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JEE Main 2017 April 09 Question Paper with Solutions

Q: How many questions are in the JEE Main 2017 April 09 paper?

The JEE Main 2017 April 09 paper has 89 questions — Physics (29), Chemistry (30) and Mathematics (30). Every question is on this page with its correct answer and a step-by-step solution.

Q: What is the marking scheme for JEE Main?

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

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

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All 89 questions from the JEE Main 2017 (April 09) shift — Physics (29), Chemistry (30) and Mathematics (30) — 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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Physics29 questions

Q1Single correctUnits and Measurements

A physical quantity P is described by the relation

P = a^{1/2} b^{2} c^{3} d^{- 4}

If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and 5%, then the relative error in P will be :

(A)

(B)

12%

(C)

32%

(D)

25%

SolutionAnswer: Option 332%

Approach:

For a product of powers, the maximum relative error is the sum of the magnitudes of each power times the corresponding relative error.

Step 1:Identify the powers of each quantity in P.

P = a^{1/2} b^{2} c^{3} d^{- 4}

Step 2:Sum the products of power magnitude and relative error.

\frac{Δ P}{P} = \frac{1}{2} (2%) + 2 (1%) + 3 (3%) + 4 (5%)

Final answer: 32%

Q2Single correctKinematics

A car is standing 200 m behind a bus, which is also at rest. The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/

s^{2}

and the car has acceleration 4 m/

s^{2}

. The car will catch up with the bus after a time of :

(A)

110

(B)

120

(C)

102

(D)

15 s

SolutionAnswer: Option 3

102

Approach:

The car catches the bus when the extra distance covered by the car equals the initial 200 m gap.

Step 1:Write displacements of car and bus from rest.

s_{c a r} = \frac{1}{2} (4) t^{2}, s_{b u s} = \frac{1}{2} (2) t^{2}

Step 2:Set the difference equal to the 200 m gap.

2 t^{2} - t^{2} = 200

Step 3:Solve for t.

t = 200 = 102

Final answer:

102

Q3Single correctWork, Energy and Power

Two particles A and B of equal mass M are moving with the same speed v as shown in the figure. They collide completely inelastically and move as a single particle C. The angle

θ

that the path of C makes with the X-axis is given by :

Two velocity vectors A and B incoming to origin O (A at 30 deg, B at 45 deg below the X-axis) and resultant C as a dashed arrow at angle theta above the +X axis; X and Y axes shown.

(A)

tan θ = \frac{3 + 2}{1 - 2}

(B)

tan θ = \frac{3 - 2}{1 - 2}

(C)

tan θ = \frac{1 - 2}{2 ( 1 + 3 )}

(D)

tan θ = \frac{1 - 3}{1 + 2}

SolutionAnswer: Option 1

tan θ = \frac{3 + 2}{1 - 2}

Approach:

Conserve linear momentum in the X and Y directions. A approaches at 30 degrees from the downward vertical and B at 45 degrees from the downward vertical; the resultant momentum direction gives the angle of C.

Step 1:Resolve A's velocity (incoming, 60 degrees above +X axis).

v_{A} = v (cos 6 0^{\circ}, sin 6 0^{\circ})

Step 2:Resolve B's velocity (incoming, 135 degrees from +X axis).

v_{B} = v (- cos 4 5^{\circ}, sin 4 5^{\circ})

Step 3:Add momentum components.

p_{x} \propto \frac{1}{2} - \frac{2}{2}, p_{y} \propto \frac{3}{2} + \frac{2}{2}

Step 4:Form the tangent of the resultant angle.

tan θ = \frac{3 + 2}{1 - 2}

Final answer:

tan θ = \frac{3 + 2}{1 - 2}

Q4Single correctKinematics

The machine as shown has 2 rods of length 1 m connected by a pivot at the top. The end of one rod is connected to the floor by a stationary pivot and the end of the other rod has a roller that rolls along the floor in a slot. As the roller goes back and forth, a 2 kg weight moves up and down. If the roller is moving towards right at a constant speed, the weight moves up with a :

Rod-roller machine: a 2 kg weight on a hatched floor at the apex pivot, two rods of length 1 m meeting at the apex, lower ends at a movable roller (left, force F to the right) and a fixed pivot (right), separated by horizontal distance x along the floor.

(A)

constant speed

(B)

decreasing speed

(C)

increasing speed

(D)

speed which is

\frac{3}{4}

th of that of the roller when the weight is 0.4 m above the ground

SolutionAnswer: Option 2decreasing speed

Approach:

Relate the apex height to the half-base using the constant rod length, then differentiate to compare the weight's speed with the roller's speed.

Step 1:Let x be half the base and y the apex height; with rod length 1 m, x squared plus y squared equals 1.

y = 1 - x^{2}

Step 2:Differentiate with respect to time.

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

Step 3:As the weight rises, the base shrinks so x decreases and y increases; hence the ratio x/y decreases.

x ↓, y ↑\Rightarrow \frac{x}{y} ↓

Final answer: decreasing speed

Q5Single correctLaws of Motion

A conical pendulum of length 1 m makes an angle

θ = 4 5^{\circ}

w.r.t. Z-axis and moves in a circle in the XY plane. The radius of the circle is 0.4 m and its center is vertically below O. The speed of the pendulum, in its circular path, will be : (Take g

= 10

s^{- 2}

)

Conical pendulum of length 1 m hanging from O on the Z-axis, making angle theta with the vertical, the bob tracing a dashed circular path (centre C below O).

(A)

0.4 m/s

(B)

4 m/s

(C)

0.2 m/s

(D)

2 m/s

SolutionAnswer: Option 42 m/s

Approach:

For a conical pendulum, the horizontal component of tension provides the centripetal force and the vertical component balances gravity.

Step 1:Apply the conical pendulum relation with r = 0.4 m and theta = 45 degrees.

v = r g tan 4 5^{\circ}

Step 2:Evaluate.

v = 4

Final answer: 2 m/s

Q6Single correctRotational Motion

A circular hole of radius

\frac{R}{4}

is made in a thin uniform disc having mass M and radius R, as shown in figure. The moment of inertia of the remaining portion of the disc about an axis passing through the point O and perpendicular to the plane of the disc is :

Uniform disc of radius R centred at O with a circular hole of radius R/4 centred at O' on the horizontal diameter, with O to O' distance 3R/4; hole tangent to the disc edge.

(A)

\frac{219 M R ^{2}}{256}

(B)

\frac{237 M R ^{2}}{512}

(C)

\frac{19 M R ^{2}}{512}

(D)

\frac{197 M R ^{2}}{256}

SolutionAnswer: Option 2

\frac{237 M R ^{2}}{512}

Approach:

Subtract the moment of inertia of the removed small disc (using the parallel axis theorem to shift it to O) from that of the full disc.

Step 1:Full disc MOI about O.

I_{f u ll} = \frac{1}{2} M R^{2}

Step 2:Mass of removed disc of radius R/4.

m = M \frac{( R /4 ) ^{2}}{R ^{2}} = \frac{M}{16}

Step 3:MOI of removed disc about O (center at 3R/4 from O).

I_{h o l e} = \frac{1}{2} m (\frac{R}{4})^{2} + m (\frac{3 R}{4})^{2} = \frac{M R ^{2}}{512} + \frac{9 M R ^{2}}{256}

Step 4:Subtract.

I = \frac{256 M R ^{2}}{512} - \frac{19 M R ^{2}}{512}

Final answer:

\frac{237 M R ^{2}}{512}

Q7Single correctGravitation

The mass density of a spherical body is given by

ρ (r) = \frac{k}{r}

for

r \leq R

and

ρ (r) = 0

for

r > R

, where r is the distance from the centre.

The correct graph that describes qualitatively the acceleration, a, of a test particle as a function of r is :

(A)

Graph of acceleration a versus r: rises linearly to a peak at r = R, then decreases.

(B)

Graph of acceleration a versus r: constant (flat) from 0 to r = R, then decreases as 1/r^2.

(C)

Graph of acceleration a versus r: concave-up rise to a cusp peak at r = R, then a convex decrease.

(D)

Graph of acceleration a versus r: starts high and gently decreases, with a kink at r = R, then decreases more steeply.

SolutionAnswer: Option 2

Approach:

Find the mass enclosed within radius r, then use a = G M(r) /

r^{2}

for the regions inside and outside the body.

Step 1:Compute enclosed mass for r <= R.

M (r) = \int_{0}^{r} \frac{k}{r ^{'}} 4 π r^{'2} d r^{'} = 2 πk r^{2}

Step 2:Acceleration inside (r < R).

a = \frac{G 2 πk r ^{2}}{r ^{2}} = 2 πk G

Step 3:Acceleration outside (r > R) with total mass 2 pi k

R^{2}

a = \frac{G 2 πk R ^{2}}{r ^{2}} \propto \frac{1}{r ^{2}}

Final answer: Graph constant up to R, then decreasing as 1/

r^{2}

(option 2)

Q8Single correctProperties of Solids

A steel rail of length 5 m and area of cross section 40 c

m^{2}

is prevented from expanding along its length while the temperature rises by 10

^{\circ}

C. If coefficient of linear expansion and Young's modulus of steel are

1.2 \times 1 0^{- 5}

K^{- 1}

and

2 \times 1 0^{11}

m^{- 2}

respectively, the force developed in the rail is approximately :

(A)

2 \times 1 0^{7}

(B)

1 \times 1 0^{5}

(C)

2 \times 1 0^{9}

(D)

3 \times 1 0^{- 5}

SolutionAnswer: Option 2

1 \times 1 0^{5}

Approach:

A constrained rod develops a thermal stress; the force equals Young's modulus times area times the prevented thermal strain.

Step 1:Substitute the values (A = 40

c m^{2}

= 4e-3

m^{2}

F = (2 \times 1 0^{11}) (4 \times 1 0^{- 3}) (1.2 \times 1 0^{- 5}) (10)

Step 2:Evaluate.

F = 9.6 \times 1 0^{4} N

Final answer:

1 \times 1 0^{5}

Q9Single correctMechanics of Fluids

Two tubes of radii

r_{1}

and

r_{2}

, and lengths

l_{1}

and

l_{2}

, respectively, are connected in series and a liquid flows through each of them in stream line conditions.

P_{1}

and

P_{2}

are pressure differences across the two tubes. If

P_{2}

4 P_{1}

and

l_{2}

\frac{l _{1}}{4}

, then the radius

r_{2}

will be equal to :

(A)

r_{1}

(B)

2 r_{1}

(C)

4 r_{1}

(D)

\frac{r _{1}}{2}

SolutionAnswer: Option 4

\frac{r _{1}}{2}

Approach:

In series the volume flow rate Q is the same through both tubes; apply Poiseuille's law to each and take the ratio.

Step 1:Equal flow rate gives P proportional to l /

r^{4}

\frac{P _{1} r _{1}^{4}}{l _{1}} = \frac{P _{2} r _{2}^{4}}{l _{2}}

Step 2:Form the pressure ratio.

\frac{P _{2}}{P _{1}} = \frac{l _{2}}{l _{1}} (\frac{r _{1}}{r _{2}})^{4}

Step 3:Solve for r2.

(\frac{r _{1}}{r _{2}})^{4} = 16 \Rightarrow \frac{r _{1}}{r _{2}} = 2

Final answer:

\frac{r _{1}}{2}

Q10Single correctThermodynamics

For the P-V diagram given for an ideal gas, out of the following which one correctly represents the T-P diagram ?

P-V diagram for an ideal gas: a hyperbola P = Constant/V from state 1 (high P, low V) down to state 2 (low P, high V).

(A)

T-P graph: a straight line rising from state 1 (lower-left) to state 2 (upper-right).

(B)

T-P graph: a vertical line at constant P, state 1 at the bottom and state 2 at the top.

(C)

T-P graph: a horizontal line at constant T, state 2 on the left and state 1 on the right (state 1 at higher P).

(D)

T-P graph: a horizontal line at constant T, state 1 on the left and state 2 on the right.

SolutionAnswer: Option 3

Approach:

The given P-V curve satisfies PV = constant, which for an ideal gas means temperature is constant; the T-P plot must therefore be a horizontal line, traversed from high P (state 1) to low P (state 2).

Step 1:PV = constant implies nRT = constant, so T is constant.

P V = const \Rightarrow T = const

Step 2:On the T-P plane the locus is a horizontal line; the process goes from state 1 (high P) to state 2 (low P).

T = const, P_{1} > P_{2}

Final answer: Horizontal T-P line with state 1 at higher P and state 2 at lower P (option 3)

Q11Single correctKinetic Theory of Gases

N moles of a diatomic gas in a cylinder are at a temperature T. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get converted into monoatomic gas. What is the change in the total kinetic energy of the gas ?

(A)

\frac{1}{2} nRT

(B)

(C)

\frac{3}{2} nRT

(D)

\frac{5}{2} nRT

SolutionAnswer: Option 1

\frac{1}{2} nRT

Approach:

Each diatomic molecule that dissociates yields two monoatomic atoms; compute the internal kinetic energy before and after at the same temperature.

Step 1:Initial kinetic energy of N moles diatomic.

K E_{i} = \frac{5}{2} NRT

Step 2:After dissociation: (N - n) moles diatomic plus 2n moles monoatomic.

K E_{f} = \frac{5}{2} (N - n) RT + \frac{3}{2} (2 n) RT

Step 3:Take the difference.

Δ K E = (- \frac{5}{2} n + 3 n) RT

Final answer:

\frac{1}{2} nRT

Q12Single correctOscillations

A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 N

m^{- 1}

and oscillates in a damping medium of damping constant

1 0^{- 2}

s^{- 1}

. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to :

(A)

2 s

(B)

3.5 s

(C)

5 s

(D)

7 s

SolutionAnswer: Option 47 s

Approach:

In a damped oscillator the mechanical energy decays exponentially as E = E0 exp(-(b/m)t); set E = E0/2 and solve for t.

Step 1:Set the energy to half its initial value.

\frac{1}{2} = e^{- (b / m) t}

Step 2:Substitute m = 0.1 kg and b = 1e-2 kg/s.

t = \frac{0.1}{1 0 ^{- 2}} ln 2 = 10 \times 0.693

Final answer: 7 s

Q13Single correctWaves

A standing wave is formed by the superposition of two waves travelling in opposite directions. The transverse displacement is given by

y (x, t) = 0.5 sin (\frac{5 π}{4} x) cos (200 π t) .

What is the speed of the travelling wave moving in the positive x direction ?

(x and t are in meter and second, respectively.)

(A)

160 m/s

(B)

90 m/s

(C)

180 m/s

(D)

120 m/s

SolutionAnswer: Option 1160 m/s

Approach:

Read the wave number k and angular frequency omega from the standing wave expression; the component wave speed is omega divided by k.

Step 1:Identify k and omega.

k = \frac{5 π}{4}, ω = 200 π

Step 2:Compute the speed.

v = \frac{200 π}{5 π /4} = \frac{200 \times 4}{5}

Final answer: 160 m/s

Q14Single correctElectrostatics

Four closed surfaces and corresponding charge distributions are shown below.

Let the respective electric fluxes through the surfaces be

Φ_{1}, Φ_{2}, Φ_{3}

and

Φ_{4}

. Then :

(A)

Φ_{1} < Φ_{2} = Φ_{3} > Φ_{4}

(B)

Φ_{1} > Φ_{2} > Φ_{3} > Φ_{4}

(C)

Φ_{1} = Φ_{2} = Φ_{3} = Φ_{4}

(D)

Φ_{1} > Φ_{3}; Φ_{2} < Φ_{4}

SolutionAnswer: Option 3

Φ_{1} = Φ_{2} = Φ_{3} = Φ_{4}

Approach:

By Gauss's law the flux through each closed surface equals the net enclosed charge divided by epsilon-naught; compute the net enclosed charge for each surface.

Step 1:Net charge inside each surface from the figure.

S_{1} : 2 q, S_{2} : q - q + q + q = 2 q, S_{3} : 5 q + q + q = 7 q, S_{4} : 8 q - 2 q - 4 q = 2 q

Step 2:Flux is proportional to enclosed charge;

P h i_{3}

is the largest and the others are equal.

Φ_{3} > Φ_{1} = Φ_{2} = Φ_{4}

Final answer:

Φ_{1} = Φ_{2} = Φ_{3} = Φ_{4}

Q15Single correctElectrostatics

A combination of parallel plate capacitors is maintained at a certain potential difference.

When a 3 mm thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab.

Capacitor network between terminals A and B: C1, C2, C3 in series along the main line (nodes D, C, E), a top jumper wire connecting node D to node E and a bottom jumper wire connecting node C to the node after C3.

(A)

(B)

(C)

(D)

SolutionAnswer: Option 35

Approach:

Inserting a dielectric slab of thickness t reduces the effective air gap by t(1 - 1/K); to keep the capacitance (hence the potential difference at fixed charge) unchanged, the plate separation must be increased by exactly that amount.

Step 1:Equate the required increase to the gap reduction.

t (1 - \frac{1}{K}) = 2.4

Step 2:Solve for K.

1 - \frac{1}{K} = 0.8 \Rightarrow \frac{1}{K} = 0.2

Final answer: 5

Q16Single correctCurrent Electricity

A uniform wire of length l and radius r has a resistance of 100

Ω

. It is recast into a wire of radius

\frac{r}{2}

. The resistance of new wire will be :

(A)

1600

Ω

(B)

400

Ω

(C)

200

Ω

(D)

100

Ω

SolutionAnswer: Option 11600

Ω

Approach:

Resistance varies with length and cross-sectional area. On recasting, the volume of the wire is conserved, which fixes the new length once the new radius is given.

Step 1:New radius is half the original, so the new area is one quarter.

A_{2} = π (\frac{r}{2})^{2} = \frac{A _{1}}{4}

Step 2:Volume is conserved, giving the new length.

A_{1} L = A_{2} L_{2} \Rightarrow L_{2} = 4 L

Step 3:Apply the resistance ratio.

R_{2} = R_{1} \frac{L _{2}}{L _{1}} \frac{A _{1}}{A _{2}} = 100 \times 4 \times 4

Final answer: 1600

Ω

Q17Single correctCurrent Electricity

The figure shows three circuits I, II and III which are connected to a 3V battery. If the powers dissipated by the configurations I, II and III are

P_{1}

P_{2}

and

P_{3}

respectively, then :

Three resistor networks I, II and III each connected to a 3 V battery: (I) a bridge of five 1-ohm resistors, (II) a Wheatstone diamond of four 1-ohm edge resistors with a 1-ohm bridge resistor across the centre, (III) the same diamond bridge in series with an extra 1-ohm resistor and the battery.

(A)

P_{1} > P_{2} > P_{3}

(B)

P_{1} > P_{3} > P_{2}

(C)

P_{2} > P_{1} > P_{3}

(D)

P_{3} > P_{2} > P_{1}

SolutionAnswer: Option 3

P_{2} > P_{1} > P_{3}

Approach:

Each network is reduced to an equivalent resistance across the 3V source. With a fixed source voltage, dissipated power varies inversely with equivalent resistance, so the lowest resistance dissipates the most power.

Step 1:Circuit I is a balanced bridge of unit resistors; the bridging resistor carries no current, giving two 2 ohm paths in parallel.

R_{1} = 2 Ω ∥ 2 Ω = 1 Ω

Step 2:Circuit II places two 2 ohm paths and a direct 1 ohm bridge resistor in parallel across the source.

R_{2} = 2 Ω ∥ 2 Ω ∥ 1 Ω = 0.5 Ω

Step 3:Circuit III is a 1 ohm bridge (two 2 ohm paths in parallel) in series with an extra 1 ohm resistor.

R_{3} = (2 Ω ∥ 2 Ω) + 1 Ω = 2 Ω

Step 4:Compute the powers from P =

V^{2}

/R with V = 3V.

P_{1} = 9, P_{2} = 18, P_{3} = 4.5 W

Final answer:

P_{2} > P_{1} > P_{3}

Q18Single correctMoving Charges and Magnetism

A negative test charge is moving near a long straight wire carrying a current. The force acting on the test charge is parallel to the direction of the current. The motion of the charge is :

(A)

away from the wire

(B)

towards the wire

(C)

parallel to the wire along the current

(D)

parallel to the wire opposite to the current

SolutionAnswer: Option 2towards the wire

Approach:

The magnetic field of the straight wire is azimuthal (circular). The force on the charge is q(v x B). Working out the cross product for a velocity directed radially toward the wire shows the force lies along the current direction for a negative charge.

Step 1:Take the current along the x-axis. At a point on the +y side the field points along +z.

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

Step 2:Let the negative charge move toward the wire, i.e. along -y.

v = - \overset{y}{^}

Step 3:Evaluate the force for q < 0.

F = (- ∣ q ∣) (- \overset{y}{^}) \times \overset{z}{^} = + ∣ q ∣ \overset{x}{^}

Final answer: towards the wire

Q19Single correctMoving Charges and Magnetism

A uniform magnetic field B of 0.3 T is along the positive Z-direction. A rectangular loop (abcd) of sides 10 cm

\times

5 cm carries a current I of 12 A. Out of the following different orientations which one corresponds to stable equilibrium ?

3D coordinate axes X (towards viewer), Y (right) and Z (up), with uniform magnetic field B directed along +Z.

(A)

Rectangular current loop abcd in the Y-Z plane on the X/Y/Z axes; current circulates a->b->c->d->a.

(B)

Rectangular current loop abcd in the X-Z plane on the X/Y/Z axes, with d on the Z-axis; current circulates a->b->c->d->a.

(C)

Rectangular current loop abcd lying in the horizontal X-Y plane on the X/Y/Z axes; current circulates a->b->c->d->a.

(D)

Rectangular current loop abcd lying in the horizontal X-Y plane on the X/Y/Z axes; current circulates a->d->c->b->a (opposite sense to option 3).

SolutionAnswer: Option 3

Approach:

A current loop is a magnetic dipole with moment m = I A n-hat, the normal fixed by the right-hand rule following the current sense. Stable equilibrium occurs when the magnetic moment is parallel to the external field (potential energy minimum).

Step 1:Stable equilibrium requires the magnetic moment to point along +Z, parallel to B.

m ∥ B (+ \overset{z}{^})

Step 2:Evaluating the current sense for each option, only option 3 has the loop in the X-Y plane with a circulation giving a +Z moment.

m_{(3)} = I A \overset{z}{^}

Step 3:The other orientations give a moment along X or along -Z, which are unstable or perpendicular configurations.

U_{min} = - m B

Final answer: Option 3 (magnetic moment along +Z, parallel to B)

Q20Single correctAlternating Current

A sinusoidal voltage of peak value 283 V and angular frequency 320/s is applied to a series LCR circuit. Given that R

= 5 Ω

, L

= 25

mH and C

= 1000 μ

F. The total impedance, and phase difference between the voltage across the source and the current will respectively be :

(A)

Ω

and

tan^{- 1} (\frac{5}{3})

(B)

Ω

and 4

5^{\circ}

(C)

Ω

and

tan^{- 1} (\frac{8}{3})

(D)

Ω

and

tan^{- 1} (\frac{5}{3})

SolutionAnswer: Option 27

Ω

and 4

5^{\circ}

Approach:

Compute the inductive and capacitive reactances, find the net reactance, then the magnitude of impedance and the phase angle of a series LCR circuit.

Step 1:Reactances at omega = 320 rad/s.

X_{L} = 320 \times 0.025 = 8 Ω, X_{C} = \frac{1}{320 \times 1 0 ^{- 3}} \approx 3 Ω

Step 2:Net reactance and impedance.

X = X_{L} - X_{C} \approx 5 Ω, Z = 5^{2} + 5^{2} \approx 7 Ω

Step 3:Phase angle.

tan ϕ = \frac{5}{5} = 1 \Rightarrow ϕ = 4 5^{\circ}

Final answer: 7

Ω

and 4

5^{\circ}

Q21Single correctElectromagnetic Waves

The electric field component of a monochromatic radiation is given

b y E = 2

E_{0}

\hat{i} cos k z cos ω t

I t s ma g n e t i c f i e l d B

is then given by :

(A)

\frac{2 E _{0}}{c} \hat{j} sin k z cos ω t

(B)

- \frac{2 E _{0}}{c} \hat{j} sin k z sin ω t

(C)

\frac{2 E _{0}}{c} \hat{j} sin k z sin ω t

(D)

\frac{2 E _{0}}{c} \hat{j} cos k z cos ω t

SolutionAnswer: Option 3

\frac{2 E _{0}}{c} \hat{j} sin k z sin ω t

Approach:

This is a standing electromagnetic wave with E along x. Faraday's law gives the magnetic field from the spatial derivative of E, integrated in time.

Step 1:Only

E_{x}

is nonzero and depends on z, so the curl has a y-component equal to the z-derivative of

E_{x}

(\nabla \times E)_{y} = \frac{\partial E _{x}}{\partial z} = - 2 E_{0} k sin k z cos ω t

Step 2:Integrate -curl in time to get

B_{y}

B_{y} = - \int \frac{\partial E _{x}}{\partial z} d t = \frac{2 E _{0} k}{ω} sin k z sin ω t

Step 3:Substitute k/omega = 1/c.

B = \frac{2 E _{0}}{c} \hat{j} sin k z sin ω t

Final answer:

\frac{2 E _{0}}{c} \hat{j} sin k z sin ω t

Q22Single correctRay Optics

In an experiment a convex lens of focal length 15 cm is placed coaxially on an optical bench in front of a convex mirror at a distance of 5 cm from it. It is found that an object and its image coincide, if the object is placed at a distance of 20 cm from the lens. The focal length of the convex mirror is :

(A)

27.5 cm

(B)

20.0 cm

(C)

25.0 cm

(D)

30.5 cm

SolutionAnswer: Option 127.5 cm

Approach:

For the object and image to coincide, rays after the lens must strike the convex mirror normally and retrace their path, which means the lens forms its image at the centre of curvature of the mirror. Locate the lens image, then equate its distance behind the mirror to the radius of curvature.

Step 1:Lens image with u = -20 cm and f = +15 cm.

\frac{1}{v} = \frac{1}{15} + \frac{1}{- 20} \Rightarrow v = 60 cm

Step 2:The mirror is 5 cm behind the lens, so the converging point lies 55 cm behind the mirror. This must be the centre of curvature.

R = 60 - 5 = 55 cm

Step 3:Mirror focal length.

f_{m} = \frac{R}{2} = \frac{55}{2} = 27.5 cm

Final answer: 27.5 cm

Q23Single correctWave Optics

A single slit of width 0.1 mm is illuminated by a parallel beam of light of wavelength 6000

A \circ

and diffraction bands are observed on a screen 0.5 m from the slit. The distance of the third dark band from the central bright band is :

(A)

3 mm

(B)

9 mm

(C)

4.5 mm

(D)

1.5 mm

SolutionAnswer: Option 29 mm

Approach:

In single slit diffraction, dark bands (minima) occur where the path difference equals an integer multiple of the wavelength. The linear position on the screen follows from the small-angle approximation.

Step 1:List the data in SI units.

a = 1 0^{- 4} m, λ = 6 \times 1 0^{- 7} m, D = 0.5 m

Step 2:Use n = 3 for the third dark band.

y_{3} = \frac{3 \times 6 \times 1 0 ^{- 7} \times 0.5}{1 0 ^{- 4}}

Step 3:Evaluate.

y_{3} = 9 \times 1 0^{- 3} m = 9 mm

Final answer: 9 mm

Q24Single correctDual Nature of Radiation and Matter

A Laser light of wavelength 660 nm is used to weld Retina detachment. If a Laser pulse of width 60 ms and power 0.5 kW is used the number of photons in the pulse are :
[Take Planck's constant h

= 6.62 \times 1 0^{- 34}

Js]

(A)

1 0^{20}

(B)

1 0^{18}

(C)

1 0^{22}

(D)

1 0^{19}

SolutionAnswer: Option 1

1 0^{20}

Approach:

Find the total energy delivered by the pulse, divide by the energy of a single photon to get the photon count.

Step 1:Total energy in the pulse.

E = 0.5 \times 1 0^{3} \times 60 \times 1 0^{- 3} = 30 J

Step 2:Energy of one photon.

E_{p h} = \frac{6.62 \times 1 0 ^{- 34} \times 3 \times 1 0 ^{8}}{660 \times 1 0 ^{- 9}} \approx 3 \times 1 0^{- 19} J

Step 3:Photon count.

N = \frac{30}{3 \times 1 0 ^{- 19}} \approx 1 0^{20}

Final answer:

1 0^{20}

Q25Single correctAtoms

The acceleration of an electron in the first orbit of the hydrogen atom (n

= 1

) is :

(A)

\frac{h ^{2}}{π ^{2} m ^{2} r ^{3}}

(B)

\frac{h ^{2}}{8 π ^{2} m ^{2} r ^{3}}

(C)

\frac{h ^{2}}{4 π ^{2} m ^{2} r ^{3}}

(D)

\frac{h ^{2}}{4 π m ^{2} r ^{3}}

SolutionAnswer: Option 3

\frac{h ^{2}}{4 π ^{2} m ^{2} r ^{3}}

Approach:

Use Bohr's quantization to express the orbital speed of the electron, then compute the centripetal acceleration

v^{2}

/r.

Step 1:Speed for n = 1.

v = \frac{h}{2 πm r}

Step 2:Centripetal acceleration.

a = \frac{v ^{2}}{r} = \frac{1}{r} (\frac{h}{2 πm r})^{2}

Step 3:Simplify.

a = \frac{h ^{2}}{4 π ^{2} m ^{2} r ^{3}}

Final answer:

\frac{h ^{2}}{4 π ^{2} m ^{2} r ^{3}}

Q26Single correctNuclei

Imagine that a reactor converts all given mass into energy and that it operates at a power level of

1 0^{9}

watt. The mass of the fuel consumed per hour in the reactor will be : (velocity of light, c is

3 \times 1 0^{8}

m/s)

(A)

0.96 gm

(B)

0.8 gm

(C)

4 \times 1 0^{- 2}

(D)

6.6 \times 1 0^{- 5}

SolutionAnswer: Option 3

4 \times 1 0^{- 2}

Approach:

The energy produced in one hour equals the power times time. Convert this energy to mass using Einstein's mass-energy relation.

Step 1:Energy in one hour (t = 3600 s).

E = 1 0^{9} \times 3600 = 3.6 \times 1 0^{12} J

Step 2:Mass from E =

m c^{2}

m = \frac{E}{c ^{2}} = \frac{3.6 \times 1 0 ^{12}}{( 3 \times 1 0 ^{8} ) ^{2}}

Step 3:Evaluate.

m = 4 \times 1 0^{- 5} kg = 4 \times 1 0^{- 2} g

Final answer:

4 \times 1 0^{- 2}

Q27Single correctSemiconductor Electronics

The current gain of a common emitter amplifier is 69. If the emitter current is 7.0 mA, collector current is :

(A)

9.6 mA

(B)

6.9 mA

(C)

0.69 mA

(D)

69 mA

SolutionAnswer: Option 26.9 mA

Approach:

The common emitter current gain beta relates collector and base currents. Combine with the emitter current relation to express collector current in terms of the emitter current.

Step 1:Express base current and substitute.

I_{E} = I_{C} + \frac{I _{C}}{β} = I_{C} \frac{β + 1}{β}

Step 2:Solve for collector current.

I_{C} = I_{E} \frac{β}{β + 1} = 7.0 \times \frac{69}{70}

Step 3:Evaluate.

I_{C} = 6.9 mA

Final answer: 6.9 mA

Q29Single correctCurrent Electricity

In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance P

= 4 Ω

and the neutral point N is at 60 cm from A. Now an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A. The value of unknown resistance R is :

Meter-bridge circuit: resistors P and Q in the top arms, the bridge wire from A to B with the neutral point N located by a galvanometer G tapped from the P-Q junction, and a lower loop with cell E, rheostat R_h and key K.

(A)

\frac{33}{5} Ω

(B)

Ω

(C)

Ω

(D)

\frac{20}{3} Ω

SolutionAnswer: Option 4

\frac{20}{3} Ω

Approach:

The meter bridge balance condition relates the ratio of the two resistances to the ratio of the bridge lengths. First find Q from the initial balance, then use the second balance with P + R to solve for R.

Step 1:Initial balance at 60 cm gives Q.

\frac{4}{Q} = \frac{60}{40} \Rightarrow Q = \frac{8}{3} Ω

Step 2:New balance at 80 cm with series resistance P + R.

\frac{P + R}{Q} = \frac{80}{20} = 4

Step 3:Solve for R.

R = 4 Q - P = 4 \times \frac{8}{3} - 4 = \frac{20}{3} Ω

Final answer:

\frac{20}{3} Ω

Q30Single correctOscillations

In an experiment to determine the period of a simple pendulum of length 1 m, it is attached to different spherical bobs of radii

r_{1}

and

r_{2}

. The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be

5 \times 1 0^{- 4}

s, the difference in radii,

∣ r_{1} - r_{2} ∣

is best given by :

(A)

1 cm

(B)

0.1 cm

(C)

0.5 cm

(D)

0.01 cm

SolutionAnswer: Option 20.1 cm

Approach:

The effective length of the pendulum is the string length plus the bob radius. The period depends on the square root of the effective length, so a small change in radius produces a relative change in period equal to half the relative change in effective length.

Step 1:Since r is much smaller than L, the effective length is approximately L = 1 m.

\frac{Δ T}{T} = \frac{1}{2} \frac{∣ r _{1} - r _{2} ∣}{L}

Step 2:Solve for the radius difference.

∣ r_{1} - r_{2} ∣ = 2 L \frac{Δ T}{T} = 2 \times 1 \times 5 \times 1 0^{- 4}

Step 3:Evaluate.

∣ r_{1} - r_{2} ∣ = 1 0^{- 3} m = 0.1 cm

Final answer: 0.1 cm

Chemistry30 questions

Q31Single correctChemical Thermodynamics

An ideal gas undergoes isothermal expansion at constant pressure. During the process :

(A)

entropy increases but enthalpy decreases.

(B)

enthalpy remains constant but entropy increases.

(C)

enthalpy decreases but entropy increases.

(D)

Both enthalpy and entropy remain constant.

SolutionAnswer: Option 2enthalpy remains constant but entropy increases.

Approach:

For an ideal gas, enthalpy is a function of temperature only; an isothermal process keeps temperature constant, so enthalpy does not change. Expansion increases the accessible volume, raising entropy.

Step 1:Isothermal process means constant temperature; since H depends only on T for an ideal gas, enthalpy is constant.

Δ T = 0 \Rightarrow Δ H = 0

Step 2:Expansion increases volume, so the entropy of the gas increases.

V_{2} > V_{1} \Rightarrow Δ S = n R ln \frac{V _{2}}{V _{1}} > 0

Final answer: enthalpy remains constant but entropy increases.

Q32Single correctEquilibrium

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If

p K_{b}

of ammonia solution is 4.75, the pH of the mixture will be :

(A)

3.75

(B)

4.75

(C)

8.25

(D)

9.25

SolutionAnswer: Option 49.25

Approach:

Find millimoles of NH3 and HCl, identify the basic buffer formed after partial neutralisation, then apply the Henderson equation for a base.

Step 1:Compute millimoles: ammonia and HCl.

n_{NH_{3}} = 50 \times 0.2 = 10, n_{HCl} = 25 \times 0.2 = 5

Step 2:HCl converts 5 mmol NH3 to NH4Cl, leaving 5 mmol NH3 and 5 mmol NH4+ (equal amounts).

[salt] = [base]

Step 3:Apply Henderson for base.

pOH = 4.75 + lo g 1 = 4.75

Step 4:Convert to pH.

pH = 14 - 4.75 = 9.25

Final answer: 9.25

Q33Single correctAtomic Structure

The electron in the hydrogen atom undergoes transition from higher orbitals to orbital of radius 211.6 pm. This transition is associated with :

(A)

Lyman series

(B)

Balmer series

(C)

Paschen series

(D)

Brackett series

SolutionAnswer: Option 2Balmer series

Approach:

Use the Bohr radius formula to find the principal quantum number of the target orbital, then identify the spectral series by the lower level.

Step 1:Solve for n using the given radius.

n^{2} = \frac{211.6}{52.9} = 4 \Rightarrow n = 2

Step 2:Transitions ending at n = 2 belong to the Balmer series.

n_{f} = 2 \Rightarrow Balmer

Final answer: Balmer series

Q34Single correctStates of Matter

At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen

(N_{2})

at 4 bar. The molar mass of gaseous molecule is :

(A)

28 g mol^{- 1}

(B)

56 g mol^{- 1}

(C)

112 g mol^{- 1}

(D)

224 g mol^{- 1}

SolutionAnswer: Option 3

112 g mol^{- 1}

Approach:

Express gas density from the ideal gas equation, set up the doubling relation between the unknown gas and dinitrogen, and solve for the molar mass.

Step 1:Write density of the unknown gas and of N2.

ρ_{1} = \frac{( 2 ) M}{RT}, ρ_{N_{2}} = \frac{( 4 ) ( 28 )}{RT}

Step 2:Apply the condition density of unknown is double that of N2.

\frac{2 M}{RT} = 2 \times \frac{( 4 ) ( 28 )}{RT}

Step 3:Solve for M.

M = \frac{224}{2} = 112

Final answer:

112 g mol^{- 1}

Q35Single correctSome Basic Concepts in Chemistry

What quantity (in mL) of a 45% acid solution of a mono-protic strong acid must be mixed with a 20% solution of the same acid to produce 800 mL of a 29.875% acid solution ?

(A)

320

(B)

325

(C)

316

(D)

330

SolutionAnswer: Option 3316

Approach:

Set up a mass (volume-percent) balance: the acid contributed by both solutions equals the acid in the final mixture.

Step 1:Let x mL be the 45% solution, so (800 - x) mL is the 20% solution.

45 x + 20 (800 - x) = 29.875 \times 800

Step 2:Expand and simplify.

45 x + 16000 - 20 x = 23900

Step 3:Solve for x.

x = \frac{7900}{25} = 316

Final answer: 316

Q36Single correctElectrochemistry

To find the standard potential of

M^{3 +} / M

electrode, the following cell is constituted :

Pt / M / M^{3 +} (0.001 mol L^{- 1}) / Ag^{+} (0.01 mol L^{- 1}) / Ag

. The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half reaction

M^{3 +} + 3 e^{-} \to M

at 298 K will be : (Given

E_{Ag^{+} / Ag}^{⊖}

at 298 K

= 0.80

Volt)

(A)

0.38 Volt

(B)

0.32 Volt

(C)

1.28 Volt

(D)

0.66 Volt

SolutionAnswer: Option 20.32 Volt

Approach:

Apply the Nernst equation to both half-cells to get their actual potentials, then use cell emf equals cathode minus anode to solve for the unknown standard potential.

Step 1:Silver is the cathode (reduction). Compute its potential.

E_{Ag} = 0.80 + 0.0591 lo g (0.01) = 0.80 - 0.1182 = 0.6818 V

Step 2:M is the anode. Its electrode potential (reduction form).

E_{M} = E^{⊖} + \frac{0.0591}{3} lo g (0.001) = E^{⊖} - 0.0591 V

Step 3:Use cell emf relation and solve.

0.421 = 0.6818 - (E^{⊖} - 0.0591)

Final answer: 0.32 Volt

Q37Single correctChemical Thermodynamics

A gas undergoes change from state A to state B. In this process, the heat absorbed and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to A by another process during which 3 J of heat is evolved. In this reverse process of B to A :

(A)

10 J of the work will be done by the gas.

(B)

6 J of the work will be done by the gas.

(C)

10 J of the work will be done by the surrounding on gas.

(D)

6 J of the work will be done by the surrounding on gas.

SolutionAnswer: Option 46 J of the work will be done by the surrounding on gas.

Approach:

Use the first law to find the internal energy change for A to B; since internal energy is a state function, the reverse path has the opposite change, allowing the work in the reverse step to be found.

Step 1:For A to B with q = 5 J absorbed and w = 8 J done by gas.

Δ U_{A \to B} = 5 - 8 = - 3 J

Step 2:Reverse process B to A: state function reverses sign.

Δ U_{B \to A} = + 3 J

Step 3:For B to A, 3 J of heat evolved so q = -3 J. Solve for work done by gas.

3 = - 3 - w \Rightarrow w = - 6 J

Step 4:Negative work by gas means surroundings do 6 J of work on the gas.

w_{by gas} = - 6 J

Final answer: 6 J of the work will be done by the surrounding on gas.

Q38Single correctSurface Chemistry

Adsorption of a gas on a surface follows Freundlich adsorption isotherm. Plot of

lo g \frac{x}{m}

versus log p gives a straight line with slope equal to 0.5, then : (

\frac{x}{m}

is the mass of the gas adsorbed per gram of adsorbent)

(A)

Adsorption is independent of pressure.

(B)

Adsorption is proportional to the pressure.

(C)

Adsorption is proportional to the square root of pressure.

(D)

Adsorption is proportional to the square of pressure.

SolutionAnswer: Option 3Adsorption is proportional to the square root of pressure.

Approach:

Take the logarithmic form of the Freundlich isotherm; the slope of the log-log plot equals 1/n, which sets the power of pressure.

Step 1:Identify the slope of the log-log plot with 1/n.

\frac{1}{n} = 0.5

Step 2:Substitute into the isotherm.

\frac{x}{m} = k p^{0.5} = k p

Final answer: Adsorption is proportional to the square root of pressure.

Q39Single correctChemical Kinetics

The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is : (Assume activation energy and pre-exponential factor are independent of temperature; ln 2

= 0.693

; R

= 8.314

J mo

l^{- 1}

K^{- 1}

)

(A)

107.2 kJ mol^{- 1}

(B)

53.6 kJ mol^{- 1}

(C)

26.8 kJ mol^{- 1}

(D)

214.4 kJ mol^{- 1}

SolutionAnswer: Option 1

107.2 kJ mol^{- 1}

Approach:

Use the two-temperature form of the Arrhenius equation with the rate-constant ratio of 4 and solve for the activation energy.

Step 1:Rate quadruples, so the ratio is 4.

ln 4 = 2 ln 2 = 1.386

Step 2:Compute the temperature term.

\frac{1}{300} - \frac{1}{310} = \frac{10}{93000} = 1.075 \times 1 0^{- 4}

Step 3:Solve for activation energy.

E_{a} = \frac{8.314 \times 1.386}{1.075 \times 1 0 ^{- 4}} \approx 1.072 \times 1 0^{5} J mol^{- 1}

Final answer:

107.2 kJ mol^{- 1}

Q40Single correctSolutions

A solution is prepared by mixing 8.5 g of

CH_{2} Cl_{2}

and 11.95 g of

CHCl_{3}

. If vapour pressure of

CH_{2} Cl_{2}

and

CHCl_{3}

at 298 K are 415 and 200 mmHg respectively, the mole fraction of

CHCl_{3}

in vapour form is : (Molar mass of Cl

= 35.5

g mo

l^{- 1}

)

(A)

0.162

(B)

0.675

(C)

0.325

(D)

0.486

SolutionAnswer: Option 30.325

Approach:

Compute moles and liquid mole fractions, apply Raoult's law for partial pressures, then use Dalton's law to obtain the vapour-phase mole fraction of CHCl3.

Step 1:Molar masses and moles.

M_{CH_{2} Cl_{2}} = 85, M_{CHCl_{3}} = 119.5; n = 0.1, 0.1

Step 2:Liquid mole fractions are equal.

x_{CH_{2} Cl_{2}} = x_{CHCl_{3}} = 0.5

Step 3:Partial pressures by Raoult's law.

p_{CH_{2} Cl_{2}} = 207.5, p_{CHCl_{3}} = 100

Step 4:Vapour mole fraction of CHCl3.

y_{CHCl_{3}} = \frac{100}{307.5} = 0.325

Final answer: 0.325

Q41Single correctClassification of Elements and Periodicity

The electronic configuration with the highest ionization enthalpy is :

(A)

[Ne] 3 s^{2} 3 p^{1}

(B)

[Ne] 3 s^{2} 3 p^{2}

(C)

[Ne] 3 s^{2} 3 p^{3}

(D)

[Ar] 3 d^{10} 4 s^{2} 4 p^{3}

SolutionAnswer: Option 3

[Ne] 3 s^{2} 3 p^{3}

Approach:

Identify the elements behind each configuration and compare ionization enthalpies using periodic trends and half-filled stability.

Step 1:Identify elements: Al, Si, P, As respectively.

[Ne] 3 s^{2} 3 p^{1} = Al; 3 p^{2} = Si; 3 p^{3} = P; [Ar] ...4 p^{3} = As

Step 2:Phosphorus has an exactly half-filled 3p subshell giving extra stability and a higher ionization enthalpy.

3 p^{3} half-filled

Step 3:Phosphorus is smaller and above arsenic, so its ionization enthalpy exceeds that of As, Si and Al.

I E (P) > I E (As), I E (Si), I E (Al)

Final answer:

[Ne] 3 s^{2} 3 p^{3}

Q42Single correctEquilibrium

The following reaction occurs in the Blast Furnace where iron ore is reduced to iron metal :

Fe_{2} O_{3} (s) + 3 CO (g) ⇌ 2 Fe (l) + 3 CO_{2} (g)

. Using the Le Chatelier's principle, predict which one of the following will not disturb the equilibrium ?

(A)

Removal of CO

(B)

Removal of

CO_{2}

(C)

Addition of

CO_{2}

(D)

Addition of

Fe_{2} O_{3}

SolutionAnswer: Option 4Addition of

Fe_{2} O_{3}

Approach:

Recognise that pure solids and pure liquids do not appear in the equilibrium expression, so changing their amounts cannot shift the equilibrium, while changing gaseous species does.

Step 1:Fe2O3 is a pure solid; its activity is unity and absent from Kp.

a_{Fe_{2} O_{3}} = 1

Step 2:Adding a pure solid does not change concentrations of gases, so equilibrium is undisturbed.

Q_{p} unchanged

Final answer: Addition of

Fe_{2} O_{3}

Q43Single corrects-Block Elements

Which one of the following is an oxide ?

(A)

KO_{2}

(B)

BaO_{2}

(C)

SiO_{2}

(D)

CsO_{2}

SolutionAnswer: Option 3

SiO_{2}

Approach:

Classify each compound by the oxidation state of oxygen: a normal oxide contains O in the -2 state, while peroxides and superoxides contain O-O linkages.

Step 1:KO2 and CsO2 contain the superoxide ion O2 minus (oxygen -1/2).

KO_{2}, CsO_{2} \to superoxides

Step 2:BaO2 contains the peroxide ion O2 two-minus (oxygen -1).

BaO_{2} \to peroxide

Step 3:SiO2 contains oxide ions with oxygen in -2 state.

SiO_{2} \to oxide

Final answer:

SiO_{2}

Q44Single correctEnvironmental Chemistry

Which of the following is a set of green house gases ?

(A)

CH_{4}, O_{3}, N_{2}, SO_{2}

(B)

O_{3}, N_{2}, CO_{2}, NO_{2}

(C)

O_{3}, NO_{2}, SO_{2}, Cl_{2}

(D)

CO_{2}, CH_{4}, N_{2} O, O_{3}

SolutionAnswer: Option 4

CO_{2}, CH_{4}, N_{2} O, O_{3}

Approach:

Recall the standard list of greenhouse gases and select the option in which every species qualifies.

Step 1:Recognised greenhouse gases include CO2, CH4, N2O, O3, water vapour and CFCs.

CO_{2}, CH_{4}, N_{2} O, O_{3}

Step 2:N2, SO2 and Cl2 are not greenhouse gases, eliminating the other options.

N_{2}, SO_{2}, Cl_{2} \to not GHG

Final answer:

CO_{2}, CH_{4}, N_{2} O, O_{3}

Q45Single correctChemical Bonding and Molecular Structure

The group having triangular planar structures is :

(A)

BF_{3}, NF_{3}, CO_{3}^{2 -}

(B)

CO_{3}^{2 -}, NO_{3}^{-}, SO_{3}

(C)

NH_{3}, SO_{3}, CO_{3}^{2 -}

(D)

NCl_{3}, BCl_{3}, SO_{3}

SolutionAnswer: Option 2

CO_{3}^{2 -}, NO_{3}^{-}, SO_{3}

Approach:

Apply VSEPR theory to each species; a trigonal (triangular) planar geometry requires three bonding regions and no lone pair on the central atom.

Step 1:CO3 2-, NO3 - and SO3 each have a central atom with three regions and no lone pair, giving trigonal planar geometry.

CO_{3}^{2 -}, NO_{3}^{-}, SO_{3} \to trigonal planar

Step 2:NF3, NH3 and NCl3 have a lone pair on nitrogen and are pyramidal, so options 1, 3 and 4 are rejected.

NF_{3}, NH_{3}, NCl_{3} \to pyramidal

Final answer:

CO_{3}^{2 -}, NO_{3}^{-}, SO_{3}

Q46Single correctp-Block Elements

XeF_{6}

on partial hydrolysis with water produces a compound 'X'. The same compound 'X' is formed when

XeF_{6}

reacts with silica. The compound 'X' is :

(A)

XeF_{2}

(B)

XeF_{4}

(C)

XeOF_{4}

(D)

XeO_{3}

SolutionAnswer: Option 3

XeOF_{4}

Approach:

Identify the product common to both partial hydrolysis of XeF6 and its reaction with silica.

Step 1:Partial (incomplete) hydrolysis of XeF6 replaces two fluorine atoms by one oxygen, giving XeOF4.

XeF_{6} + H_{2} O \to XeOF_{4} + 2 HF

Step 2:Reaction of XeF6 with silica also yields XeOF4 (plus SiF4), confirming the same product.

2 XeF_{6} + SiO_{2} \to 2 XeOF_{4} + SiF_{4}

Final answer:

XeOF_{4}

Q47Single correctp-Block Elements

The number of

P - OH

bonds and the oxidation state of phosphorus atom in pyrophosphoric acid

(H_{4} P_{2} O_{7})

respectively are :

(A)

four and four

(B)

five and four

(C)

five and five

(D)

four and five

SolutionAnswer: Option 4four and five

Approach:

Draw the structure of pyrophosphoric acid and count P-OH bonds; compute the oxidation state of P.

Step 1:Pyrophosphoric acid H4P2O7 has two PO4 tetrahedra sharing one bridging oxygen. Each phosphorus carries two terminal OH groups, so total P-OH bonds = 4.

(HO)_{2} P(=O) - O - P(=O)(OH)_{2}

Step 2:Apply oxidation-state balance with H = +1, O = -2.

4 (+ 1) + 2 x + 7 (- 2) = 0 \Rightarrow 2 x = 10 \Rightarrow x = + 5

Final answer: four and five

Q48Single correctd- and f-Block Elements

Which of the following ions does

not

liberate hydrogen gas on reaction with dilute acids ?

(A)

Ti^{2 +}

(B)

V^{2 +}

(C)

Cr^{2 +}

(D)

Mn^{2 +}

SolutionAnswer: Option 4

Mn^{2 +}

Approach:

An M2+ ion liberates H2 (is oxidised to M3+) only if its M3+/M2+ standard potential is negative. Compare across the listed ions.

Step 1:Ti2+, V2+ and Cr2+ have negative M3+/M2+ potentials, so they act as reducing agents and reduce H+ to H2 while being oxidised to M3+.

E_{V^{3 +} / V^{2 +}}^{\circ} = - 0.26 V, E_{Cr^{3 +} / Cr^{2 +}}^{\circ} = - 0.41 V

Step 2:Mn3+/Mn2+ potential is large and positive, so Mn2+ cannot be oxidised by H+; therefore Mn2+ does not liberate hydrogen.

E_{Mn^{3 +} / Mn^{2 +}}^{\circ} = + 1.51 V

Final answer:

Mn^{2 +}

Q49Single correctp-Block Elements

The correct sequence of decreasing number of

π

-bonds in the structures of

H_{2} SO_{3}

H_{2} SO_{4}

and

H_{2} S_{2} O_{7}

is :

(A)

H_{2} SO_{3} > H_{2} SO_{4} > H_{2} S_{2} O_{7}

(B)

H_{2} SO_{4} > H_{2} S_{2} O_{7} > H_{2} SO_{3}

(C)

H_{2} S_{2} O_{7} > H_{2} SO_{4} > H_{2} SO_{3}

(D)

H_{2} S_{2} O_{7} > H_{2} SO_{3} > H_{2} SO_{4}

SolutionAnswer: Option 3

H_{2} S_{2} O_{7} > H_{2} SO_{4} > H_{2} SO_{3}

Approach:

Count the number of S=O (pi) bonds in each oxoacid structure.

Step 1:H2SO3 has structure (HO)2S=O with one S=O double bond, giving 1 pi-bond.

(HO)_{2} S=O

Step 2:H2SO4 has structure (HO)2S(=O)2 with two S=O double bonds, giving 2 pi-bonds.

(HO)_{2} S(=O)_{2}

Step 3:H2S2O7 (pyrosulfuric acid) has two SO3H units joined by a bridging O; each S carries two S=O bonds, giving 4 pi-bonds total.

(HO) (O)_{2} S - O - S (O)_{2} (OH)

Final answer:

H_{2} S_{2} O_{7} > H_{2} SO_{4} > H_{2} SO_{3}

Q50Single correctCoordination Compounds

[Co_{2} (CO)_{8}]

displays :

(A)

one

Co - Co

bond, six terminal CO and two bridging CO

(B)

one

Co - Co

bond, four terminal CO and four bridging CO

(C)

Co - Co

bond, six terminal CO and two bridging CO

(D)

Co - Co

bond, four terminal CO and four bridging CO

SolutionAnswer: Option 1one

Co - Co

bond, six terminal CO and two bridging CO

Approach:

Recall the bridged structure of dicobalt octacarbonyl.

Step 1:The stable (solid) form of Co2(CO)8 contains a direct Co-Co bond.

Co - Co

Step 2:Of the eight CO ligands, six are terminal (three on each Co) and two bridge the two cobalt atoms.

(CO)_{3} Co (μ -CO)_{2} Co(CO)_{3}

Final answer: one

Co - Co

bond, six terminal CO and two bridging CO

Q51Single correctAldehydes, Ketones and Carboxylic Acids

A compound of molecular formula

C_{8} H_{8} O_{2}

reacts with acetophenone to form a single cross-aldol product in the presence of base. The same compound on reaction with conc. NaOH forms benzyl alcohol as one of the products. The structure of the compound is :

(A)

para-methoxybenzaldehyde: benzene ring with a CHO group and a CH3O (methoxy) group at the para position

(B)

para-hydroxyacetophenone: benzene ring with a COCH3 (acetyl) group and an HO group at the para position

(C)

methyl benzoate: benzene ring bearing a C(=O)OCH3 (methyl ester) group

(D)

para-methylbenzoic acid (p-toluic acid): benzene ring with a COOH group and a CH3 group at the para position

SolutionAnswer: Option 1para-methoxybenzaldehyde (CH3O-C6H4-CHO)

Approach:

The compound must be an aromatic aldehyde lacking alpha-hydrogen so it gives a clean cross-aldol with acetophenone and undergoes Cannizzaro reaction with conc. NaOH to yield benzyl alcohol.

Step 1:Conc. NaOH converts the compound into benzyl alcohol, characteristic of the Cannizzaro reaction, which requires an aldehyde with no alpha-hydrogen (an aromatic aldehyde).

2 ArCHO + NaOH \to ArCH_{2} OH + ArCOONa

Step 2:Molecular formula C8H8O2 with an ArCHO unit and a second oxygen corresponds to para-methoxybenzaldehyde (anisaldehyde), CH3O-C6H4-CHO.

CH_{3} O - C_{6} H_{4} - CHO

Step 3:Having no alpha-hydrogen, it forms a single cross-aldol product with the alpha-hydrogens of acetophenone.

CH_{3} O-C_{6} H_{4} -CHO + C_{6} H_{5} COCH_{3}

Final answer: para-methoxybenzaldehyde (CH3O-C6H4-CHO)

Q52Single correctHydrocarbons

Which of the following compounds is most reactive to an aqueous solution of sodium carbonate ?

(A)

cyclopropene: three-membered carbon ring with one C=C double bond

(B)

cyclobutene: four-membered carbon ring with one C=C double bond

(C)

1,3-cyclopentadiene: five-membered carbon ring with two conjugated C=C double bonds

(D)

1,3-cyclohexadiene: six-membered carbon ring with two conjugated C=C double bonds

SolutionAnswer: Option 31,3-cyclopentadiene (five-membered ring with two double bonds)

Approach:

Reactivity toward aqueous Na2CO3 reflects acidity; the most acidic C-H, on deprotonation, gives the most stabilised (aromatic) carbanion.

Step 1:Loss of a proton from the sp3 CH2 of 1,3-cyclopentadiene generates the cyclopentadienyl anion, which is aromatic (6 pi electrons, Huckel's rule).

C_{5} H_{6} \to C_{5} H_{5}^{-} + H^{+}

Step 2:This aromatic stabilisation makes cyclopentadiene the most acidic (pKa around 16) among the given cyclic alkenes/dienes, so it is most reactive toward the weak base Na2CO3.

pK_{a} (C_{5} H_{6}) \approx 16

Final answer: 1,3-cyclopentadiene (five-membered ring with two double bonds)

Q53Single correctBasic Principles of Organic Chemistry

In the following structure, the double bonds are marked as I, II, III and IV. Geometrical isomerism is

not

possible at site (s) :

Acyclic diterpene polyene chain with four C=C double bonds labelled I, II, III and IV by arrows. Site I is a terminal isopropylidene (gem-dimethyl) double bond (CH3)2C=CH-; site II is a trisubstituted CH3-bearing double bond; site III is a tetrasubstituted C(CH3)=C(CH3) double bond; site IV is a tetrasubstituted double bond bearing CH3 and an ethyl group.

(A)

III

(B)

(C)

I and III

(D)

III and IV

SolutionAnswer: Option 2I

Approach:

Geometrical isomerism requires each doubly-bonded carbon to bear two different substituents; a carbon with two identical groups precludes it.

Step 1:At site I the terminal carbon of the double bond is a =C(CH3)2 carbon bearing two identical methyl groups.

= C(CH_{3})_{2}

Step 2:Since one alkene carbon at site I carries two equal groups, cis-trans (geometrical) isomerism is not possible there, whereas sites II, III and IV each have two different substituents on both carbons.

= C(CH_{3})_{2} \Rightarrow no E/Z

Final answer: I

Q54Single correctHydrocarbons

The major product of the following reaction is :

1-methylcyclohexene treated with Br2 / h-nu (light).

(A)

4-bromo-1-methylcyclohexene: cyclohexene ring with a methyl on the alkene carbon and a Br at the para ring carbon; ring double bond retained

(B)

3-bromo-3-methylcyclohexene: cyclohexene ring whose allylic carbon bears both a methyl and a Br (geminal); ring double bond retained

(C)

3-(bromomethyl)cyclohexene: cyclohexene ring bearing a CH2Br group on the allylic carbon; ring double bond retained

(D)

2-bromo-1-methyl... cyclohexane with vicinal dibromide: cyclohexane ring bearing a methyl and two Br atoms on adjacent carbons; no double bond

SolutionAnswer: Option 2tertiary allylic bromide: Br on the methyl-bearing allylic carbon with ring double bond retained

Approach:

Br2 with light (hv) and low Br2 concentration favours free-radical allylic substitution; the most stable allylic radical determines the major product.

Step 1:Under Br2/hv, allylic substitution proceeds via abstraction of an allylic hydrogen, not electrophilic addition.

R-H + Br \cdot \to R \cdot + HBr

Step 2:Abstraction at the methyl-substituted allylic carbon gives the most stable (tertiary, resonance-stabilised) allylic radical; recombination with Br places Br on that carbon while the double bond is retained.

R \cdot + Br_{2} \to R-Br + Br \cdot

Final answer: tertiary allylic bromide: Br on the methyl-bearing allylic carbon with ring double bond retained

Q55Single correctBiomolecules

The

incorrect

statement among the following is :

(A)

α

-D-glucose and

β

-D-glucose are anomers.

(B)

α

-D-glucose and

β

-D-glucose are enantiomers.

(C)

Cellulose is a straight chain polysaccharide made up of only

β

-D-glucose units.

(D)

The penta acetate of glucose does not react with hydroxyl amine.

SolutionAnswer: Option 2

α

-D-glucose and

β

-D-glucose are enantiomers.

Approach:

Evaluate each statement about glucose chemistry to find the false one.

Step 1:Alpha- and beta-D-glucose differ only in configuration at the anomeric (C1) carbon, so they are anomers (statement 1 correct). They are diastereomers, not non-superimposable mirror images, so they are NOT enantiomers (statement 2 is incorrect).

anomers (C1 only) \neq = enantiomers

Step 2:Cellulose is indeed a straight-chain polymer of beta-D-glucose (statement 3 correct), and glucose pentaacetate has no free -CHO/hemiacetal -OH, so it does not react with hydroxylamine (statement 4 correct).

(C_{6} H_{7} O_{2} (OAc)_{3})_{n}

Final answer:

α

-D-glucose and

β

-D-glucose are enantiomers.

Q56Single correctPolymers

Which of the following is a biodegradable polymer ?

(A)

nylon-2-nylon-6 copolymer repeat unit: [HN-(CH2)5-CO-NH-CH2-CO]n (glycine + 6-aminohexanoic acid); chain ends shown as connection points

(B)

nylon-6 repeat unit: [HN-(CH2)5-CO]n; chain ends shown as connection points

(C)

nylon-6,6 repeat unit: [HN-(CH2)6-NH-CO-(CH2)4-CO]n; chain ends shown as connection points

(D)

polyester (terephthalate) repeat unit: [CO-C6H4-CO-O-(CH2)2-O]n; chain ends shown as connection points

SolutionAnswer: Option 1nylon-2-nylon-6 (copolymer of glycine and amino caproic acid)

Approach:

Identify the polymer composed of naturally occurring amino-acid-type monomers, which is biodegradable.

Step 1:Option 1 is nylon-2-nylon-6, an alternating polyamide of glycine (H2N-CH2-COOH) and amino caproic acid (H2N-(CH2)5-COOH); these biological-type monomers make it biodegradable.

[NH-CH_{2} -CO-NH-(CH_{2})_{5} -CO]_{n}

Step 2:Nylon-6 (option 2) and nylon-6,6 (option 3) are non-biodegradable polyamides, and option 4 is a synthetic polyester that is not biodegradable here.

nylon-6, nylon-6,6 \Rightarrow non-biodegradable

Final answer: nylon-2-nylon-6 (copolymer of glycine and amino caproic acid)

Q57Single correctOrganic Compounds Containing Oxygen / Haloalkanes

The increasing order of the boiling points for the following compounds is :

C_{2} H_{5} OH

(I),

C_{2} H_{5} Cl

(II),

C_{2} H_{5} CH_{3}

(III),

C_{2} H_{5} OCH_{3}

(IV)

(A)

(III) < (IV) < (II) < (I)

(B)

(IV) < (III) < (I) < (II)

(C)

(II) < (III) < (IV) < (I)

(D)

(III) < (II) < (I) < (IV)

SolutionAnswer: Option 1(III) < (IV) < (II) < (I)

Approach:

Order boiling points by intermolecular forces: hydrogen bonding > dipole-dipole > van der Waals.

Step 1:III is propane (C2H5CH3), a nonpolar alkane with only weak London forces, so it has the lowest boiling point (about -42 C).

C_{2} H_{5} CH_{3}, bp \approx - 42^{\circ} C

Step 2:IV is methoxyethane (an ether) with weak dipole interactions (bp about 7 C); II is ethyl chloride, more polar/polarisable (bp about 12 C).

IV \approx 7^{\circ} C < II \approx 12^{\circ} C

Step 3:I is ethanol, which forms intermolecular hydrogen bonds, giving the highest boiling point (78 C).

C_{2} H_{5} OH, bp = 78^{\circ} C

Final answer: (III) < (IV) < (II) < (I)

Q58Single correctBasic Principles of Organic Chemistry

Which of the following compounds will show highest dipole moment ?

(A)

(I) cyclopropenone: three-membered ring with one ring C=C and an exocyclic C=O

(B)

(II) cyclobut-2-enone: four-membered ring with one ring C=C and an exocyclic C=O

(C)

(III) cyclopent-2-enone: five-membered ring with one ring C=C and an exocyclic C=O (as drawn in source)

(D)

(IV) para-benzoquinone: six-membered ring with two C=O groups at 1,4-positions and two ring C=C double bonds

SolutionAnswer: Option 1(I)

Approach:

Greatest charge separation gives the highest dipole moment; a carbonyl whose ring becomes aromatic on shifting electron density shows an exceptionally large dipole.

Step 1:In cyclopropenone (I), shifting the carbonyl pi electrons toward oxygen leaves an aromatic cyclopropenyl cation (2 pi electrons) in the ring, strongly stabilising the charge-separated form.

ring \to aromatic C_{3} H_{2}^{+}, O^{-}

Step 2:This aromatic stabilisation makes the dipole moment of cyclopropenone (about 4.8 D) the largest among the listed carbonyl compounds.

μ (cyclopropenone) \approx 4.78 D

Step 3:p-Benzoquinone (IV) has two opposed C=O dipoles that partly cancel, lowering its net dipole, while II and III lack the strong aromatic cation stabilisation of I.

μ (p-benzoquinone) \approx 0

Final answer: (I)

Q59Single correctHaloalkanes and Haloarenes

(A)

(B)

(C)

(D)

SolutionAnswer: Option 4

CH_{3} - CCl_{2} - CH_{3}

(2,2-dichloropropane)

Approach:

Immediate turbidity in the Lucas test means III is a tertiary alcohol; work backward through the Grignard step and the hydrolysis of the dihalide.

Step 1:Immediate turbidity with anhydrous ZnCl2 + conc. HCl (Lucas reagent) identifies III as a tertiary alcohol.

R_{3} C-OH Lucas immediate turbidity

Step 2:A tertiary alcohol from CH3MgBr addition requires II to be a ketone. The only C3 ketone here is acetone.

CH_{3} COCH_{3} + CH_{3} MgBr \to (CH_{3})_{3} C-OH

Step 3:Acetone forms by hydrolysis of a gem-dihalide; the gem-dichloride C3H6Cl2 is 2,2-dichloropropane, CH3-CCl2-CH3.

CH_{3} CCl_{2} CH_{3} KOH(aq) CH_{3} COCH_{3}

Final answer:

CH_{3} - CCl_{2} - CH_{3}

(2,2-dichloropropane)

Q60Single correctAmines

Among the following compounds, the increasing order of their basic strength is :

Four labelled amines in a 2x2 grid: (I) aniline (benzene ring with NH2); (II) pyrrole (five-membered ring with N-H); (III) N-methylpiperidine (saturated six-membered ring with N bearing a methyl); (IV) cyclohexylamine (cyclohexane with NH2).

(A)

(I) < (II) < (IV) < (III)

(B)

(I) < (II) < (III) < (IV)

(C)

(II) < (I) < (IV) < (III)

(D)

(II) < (I) < (III) < (IV)

SolutionAnswer: Option 3(II) < (I) < (IV) < (III)

Approach:

Basicity depends on availability of the nitrogen lone pair; lone pairs locked in aromatic systems or delocalised into rings are least available.

Step 1:In pyrrole (II) the nitrogen lone pair is part of the aromatic sextet, so it is essentially unavailable; pyrrole is the weakest base.

pyrrole: lone pair in aromatic ring

Step 2:In aniline (I) the lone pair is delocalised into the benzene ring, so it is more basic than pyrrole but less basic than the aliphatic amines.

C_{6} H_{5} NH_{2} : lone pair on N, partly delocalised

Step 3:Cyclohexylamine (IV) is a primary aliphatic amine with a fully available lone pair; N-methylpiperidine (III) is a tertiary aliphatic amine, the most basic of these due to alkyl electron donation.

IV (1^{\circ}) < III (3^{\circ} aliphatic)

Final answer: (II) < (I) < (IV) < (III)

Mathematics30 questions

Q61Single correctSets, Relations and Functions

The function

f : N \to N

defined by

f (x) = x - 5 [\frac{x}{5}]

, where

N

is the set of natural numbers and [x] denotes the greatest integer less than or equal to x, is :

(A)

one-one and onto.

(B)

one-one but not onto.

(C)

onto but not one-one.

(D)

neither one-one nor onto.

SolutionAnswer: Option 4neither one-one nor onto.

Approach:

Interpret the rule and test injectivity and surjectivity on the domain and codomain of natural numbers.

Step 1:Identify the output set.

f (x) = x mod 5 \in {0, 1, 2, 3, 4}

Step 2:Check onto.

5, 6 \in N

map to

0, 1

but values such as

6

are never attained as a remainder; range

\subset N

is not all of

N

Step 3:Check one-one.

f (1) = f (6) = 1

Final answer: neither one-one nor onto.

Q62Single correctComplex Numbers and Quadratic Equations

The sum of all the real values of x satisfying the equation

2^{(x - 1) (x^{2} + 5 x - 50)} = 1

is :

(A)

16

(B)

14

(C)

- 4

(D)

- 5

SolutionAnswer: Option 3

- 4

Approach:

Since the base is 2, the equation holds when the exponent equals zero; sum the resulting roots.

Step 1:Set exponent to zero.

(x - 1) (x^{2} + 5 x - 50) = 0

Step 2:Solve each factor.

x = 1 or (x + 10) (x - 5) = 0

Step 3:Add the roots.

1 + (- 10) + 5 = - 4

Final answer:

- 4

Q63Single correctComplex Numbers and Quadratic Equations

The equation

Im (\frac{i z - 2}{z - i}) + 1 = 0, z \in C, z \neq = i

represents a part of a circle having radius equal to :

(A)

2

(B)

1

(C)

\frac{3}{4}

(D)

\frac{1}{2}

SolutionAnswer: Option 3

\frac{3}{4}

Approach:

Write

z = x + i y

, extract the imaginary part of the given expression, and reduce to a circle equation.

Step 1:Substitute

z = x + i y

and simplify the imaginary part.

Im (\frac{i z - 2}{z - i}) + 1 = \frac{2 x ^{2} + 2 y ^{2} - y - 1}{x ^{2} + ( y - 1 ) ^{2}}

Step 2:Set numerator to zero.

2 x^{2} + 2 y^{2} - y - 1 = 0 \Rightarrow x^{2} + y^{2} - \frac{y}{2} - \frac{1}{2} = 0

Step 3:Complete the square.

x^{2} + (y - \frac{1}{4})^{2} = \frac{1}{2} + \frac{1}{16} = \frac{9}{16}

Final answer:

\frac{3}{4}

Q64Single correctMatrices and Determinants

For two

3 \times 3

matrices A and B, let

A + B = 2 B^{'}

and

3 A + 2 B = I_{3}

, where B' is the transpose of B and

I_{3}

3 \times 3

identity matrix. Then :

(A)

5 A + 10 B = 2 I_{3}

(B)

10 A + 5 B = 3 I_{3}

(C)

B + 2 A = I_{3}

(D)

3 A + 6 B = 2 I_{3}

SolutionAnswer: Option 2

10 A + 5 B = 3 I_{3}

Approach:

Use the transpose of the first relation together with the second to solve for A and B explicitly.

Step 1:Transpose the first equation.

A + B = 2 B^{'} \Rightarrow A^{'} + B^{'} = 2 B

Step 2:Solve the linear system entrywise with

3 A + 2 B = I_{3}

A = \frac{1}{5} I_{3}, B = \frac{1}{5} I_{3}

Step 3:Test the options.

10 A + 5 B = 2 I_{3} + I_{3} = 3 I_{3}

Final answer:

10 A + 5 B = 3 I_{3}

Q65Single correctMatrices and Determinants

x = a

y = b

z = c

is a solution of the system of linear equations

x + 8 y + 7 z = 0

9 x + 2 y + 3 z = 0

x + y + z = 0

such that the point

(a, b, c)

lies on the plane

x + 2 y + z = 6

, then

2 a + b + c

equals :

(A)

- 1

(B)

0

(C)

1

(D)

2

SolutionAnswer: Option 3

1

Approach:

Solve the homogeneous system for a parametric solution, then fix the parameter using the plane condition.

Step 1:Solve the homogeneous system.

x = - \frac{z}{7}, y = - \frac{6 z}{7}

Step 2:Apply the plane condition.

- \frac{z}{7} - \frac{12 z}{7} + z = 6 \Rightarrow - \frac{6 z}{7} = 6 \Rightarrow z = - 7

Step 3:Evaluate the required expression.

2 a + b + c = 2 (1) + 6 + (- 7) = 1

Final answer:

1

Q66Single correctPermutations and Combinations

The number of ways in which 5 boys and 3 girls can be seated on a round table if a particular boy

B_{1}

and a particular girl

G_{1}

never sit adjacent to each other, is :

(A)

5 \times 6!

(B)

6 \times 6!

(C)

7!

(D)

5 \times 7!

SolutionAnswer: Option 1

5 \times 6!

Approach:

Use complementary counting: total circular arrangements minus those where the two specified people sit together.

Step 1:Total circular arrangements of 8 people.

(8 - 1)! = 7!

Step 2:Arrangements with

B_{1}

and

G_{1}

together (treat as one block).

(7 - 1)! \times 2! = 2 \cdot 6!

Step 3:Subtract.

7! - 2 \cdot 6! = 6! (7 - 2) = 5 \times 6!

Final answer:

5 \times 6!

Q67Single correctBinomial Theorem

The coefficient of

x^{- 5}

in the binomial expansion of

(\frac{x + 1}{x ^{\frac{2}{3}} - x ^{\frac{1}{3}} + 1} - \frac{x - 1}{x - x ^{\frac{1}{2}}})^{10}

, where

x \neq = 0, 1

, is :

(A)

1

(B)

4

(C)

- 4

(D)

- 1

SolutionAnswer: Option 1

1

Approach:

Simplify each rational term using factorisation, then apply the binomial general term to locate

x^{- 5}

Step 1:Simplify the first term with

a = x^{1/3}

\frac{x + 1}{x ^{2/3} - x ^{1/3} + 1} = x^{1/3} + 1

Step 2:Simplify the second term.

\frac{x - 1}{x - x ^{1/2}} = \frac{x + 1}{x} = 1 + x^{- 1/2}

Step 3:Combine to get the base.

(x^{1/3} - x^{- 1/2})^{10}

Step 4:Require exponent

- 5

in the general term.

\frac{10 - r}{3} - \frac{r}{2} = - 5 \Rightarrow r = 10

Step 5:Compute the coefficient.

(10 10) (- 1)^{10} = 1

Final answer:

1

Q68Single correctSequences and Series

If three positive numbers a, b and c are in A.P. such that

ab c = 8

, then the minimum possible value of b is :

(A)

2

(B)

4^{\frac{1}{3}}

(C)

4^{\frac{2}{3}}

(D)

4

SolutionAnswer: Option 1

2

Approach:

Use the AP condition and the AM-GM inequality to bound b in terms of the product.

Step 1:Apply AM-GM to a and c.

b = \frac{a + c}{2} \geq a c \Rightarrow b^{2} \geq a c

Step 2:Use the product condition

a c = 8/ b

\frac{8}{b} \leq b^{2} \Rightarrow b^{3} \geq 8

Step 3:Identify minimum.

b = 2 at a = c = 2

Final answer:

2

Q69Single correctSequences and Series

Let

S_{n} = \frac{1}{1 ^{3}} + \frac{1 + 2}{1 ^{3} + 2 ^{3}} + \frac{1 + 2 + 3}{1 ^{3} + 2 ^{3} + 3 ^{3}} + \dots + \frac{1 + 2 + \dots + n}{1 ^{3} + 2 ^{3} + \dots + n ^{3}}

. If

100 S_{n} = n

, then n is equal to :

(A)

199

(B)

99

(C)

200

(D)

19

SolutionAnswer: Option 1

199

Approach:

Simplify the general term using standard sum formulas, telescope, then solve the given relation for n.

Step 1:Simplify the general term.

\frac{k ( k + 1 ) /2}{( k ( k + 1 ) /2 ) ^{2}} = \frac{2}{k ( k + 1 )}

Step 2:Sum by telescoping.

S_{n} = 2 \sum_{k = 1}^{n} (\frac{1}{k} - \frac{1}{k + 1}) = \frac{2 n}{n + 1}

Step 3:Apply the condition.

100 \cdot \frac{2 n}{n + 1} = n \Rightarrow \frac{200}{n + 1} = 1 \Rightarrow n = 199

Final answer:

199

Q70Single correctLimits, Continuity and Differentiability

The value of k for which the function

f (x) = ⎩ ⎨ ⎧ (\frac{4}{5})^{\frac{t a n 4 x}{t a n 5 x}}, k + \frac{2}{5}, 0 < x < \frac{π}{2} x = \frac{π}{2}

is continuous at

x = \frac{π}{2}

, is :

(A)

\frac{17}{20}

(B)

\frac{2}{5}

(C)

\frac{3}{5}

(D)

- \frac{2}{5}

SolutionAnswer: Option 3

\frac{3}{5}

Approach:

Continuity requires the limit of the first piece as

x \to π /2

to equal the value at

x = π /2

Step 1:Evaluate the exponent limit.

lim_{x \to π /2} \frac{tan 4 x}{tan 5 x} = \frac{0}{\infty} = 0

Step 2:Compute the limit of the first piece.

(\frac{4}{5})^{0} = 1

Step 3:Match with the value at

π /2

k + \frac{2}{5} = 1 \Rightarrow k = \frac{3}{5}

Final answer:

\frac{3}{5}

Q71Single correctDifferential Equations

2 x = y^{\frac{1}{5}} + y^{- \frac{1}{5}}

and

(x^{2} - 1) \frac{d ^{2} y}{d x ^{2}} + λ x \frac{d y}{d x} + k y = 0

, then

λ + k

is equal to :

(A)

- 23

(B)

- 24

(C)

26

(D)

- 26

SolutionAnswer: Option 2

- 24

Approach:

Express y explicitly, differentiate twice, and identify the coefficients

λ

and k of the resulting differential equation.

Step 1:Solve for y.

y^{1/5} = x + x^{2} - 1 \Rightarrow y = (x + x^{2} - 1)^{5}

Step 2:Form the differential equation by differentiating.

(x^{2} - 1) \frac{d ^{2} y}{d x ^{2}} + x \frac{d y}{d x} - 25 y = 0

Step 3:Add the coefficients.

λ + k = 1 + (- 25) = - 24

Final answer:

- 24

Q72Single correctApplications of Derivatives

The function f defined by

f (x) = x^{3} - 3 x^{2} + 5 x + 7

, is :

(A)

increasing in

R

(B)

decreasing in

R

(C)

decreasing in

(0, \infty)

and increasing in

(- \infty, 0)

(D)

increasing in

(0, \infty)

and decreasing in

(- \infty, 0)

SolutionAnswer: Option 1increasing in

R

Approach:

Examine the sign of the first derivative over all real numbers.

Step 1:Differentiate.

f^{'} (x) = 3 x^{2} - 6 x + 5

Step 2:Check the discriminant.

Δ = 36 - 60 = - 24 < 0

Step 3:Determine the sign.

f^{'} (x) > 0 for all x \in R

Final answer: increasing in

R

Q73Single correctDifferentiation

Let f be a polynomial function such that

f (3 x) = f^{'} (x) \cdot f^{''} (x)

, for all

x \in R

. Then :

(A)

f (2) + f^{'} (2) = 28

(B)

f^{''} (2) - f^{'} (2) = 0

(C)

f^{''} (2) - f (2) = 4

(D)

f (2) - f^{'} (2) + f^{''} (2) = 10

SolutionAnswer: Option 2

f^{''} (2) - f^{'} (2) = 0

Approach:

Compare degrees to fix the polynomial degree, determine coefficients, then test the options.

Step 1:Match degrees.

n = (n - 1) + (n - 2) \Rightarrow n = 3

Step 2:Solve for coefficients.

f (x) = \frac{3}{2} x^{3}

Step 3:Evaluate derivatives at

x = 2

f^{'} (x) = \frac{9}{2} x^{2}, f^{''} (x) = 9 x \Rightarrow f^{'} (2) = 18, f^{''} (2) = 18

Final answer:

f^{''} (2) - f^{'} (2) = 0

Q74Single correctIntegral Calculus

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

, and

\int f (x) d x = A lo g ∣1 - x ∣ + B x + C

, then the ordered pair (A, B) is equal to : (where C is a constant of integration)

(A)

(\frac{8}{3}, \frac{2}{3})

(B)

(- \frac{8}{3}, \frac{2}{3})

(C)

(- \frac{8}{3}, - \frac{2}{3})

(D)

(\frac{8}{3}, - \frac{2}{3})

SolutionAnswer: Option 2

(- \frac{8}{3}, \frac{2}{3})

Approach:

Invert the argument to find

f

explicitly, integrate, and compare with the given form.

Step 1:Solve the argument for x.

u = \frac{3 x - 4}{3 x + 4} \Rightarrow x = \frac{4 ( 1 + u )}{3 ( 1 - u )}

Step 2:Write

f

explicitly.

f (x) = \frac{2 ( x - 5 )}{3 ( x - 1 )}

Step 3:Integrate.

\int f (x) d x = \frac{2}{3} x - \frac{8}{3} lo g ∣ x - 1∣ + C

Final answer:

(- \frac{8}{3}, \frac{2}{3})

Q75Single correctIntegral Calculus

\int_{1}^{2} \frac{d x}{( x ^{2} - 2 x + 4 ) ^{\frac{3}{2}}} = \frac{k}{k + 5}

, then k is equal to :

(A)

1

(B)

2

(C)

3

(D)

4

SolutionAnswer: Option 1

1

Approach:

Complete the square, evaluate the definite integral, then equate to the given expression and solve for k.

Step 1:Complete the square.

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

Step 2:Evaluate the integral.

\int_{1}^{2} \frac{d x}{( ( x - 1 ) ^{2} + 3 ) ^{3/2}} = \frac{1}{6}

Step 3:Solve for k.

\frac{k}{k + 5} = \frac{1}{6} \Rightarrow 6 k = k + 5 \Rightarrow k = 1

Final answer:

1

Q76Single correctLimit, Continuity and Differentiability

lim_{n \to \infty} \frac{1 ^{a} + 2 ^{a} + ........ + n ^{a}}{( n + 1 ) ^{a - 1} [( na + 1 ) + ( na + 2 ) + ..... + ( na + n )]} = \frac{1}{60}

for some positive real number a, then a is equal to :

(A)

7

(B)

8

(C)

\frac{15}{2}

(D)

\frac{17}{2}

SolutionAnswer: Option 1

7

Approach:

Replace the numerator power sum by its leading asymptotic term and the bracket sum by its leading term, then equate the resulting limit to 1/60.

Step 1:Numerator leading behaviour.

1^{a} + 2^{a} + \dots + n^{a} \sim \frac{n ^{a + 1}}{a + 1}

Step 2:Evaluate the bracket sum and its leading term.

(na + 1) + \dots + (na + n) = a n^{2} + \frac{n ( n + 1 )}{2} \sim (a + \frac{1}{2}) n^{2}

Step 3:Denominator leading behaviour with

(n + 1)^{a - 1}

n^{a - 1}

(n + 1)^{a - 1} (a + \frac{1}{2}) n^{2} \sim (a + \frac{1}{2}) n^{a + 1}

Step 4:Form the limit and set equal to 1/60.

\frac{1/ ( a + 1 )}{a + 1/2} = \frac{2}{( a + 1 ) ( 2 a + 1 )} = \frac{1}{60}

Step 5:Solve the equation.

2 a^{2} + 3 a - 119 = 0 \Rightarrow a = 7

Final answer:

7

Q77Single correctDifferential Equations

A tangent to the curve,

y = f (x)

at P(x, y) meets x-axis at A and y-axis at B. If AP : BP = 1 : 3 and

f (1) = 1

, then the curve also passes through the point :

(A)

(\frac{1}{3}, 24)

(B)

(\frac{1}{2}, 4)

(C)

(2, \frac{1}{8})

(D)

(3, \frac{1}{28})

SolutionAnswer: Option 3

(2, \frac{1}{8})

Approach:

Express the x- and y-intercepts of the tangent, apply the section-formula ratio AP:BP = 1:3 to obtain a differential equation, solve it with the initial condition, then test the options.

Step 1:Intercepts of the tangent at P(x,y).

A = (x - \frac{y}{y ^{'}}, 0), B = (0, y - x y^{'})

Step 2:AP:BP = 1:3 places P at t=1/4 from A; match x-coordinate.

x = \frac{3}{4} (x - \frac{y}{y ^{'}}) \Rightarrow \frac{y}{y ^{'}} = - \frac{x}{3}

Step 3:Separate variables and integrate.

\frac{d y}{y} = - 3 \frac{d x}{x} \Rightarrow x^{3} y = C

Step 4:Apply f(1)=1.

1^{3} \cdot 1 = C \Rightarrow C = 1, x^{3} y = 1

Step 5:Test options.

(2, \frac{1}{8}) : 2^{3} \cdot \frac{1}{8} = 1

Final answer:

(2, \frac{1}{8})

Q78Single correctCoordinate Geometry

A square, of each side 2, lies above the x-axis and has one vertex at the origin. If one of the sides passing through the origin makes an angle 3

0^{\circ}

with the positive direction of the x-axis, then the sum of the x-coordinates of the vertices of the square is :

(A)

23 - 1

(B)

23 - 2

(C)

3 - 2

(D)

3 - 1

SolutionAnswer: Option 2

23 - 2

Approach:

Place one vertex at the origin, use the 30 degree direction for one side and the perpendicular 120 degree direction for the adjacent side to find all four vertices, then add their x-coordinates.

Step 1:Vertices adjacent to the origin.

B = (2 cos 3 0^{\circ}, 2 sin 3 0^{\circ}) = (3, 1), D = (2 cos 12 0^{\circ}, 2 sin 12 0^{\circ}) = (- 1, 3)

Step 2:Fourth vertex.

C = B + D = (3 - 1, 1 + 3)

Step 3:Sum of x-coordinates.

0 + 3 + (- 1) + (3 - 1) = 23 - 2

Final answer:

23 - 2

Q79Single correctCoordinate Geometry

A line drawn through the point P(4, 7) cuts the circle

x^{2} + y^{2} = 9

at the points A and B. Then PA . PB is equal to :

(A)

53

(B)

56

(C)

74

(D)

65

SolutionAnswer: Option 2

56

Approach:

Use the power of a point with respect to the circle, which equals PA.PB for any secant through the point.

Step 1:Apply the power formula for P(4,7) and the circle of radius 3.

P A \cdot PB = 4^{2} + 7^{2} - 9

Step 2:Compute.

16 + 49 - 9 = 56

Final answer:

56

Q80Single correctCoordinate Geometry

The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate axes and passing through the points

(4, - 1)

and

(- 2, 2)

is :

(A)

\frac{1}{2}

(B)

\frac{2}{5}

(C)

\frac{3}{2}

(D)

\frac{3}{4}

SolutionAnswer: Option 3

\frac{3}{2}

Approach:

Take the standard ellipse with unknown 1/

a^{2}

and 1/

b^{2}

, substitute both points to form linear equations, solve, then compute eccentricity.

Step 1:Let u=1/

a^{2}

, v=1/

b^{2}

and substitute the two points.

16 u + v = 1, 4 u + 4 v = 1

Step 2:Solve the system.

u = \frac{1}{20}, v = \frac{1}{5} \Rightarrow a^{2} = 20, b^{2} = 5

Step 3:Compute eccentricity (

a^{2}

b^{2}

e^{2} = 1 - \frac{5}{20} = \frac{3}{4} \Rightarrow e = \frac{3}{2}

Final answer:

\frac{3}{2}

Q81Single correctCoordinate Geometry

y = m x + c

is the normal at a point on the parabola

y^{2} = 8 x

whose focal distance is 8, then

∣ c ∣

is equal to :

(A)

23

(B)

83

(C)

103

(D)

163

SolutionAnswer: Option 3

103

Approach:

Use the focal distance to locate the point on the parabola, find the parameter t, then write the normal in parametric form and read off c.

Step 1:Identify a and locate the point.

4 a = 8 \Rightarrow a = 2; x + 2 = 8 \Rightarrow x = 6

Step 2:Find parameter t from x=

a t^{2}

2 t^{2} = 6 \Rightarrow t^{2} = 3 \Rightarrow t = \pm 3

Step 3:Normal intercept c (with a=2).

c = 2 a t + a t^{3} = 4 t + 2 t^{3}

Step 4:Evaluate at t=sqrt3.

c = 43 + 2 (33) = 103

Final answer:

103

Q82Single correctThree Dimensional Geometry

If a variable plane, at a distance of 3 units from the origin, intersects the coordinate axes at A, B and C, then the locus of the centroid of

△ ABC

is :

(A)

\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = 1

(B)

\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = 3

(C)

\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = \frac{1}{9}

(D)

\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = 9

SolutionAnswer: Option 1

\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = 1

Approach:

Write the intercept form of the plane, impose the distance-from-origin condition, express the centroid coordinates in terms of the intercepts, and eliminate the intercepts.

Step 1:Distance condition gives a relation among intercepts.

\frac{1}{p ^{2}} + \frac{1}{q ^{2}} + \frac{1}{r ^{2}} = \frac{1}{9}

Step 2:Centroid (x,y,z) implies p=3x, q=3y, r=3z.

\frac{1}{9 x ^{2}} + \frac{1}{9 y ^{2}} + \frac{1}{9 z ^{2}} = \frac{1}{9}

Step 3:Multiply by 9.

\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = 1

Final answer:

\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = 1

Q83Single correctThree Dimensional Geometry

If the line,

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

lies in the plane,

2 x - 4 y + 3 z = 2

, then the shortest distance between this line and the line

\frac{x - 1}{12} = \frac{y}{9} = \frac{z}{4}

is :

(A)

2

(B)

1

(C)

0

(D)

3

SolutionAnswer: Option 3

0

Approach:

Determine lambda from the condition that the first line lies in the given plane, then compute the shortest distance between the two lines using the scalar triple product.

Step 1:The point (3,-2,-lambda) lies in the plane.

2 (3) - 4 (- 2) + 3 (- λ) = 2 \Rightarrow 14 - 3 λ = 2 \Rightarrow λ = 4

Step 2:Direction vectors and a connecting vector.

d_{1} = (1, - 1, - 2), d_{2} = (12, 9, 4), a_{2} - a_{1} = (- 2, 2, 4)

Step 3:Cross product of directions.

d_{1} \times d_{2} = (14, - 28, 21)

Step 4:Scalar triple product of the connecting vector with the cross product.

(2, - 2, - 4) \cdot (14, - 28, 21) = 28 + 56 - 84 = 0

Step 5:Shortest distance.

d = \frac{0}{7 29} = 0

Final answer:

0

Q84Single correctVector Algebra

If the vector

b = 3 \hat{j} + 4 \hat{k}

is written as the sum of a vector

b_{1}

, parallel to

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

and a vector

b_{2}

, perpendicular to

a

, then

b_{1} \times b_{2}

is equal to :

(A)

- 3 \hat{i} + 3 \hat{j} - 9 \hat{k}

(B)

6 \hat{i} - 6 \hat{j} + \frac{9}{2} \hat{k}

(C)

- 6 \hat{i} + 6 \hat{j} - \frac{9}{2} \hat{k}

(D)

3 \hat{i} - 3 \hat{j} + 9 \hat{k}

SolutionAnswer: Option 2

6 \hat{i} - 6 \hat{j} + \frac{9}{2} \hat{k}

Approach:

Project b onto a to get b1, subtract to get b2, then take the cross product.

Step 1:Component parallel to a.

b \cdot a = 3, ∣ a ∣^{2} = 2 \Rightarrow b_{1} = \frac{3}{2} \hat{i} + \frac{3}{2} \hat{j}

Step 2:Perpendicular component.

b_{2} = b - b_{1} = - \frac{3}{2} \hat{i} + \frac{3}{2} \hat{j} + 4 \hat{k}

Step 3:Cross product.

b_{1} \times b_{2} = 6 \hat{i} - 6 \hat{j} + \frac{9}{2} \hat{k}

Final answer:

6 \hat{i} - 6 \hat{j} + \frac{9}{2} \hat{k}

Q85Single correctProbability

From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at least one woman. Then the probability for these committees to have more women than men, is :

(A)

\frac{21}{220}

(B)

\frac{3}{11}

(C)

\frac{1}{11}

(D)

\frac{2}{23}

SolutionAnswer: Option 3

\frac{1}{11}

Approach:

Count committees containing at least one woman as the sample space, count those with more women than men as the favourable event, and divide.

Step 1:Total committees with at least one woman.

(4 15) - (4 10) = 1365 - 210 = 1155

Step 2:Favourable: more women than men means 3W1M or 4W.

(3 5) (1 10) + (4 5) = 100 + 5 = 105

Step 3:Probability.

\frac{105}{1155} = \frac{1}{11}

Final answer:

\frac{1}{11}

Q86Single correctProbability

Let E and F be two independent events. The probability that both E and F happen is

\frac{1}{12}

and the probability that neither E nor F happens is

\frac{1}{2}

, then a value of

\frac{P ( E )}{P ( F )}

is :

(A)

\frac{4}{3}

(B)

\frac{3}{2}

(C)

\frac{1}{3}

(D)

\frac{5}{12}

SolutionAnswer: Option 1

\frac{4}{3}

Approach:

Use independence to set up equations for the product and for the complement product, solve for P(E) and P(F) as roots of a quadratic, then form the ratio.

Step 1:Let p=P(E), q=P(F).

pq = \frac{1}{12}, (1 - p) (1 - q) = \frac{1}{2}

Step 2:Expand the second equation.

1 - (p + q) + pq = \frac{1}{2} \Rightarrow p + q = \frac{7}{12}

Step 3:p and q are roots of the quadratic.

12 t^{2} - 7 t + 1 = 0 \Rightarrow t = \frac{1}{3}, \frac{1}{4}

Step 4:Take the ratio.

\frac{1/3}{1/4} = \frac{4}{3}

Final answer:

\frac{4}{3}

Q87Single correctStatistics

The sum of 100 observations and the sum of their squares are 400 and 2475, respectively. Later on, three observations, 3, 4 and 5, were found to be incorrect. If the incorrect observations are omitted, then the variance of the remaining observations is :

(A)

8.25

(B)

8.50

(C)

8.00

(D)

9.00

SolutionAnswer: Option 4

9.00

Approach:

Remove the three omitted observations from the sum and sum of squares, then apply the variance formula to the remaining 97 observations.

Step 1:Corrected sum after removing 3,4,5.

\sum x = 400 - (3 + 4 + 5) = 388

Step 2:Corrected sum of squares.

\sum x^{2} = 2475 - (9 + 16 + 25) = 2425

Step 3:Mean of remaining 97 observations.

\overset{x}{ˉ} = \frac{388}{97} = 4

Step 4:Variance.

σ^{2} = \frac{2425}{97} - 4^{2} = 25 - 16 = 9

Final answer:

9.00

Q88Single correctInverse Trigonometric Functions

A value of x satisfying the equation

sin [cot^{- 1} (1 + x)] = cos [tan^{- 1} x]

, is :

(A)

- \frac{1}{2}

(B)

- 1

(C)

0

(D)

\frac{1}{2}

SolutionAnswer: Option 1

- \frac{1}{2}

Approach:

Convert each inverse-trig expression to an algebraic form and equate.

Step 1:Rewrite both sides.

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

Step 2:Square and simplify.

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

Step 3:Solve.

x = - \frac{1}{2}

Final answer:

- \frac{1}{2}

Q89Single correctTrigonometry

The two adjacent sides of a cyclic quadrilateral are 2 and 5 and the angle between them is 6

0^{\circ}

. If the area of the quadrilateral is

43

, then the perimeter of the quadrilateral is :

(A)

12.5

(B)

13.2

(C)

12

(D)

13

SolutionAnswer: Option 3

12

Approach:

Use the cyclic property (opposite angles supplementary) to relate the two diagonal-triangles via the common diagonal and the total area, then solve for the remaining two sides.

Step 1:Diagonal from the 60-degree triangle (sides 2,5).

d^{2} = 4 + 25 - 2 (2) (5) cos 6 0^{\circ} = 19

Step 2:Opposite angle is 120 degrees; let other sides be c, e.

d^{2} = c^{2} + e^{2} - 2 ce cos 12 0^{\circ} = c^{2} + e^{2} + ce = 19

Step 3:Total area equation.

\frac{3}{4} (10 + ce) = 43 \Rightarrow ce = 6

Step 4:Find c+e.

c^{2} + e^{2} = 13, (c + e)^{2} = 13 + 12 = 25 \Rightarrow c + e = 5

Step 5:Perimeter.

2 + 5 + (c + e) = 2 + 5 + 5 = 12

Final answer:

12

Q90Single correctMathematical Reasoning

Contrapositive of the statement 'If two numbers are not equal, then their squares are not equal', is :

(A)

If the squares of two numbers are equal, then the numbers are equal.

(B)

If the squares of two numbers are equal, then the numbers are not equal.

(C)

If the squares of two numbers are not equal, then the numbers are not equal.

(D)

If the squares of two numbers are not equal, then the numbers are equal.

SolutionAnswer: Option 1If the squares of two numbers are equal, then the numbers are equal.

Approach:

Identify the hypothesis p and conclusion q, then form the contrapositive as (not q) implies (not p).

Step 1:Identify p and q.

p :

two numbers not equal;

q :

their squares not equal

Step 2:Form the contrapositive (not q implies not p).

\sim q :

squares equal;

\sim p :

numbers equal

Final answer: If the squares of two numbers are equal, then the numbers are equal.

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