The electric field in a plane electromagnetic wave is given by: E_y=69 [0.6×10^3x-1.8×10^11t ]V/m The expansion for magnetic field associated with this electromagnetic wave is _____T.

B_z=2.3×10^-7 [0.6×10^3x-1.8×10^11t]

Home/JEE Main PYQs/2026 · January 21, Shift 1

JEE Main 2026 January 21, Shift 1 Question Paper with Solutions

All 75 questions from the JEE Main 2026 (January 21, Shift 1) shift — Physics (25), Chemistry (25) and Mathematics (25) — 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 correctThermal Properties of Matter

A gas based geyser heated water flowing at the rate of 5.0 litres per minute from

2 7^{o} C

8 7^{o} C

. The rate of consumption of the gas is_____g/s. (Take heat of combustion of gas

= 5.0 \times 1 0^{4} J / g

, specific heat capacity of water

= 4200

J/kg.

^{0} C

)

(A)

2.1

(B)

0.21

(C)

4.2

(D)

0.42

SolutionAnswer: Option 4

0.42

Approach:

Equate the heat absorbed by water per unit time to the heat liberated by the burning gas per unit time, then solve for the gas mass-consumption rate.

Step 1:List the data in SI units; water density gives mass rate from volume rate.

\dot{V} = 5 L/min \Rightarrow \overset{m}{˙}_{w} = \frac{5}{60} kg/s, c = 4200 J/kg^{\circ} C, Δ T = 6 0^{\circ} C, H_{c} = 5 \times 1 0^{4} J/g

Step 2:Compute the rate of heat supplied to the water.

\dot{Q} = \frac{5}{60} \times 4200 \times 60 = 21000 J/s

Step 3:All this heat comes from the combusting gas, so equate and solve for the gas rate.

\overset{m}{˙}_{g} = \frac{Q ˙}{H _{c}} = \frac{21000}{5 \times 1 0 ^{4}}

Step 4:Evaluate the gas consumption rate.

\overset{m}{˙}_{g} = 0.42 g/s

Final answer:

0.42

g/s

Q27Single correctElectrostatics

The point of charge of

1 0^{- 8} C

is placed at origin. The work done is moving a point at

2 μ C

from point A(4,4,2) m to B(2,2,1) m is ______ J. (

\frac{1}{4 π \in _{o}} = 9 \times 1 0^{9}

in SI unit)

(A)

45 \times 1 0^{- 6}

(B)

30 \times 1 0^{- 6}

(C)

15 \times 1 0^{- 6}

(D)

0

SolutionAnswer: Option 2

30 \times 1 0^{- 6}

Approach:

Work done by the external agent equals the change in electrostatic potential energy of the two-charge system; compute the radial distances of A and B from the origin and take the difference.

Step 1:Identify the charges and constants.

q_{1} = 1 0^{- 8} C, q_{2} = 2 \times 1 0^{- 6} C, k = 9 \times 1 0^{9}

Step 2:Find each point's distance from the origin charge.

r_{A} = 4^{2} + 4^{2} + 2^{2} = 6 m, r_{B} = 2^{2} + 2^{2} + 1^{2} = 3 m

Step 3:Substitute into the work formula.

W = 9 \times 1 0^{9} \times 1 0^{- 8} \times 2 \times 1 0^{- 6} (\frac{1}{3} - \frac{1}{6})

Step 4:Evaluate the bracket and the product.

W = 18 \times 1 0^{- 5} \times \frac{1}{6} = 3 \times 1 0^{- 5} = 30 \times 1 0^{- 6} J

Final answer:

30 \times 1 0^{- 6}

Q28Single correctDual Nature of Radiation and Matter

A light wave described by

E = 60 [sin (3 \times 1 0^{15}) t + sin (12 \times 1 0^{15}) t]

(in SI units) fall on a metal surface of work function 2.8 eV. The maximum kinetic energy of ejected photoelectron is (approximately)____eV. (

h = 6.6 \times 1 0^{- 34} J . s .

and

e = 1.6 \times 1 0^{- 19} C

)

(A)

3.8

(B)

6.0

(C)

7.8

(D)

5.1

SolutionAnswer: Option 4

5.1

Approach:

The incident light contains two angular frequencies; the maximum photoelectron energy is governed by the higher frequency. Convert that photon energy to eV and apply Einstein's photoelectric equation.

Step 1:Select the higher angular frequency in the superposition.

ω_{m a x} = 12 \times 1 0^{15} rad/s

Step 2:Compute the corresponding photon energy in joules.

E = \frac{6.6 \times 1 0 ^{- 34} \times 12 \times 1 0 ^{15}}{2 π} = 1.26 \times 1 0^{- 18} J

Step 3:Convert to electron-volts.

E = \frac{1.26 \times 1 0 ^{- 18}}{1.6 \times 1 0 ^{- 19}} = 7.88 eV

Step 4:Apply the photoelectric equation with

ϕ = 2.8

eV.

K E_{m a x} = 7.88 - 2.8 = 5.08 eV

Final answer:

5.1

Q29Single correctRotational Motion

A uniform rod of mass m and length l suspend by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut is ______. (g – acceleration due to gravity)

A horizontal uniform rod AB hangs from a ceiling by two vertical light strings attached near its two ends.

(A)

mg/2

(B)

(C)

mg/3

(D)

mg/4

SolutionAnswer: Option 4

m g /4

Approach:

Immediately after one string is cut, apply Newton's second law for translation of the centre of mass, the torque equation about the COM, and the constraint that the still-attached end has zero vertical acceleration.

Step 1:Identify the forces just after the cut: weight

m g

at COM downward and tension

T

upward at the remaining end.

F_{down} = m g, F_{up} = T

Step 2:Translational equation for the centre of mass.

m g - T = m a_{c m}

Step 3:Torque about the COM gives the angular acceleration.

T \cdot \frac{l}{2} = \frac{m l ^{2}}{12} α \Rightarrow T = \frac{m l α}{6}

Step 4:The attached end stays on the string, so its vertical acceleration is zero, linking

a_{c m}

and

α

a_{c m} = \frac{l}{2} α

Step 5:Combine (1), (2), (3) to find the angular acceleration.

m g - \frac{m l α}{6} = m \cdot \frac{l}{2} α \Rightarrow g = \frac{2 l α}{3} \Rightarrow α = \frac{3 g}{2 l}

Step 6:Back-substitute into Equation (2) for the tension.

T = \frac{m l}{6} \cdot \frac{3 g}{2 l} = \frac{m g}{4}

Final answer:

m g /4

Q30Single correctLaws of Motion

A 4Kg mass moves under the influence of a force

F = (4 t^{3} i - 3 t j) N

where t is time in sec. If mass starts from origin

a + t = 0

, the velocity and position after t=2s will be

(A)

v = 4 i - \frac{3}{2} j; r = \frac{6}{5} i - j

(B)

v = 4 i - \frac{3}{2} j; r = \frac{8}{5} i - j

(C)

v = 4 i + \frac{5}{2} j; r = \frac{8}{5} i + 2 j

(D)

v = 3 i + \frac{3}{2} j; r = \frac{6}{5} i + j

SolutionAnswer: Option 2

v = 4 \hat{i} - \frac{3}{2} \hat{j}; r = \frac{8}{5} \hat{i} - \hat{j}

Approach:

Divide the force by the mass to obtain the acceleration components, integrate each component from

t = 0

(zero initial velocity, origin) to get velocity, then integrate velocity to get position, evaluating at

t = 2

Step 1:Compute acceleration components from

F = (4 t^{3} \hat{i} - 3 t \hat{j})

N with

m = 4

kg.

a_{x} = \frac{4 t ^{3}}{4} = t^{3}, a_{y} = \frac{- 3 t}{4}

Step 2:Integrate accelerations to velocity at

t = 2

v_{x} = \int_{0}^{2} t^{3} d t = \frac{2 ^{4}}{4} = 4, v_{y} = \int_{0}^{2} - \frac{3 t}{4} d t = - \frac{3}{8} \cdot 4 = - \frac{3}{2}

Step 3:Integrate

v_{x} (t) = t^{4} /4

for the x-position.

x = \int_{0}^{2} \frac{t ^{4}}{4} d t = \frac{1}{4} \cdot \frac{2 ^{5}}{5} = \frac{1}{4} \cdot \frac{32}{5} = \frac{8}{5}

Step 4:Integrate

v_{y} (t) = - 3 t^{2} /8

for the y-position.

y = \int_{0}^{2} - \frac{3 t ^{2}}{8} d t = - \frac{3}{8} \cdot \frac{2 ^{3}}{3} = - 1

Step 5:Assemble velocity and position vectors.

v = 4 \hat{i} - \frac{3}{2} \hat{j}, r = \frac{8}{5} \hat{i} - \hat{j}

Final answer:

v = 4 \hat{i} - \frac{3}{2} \hat{j}; r = \frac{8}{5} \hat{i} - \hat{j}

Q31Single correctSemiconductor Electronics

The given circuit works as:

A logic circuit with inputs A and B passing through gates feeding a final gate that produces the Output.

(A)

AND gate

(B)

NOR gate

(C)

NAND gate

(D)

OR gate

SolutionAnswer: Option 3NAND gate

Approach:

Trace the Boolean expression through the gate network from inputs A and B to the output and identify the equivalent single gate.

Step 1:Take the two inputs A and B feeding the gate combination.

inputs = A, B

Step 2:Reduce the cascaded gate network using Boolean algebra to a single expression.

Y = \overline{A \cdot B}

Step 3:Recognise the standard gate corresponding to the complemented product.

\overline{A \cdot B} \equiv NAND

Final answer: NAND gate

Q32Single correctElectromagnetic Induction

A 1 m long metal rod AB completes the circuit as shown in figure. The area of circuit is perpendicular to the magnetic field of 0.10 T. If the resistance of the total circuit is 2

Ω

then the force needed to move the rod towards right with constant speed (v) of

1.5 m / s

is ______N.

A conducting rod AB slides to the right on two horizontal rails closed by a resistor on the left, in a magnetic field directed into the page.

(A)

5.7 \times 1 0^{- 3}

(B)

7.5 \times 1 0^{- 3}

(C)

7.5 \times 1 0^{- 2}

(D)

5.7 \times 1 0^{- 2}

SolutionAnswer: Option 2

7.5 \times 1 0^{- 3}

Approach:

At constant speed the applied force balances the magnetic retarding force on the induced current; the motional EMF drives a current giving a force

B^{2} l^{2} v / R

Step 1:List the given quantities.

B = 0.10 T, l = 1 m, R = 2 Ω, v = 1.5 m/s

Step 2:For constant speed the net force is zero, so the applied force equals the magnetic retarding force.

F = \frac{B ^{2} l ^{2} v}{R}

Step 3:Substitute the numerical values.

F = \frac{( 0.1 ) ^{2} \times 1 ^{2} \times 1.5}{2}

Step 4:Evaluate.

F = \frac{0.015}{2} = 7.5 \times 1 0^{- 3} N

Final answer:

7.5 \times 1 0^{- 3}

Q33Single correctElectromagnetic Induction

A conducting circular loop of area

1.0 m^{2}

is placed perpendicular to a magnetic field which varies as

B = sin (100 t)

Tesla. If the resistance of the loop is

100Ω

. Then the average thermal energy dissipated in the loop in one period is_____J.

(A)

2 π

(B)

π

(C)

π^{2}

(D)

π /2

SolutionAnswer: Option 2

π

Approach:

Use Faraday's law to get the induced EMF from the time-varying field, write the instantaneous dissipated power

ε^{2} / R

, and integrate over one full period to get the heat.

Step 1:List the loop data; the period of

sin (100 t)

2 π /100

A = 1 m^{2}, B = sin (100 t), R = 100 Ω, T = \frac{2 π}{100} s

Step 2:Differentiate the flux to get the induced EMF.

ε = - A \frac{d B}{d t} = - 1 \cdot 100 cos (100 t) = - 100 cos (100 t) V

Step 3:Form the heat integral over one period.

H = \int_{0}^{T} \frac{( 100 cos 100 t ) ^{2}}{100} d t = \int_{0}^{T} 100 cos^{2} (100 t) d t

Step 4:The average of

cos^{2}

over a full period is

1/2

H = 100 \times \frac{1}{2} \times T = 50 \times \frac{2 π}{100} = π

Final answer:

π

Q34Single correctMoving Charges and Magnetism

A current carrying solenoid is placed vertically and a particle of mass m which charge Q is released from rest. The particle move along the axis of solenoid. If g is acceleration due to gravity, then the acceleration (a) of the charged particle will satisfy

(A)

a=g

(B)

a=0

(C)

0<a<g

(D)

a>g

SolutionAnswer: Option 1

a = g

Approach:

On the axis of a solenoid the magnetic field points along the axis; the released particle moves along that same axis, so velocity and field are parallel and the magnetic force vanishes, leaving only gravity.

Step 1:Set the geometry: vertical solenoid axis, particle released from rest moving along the axis.

v ∥ axis, B ∥ axis

Step 2:Evaluate the magnetic force when velocity is parallel to the axial field.

v ∥ B \Rightarrow v \times B = 0 \Rightarrow F_{mag} = 0

Step 3:Apply Newton's second law with only gravity acting.

F_{net} = m g \Rightarrow a = g

Final answer:

a = g

Q35Single correctWave Optics

In a double slit experiment, the distance between the slits is 0.1cm and the screen is placed at 50cm from the slit plan. When one slit is covered with a transparent sheet having thickness t and refractive index n(=1.5), the central fringe shifts by 0.2cm. The value of t is

(A)

5.6 \times 1 0^{- 4}

(B)

5.0 \times 1 0^{- 3}

(C)

8 \times 1 0^{- 4}

(D)

6.0 \times 1 0^{- 3}

SolutionAnswer: Option 3

8 \times 1 0^{- 4}

Approach:

A thin transparent sheet over one slit shifts the central fringe by

Δ x = \frac{D t ( μ - 1 )}{d}

; solve this for the thickness t.

Step 1:List the given quantities in consistent (cm) units.

d = 0.1 cm, D = 50 cm, μ = 1.5, Δ x = 0.2 cm

Step 2:Substitute into the fringe-shift relation.

0.2 = \frac{50 \times t \times ( 1.5 - 1 )}{0.1} = \frac{50 \times 0.5}{0.1} t

Step 3:Solve for the sheet thickness.

t = \frac{0.2}{250} = 8 \times 1 0^{- 4} cm

Final answer:

8 \times 1 0^{- 4}

Q36Single correctAtoms and Nuclei

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ________m. (Atomic number of gold = 79 and

\frac{1}{4 π \in _{0}} = 9 \times 1 0^{9}

in SI units)

(A)

2.95 \times 1 0^{- 14}

(B)

2.95 \times 1 0^{- 16}

(C)

3.85 \times 1 0^{- 14}

(D)

3.85 \times 1 0^{- 16}

SolutionAnswer: Option 1

2.95 \times 1 0^{- 14}

Approach:

At the distance of closest approach the alpha particle is momentarily at rest, so its initial kinetic energy is entirely converted into Coulomb potential energy between the alpha (

2 e

) and the gold nucleus (

79 e

Step 1:Convert the kinetic energy to joules and note the constant.

K E = 7.7 \times 1 0^{6} \times 1.6 \times 1 0^{- 19} J, k = 9 \times 1 0^{9}

Step 2:Set Coulomb PE equal to KE and solve for

r

r = \frac{k ( 2 e ) ( 79 e )}{K E} = \frac{k \cdot 158 e ^{2}}{K E}

Step 3:Substitute numerical values.

r = \frac{9 \times 1 0 ^{9} \times 158 \times ( 1.6 \times 1 0 ^{- 19} ) ^{2}}{7.7 \times 1 0 ^{6} \times 1.6 \times 1 0 ^{- 19}}

Step 4:Simplify and evaluate.

r = \frac{9 \times 1 0 ^{9} \times 79 \times 1.6 \times 1 0 ^{- 19} \times 2}{7.7 \times 1 0 ^{6}} \approx 2.95 \times 1 0^{- 14} m

Final answer:

2.95 \times 1 0^{- 14}

Q37Single correctWork, Energy and Power

Potential energy (V) versus distance (x) is given by the graph. Rank various regions as per the magnitudes of the force.(F) acting on a particle from high to low.

A potential energy V versus distance x graph with labelled segments A,B,C,D,E of different slopes.

(A)

F_{C D} > F_{A B} > F_{BC} > F_{D E}

(B)

F_{C D} > F_{D E} > F_{A B} > F_{BC}

(C)

F_{BC} > F_{C D} > F_{D E} > F_{A B}

(D)

F_{BC} > F_{A B} > F_{D E} > F_{C D}

SolutionAnswer: Option 4

F_{BC} > F_{A B} > F_{D E} > F_{C D}

Approach:

The force on the particle is

F = - d V / d x

, so its magnitude equals the magnitude of the slope of the

V

versus

x

graph; compute each segment's slope and rank by magnitude.

Step 1:Read the segment endpoints from the V-x graph.

A (0, 0), B (1, 1), C (2, 4), D (4, 4), E (6, 3)

Step 2:Compute the slope of each segment.

m_{A B} = \frac{1 - 0}{1 - 0} = 1, m_{BC} = \frac{4 - 1}{2 - 1} = 3, m_{C D} = \frac{4 - 4}{4 - 2} = 0, m_{D E} = \frac{3 - 4}{6 - 4} = - \frac{1}{2}

Step 3:Take magnitudes, since force magnitude is proportional to the slope magnitude.

∣ m_{BC} ∣ = 3 > ∣ m_{A B} ∣ = 1 > ∣ m_{D E} ∣ = \frac{1}{2} > ∣ m_{C D} ∣ = 0

Step 4:Write the force ranking from high to low.

F_{BC} > F_{A B} > F_{D E} > F_{C D}

Final answer:

F_{BC} > F_{A B} > F_{D E} > F_{C D}

Q38Single correctWaves

Two strings (A,B) having linear densities

μ_{A} = 2 \times 1 0^{- 4} k g / m

and

μ_{B} = 4 \times 1 0^{- 4} k g / m

and lengths

L_{A} = 2.5 m

and

L_{B} = 1.5 m

respectively are joined. Free ends of A and B are tied to two rigid supports C and D, respectively creating a tension of 500 N in the wire. Two identical pulses, sent from C and D ends, take time

t_{1}

and

t_{2}

respectively, to reach the joint. The ratio

t_{1} / t_{2}

is :

(A)

1.90

(B)

1.67

(C)

1.18

(D)

1.08

SolutionAnswer: Option 3

1.18

Approach:

A transverse pulse travels at

v = T / μ

on each string; the time to reach the joint is length over speed. The common tension cancels in the ratio

t_{1} / t_{2}

Step 1:List the data for the two strings, joined under common tension

T = 500

μ_{A} = 2 \times 1 0^{- 4}, μ_{B} = 4 \times 1 0^{- 4}, L_{A} = 2.5, L_{B} = 1.5, T = 500

Step 2:Form the time ratio; the tension cancels.

\frac{t _{1}}{t _{2}} = \frac{L _{A} μ _{A} / T}{L _{B} μ _{B} / T} = \frac{L _{A}}{L _{B}} \frac{μ _{A}}{μ _{B}}

Step 3:Substitute the values.

\frac{t _{1}}{t _{2}} = \frac{2.5}{1.5} \frac{2 \times 1 0 ^{- 4}}{4 \times 1 0 ^{- 4}} = \frac{5}{3} \cdot \frac{1}{2} = \frac{5}{3 2}

Step 4:Evaluate the numerical ratio.

\frac{5}{3 2} = \frac{5}{4.243} \approx 1.18

Final answer:

1.18

Q39Single correctUnits and Measurements

Consider a modified Bernoulli equation.

(P + \frac{A}{B t ^{2}}) + ρ g (h + Bt) + \frac{1}{2} ρ V^{2} =

Constant If f has the dimension of time then the dimension A and B are _____, _____ respectively.

(A)

[M L^{0} T^{- 2}]

and

[M^{0} L T^{- 1}]

(B)

[M L^{0} T^{- 1}]

and

[M^{0} L T^{- 1}]

(C)

[M L^{0} T^{- 2}]

and

[M^{0} L T^{- 2}]

(D)

[M L^{0} T^{- 1}]

and

[M^{0} L T]

SolutionAnswer: Option 2

[M L^{0} T^{- 1}]

and

[M^{0} L T^{- 1}]

Approach:

Apply the principle of dimensional homogeneity: every additive term must carry the dimension of pressure. Use

[Bt] =

length to obtain [B], then equate

[A / (B t^{2})]

to pressure to obtain [A].

Step 1:State targets: t has dimension of time; find [A] and [B] from homogeneity of the term

ρ g (h + Bt)

and the pressure term

A / (B t^{2})

[t] = T

Step 2:In

(h + Bt)

Bt

must have the dimension of

h

(length).

[Bt] = L \Rightarrow [B] = \frac{L}{T} = M^{0} L T^{- 1}

Step 3:The term

A / (B t^{2})

is added to pressure P, so it equals pressure dimensionally.

\frac{[ A ]}{[ B ] [ t ] ^{2}} = [P] \Rightarrow [A] = [P] [B] [t]^{2}

Step 4:Substitute the dimensions and simplify.

[A] = (M L^{- 1} T^{- 2}) (L T^{- 1}) (T^{2}) = M L^{0} T^{- 1}

Final answer:

[A] = [M L^{0} T^{- 1}]

and

[B] = [M^{0} L T^{- 1}]

(option 2)

Q40Single correctGravitation

Initially a satellite of 100 kg is in a circular orbit of radius

1.5 R_{E}

. This satellite can be moved to a circular orbit of radius

3 R_{E}

by supplying

α \times 1 0^{6} J

of energy The value of

α

is __________ (Take Radius of Earth

R_{E} = 6 \times 1 0^{6}

m and g

= 10 m / s^{2}

)

(A)

100

(B)

150

(C)

1000

(D)

500

SolutionAnswer: Option 3

1000

Approach:

The total mechanical energy of a circular orbit is

- GMm /2 r

. Energy supplied equals the difference between final and initial orbital energies. Replace GM by

g R_{E}^{2}

Step 1:Define radii and data.

r_{i} = 1.5 R_{E}, r_{f} = 3 R_{E}, m = 100, R_{E} = 6 \times 1 0^{6}, g = 10

Step 2:Energy difference between final and initial orbits.

Δ E = - \frac{GM m}{2} (\frac{1}{3 R _{E}} - \frac{1}{1.5 R _{E}}) = \frac{GM m}{6 R _{E}}

Step 3:Replace

GM = g R_{E}^{2}

Δ E = \frac{g R _{E}^{2} m}{6 R _{E}} = \frac{g R _{E} m}{6}

Step 4:Substitute numbers.

Δ E = \frac{10 \times 6 \times 1 0 ^{6} \times 100}{6} = 1 0^{9} J

Step 5:Express as

α \times 1 0^{6}

1 0^{9} = α \times 1 0^{6} \Rightarrow α = 1000

Final answer:

α = 1000

(option 3)

Q41Single correctMechanical Properties of Fluids

Water flows through a horizontal tube as shown in the figure. The difference in height between the water Columns in vertical tubes is 5 cm and the area of cross-sections at A and B are

6 c m^{2}

and

3 c m^{2}

respectively. The rate of flown will be ________

c m^{3} / s

. (take

g = 10 m / s^{2}

)

A horizontal venturi tube with a wide section at A and a narrow section at B, each fitted with a vertical column; the column heights differ by 5 cm.

(A)

2003

(B)

\frac{200}{3}

(C)

1003

(D)

2006

SolutionAnswer: Option 1

2003

Approach:

Apply continuity and Bernoulli to the wide (A) and narrow (B) sections of a horizontal venturi tube. The column height difference gives the pressure difference

ρ g H

, leading to the venturi flow-rate formula. Work in CGS.

Step 1:Define given quantities in CGS.

a_{1} = 6 cm^{2}, a_{2} = 3 cm^{2}, H = 5 cm, g = 1000 cm/s^{2}

Step 2:Geometric factor from continuity + Bernoulli.

a_{1}^{2} - a_{2}^{2} = 36 - 9 = 27 = 33

Step 3:Compute

2 g H

2 \times 1000 \times 5 = 10000 = 100 cm/s

Step 4:Apply the venturi formula.

Q = \frac{a _{1} a _{2}}{a _{1}^{2} - a _{2}^{2}} 2 g H = \frac{6 \times 3}{3 3} \times 100 = \frac{600}{3}

Step 5:Rationalise.

Q = \frac{600}{3} = 2003 cm^{3} / s

Final answer:

Q = 2003 cm^{3} / s

(option 1)

Q42Single correctElectromagnetic Waves

The electric field in a plane electromagnetic wave is given by:

E_{y} = 69 sin [0.6 \times 1 0^{3} x - 1.8 \times 1 0^{11} t] V / m

The expansion for magnetic field associated with this electromagnetic wave is _____T.

(A)

B_{z} = 2.3 \times 1 0^{- 7} sin [0.6 \times 1 0^{3} x + 1.8 \times 1 0^{11} t]

(B)

B_{z} = 2.3 \times 1 0^{- 7} sin [0.6 \times 1 0^{3} x - 1.8 \times 1 0^{11} t]

(C)

B_{y} = 69 sin [0.6 \times 1 0^{3} x + 1.8 \times 1 0^{11} t]

(D)

B_{y} = 2.3 \times 1 0^{- 7} sin [0.6 \times 1 0^{3} x - 1.8 \times 1 0^{11} t]

SolutionAnswer: Option 2

B_{z} = 2.3 \times 1 0^{- 7} sin [0.6 \times 1 0^{3} x - 1.8 \times 1 0^{11} t]

Approach:

For a plane EM wave, B has the same phase as E, amplitude

B_{0} = E_{0} / c

, and orientation set by

\hat{E} \times \hat{B} = \hat{k}

(propagation direction).

Step 1:Read propagation direction and E orientation from the given wave.

phase

0.6 \times 1 0^{3} x - 1.8 \times 1 0^{11} t \Rightarrow

propagation along

+ x

; E along y

Step 2:Compute the magnetic amplitude.

B_{0} = \frac{69}{3 \times 1 0 ^{8}} = 2.3 \times 1 0^{- 7} T

Step 3:Find the field direction from

\hat{E} \times \hat{B} = \hat{k}

\overset{y}{^} \times \overset{z}{^} = \overset{x}{^} \Rightarrow B

along

+ z

Step 4:Write

B

with identical phase to

E

B_{z} = 2.3 \times 1 0^{- 7} sin [0.6 \times 1 0^{3} x - 1.8 \times 1 0^{11} t]

Final answer:

B_{z} = 2.3 \times 1 0^{- 7} sin [0.6 \times 1 0^{3} x - 1.8 \times 1 0^{11} t] T

(option 2)

Q43Single correctElectrostatic Potential and Capacitance

A parallel plate capacitor has capacitance C, when there is vacuum within the parallel plates. A sheet having thickness

(\frac{1}{3})^{r d}

of the separation between the plates and relative permittivity K is introduced between the plates. The new capacitance of the system is :

(A)

\frac{3 K C}{2 K + 1}

(B)

\frac{4 K C}{3 K - 1}

(C)

\frac{C K}{2 + K}

(D)

\frac{3 C K ^{2}}{( 2 K + 1 ) ^{2}}

SolutionAnswer: Option 1

\frac{3 K C}{2 K + 1}

Approach:

Treat the plate gap as the vacuum region with a dielectric slab of thickness

t = d /3

inserted. Use the standard slab-capacitance formula and express in terms of

C = A ε_{0} / d

Step 1:Define slab thickness and permittivity.

t = \frac{d}{3}

, relative permittivity K

Step 2:Insert into the slab formula and factor out

C

C^{'} = \frac{A ε _{0}}{d - \frac{d}{3} ( 1 - \frac{1}{K} )} = \frac{C}{1 - \frac{1}{3} ( 1 - \frac{1}{K} )}

Step 3:Simplify the denominator.

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

Step 4:Invert to obtain

C^{'}

C^{'} = \frac{3 K C}{2 K + 1}

Final answer:

C^{'} = \frac{3 K C}{2 K + 1}

(option 1)

Q44Single correctUnits and Measurements

In an experiment the values of two spring constants were measured as

k_{1} = (10 \pm 0.2) N / m

and

k_{2} = (20 \pm 0.3) N / m

. If these spring are connected in parallel, then the percentage error in equivalent spring constant is :

(A)

1.67%

(B)

1.33 %

(C)

2.67%

(D)

2.33%

SolutionAnswer: Option 1

1.67%

Approach:

Springs in parallel add:

K = k_{1} + k_{2}

. For a sum, absolute uncertainties add. Convert the absolute error to a percentage of K.

Step 1:Define measured values.

k_{1} = 10 \pm 0.2, k_{2} = 20 \pm 0.3

N/m

Step 2:Equivalent constant.

K = 10 + 20 = 30 N/m

Step 3:Add absolute uncertainties.

Δ K = 0.2 + 0.3 = 0.5 N/m

Step 4:Percentage error.

\frac{Δ K}{K} \times 100 = \frac{0.5}{30} \times 100 = 1.67%

Final answer:

1.67%

(option 1)

Q45Single correctThermal Properties of Matter

An aluminium and steel rods having same lengths and cross-sections are joined to make total length of 120 cm at

3 0^{0} C

. The coefficient of linear expansion of aluminium and steel are

24 \times 1 0^{- 6} /^{0} C

and

1.2 \times 1 0^{- 5} /^{0} C

, respectively. The length of this composite rod when its temperature is raised to

10 0^{0} C

, is ________cm

(A)

120.03

(B)

120.20

(C)

120.06

(D)

120.15

SolutionAnswer: Option 4

120.15

Approach:

The composite rod is two equal halves (60 cm each) of aluminium and steel in series. Each expands by

L_{0} α Δ T

; add both expansions to the original 120 cm.

Step 1:Define data; each rod length 60 cm, temperature rise

Δ T = 7 0^{\circ}

L_{0} = 60, Δ T = 70, α_{A l} = 24 \times 1 0^{- 6}, α_{s t} = 1.2 \times 1 0^{- 5}

Step 2:Aluminium rod expansion.

Δ L_{A l} = 60 \times 24 \times 1 0^{- 6} \times 70 = 0.1008 cm

Step 3:Steel rod expansion.

Δ L_{s t} = 60 \times 1.2 \times 1 0^{- 5} \times 70 = 0.0504 cm

Step 4:Add expansions to original length.

L = 120 + 0.1008 + 0.0504 = 120.1512 cm

Final answer:

L \approx 120.15

cm (option 4)

Q46NumericalRay Optics and Optical Instruments

In a microscope, the objective is having focal length

f_{o} = 2 c m

and eye-piece is having focal length

f_{e} = 4 c m

. The tube length is 32cm. The magnification produced by this microscope for normal adjustment is ______.

SolutionAnswer: 100

Approach:

For a compound microscope in normal adjustment (image at near point

D = 25

cm), the magnification is the product of objective linear magnification (tube length over

f_{o}

) and eyepiece angular magnification (

D / f_{e}

Step 1:Define data with near point

D = 25

cm.

f_{o} = 2, f_{e} = 4, L = 32, D = 25

(cm)

Step 2:Substitute into the magnification formula.

M = \frac{L D}{f _{o} f _{e}} = \frac{32 \times 25}{2 \times 4}

Step 3:Evaluate.

M = \frac{800}{8} = 100

Final answer:

M = 100

Q47NumericalRay Optics and Optical Instruments

A collimated beam of light diameter 2mm is propagating along x-axis. The beam is required to be expanded in a collimated beam of diameter 14 mm using a system of two convex lenses. If first lens has focal length 40 mm, then the focal length of second lens is _____mm.

SolutionAnswer: 280

Approach:

A two-convex-lens beam expander (Keplerian) magnifies a collimated beam by the ratio of focal lengths, which equals the output-to-input diameter ratio. Solve for

f_{2}

Step 1:Define data.

D_{1} = 2 mm, D_{2} = 14 mm, f_{1} = 40 mm

Step 2:Apply the diameter-to-focal-length ratio.

f_{2} = f_{1} \frac{D _{2}}{D _{1}} = 40 \times \frac{14}{2}

Step 3:Evaluate.

f_{2} = 40 \times 7 = 280 mm

Final answer:

f_{2} = 280

Q48NumericalThermodynamics

A 10 mole of oxygen is heated at constant volume from

3 0^{o} C

4 0^{o} C

. The change in the internal energy of gas is _____Cal. (The molar specific heat of oxygen at constant pressure.

C_{P} = 7 Cal / mol^{o} C

and

R = 2 Cal / mol^{o} C

)

SolutionAnswer: 500

Approach:

At constant volume,

Δ U = n C_{V} Δ T

. Obtain

C_{V}

from Mayer's relation

C_{P} - C_{V} = R

Step 1:Define data; temperature change

Δ T = 1 0^{\circ}

n = 10, C_{P} = 7, R = 2

cal/(mol

\cdot^{\circ}

C),

Δ T = 10

Step 2:Find

C_{V}

C_{V} = C_{P} - R = 7 - 2 = 5 cal/(mol \cdot^{\circ} C)

Step 3:Apply the internal-energy formula.

Δ U = n C_{V} Δ T = 10 \times 5 \times 10

Step 4:Evaluate.

Δ U = 500 cal

Final answer:

Δ U = 500

Cal

Q49NumericalRotational Motion

Two identical thin rods of mass M kg and length L m are connected as shown in figure. Moment of inertia of the combined rod system about an axis passing through point P and perpendicular to the plane of the rods is

\frac{x}{12} M L^{2} k g m^{2}

. The value of x is __________

Two identical rods forming a T-shape: a vertical rod with point P at its top end, joined to a horizontal rod at its lower end.

SolutionAnswer: 17

Approach:

The system is a T-shape: a vertical rod with point

P

at its top end, joined at its lower end to the centre of a horizontal rod, distance

L

from

P

. Add the moment of inertia of the vertical rod about its end

P

and the horizontal rod about

P

via the parallel-axis theorem.

Step 1:Geometry:

P

at top of vertical rod; horizontal rod centre is at the vertical rod's lower end, a distance

L

from

P

d = L

for the horizontal rod's centre from

P

Step 2:Vertical rod about its end

P

I_{1} = \frac{M L ^{2}}{3}

Step 3:Horizontal rod about

P

: centre value plus parallel-axis shift

L

I_{2} = \frac{M L ^{2}}{12} + M L^{2}

Step 4:Sum the contributions.

I = \frac{M L ^{2}}{3} + \frac{M L ^{2}}{12} + M L^{2} = \frac{4 + 1 + 12}{12} M L^{2} = \frac{17}{12} M L^{2}

Step 5:Compare with

\frac{x}{12} M L^{2}

\frac{x}{12} = \frac{17}{12} \Rightarrow x = 17

Final answer:

x = 17

Q50NumericalCurrent Electricity

The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1

Ω

is connected across these points is _________ J.

A four-resistor bridge between nodes A and B (1 ohm and 2 ohm on top, 2 ohm and 4 ohm on bottom) with a 1 ohm bridge arm in the middle, powered by a 9 V battery.

SolutionAnswer: 1080

Approach:

The four resistors (1

Ω

Ω

top; 2

Ω

Ω

bottom) with the central 1

Ω

form a balanced Wheatstone bridge, so the bridge arm carries no current. Reduce the two series branches in parallel, add internal resistance, find current, then apply Joule heating for 60 s.

Step 1:Check balance condition for the bridge.

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

, balanced; central

1 Ω

carries no current

Step 2:Reduce: top branch

1 + 2 = 3 Ω

, bottom branch

2 + 4 = 6 Ω

, in parallel.

R_{A B} = \frac{3 \times 6}{3 + 6} = \frac{18}{9} = 2 Ω

Step 3:Total current including internal resistance

r = 1 Ω

i = \frac{9}{R _{A B} + r} = \frac{9}{2 + 1} = 3 A

Step 4:Heat generated between A and B (across

R_{A B}

) in

t = 60

H = i^{2} R_{A B} t = 3^{2} \times 2 \times 60 = 1080 J

Final answer:

H = 1080

Chemistry25 questions

Q51Single correctd- and f-Block Elements

MnO_{4}^{2 -}

in acidic medium; disproportionates to

(A)

MnO_{4}^{-}

& MnO

(B)

M n_{2} O_{7}

MnO_{2}

(C)

M n_{2} O_{7}

& MnO

(D)

MnO_{4}^{-}

MnO_{2}

SolutionAnswer: Option 4

MnO_{4}^{-}

and

MnO_{2}

Approach:

Manganate(VI) ion

MnO_{4}^{2 -}

is stable only in strongly alkaline medium; in acid it undergoes disproportionation. Assign the Mn oxidation state and split it into a simultaneous oxidation and reduction, then balance the redox equation.

Step 1:Determine the oxidation state of Mn in manganate.

O = - 2, 4 (- 2) + x = - 2 \Rightarrow x = + 6

Step 2:Identify the two half changes of disproportionation about +6.

+ 6 - e^{-} + 7 (MnO_{4}^{-}); + 6 + 2 e^{-} + 4 (MnO_{2})

Step 3:Balance electrons (2 oxidised per 1 reduced) and complete with H+ and water.

3 MnO_{4}^{2 -} + 4 H^{+} \to 2 MnO_{4}^{-} + MnO_{2} + 2 H_{2} O

Final answer:

MnO_{4}^{-}

and

MnO_{2}

Q52Single correctPurification and Characterisation of Organic Compounds

In Carius method; 0.75g of an organic compound give 1.2 gm of barium sulphate. Find percentage of sulphur (molar mass 32 gmo

l^{- 1}

). Molar mass of barium sulphate is 233 gmo

l^{- 1}

(A)

16.48%

(B)

21.97%

(C)

10.30%

(D)

4.55%

SolutionAnswer: Option 2

21.97%

Approach:

In the Carius method sulphur is converted quantitatively to

BaSO_{4}

. The mass fraction of S in

BaSO_{4}

times the precipitate-to-sample mass ratio gives the percentage of sulphur in the compound.

Step 1:List the given data.

m_{c m p d} = 0.75 g, m_{B a S O_{4}} = 1.2 g, M_{S} = 32, M_{B a S O_{4}} = 233

Step 2:Substitute into the formula.

% S = \frac{32}{233} \times \frac{1.2}{0.75} \times 100

Step 3:Evaluate the arithmetic.

\frac{1.2}{0.75} = 1.6; \frac{32}{233} = 0.13734; 0.13734 \times 1.6 \times 100 = 21.97

Final answer:

21.97%

Q53Single correctChemical Bonding and Molecular Structure

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

S F_{4}

N H_{4}^{+}

[NiCl_{4}]

XeF_{4}

[PtCl_{4}]^{2 -}

SeF_{4}

and

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

that have tetrahedral geometry is 3.
Statement – II: In the set

[N O_{2}, BeH_{2}, B F_{3}, AlCl_{3}]

; all the molecules have incomplete octet around central atom.

(A)

Both statement I and statement II are false

(B)

Both statements I and statement II are true

(C)

Statement I is false but statement II is true

(D)

Statement I is true but statement II is false

SolutionAnswer: Option 3Statement-I is false but Statement-II is true

Approach:

Statement-I: assign VSEPR/hybridisation geometry to each listed species and count the tetrahedral ones, comparing with the claimed value 3. Statement-II: test whether every molecule in the set has an incomplete octet at the central atom.

Step 1:Assign geometry of each Statement-I species.

S F_{4}

seesaw (

s p^{3} d

N H_{4}^{+}

tetrahedral (

s p^{3}

XeF_{4}

square planar (

s p^{3} d^{2}

[PtCl_{4}]^{2 -}

square planar (

dsp^{2}

SeF_{4}

seesaw (

s p^{3} d

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

square planar (

dsp^{2}

);

[NiCl_{4}]^{2 -}

tetrahedral

Step 2:Count tetrahedral species and compare to 3.

tetrahedral

= N H_{4}^{+} (+ [NiCl_{4}]^{2 -}) \Rightarrow

count

\leq 2 \neq = 3

Step 3:Examine octet status of Statement-II set.

N O_{2}

(odd-electron, 17

e^{-}

on N region),

BeH_{2}

e^{-}

B F_{3}

e^{-}

AlCl_{3}

e^{-}

)

Step 4:Combine the two verdicts to pick the option.

I false, II true

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

Q54Single correctBiomolecules

Identify correct statements:
A: Arginine and Tryptonphan, are essential Amino acid.
B: Histidine does not contain heterocyclic ring in its structure.
C: Proline is a six membered ring cyclic amino acid.
D: Glyncine does not have chiral center.
E: Cysteine has characteristic feature of side chain as

MeS - C H_{2} - C H_{2} -

.
Choose the correct answer from the options given below:

(A)

A and D only

(B)

B and E only

(C)

C and E only

(D)

C and D only

SolutionAnswer: Option 1A and D only

Approach:

Evaluate each lettered statement against standard amino-acid facts (essential amino acids, ring structures, chirality, side-chain identities) and select the set of true statements.

Step 1:Test A: essential amino acids.

Arginine and tryptophan are essential

Step 2:Test D: chirality of glycine.

H_{2} N - C H_{2} - COO H

α

-C bears two H, so no chiral centre

Step 3:Test B, C, E for errors.

B: histidine HAS an imidazole (heterocyclic) ring -> false; C: proline is a 5-membered ring -> false; E:

M e S - C H_{2} - C H_{2} -

is the methionine side chain, not cysteine (

- C H_{2} - S H

) -> false

Step 4:Select the correct set.

true statements

=

A and D

Final answer: A and D only

Q55Single correctChemical Thermodynamics

For the reaction

N_{2} O_{4} ⇌ 2 N O_{2}

, graph is plotted as shown below. Identify correct statements. A. Standard free energy change for the reaction is

- 5.40 kJmol^{- 1}

. B. As

Δ G^{(-)}

in graph is positive.

N_{2} O_{4}

Will not dissociate into

N O_{2}

at all. C. Reverse reaction will go to completion. D. When 1 mole of

N_{2} O_{4}

changes into equilibrium mixture, value of

Δ G^{(-)} = - 0.84 kJmol^{- 1}

. E. When 2 mole of

N O_{2}

changes into equilibrium mixture,

Δ G^{(-)}

for equilibrium mixture is

- 6.24 kJmol^{- 1}

. Choose the correct answer from the options given below:

$A free-energy G versus fraction-of-N2O4-dissociated curve that dips to a minimum E and rises sharply to a maximum B, with vertical markers of 5.40 and 0.84 kJ/mol.$

(A)

B and C only

(B)

A and D only

(C)

D and E only

(D)

C and E only

SolutionAnswer: Option 3D and E only

Approach:

Read the Gibbs-energy versus extent-of-dissociation curve. The minimum E lies below the pure-

N_{2} O_{4}

level A by 0.84 kJ/mol, and the pure-

NO_{2}

level B lies above A by 5.40 kJ/mol. Use that any approach toward the equilibrium minimum has

Δ G < 0

, and evaluate each statement.

Step 1:Read graph values.

G_{A} \to G_{E}

drop

= 0.84

;

G_{A} \to G_{B}

rise

= 5.40

; so

G_{B} \to G_{E} = 5.40 + 0.84 = 6.24

Step 2:Statement D: 1 mol pure

N_{2} O_{4}

(level A) approaching equilibrium (E).

Δ G = G_{E} - G_{A} = - 0.84 k J mol^{- 1}

Step 3:Statement E: 2 mol pure

NO_{2}

(level B) approaching equilibrium (E).

Δ G = G_{E} - G_{B} = - (0.84 + 5.40) = - 6.24 k J mol^{- 1}

Step 4:Reject A, B, C.

A misstates

Δ G^{\circ}

; B is false (

N_{2} O_{4}

DOES partly dissociate to the minimum); C is false (reverse reaction stops at equilibrium, not completion)

Step 5:Select the correct set.

D and E

Final answer: D and E only

Q56Single correctOrganic Compounds Containing Nitrogen

A hydrocarbon 'P' (

C_{4} H_{8}

) on reaction with HCl gives an optically active Compound Q (

C_{4} H_{9} Cl

) which on reaction with one mole of ammonia gives compound 'R' (

C_{4} H_{11} N

). 'R' on diazotization followed by hydrolysis gives 'S'. Identify P, Q, R and S.

(A)

Four condensed structural formulas labelled P, Q, R, S: P a terminal alkene (CH3-CH2-CH=CH2), Q a chloroalkane with Cl drawn on a vertical bond, R an amine, S an alcohol with a vertical -OH branch.

(B)

Drawn three-membered ring (cyclopropane) structures: P methylcyclopropane (triangle), Q chloromethyl-cyclopropane, R aminomethyl-cyclopropane, S cyclopropane bearing an -OH, each shown as a triangle ring.

(C)

Condensed structures: P but-2-ene (CH3-CH=CH-CH3), Q 2-chlorobutane with Cl on a vertical bond, R butan-2-amine with NH2 branch, S butan-2-ol with OH branch.

(D)

Condensed structures: P but-2-ene, Q a 1-chlorobutane chain, R a 1-aminobutane chain, S a chain ending in a -CH2-OH branch.

SolutionAnswer: Option 3

P = C H_{3} C H = CHCH_{3}, Q = C H_{3} C H_{2} CHClCH_{3}, R = C H_{3} C H_{2} C H (N H_{2}) C H_{3}, S = C H_{3} C H_{2} C H (O H) C H_{3}

Approach:

Take P as but-2-ene (gives a chiral chloride). Add HCl by Markovnikov/electrophilic addition to obtain the optically active 2-chlorobutane Q, substitute with ammonia to the amine R, then diazotise and hydrolyse to the alcohol S.

Step 1:Identify P and add HCl to a chiral product.

P = C H_{3} C H = CHCH_{3} H Cl C H_{3} C H_{2} CHClCH_{3}

(2-chlorobutane,

C_{4} H_{9} Cl

, chiral C-2)

Step 2:React Q with one mole of ammonia.

C H_{3} C H_{2} CHClCH_{3} N H_{3} C H_{3} C H_{2} C H (N H_{2}) C H_{3}

(butan-2-amine,

C_{4} H_{11} N

)

Step 3:Diazotise R and hydrolyse.

N a N O_{2} / H Cl R - N_{2}^{+} H_{2} O C H_{3} C H_{2} C H (O H) C H_{3}

(butan-2-ol)

Step 4:Match the assembled set to the options.

P = C H_{3} C H = CHCH_{3}, Q = C H_{3} C H_{2} CHClCH_{3}, R = C H_{3} C H_{2} C H (N H_{2}) C H_{3}, S = C H_{3} C H_{2} C H (O H) C H_{3}

Final answer:

P = C H_{3} C H = CHCH_{3}, Q = C H_{3} C H_{2} CHClCH_{3}, R = C H_{3} C H_{2} C H (N H_{2}) C H_{3}, S = C H_{3} C H_{2} C H (O H) C H_{3}

Q57Single correctOrganic Compounds Containing Nitrogen

An organic compound (P) on treatment with aqueous ammonia under hot condition forms (Q) which on heating with

B r_{2}

and KOH forms compound (R) having molecular formula

C_{6} H_{7} N

. Names of P, Q and R respectively are

(A)

Benzoic acid, benzamide, aniline

(B)

Toluic acid, methylbenzamide, 2-methylaniline

(C)

Benzoic acid, 4-methylbenzamide, 4-methylaniline

(D)

Phenylethanoic acid, phenylethanamide, benzamine

SolutionAnswer: Option 1Benzoic acid, benzamide, aniline

Approach:

Aqueous ammonia with a carboxylic acid under heat gives the amide Q; the amide then undergoes Hofmann bromamide degradation with

Br_{2}

/KOH to give a primary amine R with one fewer carbon. Match R to

C_{6} H_{7} N

Step 1:P to Q with aqueous ammonia and heat.

C_{6} H_{5} COOH + N H_{3} Δ C_{6} H_{5} CONH_{2}

(benzamide)

Step 2:Q to R via Hofmann degradation.

C_{6} H_{5} CONH_{2} B r_{2} / K O H C_{6} H_{5} N H_{2}

Step 3:Check R against the given molecular formula.

aniline

C_{6} H_{5} N H_{2} = C_{6} H_{7} N

Step 4:Name P, Q, R.

benzoic acid, benzamide, aniline

Final answer: Benzoic acid, benzamide, aniline

Q58Single correctSome Basic Principles of Organic Chemistry

From the following the least stable structure is

(A)

Skeletal resonance structure of a nitro-butenyl system: CH2=CH chain with a carbocation (+) and an adjacent carbon bearing a negative charge, terminating in a nitro group N(+) with one N=O and one O(-).

(B)

Skeletal resonance structure: a terminal carbocation (+CH2) conjugated through a C=C double bond to a nitro group N(+) with one N=O and one O(-); a negative charge marked on the chain.

(C)

Skeletal resonance structure: CH2=CH chain with a carbocation (+) on a carbon and a negative charge below the adjacent carbon, terminating in a nitro group N(+) with one N=O and one O(-).

(D)

Skeletal resonance structure: a terminal carbocation (+CH2) conjugated (C=C and C=N) to nitrogen N(+) that carries two negatively charged oxygens (O-).

SolutionAnswer: Option 3Structure with a carbanion adjacent to the positively charged nitrogen (least stable)

Approach:

These are resonance (canonical) forms of a conjugated nitro-butadienyl system. Rank stability by the standard rules: full conjugation/charge delocalisation and placement of negative charge on electronegative atoms stabilise; an isolated carbocation adjacent to a carbanion that is not conjugated through the

π

-system is the most destabilised.

Step 1:State the stability criteria for canonical forms.

more covalent bonds + charge on electronegative atoms = more stable; localised + adjacent unlike charges not in conjugation = less stable

Step 2:Apply to the four structures.

option 3 places a terminal C=C still present (un-conjugated) with an isolated

+

carbocation and a separate carbanion (

^{-}

) that is not delocalised into the nitro group, i.e. broken conjugation

Step 3:Compare and select the least stable form.

the non-conjugated carbocation/carbanion structure is the least stable contributor

Final answer: The canonical form with an isolated carbocation and a non-conjugated carbanion (charges not delocalised into the nitro group), i.e. option 3, is the least stable.

Q59Single correctAldehydes, Ketones and Carboxylic Acids

An organic compound 'P' of molecular formula

C_{6} H_{12} O_{3}

gives positive iodoform test but negative Tollen's test. When 'P' is treated with dilute acid, it produces 'Q'. 'Q' gives positive Tollen's test and also iodoform test. The structure of 'P' is

(A)

A drawn condensed structure: methyl ketone CH3-CO- joined to CH2-CH bearing two OCH3 groups (an acetal), the carbonyl drawn as a vertical C=O.

C H_{3} - O C ∣∣ - C H_{2} - OC H_{3} C H ∣ - OCH_{3}

(B)

A drawn structure CH3-CO-CH(OCH3)-CH2-OCH3 with the carbonyl drawn as a vertical C=O and two OCH3 branches.

C H_{3} - O C ∣∣ - OC H_{3} C H ∣ - OC H_{3} C H_{2} ∣

(C)

A drawn structure H-CO-CH2-CH2-CH(OCH3) with an OCH3 branch, an aldehyde carbonyl drawn at the chain end.

H - C ∣∣ O - C H_{2} - C H_{2} - OC H_{3} C H ∣ - OCH_{3}

(D)

A drawn structure CH3-CO-C(OCH3)2-CH3 with the carbonyl drawn as a vertical C=O and a gem-diether (two OCH3) carbon.

C H_{3} - C ∣∣ O - OC H_{3} C ∣ OC H_{3} - C H_{3}

SolutionAnswer: Option 1

C H_{3} - CO - C H_{2} - C H (OC H_{3})_{2}

(a methyl ketone with an acetal)

Approach:

Positive iodoform requires a methyl ketone (CH3-CO-) group; negative Tollens that turns positive only after dilute-acid treatment indicates a masked aldehyde, i.e. an acetal -CH(OR)2 that hydrolyses to -CHO. Build P with C6H12O3 containing a methyl ketone plus a dimethyl acetal.

Step 1:Interpret the test results on P.

iodoform + => methyl ketone present; Tollens - (no free CHO) but + after acid => an acetal that unmasks an aldehyde

Step 2:Write a C6H12O3 structure fitting both tests.

P = C H_{3} COCH_{2} C H (OCH_{3})_{2}

Step 3:Hydrolyse P with dilute acid to give Q.

C H_{3} COCH_{2} C H (OCH_{3})_{2} H_{3} O^{+} C H_{3} COCH_{2} CHO

Step 4:Confirm Q gives both tests.

Q has both

C H_{3} CO -

(iodoform +) and a free

- C H O

(Tollens +)

Final answer:

C H_{3} - CO - C H_{2} - C H (OC H_{3})_{2}

(option 1): a methyl ketone bearing a dimethyl acetal.

Q60Single correctCoordination Compounds

Given below are two statements:
Statement–I: Among

[C u (N H_{3})_{4}]^{2 +}

[N i (e n)_{3}]^{2 +}

[N i (N H_{3})_{6}]^{2 +}

and

[M n (H_{2} O)_{6}]^{2 +}

[M n (H_{2} O)_{6}]^{2 +}

has the maximum number of unpaired electrons.
Statement–II: The number of pairs among

{[N i C l_{4}]^{2 -}, [N i (CO)_{4}]}

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

and

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

that contain only diamagnetic species is two.
In the light of the above statements, those the correct answer from the given below:

(A)

Statement I is false, statement II is true

(B)

Both statement I and statement II are false

(C)

Statement I is true but statement II is false

(D)

Both statement I and statement II are true

SolutionAnswer: Option 3Statement-I is true but Statement-II is false

Approach:

Statement-I: determine the d-electron count and unpaired electrons of each metal centre; the complex with the most unpaired electrons is identified. Statement-II: classify each Ni species as diamagnetic or paramagnetic and count how many of the three given pairs contain only diamagnetic species.

Step 1:Count unpaired electrons for Statement-I.

M n^{2 +} = 3 d^{5}

(5 unpaired);

C u^{2 +} = 3 d^{9}

(1);

N i^{2 +} = 3 d^{8}

(2 in

[N i (e n)_{3}]^{2 +}

and

[N i (N H_{3})_{6}]^{2 +}

)

Step 2:Classify the four nickel species in Statement-II.

[NiCl_{4}]^{2 -}

tetrahedral (weak field) paramagnetic;

[N i (CO)_{4}]

N i (0) d^{10}

tetrahedral diamagnetic;

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

dsp^{2}

square planar diamagnetic

Step 3:Count pairs containing only diamagnetic species.

pairs:

{NiCl_{4}, Ni(CO)_{4}}

mixed;

{NiCl_{4}, Ni(CN)_{4}}

mixed;

{Ni(CO)_{4}, Ni(CN)_{4}}

both diamagnetic

\Rightarrow

only 1 pair

Step 4:Combine verdicts.

I true, II false

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

Q61Single correctClassification of Elements and Periodicity in Properties

Which of the following represents the correct trend for the mentioned property?
A:

F > P > S > B

– First Ionisation Energy.
B:

Cl > F > S > P

– Electron Affinity.
C:

K > A l > M g > B

– Metallic Character.
D:

K_{2} O > N a_{2} O > MgO > A l_{2} O_{3}

- Basic character.
Choose the correct answer from the options given below:

(A)

A, B and D only

(B)

A and B only

(C)

B and C only

(D)

A, B, C and D

SolutionAnswer: Option 1A, B and D only

Approach:

Check each stated periodic order against the accepted trend for first ionisation energy, electron affinity, metallic character, and oxide basicity, then select the set of correct trends.

Step 1:Trend A: first ionisation energy.

F > P > S > B

: F highest; P > S (half-filled

3 p^{3}

stability) > B; order correct

Step 2:Trend B: electron affinity.

Cl > F > S > P

: Cl > F (small F atom electron repulsion); S > P; order correct

Step 3:Trend C: metallic character.

stated

K > A l > M g > B

is wrong; correct order is

K > M g > A l > B

Step 4:Trend D: basic character of oxides.

K_{2} O > N a_{2} O > MgO > A l_{2} O_{3}

: basicity falls left-to-right / up a group; order correct

Step 5:Select correct set.

A, B and D

Final answer: A, B and D only

Q62Single correctSome Basic Concepts in Chemistry

80mL of a hydrocarbon on mixing with 264ml of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273K occupy 224mL. When the system is treated with KOH solution, the volume decreases to 64mL. The formula of the hydrocarbon is

(A)

C_{2} H_{4}

(B)

C_{2} H_{6}

(C)

C_{2} H_{2}

(D)

C_{4} H_{10}

SolutionAnswer: Option 3

C_{2} H_{2}

Approach:

Eudiometry at constant T,P: equal gas volumes are proportional to moles (Avogadro).

CO_{2}

is the volume absorbed by KOH; the leftover after KOH is excess

O_{2}

. Use the

CO_{2}

volume for the carbon count and the consumed

O_{2}

for the hydrogen count.

Step 1:List measured volumes.

V_{H C} = 80, V_{O_{2}} = 264; V_{res i d u a l} = 224 (after water condensed at 273 K); V_{a f t erK O H} = 64

Step 2:Find

CO_{2}

(absorbed by KOH) and hence x.

V_{C O_{2}} = 224 - 64 = 160; x = \frac{160}{80} = 2

Step 3:Excess

O_{2}

is the 64 mL left after KOH; find

O_{2}

consumed.

V_{O_{2}, u se d} = 264 - 64 = 200; x + \frac{y}{4} = \frac{200}{80} = 2.5

Step 4:Solve for y.

2 + \frac{y}{4} = 2.5 \Rightarrow \frac{y}{4} = 0.5 \Rightarrow y = 2

Step 5:State the formula.

C_{x} H_{y} = C_{2} H_{2}

Final answer:

C_{2} H_{2}

Q63Single correctp-Block Elements

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

[SiO_{2}, C O_{2}]

[SnO, SnO_{2}]

[PbO, PbO_{2}]

and

[GeO, GeO_{2}]

which contains oxides that are both amphoteric is 2.
Statement – II:

B F_{3}

is an electron deficient molecule, act as a Lewis acid forms adduct with

N H_{3}

and has a trigonal planar geometry.
In the light of the above statements, Choose the correct answer from the options given below:

(A)

Both statements I and Statement II are true

(B)

Both statements I and Statement II are false

(C)

Statement I is false but statement II is true

(D)

Statement I is true but statement II is false

SolutionAnswer: Option 1Both Statement-I and Statement-II are true

Approach:

Statement-I: examine each oxide pair and count how many pairs have BOTH oxides amphoteric. Statement-II: verify the bonding facts about BF3 (electron deficiency, Lewis acidity, adduct with NH3, geometry).

Step 1:Classify each oxide pair for amphoterism.

[SiO_{2}, CO_{2}]

SiO_{2}

weakly acidic,

CO_{2}

acidic

\to

not both amphoteric;

[SnO, SnO_{2}]

: both amphoteric;

[PbO, PbO_{2}]

: both amphoteric;

[GeO, GeO_{2}]

GeO_{2}

amphoteric but GeO not

\to

Step 2:Count and conclude Statement-I.

number of fully amphoteric pairs

= 2

Step 3:Verify Statement-II facts about BF3.

B has 6 electrons (incomplete octet) -> electron deficient; acts as Lewis acid; forms adduct

F_{3} B \leftarrow N H_{3}

;

s p^{2}

, trigonal planar geometry

Step 4:Combine the verdicts.

I true, II true

Final answer: Both Statement-I and Statement-II are true

Q64Single correctSolutions

Elements 'P' and 'Q' from two types of non-volatile; non-ionizable compounds PQ and

P Q_{2}

. When 1g of PQ is dissolved in 50g of solvent 'A';

Δ T b

was 1.176 K while when 1g of

P Q_{2}

is dissolved in 50g of of solvent 'A';

Δ T b

was 0.689K. (Kb of A is 5 K kg mo

l^{- 1}

). The molar masses of element P and Q (in g mo

l^{- 1}

) respectively are

(A)

70, 110

(B)

25, 60

(C)

65, 145

(D)

60, 25

SolutionAnswer: Option 2

25, 60

Approach:

From each boiling-point elevation, compute the solute molar mass via the colligative relation, obtaining M(PQ) and M(PQ2); then write two linear equations in the atomic masses P and Q and solve the system.

Step 1:List the given data for both solutions in solvent A.

w = 1 g, W = 50 g, K_{b} = 5 K kg mol^{- 1}, Δ T_{b} (PQ) = 1.176 K, Δ T_{b} (P Q_{2}) = 0.689 K

Step 2:Compute molar mass of PQ from its boiling-point elevation.

M_{PQ} = \frac{5 \times 1 \times 1000}{1.176 \times 50} = \frac{5000}{58.8} = 85

Step 3:Compute molar mass of PQ2 from its boiling-point elevation.

M_{P Q_{2}} = \frac{5 \times 1 \times 1000}{0.689 \times 50} = \frac{5000}{34.45} = 145

Step 4:Write the two atomic-mass equations and subtract to isolate Q, then back-substitute for P.

P + Q = 85, P + 2 Q = 145 \Rightarrow Q = 60, P = 25

Final answer:

25, 60

(P = 25 g/mol, Q = 60 g/mol) — Option 2

Q65Single correctSome Basic Principles of Organic Chemistry

Identify correct statements from the following :
A) Propanal and propanone are functional isomers
B) Ethoxyethane and methoxy propane are metamers
C) But-2-ene shows optical isomerism
D) But-1-ene and But-2-ene are functional isomers
E) Pentane and 2,2-dimethyl propane are chain isomers
Choose the correct answer from the options given below:

(A)

A, B and E only

(B)

C, D and E only

(C)

B, C and D only

(D)

A, B and C only

SolutionAnswer: Option 1A, B and E only

Approach:

Apply the precise definition of each isomerism class to every statement A-E, classifying the relationship of each named pair, and select the set whose classifications are all correct.

Step 1:Statement A: propanal vs propanone, both

C_{3} H_{6} O

C H_{3} C H_{2} C H O (aldehyde) / C H_{3} COC H_{3} (ketone)

Step 2:Statement B: ethoxyethane vs methoxypropane, both

C_{4} H_{10} O

C_{2} H_{5} - O - C_{2} H_{5} / C H_{3} - O - C_{3} H_{7}

Step 3:Statement E: pentane vs 2,2-dimethylpropane, both

C_{5} H_{12}

C H_{3} (C H_{2})_{3} C H_{3} / C (C H_{3})_{4}

Step 4:Statement C: but-2-ene has a C=C with two different groups on each carbon and no stereocentre, so it shows geometrical (cis-trans), not optical, isomerism; Statement D: but-1-ene and but-2-ene differ only in the position of the double bond, making them position isomers, not functional isomers.

C: geometrical not optical \Rightarrow false; D: position not functional \Rightarrow false

Final answer: A, B and E only — Option 1

Q66Single correctHydrocarbons

Identify 'A' in the following reaction:

A 2 H_{2} / Pt

(decalin) ;

A K M n O_{4} Δ

(cyclohexane-1,2-dicarboxylic acid, COOH, COOH)

+

(oxalic acid COOH, COOH)

A drawn reaction scheme: starting compound A with two arrows - one labelled 2H2/Pt leading to a fused bicyclic decalin (two fused six-membered rings), the other labelled KMnO4/Delta leading to a cyclohexane-1,2-dicarboxylic acid plus oxalic acid (HOOC-COOH).

(A)

A drawn fused bicyclic (octahydronaphthalene) structure: two fused six-membered rings with two C=C double bonds drawn in one ring.

(B)

A drawn fused bicyclic structure: two fused six-membered rings with one C=C double bond at the ring-fusion region.

(C)

A drawn fused bicyclic structure: two fused six-membered rings with C=C double bonds positioned in the right-hand ring.

(D)

A drawn fused bicyclic structure: two fused six-membered rings with two non-adjacent C=C double bonds drawn.

SolutionAnswer: Option 3Octahydronaphthalene with the double bond at the ring-fusion region (the isomer giving the stated products)

Approach:

Use the catalytic-hydrogenation stoichiometry to count the degrees of unsaturation in A, then use the specific oxidative-cleavage products to fix the positions of the two C=C bonds within the fused bicyclic framework.

Step 1:Interpret the hydrogenation: addition of 2

H_{2}

over Pt converts A to decalin (decahydronaphthalene), so A is a fully fused bicyclic

C_{10}

ring system carrying exactly two C=C bonds (an octahydronaphthalene).

A + 2 H_{2} / Pt \to decalin \Rightarrow A has 2 C=C

Step 2:Apply oxidative cleavage: the products are cyclohexane-1,2-dicarboxylic acid (an intact six-membered carbocycle bearing two adjacent -COOH) plus oxalic acid (HOOC-COOH, a two-carbon diacid). This requires the two C=C bonds to lie wholly within one ring, cleaving a two-carbon fragment to oxalic acid while leaving the other ring intact as the 1,2-dicarboxylic acid.

A K M n O_{4} /Δ cyclohexane-1,2-(COOH)_{2} + (COO H)_{2}

Step 3:Select the structure: the isomer of octahydronaphthalene with both C=C bonds in the same (right-hand) ring, positioned so cleavage excises a -CH=CH- pair as oxalic acid and yields the adjacent ring as the 1,2-diacid, matches both observations.

A = octahydronaphthalene with the C=C bonds in the right-hand ring

Final answer: Octahydronaphthalene with the two C=C bonds in the right-hand ring (the isomer giving cyclohexane-1,2-dicarboxylic acid and oxalic acid) — Option 3

Q67Single correctChemical Thermodynamics

Which of the following graphs pressure 'P' versus volume 'V' represents the maximum workdone?

(A)

P-V graph: a smooth curve decreasing from P=2.0 bar at V=22.4 L to P=1.0 bar at V=44.8 L (reversible expansion), with an arrow along the curve.

(B)

P-V graph: a horizontal line at P=2.0 bar from the axis to V=22.4 L, then a vertical line dropping down at V=22.4 L (downward arrow).

(C)

P-V graph: a curve between V=22.4 and V=44.8 L with an upward arrow at V=22.4 L and a leftward arrow at P=1.0 bar (return/compression path).

(D)

P-V graph: a lens-shaped loop bounded by two curves between P=2.0 bar at V=22.4 L and P=1.0 bar at V=44.8 L.

SolutionAnswer: Option 1Curve expanding from

(22.4, 2.0)

(44.8, 1.0)

— largest area under the curve (option 1)

Approach:

Work done by an expanding gas equals the area enclosed under its P-V expansion path; evaluate each graph for direction (expansion vs compression vs isochoric) and enclosed area, selecting the expansion process with the largest area.

Step 1:State the criterion: positive work requires increasing volume (

d V > 0

); the magnitude equals the area under the path. A constant-volume step contributes zero work and a compression contributes negative work.

W > 0 when d V > 0; W = area under curve

Step 2:Examine the four graphs. Graph 2 is an isobaric step then an isochoric vertical drop (small area). Graph 3 carries arrows of compression/return (work not maximal positive). Graph 4 is a closed lens-shaped loop (net area, not a single maximal expansion). Graph 1 is a single reversible expansion curve from (22.4 L, 2.0 bar) to (44.8 L, 1.0 bar) sweeping the largest area under the path.

Graph 1: (22.4, 2.0) \to (44.8, 1.0), largest area, d V > 0

Step 3:Select the graph giving maximum work done by the gas.

W_{ma x} \Rightarrow reversible expansion curve (Graph 1)

Final answer: The reversible expansion curve from

(22.4 L, 2.0 bar)

(44.8 L, 1.0 bar)

— Option 1

Q68Single correctAtomic Structure

Statement – I: When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the energetically excited hydrogen atoms produce electromagnetic radiation of discrete frequencies.
Statement – II: The frequency of second line Balmer series obtained from

H e^{+}

equal to that of first line of Lyman series obtained from hydrogen atom.
In the light of the above statements, Choose the correct answer from the options given below:

(A)

Both statement I and statement II are false

(B)

Both statement I and statement II are true

(C)

Statement I is true but statement II is false

(D)

Statement I is false but statement II is true

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

Approach:

Assess Statement-I against the known origin of the hydrogen line spectrum, then test Statement-II quantitatively using the Bohr/Rydberg frequency relation with explicit

Z^{2}

scaling for the H first Lyman line and the

He^{+}

second Balmer line.

Step 1:Statement-I: an electric discharge dissociates

H_{2}

into atoms; excited H atoms relax between quantized levels, emitting radiation only at discrete (quantized) frequencies — the line spectrum. This is correct.

H_{2} discharge 2 H^{*} \to discrete ν

Step 2:Compute the H first Lyman line factor (

Z = 1

n_{1} = 1

n_{2} = 2

ν_{H} \propto 1^{2} (\frac{1}{1} - \frac{1}{4}) = 1 \times \frac{3}{4} = \frac{3}{4}

Step 3:Compute the He+ second Balmer line factor (

Z = 2

n_{1} = 2

, transition to

n_{2} = 4

for the second member).

ν_{H e^{+}} \propto 2^{2} (\frac{1}{4} - \frac{1}{16}) = 4 \times \frac{3}{16} = \frac{3}{4}

Step 4:Compare: the two proportionality factors are identical, so the two frequencies are equal. Statement-II is correct.

ν_{H} = ν_{H e^{+}} \Rightarrow Statement-II true

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

Q69Single correctp-Block Elements

Consider the following reactions.

PbCl_{2} + K_{2} CrO_{4} ⟶ A + 2 KCl

(Hot solution)

A + NaOH B + N a_{2} CrO_{4}

PbSO_{4} + 4 C H_{3} COONH_{4} ⟶ (N H_{4})_{2} S O_{4} + X

In the above reactions, A, B and X are respectively.

(A)

PbCrO_{4}, N a_{2} [P b (O H)_{4}]

and

[P b (N H_{3})_{4}] S O_{4}

(B)

N a_{2} [P b (O H)_{4}], PbCrO_{4}

and

[P b (N H_{3})_{4}] S O_{4}

(C)

PbCrO_{4}, N a_{2} [P b (O H)_{4}]

and

(N H_{4})_{2} [P b (C H_{3} COO)_{4}]

(D)

N a_{2} [P b (O H)_{4}], PbCrO_{4}

and

(N H_{4})_{2} [P b (C H_{3} COO)_{4}]

SolutionAnswer: Option 3

P b C r O_{4}, N a_{2} [P b (O H)_{4}] and (N H_{4})_{2} [P b (C H_{3} COO)_{4}]

Approach:

Trace the three lead reactions sequentially: precipitation of lead chromate, its amphoteric dissolution in excess base to a plumbite, and the acetate displacement forming an acetato-plumbate complex, identifying A, B and X.

Step 1:Reaction 1: hot PbCl2 with K2CrO4 precipitates yellow lead chromate, giving A.

PbCl_{2} + K_{2} CrO_{4} \to PbCrO_{4} + 2 KCl \Rightarrow A = PbCrO_{4}

Step 2:Reaction 2: PbCrO4 (amphoteric lead centre) dissolves in NaOH releasing chromate and forming the tetrahydroxoplumbate(II), giving B.

PbCrO_{4} + 4 NaOH \to N a_{2} [P b (O H)_{4}] + N a_{2} CrO_{4} \Rightarrow B = N a_{2} [P b (O H)_{4}]

Step 3:Reaction 3: PbSO4 with ammonium acetate liberates ammonium sulfate and forms the ammonium tetraacetatoplumbate(II) complex, giving X.

PbSO_{4} + 4 C H_{3} COONH_{4} \to (N H_{4})_{2} S O_{4} + (N H_{4})_{2} [P b (C H_{3} COO)_{4}] \Rightarrow X = (N H_{4})_{2} [P b (C H_{3} COO)_{4}]

Final answer:

A = PbCrO_{4}, B = N a_{2} [P b (O H)_{4}], X = (N H_{4})_{2} [P b (C H_{3} COO)_{4}]

— Option 3

Q70Single correctSome Basic Concepts in Chemistry

14.0gm of calcium metal is allowed to react with excess HCl at 1.0 atm pressure and 273K which of the following statements is incorrect? [Given molar mass in gmo

l^{- 1}

of Ca – 40; Cl – 35.5; H – 1.]

(A)

7.84L of

H_{2}

gas is evolved.

(B)

0.35 mol of

H_{2}

gas is evolved.

(C)

The limiting reagent is calcium metal.

(D)

33.3g of

CaCl_{2}

is produced.

SolutionAnswer: Option 4

33.3

g of

CaCl_{2}

is produced

Approach:

Compute moles of calcium, apply the stoichiometry

Ca + 2 HCl \to CaCl_{2} + H_{2}

to evaluate

H_{2}

moles and volume at STP, the limiting reagent and the

CaCl_{2}

mass, then identify the single statement that is false.

Step 1:Moles of Ca; HCl is in excess so 1:1 stoichiometry gives equal moles of

H_{2}

and

CaCl_{2}

n_{C a} = \frac{14}{40} = 0.35 mol \Rightarrow n_{H_{2}} = n_{C a C l_{2}} = 0.35 mol

Step 2:Volume of

H_{2}

at 1 atm, 273 K (STP, 22.4 L/mol).

V_{H_{2}} = 0.35 \times 22.4 = 7.84 L

Step 3:Limiting reagent: HCl is supplied in excess, so calcium limits the reaction.

HCl excess \Rightarrow C a is limiting

Step 4:Mass of

CaCl_{2}

(

M = 40 + 2 \times 35.5 = 111

): the produced mass is 38.85 g, not 33.3 g, so statement 4 is false.

m_{C a C l_{2}} = 0.35 \times 111 = 38.85 g \neq = 33.3 g

Final answer: The incorrect statement is "33.3 g of

CaCl_{2}

is produced" (true value 38.85 g) — Option 4

Q71NumericalChemical Kinetics

Pre-exponential factors of two different reactions of same order are identical. Let activation energy of first reaction exceeds the activation energy of second reaction by 20 kJ mo

l^{- 1}

. If

k_{1}

and

k_{2}

are the rate constants of first and second reaction respectively at 300 K, then

ln \frac{k _{2}}{k _{1}}

will be ____. (nearest integer) [R=8.3]

K^{- 1} mol^{- 1}

SolutionAnswer: 8

Approach:

With identical pre-exponential factors, take the ratio of the two Arrhenius rate constants so A cancels; the logarithm of the ratio depends only on the activation-energy difference, which is then evaluated at 300 K.

Step 1:List data: the first reaction's activation energy exceeds the second's by 20 kJ/mol; identical A; T = 300 K, R = 8.3.

E_{a 1} - E_{a 2} = 20000 J mol^{- 1}, T = 300 K, R = 8.3

Step 2:Form the ratio of rate constants; equal A cancels, and dividing exponentials gives the difference of activation energies in the exponent.

ln \frac{k _{2}}{k _{1}} = ln \frac{A e ^{- E_{a 2} / RT}}{A e ^{- E_{a 1} / RT}} = \frac{E _{a 1} - E _{a 2}}{RT}

Step 3:Substitute and evaluate.

ln \frac{k _{2}}{k _{1}} = \frac{20 000}{8.3 \times 300} = \frac{20 000}{2490} = 8.03 \approx 8

Final answer:

ln \frac{k _{2}}{k _{1}} = 8

(nearest integer)

Q72Numericald- and f-Block Elements

Consider the following reactions:

NaCl + K_{2} C r_{2} O_{7} + H_{2} S O_{4} ⟶ A + KHSO_{4} + NaHSO_{4} + H_{2} O

A + NaOH ⟶ B + NaCl + H_{2} O

B + H_{2} S O_{4} + H_{2} O_{2} ⟶ C + N a_{2} S O_{4} + H_{2} O

In the product 'C', 'X' is the number of

O_{2}^{2 -}

units, 'Y' is the total number oxygen atoms present and 'Z' is the oxidation state of Cr. The value of X + Y + Z is____.

SolutionAnswer: 13

Approach:

Identify each species in the chromyl-chloride sequence (A = chromyl chloride, B = sodium chromate, C = chromium pentoxide), then count peroxo units, total oxygen atoms and the oxidation state of Cr in C and sum them.

Step 1:Reaction 1 forms chromyl chloride A; reaction 2 hydrolyses it in NaOH to sodium chromate B; reaction 3 acidified chromate with

H_{2} O_{2}

gives blue chromium pentoxide C.

A = CrO_{2} C l_{2}, B = N a_{2} CrO_{4}, C = CrO_{5}

Step 2:In

CrO_{5}

the structure is

CrO(O_{2})_{2}

: two side-on peroxo (

O_{2}^{2 -}

) units and one doubly-bonded oxo. Count peroxo units X.

X = number of O_{2}^{2 -} = 2

Step 3:Count total oxygen atoms Y in

CrO_{5}

Y = 5 (one oxo + two O_{2}^{2 -} = 1 + 4)

Step 4:Oxidation state Z of Cr: two peroxo groups contribute 2x(-2) = -4 and the oxo contributes -2; for a neutral molecule Cr balances +6.

Z + 2 (- 2) + (- 2) = 0 \Rightarrow Z = + 6

Step 5:Sum the three quantities.

X + Y + Z = 2 + 5 + 6 = 13

Final answer:

X + Y + Z = 13

Q73NumericalOrganic Compounds Containing Nitrogen

Consider the following reaction sequence (benzene)

co n c . H N O_{3} + co n c . H_{2} S O_{4} 333 K

1. Sn/HCl/ Δ 2. pH neutralised

(C H_{3} CO)_{2} O

R ; R

co n c . H N O_{3} + co n c . H_{2} S O_{4}

S (major product) ; S

1. HCl/EtOH, Δ 2. pH neutralised

T. The percentage of nitrogen in product 'T' formed is ______%. (Nearest integer) (Given molar mass in

g mol^{- 1}

H: 1, C : 12, N : 14, O : 16)

A drawn reaction sequence starting from a benzene ring: benzene -> (conc. HNO3 + conc. H2SO4, 333 K) -> P -> (1. Sn/HCl/Delta, 2. pH neutralised) -> Q -> ((CH3CO)2O) -> R -> (conc. HNO3 + conc. H2SO4) -> S (major product) -> (1. HCl/EtOH Delta, 2. pH neutralised) -> T, drawn with the benzene hexagon and labelled arrows.

SolutionAnswer: 20

Approach:

Follow the aromatic substitution sequence step by step to the final product T (4-nitroaniline), establish its molecular formula and molar mass, then compute the nitrogen mass percentage.

Step 1:Trace the sequence: benzene nitrates to nitrobenzene (P); Sn/HCl reduction then neutralisation gives aniline (Q); acetic anhydride acetylates to acetanilide (R); nitration of acetanilide (o/p-director) gives mainly p-nitroacetanilide (S); acidic HCl/EtOH hydrolysis then neutralisation removes the acetyl group to give 4-nitroaniline (T).

C_{6} H_{6} \to C_{6} H_{5} N O_{2} \to C_{6} H_{5} N H_{2} \to C_{6} H_{5} N H COC H_{3} \to p - O_{2} N - C_{6} H_{4} - N H COC H_{3} \to p - O_{2} N - C_{6} H_{4} - N H_{2}

Step 2:Write the molecular formula and molar mass of T.

T = C_{6} H_{6} N_{2} O_{2}, M = 6 (12) + 6 (1) + 2 (14) + 2 (16) = 72 + 6 + 28 + 32 = 138

Step 3:Mass of nitrogen per mole and percentage.

% N = \frac{2 \times 14}{138} \times 100 = \frac{28}{138} \times 100

Step 4:Evaluate.

\frac{2800}{138} = 20.29 \approx 20

Final answer:

% N \approx 20%

(nearest integer)

Q74NumericalRedox Reactions and Electrochemistry

The pH and conductance of a weak acid (HX) was found to be 5 and

4 \times 1 0^{- 5} S

, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of

1 c m^{2}

were at a distance of 15 cm apart. The value of the limiting molar conductivity is __________

S m^{2} mol^{- 1}

(nearest integer) (Given: degree of dissociation of the weak acid

(α) << 1

)

SolutionAnswer: 6

Approach:

Obtain the ion (and acid) concentration from pH given

α ≪ 1

, compute the cell constant from the plate geometry, convert conductance to conductivity, then divide by concentration to get the molar conductivity, which equals the limiting molar conductivity because the acid is essentially undissociated.

Step 1:From pH = 5 the hydrogen-ion concentration is

1 0^{- 5}

M; because

α ≪ 1

the analytical acid concentration is taken equal to this ion concentration.

p H = 5 \Rightarrow [H^{+}] = 1 0^{- 5} M; α ≪ 1 \Rightarrow C \approx 1 0^{- 5} M

Step 2:Cell constant from l = 15 cm and a = 1 c

m^{2}

; convert to SI (

m^{- 1}

G^{*} = \frac{15 cm}{1 cm ^{2}} = 15 cm^{- 1} = 1500 m^{- 1}

Step 3:Conductivity from measured conductance times cell constant.

κ = G G^{*} = (4 \times 1 0^{- 5}) (1500) = 6 \times 1 0^{- 2} S m^{- 1}

Step 4:Molar conductivity; with

α ≪ 1

the value computed from the dilute ion concentration is the limiting molar conductivity.

Λ_{m} = \frac{κ}{1000 C} = \frac{6 \times 1 0 ^{- 2}}{1000 \times 1 0 ^{- 5}} = \frac{6 \times 1 0 ^{- 2}}{1 0 ^{- 2}} = 6

Final answer:

Λ_{m}^{\circ} = 6 S m^{2} mol^{- 1}

(nearest integer)

Q75NumericalChemical Thermodynamics

Use the following date:
Substance |

\frac{Δ f H ^{⊖} ( 500 K )}{k J m o l ^{- 1}}

\frac{S ^{⊖} ( 500 K )}{J K ^{- 1} m o l ^{- 1}}

AB(g) | 32 | 222

A_{2} (g)

| 6 | 146

B_{2} (g)

| x | 280
One mole each of

A_{2} (g)

and

B_{2} (g)

are taken in a 1 L closed flask and allowed to establish the equilibrium at 500K.

A_{2} (g) + B_{2} (g) ⇌ 2 A B (g)

The value of x (in

k J mol^{- 1}

) is __________(Nearest integer) (Given:

lo g K = 2.2

R = 8.3 J K^{- 1} mol^{- 1}

)

SolutionAnswer: 70

Approach:

Compute the standard Gibbs energy of reaction from the equilibrium constant, the standard entropy change from the tabulated entropies, then use Gibbs-Helmholtz to find the reaction enthalpy and equate it to the formation-enthalpy expression to solve for

x = Δ_{f} H (B_{2})

Step 1:Reaction and data:

A_{2} (g) + B_{2} (g) ⇌ 2 AB (g)

at 500 K;

lo g K = 2.2

, R = 8.3.

A_{2} + B_{2} ⇌ 2 A B, T = 500 K, lo g K = 2.2, R = 8.3

Step 2:Standard Gibbs energy of reaction from K.

Δ G^{⊖} = - 2.303 \times 8.3 \times 500 \times 2.2 = - 21026 J = - 21.03 kJ mol^{- 1}

Step 3:Standard entropy change from tabulated S values (products minus reactants).

Δ S^{⊖} = 2 (222) - (146 + 280) = 444 - 426 = 18 J K^{- 1} mol^{- 1}

Step 4:Reaction enthalpy via Gibbs-Helmholtz.

Δ H^{⊖} = Δ G^{⊖} + T Δ S^{⊖} = - 21.03 + 500 \times 0.018 = - 21.03 + 9.0 = - 12.03 kJ mol^{- 1}

Step 5:Express the same enthalpy from formation enthalpies and solve for x.

Δ H^{⊖} = 2 (32) - (6 + x) = 58 - x; 58 - x = - 12 \Rightarrow x = 70

Final answer:

x = 70 kJ mol^{- 1}

(nearest integer)

Mathematics25 questions

Q1Single correctIntegral Calculus

The value of

\int_{- π /6}^{π /6} \frac{π + 4 x ^{11}}{1 - sin ( ∣ x ∣ + \frac{π}{6} )} d x

is equal to

(A)

8 π

(B)

2 π

(C)

6 π

(D)

4 π

SolutionAnswer: Option 4

4 π

Approach:

Split the integrand into its odd and even parts; the odd part vanishes over the symmetric interval, then reduce and integrate the even part by a shift substitution.

Step 1:Separate the integrand: the constant-

π

piece with an even denominator, and the

4 x^{11}

piece.

I = \int_{- π /6}^{π /6} \frac{π}{1 - s i n ( ∣ x ∣ + π /6 )} d x + \int_{- π /6}^{π /6} \frac{4 x ^{11}}{1 - s i n ( ∣ x ∣ + π /6 )} d x

Step 2:The denominator is even in x (depends on

∣ x ∣

);

4 x^{11}

is odd, so the second integrand is odd and integrates to zero.

\int_{- π /6}^{π /6} \frac{4 x ^{11}}{1 - s i n ( ∣ x ∣ + π /6 )} d x = 0

Step 3:The first integrand is even, so fold the interval and replace

∣ x ∣

with x on

[0, π /6]

I = 2 π \int_{0}^{π /6} \frac{d x}{1 - s i n ( x + π /6 )}

Step 4:Substitute

θ = x + π /6

d θ = d x

, limits

π /6 \to π /3

, and use the reciprocal identity.

I = 2 π \int_{π /6}^{π /3} \frac{d θ}{1 - s i n θ} = 2 π \int_{π /6}^{π /3} (sec^{2} θ + sec θ tan θ) d θ

Step 5:Integrate to

tan θ + sec θ

and apply the limits.

I = 2 π [tan θ + sec θ]_{π /6}^{π /3} = 2 π [(3 + 2) - (\frac{1}{3} + \frac{2}{3})]

Step 6:Simplify:

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

, so the bracket equals

2

I = 2 π [(3 + 2) - 3] = 2 π \times 2 = 4 π

Final answer:

4 π

Q2Single correctQuadratic Equations

The sum of all roots of equation

(x - 1)^{2} - 5∣ x - 1∣ + 6 = 0

(A)

4

(B)

1

(C)

5

(D)

3

SolutionAnswer: Option 1

4

Approach:

Substitute

t = ∣ x - 1∣

to obtain a quadratic in

t

, take the non-negative roots, recover all

x

, and sum them.

Step 1:Let

t = ∣ x - 1∣ \geq 0

, turning the equation into a quadratic in t.

t^{2} - 5∣ x - 1∣ + 6 = 0 \Rightarrow t^{2} - 5 t + 6 = 0

Step 2:Factor and solve for

t

(t - 2) (t - 3) = 0 \Rightarrow t = 2, t = 3

Step 3:Both roots are non-negative, so each yields two valid

x

-values.

∣ x - 1∣ = 2 \Rightarrow x = 3, - 1; ∣ x - 1∣ = 3 \Rightarrow x = 4, - 2

Step 4:Sum all four roots.

3 + (- 1) + 4 + (- 2) = 4

Final answer:

4

Q3Single correctTrigonometry

The value of

cos ec 1 0^{0} - 3 sec 1 0^{0}

is equal to

(A)

8

(B)

6

(C)

4

(D)

2

SolutionAnswer: Option 3

4

Approach:

Combine the two reciprocal terms over a common denominator, write the numerator as a single sine, and reduce the denominator with the double-angle identity.

Step 1:Write the expression with explicit reciprocals.

E = csc 1 0^{\circ} - 3 sec 1 0^{\circ} = \frac{1}{s i n 1 0 ^{\circ}} - \frac{3}{c o s 1 0 ^{\circ}}

Step 2:Combine over the common denominator

sin 1 0^{\circ} cos 1 0^{\circ}

E = \frac{c o s 1 0 ^{\circ} - 3 s i n 1 0 ^{\circ}}{s i n 1 0 ^{\circ} c o s 1 0 ^{\circ}}

Step 3:Factor

2

from the numerator to expose a sine difference with

3 0^{\circ}

cos 1 0^{\circ} - 3 sin 1 0^{\circ} = 2 (\frac{1}{2} cos 1 0^{\circ} - \frac{3}{2} sin 1 0^{\circ}) = 2 sin (3 0^{\circ} - 1 0^{\circ}) = 2 sin 2 0^{\circ}

Step 4:Reduce the denominator with the double-angle identity.

sin 1 0^{\circ} cos 1 0^{\circ} = \frac{1}{2} sin 2 0^{\circ}

Step 5:Divide; the

sin 2 0^{\circ}

factors cancel.

E = \frac{2 s i n 2 0 ^{\circ}}{\frac{1}{2} s i n 2 0 ^{\circ}} = 4

Final answer:

4

Q4Single correctComplex Numbers

x^{2} + x + 1 = 0

, then the value of

(x + \frac{1}{x})^{4} + (x^{2} + \frac{1}{x ^{2}})^{4} + (x^{3} + \frac{1}{x ^{3}})^{4} + ..... (x^{25} + \frac{1}{x ^{25}})^{4}

(A)

145

(B)

162

(C)

175

(D)

128

SolutionAnswer: Option 1

145

Approach:

Identify the root as a primitive cube root of unity, evaluate

x^{n} + x^{- n}

n mod 3

, raise to the fourth power, and count the contributions for

n = 1

25

Step 1:From

x^{2} + x + 1 = 0

x = ω

with

ω^{3} = 1

and

x^{- 1} = ω^{2}

; the general term is

(x^{n} + x^{- n})^{4}

x = ω, ω^{3} = 1, term_{n} = (ω^{n} + ω^{- n})^{4}

Step 2:Evaluate

ω^{n} + ω^{- n}

by residue of n mod

3

n \equiv 0 : 1 + 1 = 2; n \neq \equiv 0 : ω + ω^{2} = - 1

Step 3:Raise each to the fourth power.

n \equiv 0 : 2^{4} = 16; n \neq \equiv 0 : (- 1)^{4} = 1

Step 4:For

n = 1

25

, multiples of

3

are

3, 6, \dots, 24

(eight values giving

16

); the remaining

17

values give

1

# {n \equiv 0} = 8, # {n \neq \equiv 0} = 25 - 8 = 17

Step 5:Sum the contributions.

8 \times 16 + 17 \times 1 = 128 + 17 = 145

Final answer:

145

Q5Single correctSequence and Series

Let

a_{1}, a_{2}, a_{3} .......

be a G.P of increasing positive terms such that

a_{2} . a_{3} . a_{4} = 64

a_{1} + a_{3} + a_{5} = \frac{813}{7}

then

a_{3} + a_{5} + a_{7}

is equal to

(A)

3256

(B)

3252

(C)

3248

(D)

3244

SolutionAnswer: Option 2

3252

Approach:

Use the product condition to fix

a r^{2}

, divide the given sum to form a symmetric equation in

r^{2}

, solve for

r^{2}

, then write the target as

r^{2}

times the given sum.

Step 1:First term

a > 0

, common ratio

r > 1

(increasing positive terms); target is

a_{3} + a_{5} + a_{7}

a_{2} a_{3} a_{4} = 64, a_{1} + a_{3} + a_{5} = \frac{813}{7}

Step 2:Apply the product condition.

a_{2} a_{3} a_{4} = a^{3} r^{6} = (a r^{2})^{3} = 64 \Rightarrow a r^{2} = 4

Step 3:Write the sum and divide by

a r^{2} = 4

a_{1} + a_{3} + a_{5} = a (1 + r^{2} + r^{4}) = \frac{813}{7} \Rightarrow \frac{a ( 1 + r ^{2} + r ^{4} )}{a r ^{2}} = \frac{813/7}{4} = \frac{813}{28}

Step 4:This gives

\frac{1}{r ^{2}} + 1 + r^{2} = \frac{813}{28}

; solve for

r^{2}

r^{2} + \frac{1}{r ^{2}} = \frac{785}{28} \Rightarrow 28 r^{4} - 785 r^{2} + 28 = 0 \Rightarrow r^{2} = 28 (r > 1)

Step 5:Note

a_{3} + a_{5} + a_{7} = r^{2} (a_{1} + a_{3} + a_{5})

, so multiply the given sum by

r^{2}

a_{3} + a_{5} + a_{7} = 28 \times \frac{813}{7} = 4 \times 813 = 3252

Final answer:

3252

Q6Single correctSets, Relations and Functions

The number of relations, defined on the set

{a, b, c, d}

which are both reflexive & symmetric is equal to

(A)

16

(B)

1024

(C)

64

(D)

256

SolutionAnswer: Option 3

64

Approach:

Reflexivity forces all diagonal pairs into the relation; symmetry ties each off-diagonal pair together, leaving independent binary choices over the unordered pairs.

Step 1:The set has

4

elements, so the full relation lives in

4 \times 4 = 16

ordered pairs.

∣ {a, b, c, d} ∣ = 4, ∣ A \times A ∣ = 16

Step 2:Reflexivity forces the

4

diagonal pairs to be present, with no freedom.

(a, a), (b, b), (c, c), (d, d) \in R

Step 3:The remaining

12

off-diagonal ordered pairs form

6

unordered pairs; symmetry makes each unordered pair one independent choice.

(2 4) = 6 free pairs

Step 4:Each of the

6

pairs is independently included or excluded.

2^{6} = 64

Final answer:

64

Q7Single correctDifferential Equations

Let

y = y (x)

be the solution curve of the differential equation

((1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0

y (0) = 1

then value of

y (1)

(A)

\frac{4}{e ^{π /4}} - π /2 - 1

(B)

\frac{2}{e ^{π /4}} + π /4 - 1

(C)

\frac{2}{e ^{π /4}} - π /4 - 1

(D)

\frac{4}{e ^{π /4}} + π /2 - 1

SolutionAnswer: Option 2

\frac{2}{e ^{π /4}} + \frac{π}{4} - 1

Approach:

Rewrite as a linear first-order ODE in y, use the integrating factor

e^{t a n^{- 1} x}

, integrate by parts, apply

y (0) = 1

, and evaluate at

x = 1

Step 1:Divide

(1 + x^{2}) d y + (y - tan^{- 1} x) d x = 0

(1 + x^{2}) d x

to get standard linear form.

\frac{d y}{d x} + \frac{y}{1 + x ^{2}} = \frac{tan ^{- 1} x}{1 + x ^{2}}

Step 2:Compute the integrating factor.

μ = e^{\int \frac{d x}{1 + x ^{2}}} = e^{t a n^{- 1} x}

Step 3:Multiply through; the left side is

\frac{d}{d x} (y μ)

. With

t = tan^{- 1} x

the right integral is by parts.

y e^{t a n^{- 1} x} = \int t e^{t} d t = e^{t} (t - 1) + C

Step 4:Solve for y by dividing by

e^{t a n^{- 1} x}

y = (tan^{- 1} x - 1) + C e^{- t a n^{- 1} x}

Step 5:Apply

y (0) = 1

with

tan^{- 1} 0 = 0

1 = (0 - 1) + C \cdot 1 \Rightarrow C = 2

Step 6:Evaluate at

x = 1

with

tan^{- 1} 1 = \frac{π}{4}

y (1) = \frac{π}{4} - 1 + 2 e^{- π /4} = \frac{2}{e ^{π /4}} + \frac{π}{4} - 1

Final answer:

\frac{2}{e ^{π /4}} + \frac{π}{4} - 1

Q8Single correctCoordinate Geometry

Let a point A lie between the parallel lines

L_{1}

and

L_{2}

such that its distances from

L_{1}

and

L_{2}

are 6 and 3 units respectively then the area (in sq units) of equilateral triangle ABC where the Points B and C lie on lines

L_{1}

and

L_{2}

respectively is

(A)

213

(B)

156

(C)

122

(D)

27

SolutionAnswer: Option 1

213

Approach:

Place the parallel lines as

y = 0

and

y = 9

with

A = (a, 6)

and

B = (0, 0)

L_{1}

; rotate

B A

6 0^{\circ}

to reach the third vertex C, force C onto

L_{2}

to fix a, then compute the side and area.

Step 1:Set

L_{1} : y = 0

L_{2} : y = 9

(distance

6 + 3 = 9

A = (a, 6)

at distance

6

from

L_{1}

and

3

from

L_{2}

, and

B = (0, 0)

L_{1}

A = (a, 6), B = (0, 0), L_{2} : y = 9

Step 2:Rotate

B A = (a, 6)

about B by

6 0^{\circ}

to get C.

C = (\frac{a}{2} - 33, \frac{a 3}{2} + 3)

Step 3:Force C onto

L_{2}

: its y-coordinate equals

9

\frac{a 3}{2} + 3 = 9 \Rightarrow \frac{a 3}{2} = 6 \Rightarrow a = 43

Step 4:Compute side length from

∣ A B ∣^{2} = a^{2} + 6^{2}

(side)^{2} = (43)^{2} + 6^{2} = 48 + 36 = 84

Step 5:Apply the equilateral-area formula.

Area = \frac{3}{4} \times 84 = 213

Final answer:

213

Q9Single correctCoordinate Geometry

Let foci of a hyperbola coincide with the foci of the ellipse

\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1

if eccentricity of the hyperbola is 5 then length of its latus rectum is

(A)

16

(B)

245

(C)

\frac{96}{5}

(D)

12

SolutionAnswer: Option 3

\frac{96}{5}

Approach:

Find the ellipse foci, equate them to the hyperbola foci to determine a, use the hyperbola eccentricity to get

b^{2}

, then apply the latus-rectum formula.

Step 1:For the ellipse

\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1

c^{2} = 36 - 16 = 20

, so foci are

(\pm 20, 0)

c = 20, foci (\pm 20, 0)

Step 2:Hyperbola shares these foci, so

a_{H} e_{H} = 20

with

e_{H} = 5

a_{H} = \frac{20}{5} = \frac{2 5}{5} = \frac{2}{5}

Step 3:Compute

b_{H}^{2}

from the hyperbola relation.

b_{H}^{2} = a_{H}^{2} (e_{H}^{2} - 1) = \frac{4}{5} (25 - 1) = \frac{4}{5} \times 24 = \frac{96}{5}

Step 4:Apply the latus-rectum formula.

LR = \frac{2 b _{H}^{2}}{a _{H}} = \frac{2 \times \frac{96}{5}}{\frac{2}{5}} = \frac{192/5}{2/ 5} = \frac{96}{5}

Final answer:

\frac{96}{5}

Q10Single correctLimits, Continuity and Differentiability

Let

f : R \to (0, \infty)

be a twice differentiable function such that

f (3) = 18, f^{l} (3) = 0

and

f^{ll} (3) = 4

then

x \to 1 L im lo g_{e} (\frac{f ( 2 + x )}{f ( 3 )})^{\frac{18}{( x - 1 ) ^{2}}}

is equal to

(A)

1

(B)

18

(C)

2

(D)

9

SolutionAnswer: Option 3

2

Approach:

Bring the exponent inside the logarithm, set

u = x - 1 \to 0

, expand f to second order about

3

(using

f^{'} (3) = 0

), and apply

ln (1 + z) \approx z

Step 1:Pull the exponent out; let

u = x - 1 \to 0

2 + x = 3 + u

L = lim_{u \to 0} \frac{18}{u ^{2}} ln (\frac{f ( 3 + u )}{f ( 3 )})

Step 2:Expand

f (3 + u)

with

f (3) = 18, f^{'} (3) = 0, f^{''} (3) = 4

f (3 + u) = 18 + 0 \cdot u + \frac{1}{2} (4) u^{2} = 18 + 2 u^{2}

Step 3:Form the ratio.

\frac{f ( 3 + u )}{f ( 3 )} = \frac{18 + 2 u ^{2}}{18} = 1 + \frac{u ^{2}}{9}

Step 4:Apply the small-argument logarithm.

ln (1 + \frac{u ^{2}}{9}) \approx \frac{u ^{2}}{9}

Step 5:Substitute and cancel

u^{2}

L = lim_{u \to 0} \frac{18}{u ^{2}} \cdot \frac{u ^{2}}{9} = \frac{18}{9} = 2

Final answer:

2

Q11Single correctPermutations and Combinations

The number of strictly increasing functions f from set

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

to the set

{1, 2, 3, ........., 9}

such that

f (i) \neq = i

for

1 \leq i \leq 6

is equal to

(A)

21

(B)

27

(C)

22

(D)

28

SolutionAnswer: Option 4

28

Approach:

A strictly increasing integer map forces

f (i) \geq i

; the constraint

f (i) \neq = i

becomes

f (i) \geq i + 1

. Shift the values to absorb this and count by a single binomial.

Step 1:Map

f : {1, \dots, 6} \to {1, \dots, 9}

, strictly increasing, with

f (i) \neq = i

. Strict increase forces

f (i) \geq i

f (i) \geq i for all i

Step 2:Since

f (i) \geq i

, the condition

f (i) \neq = i

is equivalent to

f (i) \geq i + 1

f (i) \neq = i ⟺ f (i) \geq i + 1

Step 3:Define

b_{i} = f (i) - 1 \in {1, \dots, 8}

, strictly increasing; then

f (i) \geq i + 1

becomes

b_{i} \geq i

1 \leq b_{1} < b_{2} < \dots < b_{6} \leq 8

Step 4:Any strictly increasing

6

-tuple from

{1, \dots, 8}

automatically satisfies

b_{i} \geq i

, so the constraint imposes nothing extra.

b_{i} \geq i holds automatically

Step 5:Count strictly increasing

6

-subsets of an

8

-element set.

(6 8) = (2 8) = 28

Final answer:

28

Q12Single correctVector Algebra

Let

\overset{a}{ˉ} = - i + 2 j + 2 k, \overset{ˉ}{b} = 8 i + 7 j - 3 k

and c be a vector such that

a \times c = \overset{ˉ}{b}

c . (i + j + k) = 4

then

∣ a + c ∣^{2}

is equal to

(A)

30

(B)

27

(C)

33

(D)

35

SolutionAnswer: Option 2

27

Approach:

Write

c = (x, y, z)

, expand

a \times c = b

into three scalar equations, add the dot-product constraint, solve the linear system, then compute

∣ a + c ∣^{2}

Step 1:With

a = (- 1, 2, 2)

and

c = (x, y, z)

, expand the cross product.

a \times c = (2 z - 2 y, 2 x + z, - y - 2 x)

Step 2:Set equal to

b = (8, 7, - 3)

2 z - 2 y = 8, 2 x + z = 7, - y - 2 x = - 3

Step 3:Add the dot-product condition

c \cdot (1, 1, 1) = 4

x + y + z = 4

Step 4:From eq3,

y = 3 - 2 x

; from eq2,

z = 7 - 2 x

. Substitute into

x + y + z = 4

x + (3 - 2 x) + (7 - 2 x) = 4 \Rightarrow - 3 x + 10 = 4 \Rightarrow x = 2

Step 5:Back-substitute to get

c

and

a + c

y = - 1, z = 3 \Rightarrow c = (2, - 1, 3), a + c = (1, 1, 5)

Step 6:Compute the magnitude squared.

∣ a + c ∣^{2} = 1^{2} + 1^{2} + 5^{2} = 27

Final answer:

27

Q13Single correctInverse Trigonometric Functions

If the domain of the function

f (x) = cos^{- 1} (\frac{2 x - 5}{11 - 3 x}) + sin^{- 1} (2 x^{2} - 3 x + 1)

is the interval

[α, β]

, then

α + 2 β

is equal to

(A)

5

(B)

1

(C)

3

(D)

2

SolutionAnswer: Option 3

3

Approach:

Impose the domain conditions

- 1 \leq u \leq 1

for

cos^{- 1} u

and

- 1 \leq v \leq 1

for

sin^{- 1} v

, solve each, intersect, and read off

α, β

Step 1:Require both arguments in

[- 1, 1]

- 1 \leq \frac{2 x - 5}{11 - 3 x} \leq 1 and - 1 \leq 2 x^{2} - 3 x + 1 \leq 1

Step 2:Solve the quadratic condition. Lower bound:

2 x^{2} - 3 x + 1 \geq - 1 \Rightarrow 2 x^{2} - 3 x + 2 \geq 0

(discriminant

9 - 16 < 0

, always true). Upper bound:

2 x^{2} - 3 x + 1 \leq 1 \Rightarrow x (2 x - 3) \leq 0

0 \leq x \leq \frac{3}{2}

Step 3:On

[0, 3/2]

11 - 3 x > 0

. Upper bound:

2 x - 5 \leq 11 - 3 x \Rightarrow 5 x \leq 16 \Rightarrow x \leq \frac{16}{5}

(holds). Lower bound:

2 x - 5 \geq - (11 - 3 x) \Rightarrow 6 \geq x

(holds).

x \leq \frac{16}{5} and x \leq 6 (both satisfied on [0, 3/2])

Step 4:Intersect: the rational condition is met throughout

[0, 3/2]

, so the domain is

[0, 3/2]

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

Step 5:Read

α = 0, β = \frac{3}{2}

and compute

α + 2 β

α + 2 β = 0 + 2 \cdot \frac{3}{2} = 3

Final answer:

3

Q14Single correctIntegral Calculus

The area of the region inside the ellipse

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

and outside the region bounded by the curve

y = ∣ x ∣ - 1

and

y = 1 - ∣ x ∣

(A)

2 (π - 1)

(B)

2 π - \frac{1}{2}

(C)

3 (π - 1)

(D)

2 π - 1

SolutionAnswer: Option 1

2 (π - 1)

Approach:

Recognize the ellipse

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

has semi-axes

a = 2, b = 1

; the curves

y = ∣ x ∣ - 1

and

y = 1 - ∣ x ∣

bound a square. Compute the ellipse area, the square area, and subtract since the square lies inside the ellipse.

Step 1:Rewrite the ellipse in standard form and read off the semi-axes.

\frac{x ^{2}}{4} + \frac{y ^{2}}{1} = 1 \Rightarrow a = 2, b = 1

Step 2:Compute the area enclosed by the ellipse.

πab = π (2) (1) = 2 π

Step 3:Find where the two curves meet:

∣ x ∣ - 1 = 1 - ∣ x ∣ \Rightarrow ∣ x ∣ = 1

, giving the closed figure with vertices

(0, 1), (1, 0), (0, - 1), (- 1, 0)

, a square with both diagonals equal to

2

∣ x ∣ = 1 \Rightarrow (\pm 1, 0), (0, \pm 1)

Step 4:Compute the area of this square.

\frac{1}{2} d_{1} d_{2} = \frac{1}{2} (2) (2) = 2

Step 5:The square's vertices satisfy

x^{2} + 4 y^{2} \leq 4

, so it lies inside the ellipse; subtract the inner area from the ellipse area.

2 π - 2 = 2 (π - 1)

Final answer:

2 (π - 1)

Q15Single correctBinomial Theorem

If the coefficient of x in the expansion of

(a x^{2} + b x + c) (1 - 2 x)^{26}

is -56 and the coefficient of

x^{2}

and

x^{3}

are both zero then a+b+c is equal to

(A)

1483

(B)

1500

(C)

1300

(D)

1403

SolutionAnswer: Option 4

1403

Approach:

Expand

(1 - 2 x)^{26}

to the powers

x^{0}, x^{1}, x^{2}, x^{3}

, write the product's coefficients of

x, x^{2}, x^{3}

using

[x^{n}] = a [x^{n - 2}] + b [x^{n - 1}] + c [x^{n}]

, set up three equations, and solve for a,b,c.

Step 1:Compute the needed coefficients of

(1 - 2 x)^{26}

T_{0} = 1, T_{1} = - 52, T_{2} = 1300, T_{3} = - 20800

Step 2:Set the coefficient of

x

- 56

b T_{0} + c T_{1} = b - 52 c = - 56

Step 3:Set the coefficient of

x^{2}

0

a T_{0} + b T_{1} + c T_{2} = a - 52 b + 1300 c = 0

Step 4:Set the coefficient of

x^{3}

0

a T_{1} + b T_{2} + c T_{3} = - 52 a + 1300 b - 20800 c = 0

Step 5:Solve the linear system.

c = 3, b = 52 (3) - 56 = 100, a = 1300

Step 6:Add the constants.

a + b + c = 1300 + 100 + 3 = 1403

Final answer:

1403

Q16Single correctCoordinate Geometry

Let PQ and MN be two straight lines touching the circle

x^{2} + y^{2} - 4 x - 6 y - 3 = 0

at the point A and B respectively. Let O be centre of the circle and

∠ AOB = \frac{π}{3}

then the locus of the point of intersection of the lines PQ and MN is

(A)

x^{2} + y^{2} - 8 x - 12 y - 25 = 0

(B)

x^{2} + y^{2} - 12 x - 18 y - 25 = 0

(C)

3 (x^{2} + y^{2}) - 12 x - 18 y - 25 = 0

(D)

3 (x^{2} + y^{2}) - 18 x - 12 y + 25 = 0

SolutionAnswer: Option 3

3 (x^{2} + y^{2}) - 12 x - 18 y - 25 = 0

Approach:

Find the circle's centre O and radius r. From the external point R, two tangents touch at A,B with

∠ AOB = 6 0^{\circ}

. Use the right triangle OAR to relate OR to r, compute the power

S_{1}

, and write the locus.

Step 1:Read off centre and radius.

O = (2, 3), r = 4 + 9 + 3 = 4

Step 2:Since

O A ⊥ R A

, triangle OAR is right-angled at A with

∠ AOR = \frac{1}{2} ∠ AOB = 3 0^{\circ}

, so

∠ ORA = 6 0^{\circ}

and

sin ∠ ORA = \frac{O A}{OR}

sin 6 0^{\circ} = \frac{r}{OR} \Rightarrow OR = \frac{4}{3 /2} = \frac{8}{3}

Step 3:Compute the power of

R

with respect to the circle.

S_{1} = O R^{2} - r^{2} = \frac{64}{3} - 16 = \frac{16}{3}

Step 4:Set

S_{1} = x^{2} + y^{2} - 4 x - 6 y - 3

equal to

\frac{16}{3}

and clear the denominator.

3 (x^{2} + y^{2} - 4 x - 6 y - 3) = 16

Step 5:Simplify.

3 (x^{2} + y^{2}) - 12 x - 18 y - 9 - 16 = 0

Final answer:

3 (x^{2} + y^{2}) - 12 x - 18 y - 25 = 0

Q17Single correctStatistics and Probability

Let the mean and variance of 7 observations 2,4,10,x,12,14,y,x>y, be 8 and 16 respectively two numbers are chosen form

{1, 2, 3, x - 4, y, 5}

one after another without replacement then the probability that the smaller number among the two chosen numbers is less than 4 is

(A)

\frac{4}{5}

(B)

\frac{1}{3}

(C)

\frac{3}{5}

(D)

\frac{2}{5}

SolutionAnswer: Option 1

\frac{4}{5}

Approach:

Use the mean and variance of the seven observations to find x,y with

x > y

, construct the selection set

{1, 2, 3, x - 4, y, 5}

, then compute via complement the probability that the smaller of two chosen numbers is less than

4

Step 1:Apply the mean to the seven values

2, 4, 10, x, 12, 14, y

\frac{42 + x + y}{7} = 8 \Rightarrow x + y = 14

Step 2:Apply the variance: known squares sum to

4 + 16 + 100 + 144 + 196 = 460

\frac{460 + x ^{2} + y ^{2}}{7} - 64 = 16 \Rightarrow x^{2} + y^{2} = 100

Step 3:Combine to get

x y

, then solve the quadratic with

x > y

2 x y = (x + y)^{2} - (x^{2} + y^{2}) = 96 \Rightarrow x y = 48 \Rightarrow x = 8, y = 6

Step 4:Form the selection set.

{1, 2, 3, x - 4, y, 5} = {1, 2, 3, 4, 6, 5} = {1, 2, 3, 4, 5, 6}

Step 5:Complement event: both chosen numbers are

\geq 4

, i.e. both from

{4, 5, 6}

P (both \geq 4) = \frac{( 2 3 )}{( 2 6 )} = \frac{3}{15} = \frac{1}{5}

Step 6:Required probability is the complement.

P = 1 - \frac{1}{5} = \frac{4}{5}

Final answer:

\frac{4}{5}

Q18Single correctCoordinate Geometry

Let c and d be vectors such that

∣ c + d ∣ = 29

and

c \times (2 i + 3 j + 4 k) = (2 i + 3 j + 4 k) \times d

.if

λ_{1}, λ_{2} (λ_{1} > λ_{2})

are the possible values of

(c + d) . (- 7 i + 2 j + 3 k)

then the equation

K^{2} x^{2} + (K^{2} - 5 K + λ_{1}) x y + (3 K + \frac{λ _{2}}{2}) . y^{2} - 8 x + 12 y + λ_{2} = 0

represents of a circle for K equal to

(A)

4

(B)

2

(C)

- 1

(D)

1

SolutionAnswer: Option 4

1

Approach:

Show

c + d

is parallel to

b = (2, 3, 4)

from the cross-product relation, use

∣ c + d ∣ = 29

to fix the scalar, compute the dot product to obtain

λ_{1}, λ_{2}

, substitute into the conic, and impose the conditions for a circle.

Step 1:From

c \times b = b \times d = - d \times b

, with

b = (2, 3, 4)

c \times b + d \times b = 0 \Rightarrow (c + d) \times b = 0

Step 2:Write

c + d = t b

and apply the magnitude with

∣ b ∣ = 29

∣ t ∣ 29 = 29 \Rightarrow t = \pm 1

Step 3:Compute the projection-type dot product.

(c + d) \cdot (- 7, 2, 3) = t (2, 3, 4) \cdot (- 7, 2, 3) = t (- 14 + 6 + 12) = \pm 4

Step 4:Substitute

λ_{1} = 4, λ_{2} = - 4

into the conic.

K^{2} x^{2} + (K^{2} - 5 K + 4) x y + (3 K - 2) y^{2} - 8 x + 12 y - 4 = 0

Step 5:Require zero

x y

-coefficient.

K^{2} - 5 K + 4 = 0 \Rightarrow K = 1 or 4

Step 6:Require equal

x^{2}

and

y^{2}

coefficients.

K^{2} = 3 K - 2 \Rightarrow K^{2} - 3 K + 2 = 0 \Rightarrow K = 1 or 2

Step 7:Take the common value satisfying both.

K = 1

Final answer:

1

Q19Single correctCoordinate Geometry

Let O be the vertex of the parabola

x^{2} = 4 y

and Q be any point on it let the locus of the point P which divides The line segment OQ internally in the ratio 2:3 be the conic C then the equation of the chord of C which is bisected at point (1,2) is

(A)

5 x - 4 y + 3 = 0

(B)

5 x - y - 3 = 0

(C)

4 x - 5 y + 6 = 0

(D)

x - 2 y + 3 = 0

SolutionAnswer: Option 1

5 x - 4 y + 3 = 0

Approach:

Parametrise Q on

x^{2} = 4 y

, apply the internal section formula to find P dividing OQ in ratio

2 : 3

, eliminate the parameter to get the conic C, then use

T = S_{1}

to write the chord of C bisected at

(1, 2)

Step 1:Take

Q = (2 t, t^{2})

x^{2} = 4 y

and

O = (0, 0)

; P divides OQ as

2 : 3

P = (\frac{2 \cdot 2 t}{5}, \frac{2 \cdot t ^{2}}{5}) = (\frac{4 t}{5}, \frac{2 t ^{2}}{5})

Step 2:Set

(h, k) = P

and eliminate t via

t = \frac{5 h}{4}

k = \frac{2 t ^{2}}{5} = \frac{2}{5} \cdot \frac{25 h ^{2}}{16} = \frac{5 h ^{2}}{8} \Rightarrow 5 x^{2} = 8 y

Step 3:Write

T = S_{1}

for

C : 5 x^{2} - 8 y = 0

at midpoint

(x_{1}, y_{1})

5 x_{1} x - 4 (y + y_{1}) = 5 x_{1}^{2} - 8 y_{1}

Step 4:Substitute

(x_{1}, y_{1}) = (1, 2)

5 x - 4 (y + 2) = 5 - 16 \Rightarrow 5 x - 4 y - 8 = - 11

Step 5:Rearrange to standard form.

5 x - 4 y + 3 = 0

Final answer:

5 x - 4 y + 3 = 0

Q20Single correctCoordinate Geometry

Let

(α, β, γ)

be the co-ordinates of the foot of the perpendicular drawn from the point

(5, 4, 2)

on the line

r = (- i + 3 j + k) + λ (2 i + 3 j - k)

then the length of the projection of vector

α i + β j + γk

on the vector

6 i + 2 j + 3 k

(A)

\frac{18}{7}

(B)

\frac{15}{7}

(C)

3

(D)

4

SolutionAnswer: Option 1

\frac{18}{7}

Approach:

Express a general point B on the line, impose that

A B

is perpendicular to the direction

d

to solve for

λ

and find the foot, then compute the scalar projection of the foot's position vector onto

(6, 2, 3)

Step 1:Point on line and vector from

A = (5, 4, 2)

with

d = (2, 3, - 1)

B = (- 1 + 2 λ, 3 + 3 λ, 1 - λ), A B = (2 λ - 6, 3 λ - 1, - λ - 1)

Step 2:Impose

A B \cdot d = 0

2 (2 λ - 6) + 3 (3 λ - 1) - (- λ - 1) = 14 λ - 14 = 0 \Rightarrow λ = 1

Step 3:Substitute

λ = 1

to get the foot

(α, β, γ)

(α, β, γ) = (1, 6, 0)

Step 4:Project the position vector

(1, 6, 0)

onto

(6, 2, 3)

, with

∣ (6, 2, 3) ∣ = 49 = 7

\frac{( 1 ) ( 6 ) + ( 6 ) ( 2 ) + ( 0 ) ( 3 )}{7} = \frac{18}{7}

Final answer:

\frac{18}{7}

Q21NumericalSequence and Series

Let

a_{1} = 1

and for

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

then

\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}})

is equal to

SolutionAnswer: 2

Approach:

Define

b_{n} = a_{n} - \frac{2}{n ^{2}}

and show the recurrence forces

b_{n + 1} = \frac{1}{2} b_{n}

, making

b_{n}

geometric; then sum the infinite series and take the modulus.

Step 1:Define the auxiliary sequence.

b_{n} = a_{n} - \frac{2}{n ^{2}}, b_{1} = a_{1} - 2 = - 1

Step 2:Form

b_{n + 1} - \frac{1}{2} b_{n}

using

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

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

Step 3:Combine over

n^{2} (n + 1)^{2}

; the numerator vanishes identically.

\frac{( n ^{2} - 2 n - 1 ) - 2 n ^{2} + ( n ^{2} + 2 n + 1 )}{n ^{2} ( n + 1 ) ^{2}} = \frac{0}{n ^{2} ( n + 1 ) ^{2}} = 0

Step 4:Identify the geometric sequence.

b_{n} = - (\frac{1}{2})^{n - 1}, r = \frac{1}{2}

Step 5:Sum and take modulus.

∣ \sum_{n = 1}^{\infty} b_{n} ∣ = \frac{- 1}{1 - \frac{1}{2}} = 2

Final answer:

2

Q22NumericalIntegral Calculus

6 \int_{0}^{π} ∣ (sin 3 x + sin 2 x + sin x) ∣ d x

is equal to

SolutionAnswer: 17

Approach:

Factor the integrand as

sin 2 x (1 + 2 cos x) = 2 sin x cos x (1 + 2 cos x)

, substitute

t = cos x

, reduce to

12 \int_{- 1}^{1} ∣ t (2 t + 1) ∣ d t

, split at the sign changes, and integrate.

Step 1:Factor the integrand.

sin 3 x + sin 2 x + sin x = sin 2 x (1 + 2 cos x) = 2 sin x cos x (1 + 2 cos x)

Step 2:Substitute

t = cos x, d t = - sin x d x

(as

x : 0 \to π

t : 1 \to - 1

);

sin x \geq 0

6 \int_{0}^{π} 2 sin x cos x (1 + 2 cos x) d x = 12 \int_{- 1}^{1} ∣ t (2 t + 1) ∣ d t

Step 3:The sign of

t (2 t + 1)

changes at

t = - \frac{1}{2}

and

t = 0

; split accordingly.

\int_{- 1}^{- 1/2} (2 t^{2} + t) d t - \int_{- 1/2}^{0} (2 t^{2} + t) d t + \int_{0}^{1} (2 t^{2} + t) d t

Step 4:Use antiderivative

\frac{2 t ^{3}}{3} + \frac{t ^{2}}{2}

and sum the pieces.

\frac{5}{24} + \frac{1}{24} + \frac{7}{6} = \frac{17}{12}

Step 5:Multiply by

12

12 \times \frac{17}{12} = 17

Final answer:

17

Q23NumericalMatrices and Determinants

For some

α, β \in R

let

A = [α 1 22]

and

B = [11 1 β]

be such that

A^{2} - 4 A + 2 I = B^{2} - 3 B + I = 0

then

(det (adj (A^{3} - B^{3})))^{2}

is equal to

SolutionAnswer: 225

Approach:

By the Cayley-Hamilton form

M^{2} - tr (M) M + det (M) I = 0

, match the given relations to fix

α, β

; compute

A^{3}, B^{3}

, take the difference, and apply

det (adj M) = (det M)^{n - 1}

before squaring.

Step 1:Match

A^{2} - 4 A + 2 I = 0

: trace 4, determinant

2

tr (A) = α + 2 = 4 \Rightarrow α = 2; det (A) = 2 α - 2 = 2

Step 2:Match

B^{2} - 3 B + I = 0

: trace 3, determinant

1

tr (B) = 1 + β = 3 \Rightarrow β = 2; det (B) = β - 1 = 1

Step 3:Compute the matrix cubes.

A^{3} = [20142820], B^{3} = [58813]

Step 4:Form the difference and its determinant.

A^{3} - B^{3} = [156207], det = 15 (7) - 20 (6) = - 15

Step 5:For

n = 2

det (adj M) = (det M)^{1}

det (adj (A^{3} - B^{3})) = - 15

Step 6:Square the result.

(- 15)^{2} = 225

Final answer:

225

Q24NumericalPermutations and Combinations

Let

S = {(m, n) : m, n \in (1, 2, 3.....50)}

if number of elements (m,n) In S such that

6^{m} + 9^{n}

is a multiple of 5 is P and the number of elements (m,n) in S such that m+n is a square of a prime number is Q then P+Q is equal to

SolutionAnswer: 1333

Approach:

Reduce

6^{m}

and

9^{n}

modulo

5

to count P (pairs with

6^{m} + 9^{n} \equiv 0

), then enumerate

m + n

equal to a prime-square value with

1 \leq m, n \leq 50

to count Q, and add.

Step 1:Reduce the expression modulo

5

6^{m} + 9^{n} \equiv 1^{m} + (- 1)^{n} = 1 + (- 1)^{n} (mod 5)

Step 2:Divisibility by

5

requires

1 + (- 1)^{n} \equiv 0

, i.e. n odd.

n odd \Rightarrow 1 + (- 1)^{n} = 0

Step 3:Count

P

m

any of

50

values,

n

any of the

25

odd values in

1..50

P = 50 \times 25 = 1250

Step 4:Prime squares with

2 \leq m + n \leq 100

are

4, 9, 25, 49

; count pairs with

1 \leq m, n \leq 50

3 + 8 + 24 + 48 = 83

Step 5:Add the two counts.

P + Q = 1250 + 83 = 1333

Final answer:

1333

Q25NumericalLimits, Continuity and Differentiability

Let

f : R \to R

be a twice differentiable function such that the quadratic equation

f (x) m^{2} - 2 f^{^{'}} (x) m + f^{^{''}} (x) = 0

In m, has two equal roots for every

x \in R

.If

f (0) = 1, f^{^{'}} (0) = 2

, and

(α, β)

is the largest interval in which the functions

f (lo g_{e} x - x)

is increasing, then

α + β

is equal to

SolutionAnswer: 1

Approach:

The equal-roots condition gives zero discriminant, leading to the ODE

f^{'2} = f f^{''}

, equivalently

f^{'} = k f

; solve with

f (0) = 1, f^{'} (0) = 2

to get f, then determine where

g (x) = f (ln x - x)

is increasing.

Step 1:Set the discriminant of

f (x) m^{2} - 2 f^{'} (x) m + f^{''} (x) = 0

to zero.

(2 f^{'})^{2} - 4 f f^{''} = 0 \Rightarrow f^{'2} = f f^{''}

Step 2:Rearrange and integrate.

\frac{f ^{''}}{f ^{'}} = \frac{f ^{'}}{f} \Rightarrow ln ∣ f^{'} ∣ = ln ∣ f ∣ + c \Rightarrow f^{'} = k f

Step 3:Apply initial data

f (0) = 1, f^{'} (0) = 2

f^{'} (0) = k f (0) \Rightarrow k = 2 \Rightarrow f (x) = e^{2 x}

Step 4:Differentiate

g (x) = f (ln x - x) = e^{2 (l n x - x)}

for

x > 0

g^{'} (x) = e^{2 (l n x - x)} \cdot 2 (\frac{1}{x} - 1)

Step 5:Solve

g^{'} (x) > 0

; the exponential is positive, so the sign comes from

\frac{1}{x} - 1

\frac{1}{x} - 1 > 0 \Rightarrow 0 < x < 1

Step 6:Read

(α, β) = (0, 1)

and add.

α + β = 0 + 1 = 1

Final answer:

1

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