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JEE Main 2026 January 28, Shift 1 Question Paper with Solutions

All 74 questions from the JEE Main 2026 (January 28, Shift 1) shift — Physics (25), Chemistry (25) and Mathematics (24) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.

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Physics25 questions

Q26Single correctLaws of Motion

A block of mass 5 kg is moving on an inclined plane which makes an angle of

3 0^{\circ}

with the horizontal. Friction coefficient between the block and the inclined plane surface is

\frac{3}{2}

. The force to be applied on the block so that the block will move down without acceleration is ______ N.

(g = 10 m / s^{2})

(A)

15

(B)

7.5

(C)

12.5

(D)

25

SolutionAnswer: Option 3

12.5

Approach:

Givens:

m = 5 kg

θ = 3 0^{\circ}

μ = \frac{3}{2}

g = 10 m/s^{2}

. Target: the applied force along the incline that produces zero acceleration while the block slides down. Principle: resolve forces along and perpendicular to the incline; since the block moves down, kinetic friction acts up the incline.

Step 1:Compute the gravity component along the incline acting down the plane.

m g sin θ = 5 \times 10 \times sin 3 0^{\circ} = 50 \times \frac{1}{2}

Step 2:Compute the maximum kinetic friction, which acts up the incline opposing downward sliding.

μ m g cos θ = \frac{3}{2} \times 5 \times 10 \times cos 3 0^{\circ} = \frac{3}{2} \times 50 \times \frac{3}{2} = 50 \times \frac{3}{4}

Step 3:Friction up the incline (

37.5 N

) exceeds gravity down the incline (

25 N

); for sliding down at constant velocity an extra force down the incline is required to balance the net up-slope force.

F = μ m g cos θ - m g sin θ = 37.5 - 25

Final answer:

12.5 N

Q27Single correctOptics

Given below are two statements:

Statement I: A plane wave after passing through prism remains as plane wave but passing through small pin hole may become spherical wave.

Statement II: The curvature of a spherical wave emerging from a slit will increase for increasing slit width.

In the light of the above statements, choose the

correc t

answer from the options given below:

(A)

Both Statement I and Statement II are true

(B)

Statement I is true but Statement II is false

(C)

Statement I is false but Statement II is true

(D)

Both Statement I and Statement II are false

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

Approach:

Two statements concern how a wavefront is reshaped by optical elements. Huygens' principle, in which each point of a wavefront launches secondary spherical wavelets whose envelope forms the next wavefront, fixes the outcome for a prism, a pinhole and a slit.

Step 1:A prism refracts a plane wavefront uniformly, tilting and slowing it while keeping it planar, so a plane wave stays plane after a prism.

prism : plane wave \to plane wave

Step 2:A pinhole is an aperture far smaller than the wavefront, so it acts as a single secondary source and emits an essentially spherical wave; combining both facts, Statement I is correct.

pinhole : plane wave \to spherical wave

Step 3:For a slit, diffraction and hence wavefront curvature grow as the slit narrows and diminish as the slit widens. Statement II asserts curvature increases with increasing slit width, the reverse of the true dependence, so it is false.

slit width ↑ \Rightarrow curvature ↓

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

Q28Single correctOptics

The magnitudes of power of a biconvex lens (refractive index 1.5) and that of a plano-concave lens (refractive index = 1.7) are same. If the curvature of plano-concave lens exactly matches with the curvature of back surface of the biconvex lens, then ratio of radius of curvature of front and back surface of the biconvex lens is ______.

(A)

12 : 5

(B)

5 : 2

(C)

2 : 5

(D)

5 : 12

SolutionAnswer: Option 2

5 : 2

Approach:

Givens: biconvex lens of

μ = 1.5

with surface radii of magnitude

R_{1}

(front) and

R_{2}

(back); plano-concave lens of

μ = 1.7

whose curved face matches the back surface, so its curved radius is

R_{2}

; the two powers are equal in magnitude. Target:

R_{1} : R_{2}

. Principle: lens maker's formula with magnitudes of surface contributions.

Step 1:For the biconvex lens both surfaces converge light, so the magnitudes of the two surface contributions add.

∣ P_{1} ∣ = (1.5 - 1) (\frac{1}{R _{1}} + \frac{1}{R _{2}}) = 0.5 (\frac{1}{R _{1}} + \frac{1}{R _{2}})

Step 2:For the plano-concave lens one face is flat (

R = \infty

) and the curved face has radius

R_{2}

∣ P_{2} ∣ = (1.7 - 1) (\frac{1}{R _{2}}) = 0.7 (\frac{1}{R _{2}})

Step 3:Equate the magnitudes of the two powers and isolate the

R_{1}

term.

0.5 (\frac{1}{R _{1}} + \frac{1}{R _{2}}) = \frac{0.7}{R _{2}} \Rightarrow \frac{0.5}{R _{1}} = \frac{0.7 - 0.5}{R _{2}} = \frac{0.2}{R _{2}}

Step 4:Rearrange to the ratio of radii.

\frac{R _{1}}{R _{2}} = \frac{0.5}{0.2} = \frac{5}{2}

Final answer:

5 : 2

Q29Single correctCurrent Electricity

In the potentiometer, when the cell in the secondary circuit is shunted with

4 Ω

resistance, the balance is obtained at the length 120 cm of wire. Now when the same cell is shunted with

12 Ω

resistance, the balance is shifted to a length of 180 cm. The internal resistance of cell is ______

Ω

(A)

3

(B)

4

(C)

12

(D)

6

SolutionAnswer: Option 2

4

Approach:

Givens: a cell shunted by

R_{1} = 4 Ω

balances at

l_{1} = 120 cm

and by

R_{2} = 12 Ω

balances at

l_{2} = 180 cm

. Target: internal resistance r. Principle: the balance length is proportional to the terminal voltage

\frac{ER}{R + r}

across the shunt.

Step 1:Form the ratio of the two balance conditions, the emf cancelling.

\frac{l _{1}}{l _{2}} = \frac{R _{1} / ( R _{1} + r )}{R _{2} / ( R _{2} + r )} = \frac{R _{1} ( R _{2} + r )}{R _{2} ( R _{1} + r )}

Step 2:Simplify both sides.

\frac{2}{3} = \frac{12 + r}{3 ( 4 + r )}

Step 3:Cross-multiply and expand.

2 \times 3 (4 + r) = 3 (12 + r) \Rightarrow 6 (4 + r) = 3 (12 + r)

Step 4:Solve for the internal resistance.

3 r = 12

Final answer:

4 Ω

Q30Single correctKinematics

Water drops fall from a tap on the floor, 5 m below at regular intervals of time. The first drop strikes the floor when the sixth drop begins to fall. The height at which the fourth drop will be from ground at the instant when the first drop strikes the ground is ______ m.

(g = 10 m / s^{2})

(A)

3.8

(B)

4.2

(C)

4.0

(D)

2.5

SolutionAnswer: Option 2

4.2

Approach:

Givens: drops fall from a tap

5 m

above the floor at equal time intervals

τ

; the first drop strikes the floor as the sixth begins to fall;

g = 10 m/s^{2}

. Target: the height of the fourth drop above the ground at the instant the first drop lands. Principle: distance in free fall is proportional to the square of the elapsed time.

Step 1:Five equal intervals separate drop 1 and drop 6, so when drop 1 lands it has fallen for

5 τ

and the fourth drop (released two intervals after the first) has fallen for

2 τ

t_{1} = 5 τ, t_{4} = 2 τ

Step 2:Drop 1 covers the full

5 m

in time

5 τ

, fixing the relation

\frac{1}{2} g (5 τ)^{2} = 5

\frac{1}{2} g (5 τ)^{2} = 5

Step 3:Distance fallen by the fourth drop in time

2 τ

s_{4} = \frac{1}{2} g (2 τ)^{2} = 4 \times \frac{1}{2} g τ^{2} = 4 \times 0.2

Step 4:Height above the ground equals total height minus distance fallen.

h_{4} = 5 - 0.8

Final answer:

4.2 m

Q31Single correctElectromagnetic Waves

The electric field of an electromagnetic wave travelling through a medium is given by

E (x, t) = 25 sin (2.0 \times 1 0^{7} t - 1 0^{- 1} x) \overset{n}{^}

. Then the refractive index of the medium is ______. (All given measurements are in SI units)

(A)

2.0

(B)

1.5

(C)

1.7

(D)

1.2

SolutionAnswer: Option 2

1.5

Approach:

Givens: electric field

E (x, t) = 25 sin (2.0 \times 1 0^{7} t - 1 0^{- 1} x) \overset{n}{^}

in SI units. Target: refractive index of the medium. Principle: the phase velocity follows from

ω

and k, and the refractive index is the ratio of the vacuum light speed to this velocity.

Step 1:Compare with

E = E_{0} sin (ω t - k x)

to read the angular frequency and wave number.

ω = 2.0 \times 1 0^{7} rad/s, k = 1 0^{- 1} m^{- 1}

Step 2:Compute the wave speed in the medium.

v = \frac{ω}{k} = \frac{2.0 \times 1 0 ^{7}}{1 0 ^{- 1}}

Step 3:Divide the vacuum speed of light by the medium speed.

μ = \frac{c}{v} = \frac{3 \times 1 0 ^{8}}{2 \times 1 0 ^{8}}

Final answer:

1.5

Q32Single correctThermodynamics

In the following p-V diagram the equation of state along the curved path is given by

(V - 2)^{2} = 4 a p

, where a is a constant. The total work done in the closed path is ______.

A pressure versus volume diagram with the vertical axis labelled p (N/m^2) and the horizontal axis labelled V (m^3), with volume gridlines marked at 1, 2 and 3. A closed cycle has three labelled corner points: A at the top left (at volume 1, raised pressure), C at the top right (at volume 3, raised pressure) and B at the bottom centre (at volume 2, low pressure on the axis). A downward-opening parabolic curve runs from A through C, and the curve dips down to the point B; straight segments close the loop so the path forms the closed region bounded by the parabola and the lines through A, B and C.

(A)

- \frac{1}{3 a}

(B)

- \frac{1}{a}

(C)

\frac{3}{2 a}

(D)

+ \frac{1}{3 a}

SolutionAnswer: Option 1

- \frac{1}{3 a}

Approach:

Givens: closed cycle on a p-V diagram with the curved part obeying

(V - 2)^{2} = 4 a p

; the parabola has its vertex (point B) at

V = 2, p = 0

and rises to

A (V = 1)

and

C (V = 3)

, which are joined by a horizontal segment CA at the top. Target: net work over the closed path. Principle: work in a cycle equals the signed area enclosed,

W = \oint p d V

Step 1:Evaluate the pressure at the end volumes

V = 1

and

V = 3

; both give the top level of the loop.

p (1) = p (3) = \frac{( 1 - 2 ) ^{2}}{4 a} = \frac{1}{4 a}

Step 2:Area under the parabolic path from

V = 1

V = 3

(the lower boundary

A \to B \to C

\int_{1}^{3} \frac{( V - 2 ) ^{2}}{4 a} d V = \frac{1}{4 a} \int_{- 1}^{1} u^{2} d u = \frac{1}{4 a} \cdot \frac{2}{3}

Step 3:Area under the horizontal top segment CA from

V = 1

V = 3

p = \frac{1}{4 a}

\frac{1}{4 a} \times (3 - 1)

Step 4:Traversal

A \to B \to C

increases V along the lower parabola, and

C \to A

decreases V along the upper line; this counterclockwise sense makes the net work the lower-area contribution minus the upper-area contribution.

W = \frac{1}{6 a} - \frac{1}{2 a} = \frac{1 - 3}{6 a}

Final answer:

- \frac{1}{3 a}

Q33Single correctProperties of Solids and Liquids

Which of the following best represents the temperature versus heat supplied graph for water, in the range of -2

0^{\circ}

C to 12

0^{\circ}

(A)

Graph with vertical axis T (degrees C) marked at 0 and 100, and horizontal axis labelled Heat. The curve starts at the origin level (0 degrees C), rises with a slope, then has a horizontal plateau at an intermediate temperature, rises again, has a second horizontal plateau, and finally rises again; the plateaus are placed at intermediate levels rather than exactly at 0 and 100 degrees C, and the curve does not start below 0.

(B)

Graph with vertical axis T (degrees C) marked at 0 and 100 (with label T near 100), and horizontal axis labelled Heat. A single straight line rises continuously from the origin with constant positive slope, showing no horizontal plateaus.

(C)

Graph with vertical axis T (degrees C) marked at -20, 100 and 120, and horizontal axis labelled Heat. The curve starts at -20 degrees C, rises in a straight slope up to 100 degrees C without any plateau at 0 degrees C, then has a horizontal plateau at 100 degrees C, and finally rises to 120 degrees C; the 0 degrees C melting plateau is missing.

(D)

Graph with vertical axis T (degrees C) marked at 0 and 100, and horizontal axis labelled Heat. The curve starts below 0 (at -20 degrees C), rises to 0 degrees C, has a horizontal plateau at 0 degrees C (melting), rises to 100 degrees C, has a horizontal plateau at 100 degrees C (boiling), and finally rises beyond 100 degrees C up to 120 degrees C; this is the physically correct heating curve with plateaus at both 0 and 100 degrees C.

SolutionAnswer: Option 4Heating curve starting at -20°C with horizontal plateaus at 0°C and 100°C, ending at 120°C (option 4)

Approach:

Givens: water heated from

- 2 0^{\circ} C

12 0^{\circ} C

; option graphs of temperature versus heat supplied. Target: the correct heating curve. Principle: temperature rises during single-phase heating (

Q = m c Δ T

) and stays constant during a phase change (

Q = m L

Step 1:From

- 2 0^{\circ} C

0^{\circ} C

the ice warms, so temperature rises with heat.

- 2 0^{\circ} C \to 0^{\circ} C : T increases

Step 2:At

0^{\circ} C

the ice melts at constant temperature while heat is absorbed.

at 0^{\circ} C : T constant

Step 3:From

0^{\circ} C

10 0^{\circ} C

the liquid warms, temperature rising with heat.

0^{\circ} C \to 10 0^{\circ} C : T increases

Step 4:At

10 0^{\circ} C

the water boils at constant temperature while heat is absorbed.

at 10 0^{\circ} C : T constant

Step 5:From

10 0^{\circ} C

12 0^{\circ} C

the steam warms, temperature rising again.

10 0^{\circ} C \to 12 0^{\circ} C : T increases

Final answer: Heating curve starting at

- 2 0^{\circ} C

with horizontal plateaus at

0^{\circ} C

(melting) and

10 0^{\circ} C

(boiling), then rising to

12 0^{\circ} C

Q34Single correctMagnetic Effects of Current and Magnetism

The magnetic field at the centre of a current carrying circular loop of radius R is 16

μ

T. The magnetic field at a distance

x = 3 R

on its axis from the centre is ______

μ

(A)

2

(B)

4

(C)

8

(D)

22

SolutionAnswer: Option 1

2

Approach:

Givens: field at the centre of a circular current loop

B_{0} = 16 μ T

; axial point at

x = 3 R

. Target: axial field B. Principle: the axial field of a loop scales as

(R^{2} / (R^{2} + x^{2}))^{3/2}

relative to the centre value.

Step 1:Form the ratio of the axial field to the central field.

\frac{B}{B _{0}} = (\frac{R ^{2}}{R ^{2} + x ^{2}})^{3/2}

Step 2:Substitute

x = 3 R

, giving

R^{2} + x^{2} = 4 R^{2}

B = 16 (\frac{R ^{2}}{4 R ^{2}})^{3/2} = 16 (\frac{1}{4})^{3/2}

Step 3:Evaluate the product.

B = 2 μ T

Final answer:

2 μ T

Q35Single correctMagnetic Effects of Current and Magnetism

Three long straight wires carrying current are arranged mutually parallel as shown in the figure. The force experienced by 15 cm length of wire Q is ______.

(μ_{0} = 4 π \times 1 0^{- 7} T.m/A)

Three long straight vertical wires labelled P, Q and R from left to right with separations 3 cm (P-Q) and 2 cm (Q-R). Wire P carries 3 A upward (upward arrowhead). Wire Q carries 1 A downward (downward arrowhead). Wire R carries 2 A downward (downward arrowhead).

(A)

6 \times 1 0^{- 7} N towards R

(B)

6 \times 1 0^{- 6} N towards R

(C)

6 \times 1 0^{- 7} N towards P

(D)

6 \times 1 0^{- 6} N towards P

SolutionAnswer: Option 2

6 \times 1 0^{- 6} N towards R

Approach:

Wire Q (current

1

A directed downward) lies between wire P (

3

A directed upward,

3

cm to its left) and wire R (

2

A directed downward,

2

cm to its right). Antiparallel currents repel and parallel currents attract; applying the standard force per unit length to each neighbour and combining the two contributions gives the net force on a

15

cm length of Q.

Step 1:List the data in SI units.

I_{P} = 3 A (up), I_{Q} = 1 A (down), I_{R} = 2 A (down), d_{PQ} = 0.03 m, d_{QR} = 0.02 m, L = 0.15 m

Step 2:P and Q are antiparallel, so they repel; the force on Q from P pushes Q away from P, i.e. toward R. Evaluate its magnitude on the

15

cm length.

F_{P} = \frac{μ _{0} I _{P} I _{Q}}{2 π d _{PQ}} L = \frac{( 4 π \times 1 0 ^{- 7} ) ( 3 ) ( 1 )}{2 π ( 0.03 )} (0.15) = 3 \times 1 0^{- 6} N

Step 3:Q and R are parallel (both downward), so they attract; the force on Q from R pulls Q toward R. Compute its magnitude.

F_{R} = \frac{μ _{0} I _{R} I _{Q}}{2 π d _{QR}} L = \frac{( 4 π \times 1 0 ^{- 7} ) ( 2 ) ( 1 )}{2 π ( 0.02 )} (0.15) = 3 \times 1 0^{- 6} N

Step 4:Both contributions point toward R, so they add directly to give the net force on Q.

F = F_{P} + F_{R} = 3 \times 1 0^{- 6} + 3 \times 1 0^{- 6} = 6 \times 1 0^{- 6} N toward R

Final answer:

6 \times 1 0^{- 6} N

towards R

Q36Single correctAtoms and Nuclei

An atom

_{3}^{8} X

is bombarded by shower of fundamental particles and in 10 s this atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the surface area of the nucleons is recorded by: ______

(A)

900%

(B)

225%

(C)

150%

(D)

250%

SolutionAnswer: Option 2

225%

Approach:

Givens: nucleus

_{3}^{8} X

(

A_{i} = 8

) absorbs 10 electrons, 10 protons and 9 neutrons. Target: the percentage change in nuclear surface area as reported. Principle: nuclear radius scales as

A^{1/3}

, so surface area scales as

A^{2/3}

; only nucleons (protons, neutrons) change the mass number.

Step 1:Electrons do not contribute to the mass number; added protons and neutrons raise it.

A_{f} = 8 + 10 + 9 = 27

Step 2:Form the surface-area ratio.

\frac{S _{f}}{S _{i}} = (\frac{A _{f}}{A _{i}})^{2/3} = (\frac{27}{8})^{2/3} = (\frac{3}{2})^{2}

Step 3:Express the surface-area ratio as a percentage, as reported in the paper.

\frac{S _{f}}{S _{i}} \times 100 = \frac{9}{4} \times 100

Final answer:

225%

Q37Single correctElectromagnetic Induction and Alternating Currents

The electric current in the circuit is given as

i = i_{0} (t / T)

. The r.m.s current for the period

t = 0

t = T

is ______.

(A)

\frac{i _{0}}{3}

(B)

\frac{i _{0}}{2}

(C)

\frac{i _{0}}{6}

(D)

i_{0}

SolutionAnswer: Option 1

\frac{i _{0}}{3}

Approach:

Givens: current

i = i_{0} (t / T)

over

t = 0

t = T

. Target: the r.m.s. current over this interval. Principle: the r.m.s. value is the square root of the time-averaged square of the current.

Step 1:Substitute the current into the mean-square expression.

\frac{1}{T} \int_{0}^{T} (\frac{i _{0} t}{T})^{2} d t = \frac{i _{0}^{2}}{T ^{3}} \int_{0}^{T} t^{2} d t

Step 2:Evaluate the integral.

\frac{i _{0}^{2}}{T ^{3}} [\frac{t ^{3}}{3}]_{0}^{T} = \frac{i _{0}^{2}}{T ^{3}} \cdot \frac{T ^{3}}{3}

Step 3:Take the square root of the mean square.

i_{r m s} = \frac{i _{0}^{2}}{3}

Final answer:

\frac{i _{0}}{3}

Q38Single correctExperimental Skills

When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main scale.

4^{t h}

mark on vernier scale coincides with certain mark on the main scale. While measuring the length of a cylinder, observer observes 15 divisions on main scale and

5^{t h}

division on vernier scale coincides with a main scale division. Measured length of cylinder is ______ mm. (Least count of Vernier calliper = 0.1 mm)

(A)

15.5

(B)

15.1

(C)

15.4

(D)

15.9

SolutionAnswer: Option 2

15.1

Approach:

Givens: with jaws touching, the vernier zero lies to the right of the main-scale zero and the 4th vernier division coincides (positive zero error); least count

LC = 0.1 mm

; while measuring, 15 main-scale divisions are read and the 5th vernier division coincides. Target: corrected measured length. Principle: corrected reading equals observed reading minus the positive zero error.

Step 1:The vernier zero is to the right of the main-scale zero with the 4th division coinciding, giving a positive zero error.

zero error = + 4 \times 0.1 = + 0.4 mm

Step 2:Compute the observed reading from 15 main-scale divisions (each

1 mm

) and the coinciding 5th vernier division.

observed = 15 + 5 \times 0.1 = 15 + 0.5

Step 3:Subtract the positive zero error from the observed reading.

length = 15.5 - 0.4

Final answer:

15.1 mm

Q39Single correctProperties of Solids and Liquids

Two wires A and B made of different materials of lengths 6.0 cm and 5.4 cm, respectively and area of cross sections

3.0 \times 1 0^{- 5}

m^{2}

and

4.5 \times 1 0^{- 5}

m^{2}

, respectively are stretched by the same magnitude under a given load. The ratio of the Young's modulus of A to that of B is x : 3. The value of x is ______.

(A)

10

(B)

2

(C)

5

(D)

4

SolutionAnswer: Option 3

5

Approach:

Given wire A of length

l_{A} = 6.0

cm and area

A_{A} = 3.0 \times 1 0^{- 5}

m^{2}

, wire B of length

l_{B} = 5.4

cm and area

A_{B} = 4.5 \times 1 0^{- 5}

m^{2}

, both stretched by the same elongation under the same load. Target: the value of x where

Y_{A} : Y_{B} = x : 3

. Principle: Young's modulus relates load, length, area and elongation.

Step 1:With identical load F and identical elongation

Δ l

for both wires, the ratio of moduli depends only on length and area.

\frac{Y _{A}}{Y _{B}} = \frac{F l _{A} / ( A _{A} Δ l )}{F l _{B} / ( A _{B} Δ l )} = \frac{l _{A} A _{B}}{l _{B} A _{A}}

Step 2:Substituting the lengths in cm and the areas in

m^{2}

(the common factors cancel).

\frac{Y _{A}}{Y _{B}} = \frac{( 6.0 ) ( 4.5 \times 1 0 ^{- 5} )}{( 5.4 ) ( 3.0 \times 1 0 ^{- 5} )} = \frac{27}{16.2}

Step 3:Matching the reduced ratio with the stated form

x : 3

\frac{x}{3} = \frac{5}{3}

Final answer:

5

Q40Single correctProperties of Solids and Liquids

10 kg of ice at

- 10^{\circ}

C is added to 100 kg of water to lower its temperature from 25

^{\circ}

C. Consider no heat exchange to surroundings. The decrement to the temperature of water is ______

^{\circ}

C.
(Specific heat of ice = 2100 J/Kg.

^{\circ}

C, specific heat of water = 4200 J/Kg.

^{\circ}

C, latent heat of fusion of ice =

3.36 \times 1 0^{5}

J/Kg)

(A)

11.6

(B)

10

(C)

6.67

(D)

15

SolutionAnswer: Option 2

10

Approach:

Given 10 kg ice at

- 10^{\circ}

C added to 100 kg water at

25^{\circ}

C with no heat loss;

c_{i ce} = 2100

J/kg

\cdot^{\circ}

c_{w a t er} = 4200

J/kg

\cdot^{\circ}

L_{f} = 3.36 \times 1 0^{5}

J/kg. Target: the decrement of the water temperature. Principle: conservation of heat (heat lost equals heat gained).

Step 1:Heat absorbed by the ice equals warming it from

- 10^{\circ}

C to

0^{\circ}

C, melting it, and warming the melt water from

0^{\circ}

C to the final temperature T.

Q_{g ain} = (10) (2100) (10) + (10) (3.36 \times 1 0^{5}) + (10) (4200) T

Step 2:Heat released by the 100 kg of water cooling from

25^{\circ}

C to T.

Q_{l oss} = (100) (4200) (25 - T)

Step 3:Equating heat lost to heat gained and solving for the final temperature

T

420000 (25 - T) = 3570000 + 42000 T \Rightarrow 6930000 = 462000 T

Step 4:The decrement is the drop in the water temperature from its initial value to

T

Δ T = 25 - 15

Final answer:

10^{\circ} C

Q41Single correctRotational Motion

Two circular discs of radius each 10 cm are joined at their centres by a rod of length 30 cm and mass 600 gm as shown in figure. If the mass of each disc is 600 gm and applied torque between two discs is

43 \times 1 0^{5}

dyne.cm, the angular acceleration of the discs about the given axis AB is ______ rad/

s^{2}

A vertical rotation axis labelled A at the top and B at the bottom passes through the centre of a horizontal connecting rod. Two identical circular discs, drawn as vertical ellipses, hang one on the left and one on the right end of the rod, each with radius marked 10 cm. The horizontal rod connecting the disc centres has its central segment marked 20 cm, with the lower left part of the left disc also marked 10 cm; the discs face each other across the rod which lies symmetrically about the AB axis.

(A)

11

(B)

27

(C)

22

(D)

100

SolutionAnswer: Option 1

11

Approach:

Two identical discs (each mass

m = 0.6

kg, radius

R = 0.1

m) sit at the ends of a rod (mass

0.6

kg, length

L = 0.30

m) and the assembly turns about the axis

A B

through the rod's centre, perpendicular to the rod. The total moment of inertia about

A B

is built from the two discs (parallel-axis theorem) and the rod, then the applied torque is divided by it.

Step 1:Convert the applied torque to SI units.

τ = 43 \times 1 0^{5} dyne \cdot cm = 43 \times 1 0^{5} \times 1 0^{- 7} N \cdot m = 0.43 N \cdot m

Step 2:Each disc centre lies at half the rod length from the axis,

l = L /2 = 0.15

m. Combine the disc's central inertia with the parallel-axis term.

I_{disc} = \frac{1}{2} (0.6) (0.1)^{2} + (0.6) (0.15)^{2} = 0.003 + 0.0135 = 0.0165 kg \cdot m^{2}

Step 3:Compute the rod's moment of inertia about the perpendicular axis through its centre.

I_{rod} = \frac{1}{12} (0.6) (0.30)^{2} = 0.0045 kg \cdot m^{2}

Step 4:Sum the contributions of both discs and the rod.

I = 2 (0.0165) + 0.0045 = 0.0375 kg \cdot m^{2}

Step 5:Divide the torque by the total moment of inertia.

α = \frac{τ}{I} = \frac{0.43}{0.0375} = 11.5 rad/s^{2}

Final answer:

11 rad/s^{2}

Q42Single correctCurrent Electricity

For the two cells having same EMF E and internal resistance r, the current passing through the external resistor 6

Ω

is same when both the cells are connected either in parallel or in series. The value of internal resistance r is ______

Ω

(A)

3

(B)

6

(C)

12

(D)

9

SolutionAnswer: Option 2

6

Approach:

Given two cells each of EMF E and internal resistance r feeding an external resistor

R = 6 Ω

, with the external current equal in the series and parallel configurations. Target: the value of r. Principle: the loop current equals equivalent EMF divided by total resistance.

Step 1:In parallel the equivalent EMF is

E

and the equivalent internal resistance is

r /2

, giving the external current.

I_{p a r a ll e l} = \frac{E}{R + \frac{r}{2}} = \frac{E}{6 + \frac{r}{2}}

Step 2:In series the equivalent EMF is

2 E

and the equivalent internal resistance is

2 r

I_{ser i es} = \frac{2 E}{R + 2 r} = \frac{2 E}{6 + 2 r}

Step 3:Setting the two external currents equal and cancelling

2 E

\frac{1}{12 + r} = \frac{1}{6 + 2 r} \Rightarrow 6 + 2 r = 12 + r

Final answer:

6 Ω

Q43Single correctElectrostatics

Two point charges of 1 nC and 2 nC are placed at the two corners of equilateral triangle of side 3 cm. The work done in bringing a charge of 3 nC from infinity to the third corner of the triangle is ______

μ

\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} N . m^{2} / C^{2}

(A)

5.4

(B)

2.7

(C)

27

(D)

3.3

SolutionAnswer: Option 2

2.7

Approach:

Given charges

q_{1} = 1

nC and

q_{2} = 2

nC fixed at two corners of an equilateral triangle of side

a = 3

= 0.03

m, and a charge

q = 3

nC brought from infinity to the third corner, with

\frac{1}{4 π ε _{0}} = 9 \times 1 0^{9} N \cdot m^{2} / C^{2}

. Target: the work done. Principle: work equals the change in electrostatic potential energy.

Step 1:The third corner is at distance a from each fixed charge, so the work equals the sum of the two interaction energies of q with

q_{1}

and

q_{2}

W = \frac{1}{4 π ε _{0}} \frac{q q _{1}}{a} + \frac{1}{4 π ε _{0}} \frac{q q _{2}}{a} = \frac{1}{4 π ε _{0}} \frac{q ( q _{1} + q _{2} )}{a}

Step 2:Substituting

q = 3 \times 1 0^{- 9}

q_{1} + q_{2} = 3 \times 1 0^{- 9}

a = 0.03

m and the Coulomb constant.

W = (9 \times 1 0^{9}) \frac{( 3 \times 1 0 ^{- 9} ) ( 3 \times 1 0 ^{- 9} )}{3 \times 1 0 ^{- 2}}

Step 3:Evaluating the arithmetic.

W = \frac{8.1 \times 1 0 ^{- 8}}{3 \times 1 0 ^{- 2}} = 2.7 \times 1 0^{- 6} J

Final answer:

2.7 μ J

Q44Single correctProperties of Solids and Liquids

Particle of mass m falls from rest through a resistive medium having resistive force

F = - k v

, where v is the velocity of the particle and k is a constant. Which of the following graphs represents velocity (v) versus time (t)?

(A)

v versus t graph: a curve starting high on the v-axis and decaying toward the t-axis (exponential decay).

(B)

v versus t graph: a curve rising from the origin and saturating to a horizontal asymptote (terminal-velocity curve).

(C)

v versus t graph: a straight line rising from the origin with constant positive slope.

(D)

v versus t graph: a concave-up curve rising from the origin with increasing slope.

SolutionAnswer: Option 2

v = \frac{m g}{k} (1 - e^{- k t / m})

Approach:

Given a particle of mass

m

released from rest in a medium with resistive force

F = - k v

(

k

constant). Target: identify the

v

versus

t

graph. Principle: Newton's second law produces a first-order linear differential equation whose solution saturates to a terminal velocity.

Step 1:The net downward force is gravity reduced by the resistive force, giving the equation of motion.

m \frac{d v}{d t} = m g - k v \Rightarrow \frac{d v}{d t} = g - \frac{k}{m} v

Step 2:Separating variables and integrating from

v = 0

t = 0

\int_{0}^{v} \frac{d v}{g - \frac{k}{m} v} = \int_{0}^{t} d t

Step 3:At

t = 0

the velocity is zero, and as

t \to \infty

it approaches the terminal velocity, so the curve rises from the origin and flattens to a horizontal asymptote.

v \to \frac{m g}{k} as t \to \infty

Final answer:

v = \frac{m g}{k} (1 - e^{- k t / m})

Q45Single correctElectronic Devices

Assuming in forward bias condition there is a voltage drop of 0.7 V across a silicon diode, the current through diode

D_{1}

in the circuit is ______ mA. (Assume all diodes in the given circuit are identical)

A 12 V battery in series with a resistor R1 of 0.3 k ohm feeds a node from which three silicon diodes hang in parallel branches to the bottom rail. The left diode D1 is forward biased (arrow pointing down toward the bottom rail, labelled Si), the middle diode D2 is connected with reversed polarity (arrow pointing up, labelled Si), and the right diode D3 is forward biased (arrow pointing down, labelled Si).

(A)

20.15

(B)

11.7

(C)

17.6

(D)

18.8

SolutionAnswer: Option 4

18.8

Approach:

A 12 V source feeds three identical silicon diodes (

V_{D} = 0.7

V) through a series resistor

R_{1} = 0.3 k Ω

. The diode orientations in the figure decide which branches conduct, after which the series current and its division give the current through

D_{1}

Step 1:From the circuit,

D_{1}

and

D_{3}

are forward biased while

D_{2}

is reverse biased, so

D_{2}

blocks and only the two forward branches carry current.

D_{2} reverse biased \Rightarrow i_{D_{2}} = 0

Step 2:The total current from the source flows through

R_{1}

after a single forward diode drop of

0.7

i = \frac{12 - 0.7}{0.3 \times 1 0 ^{3}} = \frac{11.3}{300} = 37.67 mA

Step 3:The two identical conducting diodes are in parallel across the same node pair, so they share the total current equally; the current through

D_{1}

is half.

i_{D_{1}} = \frac{i}{2} = \frac{37.67}{2} = 18.8 mA

Final answer:

18.8 mA

Q46NumericalOscillations and Waves

The displacement of a particle, executing simple harmonic motion with time period T, is expressed as

x (t) = A sin ω t

, where A is the amplitude. The maximum value of potential energy of this oscillator is found at

t = T /2 β

. The value of

β

is ______.

SolutionAnswer: 2

Approach:

For an oscillator

x (t) = A sin ω t

the potential energy is proportional to

x^{2}

, so it peaks where the displacement is largest. Equating the first such instant to the given form

t = T / (2 β)

fixes

β

Step 1:Potential energy is maximum when the displacement reaches the amplitude,

∣ x ∣ = A

A sin ω t = A \Rightarrow sin ω t = 1 \Rightarrow ω t = \frac{π}{2}

Step 2:Substitute

ω = 2 π / T

to obtain the first instant of maximum potential energy.

\frac{2 π}{T} t = \frac{π}{2} \Rightarrow t = \frac{T}{4}

Step 3:Match this instant to the given expression

t = T / (2 β)

and solve.

\frac{T}{2 β} = \frac{T}{4} \Rightarrow 2 β = 4 \Rightarrow β = 2

Final answer:

2

Q47NumericalCurrent Electricity

The equivalent resistance between the points A and B in the following circuit is

\frac{x}{5} Ω

. The value of x is ______.

A bridge resistor network between terminals A and B. From A the top branch has a 6 ohm resistor followed by a 3 ohm resistor to the top-right node. From A the bottom branch has a 3 ohm resistor followed by a 6 ohm resistor to B at the bottom-right. A 3 ohm resistor bridges vertically between the junction of the top two resistors and the junction of the bottom two resistors.

SolutionAnswer: 21

Approach:

From the figure: terminal A at top-left, terminal B at bottom-right. Top path A

- 6 Ω - C - 3 Ω -

B; bottom path A

- 3 Ω - D - 6 Ω -

B; a

3 Ω

resistor bridges C to D. Target: the value of x where

R_{A B} = x /5 Ω

. Principle: this is an unbalanced Wheatstone bridge, solved by nodal analysis (or delta-star reduction).

Step 1:Check balance: the arm ratios

\frac{R _{A C}}{R _{CB}} = \frac{6}{3} = 2

and

\frac{R _{A D}}{R _{D B}} = \frac{3}{6} = 0.5

are unequal, so the bridge is unbalanced and the

3 Ω

bridge arm carries current.

\frac{6}{3} \neq = \frac{3}{6}

Step 2:Inject

1

A at A, take it out at B, and set

V_{B} = 0

. Writing KCL at A, C and D gives three equations in

V_{A}, V_{C}, V_{D}

\frac{V _{A} - V _{C}}{6} + \frac{V _{A} - V _{D}}{3} = 1; \frac{V _{C} - V _{A}}{6} + \frac{V _{C}}{3} + \frac{V _{C} - V _{D}}{3} = 0; \frac{V _{D} - V _{A}}{3} + \frac{V _{D}}{6} + \frac{V _{D} - V _{C}}{3} = 0

Step 3:With

1

A injected and

V_{B} = 0

, the equivalent resistance equals

V_{A}

R_{A B} = \frac{V _{A} - V _{B}}{I} = \frac{4.2}{1} = \frac{21}{5} Ω

Step 4:Matching with the stated form

x /5

\frac{x}{5} = \frac{21}{5}

Final answer:

21

Q48NumericalDual Nature of Matter and Radiation

The ratio of de Broglie wavelength of a deuteron with kinetic energy E to that of an alpha particle with kinetic energy 2E, is n : 1. The value of n is ______. (Assume mass of proton = mass of neutron).

SolutionAnswer: 2

Approach:

Given a deuteron (mass

2 u

) with kinetic energy E and an alpha particle (mass 4u) with kinetic energy 2E, with proton and neutron masses taken equal. Target: the value of n where

λ_{d} : λ_{α} = n : 1

. Principle: the de Broglie wavelength relates to mass and kinetic energy through momentum

p = 2 m E

Step 1:Forming the ratio of the deuteron wavelength to the alpha particle wavelength with their respective masses and kinetic energies.

\frac{λ _{d}}{λ _{α}} = \frac{h / 2 m _{d} E}{h / 2 m _{α} ( 2 E )} = \frac{2 m _{α} ( 2 E )}{2 m _{d} E} = \frac{2 m _{α}}{m _{d}}

Step 2:Substituting

m_{d} = 2 u

and

m_{α} = 4 u

\frac{λ _{d}}{λ _{α}} = \frac{2 \times 4 u}{2 u} = 4

Step 3:Expressing the ratio in the form

n : 1

n : 1 = 2 : 1

Final answer:

2

Q49NumericalRotational Motion

A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration of this axis is

n

cm. The value of n is ______.

SolutionAnswer: 265

Approach:

Given a solid sphere of radius

R = 10

cm rotating about an axis at distance

d = 15

cm from its centre, with radius of gyration

K = n

cm. Target: the value of n. Principle: the moment of inertia about the external axis follows from the parallel-axis theorem and equals

m K^{2}

Step 1:Applying the parallel-axis theorem and equating to

m K^{2}

, then cancelling the mass.

m K^{2} = \frac{2}{5} m R^{2} + m d^{2} \Rightarrow K^{2} = \frac{2}{5} R^{2} + d^{2}

Step 2:Substituting

R = 10

cm and

d = 15

cm (lengths in cm so

K^{2}

is in c

m^{2}

K^{2} = \frac{2}{5} (10)^{2} + (15)^{2} = 40 + 225

Step 3:Matching with

K = n

cm.

K = 265 cm

Final answer:

265

Q50NumericalOptics

A convex lens of refractive index 1.5 and focal length f = 18 cm is immersed in water. The difference in focal lengths of the given lens when it is in water and in air is

n \times f

. The value of n is ______. (refractive index of water = 4/3)

SolutionAnswer: 3

Approach:

Given a convex lens of refractive index

n_{g} = 1.5

with focal length

f = 18

cm in air, immersed in water (

n_{w} = 4/3

); the focal-length difference is

n \times f

. Target: the value of n. Principle: the lens maker formula links focal length to the relative refractive index of the lens material with respect to its surroundings.

Step 1:In air the surrounding index is 1, giving the air focal length in terms of the surface term.

\frac{1}{18} = (1.5 - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}}) = 0.5 (\frac{1}{R _{1}} - \frac{1}{R _{2}})

Step 2:In water the surrounding index is

4/3

, giving the water focal length with the same surface term

S

\frac{1}{f _{w}} = (\frac{1.5}{4/3} - 1) S = (1.125 - 1) S = 0.125 S

Step 3:Dividing the two relations eliminates

S

and gives the water focal length.

\frac{f _{w}}{f} = \frac{0.5}{0.125} = 4 \Rightarrow f_{w} = 4 \times 18

Step 4:Expressing the difference of focal lengths as a multiple of the air focal length.

f_{w} - f = 72 - 18 = 54 = n \times 18

Final answer:

3

Chemistry25 questions

Q51Single correctChemical Thermodynamics

20.0 d m^{3}

of an ideal gas 'X' at 600 K and 0.5 MPa undergoes isothermal reversible expansion until pressure of the gas is 0.2 MPa. Which of the following option is correct?
(Given:

lo g 2 = 0.3010

and

lo g 5 = 0.6989

)

(A)

w = - 3.9 k J, Δ U = 0, Δ H = 0; q = 3.9 k J

(B)

w = + 4.1 k J, Δ U = 0, Δ H = 0; q = - 4.1 k J

(C)

w = 9.1 J, Δ U = 9.1 J, Δ H = 0; q = 0

(D)

w = - 9.1 J, Δ U = 0, Δ H = 0, q = 9.1 k J

SolutionAnswer: Option 4

w = - 9.1 J, Δ U = 0, Δ H = 0, q = 9.1 k J

Approach:

An ideal gas at 600 K expands isothermally and reversibly from 0.5 MPa to 0.2 MPa, starting at 20.0

d m^{3}

. For an isothermal change of an ideal gas the internal energy and enthalpy depend only on temperature, so both are zero. The reversible work is evaluated from the isothermal work expression and the heat follows from the first law.

Step 1:For an ideal gas at constant temperature, internal energy and enthalpy are functions of temperature alone and therefore do not change.

Δ U = 0, Δ H = 0

Step 2:The product nRT equals the initial pressure times volume. With

P_{1} = 0.5 MPa = 0.5 \times 1 0^{6} P a

and

V_{1} = 20.0 d m^{3} = 20.0 \times 1 0^{- 3} m^{3}

, this gives the energy scale of the process and corresponds to about two moles of gas.

nRT = P_{1} V_{1} = 0.5 \times 1 0^{6} \times 20.0 \times 1 0^{- 3} = 1.0 \times 1 0^{4} J

Step 3:The pressure ratio for the reversible isothermal work is

P_{1} / P_{2} = 0.5/0.2 = 2.5

, and

lo g 2.5 = lo g 5 - lo g 2 = 0.6989 - 0.3010 = 0.3979

ln \frac{P _{1}}{P _{2}} = 2.303 lo g 2.5 = 2.303 \times 0.3979 = 0.9163

Step 4:Insert the values into the reversible isothermal work expression. The gas expands, so it performs work on the surroundings and the work done on the gas is negative.

w = - nRT ln \frac{P _{1}}{P _{2}} = - 1.0 \times 1 0^{4} \times 0.9163 = - 9.16 \times 1 0^{3} J

Step 5:Since the internal energy is unchanged, the first law gives heat equal in magnitude and opposite in sign to the work; heat is absorbed during the expansion.

q = - w = + 9.1 k J

Final answer:

w \approx - 9.1 k J, Δ U = 0, Δ H = 0, q \approx + 9.1 k J

Q52Single correctEquilibrium

Consider a weak base 'B' of

p K_{b} = 5.699

. 'x' mL of 0.02 M HCl and 'y' mL of 0.02 M weak base 'B' are mixed to make 100 mL of a buffer of pH 9 at 25

^{\circ} C

. The values of 'x' and 'y' respectively are:
(Given:

lo g 2 = 0.3010

lo g 3 = 0.4771

lo g 5 = 0.699

)

(A)

x 11.1 y 88.9

(B)

x 14.3 y 85.7

(C)

x 42.7 y 57.3

(D)

x 85.7 y 14.3

SolutionAnswer: Option 2

x 14.3 y 85.7

Approach:

A weak base B (pKb = 5.699) is partly neutralised by strong acid HCl to form a basic buffer of the conjugate acid BH+ and the leftover base B in a total volume of 100 mL. The target pH 9 fixes pOH = 5, the Henderson equation for the base relates the salt-to-base ratio, and the volume balance x + y = 100 closes the system.

Step 1:Convert the target pH to pOH for the basic buffer.

pO H = 14 - 9 = 5

Step 2:HCl (0.02x mmol) neutralises an equal amount of base to form the conjugate acid salt BH+; the unreacted base is the original base minus the neutralised portion. Both salt and base share the same 100 mL volume, so the concentration ratio reduces to the mole ratio.

[salt] \propto 0.02 x, [base] \propto 0.02 (y - x)

Step 3:Apply the Henderson equation and isolate the logarithm.

5 = 5.699 + lo g \frac{x}{y - x}

Step 4:Since log 5 = 0.699, the antilog of -0.699 is 1/5; solve the ratio for the relation between y and x.

\frac{x}{y - x} = 1 0^{- 0.699} = \frac{1}{5} = 0.2

Step 5:Substitute into the volume balance and solve.

x + 6 x = 100 \Rightarrow 7 x = 100

Final answer:

x 14.3 y 85.7

Q53Single correctAtomic Structure

Which of the following point in Figure 2 most accurately represents the nodal surface as shown in Figure 1?

(A)

C

(B)

B

(C)

D

(D)

A

SolutionAnswer: Option 2

B

Approach:

The spherical nodal surface of the 2s orbital is its radial node, the radius at which the wave function

ψ_{2 s}

passes through zero and reverses sign. Locating that zero crossing on the Figure 2 plot identifies the labelled point.

Step 1:The 2s orbital has exactly one radial node; the spherical nodal surface in Figure 1 marks the radius where the electron probability density falls to zero.

2s : one radial node

Step 2:On the Figure 2 curve, point A is the maximum near the

ψ

axis and point C is the negative minimum, so the curve crosses zero somewhere between them.

A : ψ_{2 s} > 0, C : ψ_{2 s} < 0

Step 3:The point where

ψ_{2 s} = 0

and the sign changes is labelled B, so B represents the nodal surface.

ψ_{2 s} = 0 at point B

Final answer:

B

Q54Single correctBiomolecules

In the given pentapeptide, find out an essential amino acid (Y) and the sequence present in the pentapeptide:

(A)

(Y) Serine (Sequence) Thr - Ser - Asp - Ala - Gly

(B)

(Y) Threonine (Sequence) Thr - Ser - Asp - Gly - Ala

(C)

(Y) Serine (Sequence) Ser - Asp - Thr - Ala - Gly

(D)

(Y) Threonine (Sequence) Ser - Thr - Asp - Gly - Ala

SolutionAnswer: Option 2

(Y) Threonine (Sequence) Thr - Ser - Asp - Gly - Ala

Approach:

Each residue of the pentapeptide is named from its side chain read from the N-terminal H2N end to the C-terminal COOH end; the essential amino acid among them is identified, and the residues are listed in order to obtain the sequence.

Step 1:From the N-terminal H2N end, the first alpha-carbon carries a side chain bearing both an OH and a CH3 on the beta carbon, the threonine side chain; the second residue carries -CH2OH (serine); the third carries -CH2COOH (aspartic acid); the fourth carries -H (glycine); the fifth, ending in COOH, carries -CH3 (alanine).

Thr - Ser - Asp - Gly - Ala

Step 2:Among these five residues, threonine is the essential amino acid (it cannot be synthesised by the body), so Y is threonine.

Y : H_{3} C - C H (O H) - (threonine)

Step 3:Reading the chain from the free amino end to the free carboxyl end gives the primary sequence.

Thr - Ser - Asp - Gly - Ala

Final answer:

(Y) Threonine (Sequence) Thr - Ser - Asp - Gly - Ala

Q55Single correctClassification of Elements and Periodicity in Properties

In period 4 of the periodic table, the elements with highest and lowest atomic radii are respectively.

(A)

K\ &\ Br

(B)

Rb\ &\ Br

(C)

Na\ &\ Cl

(D)

K\ &\ Se

SolutionAnswer: Option 1K\ &\ Br

Approach:

Atomic radius across a period is set by the competition between rising effective nuclear charge and electrons filling the same principal shell. Applying this trend across period 4 identifies the largest and smallest atoms.

Step 1:Across a period the effective nuclear charge rises while electrons enter the same shell, so atomic radius decreases from left to right.

period 4 : radius decreases K \to Kr

Step 2:Potassium opens period 4 as the group-1 element with one loosely held 4s electron, giving the largest atomic radius of the period.

K : largest radius (\sim 235 pm)

Step 3:Toward the right the radius is smallest at bromine among the representative elements, whose high effective nuclear charge contracts the 4p shell.

B r : smallest representative radius (\sim 114 pm)

Final answer: K\ &\ Br

Q56Single correctCoordination Compounds

The correct statement among the following is:

(A)

N i (CO)_{4}

is diamagnetic and

[N i Cl_{4}]^{2 -}

and

[N i (CN)_{4}]^{2 -}

are paramagnetic

(B)

[N i (CN)_{4}]^{2 -}

and

[N i Cl_{4}]^{2 -}

are diamagnetic and

N i (CO)_{4}

is paramagnetic

(C)

N i (CO)_{4}

and

[N i Cl_{4}]^{2 -}

are diamagnetic and

[N i (CN)_{4}]^{2 -}

is paramagnetic

(D)

N i (CO)_{4}

and

[N i (CN)_{4}]^{2 -}

are diamagnetic and

[N i Cl_{4}]^{2 -}

is paramagnetic

SolutionAnswer: Option 4

N i (CO)_{4}

and

[N i (CN)_{4}]^{2 -}

are diamagnetic and

[N i Cl_{4}]^{2 -}

is paramagnetic

Approach:

The oxidation state of nickel and the field strength of the ligand together fix the d-electron count, the geometry, and the number of unpaired electrons, which determine whether each complex is diamagnetic or paramagnetic.

Step 1:In Ni(CO)4 nickel is in the zero oxidation state with a 3d10 configuration after rearrangement; CO is a strong-field ligand and the complex is sp3 tetrahedral with every electron paired, hence diamagnetic.

N i (0) : 3 d^{10}, s p^{3} \Rightarrow 0 unpaired

Step 2:In [Ni(CN)4]2- nickel is Ni2+ (d8); the strong-field cyanide ligand forces pairing, giving a dsp2 square-planar complex with no unpaired electrons, hence diamagnetic.

N i^{2 +} : d^{8}, d s p^{2} (square planar) \Rightarrow 0 unpaired

Step 3:In [NiCl4]2- nickel is again Ni2+ (d8); chloride is a weak-field ligand, so the complex is sp3 tetrahedral with two unpaired electrons, hence paramagnetic.

N i^{2 +} : d^{8}, s p^{3} (tetrahedral) \Rightarrow 2 unpaired

Final answer:

N i (CO)_{4}

and

[N i (CN)_{4}]^{2 -}

are diamagnetic and

[N i Cl_{4}]^{2 -}

is paramagnetic

Q57Single correctChemical Kinetics

An organic compound undergoes first order decomposition. The time taken for decomposition to

(\frac{1}{8})^{t h}

and

(\frac{1}{10})^{t h}

of its initial concentration are

t_{1/8}

and

t_{1/10}

respectively. What is the value of

\frac{t _{1/8}}{t _{1/10}} \times 10

?
(log 2 = 0.3)

(A)

0.9

(B)

30

(C)

3

(D)

9

SolutionAnswer: Option 4

9

Approach:

For a first-order decomposition the time to reach a given fraction of the initial concentration is proportional to the logarithm of the ratio of initial to final concentration. Taking the ratio of the two times cancels the rate constant, leaving a ratio of logarithms.

Step 1:Decomposition to one-eighth leaves [A] = [A0]/8, so the logarithmic factor is log 8; decomposition to one-tenth leaves [A] = [A0]/10, so the factor is log 10.

t_{1/8} = \frac{2.303}{k} lo g 8, t_{1/10} = \frac{2.303}{k} lo g 10

Step 2:Form the ratio; the constant 2.303/k cancels, and log 8 = 3 log 2 = 3(0.3) = 0.9 while log 10 = 1.

\frac{t _{1/8}}{t _{1/10}} = \frac{lo g 8}{lo g 10} = \frac{3 \times 0.3}{1} = 0.9

Step 3:Multiply by ten as required by the question.

\frac{t _{1/8}}{t _{1/10}} \times 10 = 0.9 \times 10

Final answer:

9

Q58Single correctOrganic Compounds Containing Oxygen

Given below are two statements for the following reaction sequence.

(A)

Statement I is true but Statement II is false

(B)

Both Statement I and Statement II are false

(C)

Statement I is false but Statement II is true

(D)

Both Statement I and Statement II are true

SolutionAnswer: Option 4Both Statement I and Statement II are true

Approach:

Compound X (C3H6Cl2) on double dehydrohalogenation with excess NaNH2 yields propyne (Y). Markovnikov hydration of propyne with dil. H2SO4 and Hg2+ gives acetone (Z), tested against the iodoform reaction. Cyclic trimerisation of propyne over a red-hot iron tube gives mesitylene (Q, C9H12), whose hydrogen environments are counted.

Step 1:X is a dichloropropane (C3H6Cl2). Excess sodamide removes two molecules of HCl to build a carbon-carbon triple bond, giving propyne as compound Y.

C_{3} H_{6} Cl_{2} excess N a N H_{2} C H_{3} - C \equiv C H (Y)

Step 2:Propyne undergoes Markovnikov hydration with dil. H2SO4 and Hg2+; water adds with OH on the more substituted carbon, and the resulting enol tautomerises to acetone, a methyl ketone (Z). A methyl ketone responds to the iodoform reaction with NaOI to give the yellow precipitate of iodoform.

C H_{3} - C \equiv C H H_{2} S O_{4}, H g^{2 +} C H_{3} - CO - C H_{3} (Z)

Step 3:Three molecules of propyne trimerise over a red-hot iron tube to 1,3,5-trimethylbenzene (mesitylene), C9H12. It carries 3 aromatic H on the ring and 9 aliphatic H in the three methyl groups, an aromatic-to-aliphatic ratio of 3 : 9 = 1 : 3.

3 C H_{3} - C \equiv C H red hot Fe C_{9} H_{12} (Q)

Final answer: Both Statement I and Statement II are true

Q59Single correctOrganic Compounds Containing Oxygen

Consider the following reaction sequence

(A)

Both compounds x and y will dissolve in NaOH

(B)

Compound y will dissolve in

NaHCO_{3}

and evolve a gas.

(C)

Both compounds x and y will burn with sooty flame

(D)

Compound x is more acidic than compound y.

SolutionAnswer: Option 4Compound x is more acidic than compound y.

Approach:

The elemental data and the colour with neutral FeCl3 identify compound x as phenol. The Kolbe-Schmitt reaction (CO2/NaOH at high pressure, then acidification) converts phenol to salicylic acid (y). Each statement is evaluated and the false one is selected.

Step 1:Vapour density 47 gives molar mass 94, matching C6H5OH; carbon 76.6% and hydrogen 6.38% fit C6H6O, and a characteristic colour with neutral FeCl3 is diagnostic of a phenol. Thus x is phenol.

M = 2 \times 47 = 94 \Rightarrow C_{6} H_{5} O H

Step 2:Phenol treated with CO2 and NaOH at about 120 C under high pressure, followed by acidification, undergoes the Kolbe-Schmitt reaction to give salicylic acid (2-hydroxybenzoic acid), bearing an OH and an ortho COOH; salicylic acid is y and also gives a colour with neutral FeCl3 owing to its phenolic OH.

y = salicylic acid (2-hydroxybenzoic acid)

Step 3:Statement 1: both phenol and salicylic acid have acidic OH/COOH groups and dissolve in NaOH (true). Statement 2: salicylic acid contains a carboxylic acid group, so it dissolves in NaHCO3 with evolution of CO2 (true). Statement 3: both are aromatic with high carbon-to-hydrogen content and burn with a sooty flame (true). Statement 4: salicylic acid (a carboxylic acid) is more acidic than phenol, so the claim that phenol is more acidic is false.

acidity: salicylic acid (y) > phenol (x)

Final answer: Compound x is more acidic than compound y.

Q60Single correctSome Basic Principles of Organic Chemistry

CORRECT order of stability for the following is

C H_{2} = C H^{-}, C H_{3} - C H_{2}^{-}, C H \equiv C^{-}

(A)

C H_{3} - C H_{2}^{-} > C H_{2} = C H^{-} > C H \equiv C^{-}

(B)

C H_{2} = C H^{-} > C H \equiv C^{-} > C H_{3} - C H_{2}^{-}

(C)

C H \equiv C^{-} > C H_{3} - C H_{2}^{-} > C H_{2} = C H^{-}

(D)

C H \equiv C^{-} > C H_{2} = C H^{-} > C H_{3} - C H_{2}^{-}

SolutionAnswer: Option 4

C H \equiv C^{-} > C H_{2} = C H^{-} > C H_{3} - C H_{2}^{-}

Approach:

Carbanion stability tracks the s-character of the orbital holding the lone pair: greater s-character keeps the negative charge closer to the nucleus and lowers its energy. Assigning hybridisation to each carbanion carbon orders the stabilities.

Step 1:Assign the hybridisation of the carbon bearing the negative charge:

C H_{3} - C H_{2}^{-}

s p^{3}

C H_{2} = C H^{-}

s p^{2}

, and

C H \equiv C^{-}

is sp.

C H_{3} C H_{2}^{-} : s p^{3}, C H_{2} = C H^{-} : s p^{2}, C H \equiv C^{-} : s p

Step 2:Order the s-character, since a lone pair in an orbital of higher s-character is held closer to the nucleus and is more stable.

s p^{3} (25%) < s p^{2} (33%) < s p (50%)

Step 3:Greater s-character gives greater stability, so the stability order follows the s-character order.

C H \equiv C^{-} > C H_{2} = C H^{-} > C H_{3} - C H_{2}^{-}

Final answer:

C H \equiv C^{-} > C H_{2} = C H^{-} > C H_{3} - C H_{2}^{-}

Q61Single correctAtomic Structure

The wave numbers of three spectral lines of H atom are considered. Identify the set of spectral belonging to Balmer series.
(R = Rydberg constant)

(A)

\frac{3 R}{4}, \frac{3 R}{16}, \frac{7 R}{144}

(B)

\frac{7 R}{144}, \frac{3 R}{16}, \frac{16 R}{255}

(C)

\frac{5 R}{36}, \frac{8 R}{9}, \frac{15 R}{16}

(D)

\frac{5 R}{36}, \frac{3 R}{16}, \frac{21 R}{100}

SolutionAnswer: Option 4

\frac{5 R}{36}, \frac{3 R}{16}, \frac{21 R}{100}

Approach:

The Balmer series of hydrogen arises from electronic transitions terminating at n = 2. The Rydberg formula with the final level fixed at 2 and the initial level taken as 3, 4 and 5 generates the wave numbers, which are matched to the option set.

Step 1:Transition from n = 3 to n = 2.

\overset{ν}{ˉ} = R (\frac{1}{4} - \frac{1}{9}) = R \cdot \frac{9 - 4}{36} = \frac{5 R}{36}

Step 2:Transition from n = 4 to n = 2.

\overset{ν}{ˉ} = R (\frac{1}{4} - \frac{1}{16}) = R \cdot \frac{4 - 1}{16} = \frac{3 R}{16}

Step 3:Transition from n = 5 to n = 2.

\overset{ν}{ˉ} = R (\frac{1}{4} - \frac{1}{25}) = R \cdot \frac{25 - 4}{100} = \frac{21 R}{100}

Final answer:

\frac{5 R}{36}, \frac{3 R}{16}, \frac{21 R}{100}

Q62Single correctd- and f-Block Elements

Given below are two statements:
Statement I: The number of pairs, from the following, in which both the ions are coloured in aqueous solutions is 3.

[S c^{3 +}, T i^{3 +}], [M n^{2 +}, C r^{2 +}], [C u^{2 +}, Z n^{2 +}]

and

[N i^{2 +}, T i^{4 +}]

Statement II:

T h^{4 +}

is the strongest reducing agent among

T h^{4 +}, C e^{4 +}, G d^{3 +}

and

E u^{2 +}

.
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)

Both Statement I and Statement II are false

(D)

Statement I is true but Statement II is false

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

Approach:

Colour of an aqueous transition-metal ion arises from d-d transitions, which require a partially filled d subshell (d1 to d9); d0 and d10 ions are colourless. Each pair is examined to count those in which both ions are coloured. For the f-block ions, reducing strength is judged from the tendency of the ion to be oxidised to a more stable configuration.

Step 1:Evaluate each pair. [Sc3+, Ti3+]: Sc3+ is d0 (colourless), Ti3+ is d1 (coloured) - not both coloured. [Mn2+, Cr2+]: Mn2+ is d5 and Cr2+ is d4, both coloured. [Cu2+, Zn2+]: Cu2+ is d9 (coloured), Zn2+ is d10 (colourless) - not both coloured. [Ni2+, Ti4+]: Ni2+ is d8 (coloured), Ti4+ is d0 (colourless) - not both coloured.

S c^{3 +}, T i^{4 +} (d^{0}), Z n^{2 +} (d^{10}) colourless

Step 2:The number of pairs with both ions coloured is one, not three, so Statement I is false.

coloured pairs = 1 \neq = 3

Step 3:Among Th4+, Ce4+, Gd3+ and Eu2+, the strongest reducing agent is the one most readily oxidised. Eu2+ is readily oxidised to the stable half-filled Eu3+ (4f7) and is the strongest reducing agent; Th4+ has the inert [Rn] core and is neither a strong reductant nor oxidant. Hence Statement II, which names Th4+, is false.

E u^{2 +} \to E u^{3 +} (4 f^{7}) strongest reductant, not T h^{4 +}

Final answer: Both Statement I and Statement II are false

Q63Single correctChemical Bonding and Molecular Structure

Given below are two statements:
Statement I: The number of species among

B F_{4}^{-}, SiF_{4}, XeF_{4}

and

S F_{4}

, that have unequal E-F bond lengths is two. Here, E is the central atom.
Statement II: Among

O_{2}^{-}, O_{2}^{2 -}, F_{2}

and

O_{2}^{+}, O_{2}^{-}

has the highest bond order
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 false

(C)

Both Statement I and Statement II are true

(D)

Statement I is true but Statement II is false

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

Approach:

The geometry of each fluoride is obtained from VSEPR/hybridisation to decide whether all E-F bonds are equivalent. Bond orders of the dioxygen species are computed from molecular orbital theory to find the largest, and each statement is judged.

Step 1:BF4- (sp3) and SiF4 (sp3) are regular tetrahedra, so all E-F bonds are equal. XeF4 (sp3d2, two lone pairs) is square planar with four equivalent Xe-F bonds, all equal. SF4 (sp3d, one lone pair) is a see-saw shape with two axial bonds longer than two equatorial bonds, so its bonds are unequal.

B F_{4}^{-}, S i F_{4} (tetrahedral), X e F_{4} (square planar) \Rightarrow equal; S F_{4} (see-saw) \Rightarrow unequal

Step 2:The number of species with unequal E-F bond lengths is one, not two, so Statement I is false.

unequal-bond species = 1 \neq = 2

Step 3:From MO theory the bond orders are O2+ = 2.5 (15 electrons), O2- = 1.5 (17 electrons),

O 2^{2}

- = 1.0 (18 electrons), and F2 = 1.0 (18 electrons). The highest bond order is that of O2+, not O2-, so Statement II is false.

B.O. (O_{2}^{+}) = 2.5 > B.O. (O_{2}^{-}) = 1.5 > B.O. (O_{2}^{2 -}) = B.O. (F_{2}) = 1

Final answer: Both Statement I and Statement II are false

Q64Single correctSolutions

At T(K), 2 moles of liquid A and 3 moles of liquid B are mixed. The vapour pressure of ideal solution formed is 320 mm Hg. At this stage, one mole of A and one mole of B added to the solution. The vapour pressure is now measured as 328.6 mm Hg. The vapour pressre (in mm Hg) of A and B are respectively:

(A)

300, 200

(B)

600, 400

(C)

500, 200

(D)

400, 300

SolutionAnswer: Option 3

500, 200

Approach:

Given two ideal-solution vapour pressures at different compositions, the pure-component vapour pressures are required. Raoult's law writes the total pressure as the mole-fraction-weighted sum of pure vapour pressures. Applying it to the 2:3 mixture and to the 3:4 mixture obtained after adding one mole of each component gives two linear equations in

P_{A}^{\circ}

and

P_{B}^{\circ}

, solved simultaneously.

Step 1:For the initial mixture of 2 mol A and 3 mol B (total 5), the mole fractions are

x_{A} = 2/5

and

x_{B} = 3/5

; applying Raoult's law to the measured 320 mm Hg gives the first equation.

320 = \frac{2}{5} P_{A}^{\circ} + \frac{3}{5} P_{B}^{\circ}

Step 2:Adding one mole of A and one mole of B changes the composition to 3 mol A and 4 mol B (total 7), with

x_{A} = 3/7

and

x_{B} = 4/7

; the measured 328.6 mm Hg gives the second equation.

328.6 = \frac{3}{7} P_{A}^{\circ} + \frac{4}{7} P_{B}^{\circ}

Step 3:Eliminate

P_{A}^{\circ}

by computing

3 \times (eq.1) - 2 \times (eq.2)

3 (2 P_{A}^{\circ} + 3 P_{B}^{\circ}) - 2 (3 P_{A}^{\circ} + 4 P_{B}^{\circ}) = 3 (1600) - 2 (2300.2)

Step 4:Substitute

P_{B}^{\circ} = 200

into the first equation to obtain

P_{A}^{\circ}

2 P_{A}^{\circ} = 1600 - 3 (200) = 1000

Final answer:

P_{A}^{\circ} = 500, P_{B}^{\circ} = 200

mm Hg

Q65Single correctPurification and Characterisation of Organic Compounds

Method used for separation of mixture of products (B and C) obtained in the following reaction is

Benzene reacts with Br2 in presence of FeBr3 to give A (bromobenzene), which on treatment with concentrated HNO3 and concentrated H2SO4 gives products B and C (ortho- and para-bromonitrobenzene).

(A)

Fractional distillation

(B)

simple distillation

(C)

steam distillation

(D)

sublimation

SolutionAnswer: Option 1

Fractional distillation

Approach:

Benzene is brominated to bromobenzene (A), whose halogen is ortho/para directing; nitration then gives a mixture of two positional isomers, B and C. The required separation method follows from the physical nature of the two isomeric products.

Step 1:Electrophilic bromination of benzene with Br2 and FeBr3 yields bromobenzene; the C-Br bond makes the halogen an ortho/para director.

C_{6} H_{6} + Br_{2} F e Br_{3} C_{6} H_{5} B r (A)

Step 2:Nitration of bromobenzene with conc. HNO3 and conc. H2SO4 substitutes a nitro group at the ortho and para positions, producing the two products B and C.

C_{6} H_{5} B r H N O_{3} / H_{2} S O_{4} o - B r - C_{6} H_{4} - N O_{2} (B) + p - B r - C_{6} H_{4} - N O_{2} (C)

Step 3:The two products are mutually miscible liquids that differ moderately in boiling point; such a mixture of miscible liquids with close boiling points is separated by repeated vaporisation-condensation, i.e. fractional distillation. Simple distillation fails for close boiling points, steam distillation suits immiscible volatile solids, and sublimation suits compounds that pass directly to vapour.

miscible isomers, close b.p. \Rightarrow fractional distillation

Final answer: Fractional distillation

Q66Single correctp-Block Elements

Regarding the hydrides of group 15 elements

E H_{3}

(E = N, P, As, Sb), select the correct statement from the following:
A) The stability of hydrides decreases down the group
B) The basicity of hydrides decreases down the group
C) The reducing character increases down the group
D) The boiling point increase down the group
Choose the correct answer from the options given below:

(A)

A, B & C only

(B)

A,\ B,\ C\ &\ D

(C)

A & D only

(D)

B & C only

SolutionAnswer: Option 1

A, B & C only

Approach:

The four statements about the group 15 hydrides

E H_{3}

are each tested against established periodic trends in thermal stability, basicity, reducing character and boiling point as the central atom increases in size down the group.

Step 1:Statement A: as the central atom grows down the group the E-H bond weakens, so thermal stability falls in the order shown. A is correct.

N H_{3} > P H_{3} > AsH_{3} > SbH_{3}

Step 2:Statement B: the lone pair becomes more diffuse and less available for donation as size increases, so basicity decreases down the group (

N H_{3} > P H_{3} > AsH_{3} > SbH_{3}

). B is correct.

basicity ↓ down the group

Step 3:Statement C: weaker E-H bonds make the hydrides more readily oxidised, so the reducing character increases down the group. C is correct.

reducing character ↑ down the group

Step 4:Statement D: boiling point does not increase monotonically because

N H_{3}

is anomalously high owing to hydrogen bonding, giving the irregular order

P H_{3} < AsH_{3} < N H_{3} < SbH_{3} < BiH_{3}

. D is incorrect.

P H_{3} < AsH_{3} < N H_{3} < SbH_{3} < BiH_{3}

Final answer: A, B & C only

Q67Single correctOrganic Compounds Containing Oxygen

Given below are the four isomeric compounds (P, Q, R, S)

(A)

A, C and E only

(B)

C and E only

(C)

A and E only

(D)

A, B, D and E only

SolutionAnswer: Option 1

A, C and E only

Approach:

The four

C_{9} H_{10} O

isomers are P (1-phenylprop-2-en-1-ol,

P h - C H (O H) - C H = C H_{2}

, allylic-benzylic alcohol with a vinyl group), Q (3-phenylpropanal,

P h - C H_{2} - C H_{2} - C H O

, aldehyde), R (1-phenylpropan-2-one,

P h - C H_{2} - CO - C H_{3}

, methyl ketone) and S (1-phenylpropan-1-one,

P h - CO - C H_{2} - C H_{3}

, ethyl ketone). Each diagnostic test is applied to every structure to judge statements A-E.

Step 1:Statement A: Q (aldehyde), R (ketone) and S (ketone) each contain a carbonyl group and give an orange precipitate with 2,4-DNP; P is an alcohol and gives none. A is correct.

Q, R, S 2, 4 -DNP precipitate; P \to no precipitate

Step 2:Statement B: Baeyer's reagent (cold dilute alkaline

KMnO_{4}

) tests for an alkene C=C. P bears a vinyl group and decolourises it, but Q (

P h - C H_{2} - C H_{2} - C H O

) has no carbon-carbon double bond, so the pair 'P and Q' is not jointly positive. B is incorrect.

P \to positive (C=C); Q \to no C=C

Step 3:Statement C: Q and R are aromatic compounds with high carbon-to-hydrogen ratio and burn with a sooty (luminous) flame; the statement is a true observation. C is correct.

Q, R \to sooty flame

Step 4:Statement D: the iodoform test needs a

C H_{3} CO -

C H_{3} C H (O H) -

unit. R (

P h - C H_{2} - CO - C H_{3}

) has

C H_{3} CO -

and is positive, but S (

P h - CO - C H_{2} - C H_{3}

) is an ethyl ketone lacking that unit and is negative, so the pair 'R and S' is not jointly positive. D is incorrect.

R \to C H_{3} CO - (positive); S \to no C H_{3} CO - (negative)

Step 5:Statement E: only an aldehyde reduces Tollens' reagent; Q is the sole aldehyde, so Q alone deposits silver. E is correct.

Q Tollens’ A g ↓; P, R, S \to no reaction

Final answer: A, C and E only

Q68Single correctOrganic Compounds Containing Nitrogen

Consider the following reactions giving major product. Identify the correct reaction

(A)

Benzanilide (C6H5-NH-CO-C6H5) on treatment with HNO3 : H2SO4 (1:2) drawn to give a product in which the NHCO bridge connects the two rings and a NO2 group is at the para position of the acyl (benzoyl) ring.

(B)

Propanamide H3CCH2-C(=O)-NH2 plus Br2 plus 4KOH (alc) on heating gives ethylamine H3CCH2NH2 plus 2KBr plus K2CO3 plus 2H2O (Hofmann bromamide degradation).

(C)

Benzylamine C6H5-CH2NH2 with CHCl3/KOH and heat gives benzyl isocyanide C6H5-CH2-N(+)≡C(-), which on (i) Sn/HCl/heat (ii) pH adjusted is drawn returning to benzylamine C6H5-CH2NH2.

(D)

Potassium phthalimide (cyclic imide of benzene with two C=O groups and N(-)K(+)) plus C6H5-X on (i) Heat (ii) Hydrolysis drawn to give aniline C6H5-NH2.

SolutionAnswer: Option 2

Hofmann bromamide degradation: H_{3} CC H_{2} CON H_{2} + Br_{2} + 4 K O H (a l c) Δ H_{3} CC H_{2} N H_{2} + 2 K B r + K_{2} C O_{3} + 2 H_{2} O

Approach:

Each drawn scheme is judged for whether its major product is the genuine outcome. Option 2 is the Hofmann bromamide degradation, the standard conversion of a primary amide to a primary amine with one fewer carbon; its balanced stoichiometry is checked. The remaining schemes are tested for the validity of their drawn products.

Step 1:Option 2: propanamide is degraded by

Br_{2}

and alkali, losing the carbonyl carbon to give ethylamine (a primary amine with one fewer carbon) together with

2 KBr

K_{2} C O_{3}

and water.

H_{3} CC H_{2} CON H_{2} + Br_{2} + 4 K O H (a l c) Δ H_{3} CC H_{2} N H_{2} + 2 K B r + K_{2} C O_{3} + 2 H_{2} O

Step 2:Balance check of option 2: C

3 = 2 + 1

, N

1 = 1

, Br

2 = 2

, K

4 = 2 + 2

, O

5 = 3 + 2

, H

11 = 7 + 4

, all conserved, so the drawn equation is correct.

C,H,N,O,Br,K all balanced

Step 3:Option 1 (benzanilide nitration) places the nitro group incorrectly on the acyl ring rather than where the strongly activating -NH-CO- amide nitrogen directs; option 3 does not regenerate benzylamine through the carbylamine intermediate as drawn; option 4 (Gabriel synthesis) cannot give aniline because aryl halides do not undergo nucleophilic substitution with potassium phthalimide. These schemes are wrong.

options 1, 3, 4 drawn products incorrect

Final answer: Hofmann bromamide degradation:

H_{3} CC H_{2} CON H_{2} + Br_{2} + 4 K O H (a l c) Δ H_{3} CC H_{2} N H_{2} + 2 K B r + K_{2} C O_{3} + 2 H_{2} O

Q69Single correctHydrocarbons

P h - C H = C H_{2} (P h COO)_{2} H B r Product

Consider the above reaction
A. The reaction proceeds through a more stable radical intermediate.
B. The role of peroxide is to generate

H^{\cdot}

(hydrogen radical).
C. During this reaction, benzene is formed as a byproduct.
D. 1-Bromo-2-phenylethane is formed as the minor product.
E. The same reaction in absence of peroxide proceeds via carbocation intermediate.
Identify the correct statements. Choose the correct answer from the options given below.

(A)

C, D & E only

(B)

A & E only

(C)

A, B & D only

(D)

A, C & E only

SolutionAnswer: Option 4

A, C & E only

Approach:

Addition of HBr to styrene with benzoyl peroxide follows the anti-Markovnikov (peroxide/Kharasch) free-radical pathway, while without peroxide it follows the ionic Markovnikov pathway through a carbocation. Each statement is judged against the mechanism and the stability of the intermediates.

Step 1:Statement A: in the chain step a bromine radical adds to the terminal

C H_{2}

, generating the resonance-stabilised benzylic radical

P h \dot{C} H - C H_{2} B r

rather than the less stable primary radical, so the reaction proceeds through the more stable radical. A is correct.

B r^{\cdot} + P h - C H = C H_{2} \to P h \dot{C} H - C H_{2} B r

Step 2:Statement B: the peroxide initiates the chain by decomposing into radicals that abstract H from HBr to generate bromine radicals (

B r^{\cdot}

), which are the chain carriers; it does not generate hydrogen radicals. B is incorrect.

P h^{\cdot} + HBr \to PhH + B r^{\cdot}

Step 3:Statement C: benzoyl peroxide decomposes via

PhCOO^{\cdot}

to phenyl radicals that abstract hydrogen from HBr, forming benzene (

C_{6} H_{6}

) as a by-product of initiation. C is correct.

P h^{\cdot} + HBr \to C_{6} H_{6} + B r^{\cdot}

Step 4:Statement D: the benzylic radical abstracts H from HBr to give

P h - C H_{2} - C H_{2} B r

(1-bromo-2-phenylethane) as the major anti-Markovnikov product, so calling it the minor product is wrong. D is incorrect.

P h \dot{C} H - C H_{2} B r + H B r \to P h - C H_{2} - C H_{2} B r + B r^{\cdot} (major)

Step 5:Statement E: without peroxide, electrophilic addition of HBr proceeds by protonation to the more stable benzylic carbocation

P h C + H - C H_{3}

(Markovnikov), so the no-peroxide route is via a carbocation. E is correct.

P h - C H = C H_{2} + H^{+} \to P h C + H - C H_{3} B r^{-} P h - C H B r - C H_{3}

Final answer: A, C & E only

Q70Single correctPrinciples Related to Practical Chemistry

Given below are two statements:
Statement I: Griss-Ilosvay test is used for the detection of nitrite ion, which involves the use of sulphanilic acid and

α

-naphthylamine reagent.
Statement II: In the above test, sulphanilic acid is diazotized by the acidified nitrite ion, which on further coupling with

α

-naphthylamine forms an azo-dye.
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)

Both Statement I and Statement II are false

(D)

Statement I is true but Statement II is false

SolutionAnswer: Option 2

Both Statement I and Statement II are true

Approach:

The Griess-Ilosvay test for nitrite ion is examined for both its overall identity (reagents and purpose) and the detailed sequence (which species is diazotised and which couples). The standard mechanism is that acidified nitrite diazotises sulphanilic acid and the resulting diazonium salt couples with

α

-naphthylamine to give a red azo dye; each statement is checked against this sequence.

Step 1:Statement I identifies the Griess-Ilosvay test as the detection of nitrite ion using sulphanilic acid and

α

-naphthylamine, which is the correct description of the test. Statement I is true.

detection of N O_{2}^{-} using sulphanilic acid and α -naphthylamine

Step 2:Statement II describes the sequence: the acidified nitrite (as

HNO_{2}

) diazotises sulphanilic acid, and the diazonium salt couples with

α

-naphthylamine to form an azo dye. This is the established mechanism of the test, so Statement II is also true.

sulphanilic acid H^{+}, N O_{2}^{-} diazonium α -naphthylamine azo dye

Step 3:Both statements are correct and Statement II gives the mechanistic explanation underlying Statement I.

I true, II true

Final answer: Both Statement I and Statement II are true

Q71NumericalEquilibrium

Consider the dissociation equilibrium of the following weak acid:

H A ⇌ H^{+} (a q) + A^{-} (a q)

If the pKa of the acid is 4, then the pH of 10 mM HA solution is ______.(Nearest integer)
Given: [The degree of dissociation can be neglected with respect to unity.]

SolutionAnswer: 3

Approach:

For a weak monoprotic acid where the degree of dissociation is negligible compared with unity, the hydrogen-ion concentration is the geometric mean of

K_{a}

and the initial concentration C. The pH follows from the half-relation in

p K_{a}

and

lo g C

. Given

p K_{a} = 4

and

C = 10

mM.

Step 1:Convert the data:

p K_{a} = 4

gives

K_{a} = 1 0^{- 4}

, and 10 mM gives

C = 10 \times 1 0^{- 3} = 1 0^{- 2}

K_{a} = 1 0^{- 4}, C = 1 0^{- 2} M

Step 2:Compute the hydrogen-ion concentration with dissociation neglected against unity.

[H^{+}] = 1 0^{- 4} \times 1 0^{- 2} = 1 0^{- 6} = 1 0^{- 3} M

Step 3:Take the negative logarithm to obtain the pH.

p H = - lo g (1 0^{- 3}) = 3

Final answer:

3

Q72NumericalRedox Reactions and Electrochemistry

Consider the following redox reaction taking place in acidic medium:

B H_{4}^{-} (a q) + ClO_{3}^{-} (a q) \to H_{2} B O_{3}^{-} (a q) + C l^{-} (a q)

If the Nernst equation for the above balanced reaction is

E_{ce ll} = E_{ce ll}^{\circ} - \frac{RT}{n F} ln Q,

then the value of n is ____. (Nearest integer)

SolutionAnswer: 24

Approach:

The n in the Nernst equation equals the total electrons transferred in the balanced overall reaction. Oxidation-state changes give the n-factor of the oxidant (

ClO_{3}^{-}

) and of the reductant (

B H_{4}^{-}

); balancing electrons gives the smallest whole-number coefficients, and the total electron transfer is the common multiple.

Step 1:Oxidation states: in

B H_{4}^{-}

the hydride H is

- 1

and B is

+ 3

; in

H_{2} B O_{3}^{-}

B remains

+ 3

while the four hydrogens go from

- 1

+ 1

. Each

B H_{4}^{-}

therefore loses

4 \times 2 = 8

electrons (n-factor 8).

H : - 1 \to + 1 (\times 4) \Rightarrow 8 e^{-} lost per B H_{4}^{-}

Step 2:In

ClO_{3}^{-}

chlorine is

+ 5

and in

C l^{-}

it is

- 1

, so each

ClO_{3}^{-}

gains

6

electrons (n-factor 6).

Cl : + 5 \to - 1 \Rightarrow 6 e^{-} gained per Cl O_{3}^{-}

Step 3:Equate electrons lost and gained using the least common multiple of 8 and 6, which is 24, giving 3

B H_{4}^{-}

(

3 \times 8 = 24

) and 4

ClO_{3}^{-}

(

4 \times 6 = 24

). The total electron transfer is 24.

3 B H_{4}^{-} + 4 ClO_{3}^{-} \to 3 H_{2} B O_{3}^{-} + 4 C l^{-} + 3 H_{2} O

Final answer:

24

Q73NumericalCoordination Compounds

X is the number of geometrical isomers exhibited by

[Pt (N H_{3}) (H_{2} O) BrCl]

.
Y is the number of optically inactive isomer(s) exhibited by

[C r Cl_{2} (o x)_{2}]^{3 -}

.
Z is the number of geometrical isomers exhibited by

[C o (N H_{3})_{3} (N O_{2})_{3}]

.
The value of X+Y+Z is ______.

SolutionAnswer: 6

Approach:

Each coordination entity is analysed for its required isomer count from its geometry and ligand set: a square-planar [Mabcd] Pt complex (X, geometrical isomers), an octahedral

[M (AA)_{2} b_{2}]

chromium complex (Y, optically inactive isomers) and an octahedral

[M a_{3} b_{3}]

cobalt complex (Z, geometrical isomers). The three counts are then summed.

Step 1:

[Pt (N H_{3}) (H_{2} O) BrCl]

is square planar of type Mabcd (four different monodentate ligands); such a complex shows three geometrical isomers, defined by which ligand sits trans to each chosen one. Hence X = 3.

X = 3

Step 2:

[C r Cl_{2} (o x)_{2}]^{3 -}

is octahedral of type

M (AA)_{2} b_{2}

with two bidentate oxalate ligands and two chlorides; it exists as cis and trans forms. The cis form is chiral (optically active,

d / l

pair) while the trans form has a plane of symmetry and is optically inactive. The number of optically inactive isomers is therefore one (the trans). Hence Y = 1.

Y = 1 (trans, achiral)

Step 3:

[C o (N H_{3})_{3} (N O_{2})_{3}]

is octahedral of type

M a_{3} b_{3}

; the three identical ligands can occupy one face (facial, fac) or a meridian (meridional, mer), giving two geometrical isomers. Hence Z = 2.

Z = 2 (fac and mer)

Step 4:Add the three counts.

X + Y + Z = 3 + 1 + 2

Final answer:

6

Q74NumericalPurification and Characterisation of Organic Compounds

0.53 g of an organic compound (x) when heated with excess of nitric acid (concentrated) and then with silver nitrate gave 0.75 g of silver bromide precipitate. 1.0 g of (x) gave 1.32 g of

C O_{2}

gas on combustion. The percentage of hydrogen in the compound (x) is ____%. [Nearest Integer]
[Given: Molar mass in

g mol^{- 1}

H : 1, C : 12, Br : 80, Ag : 108, O : 16; Compound (x)

C_{x} H_{y} Br_{z}

]

SolutionAnswer: 4

Approach:

The compound contains only C, H and Br. The carbon percentage is found from the

C O_{2}

produced per gram of sample, the bromine percentage from the AgBr precipitate per gram of sample (Carius method), and the hydrogen percentage is obtained by difference.

Step 1:From 1.0 g of compound giving 1.32 g

C O_{2}

, the mass of carbon is

\frac{12}{44} \times 1.32 = 0.36

g, so the carbon percentage is 36%.

% C = \frac{12}{44} \times \frac{1.32}{1.0} \times 100 = 36%

Step 2:Molar mass of AgBr is

108 + 80 = 188

. From 0.53 g of compound giving 0.75 g AgBr, the mass of bromine is

\frac{80}{188} \times 0.75 = 0.3191

m_{B r} = \frac{80}{188} \times 0.75 = 0.3191 g

Step 3:The bromine percentage uses the 0.53 g sample.

% B r = \frac{0.3191}{0.53} \times 100 = 60.2%

Step 4:Hydrogen is the remainder after subtracting carbon and bromine.

% H = 100 - (36 + 60.2) = 3.8

Final answer:

4

Q75NumericalRedox Reactions and Electrochemistry

500 mL of 1.2 M KI solution is mixed with 500 mL of 0.2 M

KMnO_{4}

solution in basic medium. The liberated iodine was titrated with standard 0.1 M

Na_{2} S_{2} O_{3}

solution in presence of starch indicator till the blue color disappeared. The volume (in L) of

Na_{2} S_{2} O_{3}

consumed is ______. (Nearest integer)

SolutionAnswer: 3

Approach:

In basic medium permanganate is reduced from

+ 7

+ 4

(n-factor 3) while oxidising iodide to iodine. The liberated iodine is then reduced by thiosulphate (n-factor 1). By electron conservation through the iodine intermediate, the equivalents of permanganate equal the equivalents of thiosulphate, which gives the required thiosulphate volume.

Step 1:Moles of

KMnO_{4} = 0.2 \times 0.5 = 0.1

mol; in basic medium

MnO_{4}^{-}

goes to

MnO_{2}

(

+ 7 \to + 4

), n-factor 3, so its equivalents are

0.1 \times 3 = 0.3

. Iodide (0.6 mol available) is in excess, so permanganate is fully consumed.

eq of K M n O_{4} = 0.2 \times \frac{500}{1000} \times 3 = 0.3

Step 2:The electrons captured from iodide reappear when the liberated iodine is titrated by thiosulphate (n-factor 1), so equivalents of thiosulphate equal 0.3; set

0.1 \times V \times 1 = 0.3

0.1 \times V \times 1 = 0.3

Step 3:Solve for the volume in litres.

V = \frac{0.3}{0.1}

Final answer:

3

Mathematics24 questions

Q1Single correctMatrices and Determinants

Let A,B and C be three

2 \times 2

matrices with real entries such that

B = (I + A)^{- 1}

and

A + C = I

. If

BC = (1 - 1 - 5 2)

and

CB (x_{1} x_{2}) = (12 - 6)

, then

x_{1} + x_{2}

(A)

2

(B)

- 2

(C)

4

(D)

0

SolutionAnswer: Option 4

0

Approach:

Given

B = (I + A)^{- 1}

A + C = I

, and the product BC, the objective is

x_{1} + x_{2}

where

CB (x_{1} x_{2}) = (12 - 6)

. The relations are used to express BC and CB as the same matrix, after which the linear system is solved.

Step 1:The data fix the matrices A,B,C with

B = (I + A)^{- 1}

C = I - A

BC = (1 - 1 - 5 2)

, and the target is the sum of components of the solution vector of

CB x = (12 - 6)

. Expand

B (I + A) = I

B + B A = I

Step 2:Substitute

C = I - A

into

BC

and apply

B A = I - B

BC = B (I - A) = B - B A = B - (I - B)

Step 3:Expand

(I + A) B = I

to obtain the companion relation for

A B

B + A B = I

Step 4:Substitute

C = I - A

into

CB

and apply

A B = I - B

, giving the same matrix as

BC

CB = (I - A) B = B - A B = B - (I - B) = 2 B - I = BC

Step 5:Form the linear system from

CB x = (12 - 6)

and eliminate by addition.

x_{1} - 5 x_{2} = 12, - x_{1} + 2 x_{2} = - 6

Step 6:Add the two components.

x_{1} + x_{2} = 2 + (- 2)

Final answer:

0

Q2Single correctStatistics and Probability

The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are 2,3,5,10,11,13,15,21, then the mean deviation about the median of all the10 observations is

(A)

7

(B)

4

(C)

6

(D)

5

SolutionAnswer: Option 4

5

Approach:

Ten observations have mean

9

and variance

34.2

with eight values listed. The two missing values are recovered from the sum and sum-of-squares conditions, all ten values are ordered, and the mean deviation about the median is computed.

Step 1:The known eight values

2, 3, 5, 10, 11, 13, 15, 21

sum to

80

. With mean

9

over ten observations the total is

90

, so the two missing values

a, b

satisfy the sum condition.

80 + a + b = 90

Step 2:Variance gives

\sum x_{i}^{2} = 10 (34.2 + 9^{2}) = 10 (115.2) = 1152

. The known eight squares sum to

4 + 9 + 25 + 100 + 121 + 169 + 225 + 441 = 1094

a^{2} + b^{2} = 1152 - 1094

Step 3:Use the identity

(a + b)^{2} = a^{2} + b^{2} + 2 ab

to find ab, then solve the pair.

ab = \frac{( a + b ) ^{2} - ( a ^{2} + b ^{2} )}{2} = \frac{100 - 58}{2} = 21

Step 4:Order all ten values and take the median as the mean of the fifth and sixth terms.

2, 3, 3, 5, 7, 10, 11, 13, 15, 21

Step 5:Sum the absolute deviations from the median and divide by ten.

\sum ∣ x_{i} - M ∣ = 6.5 + 5.5 + 5.5 + 3.5 + 1.5 + 1.5 + 2.5 + 4.5 + 6.5 + 12.5 = 50

Final answer:

5

Q3Single correctCo-ordinate Geometry

Let

y = x

be the equation of a chord of the circle

C_{1}

(in the closed half - plane

x \geq 0

) of diameter 10 passing through the origin. Let

C_{2}

be another circle described on the given chord as its diameter. If the equation of the chord of the circle

C_{2}

, which passes through the point

(2, 3)

and is farthest from the center of

C_{2}

, is

x + a y + b = 0,

then

a - b

is equal to

(A)

- 2

(B)

- 6

(C)

6

(D)

10

SolutionAnswer: Option 1

- 2

Approach:

The chord

y = x

through the origin meets circle

C_{1} : (x - 5)^{2} + y^{2} = 25

(diameter

10

) at two points that form the diameter of

C_{2}

. Its centre is found, then the chord through

(2, 3)

farthest from that centre, being perpendicular to the line joining the centre and the point, is determined and matched to

x + a y + b = 0

Step 1:The circle

C_{1}

of diameter

10

in the half-plane

x \geq 0

through the origin is

(x - 5)^{2} + y^{2} = 25

. Intersect with

y = x

(x - 5)^{2} + x^{2} = 25 \Rightarrow 2 x^{2} - 10 x = 0

Step 2:Circle

C_{2}

has this chord as diameter, so its centre is the midpoint of

(0, 0)

and

(5, 5)

P = (\frac{0 + 5}{2}, \frac{0 + 5}{2})

Step 3:A chord through a fixed point

Q

is farthest from the centre when perpendicular to

PQ

. Compute the slope of

PQ

with

Q (2, 3)

m_{PQ} = \frac{3 - \frac{5}{2}}{2 - \frac{5}{2}} = \frac{\frac{1}{2}}{- \frac{1}{2}} = - 1

Step 4:Write the chord of slope

1

through

(2, 3)

and match with

x + a y + b = 0

y - 3 = 1 \cdot (x - 2) \Rightarrow x - y + 1 = 0

Final answer:

- 2

Q4Single correctVector Algebra

For three unit vectors a,b,c satisfying

∣ a - b ∣^{2} + ∣ b - c ∣^{2} + ∣ c - a ∣^{2} = 9

and

∣2 a + kb + k c ∣ = 3

, the positive value of k is

(A)

3

(B)

5

(C)

4

(D)

6

SolutionAnswer: Option 2

5

Approach:

For unit vectors a,b,c the first condition fixes the sum of pairwise dot products; this forces

a + b + c = 0

, which reduces the second magnitude condition to a one-vector equation solved for the positive k.

Step 1:Expand the first condition using

∣ a ∣ = ∣ b ∣ = ∣ c ∣ = 1

∣ a - b ∣^{2} + ∣ b - c ∣^{2} + ∣ c - a ∣^{2} = 6 - 2 (a \cdot b + b \cdot c + c \cdot a) = 9

Step 2:Substitute this into the norm of the vector sum.

∣ a + b + c ∣^{2} = 3 + 2 (- \frac{3}{2}) = 0

Step 3:Rewrite the second condition with

b + c = - a

∣2 a + kb + k c ∣ = ∣2 a + k (b + c) ∣ = ∣2 a - ka ∣ = ∣ (2 - k) a ∣

Step 4:Solve the absolute-value equation.

2 - k = \pm 3

Step 5:Select the positive root.

k > 0

Final answer:

5

Q5Single correctSequence and Series

The value of

\sum_{k = 1}^{\infty} (- 1)^{k + 1} (\frac{k ( k + 1 )}{k !})

(A)

2/ e

(B)

e /2

(C)

e

(D)

1/ e

SolutionAnswer: Option 4

1/ e

Approach:

The summand

\frac{k ( k + 1 )}{k !}

is split via

k (k + 1) = k (k - 1) + 2 k

into two factorial series; each is re-indexed to the expansion of

e^{- 1}

and the contributions combined.

Step 1:Split the numerator

k (k + 1) = k (k - 1) + 2 k

and apply factorial reduction.

\sum_{k = 1}^{\infty} (- 1)^{k + 1} \frac{k ( k + 1 )}{k !} = \sum_{k = 2}^{\infty} (- 1)^{k + 1} \frac{1}{( k - 2 )!} + 2 \sum_{k = 1}^{\infty} (- 1)^{k + 1} \frac{1}{( k - 1 )!}

Step 2:Re-index the first series with

m = k - 2

, so

(- 1)^{k + 1} = (- 1)^{m + 1}

\sum_{m = 0}^{\infty} (- 1)^{m + 1} \frac{1}{m !} = - \sum_{m = 0}^{\infty} \frac{( - 1 ) ^{m}}{m !} = - e^{- 1}

Step 3:Re-index the second series with

m = k - 1

, so

(- 1)^{k + 1} = (- 1)^{m}

2 \sum_{m = 0}^{\infty} (- 1)^{m} \frac{1}{m !} = 2 e^{- 1}

Step 4:Add the two contributions.

- e^{- 1} + 2 e^{- 1}

Final answer:

1/ e

Q6Single correctPermutations and Combinations

Let S={1,2,3,4,5,6,7,8,9}. Let

x

be the number of 9-digit numbers formed using the digits of the set

S

such that only one digit is repeated and it is repeated exactly twice. Let

y

be the number of 9-digit numbers formed using the digits of the set S such that only two digits are repeated and each of these is repeated exactly twice. Then:

(A)

45 x = 7 y

(B)

21 x = 4 y

(C)

56 x = 9 y

(D)

29 x = 5 y

SolutionAnswer: Option 2

21 x = 4 y

Approach:

From

S = {1, \dots, 9}

, x counts nine-digit numbers with exactly one digit repeated twice and y counts those with exactly two digits each repeated twice. Both counts use selection of the repeated and single digits times the multinomial arrangement; their ratio gives the relation.

Step 1:For

x

: a nine-digit number with one digit repeated twice uses eight distinct digits from

S

. Choose the eight distinct digits, choose which one is doubled, then arrange nine positions with one identical pair.

x = (8 9) (1 8) \frac{9 !}{2 !} = 9 \cdot 8 \cdot \frac{9 !}{2}

Step 2:For

y

: two digits each repeated twice use seven distinct digits. Choose the seven distinct digits, choose the two that are doubled, then arrange with two identical pairs.

y = (7 9) (2 7) \frac{9 !}{2 ! 2 !} = 36 \cdot 21 \cdot \frac{9 !}{4}

Step 3:Form the ratio of the counts.

\frac{x}{y} = \frac{36 \cdot 9 !}{189 \cdot 9 !} = \frac{36}{189} = \frac{4}{21}

Step 4:Cross-multiply to obtain the relation.

21 x = 4 y

Final answer:

21 x = 4 y

Q7Single correctStatistics and Probability

A bag contains 10 balls out of which k are red and

(10 - k)

are black, where

0 \leq k \leq 10

. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

(A)

\frac{14}{55}

(B)

\frac{7}{11}

(C)

\frac{7}{55}

(D)

\frac{7}{110}

SolutionAnswer: Option 1

\frac{14}{55}

Approach:

Each composition

k = 0, \dots, 10

is treated as equally likely a priori. The likelihood of drawing three black balls is hypergeometric, and Bayes' theorem gives the posterior probability of the composition with one red and nine black (

k = 1

Step 1:With eleven equally likely compositions

k = 0, 1, \dots, 10

, the prior is

\frac{1}{11}

and the likelihood of three black balls when

(10 - k)

are black is

(3 10 - k) / (3 10)

. The common factors

\frac{1}{11}

and

1/ (3 10)

cancel, leaving counts proportional to

(3 10 - k)

P (3B ∣ k) \propto (3 10 - k)

Step 2:The composition one red, nine black corresponds to

k = 1

with weight

(3 9)

(3 9) = 84

Step 3:Sum the weights over all

k

, equivalently over black count

j = 10 - k

from

3

10

, using the hockey-stick identity.

\sum_{j = 3}^{10} (3 j) = (4 11) = 330

Step 4:Apply Bayes' theorem and simplify.

P = \frac{84}{330} = \frac{14}{55}

Final answer:

\frac{14}{55}

Q8Single correctCo-ordinate Geometry

Let ABC be an equilateral triangle with orthocenter at the origin and the side BC on the line

x + 22 y = 4

. If the co-ordinates of the vertex A are

(α, β)

, then the greatest integer less than or equal to

∣ α + 2 β ∣

is:

(A)

4

(B)

2

(C)

5

(D)

3

SolutionAnswer: Option 1

4

Approach:

For an equilateral triangle the orthocentre, centroid and circumcentre coincide at the origin. The centroid divides the median from A to the midpoint M of BC in ratio

2 : 1

, with A and M on opposite sides of the origin along the perpendicular to BC. The vertex

A = (α, β)

is located and

⌊ ∣ α + 2 β ∣ ⌋

evaluated.

Step 1:The orthocentre O at the origin coincides with the centroid. The foot M of the perpendicular from O to

BC : x + 22 y - 4 = 0

is at distance OM.

OM = \frac{∣0 + 0 - 4∣}{1 + 8} = \frac{4}{3}

Step 2:The unit normal to BC is

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

. Since the line value at O is

- 4 < 0

, M lies in the

+

normal direction,

M = \frac{4}{3} (\frac{1}{3}, \frac{2 2}{3}) = (\frac{4}{9}, \frac{8 2}{9})

A = 3 G - 2 M = - 2 M

Step 3:Confirm A lies on the perpendicular from O to BC (slope

22

β = 22 α

gives

- \frac{16 2}{9} = 22 (- \frac{8}{9})

, which holds. Compute

α + 2 β

α + 2 β = - \frac{8}{9} + 2 (- \frac{16 2}{9}) = - \frac{8}{9} - \frac{32}{9}

Step 4:Take the absolute value and apply the greatest-integer function.

∣ α + 2 β ∣ = \frac{40}{9} \approx 4.44

Final answer:

4

Q9Single correctComplex Numbers and Quadratic Equations

α, β

, where

α < β

, are the roots of the equation

λ x^{2} - (λ + 3) x + 3 = 0

such that

\frac{1}{α} - \frac{1}{β} = \frac{1}{3}

, then the sum of all possible values of

λ

(A)

4

(B)

8

(C)

2

(D)

6

SolutionAnswer: Option 4

6

Approach:

The quadratic

λ x^{2} - (λ + 3) x + 3 = 0

factors with roots

1

and

3/ λ

. Imposing

\frac{1}{α} - \frac{1}{β} = \frac{1}{3}

for both orderings consistent with

α < β

yields the admissible values of

λ

, whose sum is required.

Step 1:Factor the quadratic to read off the roots.

(λ x - 3) (x - 1) = 0

Step 2:For the ordering

α = 1 < β = \frac{3}{λ}

(requires

\frac{3}{λ} > 1

), impose the condition.

\frac{1}{1} - \frac{1}{3/ λ} = 1 - \frac{λ}{3} = \frac{1}{3}

Step 3:For the ordering

α = \frac{3}{λ} < β = 1

(requires

\frac{3}{λ} < 1

), impose the condition.

\frac{1}{3/ λ} - \frac{1}{1} = \frac{λ}{3} - 1 = \frac{1}{3}

Step 4:Add the admissible values.

2 + 4

Final answer:

6

Q10Single correctIntegral Calculus

\int (\frac{1 - 5 cos ^{2} x}{sin ^{5} x cos ^{2} x}) d x = f (x) + C

, where C is the constant of integration, then

f (\frac{π}{6}) - f (\frac{π}{4})

is equal to

(A)

\frac{1}{3} (26 - 3)

(B)

\frac{2}{3} (4 + 6)

(C)

\frac{4}{3} (8 - 6)

(D)

\frac{1}{3} (26 + 3)

SolutionAnswer: Option 3

\frac{4}{3} (8 - 6)

Approach:

The integrand

\frac{1 - 5 cos ^{2} x}{sin ^{5} x cos ^{2} x}

is the exact derivative of

\frac{tan x}{sin ^{5} x}

, so

f (x) = \frac{tan x}{sin ^{5} x}

; evaluation at

\frac{π}{6}

and

\frac{π}{4}

gives the required difference.

Step 1:Differentiating

\frac{tan x}{sin ^{5} x}

reproduces the integrand, so the antiderivative is identified.

f (x) = \frac{tan x}{sin ^{5} x}

Step 2:Evaluate f at

\frac{π}{6}

f (\frac{π}{6}) = \frac{1/ 3}{( 1/2 ) ^{5}} = \frac{32}{3}

Step 3:Evaluate f at

\frac{π}{4}

f (\frac{π}{4}) = \frac{1}{( 1/ 2 ) ^{5}} = (2)^{5} \cdot \frac{1}{4} = 42

Step 4:Form the difference and factor by

\frac{4}{3}

\frac{32}{3} - 42 = \frac{4}{3} (8 - 3 \cdot 2) = \frac{4}{3} (8 - 6)

Final answer:

\frac{4}{3} (8 - 6)

Q11Single correctComplex Numbers and Quadratic Equations

Let

S = {x^{3} + a x^{2} + b x + c : a, b, c \in N and a, b, c \leq 20}

be a set of polynomials. Then the number of polynomials in S, which are divisible by

x^{2} + 2

, is

(A)

20

(B)

6

(C)

120

(D)

10

SolutionAnswer: Option 4

10

Approach:

Divisibility of the monic cubic

x^{3} + a x^{2} + b x + c

x^{2} + 2

forces it to equal

(x^{2} + 2) (x + q)

. Coefficient comparison fixes b and ties c to a; the natural-number triples within the bound are then counted.

Step 1:Since the cubic is monic, the quotient on division by

x^{2} + 2

x + q

. Expand the product.

(x^{2} + 2) (x + q) = x^{3} + q x^{2} + 2 x + 2 q

Step 2:Match coefficients with

x^{3} + a x^{2} + b x + c

a = q, b = 2, c = 2 q

Step 3:Apply

a, b, c \in N

with

a, b, c \leq 20

. The value

b = 2

is admissible, and

c = 2 a \leq 20

bounds a.

1 \leq a \leq 10

Step 4:Each admissible

a

gives one valid triple

(a, 2, 2 a)

; count them.

# {a : 1 \leq a \leq 10} = 10

Final answer:

10

Q12Single correctDifferential Equations

Let

y = y (x)

be the solution of the differential equation

x \frac{d y}{d x} - sin 2 y = x^{3} (2 - x^{3}) cos^{2} y, x \neq = 0

. If

y (2) = 0

, then

tan (y (1))

is equal to

(A)

\frac{3}{4}

(B)

\frac{7}{4}

(C)

- \frac{7}{4}

(D)

- \frac{3}{4}

SolutionAnswer: Option 2

\frac{7}{4}

Approach:

Dividing by

cos^{2} y

and substituting

t = tan y

converts the equation into a first-order linear ODE in t, solved by an integrating factor; the condition

y (2) = 0

fixes the constant and

tan (y (1))

is read off.

Step 1:Divide the equation by

x cos^{2} y

and use

\frac{sin 2 y}{cos ^{2} y} = 2 tan y

to obtain a relation in

t = tan y

sec^{2} y \frac{d y}{d x} - \frac{2 tan y}{x} = x^{2} (2 - x^{3})

Step 2:Compute the integrating factor.

I . F = e^{\int - \frac{2}{x} d x} = e^{- 2 l n x} = \frac{1}{x ^{2}}

Step 3:Multiply through and integrate the exact derivative.

\frac{d}{d x} (\frac{t}{x ^{2}}) = 2 - x^{3} \Rightarrow \frac{t}{x ^{2}} = 2 x - \frac{x ^{4}}{4} + c

Step 4:Apply

y (2) = 0

, so

tan y = 0

x = 2

0 = 2 (8) - \frac{64}{4} + 4 c = 16 - 16 + 4 c

Step 5:Evaluate at

x = 1

tan y (1) = 2 (1) - \frac{1}{4}

Final answer:

\frac{7}{4}

Q13Single correctIntegral Calculus

The area of the region

R = {(x, y) : x y \leq 8, 1 \leq y \leq x^{2}, x \geq 0}

(A)

\frac{1}{3} (40 lo g_{e} (2) + 27)

(B)

\frac{1}{3} (49 lo g_{e} (2) - 15)

(C)

\frac{2}{3} (24 lo g_{e} (2) - 7)

(D)

\frac{2}{3} (20 lo g_{e} (2) + 9)

SolutionAnswer: Option 3

\frac{2}{3} (24 lo g_{e} (2) - 7)

Approach:

Givens: the region bounded by

y = x^{2}

y = 1

, the hyperbola

x y = 8

and

x \geq 0

. Target: its area. Method: determine the upper boundary as

min (x^{2}, 8/ x)

above the line

y = 1

, split the x-integration at the crossover, and integrate.

Step 1:The lower boundary is

y = 1

. The upper boundary is the smaller of

y = x^{2}

and

y = 8/ x

. Equating

x^{2} = 8/ x

gives

x^{3} = 8

, so the curves cross at

x = 2

x^{2} = \frac{8}{x} \Rightarrow x^{3} = 8 \Rightarrow x = 2

Step 2:Determine the x-range. The lower curve

y = x^{2}

meets

y = 1

x = 1

, and the hyperbola

y = 8/ x

meets

y = 1

x = 8

. For

1 \leq x \leq 2

the top is

y = x^{2}

; for

2 \leq x \leq 8

the top is

y = 8/ x

x \in [1, 2] : y_{top} = x^{2}; x \in [2, 8] : y_{top} = \frac{8}{x}

Step 3:Integrate the first part from

x = 1

x = 2

with top

y = x^{2}

and bottom

y = 1

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

Step 4:Integrate the second part from

x = 2

x = 8

with top

y = 8/ x

and bottom

y = 1

\int_{2}^{8} (\frac{8}{x} - 1) d x = [8 ln x - x]_{2}^{8} = (8 ln 8 - 8) - (8 ln 2 - 2)

Step 5:Add the two contributions and factor.

A = \frac{4}{3} + 16 ln 2 - 6 = 16 ln 2 - \frac{14}{3}

Final answer:

\frac{2}{3} (24 lo g_{e} (2) - 7)

Q14Single correctSets, Relations and Functions

g (x) = 3 x^{2} + 2 x - 3

f (0) = - 3

and

4 g (f (x)) = 3 x^{2} - 32 x + 72

, then

f (g (2))

(A)

- \frac{25}{6}

(B)

\frac{25}{6}

(C)

\frac{7}{2}

(D)

- \frac{7}{2}

SolutionAnswer: Option 3

\frac{7}{2}

Approach:

Givens:

g (x) = 3 x^{2} + 2 x - 3

f (0) = - 3

, and

4 g (f (x)) = 3 x^{2} - 32 x + 72

. Target:

f (g (2))

. Method: take f linear, match coefficients to fix it, then evaluate the composition.

Step 1:Since the right side

3 x^{2} - 32 x + 72

is quadratic and g is quadratic, f is linear. Apply

f (0) = - 3

f (x) = p x + q, f (0) = q = - 3

Step 2:Substitute

f (x) = p x - 3

into

4 g (f (x)) = 4 [3 (p x - 3)^{2} + 2 (p x - 3) - 3]

and expand.

4 [3 (p^{2} x^{2} - 6 p x + 9) + 2 p x - 6 - 3] = 12 p^{2} x^{2} - 64 p x + 72

Step 3:Match the coefficient of

x^{2}

and the coefficient of x.

12 p^{2} = 3 \Rightarrow p = \pm \frac{1}{2}; - 64 p = - 32 \Rightarrow p = \frac{1}{2}

Step 4:Write

f

and evaluate

g (2)

f (x) = \frac{x}{2} - 3, g (2) = 3 (4) + 2 (2) - 3 = 13

Step 5:Compute

f (g (2)) = f (13)

f (13) = \frac{13}{2} - 3 = \frac{13 - 6}{2}

Final answer:

\frac{7}{2}

Q15Single correctSequence and Series

The common difference of the A.P.:

a_{1}, a_{2}, \dots\dots a_{m}

is 13 more than the common difference of the A.P.:

b_{1}, b_{2}, \dots\dots b_{n}

. If

b_{31} = - 277

b_{43} = - 385

, and

a_{78} = 327

, then

a_{1}

is equal to:

(A)

19

(B)

21

(C)

16

(D)

24

SolutionAnswer: Option 1

19

Approach:

Givens:

d_{a} = d_{b} + 13

b_{31} = - 277

b_{43} = - 385

a_{78} = 327

. Target:

a_{1}

. Method: extract

d_{b}

from two b-terms, get

d_{a}

, then back out

a_{1}

Step 1:Use

b_{43} - b_{31} = (43 - 31) d_{b}

to find the common difference of the b-series.

12 d_{b} = - 385 - (- 277) = - 108

Step 2:The common difference of the a-series is

13

more than

d_{b}

d_{a} = d_{b} + 13 = - 9 + 13

Step 3:Express

a_{1}

from

a_{78} = a_{1} + 77 d_{a}

a_{1} = 327 - 77 (4) = 327 - 308

Final answer:

19

Q16Single correctLimit, Continuity and Differentiability

The Value of

x \to 0 lim \frac{lo g _{e} ( sec ( e x ) \cdot sec ( e ^{2} x ) \dots\dots sec ( e ^{10} x ) )}{e ^{2} - e ^{2 c o s x}}

(A)

\frac{( e ^{10} - 1 )}{2 e ^{2} ( e ^{2} - 1 )}

(B)

\frac{( e ^{20} - 1 )}{2 e ^{2} ( e ^{2} - 1 )}

(C)

\frac{( e ^{10} - 1 )}{2 ( e ^{2} - 1 )}

(D)

\frac{( e ^{20} - 1 )}{2 ( e ^{2} - 1 )}

SolutionAnswer: Option 4

\frac{( e ^{20} - 1 )}{2 ( e ^{2} - 1 )}

Approach:

Givens: the limit of

\frac{lo g _{e} \prod _{k = 1}^{10} sec ( e ^{k} x )}{e ^{2} - e ^{2 c o s x}}

x \to 0

. Target: the limit value. Method: expand numerator and denominator to leading order in

x^{2}

and take the ratio of coefficients.

Step 1:Convert the logarithm of the product into a sum and apply

lo g_{e} sec (e^{k} x) \to \frac{1}{2} e^{2 k} x^{2}

\sum_{k = 1}^{10} lo g_{e} sec (e^{k} x) \sim \frac{x ^{2}}{2} \sum_{k = 1}^{10} e^{2 k}

Step 2:Expand the denominator using

cos x = 1 - \frac{x ^{2}}{2} + O (x^{4})

e^{2} - e^{2 c o s x} = e^{2} - e^{2 - x^{2}} = e^{2} (1 - e^{- x^{2}})

Step 3:Form the ratio of leading

x^{2}

terms.

\frac{\frac{x ^{2}}{2} e ^{2} \frac{e ^{20} - 1}{e ^{2} - 1}}{e ^{2} x ^{2}} = \frac{1}{2} \cdot \frac{e ^{20} - 1}{e ^{2} - 1}

Final answer:

\frac{( e ^{20} - 1 )}{2 ( e ^{2} - 1 )}

Q17Single correctTrigonometry

\frac{tan ( A - B )}{tan A} + \frac{sin ^{2} C}{sin ^{2} A} = 1

A, B, C \in (0, \frac{π}{2})

, then

(A)

tan A, tan B, tan C are in G.P.

(B)

tan A, tan B, tan C are in A.P.

(C)

tan A, tan C, tan B are in A.P.

(D)

tan A, tan C, tan B are in G.P

SolutionAnswer: Option 4tan A, tan C, tan B are in G.P

Approach:

Givens:

\frac{tan ( A - B )}{tan A} + \frac{sin ^{2} C}{sin ^{2} A} = 1

with

A, B, C \in (0, \frac{π}{2})

. Target: the progression relating

tan A, tan B, tan C

. Method: reduce the relation to a product condition on the tangents.

Step 1:Isolate the first term.

\frac{tan ( A - B )}{tan A} = 1 - \frac{sin ^{2} C}{sin ^{2} A} = \frac{sin ^{2} A - sin ^{2} C}{sin ^{2} A}

Step 2:Multiply both sides by

sin^{2} A

and use

sin A cos A

on the left and the product form on the right.

tan (A - B) sin A cos A = sin (A + C) sin (A - C)

Step 3:Reducing the trigonometric identity yields a symmetric product relation among the tangents.

tan^{2} C = tan A tan B

Step 4:Since

tan C

is the geometric mean of

tan A

and

tan B

, the ordering

tan A, tan C, tan B

is geometric.

\frac{tan C}{tan A} = \frac{tan B}{tan C}

Final answer: tan A, tan C, tan B are in G.P

Q18Single correctComplex Numbers and Quadratic Equations

Let z be a complex number such that

∣ z - 6∣ = 5

and

∣ z + 2 - 6 i ∣ = 5

. Then the value of

z^{3} + 3 z^{2} - 15 z + 141

is equal to

(A)

37

(B)

61

(C)

50

(D)

42

SolutionAnswer: Option 3

50

Approach:

Givens:

∣ z - 6∣ = 5

and

∣ z + 2 - 6 i ∣ = 5

. Target:

z^{3} + 3 z^{2} - 15 z + 141

. Method: identify the two circles, locate their unique common point, and evaluate the polynomial there.

Step 1:Read off the two circles: centre

A (6, 0)

radius

5

, and centre

B (- 2, 6)

radius

5

∣ z - 6∣ = 5, ∣ z - (- 2 + 6 i) ∣ = 5

Step 2:Compute the distance between the centres.

A B = (6 + 2)^{2} + (0 - 6)^{2} = 64 + 36 = 10

Step 3:Since

A B

equals the sum of radii, the circles touch externally at the midpoint of

A B

z = \frac{( 6 , 0 ) + ( - 2 , 6 )}{2} = (2, 3)

Step 4:Compute

z^{2}

and

z^{3}

z^{2} = (2 + 3 i)^{2} = - 5 + 12 i, z^{3} = (2 + 3 i) (- 5 + 12 i) = - 46 + 9 i

Step 5:Assemble the polynomial.

(- 46 + 9 i) + 3 (- 5 + 12 i) - 15 (2 + 3 i) + 141

Final answer:

50

Q19Single correctThree Dimensional Geometry

If the distances of the point

(1, 2, a)

from the line

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

along the lines

L_{1} : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b}

and

L_{2} : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c}

are equal, then

a + b + c

is equal to

(A)

4

(B)

7

(C)

5

(D)

6

SolutionAnswer: Option 2

7

Approach:

Givens: point

P (1, 2, a)

, the fixed line

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

, and lines

L_{1}, L_{2}

through P whose distances from P to the fixed line measured along

L_{1}

and

L_{2}

are equal. Target:

a + b + c

. Method: find where

L_{1}

and

L_{2}

meet the fixed line, impose equal distances, and read off a,b,c.

Step 1:Intersect

L_{1}

with the fixed line. A point on

L_{1}

(1 + 3 s, 2 + 4 s, a + b s)

; matching x and y with

(1 + t, 2 t, 1 + t)

gives

t = 3 s

and

2 + 4 s = 6 s

, so

s = 1, t = 3

. The meeting point on the fixed line is

Q (4, 6, 4)

2 + 4 s = 6 s \Rightarrow s = 1, t = 3

Step 2:Intersect

L_{2}

with the fixed line. A point on

L_{2}

(1 + s^{'}, 2 + 4 s^{'}, a + c s^{'})

; matching gives

t = s^{'}

and

2 + 4 s^{'} = 2 s^{'}

, so

s^{'} = - 1, t = - 1

. The meeting point on the fixed line is

R (0, - 2, 0)

2 + 4 s^{'} = 2 s^{'} \Rightarrow s^{'} = - 1, t = - 1

Step 3:Impose equal distances

PQ = PR

from

P (1, 2, a)

, taking the

z

-coordinates of

Q

and

R

on the fixed line as

4

and

0

9 + 16 + (4 - a)^{2} = 1 + 16 + a^{2}

Step 4:Use

a + b = 4

and

a - c = 0

with

a = 3

b = 4 - 3 = 1, c = a = 3

Step 5:Add the three values.

a + b + c = 3 + 1 + 3

Final answer:

7

Q20Single correctIntegral Calculus

Let f be a polynomial function such that

f (x^{2} + 1) = x^{4} + 5 x^{2} + 2

, for all

x \in R

. Then

\int_{0}^{3} f (x) d x

is equal to

(A)

\frac{33}{2}

(B)

\frac{27}{2}

(C)

\frac{41}{3}

(D)

\frac{5}{3}

SolutionAnswer: Option 1

\frac{33}{2}

Approach:

Givens:

f (x^{2} + 1) = x^{4} + 5 x^{2} + 2

. Target:

\int_{0}^{3} f (x) d x

. Method: recover f via the substitution

t = x^{2} + 1

, then integrate the explicit polynomial.

Step 1:Express the right side in

x^{2}

and substitute

x^{2} = t - 1

x^{4} + 5 x^{2} + 2 = (x^{2})^{2} + 5 x^{2} + 2 = (t - 1)^{2} + 5 (t - 1) + 2

Step 2:Integrate

f (x) = x^{2} + 3 x - 2

from

0

3

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

Step 3:Simplify.

3 + \frac{27}{2} = \frac{6 + 27}{2}

Final answer:

\frac{33}{2}

Q22NumericalTrigonometry

k = tan (\frac{π}{4} + \frac{1}{2} cos^{- 1} (\frac{2}{3})) + tan (\frac{1}{2} sin^{- 1} (\frac{2}{3}))

, then the number of solutions of the equation

sin^{- 1} (k x - 1) = sin^{- 1} x - cos^{- 1} x

SolutionAnswer: 1

Approach:

Givens:

k = tan (\frac{π}{4} + \frac{1}{2} cos^{- 1} \frac{2}{3}) + tan (\frac{1}{2} sin^{- 1} \frac{2}{3})

and the equation

sin^{- 1} (k x - 1) = sin^{- 1} x - cos^{- 1} x

. Target: the number of solutions. Method: simplify k via complementary angles, reduce to an algebraic equation, and count roots inside the valid domain.

Step 1:Set

θ = \frac{1}{2} cos^{- 1} \frac{2}{3}

. Since

sin^{- 1} \frac{2}{3} = \frac{π}{2} - cos^{- 1} \frac{2}{3}

, the second angle is

\frac{1}{2} sin^{- 1} \frac{2}{3} = \frac{π}{4} - θ

, so

k = tan (\frac{π}{4} + θ) + tan (\frac{π}{4} - θ)

k = \frac{2}{cos 2 θ} = \frac{2}{cos ( cos ^{- 1} \frac{2}{3} )}

Step 2:Substitute

k = 3

and rewrite the right side using the identity.

sin^{- 1} (3 x - 1) = 2 sin^{- 1} x - \frac{π}{2}

Step 3:Take sine of both sides;

sin (2 sin^{- 1} x - \frac{π}{2}) = - cos (2 sin^{- 1} x) = - (1 - 2 x^{2})

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

Step 4:The roots are

x = 0

and

x = \frac{3}{2}

; the root

x = \frac{3}{2}

lies outside the domain

[0, \frac{2}{3}]

and is discarded.

x = 0 (admissible), x = \frac{3}{2} (rejected)

Step 5:Confirm

x = 0

in the original equation.

sin^{- 1} (- 1) = sin^{- 1} 0 - cos^{- 1} 0 = 0 - \frac{π}{2} = - \frac{π}{2}

Final answer:

1

Q23NumericalSequence and Series

In a G.P., if the product of the first three terms is 27 and the set of all possible values for the sum of its first three terms is

R - (a, b)

, then

a^{2} + b^{2}

is equal to

SolutionAnswer: 90

Approach:

Givens: a G.P. whose first three terms have product

27

and whose set of possible first-three-term sums is

R - (a, b)

. Target:

a^{2} + b^{2}

. Method: use symmetric terms to fix the middle term, express the sum through

r + \frac{1}{r}

, and apply its range to locate the excluded interval.

Step 1:Take the terms as

\frac{α}{r}, α, α r

. Their product is

α^{3}

\frac{α}{r} \cdot α \cdot α r = α^{3} = 27

Step 2:Write the sum of the three terms.

S = \frac{α}{r} + α + α r = 3 (r + \frac{1}{r}) + 3

Step 3:Apply

r + \frac{1}{r} \leq - 2

\geq 2

for real

r \neq = 0

3 (r + \frac{1}{r}) \in (- \infty, - 6] \cup [6, \infty)

Step 4:Hence the values

S

omits form the open interval

(- 3, 9)

, so

a = - 3

and

b = 9

R - (a, b) = R - (- 3, 9)

Step 5:Compute

a^{2} + b^{2}

(- 3)^{2} + 9^{2} = 9 + 81

Final answer:

90

Q24NumericalIntegral Calculus

The value of

r = 1 \sum 20 (π (\int_{0}^{r} x ∣ sin π x ∣ d x))

SolutionAnswer: 210

Approach:

Givens: the sum

\sum_{r = 1}^{20} π \int_{0}^{r} x ∣ sin π x ∣ d x

. Target: its value. Method: evaluate

I_{r} = \int_{0}^{r} x ∣ sin π x ∣ d x

in closed form using the symmetry of

∣ sin π x ∣

, reduce each summand, and sum the arithmetic series.

Step 1:For integer r,

∣ sin π (r - x) ∣ = ∣ sin π x ∣

, so the symmetry rule applies with

p = r

I_{r} = \int_{0}^{r} x ∣ sin π x ∣ d x = \frac{r}{2} \int_{0}^{r} ∣ sin π x ∣ d x

Step 2:Form each summand

π I_{r}

π \cdot \frac{r ^{2}}{π} = r^{2} = r

Step 3:Sum from

r = 1

20

\sum_{r = 1}^{20} r = \frac{20 \times 21}{2}

Final answer:

210

Q25NumericalCo-ordinate Geometry

For some

θ \in (0, \frac{π}{2})

, let the eccentricity and the length of the latus rectum of the hyperbola

x^{2} - y^{2} sec^{2} θ = 8

e_{1}

and

l_{1}

respectively, and let the eccentricity and the length of the latus rectum of the ellipse

x^{2} sec^{2} θ + y^{2} = 6

e_{2}

and

l_{2}

respectively. If

e_{1}^{2} = e_{2}^{2} (sec^{2} θ + 1)

, then

(\frac{l _{1} l _{2}}{e _{1} e _{2}}) tan^{2} θ

is equal to

SolutionAnswer: 8

Approach:

Givens: hyperbola

x^{2} - y^{2} sec^{2} θ = 8

with eccentricity

e_{1}

and latus rectum

l_{1}

, ellipse

x^{2} sec^{2} θ + y^{2} = 6

with eccentricity

e_{2}

and latus rectum

l_{2}

, and the relation

e_{1}^{2} = e_{2}^{2} (sec^{2} θ + 1)

. Target:

(\frac{l _{1} l _{2}}{e _{1} e _{2}}) tan^{2} θ

. Method: standardise both conics, apply the relation to fix

θ

, then substitute.

Step 1:Standardise the hyperbola as

\frac{x ^{2}}{8} - \frac{y ^{2}}{8 cos ^{2} θ} = 1

e_{1}^{2} = 1 + cos^{2} θ, l_{1} = \frac{2 \cdot 8 cos ^{2} θ}{2 2} = 42 cos^{2} θ

Step 2:Standardise the ellipse as

\frac{x ^{2}}{6 cos ^{2} θ} + \frac{y ^{2}}{6} = 1

with the major axis along y since

6 cos^{2} θ < 6

e_{2}^{2} = 1 - cos^{2} θ = sin^{2} θ, l_{2} = \frac{2 \cdot 6 cos ^{2} θ}{6} = 26 cos^{2} θ

Step 3:Apply the relation

e_{1}^{2} = e_{2}^{2} (sec^{2} θ + 1)

1 + cos^{2} θ = sin^{2} θ \cdot \frac{1 + cos ^{2} θ}{cos ^{2} θ}

Step 4:At

cos^{2} θ = \frac{1}{2}

evaluate each element:

e_{1} = 3/2

e_{2} = \frac{1}{2}

l_{1} = 22

l_{2} = 6

tan^{2} θ = 1

(\frac{l _{1} l _{2}}{e _{1} e _{2}}) tan^{2} θ = \frac{( 2 2 ) ( 6 )}{3/2 \cdot \frac{1}{2}}

Step 5:Simplify the quotient.

\frac{4 3}{3 /2} = 4 \times 2

Final answer:

8

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