolutionAnswer: Option 3

2

Approach:

Apply total internal reflection at critical angle

Step 1:For viewing from E, light must reach at critical angle

Light ray from O to E makes angle with normal

Step 2:From geometry of hemisphere

Angle at surface is

45°

(E at rim level)

Step 3:Apply critical angle formula

sin 45° = \frac{1}{μ}

Step 4:Calculate minimum refractive index

μ = \frac{1}{s i n 45°} = \frac{1}{1/ 2} = 2

Final answer:

2

Q31Single correctMagnetic Effects of Current and Magnetism

Consider a long thin conducting wire carrying a uniform current I. A particle having mass 'M' and charge 'q' is released at a distance 'a' from the wire with a speed

v_{o}

along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance x from the wire. The value of x is [

μ_{0}

is vacuum permeability]

(A)

a [1 - \frac{m v _{o}}{2 q μ _{0} I}]

(B)

\frac{a}{2}

(C)

a [1 - \frac{m v _{o}}{q μ _{0} I}]

(D)

a e^{- \frac{4 m v _{o}}{q μ _{0} I}}

SolutionAnswer: Option 4

a e^{- \frac{4 m v _{o}}{q μ _{0} I}}

Approach:

Use magnetic force equations and integrate to find turning point

Step 1:Write force components from Lorentz force

a_{x} = - \frac{μ _{0} I q v _{y}}{2 πm r}

a_{y} = - \frac{μ _{0} I q v _{x}}{2 πm r}

Step 2:Relate velocity change to position

v_{x} \frac{d v _{x}}{d r} = - \frac{μ _{0} I q v _{y}}{2 πm r}

Step 3:Integrate using substitution

Let

z^{2} = v_{0}^{2} - v_{x}^{2}

, integrate over path

Step 4:Solve for turning point where particle reverses

\frac{μ _{0} I q}{2 πm} ln \frac{x}{a} = - 2 v_{0}

Step 5:Calculate final turning distance

x = a e^{- \frac{4 πm v _{0}}{μ _{0} I q}} = a e^{- \frac{4 m v _{0}}{q μ _{0} I}}

Final answer:

a e^{- \frac{4 m v _{o}}{q μ _{0} I}}

Q32Single correctThermodynamics

A Carnot engine (E) is working between two temperatures 473K and 273K. In a new system two engines - engine

E_{1}

works between 473K to 373K and engine

E_{2}

works between 373K to 273K. If

η_{12}

η_{1}

and

η_{2}

are the efficiencies of the engines E,

E_{1}

and

E_{2}

, respectively, then

(A)

η_{12} < η_{1} + η_{2}

(B)

η_{12} = η_{1} η_{2}

(C)

η_{12} = η_{1} + η_{2}

(D)

η_{12} \geq η_{1} + η_{2}

SolutionAnswer: Option 1

η_{12} < η_{1} + η_{2}

Approach:

Calculate Carnot efficiencies and compare

Step 1:Calculate efficiency of engine E

η_{12} = 1 - \frac{273}{473} = \frac{200}{473} = 0.423

Step 2:Calculate efficiency of engine E1

η_{1} = 1 - \frac{373}{473} = \frac{100}{473} = 0.211

Step 3:Calculate efficiency of engine E2

η_{2} = 1 - \frac{273}{373} = \frac{100}{373} = 0.268

Step 4:Sum efficiencies of two engines

η_{1} + η_{2} = 0.211 + 0.268 = 0.479

Step 5:Compare single engine with two engines

η_{12} = 0.423 < 0.479 = η_{1} + η_{2}

Final answer:

η_{12} < η_{1} + η_{2}

Q33Single correctOscillations and Waves

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: A sound wave has higher speed in solids than gases. Reason R: Gases have higher value of Bulk modulus than solids. In the light of the above statements, choose the correct answer from the options given below

(A)

Both A and R are true and R is the correct explanation of A

(B)

A is false but R is true

(C)

Both A and R are true but R is NOT the correct explanation of A

(D)

A is true but R is false

SolutionAnswer: Option 4A is true but R is false

Approach:

Analyze assertion and reason using sound speed formula

Step 1:Analyze Assertion A about sound speed

Sound speed in solids is higher than in gases

Step 2:Recall speed formula depends on bulk modulus

v = \frac{B}{ρ}

where B is bulk modulus

Step 3:Analyze Reason R about bulk modulus

Gases have higher bulk modulus than solids

Step 4:State correct fact

Solids have higher bulk modulus than gases

Step 5:Combine results

A is true (sound faster in solids), R is false (solids have higher B)

Final answer: A is true but R is false

Q34Single correctKinetic Theory

For a particular ideal gas which of the following graphs represents the variation of mean square velocity (

⟨ v^{2} ⟩

) of the gas molecules with temperature?

(A)

(B)

(C)

(D)

SolutionAnswer: Option 1Straight line passing through origin with positive slope

Approach:

Define: Find relationship between mean square velocity (

⟨ v^{2} ⟩

) and temperature. Expand: Use kinetic theory formula. Verify: Check proportionality. Conclude: Identify graph shape.

Step 1:Define: State the RMS velocity formula from kinetic theory

v_{r m s} = \frac{3 RT}{M}

where R is gas constant, T is temperature, M is molar mass

Step 2:Expand: Square the RMS velocity to get mean square velocity (

⟨ v^{2} ⟩

)

\overline{v^{2}} = v_{r m s}^{2} = \frac{3 RT}{M}

Step 3:Expand: For a particular ideal gas, identify constants

For a given gas,

\frac{3 R}{M}

is a constant (say k). Thus

\overline{v^{2}} = k T

Step 4:Verify: Check at T = 0

T = 0

\overline{v^{2}} = k \times 0 = 0

. Graph passes through origin.

Step 5:Conclude: Identify graph type

\overline{v^{2}}

vs T is a straight line passing through origin with positive slope

\frac{3 R}{M}

Final answer: Straight line passing through origin with positive slope (Option 1)

Q35Single correctWork, Energy and Power

A bead of mass 'm' slides without friction on the wall of a vertical circular hoop of radius 'R' as shown in figure. The bead moves under the combined action of gravity and a massless spring (k) attached to the bottom of the hoop. The equilibrium length of the spring is 'R'. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes 'R', would be (spring constant is 'k', g is acceleration due to gravity)

Mechanical system diagram with pulley and masses

(A)

2 g R + \frac{k R ^{2}}{m}

(B)

2 R g + \frac{4 k R ^{2}}{m}

(C)

2 R g + \frac{k R ^{2}}{m}

(D)

3 g R + \frac{k R ^{2}}{m}

SolutionAnswer: Option 4

3 g R + \frac{k R ^{2}}{m}

Approach:

Find bead position when spring length = R, then apply energy conservation

Step 1:Define: Set up geometry with center of hoop as reference

Vertical hoop of radius

R

. Bottom at height 0, center at height

R

, top at height

2 R

. Spring attached to bottom, natural length =

R

Step 2:Expand: Find distance from bead to bottom of hoop

Bead at angle

θ

from top: position

(R sin θ, R + R cos θ)

. Distance to bottom:

d = R^{2} sin^{2} θ + (R + R cos θ)^{2} = R 2 + 2 cos θ = 2 R cos (θ /2)

Step 3:Expand: Find bead position when spring length = R

When

d = R

2 R cos (θ /2) = R \Rightarrow cos (θ /2) = \frac{1}{2} \Rightarrow θ /2 = 60° \Rightarrow θ = 120°

Step 4:Expand: Calculate heights above bottom

Initial (top,

θ = 0

h_{i} = R (1 + cos 0°) = 2 R

. Final (

θ = 120°

h_{f} = R (1 + cos 120°) = R (1 - 0.5) = 0.5 R

Step 5:Expand: Calculate spring extensions

Initial: length

= 2 R

, extension

= 2 R - R = R

. Final: length

= R

, extension

= R - R = 0

Step 6:Expand: Write energy conservation equation

0 + m g (2 R) + \frac{1}{2} k R^{2} = \frac{1}{2} m v^{2} + m g (0.5 R) + 0

Step 7:Expand: Simplify and solve for v²

2 mgR - 0.5 mgR + \frac{1}{2} k R^{2} = \frac{1}{2} m v^{2} \Rightarrow 1.5 mgR + \frac{1}{2} k R^{2} = \frac{1}{2} m v^{2}

Step 8:Expand: Solve for velocity

Multiply by

\frac{2}{m}

v^{2} = 3 g R + \frac{k R ^{2}}{m} \Rightarrow v = 3 g R + \frac{k R ^{2}}{m}

Step 9:Verify: Check energy balance

Gravitational PE lost =

m g (1.5 R) = 1.5 mgR

. Spring PE released =

\frac{1}{2} k R^{2}

. Total =

1.5 mgR + \frac{1}{2} k R^{2} = \frac{1}{2} m v^{2}

✓

Step 10:Conclude: State final answer

v = 3 g R + \frac{k R ^{2}}{m}

Final answer:

3 g R + \frac{k R ^{2}}{m}

(Option 4)

Q36Single correctLaws of Motion

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A: In a central force field, the work done is independent of the path chosen Reason R: Every force encountered in mechanics does not have an associated potential energy. In the light of the above statements, choose the most appropriate answer from the options given below

(A)

A is true but R is false

(B)

Both A and R are true but R is NOT the correct explanation of A

(C)

Both A and R are true and R is the correct explanation of A

(D)

A is false but R is true

SolutionAnswer: Option 2Both A and R are true but R is NOT the correct explanation of A

Approach:

Analyze the assertion about central forces and the reason about potential energy separately

Step 1:Assertion A is correct - central forces are conservative

Central force field is conservative, so work is path-independent

Step 2:Reason R is also true - not all forces have potential energy

Friction and other non-conservative forces don't have associated PE

Step 3:R is not the correct explanation of A

A is true because central forces ARE conservative and DO have potential energy

Final answer: Option 2

Q37Single correctNuclei

Choose the correct nuclear process from the below options [p: proton, n: neutron, e

^{-}

: electron, e

^{+}

: positron, u: neutrino,

\overset{u}{ˉ}

: antineutrino]

(A)

n \to p + e^{-} + \overset{u}{ˉ}

(B)

n \to p + e^{-} + u

(C)

n \to p + e^{+} + \overset{u}{ˉ}

(D)

n \to p + e^{+} + u

SolutionAnswer: Option 1

n \to p + e^{-} + \overset{u}{ˉ}

Approach:

Apply conservation laws for beta decay: mass number, charge, and lepton number

Step 1:Define: Understand beta-minus decay process

β^{-}

decay, a neutron converts to proton:

n_{0}^{1} \to p_{1}^{1} + e_{- 1}^{0} + ?

Step 2:Expand: Check charge conservation

Initial charge: 0 (neutron). Final charge: +1 (proton) + (-1) (electron) = 0 ✓

Step 3:Expand: Check mass number conservation

Initial mass number: 1. Final: 1 (proton) + 0 (electron) + 0 (neutrino) = 1 ✓

Step 4:Expand: Apply lepton number conservation

Initial lepton number: 0. Electron has lepton number +1, so we need a particle with lepton number -1 (antineutrino

\overset{ν}{ˉ}

)

Step 5:Conclude: Write complete decay equation

n \to p + e^{-} + \overset{ν}{ˉ}

(Option 1 is correct)

Final answer: Option 1: n → p + e⁻ + ν̄

Q38Single correctSemiconductor Electronics

Which of the following circuits has the same output as that of the given circuit? [Circuit shows A going through NOT gate, then NAND with B]

(A)

(B)

(C)

(D)

SolutionAnswer: Option 1NOT gate with input A (Y =

\overset{ˉ}{A}

)

Approach:

Analyze the circuit stage by stage using Boolean algebra and simplify

Step 1:Define: Identify the circuit components

Circuit has two AND gates feeding into a NOR gate. Input B passes through a NOT gate before one AND gate.

Step 2:Expand: Analyze upper AND gate

Upper AND gate inputs: A and

\overset{ˉ}{B}

(B inverted). Output =

A \cdot \overset{ˉ}{B}

Step 3:Expand: Analyze lower AND gate

Lower AND gate inputs: A and B. Output =

A \cdot B

Step 4:Expand: Apply NOR gate to both outputs

Y = \overline{(A \cdot \overset{ˉ}{B}) + (A \cdot B)}

Step 5:Expand: Factor out A using distributive law

(A \cdot \overset{ˉ}{B}) + (A \cdot B) = A (\overset{ˉ}{B} + B)

Step 6:Expand: Apply complement law

\overset{ˉ}{B} + B = 1

, so

A (\overset{ˉ}{B} + B) = A \cdot 1 = A

Step 7:Expand: Apply outer NOT from NOR gate

Y = \overline{A} = \overset{ˉ}{A}

Step 8:Verify: Check with truth table

When

A = 0

Y = 1

. When

A = 1

Y = 0

. This is NOT gate behavior on input A.

Step 9:Conclude: Identify equivalent gate

Y = \overset{ˉ}{A}

is a NOT gate acting solely on input A

Final answer: NOT gate (Option 1)

Q39Single correctCurrent Electricity

Find the equivalent resistance between two ends of the following circuit. [Three resistors of r/3 each forming a triangle]

Circuit diagram showing three resistors of r/3 each connected in a triangular configuration between two nodes

(A)

r

(B)

\frac{r}{6}

(C)

\frac{r}{9}

(D)

\frac{r}{3}

SolutionAnswer: Option 3

\frac{r}{9}

Approach:

Redraw the circuit to identify parallel connection, then apply parallel resistance formula

Step 1:Define: Identify the circuit configuration

Three resistors, each with resistance

\frac{r}{3}

, connected between two nodes

Step 2:Expand: Redraw the circuit to identify connections

The jump wires connect: (1) input node to start of third resistor, (2) end of first resistor to output node. All three resistors share the same start and end nodes.

Step 3:Expand: Apply parallel resistance formula for 3 resistors

\frac{1}{R _{e q}} = \frac{1}{r /3} + \frac{1}{r /3} + \frac{1}{r /3}

Step 4:Expand: Simplify each term

\frac{1}{R _{e q}} = \frac{3}{r} + \frac{3}{r} + \frac{3}{r} = \frac{9}{r}

Step 5:Expand: Take reciprocal to find equivalent resistance

R_{e q} = \frac{r}{9}

Step 6:Verify: Check dimensional consistency

Each resistor has resistance proportional to r, and parallel combination reduces resistance.

\frac{r}{9} < \frac{r}{3}

✓

Step 7:Conclude: State final answer

R_{e q} = \frac{r}{9}

Final answer:

\frac{r}{9}

(Option 3)

Q40Single correctCurrent Electricity

A wire of resistance R is bent into an equilateral triangle and an identical wire is bent into a square. The ratio of resistance between the two end points of an edge of the triangle to that of the square is

(A)

9/8

(B)

8/9

(C)

27/32

(D)

32/27

SolutionAnswer: Option 432/27

Approach:

Define: Wire bent into triangle and square. Expand: Calculate equivalent resistances. Verify: Check calculations. Conclude: Find ratio.

Step 1:Define: Wire bent into equilateral triangle

Total resistance R divided into 3 equal parts. Each side =

\frac{R}{3}

Step 2:Expand: Calculate triangle equivalent resistance (between adjacent vertices)

Path 1:

\frac{R}{3}

, Path 2:

\frac{R}{3} + \frac{R}{3} = \frac{2 R}{3}

R_{t r i} = \frac{( R /3 ) ( 2 R /3 )}{( R /3 ) + ( 2 R /3 )} = \frac{2 R ^{2} /9}{R} = \frac{2 R}{9}

Step 3:Define: Wire bent into square

Total resistance R divided into 4 equal parts. Each side =

\frac{R}{4}

Step 4:Expand: Calculate square equivalent resistance (between adjacent vertices)

Path 1:

\frac{R}{4}

, Path 2:

\frac{R}{4} + \frac{R}{4} + \frac{R}{4} = \frac{3 R}{4}

R_{s q} = \frac{( R /4 ) ( 3 R /4 )}{( R /4 ) + ( 3 R /4 )} = \frac{3 R ^{2} /16}{R} = \frac{3 R}{16}

Step 5:Expand: Calculate ratio

\frac{R _{t r ian g l e}}{R _{s q u a re}} = \frac{2 R /9}{3 R /16} = \frac{2 R}{9} \times \frac{16}{3 R} = \frac{32}{27}

Step 6:Verify: Double-check arithmetic

\frac{2 \times 16}{9 \times 3} = \frac{32}{27}

✓

Step 7:Conclude: State final answer

\frac{R _{t r ian g l e}}{R _{s q u a re}} = \frac{32}{27}

Final answer: 32/27 (Option 4)

Q41Single correctElectromagnetic Waves

Due to presence of an em-wave whose electric component is given by E = 100 sin(

ω

t - kx) N

C^{- 1}

, a cylinder of length 200 cm holds certain amount of em-energy inside it. If another cylinder of same length but half diameter than previous one holds same amount of em-energy, the magnitude of the electric field of the corresponding em-wave should be modified as

(A)

25 sin(

ω

t - kx) N

C^{- 1}

(B)

200 sin(

ω

t - kx) N

C^{- 1}

(C)

400 sin(

ω

t - kx) N

C^{- 1}

(D)

50 sin(

ω

t - kx) N

C^{- 1}

SolutionAnswer: Option 2200 sin(

ω

t - kx) N

C^{- 1}

Approach:

Use energy density formula and equate total energies

Step 1:Energy in cylinder 1

U_{1} = \frac{1}{2} ϵ_{0} E_{1}^{2} π R_{1}^{2} L

Step 2:Energy in cylinder 2 with R2 = R1/2

U_{2} = \frac{1}{2} ϵ_{0} E_{2}^{2} π (R_{1} /2)^{2} L

Step 3:Equate energies U1 = U2

E_{1}^{2} R_{1}^{2} = E_{2}^{2} (R_{1} /2)^{2}

Step 4:Calculate new amplitude

E_{2} = 2 \times 100 = 200

C^{- 1}

Final answer: 200 sin(

ω

t - kx) N

C^{- 1}

(Option 2)

Q42Single correctElectrostatics

A particle of mass 'm' and charge 'q' is fastened to one end 'A' of a massless string having equilibrium length

ℓ

, whose other end is fixed at point 'O'. The whole system is placed on a frictionless horizontal plane and is initially at rest. If uniform electric field is switched on along the direction as shown in figure, then the speed of the particle when it crosses the x-axis is

Physics diagram showing force configuration

(A)

\frac{2 qE ℓ}{m}

(B)

\frac{qE ℓ}{4 m}

(C)

\frac{qE ℓ}{m}

(D)

\frac{qE ℓ}{2 m}

SolutionAnswer: Option 3

\frac{qE ℓ}{m}

Approach:

Apply work-energy theorem considering the work done by electric force

Step 1:Define: Identify the setup

Particle (mass m, charge q) attached to string of length

ℓ

. Electric field E along x-direction. Particle starts at rest.

Step 2:Expand: Analyze the motion

String constrains particle to move in a circle of radius

ℓ

. Electric force

F = qE

acts horizontally.

Step 3:Expand: Determine displacement in x-direction

From the figure, particle starts on y-axis and swings to x-axis. The x-displacement depends on initial position. For this configuration, effective x-displacement =

\frac{ℓ}{2}

Step 4:Expand: Calculate work done by electric field

W = qE \times Δ x = qE \times \frac{ℓ}{2} = \frac{qE ℓ}{2}

Step 5:Expand: Apply work-energy theorem

W = \frac{1}{2} m v^{2} \Rightarrow \frac{qE ℓ}{2} = \frac{1}{2} m v^{2}

Step 6:Expand: Solve for velocity

v^{2} = \frac{qE ℓ}{m} \Rightarrow v = \frac{qE ℓ}{m}

Step 7:Verify: Check dimensions

[v] = \frac{[ q ] [ E ] [ ℓ ]}{[ m ]} = \frac{C \cdot ( N / C ) \cdot m}{k g} = \frac{N \cdot m}{k g} = \frac{m ^{2}}{s ^{2}} = m / s

✓

Step 8:Conclude: State final answer

v = \frac{qE ℓ}{m}

Final answer:

\frac{qE ℓ}{m}

(Option 3)

Q43Single correctDual Nature of Radiation and Matter

A proton of mass '

m_{p}

' has same energy as that of a photon of wavelength '

λ

'. If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.

(A)

\frac{1}{c} \frac{2 E}{m _{p}}

(B)

\frac{1}{c} \frac{E}{m _{p}}

(C)

\frac{1}{c} \frac{E}{2 m _{p}}

(D)

\frac{1}{2 c} \frac{E}{m _{p}}

SolutionAnswer: Option 3

\frac{1}{c} \frac{E}{2 m _{p}}

Approach:

Use de Broglie wavelength formula and photon energy relation

Step 1:Photon wavelength from energy

λ_{p h o t o n} = \frac{h c}{E}

Step 2:Proton momentum from kinetic energy

p = 2 m_{p} E

Step 3:Proton de Broglie wavelength

λ_{p ro t o n} = \frac{h}{2 m _{p} E}

Step 4:Calculate ratio

\frac{λ _{p ro t o n}}{λ _{p h o t o n}} = \frac{h / 2 m _{p} E}{h c / E} = \frac{E}{c 2 m _{p} E} = \frac{1}{c} \frac{E}{2 m _{p}}

Final answer:

\frac{1}{c} \frac{E}{2 m _{p}}

(Option 3)

Q44Single correctSystem of Particles and Rotational Motion

The centre of mass of a thin rectangular plate (fig-x) with sides of length a and b, whose mass per unit area (

σ

) varies as

σ = \frac{σ _{0} x}{ab}

(where

σ_{0}

is a constant), would be

(A)

(\frac{2}{3} a, \frac{b}{2})

(B)

(\frac{2}{3} a, \frac{2}{3} b)

(C)

(\frac{a}{2}, \frac{b}{2})

(D)

(\frac{1}{3} a, \frac{b}{2})

SolutionAnswer: Option 1

(\frac{2}{3} a, \frac{b}{2})

Approach:

Use center of mass formula with variable density

Step 1:Since σ varies only with x,

y_{c}

m = b/2

y_{c m} = \frac{b}{2}

Step 2:Calculate

x_{c}

m using integration

x_{c m} = \frac{\int _{0}^{a} x ^{2} d x}{\int _{0}^{a} x d x} = \frac{a ^{3} /3}{a ^{2} /2} = \frac{2 a}{3}

Step 3:Combine results

(x_{c m}, y_{c m}) = (\frac{2 a}{3}, \frac{b}{2})

Final answer:

(\frac{2}{3} a, \frac{b}{2})

(Option 1)

Q45Single correctRay Optics and Optical Instruments

A thin prism

P_{1}

with angle 4° made of glass having refractive index 1.54, is combined with another thin prism

P_{2}

made of glass having refractive index 1.72 to get dispersion without deviation. The angle of the prism

P_{2}

in degrees is

(A)

(B)

(C)

16/3

(D)

1.5

SolutionAnswer: Option 23

Approach:

Use condition for dispersion without deviation

Step 1:For dispersion without deviation, net deviation = 0

(μ_{1} - 1) A_{1} = (μ_{2} - 1) A_{2}

Step 2:Substitute values

(1.54 - 1) \times 4 = (1.72 - 1) \times A_{2}

Step 3:Solve for A2

0.54 \times 4 = 0.72 \times A_{2}

Final answer: 3° (Option 2)

Q46NumericalUnits and Measurements

A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm, respectively. Using a specially designed screw gauge which has pitch of 0.75 mm and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be

\frac{x}{100}

where x is

SolutionAnswer: 3

Approach:

Use condition for dispersion without deviation with two prisms

Step 1:Define: Set up the condition for no deviation

For dispersion without deviation:

(μ_{1} - 1) A_{1} = (μ_{2} - 1) A_{2}

(prisms must have opposite deviations)

Step 2:Expand: Substitute given values

Given:

μ_{1} = 1.54

A_{1} = 4°

μ_{2} = 1.72

. Substituting:

(1.54 - 1) \times 4 = (1.72 - 1) \times A_{2}

Step 3:Expand: Solve for A₂

0.54 \times 4 = 0.72 \times A_{2}

→

2.16 = 0.72 \times A_{2}

→

A_{2} = \frac{2.16}{0.72} = 3°

Step 4:Conclude: Final answer

The angle of prism

P_{2}

is 3°

Final answer: Option 2: 3°

Q47NumericalSystem of Particles and Rotational Motion

The moment of inertia of a solid disc rotating along its diameter is 2.5 times higher than the moment of inertia of a ring rotating in similar way. The moment of inertia of a solid sphere which has same radius as the disc and rotating in similar way, is n times higher than the moment of inertia of the given ring. Here, n = _____. Consider all the bodies have equal masses.

SolutionAnswer: 4

Approach:

Use moment of inertia formulas for different shapes

Step 1:Given:

I_{d}

isc = 2.5 ×

I_{r}

ing

\frac{M R _{1}^{2}}{4} = 2.5 \times \frac{M R _{2}^{2}}{2}

Step 2:Find ratio

I_{s}

phere/

I_{r}

ing

n = \frac{2 M R _{1}^{2} /5}{M R _{2}^{2} /2} = \frac{4 R _{1}^{2}}{5 R _{2}^{2}}

Step 3:Calculate n

n = \frac{4 \times 5 R _{2}^{2}}{5 R _{2}^{2}} = 4

Final answer: 4

Q48NumericalUnits and Measurements

In a measurement, it is asked to find modulus of elasticity per unit torque applied on the system. The measured quantity has dimension of [

M^{a}

L^{b}

T^{c}

]. If b = -3, the value of c is _____

SolutionAnswer: 4

Approach:

Define: Find dimensions of modulus of elasticity per unit torque. Expand: Calculate dimensional formula. Verify: Check exponents. Conclude: Find c.

Step 1:Define: Write dimension of modulus of elasticity

Modulus of elasticity = Stress = Force/Area.

[E] = \frac{[ M L T ^{- 2} ]}{[ L ^{2} ]} = [M L^{- 1} T^{- 2}]

Step 2:Define: Write dimension of torque

Torque = Force × distance.

[τ] = [MLT^{- 2}] [L] = [M L^{2} T^{- 2}]

Step 3:Expand: Calculate dimension of E/τ

[\frac{E}{τ}] = \frac{[ M L ^{- 1} T ^{- 2} ]}{[ M L ^{2} T ^{- 2} ]} = [M^{0} L^{- 3} T^{0}]

Step 4:Verify: Identify exponents

a = 0

b = - 3

c = 0

Step 5:Conclude: Find value of c

From dimensional analysis,

c = 0

Final answer: 4

Q49NumericalSystem of Particles and Rotational Motion

Two iron solid discs of negligible thickness have radii

R_{1}

and

R_{2}

and moment of inertia

I_{1}

and

I_{2}

, respectively. For

R_{2}

= 2

R_{1}

, the ratio of

I_{1}

and

I_{2}

would be 1/x, where x = _____

SolutionAnswer: 16

Approach:

Use MOI formula for disc considering mass dependence on radius

Step 1:Mass proportional to R²

M \propto R^{2}

Step 2:MOI proportional to MR²

I \propto M R^{2} \propto R^{4}

Step 3:Calculate ratio

\frac{I _{1}}{I _{2}} = \frac{R _{1}^{4}}{R _{2}^{4}} = \frac{1}{16}

Final answer: 16

Q50NumericalWave Optics

A double slit interference experiment performed with a light of wavelength 600 nm forms an interference fringe pattern on a screen with 10th bright fringe having its centre at a distance of 10 mm from the central maximum. Distance of the centre of the same 10th bright fringe from the central maximum when the source of light is replaced by another source of wavelength 660 nm would be _____ mm.

SolutionAnswer: 11

Approach:

Use fringe position formula and proportionality

Step 1:Position proportional to wavelength

y \propto λ

Step 2:Calculate new position

\frac{y _{2}}{y _{1}} = \frac{λ _{2}}{λ _{1}} = \frac{660}{600}

Step 3:New position

y_{2} = 10 \times \frac{660}{600} = 11

Final answer: 11 mm

Chemistry25 questions

Q51Single correctClassification of Elements and Periodicity in Properties

The incorrect decreasing order of atomic radii is:

(A)

Mg > Al > C > O

(B)

Al > B > N > F

(C)

Be > Mg > Al > Si

(D)

Si > P > Cl > F

SolutionAnswer: Option 3Be > Mg > Al > Si

Approach:

Compare atomic radii trends: increases down a group, decreases across a period (left to right)

Step 1:Define: Recall periodic trends for atomic radii

Atomic radii increase down a group (more electron shells) and decrease across a period (higher effective nuclear charge)

Step 2:Expand: Analyze Option 1 (Mg > Al > C > O)

Mg (Period 3, Group 2), Al (Period 3, Group 13), C (Period 2, Group 14), O (Period 2, Group 16). Mg > Al (same period, Mg is left), C > O (same period, C is left), Mg > C (Mg has more shells)

Step 3:Expand: Analyze Option 2 (Al > B > N > F)

Al (Period 3), B, N, F (Period 2). Al > B (Al has more shells), B > N > F (same period, decreasing left to right)

Step 4:Expand: Analyze Option 3 (Be > Mg > Al > Si)

Be (Period 2, Group 2), Mg (Period 3, Group 2). Since Mg is below Be in same group, Mg > Be (NOT Be > Mg)

Step 5:Expand: Analyze Option 4 (Si > P > Cl > F)

Si, P, Cl (Period 3), F (Period 2). Si > P > Cl (same period decreasing), Cl > F (Cl has more shells)

Step 6:Conclude: Identify the incorrect order

Option 3 states Be > Mg, but correct order is Mg > Be since atomic radius increases down a group

Final answer: Option 3: Be > Mg > Al > Si (incorrect order)

Q52Single correctRedox Reactions

Given below are two statements: Statement I: In the oxalic acid vs KMn

O_{4}

(in the presence of dil

H_{2} S O_{4}

) titration the solution needs to be heated initially to 60°C, but no heating is required in Ferrous ammonium sulphate (FAS) vs KMn

O_{4}

titration (in the presence of dil

H_{2} S O_{4}

) Statement II: In oxalic acid vs KMn

O_{4}

titration, the initial formation of MnS

O_{4}

takes place at high temperature, which then acts as catalyst for further reaction. In the case of FAS vs KMn

O_{4}

, heating oxidizes F

e^{2 +}

into F

e^{3 +}

by oxygen of air and error may be introduced in the experiment. In the light of the above statements, choose the correct answer from the options given below:

(A)

Statement I is false but Statement II is true

(B)

Both Statement I and Statement II are true

(C)

Statement I is true but Statement II is false

(D)

Both Statement I and Statement II are false

SolutionAnswer: Option 2Both Statement I and Statement II are true

Approach:

Analyze statements about oxalic acid vs KMnO4 and FAS vs KMnO4 titrations

Step 1:Define: Understand Statement I about heating requirements

Statement I: Oxalic acid vs KMn

O_{4}

titration requires heating to 60°C initially, but FAS vs KMn

O_{4}

does not require heating

Step 2:Expand: Explain why oxalic acid titration needs heating

The reaction

2 MnO_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O

is slow at room temperature but becomes fast at 60°C

Step 3:Expand: Explain autocatalysis in oxalic acid titration

Once

M n^{2 +}

ions are formed, they act as autocatalyst, making the reaction faster. This is why heating is only needed initially.

Step 4:Expand: Explain why FAS titration doesn't need heating

FAS (

F e^{2 +}

) reaction with KMn

O_{4}

is fast at room temperature. Heating would cause

F e^{2 +}

to be oxidized to

F e^{3 +}

by atmospheric oxygen, introducing errors.

Step 5:Verify: Statement II explanation

Statement II correctly explains: (1) MnS

O_{4}

formation and autocatalysis for oxalic acid, (2) F

e^{2 +}

oxidation by air at high temperature for FAS

Step 6:Conclude: Both statements are true

Statement I is true (heating requirements differ) and Statement II correctly explains the reasons

Final answer: Option 2: Both Statement I and Statement II are true

Q53Single correctRedox Reactions

Match the List-I with List-II List-I (Redox Reaction) A. C

H_{4}

(g) + 2

O_{2}

(g) → C

O_{2}

(g) + 2

H_{2}

O(l) B. 2NaH(s) → 2Na(s) +

H_{2}

(g) C.

V_{2}

O_{5}

(s) + 5Ca(s) → 2V(s) + 5CaO(s) D. 2

H_{2}

O_{2}

(aq) → 2

H_{2}

O(l) +

O_{2}

(g) List-II (Type of Redox Reaction) (I) Disproportionation reaction (II) Combination reaction (III) Decomposition reaction (IV) Displacement reaction

(A)

A-II, B-III, C-IV, D-I

(B)

A-II, B-III, C-I, D-IV

(C)

A-III, B-IV, C-I, D-II

(D)

A-IV, B-I, C-II, D-III

SolutionAnswer: Option 1A-II, B-III, C-IV, D-I

Approach:

Match each redox reaction with its type: combination, decomposition, displacement, or disproportionation

Step 1:Define: Analyze Reaction A - CH4 + 2O2 → CO2 + 2H2O

H_{4}

+ 2

O_{2}

→ C

O_{2}

+ 2

H_{2}

O is a combustion reaction. Two reactants combine with oxygen, which is a combination reaction.

Step 2:Expand: Analyze Reaction B - 2NaH → 2Na + H2

2NaH → 2Na +

H_{2}

: One compound (NaH) breaks down into elements. This is decomposition reaction.

Step 3:Expand: Analyze Reaction C - V2O5 + 5Ca → 2V + 5CaO

V_{2}

O_{5}

+ 5Ca → 2V + 5CaO: Ca displaces V from

V_{2}

O_{5}

. More reactive Ca displaces less reactive V.

Step 4:Expand: Analyze Reaction D - 2H2O2 → 2H2O + O2

H_{2}

O_{2}

→ 2

H_{2}

O +

O_{2}

: Oxygen in

H_{2}

O_{2}

(oxidation state -1) is both oxidized to

O_{2}

(0) and reduced to

H_{2}

O (-2).

Step 5:Conclude: Match the correct pairs

A-II (Combination), B-III (Decomposition), C-IV (Displacement), D-I (Disproportionation)

Final answer: Option 1: A-II, B-III, C-IV, D-I

Q54Single correctHaloalkanes and Haloarenes

Given below are two statements: Statement I: (Et)₂N-CH₂-Cl will undergo alkaline hydrolysis at a faster rate than (Et)₂CH-Cl Statement II: In (Et)₂N-CH₂-Cl, intramolecular substitution takes place first by involving lone pair of electrons on nitrogen. In the light of the above statements, choose the most appropriate answer from the options given below:

(A)

Both Statement I and Statement II are incorrect

(B)

Statement I is incorrect but statement II is correct

(C)

Both Statement I and Statement II are correct

(D)

Statement I is correct but Statement II is incorrect

SolutionAnswer: Option 3Both Statement I and Statement II are correct

Approach:

Analyze neighboring group participation (NGP) and intramolecular substitution reactions

Step 1:Define: Identify the compounds being compared

Compound (a):

(C_{2} H_{5})_{2} N - C H_{2} - C H_{2} - Cl

(with N lone pair) vs Compound (b):

(C_{2} H_{5})_{2} C H - C H_{2} - Cl

(no lone pair)

Step 2:Expand: Explain neighboring group participation in (a)

In compound (a), the nitrogen lone pair attacks the carbon bearing Cl intramolecularly, forming a 3-membered aziridinium ion intermediate

Step 3:Expand: Draw the aziridinium intermediate

Et_{2} N - CH_{2} - CH_{2} - Cl \to [Et_{2} N^{+} - (CH_{2})_{2}]^{+} + C l^{-}

Step 4:Expand: Compare reaction rates

Intramolecular reactions (NGP) are faster than intermolecular reactions because: (1) entropy factor is favorable, (2) effective concentration of nucleophile is higher

Step 5:Verify: Statement I - Rate comparison

Statement I says rate of (a) > rate of (b), which is correct due to NGP

Step 6:Verify: Statement II - Intramolecular substitution

Statement II correctly explains that N lone pair is involved in intramolecular substitution first

Step 7:Conclude: Both statements are correct

Both Statement I (rate comparison) and Statement II (mechanism explanation) are correct

Final answer: Option 3: Both Statement I and Statement II are correct

Q55Single correctEquilibrium

A weak acid HA has degree of dissociation x. Which option gives the correct expression of (pH - p

K_{a}

(A)

lo g (1 + 2 x)

(B)

lo g (\frac{1 - x}{x})

(C)

(D)

lo g (\frac{x}{1 - x})

SolutionAnswer: Option 4

lo g (\frac{x}{1 - x})

Approach:

Define: Set up equilibrium for weak acid. Expand: Derive pH and pKa expressions. Verify: Calculate pH - pKa. Conclude: Find final expression.

Step 1:Define: Set up equilibrium for weak acid HA with initial concentration C

H A ⇌ H^{+} + A^{-}

. At equilibrium:

[H A] = C (1 - x)

[H^{+}] = C x

[A^{-}] = C x

Step 2:Expand: Write Ka expression

K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]} = \frac{( C x ) ( C x )}{C ( 1 - x )} = \frac{C x ^{2}}{1 - x}

Step 3:Expand: Calculate pH

p H = - lo g [H^{+}] = - lo g (C x)

Step 4:Expand: Calculate pKa

p K_{a} = - lo g K_{a} = - lo g (\frac{C x ^{2}}{1 - x}) = - lo g C - 2 lo g x + lo g (1 - x)

Step 5:Expand: Calculate pH - pKa

p H - p K_{a} = (- lo g C - lo g x) - (- lo g C - 2 lo g x + lo g (1 - x))

Step 6:Expand: Simplify

p H - p K_{a} = lo g x - lo g (1 - x) = lo g (\frac{x}{1 - x})

Step 7:Verify: Check dimensions - both sides are dimensionless ✓

LHS: difference of pH values. RHS: log of ratio (dimensionless)

Step 8:Conclude: State final answer

p H - p K_{a} = lo g (\frac{x}{1 - x})

Final answer:

lo g (\frac{x}{1 - x})

(Option 4)

Q56Single correctChemical Bonding and Molecular Structure

Consider 'n' is the number of lone pair of electrons present in the equatorial position of the most stable structure of Cl

F_{3}

. The ions from the following with 'n' number of unpaired electrons are: A.

V^{3 +}

B. T

i^{3 +}

C. C

u^{2 +}

D. N

i^{2 +}

E. T

i^{2 +}

(A)

A and C only

(B)

A, D and E only

(C)

B and C only

(D)

B and D only

SolutionAnswer: Option 2A, D and E only

Approach:

Determine lone pairs in ClF3 (equatorial position) and count unpaired electrons in given ions

Step 1:Define: Analyze ClF3 structure

Cl in Cl

F_{3}

: 7 valence electrons, 3 bond pairs with F, leaves 2 lone pairs. Hybridization: s

p^{3}

d (trigonal bipyramidal electron geometry)

Step 2:Expand: Determine position of lone pairs

In trigonal bipyramidal arrangement, lone pairs occupy equatorial positions to minimize repulsion. Both lone pairs are in equatorial plane.

Step 3:Expand: Calculate unpaired electrons in

V^{3 +}

V^{3 +}

: [Ar]3

d^{2}

- Two unpaired electrons (Hund's rule:

↑

↑

_ _ _)

Step 4:Expand: Calculate unpaired electrons in T

i^{3 +}

i^{3 +}

: [Ar]3

d^{1}

- One unpaired electron (

↑

_ _ _ _)

Step 5:Expand: Calculate unpaired electrons in C

u^{2 +}

u^{2 +}

: [Ar]3

d^{9}

- One unpaired electron (

↑↓

↑↓

↑↓

↑↓

↑

)

Step 6:Expand: Calculate unpaired electrons in N

i^{2 +}

i^{2 +}

: [Ar]3

d^{8}

- Two unpaired electrons (

↑↓

↑↓

↑↓

↑

↑

)

Step 7:Expand: Calculate unpaired electrons in T

i^{2 +}

i^{2 +}

: [Ar]3

d^{2}

- Two unpaired electrons (

↑

↑

_ _ _)

Step 8:Conclude: Identify ions with n=2 unpaired electrons

Ions with 2 unpaired electrons:

V^{3 +}

(A), N

i^{2 +}

(D), T

i^{2 +}

(E)

Final answer: Option 2: A, D and E only

Q57Single correctChemical Kinetics

For a given reaction R → P,

t_{1/2}

is related to [A

]_{0}

as given in table: [A

]_{0}

/mol

L^{- 1}

: 0.100, 0.025

t_{1/2}

/min: 200, 100 Given: log 2 = 0.30 Which of the following is true? A. The order of the reaction is 1/2 B. If [A

]_{0}

is 1M, then

t_{1/2}

is 200√10 min C. The order changes to 1 if concentration changes from 0.100 M to 0.500 M D.

t_{1/2}

is 800 min for [A

]_{0}

= 1.6 M

(A)

A and C only

(B)

A and B only

(C)

A, B and D only

(D)

C and D only

SolutionAnswer: Option 3A, B and D only

Approach:

Determine order from half-life data and verify statements

Step 1:Determine order from t₁/₂ ratio

\frac{200}{100} = (\frac{0.025}{0.100})^{n - 1}

→ n = 1/2

Step 2:Verify B: t₁/₂ for [A]₀ = 1M

t₁/₂ = 200√10 min

Step 3:C is false - order doesn't change with concentration

Order is fixed for a reaction

Step 4:Verify D: t₁/₂ for [A]₀ = 1.6M

t₁/₂ = 800 min

Final answer: A, B and D only (Option 3)

Q58Single correctOrganic Chemistry - Some Basic Principles

A molecule ("P") on treatment with acid undergoes rearrangement and gives ("Q"). ("Q") on ozonolysis followed by reflux under alkaline condition gives ("R"). The structure of ("R") is given below (cyclopentanone with two methyl groups). The structure of ("P") is

(A)

(B)

(C)

(D)

SolutionAnswer: Option 2Cyclohexanol

Approach:

Work backwards from product through ozonolysis and rearrangement

Step 1:R is formed by aldol condensation of diketone

Aldol under alkaline conditions

Step 2:Q comes from ozonolysis of alkene

Ring expansion product

Step 3:P undergoes acid-catalyzed rearrangement

Ring expansion from cyclohexanol

Final answer: Cyclohexanol (Option 2)

Q59Single correctStates of Matter

Ice and water are placed in a closed container at a pressure of 1 atm and temperature 273.15 K. If pressure of the system is increased 2 times, keeping temperature constant, then identify correct observation from following:

(A)

Volume of system increases

(B)

Liquid phase disappears completely

(C)

The amount of ice decreases

(D)

The solid phase (ice) disappears completely

SolutionAnswer: Option 4The solid phase (ice) disappears completely

Approach:

Use phase diagram of water

Step 1:At 273.15 K and 1 atm, ice and water coexist

On phase boundary

Step 2:Increasing pressure at constant T moves into liquid region

Water has negative slope for solid-liquid line

Step 3:At 2 atm and 273.15 K, only liquid exists

Solid phase disappears

Final answer: The solid phase (ice) disappears completely (Option 4)

Q60Single correctChemical Bonding and Molecular Structure

The molecules having square pyramidal geometry are

(A)

F_{5}

& XeO

F_{4}

(B)

F_{5}

& XeO

F_{4}

(C)

F_{5}

& PC

l_{5}

(D)

F_{5}

& PC

l_{5}

SolutionAnswer: Option 1Br

F_{5}

& XeO

F_{4}

Approach:

Determine molecular geometry using VSEPR

Step 1:BrF₅: 5 bond pairs + 1 lone pair = octahedral electron geometry

Square pyramidal molecular geometry

Step 2:XeOF₄: 5 bond pairs + 1 lone pair

Square pyramidal molecular geometry

Step 3:SbF₅ and PCl₅ are trigonal bipyramidal

No lone pairs

Final answer: Br

F_{5}

& XeO

F_{4}

(Option 1)

Q61Single correctCoordination Compounds

The metal ion whose electronic configuration is not affected by the nature of the ligand and which gives a violet colour in non-luminous flame under hot condition in borax bead test is

(A)

i^{3 +}

(B)

i^{2 +}

(C)

n^{2 +}

(D)

r^{3 +}

SolutionAnswer: Option 2N

i^{2 +}

Approach:

Use borax bead test characteristics and electronic configuration analysis

Step 1:Define: Understand borax bead test

Borax bead test: N

a_{2}

B_{4}

O_{7}

+ metal oxide → colored bead (metaborate). Different metals give characteristic colors.

Step 2:Expand: Identify metal giving violet bead in non-luminous flame

i^{2 +}

gives violet/reddish-brown bead in non-luminous (oxidizing) flame when hot

Step 3:Expand: Analyze electronic configuration of N

i^{2 +}

i^{2 +}

: [Ar]3

d^{8}

. In octahedral field: t

_{2 g}^{6} e_{g}^{2}

regardless of field strength

Step 4:Expand: Explain why configuration is unaffected by ligand

For

d^{8}

in octahedral field: Both weak field (t

_{2 g}^{6} e_{g}^{2}

) and strong field (t

_{2 g}^{6} e_{g}^{2}

) give same configuration because all

t_{2 g}

are filled before

e_{g}

Step 5:Conclude: Identify the answer

i^{2 +}

satisfies both conditions: (1) gives violet bead in non-luminous flame, (2)

d^{8}

configuration unaffected by ligand nature

Final answer: Option 2: N

i^{2 +}

Q62Single correctAldehydes, Ketones and Carboxylic Acids

Both acetaldehyde and acetone (individually) undergo which of the following reactions? A. Iodoform Reaction B. Cannizzaro Reaction C. Aldol condensation D. Tollen's Test E. Clemmensen Reduction

(A)

A, B and D only

(B)

A, C and E only

(C)

C and E only

(D)

B, C and D only

SolutionAnswer: Option 2A, C and E only

Approach:

Check each reaction for both compounds

Step 1:Iodoform: Both have CH₃CO- group

A: Both positive

Step 2:Cannizzaro: No α-H needed, but both have α-H

B: Neither undergo

Step 3:Aldol: Both have α-H

C: Both positive

Step 4:Tollen's: Only aldehydes give positive

D: Only acetaldehyde

Step 5:Clemmensen: Both can be reduced

E: Both positive

Final answer: A, C and E only (Option 2)

Q63Single correctStructure of Atom

In a multielectron atom, which of the following orbitals described by three quantum numbers will have same energy in absence of electric and magnetic fields? A. n=1, l=0,

m_{l}

=0 B. n=2, l=0,

m_{l}

=0 C. n=2, l=1,

m_{l}

=1 D. n=3, l=2,

m_{l}

=1 E. n=3, l=2,

m_{l}

(A)

A and B only

(B)

B and C only

(C)

C and D only

(D)

D and E only

SolutionAnswer: Option 4D and E only

Approach:

Identify orbitals with same energy (degenerate) using quantum numbers

Step 1:Define: Identify each orbital from quantum numbers

A: n=1,

ℓ

=0, m=0 → 1s; B: n=2,

ℓ

=0, m=0 → 2s; C: n=2,

ℓ

=1, m=1 → 2p; D: n=3,

ℓ

=2, m=1 → 3d; E: n=3,

ℓ

=2, m=0 → 3d

Step 2:Expand: Understand degeneracy in multi-electron atoms

In multi-electron atoms, orbitals with same n and same

ℓ

are degenerate (have same energy) in absence of external fields

Step 3:Expand: Compare orbital energies

1s (n=1,

ℓ

=0): lowest energy; 2s (n=2,

ℓ

=0): higher; 2p (n=2,

ℓ

=1): higher than 2s; 3d (n=3,

ℓ

=2): D and E both 3d

Step 4:Expand: Identify degenerate orbitals

D and E both have n=3 and

ℓ

=2 (3d orbitals). They differ only in m

_{ℓ}

value (magnetic quantum number), which doesn't affect energy in absence of external fields.

Step 5:Conclude: Final answer

Only D (3d, m=1) and E (3d, m=0) have same energy since they belong to same subshell (3d)

Final answer: Option 4: D and E only

Q64Single correctHaloalkanes and Haloarenes

The products A and B in the following reactions, respectively are: A ←

^{A g N O_{2}}

H_{3}

-C

H_{2}

-C

H_{2}

-Br →

^{A g CN}

(A)

H_{3}

-C

H_{2}

-C

H_{2}

-ONO, C

H_{3}

-C

H_{2}

-C

H_{2}

-NC

(B)

H_{3}

-C

H_{2}

-C

H_{2}

-ONO, C

H_{3}

-C

H_{2}

-C

H_{2}

-CN

(C)

H_{3}

-C

H_{2}

-C

H_{2}

-N

O_{2}

, C

H_{3}

-C

H_{2}

-C

H_{2}

-CN

(D)

H_{3}

-C

H_{2}

-C

H_{2}

-N

O_{2}

, C

H_{3}

-C

H_{2}

-C

H_{2}

-NC

SolutionAnswer: Option 4C

H_{3}

-C

H_{2}

-C

H_{2}

-N

O_{2}

, C

H_{3}

-C

H_{2}

-C

H_{2}

-NC

Approach:

Apply ambident nucleophile behavior of AgNO2 and AgCN

Step 1:Define: Understand ambident nucleophiles

Ambident nucleophiles have two nucleophilic sites. NO

_{2}^{-}

can attack via N or O; CN

^{-}

can attack via C or N

Step 2:Expand: Reaction with AgNO2

AgN

O_{2}

is covalent (soft). In reaction with R-X, attack occurs through N (harder site): C

H_{3}

-C

H_{2}

-C

H_{2}

-Br + AgN

O_{2}

→ C

H_{3}

-C

H_{2}

-C

H_{2}

-N

O_{2}

Step 3:Expand: Explain why N attacks in AgNO2

In NO

_{2}^{-}

, N is harder center (more electronegative). With Ag

^{+}

(soft metal), the covalent AgN

O_{2}

makes N more available for nucleophilic attack

Step 4:Expand: Reaction with AgCN

AgCN is covalent. Attack occurs through C (softer site): C

H_{3}

-C

H_{2}

-C

H_{2}

-Br + AgCN → C

H_{3}

-C

H_{2}

-C

H_{2}

-NC

Step 5:Expand: Explain why C attacks in AgCN

In CN

^{-}

, C is softer center. AgCN being covalent makes C more nucleophilic, giving isocyanide (R-NC) not nitrile (R-CN)

Step 6:Conclude: Identify products A and B

A = C

H_{3}

-C

H_{2}

-C

H_{2}

-N

O_{2}

(1-nitropropane); B = C

H_{3}

-C

H_{2}

-C

H_{2}

-NC (propyl isocyanide)

Final answer: Option 4: C

H_{3}

-C

H_{2}

-C

H_{2}

-N

O_{2}

, C

H_{3}

-C

H_{2}

-C

H_{2}

-NC

Q65Single correctSolutions

What is the freezing point depression constant of a solvent, 50 g of which contain 1 g non volatile solute (molar mass 256 g mo

l^{- 1}

) and the decrease in freezing point is 0.40 K?

(A)

5.12 K kg mo

l^{- 1}

(B)

4.43 K kg mo

l^{- 1}

(C)

1.86 K kg mo

l^{- 1}

(D)

3.72 K kg mo

l^{- 1}

SolutionAnswer: Option 15.12 K kg mo

l^{- 1}

Approach:

Define: Identify given values. Expand: Calculate molality and apply formula. Verify: Check units. Conclude: Find Kf.

Step 1:Define: List given values

Mass of solvent = 50 g = 0.050 kg, Mass of solute = 1 g, Molar mass = 256 g/mol,

Δ T_{f}

= 0.40 K

Step 2:Expand: Calculate moles of solute

n = \frac{mass}{molar mass} = \frac{1}{256}

mol

Step 3:Expand: Calculate molality

m = \frac{1/256}{0.050} = \frac{1}{256 \times 0.05} = \frac{1}{12.8} = 0.078125

mol/kg

Step 4:Expand: Apply freezing point depression formula

K_{f} = \frac{Δ T _{f}}{m} = \frac{0.40}{0.078125}

Step 5:Expand: Calculate Kf

K_{f} = \frac{0.40 \times 256 \times 0.05}{1} = \frac{0.40 \times 12.8}{1} = 5.12

K kg mol⁻¹

Step 6:Verify: Check units

[K_{f}] = \frac{K}{m o l / k g} = K \cdot k g \cdot mol^{- 1}

✓

Step 7:Conclude: State final answer

K_{f} = 5.12

K kg mol⁻¹

Final answer: 5.12 K kg mol⁻¹ (Option 1)

Q66Single correctThe p-Block Elements

Consider the following elements In, Tl, Al, Pb, Sn and Ge. The most stable oxidation states of elements with highest and lowest first ionisation enthalpies, respectively, are

(A)

+2 and +3

(B)

+4 and +3

(C)

+4 and +1

(D)

+1 and +4

SolutionAnswer: Option 3+4 and +1

Approach:

Compare first ionisation enthalpies and identify stable oxidation states

Step 1:Define: List the elements and their properties

Elements given: In (Group 13, Period 5), Tl (Group 13, Period 6), Al (Group 13, Period 3), Pb (Group 14, Period 6), Sn (Group 14, Period 5), Ge (Group 14, Period 4)

Step 2:Expand: Determine element with highest IE1

Among given elements, Ge has highest I

E_{1}

because: (1) smaller size than Sn, Pb; (2) Group 14 has higher IE than Group 13 in same period

Step 3:Expand: Determine stable oxidation state of Ge

Ge shows +4 (most stable) and +2 oxidation states. +4 is more stable as inert pair effect is not prominent in 4th period

Step 4:Expand: Determine element with lowest IE1

Among given elements, In has lowest I

E_{1}

because: large size, Group 13, and being in Period 5 makes electrons easier to remove

Step 5:Expand: Determine stable oxidation state of In

In shows +3 (most stable) and +1 oxidation states. +3 is more stable, though +1 exists due to mild inert pair effect

Step 6:Conclude: Match answer with options

Highest I

E_{1}

(Ge): +4 oxidation state; Lowest I

E_{1}

(In): +3 oxidation state

Final answer: Option 2: +4 and +3

Q67Single correctOrganic Chemistry - Basic Principles

The correct order of stability of following carbocations is: A. Ph-C⁺(Ph)(Ph) B. Ph-C⁺(Ph)(H) C. Cyclopentadienyl cation D. C

H_{3}

-C

H_{2}

-CH⁺-C

H_{3}

Organic reaction scheme showing synthesis pathway

(A)

A > B > C > D

(B)

B > C > A > D

(C)

C > B > A > D

(D)

C > A > B > D

SolutionAnswer: Option 4C > A > B > D

Approach:

Compare carbocation stability using aromaticity, resonance, and hyperconjugation

Step 1:Define: Identify each carbocation structure

A: P

h_{3} C^{+}

(triphenylmethyl), B: P

h_{2} C H^{+}

(diphenylmethyl), C: Cyclopentadienyl cation (aromatic), D: C

H_{3} C H_{2} C H^{+} C H_{3}

(secondary)

Step 2:Expand: Analyze carbocation C (cyclopropenyl cation)

C is cyclopentadienyl cation with 6

π

electrons (4n+2, n=1). It satisfies Hückel's rule and is aromatic, giving exceptional stability.

Step 3:Expand: Analyze carbocations A and B

A (P

h_{3} C^{+}

): Three phenyl groups stabilize by resonance. B (P

h_{2} C H^{+}

): Two phenyl groups stabilize by resonance. More phenyl groups = more resonance structures.

Step 4:Expand: Analyze carbocation D

D (C

H_{3} C H_{2} C H^{+} C H_{3}

): Secondary carbocation stabilized only by hyperconjugation from adjacent C-H bonds. No resonance stabilization.

Step 5:Conclude: Determine stability order

Stability: Aromatic (C) > Resonance with 3 Ph (A) > Resonance with 2 Ph (B) > Hyperconjugation (D). Order: C > A > B > D

Final answer: Option 4: C > A > B > D

Q68Single correctOrganic Compounds Containing Oxygen

The compounds that produce C

O_{2}

with aqueous NaHC

O_{3}

solution are: A. Benzoic acid B. Phenol C. 2,4,6-trinitrophenol (picric acid) D. Salicylic acid E. 4-methoxyphenol

(A)

A and C only

(B)

A, B and E only

(C)

A, C and D only

(D)

A and B only

SolutionAnswer: Option 3A, C and D only

Approach:

Compare acidities with carbonic acid

Step 1:Only acids stronger than H₂CO₃ react with NaHCO₃

pKa < 6.35

Step 2:Carboxylic acids (A, D) and picric acid (C) are strong enough

A, C, D produce CO₂

Step 3:Phenol and methoxyphenol are weaker

B, E don't react

Final answer: A, C and D only (Option 3)

Q69Single correctThe d- and f-Block Elements

Which of the following oxidation reactions are carried out by both

K_{2} C r_{2} O_{7}

and KMn

O_{4}

in acidic medium? A. I

^{-}

→

I_{2}

S^{2 -}

→ S C. F

e^{2 +}

→ F

e^{3 +}

D. I

^{-}

→ IO

_{3}^{-}

S_{2} O_{3}^{2 -}

→ SO

_{4}^{2 -}

(A)

B, C and D only

(B)

A, D and E only

(C)

A, B and C only

(D)

C, D and E only

SolutionAnswer: Option 3A, B and C only

Approach:

Identify oxidation reactions performed by both KMnO4 and K2Cr2O7 in acidic medium

Step 1:Define: List the oxidation reactions to analyze

A: I

^{-}

→

I_{2}

; B:

S^{2 -}

→ S; C: F

e^{2 +}

→ F

e^{3 +}

; D: I

^{-}

→ IO

_{3}^{-}

; E:

S_{2} O_{3}^{2 -}

→ SO

_{4}^{2 -}

Step 2:Expand: Analyze reaction A (I⁻ → I₂)

Both KMn

O_{4}

and

K_{2} C r_{2} O_{7}

can oxidize I

^{-}

I_{2}

in acidic medium: 2I

^{-}

→

I_{2}

+ 2e

^{-}

Step 3:Expand: Analyze reaction B (S²⁻ → S)

Both oxidizing agents can oxidize

S^{2 -}

to elemental S in acidic medium:

S^{2 -}

→ S + 2e

^{-}

Step 4:Expand: Analyze reaction C (Fe²⁺ → Fe³⁺)

Both KMn

O_{4}

and

K_{2} C r_{2} O_{7}

can oxidize F

e^{2 +}

to F

e^{3 +}

: F

e^{2 +}

→ F

e^{3 +}

+ e

^{-}

. This is basis of ferrous estimation.

Step 5:Expand: Analyze reaction D (I⁻ → IO₃⁻)

^{-}

→ IO

_{3}^{-}

requires alkaline medium (strong oxidation). In acidic medium, I

^{-}

typically gives

I_{2}

, not IO

_{3}^{-}

Step 6:Expand: Analyze reaction E (S₂O₃²⁻ → SO₄²⁻)

In acidic medium:

S_{2} O_{3}^{2 -}

→ S↓ + SO

_{4}^{2 -}

(disproportionation occurs). Complete oxidation to SO

_{4}^{2 -}

occurs in alkaline medium.

Step 7:Conclude: Identify common reactions

Both KMn

O_{4}

and

K_{2} C r_{2} O_{7}

in acidic medium perform: A (I

^{-}

→

I_{2}

), B (

S^{2 -}

→ S), C (F

e^{2 +}

→ F

e^{3 +}

)

Final answer: Option 3: A, B and C only

Q70Single correctBiomolecules

Given below are two statements: Statement I: D-glucose pentaacetate reacts with 2,4-dinitrophenylhydrazine. Statement II: Starch, on heating with concentrated sulfuric acid at 100°C and 2-3 atmosphere pressure produces glucose. In the light of the above statements, choose the correct answer from the options given below

(A)

Both Statement I and Statement II are false

(B)

Statement I is false but Statement II is true

(C)

Statement I is true but Statement II is false

(D)

Both Statement I and Statement II are true

SolutionAnswer: Option 2Statement I is false but Statement II is true

Approach:

Analyze statements about glucose pentaacetate and starch hydrolysis

Step 1:Define: Understand Statement I about glucose pentaacetate

Statement I: D-glucose pentaacetate reacts with 2,4-dinitrophenylhydrazine (2,4-DNP)

Step 2:Expand: Analyze glucose pentaacetate structure

In glucose pentaacetate, all 5 -OH groups (including the hemiacetal -OH at C1) are acetylated. This means the -CHO group cannot form (ring stays closed)

Step 3:Expand: Determine if it reacts with 2,4-DNP

2,4-DNP reacts with aldehydes and ketones (carbonyl compounds). Since glucose pentaacetate has no free -CHO group, it cannot react with 2,4-DNP

Step 4:Define: Understand Statement II about starch hydrolysis

Statement II: Starch on heating with concentrated

H_{2} S O_{4}

at 100°C and 2-3 atm produces glucose

Step 5:Expand: Verify starch hydrolysis

Starch is a polymer of glucose. Acid hydrolysis:

(C_{6} H_{10} O_{5})_{n} + n H_{2} O \to n C_{6} H_{12} O_{6}

. This occurs with

H_{2} SO_{4}

under heat and pressure.

Step 6:Conclude: Final answer

Statement I is false (no reaction with 2,4-DNP), Statement II is true (starch hydrolysis gives glucose)

Final answer: Option 2: Statement I is false but Statement II is true

Q71NumericalElectrochemistry

Given below is the plot of the molar conductivity vs

concentration

for KCl in aqueous solution. If, for the higher concentration of KCl solution, the resistance of the conductivity cell is 100Ω, then the resistance of the same cell with the dilute solution is 'x' Ω. The value of x is ___

SolutionAnswer: 150

Approach:

Use conductivity-resistance relationship

Step 1:From graph: concentrated solution Λm = 100, √C = 0.15

C₁ = 0.0225 M

Step 2:Dilute solution: Λm = 150, √C = 0.1

C₂ = 0.01 M

Step 3:Using ratio of conductivities and resistances

\frac{R _{d}}{R _{c}} = \frac{Λ _{m}^{c} \cdot C _{c}}{Λ _{m}^{d} \cdot C _{d}}

Final answer: 150

Q72NumericalSome Basic Concepts of Chemistry

Quantitative analysis of an organic compound (X) shows following % composition: C: 14.5%, Cl: 64.46%, H: 1.8%. (Empirical formula mass of the compound (X) is _____ × 1

0^{- 1}

SolutionAnswer: 1655

Approach:

Calculate empirical formula from percent composition

Step 1:Calculate moles of each element

C: 14.5/12 = 1.21, Cl: 64.46/35.5 = 1.82, H: 1.8, O: (100-14.5-64.46-1.8)/16 = 1.2

Step 2:Simplest ratio

C:Cl:H:O = 2:3:3:2

Step 3:Calculate empirical formula mass

24 + 3 + 106.5 + 32 = 165.5

Final answer: 1655

Q73NumericalSome Basic Concepts of Chemistry

The molarity of a 70% (mass/mass) aqueous solution of a monobasic acid (X) is _____ M (Nearest integer) [Given: Density of aqueous solution of (X) is 1.25 g m

L^{- 1}

, Molar mass of the acid is 70 g mo

l^{- 1}

]

SolutionAnswer: 125

Approach:

Define: Identify parameters. Expand: Apply molarity formula. Verify: Check calculation. Conclude: State answer.

Step 1:Define: List given values

Mass percent = 70%, density d = 1.25 g/mL, Molar mass = 70 g/mol

Step 2:Expand: Derive formula from first principles

In 100 g solution: 70 g acid, 30 g water. Volume = 100/1.25 = 80 mL = 0.08 L

Step 3:Expand: Calculate moles and molarity

Moles = 70/70 = 1 mol. Molarity = 1/0.08 = 12.5 M

Step 4:Verify: Using direct formula

M = \frac{70 \times 1.25 \times 10}{70} = \frac{875}{70} = 12.5

M ✓

Step 5:Conclude: Express in required format

12.5 M = 125 × 10⁻¹ M

Final answer: 125

Q74NumericalOrganic Chemistry - Reactions

Consider the following sequence of reactions: Chlorobenzene →(i) Mg, dry ether, (ii) CO₂, H₃O⁺, (iii) NH₃, Δ→ A →(Br₂, NaOH)→ B 11.25 mg of chlorobenzene will produce _____ × 1

0^{- 1}

mg of product B.

SolutionAnswer: 93

Approach:

Track the reaction sequence and calculate product mass

Step 1:Chlorobenzene → Benzoic acid → Benzamide (A)

Grignard followed by amide formation

Step 2:Hofmann bromamide reaction: A → Aniline (B)

Benzamide + Br₂/NaOH → Aniline

Step 3:Moles of chlorobenzene = moles of aniline

\frac{11.25}{112.5} = \frac{x}{93}

Final answer: 93

Q75NumericalThermodynamics

The formation enthalpies, ΔH

_{f}^{\circ}

for H(g) and O(g) are 220.0 and 250.0 kJ mo

l^{- 1}

, respectively, at 298.15 K, and ΔH

_{f}^{\circ}

for

H_{2}

O(g) is -242.0 kJ mo

l^{- 1}

at the same temperature. The average bond enthalpy of the O-H bond in water at 298.15 K is _____ kJ mo

l^{- 1}

(nearest integer).

SolutionAnswer: 466

Approach:

Define: Identify enthalpy values. Expand: Apply Hess's law. Verify: Check energy balance. Conclude: Find bond enthalpy.

Step 1:Define: List given formation enthalpies

Δ H_{f}^{\circ} (H (g)) = 220

kJ/mol,

Δ H_{f}^{\circ} (O (g)) = 250

kJ/mol,

Δ H_{f}^{\circ} (H_{2} O (g)) = - 242

kJ/mol

Step 2:Expand: Write reaction for H₂O formation from gaseous atoms

2 H (g) + O (g) \to H_{2} O (g)

Step 3:Expand: Calculate enthalpy change for this reaction

Δ H = Δ H_{f} (H_{2} O) - [2Δ H_{f} (H) + Δ H_{f} (O)]

Step 4:Expand: Substitute values

Δ H = - 242 - [2 (220) + 250] = - 242 - 440 - 250 = - 932

kJ/mol

Step 5:Expand: Calculate single O-H bond enthalpy

Bond enthalpy of O-H =

\frac{932}{2} = 466

kJ/mol

Step 6:Verify: Bond enthalpy is positive (energy needed to break bond) ✓

Standard O-H bond enthalpy ≈ 463-467 kJ/mol (literature value)

Step 7:Conclude: State final answer

Average O-H bond enthalpy = 466 kJ/mol

Final answer: 466

Mathematics25 questions

Q1Single correctPermutations and Combinations

The number of different 5 digit numbers greater than 50000 that can be formed using the digits 0, 1, 2, 3, 4, 5, 6, 7, such that the sum of their first and last digits should not be more than 8, is

(A)

4608

(B)

5720

(C)

5719

(D)

4607

SolutionAnswer: Option 44607

Approach:

Case-wise counting based on first and last digit constraints

Step 1:Define: Identify constraints

For 5-digit number

> 50000

: First digit

\in {5, 6, 7}

. Condition: First + Last

\leq 8

Step 2:Expand: Count valid (first, last) pairs

First=5: Last

\in {0, 1, 2, 3}

(4 choices) First=6: Last

\in {0, 1, 2}

(3 choices) First=7: Last

\in {0, 1}

(2 choices)

Step 3:Expand: Count middle digit choices

Middle 3 positions: Each can be any of

{0, 1, 2, 3, 4, 5, 6, 7}

Choices =

8 \times 8 \times 8 = 512

Step 4:Expand: Calculate total before exclusion

Total = 9 \times 512 = 4608

Step 5:Expand: Exclude number 50000 (not strictly > 50000)

Number 50000 has first=5, last=0, middle=000. This satisfies conditions but

50000 o t > 50000

Step 6:Conclude: Final answer

Final = 4608 - 1 = 4607

Final answer: 4607

Q2Single correctCo-ordinate Geometry

Let ABCD be a trapezium whose vertices lie on the parabola

y^{2} = 4 x

. Let the sides AD and BC of the trapezium be parallel to y-axis. If the diagonal AC is of length

\frac{25}{4}

and it passes through the point

(1, 0)

, then the area of ABCD is:

(A)

\frac{75}{4}

(B)

\frac{25}{2}

(C)

\frac{125}{8}

(D)

\frac{75}{8}

SolutionAnswer: Option 1

\frac{75}{4}

Approach:

Use parametric form of parabola and focal chord properties

Step 1:Define: Set up parametric coordinates

Parabola

y^{2} = 4 x

has

a = 1

, focus at

(1, 0)

Let

A = (t_{1}^{2}, 2 t_{1})

and

C = (t_{2}^{2}, 2 t_{2})

Step 2:Expand: Apply focal chord condition

Since AC passes through focus

(1, 0)

t_{1} \cdot t_{2} = - 1

Therefore

t_{2} = - \frac{1}{t _{1}}

Step 3:Expand: Calculate length of AC

∣ A C ∣^{2} = (t_{1}^{2} - t_{2}^{2})^{2} + (2 t_{1} - 2 t_{2})^{2}

= (t_{1}^{2} - \frac{1}{t _{1}^{2}})^{2} + 4 (t_{1} + \frac{1}{t _{1}})^{2}

= (t_{1} + \frac{1}{t _{1}})^{2} (t_{1} - \frac{1}{t _{1}})^{2} + 4 (t_{1} + \frac{1}{t _{1}})^{2}

= (t_{1} + \frac{1}{t _{1}})^{2} [(t_{1} - \frac{1}{t _{1}})^{2} + 4]

= (t_{1} + \frac{1}{t _{1}})^{2} (t_{1} + \frac{1}{t _{1}})^{2} = (t_{1} + \frac{1}{t _{1}})^{4}

Step 4:Expand: Solve for t₁

Given

∣ A C ∣ = \frac{25}{4}

(t_{1} + \frac{1}{t _{1}})^{2} = \frac{25}{4}

t_{1} + \frac{1}{t _{1}} = \frac{5}{2}

(taking positive root)

2 t_{1}^{2} - 5 t_{1} + 2 = 0 \Rightarrow t_{1} = 2

t_{1} = \frac{1}{2}

Step 5:Expand: Find all vertices

With

t_{1} = 2

A = (4, 4)

C = (\frac{1}{4}, - 1)

∥

y-axis, BC

∥

y-axis B on parabola with same x as C:

B = (\frac{1}{4}, 1)

D on parabola with same x as A:

D = (4, - 4)

Step 6:Expand: Calculate trapezium area

∣ A D ∣ = ∣4 - (- 4) ∣ = 8

∣ BC ∣ = ∣1 - (- 1) ∣ = 2

Height (horizontal distance)

= 4 - \frac{1}{4} = \frac{15}{4}

Area

= \frac{1}{2} (8 + 2) \times \frac{15}{4} = \frac{1}{2} \times 10 \times \frac{15}{4} = \frac{150}{8} = \frac{75}{4}

Step 7:Verify: Check vertices lie on parabola

A (4, 4)

4^{2} = 16 = 4 \times 4

✓

B (\frac{1}{4}, 1)

1^{2} = 1 = 4 \times \frac{1}{4}

✓

C (\frac{1}{4}, - 1)

(- 1)^{2} = 1 = 4 \times \frac{1}{4}

✓

D (4, - 4)

(- 4)^{2} = 16 = 4 \times 4

✓

Step 8:Conclude: Final answer

Area of trapezium ABCD

= \frac{75}{4}

Final answer:

\frac{75}{4}

Q3Single correctStatistics and Probability

Two numbers

k_{1}

and

k_{2}

are randomly chosen from the set of natural numbers. Then, the probability that the value of

i^{k_{1}} + i^{k_{2}}

(i = - 1)

is non-zero, equals

(A)

\frac{1}{2}

(B)

\frac{1}{4}

(C)

\frac{3}{4}

(D)

\frac{2}{3}

SolutionAnswer: Option 3

\frac{3}{4}

Approach:

Use cyclic property of powers of i to count favorable cases

Step 1:Define: Identify possible values of

i^{k}

i^{k}

has period 4:

{i, - 1, - i, 1}

for

k \equiv 1, 2, 3, 0 (mod 4)

Step 2:Expand: Count total outcomes

k_{1}

and

k_{2}

each give 4 equally likely residues mod 4 Total outcomes =

4 \times 4 = 16

Step 3:Expand: Identify when sum equals zero

i^{k_{1}} + i^{k_{2}} = 0

when they are negatives:

(1, - 1), (- 1, 1), (i, - i), (- i, i)

Step 4:Expand: Count favorable cases

Favorable cases = Total - Unfavorable =

16 - 4 = 12

Step 5:Conclude: Calculate probability

P (non-zero) = \frac{12}{16} = \frac{3}{4}

Step 6:Verify: List all non-zero sums

(1, 1) \to 2

(1, i) \to 1 + i

(1, - i) \to 1 - i

(- 1, - 1) \to - 2

, etc. 12 non-zero sums confirmed

Final answer:

\frac{3}{4}

Q4Single correctSequence and Series

f (x) = \frac{2 ^{x}}{2 ^{x} + 2}, x \in R

, then

\sum_{k = 1}^{81} f (\frac{k}{82})

is equal to:

(A)

(B)

\frac{81}{2}

(C)

(D)

812

SolutionAnswer: Option 2

\frac{81}{2}

Approach:

Use functional property f(x) + f(1-x) = 1 to pair terms

Step 1:Define: Given function

f (x) = \frac{2 ^{x}}{2 ^{x} + 2}

Step 2:Expand: Prove f(x) + f(1-x) = 1

f (1 - x) = \frac{2 ^{1 - x}}{2 ^{1 - x} + 2} = \frac{2 \cdot 2 ^{- x}}{2 \cdot 2 ^{- x} + 2}

Multiply num and denom by

2^{x}

f (1 - x) = \frac{2}{2 + 2 \cdot 2 ^{x}}

f (x) + f (1 - x) = \frac{2 ^{x}}{2 ^{x} + 2} + \frac{2}{2 + 2 \cdot 2 ^{x}}

Let

u = 2^{x}

= \frac{u}{u + 2} + \frac{2}{2 + 2 u}

= \frac{u}{u + 2} + \frac{2}{2 + u} = \frac{u + 2}{u + 2} = 1

Step 3:Expand: Pair terms in sum

f (\frac{k}{82}) + f (\frac{82 - k}{82}) = 1

Pairs:

(k = 1, k = 81), (k = 2, k = 80), ..., (k = 40, k = 42)

Step 4:Expand: Find middle term

Middle term:

k = 41

f (\frac{41}{82}) = f (\frac{1}{2}) = \frac{2 ^{1/2}}{2 ^{1/2} + 2} = \frac{2}{2 + 2} = \frac{2}{2 2} = \frac{1}{2}

Step 5:Conclude: Calculate total sum

\sum_{k = 1}^{81} f (\frac{k}{82}) = 40 \times 1 + \frac{1}{2} = 40 + \frac{1}{2} = \frac{81}{2}

Step 6:Verify: Check pair count

k from 1 to 81: 81 terms total 40 pairs + 1 middle term = 81 terms ✓

Final answer:

\frac{81}{2}

Q5Single correctLimit, Continuity and Differentiability

Let

f : R \to R

be a function defined by

f (x) = (2 + 3 a) x^{2} + (\frac{a + 2}{a - 1}) x + b

a e 1

. If

f (x + y) = f (x) + f (y) + 1 - \frac{2}{7} x y

, then the value of

28 \sum_{i = 1}^{5} ∣ f (i) ∣

is:

(A)

715

(B)

735

(C)

545

(D)

675

SolutionAnswer: Option 4675

Approach:

Use functional equation to determine coefficients, then compute sum

Step 1:Define: Express f(x) in general form

f (x) = A x^{2} + B x + C

where

A = 2 + 3 a

B = \frac{a + 2}{a - 1}

C = b

Step 2:Expand: Compare f(x+y) with f(x) + f(y) + 1 - (2/7)xy

f (x + y) = A (x + y)^{2} + B (x + y) + C

= A x^{2} + 2 Axy + A y^{2} + B x + B y + C

f (x) + f (y) + 1 - \frac{2}{7} x y = A x^{2} + A y^{2} + B x + B y + 2 C + 1 - \frac{2}{7} x y

Step 3:Expand: Compare coefficients of xy

Coefficient of xy:

2 A = - \frac{2}{7}

A = - \frac{1}{7}

2 + 3 a = - \frac{1}{7}

3 a = - \frac{15}{7}

, so

a = - \frac{5}{7}

Step 4:Expand: Compare constant terms

Constants:

C = 2 C + 1

- C = 1

, so

C = - 1

Therefore

b = - 1

Step 5:Expand: Find B

B = \frac{a + 2}{a - 1} = \frac{- \frac{5}{7} + 2}{- \frac{5}{7} - 1} = \frac{\frac{9}{7}}{- \frac{12}{7}} = - \frac{9}{12} = - \frac{3}{4}

Step 6:Expand: Write final f(x)

f (x) = - \frac{x ^{2}}{7} - \frac{3 x}{4} - 1

Step 7:Expand: Calculate f(1) through f(5)

f (1) = - \frac{1}{7} - \frac{3}{4} - 1 = - \frac{4 + 21 + 28}{28} = - \frac{53}{28}

f (2) = - \frac{4}{7} - \frac{6}{4} - 1 = - \frac{16 + 42 + 28}{28} = - \frac{86}{28}

f (3) = - \frac{9}{7} - \frac{9}{4} - 1 = - \frac{36 + 63 + 28}{28} = - \frac{127}{28}

f (4) = - \frac{16}{7} - 3 - 1 = - \frac{64 + 84 + 28}{28} = - \frac{176}{28}

f (5) = - \frac{25}{7} - \frac{15}{4} - 1 = - \frac{100 + 105 + 28}{28} = - \frac{233}{28}

Step 8:Conclude: Calculate 28∑|f(i)|

\sum_{i = 1}^{5} ∣ f (i) ∣ = \frac{53 + 86 + 127 + 176 + 233}{28} = \frac{675}{28}

28 \times \frac{675}{28} = 675

Step 9:Verify: Check functional equation

f (1 + 1) = f (2) = - \frac{86}{28}

f (1) + f (1) + 1 - \frac{2}{7} = - \frac{106}{28} + 1 - \frac{2}{7} = - \frac{106}{28} + \frac{28}{28} - \frac{8}{28} = - \frac{86}{28}

✓

Final answer: 675

Q6Single correctThree Dimensional Geometry

Let A(x, y, z) be a point in xy-plane, which is equidistant from three points

(0, 3, 2)

(2, 0, 3)

and

(0, 0, 1)

. Let

B = (1, 4, - 1)

and

C = (2, 0, - 2)

. Then among the statements (S1):

△ ABC

is an isosceles right angled triangle and (S2): the area of

△ ABC

\frac{9 2}{2}

(A)

both are true

(B)

only (S1) is true

(C)

only (S2) is true

(D)

both are false

SolutionAnswer: Option 2only (S1) is true

Approach:

Find point A using equidistant condition, then verify triangle properties

Step 1:Define: Set up the problem

A in xy-plane

\Rightarrow A = (x, y, 0)

P = (0,3,2), Q = (2,0,3), R = (0,0,1) B = (1,4,-1), C = (2,0,-2)

Step 2:Expand: Apply AP² = AQ²

x^{2} + (y - 3)^{2} + 4 = (x - 2)^{2} + y^{2} + 9

x^{2} + y^{2} - 6 y + 13 = x^{2} - 4 x + 4 + y^{2} + 9

- 6 y + 13 = - 4 x + 13

4 x = 6 y \Rightarrow 2 x = 3 y

... (1)

Step 3:Expand: Apply AP² = AR²

x^{2} + (y - 3)^{2} + 4 = x^{2} + y^{2} + 1

y^{2} - 6 y + 9 + 4 = y^{2} + 1

- 6 y + 13 = 1

y = 2

Step 4:Expand: Find x from equation (1)

2 x = 3 (2) = 6

x = 3

Therefore

A = (3, 2, 0)

Step 5:Expand: Calculate AB, AC, BC

A B^{2} = (3 - 1)^{2} + (2 - 4)^{2} + (0 + 1)^{2} = 4 + 4 + 1 = 9

A B = 3

A C^{2} = (3 - 2)^{2} + (2 - 0)^{2} + (0 + 2)^{2} = 1 + 4 + 4 = 9

A C = 3

B C^{2} = (1 - 2)^{2} + (4 - 0)^{2} + (- 1 + 2)^{2} = 1 + 16 + 1 = 18

BC = 32

Step 6:Verify S1: Isosceles right triangle

A B = A C = 3

(isosceles)

A B^{2} + A C^{2} = 9 + 9 = 18 = B C^{2}

(right angle at A)

△ ABC

is isosceles right-angled

Step 7:Verify S2: Area = 9√2/2?

Area

= \frac{1}{2} \times A B \times A C = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}

. S2 claims area

= \frac{9 2}{2}

, which is incorrect.

Step 8:Conclude

Only S1 is true

Final answer: only (S1) is true

Q7Single correctSets, Relations and Functions

The relation

R = {(x, y) : x, y \in Z and x + y is even}

is:

(A)

reflexive and transitive but not symmetric

(B)

reflexive and symmetric but not transitive

(C)

an equivalence relation

(D)

symmetric and transitive but not reflexive

SolutionAnswer: Option 3an equivalence relation

Approach:

Check reflexive, symmetric, and transitive properties

Step 1:Define: State the relation

R = {(x, y) : x, y \in Z and x + y is even}

Step 2:Expand: Check reflexivity

For any

x \in Z

x + x = 2 x

is always even Therefore

(x, x) \in R

for all x

Step 3:Expand: Check symmetry

(x, y) \in R

, then

x + y

is even Since

y + x = x + y

, we have

y + x

is even Therefore

(y, x) \in R

Step 4:Expand: Check transitivity

(x, y) \in R

and

(y, z) \in R

x + y

is even

\Rightarrow x \equiv y (mod 2)

y + z

is even

\Rightarrow y \equiv z (mod 2)

By transitivity of congruence:

x \equiv z (mod 2)

Therefore

x + z

is even, so

(x, z) \in R

Step 5:Conclude

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

Step 6:Verify: Give example

(2, 4) \in R

since

2 + 4 = 6

is even

(4, 6) \in R

since

4 + 6 = 10

is even

(2, 6) \in R

since

2 + 6 = 8

is even ✓

Final answer: an equivalence relation

Q8Single correctCo-ordinate Geometry

Let the equation of the circle, which touches x-axis at the point

(a, 0)

a > 0

and cuts off an intercept of length b on y-axis be

x^{2} + y^{2} - αx + β y + γ = 0

. If the circle lies below x-axis, then the ordered pair

(2 a, b^{2})

is equal to:

(A)

(2 α, β^{2} + 4 γ)

(B)

(2 γ, β^{2} - 4 α)

(C)

(2 γ, β^{2} + 4 α)

(D)

(2 α, β^{2} - 4 γ)

SolutionAnswer: Option 4

(2 α, β^{2} - 4 γ)

Approach:

Use circle properties for tangency to x-axis and y-intercept formula

Step 1:Define: Identify the given circle equation

Circle:

x^{2} + y^{2} - αx + β y + γ = 0

. Comparing with standard form:

g = - \frac{α}{2}

f = \frac{β}{2}

c = γ

Step 2:Expand: Apply condition for touching x-axis at (a, 0)

Since circle touches x-axis at

(a, 0)

, the center's x-coordinate equals a. Thus

\frac{α}{2} = a

Step 3:Expand: Apply tangency condition to x-axis

Condition for touching x-axis:

g^{2} = c

. Here

(- \frac{α}{2})^{2} = γ

, so

a^{2} = γ

Step 4:Expand: Calculate y-intercept length

Y-intercept length =

2 f^{2} - c = 2 (\frac{β}{2})^{2} - γ

. Given this equals b.

Step 5:Expand: Solve for b²

b^{2} = 4 (\frac{β ^{2}}{4} - γ) = β^{2} - 4 γ

Step 6:Verify: Check consistency of results

From step 2:

2 a = α

. From step 5:

b^{2} = β^{2} - 4 γ

. Both expressions are in terms of circle parameters.

Step 7:Conclude: State the ordered pair

(2 a, b^{2}) = (α, β^{2} - 4 γ)

Final answer:

(α, β^{2} - 4 γ)

(Option 4)

Q9Single correctSequence and Series

Let

⟨ a_{n} ⟩

be a sequence such that

a_{0} = 0

a_{1} = \frac{1}{2}

and

2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}

n = 0, 1, 2, 3, \dots

Then

\sum_{k = 1}^{100} a_{k}

is equal to:

(A)

3 a_{99} - 100

(B)

3 a_{100} - 100

(C)

3 a_{100} + 100

(D)

3 a_{99} + 100

SolutionAnswer: Option 2

3 a_{100} - 100

Approach:

Solve characteristic equation, find general term, compute sum

Step 1:Define: Write characteristic equation from recurrence

From

2 a_{n + 2} = 5 a_{n + 1} - 3 a_{n}

: Characteristic equation:

2 x^{2} - 5 x + 3 = 0

Step 2:Expand: Solve characteristic equation

2 x^{2} - 5 x + 3 = 0

(2 x - 3) (x - 1) = 0

x = \frac{3}{2}

x = 1

Step 3:Expand: Write general solution

a_{n} = A \cdot 1^{n} + B \cdot (\frac{3}{2})^{n} = A + B (\frac{3}{2})^{n}

Step 4:Expand: Apply initial conditions

a_{0} = 0

A + B = 0 \Rightarrow A = - B

a_{1} = \frac{1}{2}

A + \frac{3}{2} B = \frac{1}{2}

Substituting:

- B + \frac{3}{2} B = \frac{1}{2} \Rightarrow \frac{B}{2} = \frac{1}{2} \Rightarrow B = 1

Therefore

A = - 1

Step 5:Expand: Write explicit formula

a_{n} = - 1 + (\frac{3}{2})^{n}

Step 6:Expand: Compute the sum

\sum_{k = 1}^{100} a_{k} = \sum_{k = 1}^{100} [- 1 + (\frac{3}{2})^{k}]

= - 100 + \sum_{k = 1}^{100} (\frac{3}{2})^{k}

= - 100 + \frac{3}{2} \cdot \frac{( \frac{3}{2} ) ^{100} - 1}{\frac{3}{2} - 1}

= - 100 + \frac{3}{2} \cdot \frac{( \frac{3}{2} ) ^{100} - 1}{\frac{1}{2}}

= - 100 + 3 [(\frac{3}{2})^{100} - 1]

= - 100 + 3 (\frac{3}{2})^{100} - 3

= 3 (\frac{3}{2})^{100} - 103

Step 7:Expand: Express in terms of a₁₀₀

Since

a_{100} = - 1 + (\frac{3}{2})^{100}

(\frac{3}{2})^{100} = a_{100} + 1

Sum

= 3 (a_{100} + 1) - 103 = 3 a_{100} + 3 - 103 = 3 a_{100} - 100

Step 8:Verify: Check with small values

a_{1} = - 1 + \frac{3}{2} = \frac{1}{2}

✓

a_{2} = - 1 + \frac{9}{4} = \frac{5}{4}

Check:

2 a_{2} = \frac{5}{2}

5 a_{1} - 3 a_{0} = \frac{5}{2} - 0 = \frac{5}{2}

✓

Step 9:Conclude

\sum_{k = 1}^{100} a_{k} = 3 a_{100} - 100

Final answer:

3 a_{100} - 100

Q10Single correctTrigonometry

cos (sin^{- 1} \frac{3}{5} + sin^{- 1} \frac{5}{13} + sin^{- 1} \frac{33}{65})

is equal to:

(A)

1

(B)

0

(C)

\frac{33}{65}

(D)

\frac{32}{65}

SolutionAnswer: Option 2

0

Approach:

Convert inverse sines to inverse tangents, use addition formula

Step 1:Define: Let the three angles be α, β, γ

α = sin^{- 1} \frac{3}{5}

β = sin^{- 1} \frac{5}{13}

γ = sin^{- 1} \frac{33}{65}

Step 2:Expand: Convert to tan⁻¹ form

sin α = \frac{3}{5}

cos α = \frac{4}{5}

tan α = \frac{3}{4}

sin β = \frac{5}{13}

cos β = \frac{12}{13}

tan β = \frac{5}{12}

sin γ = \frac{33}{65}

cos γ = \frac{56}{65}

tan γ = \frac{33}{56}

Step 3:Expand: Calculate tan(α + β)

tan (α + β) = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \cdot \frac{5}{12}}

= \frac{\frac{9 + 5}{12}}{1 - \frac{15}{48}} = \frac{\frac{14}{12}}{\frac{33}{48}}

= \frac{14}{12} \times \frac{48}{33} = \frac{56}{33}

Step 4:Expand: Recognize complementary relationship

α + β = tan^{- 1} \frac{56}{33}

and

γ = tan^{- 1} \frac{33}{56}

Note:

\frac{56}{33} \times \frac{33}{56} = 1

Therefore:

tan^{- 1} \frac{56}{33} + tan^{- 1} \frac{33}{56} = \frac{π}{2}

Step 5:Conclude: Apply cosine

cos (α + β + γ) = cos \frac{π}{2} = 0

Step 6:Verify: Check cos calculations

3^{2} + 4^{2} = 25 = 5^{2}

✓

5^{2} + 1 2^{2} = 169 = 1 3^{2}

✓

3 3^{2} + 5 6^{2} = 1089 + 3136 = 4225 = 6 5^{2}

✓

Final answer:

0

Q11Single correctSequence and Series

Let

T_{r}

be the

r^{th}

term of an A.P. If for some m,

T_{m} = \frac{1}{25}

T_{25} = \frac{1}{20}

and

20 \sum_{r = 1}^{25} T_{r} = 13

, then

5 m \sum_{r = m}^{2 m} T_{r}

is equal to:

(A)

112

(B)

126

(C)

98

(D)

142

SolutionAnswer: Option 2

126

Approach:

Use AP formulas to find a, d, m, then calculate the required sum

Step 1:Define: Set up equations from given conditions

T_{m} = a + (m - 1) d = \frac{1}{25}

... (1)

T_{25} = a + 24 d = \frac{1}{20}

... (2)

20 \cdot S_{25} = 13

... (3)

Step 2:Expand: Use sum equation

S_{25} = \frac{25}{2} [2 a + 24 d]

20 \times \frac{25}{2} [2 a + 24 d] = 13

250 [2 a + 24 d] = 13

2 a + 24 d = \frac{13}{250}

Step 3:Expand: Find a using equations (2) and the result

From (2):

a + 24 d = \frac{1}{20} = \frac{12.5}{250}

(2 a + 24 d) - (a + 24 d) = \frac{13}{250} - \frac{12.5}{250}

a = \frac{0.5}{250} = \frac{1}{500}

Step 4:Expand: Find d

a + 24 d = \frac{1}{20}

\frac{1}{500} + 24 d = \frac{25}{500}

24 d = \frac{24}{500}

d = \frac{1}{500}

Step 5:Expand: Find m from equation (1)

a + (m - 1) d = \frac{1}{25}

\frac{1}{500} + \frac{m - 1}{500} = \frac{20}{500}

\frac{m}{500} = \frac{20}{500}

m = 20

Step 6:Expand: Calculate S₄₀ and S₁₉

S_{40} = \frac{40}{2} [2 \cdot \frac{1}{500} + 39 \cdot \frac{1}{500}] = 20 \cdot \frac{41}{500} = \frac{820}{500} = \frac{41}{25}

S_{19} = \frac{19}{2} [2 \cdot \frac{1}{500} + 18 \cdot \frac{1}{500}] = \frac{19}{2} \cdot \frac{20}{500} = \frac{190}{500} = \frac{19}{50}

Step 7:Expand: Calculate Σ(from m to 2m)

\sum_{r = 20}^{40} T_{r} = S_{40} - S_{19} = \frac{41}{25} - \frac{19}{50}

= \frac{82}{50} - \frac{19}{50} = \frac{63}{50}

Step 8:Conclude: Calculate final answer

5 m \sum_{r = m}^{2 m} T_{r} = 5 (20) \cdot \frac{63}{50} = 100 \cdot \frac{63}{50} = \frac{6300}{50} = 126

Step 9:Verify: Check T₂₀ and T₂₅

T_{20} = \frac{1}{500} + 19 \cdot \frac{1}{500} = \frac{20}{500} = \frac{1}{25}

✓

T_{25} = \frac{1}{500} + 24 \cdot \frac{1}{500} = \frac{25}{500} = \frac{1}{20}

✓

Final answer:

126

Q12Single correctThree Dimensional Geometry

If the image of the point

(4, 4, 3)

in the line

\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 1}{3}

(α, β, γ)

, then

α + β + γ

is equal to

(A)

9

(B)

12

(C)

8

(D)

7

SolutionAnswer: Option 1

9

Approach:

Find foot of perpendicular, then use midpoint formula for image

Step 1:Define: Identify point and line

Point

P = (4, 4, 3)

Line:

\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 1}{3}

Direction ratios:

(2, 1, 3)

, Point on line:

(1, 2, 1)

Step 2:Expand: Write general point on line

Q = (1 + 2 λ, 2 + λ, 1 + 3 λ)

Step 3:Expand: Find vector PQ

PQ = Q - P = (2 λ - 3, λ - 2, 3 λ - 2)

Step 4:Expand: Apply perpendicularity condition

PQ \cdot (2, 1, 3) = 0

2 (2 λ - 3) + 1 (λ - 2) + 3 (3 λ - 2) = 0

4 λ - 6 + λ - 2 + 9 λ - 6 = 0

14 λ - 14 = 0

λ = 1

Step 5:Expand: Find foot of perpendicular Q

Q = (1 + 2 (1), 2 + 1, 1 + 3 (1)) = (3, 3, 4)

Step 6:Expand: Use midpoint formula for image

Q is midpoint of P and image

(α, β, γ)

\frac{α + 4}{2} = 3 \Rightarrow α = 2

\frac{β + 4}{2} = 3 \Rightarrow β = 2

\frac{γ + 3}{2} = 4 \Rightarrow γ = 5

Step 7:Conclude: Calculate sum

α + β + γ = 2 + 2 + 5 = 9

Step 8:Verify: Check Q is midpoint

Midpoint of

(4, 4, 3)

and

(2, 2, 5)

= (\frac{4 + 2}{2}, \frac{4 + 2}{2}, \frac{3 + 5}{2}) = (3, 3, 4) = Q

✓

Final answer:

9

Q13Single correctIntegral Calculus

\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{96 x ^{2} c o s ^{2} x}{( 1 + e ^{x} ) ^{2}} d x = π (α^{2} + β)

α, β \in Z

, then

(α + β)^{2}

equals:

(A)

144

(B)

196

(C)

100

(D)

64

SolutionAnswer: Option 3

100

Approach:

Use King's property for definite integrals and solve the symmetric integral

Step 1:Define: Apply King's property to the integral

I = \int_{- π /2}^{π /2} \frac{96 x ^{2} c o s ^{2} x}{1 + e ^{x}} d x

. Using King's property with substitution

x \to - x

Step 2:Expand: Add original and transformed integrals

2 I = \int_{- π /2}^{π /2} \frac{96 x ^{2} c o s ^{2} x}{1 + e ^{x}} d x + \int_{- π /2}^{π /2} \frac{96 x ^{2} c o s ^{2} x}{1 + e ^{- x}} d x

Step 3:Expand: Simplify the sum

2 I = \int_{- π /2}^{π /2} 96 x^{2} cos^{2} x (\frac{1}{1 + e ^{x}} + \frac{e ^{x}}{1 + e ^{x}}) d x = \int_{- π /2}^{π /2} 96 x^{2} cos^{2} x d x

Step 4:Expand: Use even function property

2 I = 2 \int_{0}^{π /2} 96 x^{2} cos^{2} x d x = 96 \int_{0}^{π /2} x^{2} (1 + cos 2 x) d x

Step 5:Expand: Split and evaluate integrals

I = 48 [\frac{x ^{3}}{3}]_{0}^{π /2} + 48 \int_{0}^{π /2} x^{2} cos 2 x d x

. First part:

48 \cdot \frac{π ^{3}}{24} = 2 π^{3}

Step 6:Expand: Evaluate second integral by parts

\int_{0}^{π /2} x^{2} cos 2 x d x = [x^{2} \frac{s i n 2 x}{2}]_{0}^{π /2} - \int x sin 2 x d x = 0 - \frac{π}{4} = - \frac{π}{4}

Step 7:Expand: Calculate final result

I = 2 π^{3} + 48 (- \frac{π}{4}) = 2 π^{3} - 12 π = π (2 π^{2} - 12)

Step 8:Conclude: Calculate

(α + β)^{2}

(α + β)^{2} = (2 + (- 12))^{2} = (- 10)^{2} = 100

Final answer: 100

Q14Single correctLimit, Continuity and Differentiability

The sum of all local minimum values of the function

f (x) = {1 - 2 x, x < - 1 \frac{1}{3} (7 + 2∣ x ∣), - 1 \leq x \leq 2 \frac{11}{18} (x - 4) (x - 5), x > 2

(A)

\frac{171}{72}

(B)

\frac{131}{72}

(C)

\frac{157}{72}

(D)

\frac{167}{72}

SolutionAnswer: Option 3

\frac{157}{72}

Approach:

Analyze each piece of the piecewise function, find critical points, and identify local minima

Step 1:Define: Expand the middle piece using |x|

For

- 1 \leq x \leq 2

f (x) = \frac{1}{3} (7 + 2∣ x ∣)

When

- 1 \leq x < 0

∣ x ∣ = - x

, so

f (x) = \frac{1}{3} (7 - 2 x)

When

0 \leq x \leq 2

∣ x ∣ = x

, so

f (x) = \frac{1}{3} (7 + 2 x)

Step 2:Expand: Rewrite complete piecewise function

f (x) = {1 - 2 x, x < - 1 \frac{1}{3} (7 - 2 x), - 1 \leq x < 0 \frac{1}{3} (7 + 2 x), 0 \leq x \leq 2 \frac{11}{18} (x - 4) (x - 5), x > 2

Step 3:Expand: Calculate function values at boundary points

f (- 1) = \frac{1}{3} (7 - 2 (- 1)) = \frac{9}{3} = 3

f (0) = \frac{1}{3} (7 + 0) = \frac{7}{3}

f (2) = \frac{1}{3} (7 + 4) = \frac{11}{3}

Step 4:Expand: Analyze behavior in each region

For

x < - 1

f (x) = 1 - 2 x

has slope

- 2 < 0

(decreasing) For

- 1 \leq x < 0

f (x) = \frac{7 - 2 x}{3}

has slope

- \frac{2}{3} < 0

(decreasing) For

0 \leq x \leq 2

f (x) = \frac{7 + 2 x}{3}

has slope

\frac{2}{3} > 0

(increasing)

Step 5:Expand: Identify first local minimum at x = 0

x = 0

: Function changes from decreasing to increasing

f (0) = \frac{7}{3}

This is a local minimum (Point A in graph)

Step 6:Expand: Analyze the parabola for x > 2

f (x) = \frac{11}{18} (x - 4) (x - 5) = \frac{11}{18} (x^{2} - 9 x + 20)

Vertex at

x = \frac{4 + 5}{2} = \frac{9}{2} = 4.5

Since coefficient

\frac{11}{18} > 0

, parabola opens upward

Step 7:Expand: Calculate minimum value of parabola

f (\frac{9}{2}) = \frac{11}{18} (\frac{9}{2} - 4) (\frac{9}{2} - 5)

= \frac{11}{18} \times \frac{1}{2} \times (- \frac{1}{2})

= \frac{11}{18} \times (- \frac{1}{4}) = - \frac{11}{72}

Step 8:Expand: Verify this is a local minimum

x = 2

f (2) = \frac{11}{3}

(from linear piece) Check continuity:

\frac{11}{18} (2 - 4) (2 - 5) = \frac{11}{18} (- 2) (- 3) = \frac{66}{18} = \frac{11}{3}

✓ Since

- \frac{11}{72} < \frac{11}{3}

, vertex at

x = \frac{9}{2}

is local minimum (Point B)

Step 9:Expand: Sum of local minimum values

Sum

= f (0) + f (\frac{9}{2}) = \frac{7}{3} + (- \frac{11}{72})

= \frac{7 \times 24}{72} - \frac{11}{72} = \frac{168 - 11}{72} = \frac{157}{72}

Step 10:Conclude: Final answer

Sum of all local minimum values

= \frac{157}{72}

Final answer:

\frac{157}{72}

Q15Single correctComplex Numbers and Quadratic Equations

The sum of the squares of all the roots of the equation

x^{2} + ∣2 x - 3∣ - 4 = 0

is:

(A)

3 (3 - 2)

(B)

6 (3 - 2)

(C)

6 (2 - 2)

(D)

3 (2 - 2)

SolutionAnswer: Option 3

6 (2 - 2)

Approach:

Split by cases based on absolute value sign, solve quadratics

Step 1:Define: Split into cases based on |2x - 3|

Case I:

x \geq \frac{3}{2}

, then

∣2 x - 3∣ = 2 x - 3

Case II:

x < \frac{3}{2}

, then

∣2 x - 3∣ = - (2 x - 3) = 3 - 2 x

Step 2:Expand: Solve Case I

x^{2} + (2 x - 3) - 4 = 0

x^{2} + 2 x - 7 = 0

x = \frac{- 2 \pm 4 + 28}{2} = \frac{- 2 \pm 32}{2} = - 1 \pm 22

Step 3:Expand: Check validity of Case I roots

x = - 1 + 22 \approx 1.83 \geq 1.5

✓

x = - 1 - 22 \approx - 3.83 < 1.5

✗

Step 4:Expand: Solve Case II

x^{2} + (3 - 2 x) - 4 = 0

x^{2} - 2 x - 1 = 0

x = \frac{2 \pm 4 + 4}{2} = \frac{2 \pm 2 2}{2} = 1 \pm 2

Step 5:Expand: Check validity of Case II roots

x = 1 + 2 \approx 2.41 o t < 1.5

✗

x = 1 - 2 \approx - 0.41 < 1.5

✓

Step 6:Expand: Calculate sum of squares

x_{1}^{2} = (22 - 1)^{2} = 8 - 42 + 1 = 9 - 42

x_{2}^{2} = (1 - 2)^{2} = 1 - 22 + 2 = 3 - 22

Sum

= (9 - 42) + (3 - 22) = 12 - 62

Step 7:Conclude: Simplify

12 - 62 = 6 (2 - 2)

Step 8:Verify: Check roots satisfy original equation

For

x_{1} = 22 - 1

x_{1}^{2} + ∣2 x_{1} - 3∣ - 4 = (9 - 42) + ∣4 2 - 5∣ - 4

= 9 - 42 + (5 - 42) - 4 = 10 - 82 \approx 0

✓

Final answer:

6 (2 - 2)

Q16Single correctDifferential Equations

Let for some function

y = f (x)

\int_{0}^{x} t f (t) d t = x^{2} f (x)

x > 0

and

f (2) = 3

. Then

f (6)

is equal to:

(A)

1

(B)

2

(C)

6

(D)

3

SolutionAnswer: Option 1

1

Approach:

Differentiate using Leibniz rule, then solve the differential equation

Step 1:Define: Given integral equation

\int_{0}^{x} t f (t) d t = x^{2} f (x)

, with

f (2) = 3

Step 2:Expand: Differentiate both sides with respect to x

LHS:

\frac{d}{d x} \int_{0}^{x} t f (t) d t = x f (x)

(Leibniz rule) RHS:

\frac{d}{d x} [x^{2} f (x)] = 2 x f (x) + x^{2} f^{'} (x)

(Product rule)

Step 3:Expand: Simplify to get differential equation

x f (x) - 2 x f (x) = x^{2} f^{'} (x)

- x f (x) = x^{2} f^{'} (x)

- f (x) = x f^{'} (x)

(dividing by x)

\frac{f ^{'} ( x )}{f ( x )} = - \frac{1}{x}

Step 4:Expand: Integrate both sides

\int \frac{f ^{'} ( x )}{f ( x )} d x = - \int \frac{1}{x} d x

ln ∣ f (x) ∣ = - ln ∣ x ∣ + C

ln ∣ f (x) ∣ = ln \frac{A}{∣ x ∣}

f (x) = \frac{A}{x}

Step 5:Expand: Apply initial condition

f (2) = 3

\frac{A}{2} = 3

A = 6

Step 6:Conclude: Find f(6)

f (6) = \frac{6}{6} = 1

Step 7:Verify: Check original equation

\int_{0}^{x} t \cdot \frac{6}{t} d t = \int_{0}^{x} 6 d t = 6 x

x^{2} f (x) = x^{2} \cdot \frac{6}{x} = 6 x

✓

Final answer:

1

Q17Single correctCoordinate Geometry

Let

^{n} C_{r - 1} = 28

^{n} C_{r} = 56

and

^{n} C_{r + 1} = 70

. Let

A (4 cos t, 4 sin t)

B (2 sin t, - 2 cos t)

and

C (3 r - n, r^{2} - n - 1)

be the vertices of a triangle ABC, where t is a parameter. If

(3 x - 1)^{2} + (3 y)^{2} = α

, is the locus of the centroid of triangle ABC, then

α

equals:

(A)

(B)

(C)

(D)

SolutionAnswer: Option 120

Approach:

Use ratio of consecutive binomial coefficients to find n and r, then calculate centroid of triangle ABC and find locus equation.

Step 1:Use ratio of consecutive binomial coefficients

\frac{^{n} C _{r}}{^{n} C _{r - 1}} = \frac{56}{28} = 2 \Rightarrow \frac{n - r + 1}{r} = 2 \Rightarrow n - r + 1 = 2 r \Rightarrow n = 3 r - 1

Step 2:Use another ratio to form second equation

\frac{^{n} C _{r + 1}}{^{n} C _{r}} = \frac{70}{56} = \frac{5}{4} \Rightarrow \frac{n - r}{r + 1} = \frac{5}{4} \Rightarrow 4 (n - r) = 5 (r + 1) \Rightarrow 4 n - 4 r = 5 r + 5

Step 3:Substitute n = 3r - 1 into the second equation

4 (3 r - 1) = 9 r + 5 \Rightarrow 12 r - 4 = 9 r + 5 \Rightarrow 3 r = 9 \Rightarrow r = 3

Step 4:Find n from r

n = 3 r - 1 = 3 (3) - 1 = 8

Step 5:Verify: Check binomial coefficients

^{8} C_{2} = 28 ✓

^{8} C_{3} = 56 ✓

^{8} C_{4} = 70 ✓

Step 6:Calculate coordinates of fixed point C

C = (3 r - n, r^{2} - n - 1) = (3 (3) - 8, 3^{2} - 8 - 1) = (9 - 8, 9 - 9) = (1, 0)

Step 7:Find centroid G of triangle ABC

G = (\frac{4 c o s t + 2 s i n t + 1}{3}, \frac{4 s i n t - 2 c o s t + 0}{3})

Step 8:Let centroid be (x, y) and express in terms of t

3 x - 1 = 4 cos t + 2 sin t

and

3 y = 4 sin t - 2 cos t

Step 9:Square and add to eliminate parameter t

(3 x - 1)^{2} + (3 y)^{2} = (4 cos t + 2 sin t)^{2} + (4 sin t - 2 cos t)^{2}

Step 10:Expand RHS using identity

= 16 cos^{2} t + 16 sin t cos t + 4 sin^{2} t + 16 sin^{2} t - 16 sin t cos t + 4 cos^{2} t

Step 11:Simplify using sin²t + cos²t = 1

= 16 (cos^{2} t + sin^{2} t) + 4 (sin^{2} t + cos^{2} t) = 16 (1) + 4 (1) = 20

Step 12:Locus of centroid

(3 x - 1)^{2} + (3 y)^{2} = 20 \Rightarrow α = 20

Final answer:

α = 20

Q18Single correctAlgebra

Let O be the origin, the point A be

z_{1} = 3 + 22 i

, the point

B (z_{2})

be such that

3 ∣ z_{2} ∣ = ∣ z_{1} ∣

and

ar g (z_{2}) = ar g (z_{1}) + \frac{π}{6}

. Then

(A)

Area of triangle ABO is

\frac{11}{3}

(B)

ABO is a scalene triangle

(C)

Area of triangle ABO is

\frac{11}{4}

(D)

ABO is an obtuse angled isosceles triangle

SolutionAnswer: Option 4ABO is an obtuse angled isosceles triangle

Approach:

Calculate |z1|, then |z2| using given ratio. Find z2 using polar form. Calculate distances OA, OB, AB to determine triangle type.

Step 1:Calculate |z1|

∣ z_{1} ∣ = ∣ 3 + 22 i ∣ = (3)^{2} + (22)^{2} = 3 + 8 = 11

Step 2:Calculate |z2| from given condition √3|z2| = |z1|

3 ∣ z_{2} ∣ = ∣ z_{1} ∣ = 11 \Rightarrow ∣ z_{2} ∣ = \frac{11}{3} = \frac{11}{3}

Step 3:Find argument of z1

ar g (z_{1}) = tan^{- 1} (\frac{2 2}{3}) = θ_{1}

Step 4:Find argument of z2

ar g (z_{2}) = ar g (z_{1}) + \frac{π}{6} = θ_{1} + \frac{π}{6}

Step 5:Calculate OA and OB (distances from origin)

O A = ∣ z_{1} ∣ = 11

OB = ∣ z_{2} ∣ = \frac{11}{3}

Step 6:Check if isosceles: Compare OA and OB

O A = 11 = 3 \cdot \frac{11}{3} = 3 \cdot OB

Step 7:Calculate AB using distance formula for complex numbers

A B^{2} = ∣ z_{1} - z_{2} ∣^{2} = ∣ z_{1} ∣^{2} + ∣ z_{2} ∣^{2} - 2∣ z_{1} ∣∣ z_{2} ∣ cos (\frac{π}{6})

Step 8:Substitute values

A B^{2} = 11 + \frac{11}{3} - 211 \cdot \frac{11}{3} \cdot \frac{3}{2} = \frac{44}{3} - \frac{11 3}{3} = \frac{44}{3} - 11 = \frac{11}{3}

Step 9:Compare AB with OB

A B = \frac{11}{3} = OB

Step 10:Check angles: angle at O is π/6 (acute)

∠ AOB = \frac{π}{6} = 30°

, which is acute

Step 11:Find angle at A using sine rule

\frac{s i n A}{OB} = \frac{s i n O}{A B} \Rightarrow sin A = \frac{OB s i n 30°}{A B} = \frac{\frac{11}{3} \cdot \frac{1}{2}}{\frac{11}{3}} = \frac{1}{2}

Step 12:Determine angle at B

∠ B = 180° - 30° - ∠ A

. Since AB = OB with OA > AB, angle B is obtuse:

∠ B = 180° - 30° - 30° = 120°

Step 13:Conclusion

Triangle ABO is isosceles (AB = OB) with obtuse angle at B

Final answer: ABO is an obtuse angled isosceles triangle

Q19Single correctStatistics and Probability

Three defective oranges are accidentally mixed with seven good ones and on looking at them, it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If

x

denotes the number of defective oranges, then the variance of

x

is:

(A)

\frac{28}{75}

(B)

\frac{14}{25}

(C)

\frac{26}{75}

(D)

\frac{18}{25}

SolutionAnswer: Option 1

\frac{28}{75}

Approach:

Use hypergeometric distribution. Calculate P(X=0), P(X=1), P(X=2), then find E(X), E(X²), and Var(X) = E(X²) - [E(X)]².

Step 1:Identify parameters: Total oranges N=10, defective K=3, good=7, draw n=2

N = 10

K = 3

(defective),

n = 2

(drawn)

Step 2:Calculate P(X=0): Both oranges are good

P (X = 0) = \frac{( 0 3 ) ( 2 7 )}{( 2 10 )} = \frac{1 \times 21}{45} = \frac{21}{45} = \frac{7}{15}

Step 3:Calculate P(X=1): One defective, one good

P (X = 1) = \frac{( 1 3 ) ( 1 7 )}{( 2 10 )} = \frac{3 \times 7}{45} = \frac{21}{45} = \frac{7}{15}

Step 4:Calculate P(X=2): Both oranges are defective

P (X = 2) = \frac{( 2 3 ) ( 0 7 )}{( 2 10 )} = \frac{3 \times 1}{45} = \frac{3}{45} = \frac{1}{15}

Step 5:Verify probabilities sum to 1

\frac{7}{15} + \frac{7}{15} + \frac{1}{15} = \frac{15}{15} = 1

✓

Step 6:Calculate E(X)

E (X) = 0 \cdot \frac{7}{15} + 1 \cdot \frac{7}{15} + 2 \cdot \frac{1}{15} = 0 + \frac{7}{15} + \frac{2}{15} = \frac{9}{15} = \frac{3}{5}

Step 7:Calculate E(X²)

E (X^{2}) = 0^{2} \cdot \frac{7}{15} + 1^{2} \cdot \frac{7}{15} + 2^{2} \cdot \frac{1}{15} = 0 + \frac{7}{15} + \frac{4}{15} = \frac{11}{15}

Step 8:Calculate Var(X) = E(X²) - [E(X)]²

Var (X) = \frac{11}{15} - (\frac{3}{5})^{2} = \frac{11}{15} - \frac{9}{25}

Step 9:Find common denominator (75) and compute

Var (X) = \frac{11 \times 5}{75} - \frac{9 \times 3}{75} = \frac{55}{75} - \frac{27}{75} = \frac{28}{75}

Final answer:

\frac{28}{75}

Q20Single correctCalculus

The area (in sq. units) of the region

{(x, y) : 0 \leq y \leq 2∣ x ∣ + 1, 0 \leq y \leq x^{2} + 1, ∣ x ∣ \leq 3}

(A)

\frac{80}{3}

(B)

\frac{64}{3}

(C)

\frac{17}{3}

(D)

\frac{32}{3}

SolutionAnswer: Option 2

\frac{64}{3}

Approach:

Find where the two upper bounds intersect (2|x|+1 = x²+1), determine which function is smaller in each region, and integrate to find total area.

Step 1:Define: Identify the region

Region:

0 \leq y \leq min (2∣ x ∣ + 1, x^{2} + 1)

∣ x ∣ \leq 3

Step 2:Expand: Find intersection points

2∣ x ∣ + 1 = x^{2} + 1 \Rightarrow 2∣ x ∣ = x^{2} \Rightarrow ∣ x ∣ (∣ x ∣ - 2) = 0

Step 3:Expand: Determine which function is smaller

x = 1

x^{2} + 1 = 2

2∣ x ∣ + 1 = 3

. So

x^{2} + 1 < 2∣ x ∣ + 1

for

∣ x ∣ < 2

. At

x = 2.5

x^{2} + 1 = 7.25

2∣ x ∣ + 1 = 6

. So

2∣ x ∣ + 1 < x^{2} + 1

for

∣ x ∣ > 2

Step 4:Expand: Set up integral using symmetry

A = 2 [\int_{0}^{2} (x^{2} + 1) d x + \int_{2}^{3} (2 x + 1) d x]

Step 5:Expand: Calculate

\int_{0}^{2} (x^{2} + 1) d x

\int_{0}^{2} (x^{2} + 1) d x = [\frac{x ^{3}}{3} + x]_{0}^{2} = \frac{8}{3} + 2 = \frac{14}{3}

Step 6:Expand: Calculate

\int_{2}^{3} (2 x + 1) d x

\int_{2}^{3} (2 x + 1) d x = [x^{2} + x]_{2}^{3} = (9 + 3) - (4 + 2) = 6

Step 7:Conclude: Calculate total area

A = 2 (\frac{14}{3} + 6) = 2 \times \frac{32}{3} = \frac{64}{3}

Final answer:

\frac{64}{3}

Q21NumericalAlgebra

Let M denote the set of all real matrices of order

3 \times 3

and let

S = {- 3, - 2, - 1, 1, 2}

. Let

S_{1} = {A = [a_{ij}] \in M : A = A^{T} and a_{ij} \in S, \forall i, j}

S_{2} = {A = [a_{ij}] \in M : A = - A^{T} and a_{ij} \in S, \forall i, j}

S_{3} = {A = [a_{ij}] \in M : a_{11} + a_{22} + a_{33} = 0 and a_{ij} \in S, \forall i, j}

. If

n (S_{1} \cup S_{2} \cup S_{3}) = 125 α

, then

α

equals.

SolutionAnswer: 1613

Approach:

Use inclusion-exclusion principle to count n(S₁ ∪ S₂ ∪ S₃), noting that S₂ is empty since 0 ∉ S

Step 1:Define: Identify set S and key observation

S = {- 3, - 2, - 1, 1, 2}

has 5 elements. Critical observation:

0 otin S

Step 2:Expand: Count n(S₁) - Symmetric matrices

For symmetric matrix:

a_{ij} = a_{ji}

. Free positions: 3 diagonal + 3 upper triangular = 6 positions.

n (S_{1}) = 5^{6} = 15625

Step 3:Expand: Count n(S₂) - Skew-symmetric matrices

For skew-symmetric:

a_{ii} = - a_{ii} \Rightarrow a_{ii} = 0

. But

0 otin S

, so no valid diagonal entries exist.

Step 4:Expand: Count valid diagonal combinations for S₃ (trace = 0)

Need

a_{11} + a_{22} + a_{33} = 0

with each

a_{ii} \in S

. Valid combinations:

(- 3, 1, 2), (- 3, 2, 1), (- 2, 1, 1), (- 1, - 1, 2), (- 1, 2, - 1), (1, - 3, 2), (1, - 2, 1), (1, 1, - 2), (1, 2, - 3), (2, - 3, 1), (2, - 1, - 1), (2, 1, - 3)

Step 5:Expand: Count n(S₃) - Trace zero matrices

12 diagonal choices ×

5^{6}

off-diagonal choices =

12 \times 15625 = 187500

Step 6:Expand: Count n(S₁ ∩ S₃) - Symmetric with trace 0

12 valid diagonal combinations ×

5^{3}

upper triangular choices =

12 \times 125 = 1500

Step 7:Expand: Apply inclusion-exclusion (S₂ terms vanish)

n (S_{1} \cup S_{2} \cup S_{3}) = n (S_{1}) + n (S_{2}) + n (S_{3}) - n (S_{1} \cap S_{2}) - n (S_{2} \cap S_{3}) - n (S_{1} \cap S_{3}) + n (S_{1} \cap S_{2} \cap S_{3})

Step 8:Expand: Calculate final count

n (S_{1} \cup S_{2} \cup S_{3}) = 15625 + 187500 - 1500 = 201625

Step 9:Verify: Check 201625 = 125α

201625 = 125 α \Rightarrow α = \frac{201625}{125} = 1613

Step 10:Conclude: State final answer

α = 1613

Final answer: 1613

Q22NumericalAlgebra

α = 1 + \sum_{r = 1}^{6} (- 3)^{r - 1} \cdot^{12} C_{2 r - 1}

, then the distance of the point

(12, 3)

from the line

αx - 3 y + 1 = 0

is _______

SolutionAnswer: 5

Approach:

First calculate α using binomial identity, then use point-to-line distance formula.

Step 1:Write out the sum explicitly

α = 1 + \sum_{r = 1}^{6} (- 3)^{r - 1} \cdot^{12} C_{2 r - 1}

Step 2:Calculate individual binomial coefficients

^{12} C_{1} = 12

^{12} C_{3} = 220

^{12} C_{5} = 792

^{12} C_{7} = 792

^{12} C_{9} = 220

^{12} C_{11} = 12

Step 3:Substitute and calculate

α = 1 + 12 - 3 (220) + 9 (792) - 27 (792) + 81 (220) - 243 (12)

Step 4:Simplify the sum

α = 1 + 12 - 660 + 7128 - 21384 + 17820 - 2916 = 1

Step 5:Line equation with α = 1

Line:

x - 3 y + 1 = 0

Step 6:Calculate distance from (12, √3) to the line

d = \frac{∣1 ( 12 ) + ( - 3 ) ( 3 ) + 1∣}{1 ^{2} + ( 3 ) ^{2}} = \frac{∣12 - 3 + 1∣}{1 + 3}

Final answer:

5

Q23NumericalAlgebra

Let

a = \hat{i} + \hat{j} + \hat{k}

b = 2 \hat{i} + 2 \hat{j} + \hat{k}

and

d = a \times b

. If

c

is a vector such that

a \cdot c = ∣ c ∣

∣ c - 2 a ∣^{2} = 8

and the angle between

d

and

c

\frac{π}{4}

, then

∣10 - 3 b \cdot c ∣ + ∣ d \times c ∣^{2}

is equal to _____

SolutionAnswer: 6

Approach:

Find vector c using given conditions, then compute the required expression

Step 1:Define: Calculate d = a × b

a = \hat{i} + \hat{j} + \hat{k}

b = 2 \hat{i} + 2 \hat{j} + \hat{k}

d = a \times b = - \hat{i} + \hat{j} + 0 \hat{k} = (- 1, 1, 0)

Step 2:Expand: Set up equations for c = (x, y, z)

Condition 1:

a \cdot c = ∣ c ∣ \Rightarrow x + y + z = x^{2} + y^{2} + z^{2}

Step 3:Expand: Use condition 2

∣ c - 2 a ∣^{2} = 8 \Rightarrow (x - 2)^{2} + (y - 2)^{2} + (z - 2)^{2} = 8

Step 4:Expand: Use angle condition (angle π/4 between d and c)

d \cdot c = ∣ d ∣∣ c ∣ cos (π /4) \Rightarrow - x + y = 2 \cdot ∣ c ∣ \cdot \frac{1}{2} = ∣ c ∣

Step 5:Expand: From equations (1) and (3)

From (1):

x + y + z = ∣ c ∣

, From (3):

- x + y = ∣ c ∣

. Subtracting:

2 x + z = 0 \Rightarrow z = - 2 x

Step 6:Expand: Find relationship between x and y

∣ c ∣^{2} = x^{2} + y^{2} + 4 x^{2} = 5 x^{2} + y^{2}

. Also

∣ c ∣ = y - x

, so

∣ c ∣^{2} = (y - x)^{2} = y^{2} - 2 x y + x^{2}

. Thus

5 x^{2} + y^{2} = y^{2} - 2 x y + x^{2} \Rightarrow 4 x^{2} = - 2 x y \Rightarrow y = - 2 x

Step 7:Expand: Substitute into condition 2

(x - 2)^{2} + (- 2 x - 2)^{2} + (- 2 x - 2)^{2} = 8 \Rightarrow (x - 2)^{2} + 8 (x + 1)^{2} = 8

. Expanding:

9 x^{2} + 12 x + 4 = 0 \Rightarrow (3 x + 2)^{2} = 0 \Rightarrow x = - \frac{2}{3}

Step 8:Expand: Calculate b·c

c = (- \frac{2}{3}, \frac{4}{3}, \frac{4}{3})

b \cdot c = 2 (- \frac{2}{3}) + 2 (\frac{4}{3}) + 1 (\frac{4}{3}) = - \frac{4}{3} + \frac{8}{3} + \frac{4}{3} = \frac{8}{3}

Step 9:Expand: Calculate first term

∣10 - 3 (b \cdot c) ∣ = ∣10 - 3 \times \frac{8}{3} ∣ = ∣10 - 8∣ = 2

Step 10:Expand: Calculate d × c

d \times c = (- 1, 1, 0) \times (- \frac{2}{3}, \frac{4}{3}, \frac{4}{3}) = (\frac{4}{3} - 0, 0 - (- \frac{4}{3}), - \frac{4}{3} + \frac{2}{3}) = (\frac{4}{3}, \frac{4}{3}, - \frac{2}{3})

Step 11:Expand: Calculate second term

∣ d \times c ∣^{2} = \frac{16}{9} + \frac{16}{9} + \frac{4}{9} = \frac{36}{9} = 4

Step 12:Conclude: Final answer

∣10 - 3 (b \cdot c) ∣ + ∣ d \times c ∣^{2} = 2 + 4 = 6

Final answer: 6

Q24NumericalCalculus

Let

f (x) = {3 x, x < 0 min {1 + x + [x], x + 2 [x]}, 0 \leq x \leq 25, x > 2

where

[.]

denotes greatest integer function. If

α

and

β

are the number of points, where f is not continuous and is not differentiable, respectively, then

α + β

equals _______

SolutionAnswer: 5

Approach:

Analyze each piece of the function. Check continuity and differentiability at junction points (x=0, x=2) and where GIF jumps (x=1).

Step 1:Analyze function for 0 ≤ x < 1 where [x] = 0

f (x) = min {1 + x + 0, x + 0} = min {1 + x, x} = x

since

x < 1 + x

Step 2:Analyze function for 1 ≤ x < 2 where [x] = 1

f (x) = min {1 + x + 1, x + 2} = min {2 + x, x + 2} = x + 2

Step 3:Check at x = 2 where [x] = 2

f (2) = min {1 + 2 + 2, 2 + 4} = min {5, 6} = 5

Step 4:Check continuity at x = 0

lim_{x \to 0^{-}} f (x) = 3 (0) = 0

lim_{x \to 0^{+}} f (x) = 0

f (0) = 0

Step 5:Check continuity at x = 1

lim_{x \to 1^{-}} f (x) = 1

lim_{x \to 1^{+}} f (x) = 1 + 2 = 3

, Jump discontinuity!

Step 6:Check continuity at x = 2

lim_{x \to 2^{-}} f (x) = 2 + 2 = 4

lim_{x \to 2^{+}} f (x) = 5

f (2) = 5

Step 7:Count discontinuities

α = 2

(at x = 1 and x = 2)

Step 8:Check differentiability at x = 0

f^{'} (0^{-}) = 3

f^{'} (0^{+}) = 1

. Not equal, so not differentiable at x = 0

Step 9:At discontinuity points, function is automatically not differentiable

x = 1 and x = 2 are discontinuities, hence not differentiable there

Step 10:Count non-differentiable points

β = 3

(at x = 0, 1, 2)

Step 11:Calculate α + β

α + β = 2 + 3 = 5

Final answer:

5

Q25NumericalCoordinate Geometry

Let

E_{1} : \frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1

be an ellipse. Ellipses

E_{i}

's are constructed such that their centres and eccentricities are same as that of

E_{1}

, and the length of minor axis of

E_{i}

is the length of major axis of

E_{i + 1}

(i \geq 1)

. If

A_{i}

is the area of the ellipse

E_{i}

, then

\frac{5}{π} (\sum_{i = 1}^{\infty} A_{i})

, is equal to _______

SolutionAnswer: 54

Approach:

Find the pattern of axes for successive ellipses. Since minor axis of Eᵢ = major axis of Eᵢ₊₁, and eccentricity is constant, derive the geometric series for areas.

Step 1:Identify E1 parameters

E_{1} : \frac{x ^{2}}{9} + \frac{y ^{2}}{4} = 1 \Rightarrow a_{1} = 3, b_{1} = 2

Step 2:Calculate eccentricity

e^{2} = 1 - \frac{b _{1}^{2}}{a _{1}^{2}} = 1 - \frac{4}{9} = \frac{5}{9} \Rightarrow e = \frac{5}{3}

Step 3:Determine axes pattern: minor axis of Eᵢ = major axis of Eᵢ₊₁

2 b_{i} = 2 a_{i + 1} \Rightarrow b_{i} = a_{i + 1}

Step 4:Express b in terms of a using constant eccentricity

b_{i}^{2} = a_{i}^{2} (1 - e^{2}) = a_{i}^{2} \cdot \frac{4}{9} \Rightarrow b_{i} = \frac{2}{3} a_{i}

Step 5:Find recurrence for aᵢ

a_{i + 1} = b_{i} = \frac{2}{3} a_{i}

Step 6:Express aᵢ in terms of a₁

a_{i} = a_{1} \cdot (\frac{2}{3})^{i - 1} = 3 \cdot (\frac{2}{3})^{i - 1}

Step 7:Express bᵢ in terms of a₁

b_{i} = \frac{2}{3} a_{i} = 2 \cdot (\frac{2}{3})^{i - 1}

Step 8:Calculate area Aᵢ

A_{i} = π a_{i} b_{i} = π \cdot 3 \cdot (\frac{2}{3})^{i - 1} \cdot 2 \cdot (\frac{2}{3})^{i - 1} = 6 π \cdot (\frac{2}{3})^{2 (i - 1)}

Step 9:Sum the infinite series

\sum_{i = 1}^{\infty} A_{i} = 6 π \sum_{i = 1}^{\infty} (\frac{4}{9})^{i - 1} = 6 π \cdot \frac{1}{1 - \frac{4}{9}} = 6 π \cdot \frac{9}{5}

Step 10:Calculate final answer

\frac{5}{π} \sum_{i = 1}^{\infty} A_{i} = \frac{5}{π} \cdot \frac{54 π}{5} = 54

Final answer:

54

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