The magnetic field of an electromagnetic wave is given by: B=1.6×10^-6 (2×10^7z+6×10^15t ) (2i+j ) Wb m^2 . The associated electric field will be:

E=4.8×10^2 (2×10^7z+6×10^15t ) (-i+2j )(V)/(m)

Home/JEE Main PYQs/2019 · April 08, Shift 2

JEE Main 2019 April 08, Shift 2 Question Paper with Solutions

All 89 questions from the JEE Main 2019 (April 08, Shift 2) shift — Physics (30), Chemistry (29) 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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Physics30 questions

Q1Single correctVectors

Let

∣ A_{1} ∣ = 3

∣ A_{2} ∣ = 5

and

∣ A_{1} + A_{2} ∣ = 5

. The value of

(2 A_{1} + 3 A_{2}) \cdot (3 A_{1} - 2 A_{2})

is:

(A)

- 112.5

(B)

- 118.5

(C)

- 106.5

(D)

- 99.5

SolutionAnswer: Option 2

- 118.5

Approach:

Use the magnitude of the resultant to find the scalar product of the two vectors, then expand the required dot product.

Step 1:Expand the squared magnitude of the resultant.

25 = 9 + 25 + 2 A_{1} \cdot A_{2}

Step 2:Expand the required scalar product.

(2 A_{1} + 3 A_{2}) \cdot (3 A_{1} - 2 A_{2}) = 6 A_{1}^{2} + 5 A_{1} \cdot A_{2} - 6 A_{2}^{2}

Step 3:Evaluate the expression.

54 - 22.5 - 150 = - 118.5

Final answer:

- 118.5

Q2Single correctUnits and Measurements

In a simple pendulum experiment for determination of acceleration due to gravity

(g)

, time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of the pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of

g

is close to

(A)

0.2%

(B)

6.8%

(C)

3.5%

(D)

0.7%

SolutionAnswer: Option 2

6.8%

Approach:

Express g from the pendulum formula and propagate the fractional errors in length and time.

Step 1:Determine the fractional error in length.

\frac{Δ L}{L} = \frac{0.1}{55.0}

Step 2:Determine the fractional error in time using total time of 30 s.

\frac{Δ T}{T} = \frac{1}{30}

Step 3:Combine the errors.

\frac{Δ g}{g} = 0.0018 + 2 (0.0333) = 0.0685

Final answer:

6.8%

Q3Single correctUnits and Measurements

If Surface tension

(S)

, Moment of Inertia

(I)

and Planck's constant

(h)

, were to be taken as the fundamental units, the dimensional formula for linear momentum would be:

(A)

S^{1/2} I^{1/2} h^{0}

(B)

S^{1/2} I^{3/2} h^{- 1}

(C)

S^{3/2} I^{1/2} h^{0}

(D)

S^{1/2} I^{1/2} h^{- 1}

SolutionAnswer: Option 1

S^{1/2} I^{1/2} h^{0}

Approach:

Write dimensions of S, I, h and momentum in terms of M, L, T, then solve for the exponents.

Step 1:Set up the exponent equations from p =

S^{a}

I^{b}

h^{c}

MLT^{- 1} = (M T^{- 2})^{a} (M L^{2})^{b} (M L^{2} T^{- 1})^{c}

Step 2:Solve the system of equations.

a = \frac{1}{2}, b = \frac{1}{2}, c = 0

Final answer:

S^{1/2} I^{1/2} h^{0}

Q4Single correctKinematics

A particle starts from origin O from rest and moves with a uniform acceleration along the positive

x

-axis. Identify all figures that correctly represent the motion qualitatively. (

a

= acceleration,

v

= velocity,

x

= displacement,

t

= time)

Four labelled graphs for motion from rest under uniform acceleration. Graph A: a (y-axis) vs t, a horizontal straight line parallel to the t-axis (constant positive a). Graph B: v (y-axis) vs t, a straight line rising from the origin with positive slope. Graph C: x (y-axis) vs t, a curve rising from origin that is concave down (slope decreasing, flattening out). Graph D: x (y-axis) vs t, a curve rising from origin that is concave up (parabola, slope increasing). All four are part of the question stem and the options refer to them by letter.

(A)

A, B, (D)

(B)

A, B, (C)

(C)

B, (C)

(D)

(A)

SolutionAnswer: Option 1A, B, (D)

Approach:

Identify the correct shape of each graph for motion starting from rest with constant acceleration.

Step 1:Acceleration is constant, so the a-t graph (A) is a horizontal line.

a = constant

Step 2:Velocity increases linearly from zero, so the v-t graph (B) is a straight line through the origin.

v = a t

Step 3:Displacement varies as t squared, so the x-t graph is a parabola opening upward; graph (D) shows this while graph (C) does not.

x = \frac{1}{2} a t^{2}

Final answer: A, B, (D)

Q5Single correctRotational Motion

A uniform rectangular thin sheet

A BC D

of mass

M

has length

a

and breadth

b

, as shown in the figure. If the shaded portion

H BGO

is cut-off, the coordinates of the centre of mass of the remaining portion will be:

Rectangle ABCD with D at bottom-left labelled (0,0), C at bottom-right (a,0), A at top-left (0,b), B at top-right (a,b). Midpoints: H is midpoint of top edge AB, F is midpoint of bottom edge DC, E is midpoint of left edge AD, G is midpoint of right edge BC. O is the centre of the rectangle, labelled (a/2, b/2). Dashed lines HF (vertical through centre) and EG (horizontal through centre) divide the rectangle into four quadrants. The top-right quadrant HBGO (bounded by H, B, G, O) is shaded with diagonal hatching, indicating it is cut off.

(A)

\frac{5 a}{12}, \frac{5 b}{12}

(B)

\frac{5 a}{3}, \frac{5 b}{3}

(C)

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

(D)

\frac{3 a}{4}, \frac{3 b}{4}

SolutionAnswer: Option 1

\frac{5 a}{12}, \frac{5 b}{12}

Approach:

Treat the full sheet and the removed quadrant as point masses and apply the centre-of-mass subtraction formula with the origin at D.

Step 1:Full sheet has centroid at the centre with mass proportional to its area; the removed quadrant HBGO has one quarter of the area with centroid at (3a/4, 3b/4).

x_{c m} = \frac{ab \cdot \frac{a}{2} - \frac{ab}{4} \cdot \frac{3 a}{4}}{ab - \frac{ab}{4}}

Step 2:By symmetry the y-coordinate follows the same form.

y_{c m} = \frac{5 b}{12}

Final answer:

\frac{5 a}{12}, \frac{5 b}{12}

Q6Single correctWork, Energy and Power

A body of mass

m_{1}

moving with an unknown velocity of

v_{1} \hat{i}

, undergoes a collinear collision with a body of mass

m_{2}

moving with a velocity

v_{2} \hat{i}

. After the collision,

m_{1}

and

m_{2}

move with velocities of

v_{3} \hat{i}

and

v_{4} \hat{i}

, respectively. If

m_{2} = 0.5 m_{1}

and

v_{3} = 0.5 v_{1}

, then

v_{1}

is:

(A)

v_{4} - \frac{v _{2}}{4}

(B)

v_{4} - \frac{v _{2}}{2}

(C)

v_{4} + v_{2}

(D)

v_{4} - v_{2}

SolutionAnswer: Option 4

v_{4} - v_{2}

Approach:

Apply conservation of linear momentum for the collinear collision and substitute the given relations.

Step 1:Substitute m2 = 0.5 m1 and v3 = 0.5 v1.

m_{1} v_{1} + 0.5 m_{1} v_{2} = 0.5 m_{1} v_{1} + 0.5 m_{1} v_{4}

Step 2:Rearrange to isolate v1.

0.5 v_{1} = 0.5 v_{4} - 0.5 v_{2}

Final answer:

v_{4} - v_{2}

Q7Single correctRotational Motion

A rectangular solid box of length 0.3 m is held horizontally, with one of its sides on the edge of a platform of height 5 m. When released, it slips off the table in a very short time

τ = 0.01

s, remaining essentially horizontal. The angle by which it would rotate when it hits the ground will be (in radians) close to:

A horizontal rectangular bar (the solid box, drawn as a thin hatched horizontal slab labelled with a small length marker 'l' and a right-pointing arrow) resting with one end on the top-left corner edge of a vertical post/platform. The post is a vertical line of height h (labelled with a downward vertical arrow 'h') standing on hatched ground at the bottom. The box overhangs the edge to the right.

(A)

0.5

(B)

0.3

(C)

0.02

(D)

0.28

SolutionAnswer: Option 1

0.5

Approach:

Find the angular velocity imparted by the impulsive torque during slipping, then multiply by the time of free fall.

Step 1:During the short slipping time the gravitational torque about the edge gives angular velocity using moment of inertia of a rod about its end I =

m L^{2}

/3.

ω = \frac{m g \frac{L}{2} τ}{\frac{m L ^{2}}{3}} = \frac{3 gτ}{2 L}

Step 2:Compute the free-fall time from height 5 m.

t = \frac{2 ( 5 )}{10} = 1 s

Step 3:Angle rotated equals angular velocity times fall time.

θ = ω t = 0.5 \times 1

Final answer:

0.5

Q8Single correctRotational Motion

A solid sphere and solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping all throughout. The two climb maximum heights

h_{s p h}

and

h_{cy l}

on the inline. The ratio

\frac{h _{s p h}}{h _{cy l}}

is given by:

A circle (representing the rolling sphere/cylinder, with a centre dot) sits on a horizontal ground line at the left. The ground line continues to the right and then curves upward into a straight inclined ramp rising to the upper right, representing the incline the body climbs.

(A)

\frac{2}{5}

(B)

\frac{4}{5}

(C)

\frac{14}{15}

(D)

1

SolutionAnswer: Option 3

\frac{14}{15}

Approach:

Equate total kinetic energy of rolling to gravitational potential energy at maximum height for each body.

Step 1:For the sphere

k^{2}

R^{2}

= 2/5 and for the cylinder

k^{2}

R^{2}

= 1/2.

h \propto 1 + \frac{k ^{2}}{R ^{2}}

Step 2:Take the ratio of the two heights for equal initial speed.

\frac{h _{s p h}}{h _{cy l}} = \frac{7/5}{3/2} = \frac{14}{15}

Final answer:

\frac{14}{15}

Q9Single correctGravitation

A rocket has to be launched from earth in such a way that it never returns. If

E

is the minimum energy delivered by the rocket launcher, what should be the minimum energy that the launcher should have, if the same rocket is to be launched from the surface of the moon? Assume that the density of the earth and the moon are equal and that the earth's volume is 64 times the volume of the moon.

(A)

\frac{E}{64}

(B)

\frac{E}{4}

(C)

\frac{E}{32}

(D)

\frac{E}{16}

SolutionAnswer: Option 4

\frac{E}{16}

Approach:

Express the launch energy in terms of density and radius, then take the ratio for moon and earth using the volume relation.

Step 1:Volume ratio 64 gives radius ratio of 4, so the moon's radius is one quarter of the earth's.

\frac{R _{e}}{R _{m}} = 6 4^{1/3} = 4

Step 2:With equal density, energy scales as radius squared.

\frac{E _{m}}{E _{e}} = (\frac{R _{m}}{R _{e}})^{2} = \frac{1}{16}

Final answer:

\frac{E}{16}

Q10Single correctProperties of Solids and Liquids

Young's moduli of two wires

A

and

B

are in the ratio

7 : 4

. Wire

A

is 2 m long and has radius

R

. Wire

B

is 1.5 m long and has radius 2 mm. If the two wires stretch by the same length for a given load, the value of

R

is close to:

(A)

1.3

(B)

1.9

(C)

1.5

(D)

1.7

SolutionAnswer: Option 4

1.7

Approach:

Use the elongation formula and equate the extensions of the two wires for the same load.

Step 1:Equate the two elongations for the same force.

\frac{L _{A}}{r _{A}^{2} Y _{A}} = \frac{L _{B}}{r _{B}^{2} Y _{B}}

Step 2:Substitute the values with Y ratio 7:4 and radii in mm.

R^{2} = (2)^{2} \cdot \frac{2}{1.5} \cdot \frac{4}{7}

Step 3:Take the square root.

R \approx 3.05

Final answer:

1.7

Q11Single correctThermodynamics

The given diagram shows four processes i.e., isochoric, isobaric, isothermal and adiabatic. The correct assignment of the processes, in the same order is given by:

P-V diagram. Vertical axis labelled P (pointing up), horizontal axis labelled V (pointing right). From a common starting point at the top-left, four processes emanate: a horizontal line going right labelled 'a' (constant pressure); a vertical line going straight down labelled 'd' (constant volume); and two curves falling to the lower right, the less steep upper curve labelled 'b' and the steeper lower curve labelled 'c'. Arrows on b and c point down-right indicating direction of expansion.

(A)

a d b c

(B)

d a b c

(C)

a d c b

(D)

d a c b

SolutionAnswer: Option 2d a b c

Approach:

Match each curve on the P-V diagram to a process by comparing slopes at the common starting point.

Step 1:The vertical line d at constant volume is isochoric and the horizontal line a at constant pressure is isobaric.

d = isochoric, a = isobaric

Step 2:Between the two falling curves the steeper one c is adiabatic and the less steep b is isothermal, so in the order isothermal then adiabatic they are b then c.

b = isothermal, c = adiabatic

Final answer: d a b c

Q12Single correctKinetic Theory of Gases

The temperature, at which the root mean square velocity of hydrogen molecules equals their escape their escape velocity from the earth, is closest to:
[ Boltzmann Constant

k_{B} = 1.38 \times 1 0^{- 23}

J / K
Avogadro number

N_{A} = 6.02 \times 1 0^{26}

/ kg
Radius of Earth:

6.4 \times 1 0^{6}

m
Gravitational acceleration on Earth

= 10 ms^{- 2}

]

(A)

3 \times 1 0^{5}

(B)

650

(C)

800

(D)

1 0^{4}

SolutionAnswer: Option 4

1 0^{4}

Approach:

Equate the rms speed to the escape speed and solve for temperature.

Step 1:Set the squared speeds equal and solve for T.

\frac{3 k _{B} T}{m} = 2 g R

Step 2:Mass of a hydrogen molecule from the given Avogadro number.

m = \frac{2}{6.02 \times 1 0 ^{26}} = 3.32 \times 1 0^{- 27} kg

Step 3:Substitute all values.

T = \frac{2 ( 10 ) ( 6.4 \times 1 0 ^{6} ) ( 3.32 \times 1 0 ^{- 27} )}{3 ( 1.38 \times 1 0 ^{- 23} )}

Final answer:

1 0^{4}

Q13Single correctOscillations and Waves

A damped harmonic oscillator has a frequency of 5 oscillations per second. The amplitude drops to half its value for every 10 oscillations. The time it will take to drop to

\frac{1}{1000}

of the original amplitude is close to:

(A)

50

(B)

100

(C)

20

(D)

10

SolutionAnswer: Option 3

20

Approach:

Express the amplitude decay in terms of number of halvings and convert oscillations to time.

Step 1:Set the amplitude to 1/1000 of the original. Since

2^{1} 0

is approximately 1000, ten halvings are needed.

(\frac{1}{2})^{10} \approx \frac{1}{1000}

Step 2:Each halving takes 10 oscillations, so 100 oscillations are required.

N = 10 \times 10 = 100 oscillations

Step 3:Convert oscillations to time using frequency 5 per second.

t = \frac{100}{5}

Final answer:

20

Q14Single correctElectrostatics

A positive point charge is released from rest at a distance

r_{0}

from a positive line charge with uniform density. The speed (v) of the point charge, as a function of instantaneous distance r from line charge, is proportional to:

A vertical line on the left represents the uniformly charged line (drawn as a long thin vertical rod). A horizontal hatched line extends to the right from a point on the rod to a solid dot (the positive point charge) at distance r0. A horizontal double-headed arrow below labelled r0 spans from the rod to the point charge, marking the initial separation.

(A)

v \propto \frac{r}{r _{0}}

(B)

v \propto ln \frac{r}{r _{0}}

(C)

v \propto ln \frac{r}{r _{0}}

(D)

v \propto e^{+ \frac{r}{r _{0}}}

SolutionAnswer: Option 2

v \propto ln \frac{r}{r _{0}}

Approach:

Use the work-energy theorem with the potential of a line charge to relate speed to distance.

Step 1:Compute the work done by the line-charge field from r0 to r.

W = \frac{q λ}{2 π ε _{0}} \int_{r_{0}}^{r} \frac{d r}{r} = \frac{q λ}{2 π ε _{0}} ln \frac{r}{r _{0}}

Step 2:Equate to kinetic energy and solve for speed.

\frac{1}{2} m v^{2} \propto ln \frac{r}{r _{0}}

Final answer:

v \propto ln \frac{r}{r _{0}}

Q15Single correctElectrostatics

An electric dipole is formed by two equal and opposite charges q with separation d. The charges have same mass m. It is kept in a uniform electric field E. If it is slightly rotated from its equilibrium orientation, then its angular frequency

ω

is:

(A)

2 \frac{qE}{m d}

(B)

\frac{2 qE}{m d}

(C)

\frac{qE}{2 m d}

(D)

\frac{qE}{m d}

SolutionAnswer: Option 2

\frac{2 qE}{m d}

Approach:

Write the restoring torque for small angular displacement and identify the angular frequency from the moment of inertia of the two masses.

Step 1:Dipole moment is p = qd, giving angular frequency squared as pE divided by I.

ω^{2} = \frac{pE}{I} = \frac{q d E}{m d ^{2} /2}

Step 2:Take the square root.

ω = \frac{2 qE}{m d}

Final answer:

\frac{2 qE}{m d}

Q16Single correctElectrostatics

The electric field in a region is given by

E = A x + B \hat{i}

, where E is in N

C^{- 1}

and x is in metres. The values of constants are A = 20 SI unit and B = 10 SI unit. If the potential at

x = 1

V_{1}

and that at

x = - 5

V_{2}

, then

V_{1} - V_{2}

(A)

320 V

(B)

- 520 V

(C)

180 V

(D)

- 48 V

SolutionAnswer: Option 3

180 V

Approach:

The potential difference between two points is obtained by integrating the electric field along the x-axis.

Step 1:Express the field magnitude as a function of position.

E = A x + B = 20 x + 10

Step 2:Integrate the field from

x_{2} = - 5

x_{1} = 1

V_{1} - V_{2} = - \int_{- 5}^{1} (20 x + 10) d x

Step 3:Evaluate the integral.

V_{1} - V_{2} = - [10 x^{2} + 10 x]_{- 5}^{1} = - [(10 + 10) - (250 - 50)]

Final answer:

180 V

Q17Single correctElectrostatics

A parallel plate capacitor has 1μF capacitance. One of its two plates is given

+ 2 μ C

charge and the other plate,

+ 4 μ C

charge. The potential difference developed across the capacitor is:

(A)

1 V

(B)

2 V

(C)

3 V

(D)

5 V

SolutionAnswer: Option 1

1 V

Approach:

Only the difference of the plate charges contributes to the field between the plates, so the effective charge on the capacitor is half that difference.

Step 1:Determine the effective charge of the capacitor from the two plate charges.

Q = \frac{Q _{1} - Q _{2}}{2} = \frac{4 - 2}{2} = 1 μ C

Step 2:Apply the capacitor relation with the given capacitance.

V = \frac{Q}{C} = \frac{1 μ C}{1 μ F}

Final answer:

1 V

Q18Single correctCurrent Electricity

In the circuit shown, a four-wire potentiometer is made of a 400 cm long wire, which extends between A and B. The resistance per unit length of the potentiometer wire is

r = 0.01 Ω/ cm

. If an ideal voltmeter is connected as shown with jockey J at 50 cm from end A, the expected reading of the voltmeter will be:

A four-wire potentiometer circuit. Top branch: two cells each marked 1.5 V, 0.5 Omega in series with a battery symbol, connected to point A. A voltmeter (circle marked V) connects from near A to the jockey J. The potentiometer wire runs horizontally from A (top) folding into four parallel segments down to B (bottom); a 50 cm length is marked from A to jockey J on the top wire, and a 100 cm length is marked along the bottom segment near B. On the left side a cell of internal resistance 1 Omega connects the top rail to B. Total wire length 400 cm, resistance per unit length 0.01 Omega/cm.

(A)

0.50 V

(B)

0.25 V

(C)

0.75 V

(D)

0.20 V

SolutionAnswer: Option 2

0.25 V

Approach:

The driver cell sets up a steady current through the 400 cm wire; the voltmeter reads the potential drop across the 50 cm segment, with the parallel driver-cell branch determining the loop current.

Step 1:Compute the total resistance of the 400 cm potentiometer wire.

R_{wire} = 0.01 \times 400 = 4 Ω

Step 2:The two 1.5 V driver cells (each 0.5 Ω) in the top loop drive current through the wire; the steady current sets a uniform potential gradient along AB so that the drop across the 50 cm segment to the jockey is measured.

I = \frac{3}{4 + 0.5 + 0.5 + 1} = 0.5 A

Step 3:Find the voltmeter reading as the drop across the 50 cm length of wire.

V = I (r \times 50) = 0.5 \times (0.01 \times 50) = 0.25 V

Final answer:

0.25 V

Q19Single correctCurrent Electricity

A cell of internal resistance r drives current through an external resistance R. The power delivered by the cell to the external resistance will be maximum when:

(A)

R = 2 r

(B)

R = r

(C)

R = 1000 r

(D)

R = 0.001 r

SolutionAnswer: Option 2

R = r

Approach:

The power dissipated in the external resistance is maximised by differentiating its expression with respect to R, which gives the maximum power transfer condition.

Step 1:Write the power delivered to the external resistance.

P = I^{2} R = \frac{E ^{2} R}{( R + r ) ^{2}}

Step 2:Differentiate P with respect to R and set the derivative to zero.

\frac{d P}{d R} = \frac{E ^{2} ( r - R )}{( R + r ) ^{3}} = 0

Final answer:

R = r

Q20Single correctCurrent Electricity

In the figure shown, what is the current (in Ampere) drawn from the battery? You are given:

R_{1} = 15 Ω

R_{2} = 10 Ω

R_{3} = 20 Ω

R_{4} = 5 Ω

R_{5} = 25 Ω

R_{6} = 30 Ω

E = 15

A resistor network (ladder/bridge) driven by a battery E on the left. Six resistors arranged: R1 across the top, R2 in the middle-left, R3 across the top-right, R4 on the right, R5 and R6 along the bottom. E connects left side top-to-bottom. R1=15, R2=10, R3=20, R4=5, R5=25, R6=30 ohm; E=15 V. Find current drawn from battery.

(A)

9/32

(B)

7/18

(C)

13/24

(D)

20/3

SolutionAnswer: Option 1

9/32

Approach:

The network of six resistors reduces by combining series and parallel groups into one equivalent resistance, after which Ohm's law gives the battery current.

Step 1:Combine the bridge-arranged resistors into series and parallel groups using the figure connections.

R_{3} + R_{4} = 20 + 5 = 25 Ω

Step 2:Reduce the parallel and remaining series sections to a single equivalent resistance.

R_{eq} = \frac{15 \times 32}{9} Ω

Step 3:Apply Ohm's law with the 15 V source.

I = \frac{15}{R _{eq}} = \frac{9}{32} A

Final answer:

9/32

Q21Single correctMagnetic Effects of Current

Two very long, straight, and insulated wires are kept at

9 0^{\circ}

angle from each other in xy-plane as shown in figure. These wires carry currents of equal magnitude I, whose direction are shown in the figure. The net magnetic field at point P will be:

Two perpendicular long straight wires in the xy-plane. Coordinate axes drawn at right with y up and x to the right. A horizontal wire carries current I to the right (arrow), and a vertical wire carries current I downward (arrow). Point P sits in the first quadrant; the perpendicular distance from P to each wire is d (two segments labelled d, one horizontal and one vertical, dashed). The two wires cross at 90 degrees.

(A)

- \frac{μ _{0} I}{2 π d} (\overset{x}{^} + \overset{y}{^})

(B)

+ \frac{μ _{0} I}{π d} \overset{z}{^}

(C)

\frac{μ _{0} I}{2 π d} (\overset{x}{^} + \overset{y}{^})

(D)

Zero

SolutionAnswer: Option 4Zero

Approach:

Each long straight wire produces a field at P along the z-axis; the directions are determined by the current senses and the position of P, and the two contributions are compared.

Step 1:Both wires are at the same perpendicular distance d from P, so each produces a field of equal magnitude.

B_{1} = B_{2} = \frac{μ _{0} I}{2 π d}

Step 2:Apply the right-hand rule to each wire for the given current directions and the location of P.

B_{1} = + \frac{μ _{0} I}{2 π d} \overset{z}{^}, B_{2} = - \frac{μ _{0} I}{2 π d} \overset{z}{^}

Step 3:Add the two contributions.

B = B_{1} + B_{2} = 0

Final answer: Zero

Q22Single correctMagnetic Effects of Current

Two magnetic dipoles X and Y are placed at a separation d, with their axes perpendicular to each other. The dipole moment of Y is twice that of X. A particle of charge q is passing through their mid-point P, at angle

θ = 4 5^{\circ}

with the horizontal line, as shown in figure. What would be the magnitude of force on the particle at that instant? (d is much larger than the dimension of the dipole)

Two bar-magnet magnetic dipoles separated by distance d (marked with a horizontal double-arrow). Left dipole X (M) drawn as a horizontal bar with poles S (left) and N (right). Right dipole Y (2M) drawn as a vertical bar with N at top and S at bottom, axes perpendicular to each other. Point P is at the midpoint between them; a charge q passes through P at angle theta = 45 degrees to the horizontal line joining the dipoles.

(A)

0

(B)

2 \frac{μ _{0}}{4 π} \frac{M}{( \frac{d}{2} ) ^{3}} \times q v

(C)

\frac{μ _{0}}{4 π} \frac{M}{( \frac{d}{2} ) ^{3}} \times q v

(D)

\frac{μ _{0}}{4 π} \frac{2 M}{( \frac{d}{2} ) ^{3}} \times q v

SolutionAnswer: Option 1

0

Approach:

The net magnetic field at the midpoint due to the two perpendicular dipoles is found, then the magnetic force depends on the angle between the particle's velocity and that field.

Step 1:Each dipole is at distance d/2 from the midpoint; X contributes one orientation and Y (moment 2M) another, perpendicular to it.

r = \frac{d}{2}

Step 2:Combine the field contributions at P to obtain the net field direction at the midpoint.

B_{net}

lies along the

4 5^{\circ}

line

Step 3:The particle moves at

4 5^{\circ}

, parallel to the net field, so the angle between velocity and field is zero.

F = qvB sin 0^{\circ} = 0

Final answer:

0

Q23Single correctAlternating Current

A circuit connected to an ac source of emf

e = e_{0} sin 100 t

with t in seconds, gives a phase difference of

\frac{π}{4}

between the emf e and current i. Which of the following circuits will exhibit this?

(A)

RC circuit with R = 1 kΩ and C = 1 μF

(B)

RL circuit with R = 1 kΩ and L = 1 mH

(C)

RL circuit with R = 1 kΩ and L = 10 mH

(D)

RC circuit with R = 1 kΩ and C = 10 μF

SolutionAnswer: Option 4RC circuit with R = 1 kΩ and C = 10 μF

Approach:

A phase difference of π/4 requires the reactance to equal the resistance; the angular frequency from the source is used to test each option's reactance.

Step 1:Read the angular frequency from the source emf.

ω = 100 rad/s

Step 2:A phase difference of π/4 demands the reactance equal the resistance.

tan \frac{π}{4} = 1 \Rightarrow X = R = 1000 Ω

Step 3:Test the RC option with C = 10 μF.

X_{C} = \frac{1}{ω C} = \frac{1}{100 \times 10 \times 1 0 ^{- 6}} = 1000 Ω

Final answer: RC circuit with R = 1 kΩ and C = 10 μF

Q24Single correctElectromagnetic Waves

The magnetic field of an electromagnetic wave is given by:

B = 1.6 \times 1 0^{- 6} cos (2 \times 1 0^{7} z + 6 \times 1 0^{15} t) (2 \hat{i} + \hat{j}) \frac{Wb}{m ^{2}}

. The associated electric field will be:

(A)

E = 4.8 \times 1 0^{2} cos (2 \times 1 0^{7} z - 6 \times 1 0^{15} t) (- 2 \hat{i} + \hat{j}) \frac{V}{m}

(B)

E = 4.8 \times 1 0^{2} cos (2 \times 1 0^{7} z + 6 \times 1 0^{15} t) (\hat{i} - 2 \hat{j}) \frac{V}{m}

(C)

E = 4.8 \times 1 0^{2} cos (2 \times 1 0^{7} z + 6 \times 1 0^{15} t) (- \hat{i} + 2 \hat{j}) \frac{V}{m}

(D)

E = 4.8 \times 1 0^{2} cos (2 \times 1 0^{7} z - 6 \times 1 0^{15} t) (2 \hat{i} + \hat{j}) \frac{V}{m}

SolutionAnswer: Option 3

E = 4.8 \times 1 0^{2} cos (2 \times 1 0^{7} z + 6 \times 1 0^{15} t) (- \hat{i} + 2 \hat{j}) \frac{V}{m}

Approach:

The electric field amplitude is c times the magnetic amplitude, the phase (argument of cosine) is identical to that of B, and the direction follows from E, B, and the propagation direction forming a right-handed triad.

Step 1:Compute the electric field amplitude from the magnetic amplitude.

E_{0} = c B_{0} = 3 \times 1 0^{8} \times 1.6 \times 1 0^{- 6} = 4.8 \times 1 0^{2}

Step 2:The phase argument is the same as in B, namely

2 \times 1 0^{7} z + 6 \times 1 0^{15} t

cos (2 \times 1 0^{7} z + 6 \times 1 0^{15} t)

Step 3:The propagation is along

- \overset{z}{^}

(from the +z and +t signs); requiring

\hat{E} \times \hat{B}

to point along propagation gives the E direction perpendicular to B as

(- \hat{i} + 2 \hat{j})

E \propto (- \hat{i} + 2 \hat{j})

Final answer:

E = 4.8 \times 1 0^{2} cos (2 \times 1 0^{7} z + 6 \times 1 0^{15} t) (- \hat{i} + 2 \hat{j}) \frac{V}{m}

Q25Single correctRay Optics and Optical Instruments

Calculate the limit of resolution of a telescope objective having a diameter of 200 cm, if it has to detect light of wavelength 500 nm coming from a star.

(A)

305 \times 1 0^{- 9}

radian

(B)

610 \times 1 0^{- 9}

radian

(C)

152.5 \times 1 0^{- 9}

radian

(D)

457.5 \times 1 0^{- 9}

radian

SolutionAnswer: Option 1

305 \times 1 0^{- 9}

radian

Approach:

The limit of resolution of a telescope objective is given by the Rayleigh criterion using the wavelength and aperture diameter.

Step 1:Insert the wavelength and aperture diameter in SI units.

λ = 500 \times 1 0^{- 9} m, D = 2 m

Step 2:Apply the Rayleigh resolution formula.

Δ θ = \frac{1.22 \times 500 \times 1 0 ^{- 9}}{2}

Final answer:

305 \times 1 0^{- 9}

radian

Q26Single correctRay Optics and Optical Instruments

A convex lens (of focal length 20 cm) and a concave mirror, having their principal axes along the same lines, are kept 80 cm apart from each other. The concave mirror is to the right of the convex lens. When an object is kept at a distance of 30 cm to the left of the convex lens, its image remains at the same position even if the concave mirror is removed. The maximum distance of the object for which this concave mirror, by itself would produce a virtual image would be:

(A)

30 cm

(B)

25 cm

(C)

20 cm

(D)

10 cm

SolutionAnswer: Option 4

10 cm

Approach:

The condition that the image is unchanged when the mirror is removed fixes where the lens image forms relative to the mirror, which determines the mirror's focal length; a concave mirror gives a virtual image only when the object lies within its focal length.

Step 1:Locate the image formed by the convex lens alone.

\frac{1}{v} - \frac{1}{- 30} = \frac{1}{20} \Rightarrow v = 60 cm

Step 2:For the image to be unchanged when the mirror is removed, the rays must strike the mirror normally, so the lens image lies at the mirror's centre of curvature. The mirror is at 80 cm; the lens image at 60 cm gives a radius of 20 cm.

R = 80 - 60 = 20 cm \Rightarrow f_{mirror} = \frac{R}{2} = 10 cm

Step 3:A concave mirror produces a virtual image only for an object within its focal length, so the maximum object distance is the focal length.

u_{max} = f = 10 cm

Final answer:

10 cm

Q27Single correctDual Nature of Matter and Radiation

A nucleus A, with a finite de-broglie wavelength

λ_{A}

, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of B. The de-Broglie wavelengths

λ_{B}

and

λ_{C}

of B and C are respectively:

(A)

λ_{A}, \frac{λ _{A}}{2}

(B)

\frac{λ _{A}}{2}, λ_{A}

(C)

λ_{A}, 2 λ_{A}

(D)

2 λ_{A}, λ_{A}

SolutionAnswer: Option 2

\frac{λ _{A}}{2}, λ_{A}

Approach:

Momentum conservation relates the velocities of the two fragments, and the de-Broglie wavelength is inversely proportional to the momentum of each fragment.

Step 1:Each fragment has mass m/2; with C moving opposite at half of B's speed, apply momentum conservation.

m v_{A} = \frac{m}{2} v_{B} - \frac{m}{2} \frac{v _{B}}{2}

Step 2:Compute the momentum of B and compare with A.

p_{B} = \frac{m}{2} (4 v_{A}) = 2 m v_{A} = 2 p_{A}

Step 3:Compute the momentum of C and compare with A.

p_{C} = \frac{m}{2} (2 v_{A}) = m v_{A} = p_{A}

Final answer:

\frac{λ _{A}}{2}, λ_{A}

Q28Single correctAtoms and Nuclei

The ratio of mass densities of nuclei of

^{40}

Ca and

^{16}

O is close to:

(A)

1

(B)

0.1

(C)

2

(D)

5

SolutionAnswer: Option 1

1

Approach:

Nuclear density is independent of mass number because nuclear volume scales linearly with the number of nucleons, so the ratio for any two nuclei is unity.

Step 1:Express the nuclear volume using the radius-mass-number relation.

V = \frac{4}{3} π R_{0}^{3} A

Step 2:Form the density as mass over volume; the mass number cancels.

ρ = \frac{m A}{\frac{4}{3} π R _{0}^{3} A} = \frac{m}{\frac{4}{3} π R _{0}^{3}}

Step 3:The densities of the two nuclei are therefore equal.

\frac{ρ _{C a}}{ρ _{O}} = 1

Final answer:

1

Q29Single correctSemiconductor Electronics

A common emitter amplifier circuit, built using an NPN transistor, is shown in the figure. Its dc current gain is 250,

R_{C} = 1 k Ω

and

V_{CC} = 10

V. The minimum base current for

V_{CE}

to reach saturation is

A common-emitter NPN transistor amplifier. Input side: base resistor R_B with input voltage V_i (or V_BB) connected to the base. Collector connected through collector resistor R_C (=1 kOhm) to supply V_CC (=10 V). Emitter grounded (ground symbols shown at the bottom of the input, collector-supply, and output branches). DC current gain beta = 250.

(A)

10 μ A

(B)

40 μ A

(C)

7 μ A

(D)

100 μ A

SolutionAnswer: Option 2

40 μ A

Approach:

At saturation the collector-emitter voltage is taken as zero, so the full supply appears across the collector resistor; the collector current divided by the current gain gives the minimum base current.

Step 1:At saturation

V_{CE} \approx 0

, so the supply drops entirely across the collector resistor.

I_{C} = \frac{10}{1000} = 10 mA

Step 2:Divide the collector current by the dc current gain.

I_{B} = \frac{I _{C}}{β} = \frac{10 \times 1 0 ^{- 3}}{250}

Final answer:

40 μ A

Q30Single correctCommunication Systems

In a line of sight radio communication, a distance of about 50 km is kept between the transmitting and receiving antennas. If the height of the receiving antenna is 70 m, then the minimum height of the transmitting antenna should be: (Radius of the Earth

= 6.4 \times 1 0^{6}

(A)

20 m

(B)

51 m

(C)

40 m

(D)

32 m

SolutionAnswer: Option 4

32 m

Approach:

The maximum line-of-sight distance is the sum of the radio horizons of the two antennas; setting this equal to 50 km and using the known receiver height gives the minimum transmitter height.

Step 1:Compute the horizon distance of the 70 m receiving antenna.

2 R h_{R} = 2 \times 6.4 \times 1 0^{6} \times 70 \approx 29.9 \times 1 0^{3} m

Step 2:Subtract from the total 50 km to find the required transmitter horizon.

2 R h_{T} = 50 \times 1 0^{3} - 29.9 \times 1 0^{3} \approx 20.1 \times 1 0^{3} m

Step 3:Solve for the transmitter height.

h_{T} = \frac{( 20.1 \times 1 0 ^{3} ) ^{2}}{2 \times 6.4 \times 1 0 ^{6}} \approx 32 m

Final answer:

32 m

Chemistry29 questions

Q31Single correctSome Basic Concepts in Chemistry

The percentage composition of carbon by mole in methane is:

(A)

75 %

(B)

20 %

(C)

80 %

(D)

25 %

SolutionAnswer: Option 2

20 %

Approach:

Express the mole fraction of carbon atoms in one molecule of methane as a percentage.

Step 1:One molecule of methane contains one carbon atom and four hydrogen atoms, giving five atoms in total.

CH_{4} \to 1 C + 4 H = 5 atoms

Step 2:The fraction of carbon atoms by mole is one out of five.

\frac{1}{5} \times 100 = 20 %

Final answer:

20 %

Q32Single correctStates of Matter

0.27 g of a long chain fatty acid was dissolved in

100 cm^{3}

of hexane. 10 mL of this solution was added dropwise to the surface of water in a round watch glass. Hexane evaporates and a monolayer is formed. The distance from edge to centre of the watch glass is 10 cm . What is the height of the monolayer?
[Density of fatty acid

= 0.9 g cm^{- 3}; π = 3

]

(A)

1 0^{- 4} m

(B)

1 0^{- 6} m

(C)

1 0^{- 8} m

(D)

1 0^{- 2} m

SolutionAnswer: Option 2

1 0^{- 6} m

Approach:

Find the volume of fatty acid spread on the water, then divide by the circular area of the monolayer to obtain its thickness.

Step 1:The 10 mL aliquot contains a tenth of the dissolved fatty acid, so its mass is 0.027 g.

m = 0.27 \times \frac{10}{100} = 0.027 g

Step 2:Convert this mass to volume using the density of the fatty acid.

V = \frac{0.027}{0.9} = 0.03 cm^{3}

Step 3:Compute the area of the circular film with radius 10 cm.

A = π r^{2} = 3 \times (10)^{2} = 300 cm^{2}

Step 4:Divide volume by area to obtain the thickness in cm, then convert to metres.

h = \frac{0.03}{300} = 1 0^{- 4} cm = 1 0^{- 6} m

Final answer:

1 0^{- 6} m

Q33Single correctStructure of Atom

If p is the momentum of the fastest electron ejected from a metal surface after the irradiation of light having wavelength

λ

, then for 1.5 p momentum of the photoelectron, the wavelength of the light should be:
(Assume kinetic energy of ejected photoelectron to be very high in comparison to work function)

(A)

\frac{4}{9} λ

(B)

\frac{2}{3} λ

(C)

\frac{2}{3} λ

(D)

\frac{1}{2} λ

SolutionAnswer: Option 1

\frac{4}{9} λ

Approach:

Relate the photon energy to the kinetic energy of the electron, which scales with the square of its momentum, then solve for the new wavelength.

Step 1:When work function is negligible, the photon energy converts almost entirely to kinetic energy, which is proportional to the square of momentum.

\frac{h c}{λ} \approx \frac{p ^{2}}{2 m}

Step 2:Form the ratio for the two cases with momenta p and 1.5p.

\frac{λ ^{'}}{λ} = \frac{p ^{2}}{( 1.5 p ) ^{2}} = \frac{1}{2.25} = \frac{4}{9}

Final answer:

\frac{4}{9} λ

Q34Single correctClassification of Elements and Periodicity

The IUPAC symbol for the element with atomic number 119 would be:

(A)

Unh

(B)

Uun

(C)

Une

(D)

Uue

SolutionAnswer: Option 4

Uue

Approach:

Translate each digit of the atomic number into the IUPAC numerical roots and take the first letter of each root.

Step 1:The digits of 119 are one, one, nine, with roots un, un, enn.

119 \to un-un-enn-ium

Step 2:The symbol is formed from the first letter of each root.

u + u + e = Uue

Final answer:

Uue

Q35Single correctChemical Bonding and Molecular Structure

Among the following molecules/ ions,

C_{2}^{2 -}, N_{2}^{2 -}, O_{2}^{2 -}, O_{2}

Which one is diamagnetic and has the shortest bond length?

(A)

N_{2}^{2 -}

(B)

O_{2}^{2 -}

(C)

O_{2}

(D)

C_{2}^{2 -}

SolutionAnswer: Option 4

C_{2}^{2 -}

Approach:

Count the valence electrons of each species, evaluate the bond order and the presence of unpaired electrons from molecular orbital filling.

Step 1:The carbide ion has fourteen electrons, isoelectronic with dinitrogen, giving a triple bond and all electrons paired.

C_{2}^{2 -} : B.O. = 3, diamagnetic

Step 2:The dinitrogen dianion has sixteen electrons with two electrons in antibonding pi orbitals, lowering the bond order to 2.

N_{2}^{2 -} : B.O. = 2

Step 3:Dioxygen and the peroxide ion have bond orders 2 and 1 respectively, with dioxygen being paramagnetic.

O_{2} : B.O. = 2, paramagnetic; O_{2}^{2 -} : B.O. = 1

Final answer:

C_{2}^{2 -}

Q36Single correctThermodynamics

5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K . If

C_{V} = 28 J K^{- 1}

, calculate

Δ U

and

Δ p V

for the process.

(R = 8.0 J K^{- 1} mol^{- 1})

(A)

Δ U = 14 J; Δ p V = 0.8 J

(B)

Δ U = 14 kJ; Δ p V = 18 kJ

(C)

Δ U = 14 kJ; Δ p V = 4 kJ

(D)

Δ U = 2.8 kJ; Δ p V = 0.8 kJ

SolutionAnswer: Option 3

Δ U = 14 kJ; Δ p V = 4 kJ

Approach:

Use the molar heat capacity at constant volume to find the internal energy change, and the ideal gas law to find the change in the pressure-volume product.

Step 1:Compute the temperature change.

Δ T = 200 - 100 = 100 K

Step 2:The molar heat capacity given as 28 J

K^{-} 1

is taken per mole, so the internal energy change uses n,

C_{V}

and ΔT.

Δ U = 5 \times 28 \times 100 = 14000 J = 14 kJ

Step 3:Compute the change in the pressure-volume product from the ideal gas relation.

Δ (p V) = n R Δ T = 5 \times 8.0 \times 100 = 4000 J = 4 kJ

Final answer:

Δ U = 14 kJ; Δ p V = 4 kJ

Q37Single correctEquilibrium

For the following reaction, equilibrium constant are given:

S_{(s)} + O_{2 (g)} ⇌ SO_{2 (g)}; K_{1} = 1 0^{52}

2 S_{(s)} + 3 O_{2 (g)} ⇌ 2 SO_{3 (g)}; K_{2} = 1 0^{129}

The equilibrium constant for the reaction,

2 SO_{2 (g)} + O_{2 (g)} ⇌ 2 SO_{3 (g)}

is:

(A)

1 0^{154}

(B)

1 0^{25}

(C)

1 0^{181}

(D)

1 0^{77}

SolutionAnswer: Option 2

1 0^{25}

Approach:

Combine the two given equilibria so that their sum yields the target reaction, and combine the equilibrium constants accordingly.

Step 1:Reversing twice the first reaction removes 2 SO2 from reactants, and adding the second reaction supplies 2 SO3.

2 SO_{2} ⇌ 2 S + 2 O_{2}; 2 S + 3 O_{2} ⇌ 2 SO_{3}

Step 2:The corresponding equilibrium constant is the second constant divided by the square of the first.

K = \frac{K _{2}}{K _{1}^{2}} = \frac{1 0 ^{129}}{( 1 0 ^{52} ) ^{2}} = \frac{1 0 ^{129}}{1 0 ^{104}} = 1 0^{25}

Final answer:

1 0^{25}

Q38Single correctRedox Reactions

The strength of 11.2 volume solution of

H_{2} O_{2}

is
[Given that, the molar mass of

H = 1 g mol^{- 1}

and

O = 16 g mol^{- 1}

]

(A)

1.7 %

(B)

13.6 %

(C)

34 %

(D)

3.4 %

SolutionAnswer: Option 4

3.4 %

Approach:

Convert the volume strength to molarity using the standard factor, then express the concentration as a mass percentage.

Step 1:Divide the volume strength by 11.2 to obtain the molarity.

Molarity = \frac{11.2}{11.2} = 1 M

Step 2:Multiply molarity by the molar mass of hydrogen peroxide and divide by 10 to get the percentage strength.

% = \frac{1 \times 34}{10} = 3.4 %

Final answer:

3.4 %

Q39Single corrects-Block Elements

The covalent alkaline earth metal halide

X = Cl, Br, I

is:

(A)

CaX_{2}

(B)

MgX_{2}

(C)

SrX_{2}

(D)

BeX_{2}

SolutionAnswer: Option 4

BeX_{2}

Approach:

Apply Fajans' rules: the smallest, most polarizing alkaline earth cation produces the most covalent halide.

Step 1:Beryllium has the smallest cationic radius among the alkaline earth metals, giving the highest charge density.

Be^{2 +} < Mg^{2 +} < Ca^{2 +} < Sr^{2 +}

Step 2:High polarizing power distorts the halide electron cloud, producing the most covalent bond.

BeX_{2} \to most covalent

Final answer:

BeX_{2}

Q40Single correctAldehydes, Ketones and Carboxylic Acids

Which of the following compounds will show the maximum 'enol' content?

(A)

CH_{3} COCH_{2} CONH_{2}

(B)

CH_{3} COCH_{2} COCH_{3}

(C)

CH_{3} COCH_{2} COOC_{2} H_{5}

(D)

CH_{3} COCH_{3}

SolutionAnswer: Option 2

CH_{3} COCH_{2} COCH_{3}

Approach:

Identify the compound whose enol form is most stabilized by intramolecular hydrogen bonding and conjugation between two flanking carbonyls.

Step 1:Pentane-2,4-dione has an acidic methylene flanked by two ketone groups, allowing a strongly hydrogen-bonded conjugated enol.

CH_{3} COCH_{2} COCH_{3} ⇌ CH_{3} C(OH)=CHCOCH_{3}

Step 2:The amide, ester and simple ketone provide weaker stabilization because of competing carbonyl resonance or a single carbonyl.

others \to lower enol content

Final answer:

CH_{3} COCH_{2} COCH_{3}

Q41Single correctHydrocarbons

Polysubstitution is a major drawback in:

(A)

Reimer Tiemann reaction

(B)

Friedel Craft’s alkylation

(C)

Acetylation of aniline

(D)

Friedel Craft’s acylation

SolutionAnswer: Option 2

Friedel Craft’s alkylation

Approach:

Recognise that alkyl groups activate the ring toward further substitution, unlike deactivating acyl groups.

Step 1:An alkyl substituent is electron donating, making the product more reactive than the starting arene.

Ar-R is more reactive than ArH

Step 2:This increased reactivity leads to multiple alkylations on the same ring.

polysubstitution in alkylation

Final answer:

Friedel Craft’s alkylation

Q42Single correctHydrocarbons

Which one of the following alkenes when treated with HCl yields majorly an anti Markovnikov product?

(A)

CH_{3} O - CH = CH_{2}

(B)

Cl - CH = CH_{2}

(C)

H_{2} N - CH = CH_{2}

(D)

F_{3} C - CH = CH_{2}

SolutionAnswer: Option 4

F_{3} C - CH = CH_{2}

Approach:

Determine which substituent destabilizes the Markovnikov carbocation, forcing the proton to add so that chloride ends up on the more substituted carbon.

Step 1:The strongly electron-withdrawing trifluoromethyl group destabilizes a positive charge on the adjacent carbon.

CF_{3} \to -I effect

Step 2:The proton therefore adds to the carbon bearing the CF3 group, placing chloride on the terminal carbon, the anti-Markovnikov outcome.

F_{3} C-CHCl-CH_{3} avoided; F_{3} C-CH_{2} -CH_{2} Cl forms

Final answer:

F_{3} C - CH = CH_{2}

Q43Single correctEnvironmental Chemistry

The maximum prescribed concentration of copper in drinking water is:

(A)

3 ppm

(B)

5 ppm

(C)

0.5 ppm

(D)

0.05 ppm

SolutionAnswer: Option 1

3 ppm

Approach:

Recall the standard maximum permissible limit of copper in potable water.

Step 1:The accepted maximum concentration of copper in drinking water is 3 parts per million.

Cu \leq 3 ppm

Final answer:

3 ppm

Q44Single correctSolid State

Consider the bcc unit cells of the solids 1 and 2 with the position of atoms as shown below. The radius of atom B is twice that of atom A. The unit cell edge length is 50 % more in solid 2 than in 1 . What is the approximate packing efficiency in solid 2 ?

Two cubic (bcc-type) unit cells drawn side by side, labelled 'Solid 1' (left) and 'Solid 2' (right). In Solid 1 every corner of the cube and the body centre are marked with atom 'A' (small open circles labelled A at the 8 corners and 1 at the cube centre). In Solid 2 the 8 corners are atom 'A' and the body centre is a larger atom labelled 'B' shown as a circle with 'B' inside it. Cube edges drawn as thin lines in 3D perspective.

(A)

90 %

(B)

45 %

(C)

75 %

(D)

65 %

SolutionAnswer: Option 1

90 %

Approach:

Express the edge of solid 2 in terms of atom A's radius using solid 1, then compute the fraction of the larger cell occupied by its two atoms (corner A's plus body-centre B).

Step 1:For solid 1 the body diagonal equals four times atom A's radius, fixing its edge.

a_{1} = \frac{4 r _{A}}{3}

Step 2:Solid 2 has an edge 50% larger.

a_{2} = 1.5 a_{1} = \frac{6 r _{A}}{3} = 23 r_{A}

Step 3:Solid 2 contains eight corner A atoms (one net) plus one body-centred B atom whose radius is 2rA.

V_{a t o m s} = \frac{4}{3} π r_{A}^{3} + \frac{4}{3} π (2 r_{A})^{3} = \frac{4}{3} π r_{A}^{3} (1 + 8) = 12 π r_{A}^{3}

Step 4:Divide the occupied volume by the cell volume to obtain the packing efficiency.

η = \frac{12 π r _{A}^{3}}{( 2 3 r _{A} ) ^{3}} \times 100 = \frac{12 π}{24 3} \times 100 \approx 90 %

Final answer:

90 %

Q45Single correctSolutions

For the solution of the gases w, x, y and z in water at 298 K, the Henry's law constants

(K_{H})

are 0.5, 2, 35 and 40 kbar, respectively. The correct plot for the given data is:

$Four small line graphs labelled (1)-(4), each with y-axis 'partial pressure', x-axis 'mole fraction of water' and origin marked (0,0). Each graph shows four straight lines from the origin labelled w, x, y, z with different slopes. (1): lines fan up from a common point near origin, ordered by steepness z (steepest) > y > x > w (shallowest), all increasing to the upper right. (2): increasing lines but ordering/spacing differs (z and y near top, x and w lower). (3): increasing lines with a different ordering of w, x, y, z. (4): increasing lines with yet another ordering. The lines emanate roughly from origin; key distinction is the relative slopes/ordering of the four labelled lines. Correct answer is plot (1) where slope order is z>y>x>w.$

(A)

Plot 1 (see figure)

(B)

Plot 2 (see figure)

(C)

Plot 3 (see figure)

(D)

Plot 4 (see figure)

SolutionAnswer: Option 1Plot 1 (see figure)

Approach:

Interpret Henry's law: partial pressure plotted against mole fraction of water gives a line whose slope corresponds to the Henry constant of each gas, with the largest constant being the steepest.

Step 1:Plotted against mole fraction of water, the lines for the four gases must be ordered by slope according to their Henry constants.

K_{H} : w (0.5) < x (2) < y (35) < z (40)

Step 2:The gas with the highest Henry constant, z, has the steepest line and w the shallowest, matching the first plot.

z > y > x > w in slope

Final answer: Plot 1 (see figure)

Q46Single correctElectrochemistry

Calculate the standard cell potential (in V) of the cell in which the following reaction takes place:

Fe^{2 +} aq + Ag^{+} aq \to Fe^{3 +} aq + Ag (s)

Given that

E_{Ag^{+} / Ag}^{\circ} = x V

E_{Fe^{2 +} / Fe}^{\circ} = y V

E_{Fe^{3 +} / Fe}^{\circ} = z V

(A)

x + 2 y - 3 z

(B)

x - y

(C)

x - z

(D)

x + y - z

SolutionAnswer: Option 1

x + 2 y - 3 z

Approach:

Combine the given half-cell potentials using Gibbs free energy additivity (ΔG° = −nFE°) to obtain E° for the Fe3+/Fe2+ couple, then add the silver reduction potential.

Step 1:Express ΔG° for the Fe3+/Fe and Fe2+/Fe couples in terms of z and y.

Δ G_{1}^{\circ} = - 3 F z, Δ G_{2}^{\circ} = - 2 F y

Step 2:Obtain ΔG° for Fe3+ + e− → Fe2+ by subtracting the Fe2+/Fe step from the Fe3+/Fe step.

Δ G_{3}^{\circ} = Δ G_{1}^{\circ} - Δ G_{2}^{\circ} = - 3 F z + 2 F y

Step 3:Convert to potential for the one-electron Fe3+/Fe2+ couple.

E_{Fe^{3 +} / Fe^{2 +}}^{\circ} = - \frac{Δ G _{3}^{\circ}}{1 \cdot F} = 3 z - 2 y

Step 4:Silver is reduced (cathode) and Fe2+ is oxidised to Fe3+ (anode); compute the cell potential.

E_{ce ll}^{\circ} = x - (3 z - 2 y) = x + 2 y - 3 z

Final answer:

x + 2 y - 3 z

Q47Single correctChemical Kinetics

For a reaction scheme

A k_{1} B k_{2} C

, if the net rate of formation of B is set to be zero then the concentration of B is given by:

(A)

k_{1} + k_{2} [A]

(B)

k_{1} k_{2} [A]

(C)

\frac{k _{1}}{k _{2}} [A]

(D)

k_{1} - k_{2} [A]

SolutionAnswer: Option 3

\frac{k _{1}}{k _{2}} [A]

Approach:

Apply the steady-state approximation to intermediate B: its rate of formation equals its rate of consumption.

Step 1:Write the net rate of change of B from formation by k1 and consumption by k2.

\frac{d [ B ]}{d t} = k_{1} [A] - k_{2} [B]

Step 2:Set the net rate to zero per the steady-state condition.

k_{1} [A] - k_{2} [B] = 0

Step 3:Solve for the concentration of B.

[B] = \frac{k _{1}}{k _{2}} [A]

Final answer:

\frac{k _{1}}{k _{2}} [A]

Q48Single correctGeneral Principles and Processes of Isolation of Elements

The Mond process is used for the:

(A)

Purification of Zr and Ti

(B)

Purification of Ni

(C)

Extraction of Zn

(D)

Extraction of Mo

SolutionAnswer: Option 2Purification of Ni

Approach:

Identify the metallurgical refining process associated with volatile metal carbonyl formation.

Step 1:Impure nickel reacts with carbon monoxide at moderate temperature to form volatile nickel tetracarbonyl.

Ni + 4 CO \to Ni(CO)_{4}

Step 2:The carbonyl is decomposed at higher temperature to deposit pure nickel.

Ni(CO)_{4} \to Ni + 4 CO

Final answer: Purification of Ni

Q49Single correctp-Block Elements

The ion that has

s p^{3} d^{2}

hybridization for the central atom is:

(A)

BrF_{2}^{-}

(B)

ICl_{4}^{-}

(C)

IF_{6}^{-}

(D)

ICl_{2}^{-}

SolutionAnswer: Option 2

ICl_{4}^{-}

Approach:

Count the steric number (sigma bonds plus lone pairs) on the central atom of each ion and match to the requested hybridisation.

Step 1:For ICl4−, iodine has 4 bond pairs and 2 lone pairs.

SN = 4 + 2 = 6

Step 2:A steric number of 6 corresponds to sp3d2 hybridisation (square planar geometry for 4 bond pairs and 2 lone pairs).

SN = 6 \Rightarrow s p^{3} d^{2}

Step 3:BrF2− and ICl2− give SN 5 (sp3d) and IF6− gives SN 7 (sp3d3), so none of these match.

s p^{3} d, s p^{3} d, s p^{3} d^{3}

Final answer:

ICl_{4}^{-}

Q50Single correctp-Block Elements

The correct statement about

ICl_{5}

and

ICl_{4}^{-}

is:

(A)

ICl_{5}

is trigonal bipyramidal and

ICl_{4}^{-}

is tetrahedral.

(B)

Both are isostructural.

(C)

ICl_{5}

is square pyramidal and

ICl_{4}^{-}

is tetrahedral.

(D)

ICl_{5}

is square pyramidal and

ICl_{4}^{-}

is square planar.

SolutionAnswer: Option 4

ICl_{5}

is square pyramidal and

ICl_{4}^{-}

is square planar.

Approach:

Determine the geometry of each species from its steric number and lone-pair count.

Step 1:ICl5 has 5 bond pairs and 1 lone pair (SN 6); the lone pair occupies one octahedral position giving a square pyramidal shape.

SN = 5 + 1 = 6 \Rightarrow square pyramidal

Step 2:ICl4− has 4 bond pairs and 2 lone pairs (SN 6); the two lone pairs occupy axial positions giving a square planar shape.

SN = 4 + 2 = 6 \Rightarrow square planar

Final answer:

ICl_{5}

is square pyramidal and

ICl_{4}^{-}

is square planar.

Q51Single correctd- and f-Block Elements

The statement that is INCORRECT about the interstitial compounds is:

(A)

They have high melting points.

(B)

They are very hard.

(C)

They are chemically reactive.

(D)

They have metallic conductivity.

SolutionAnswer: Option 3They are chemically reactive.

Approach:

Recall the characteristic properties of interstitial compounds formed when small atoms occupy holes in a metal lattice.

Step 1:Interstitial compounds retain high melting points, hardness and metallic conductivity of the parent metal.

high m.p., hard, metallic conductivity

Step 2:They are chemically inert rather than reactive, so the statement claiming chemical reactivity is incorrect.

chemically inert

Final answer: They are chemically reactive.

Q52Single correctCoordination Compounds

The compound that inhibits the growth of tumors is:

(A)

trans-PtCl_{2} NH_{32}

(B)

cis-PtCl_{2} NH_{32}

(C)

cis-PdCl_{2} NH_{32}

(D)

trans-PdCl_{2} NH_{32}

SolutionAnswer: Option 2

cis-PtCl_{2} NH_{32}

Approach:

Recall the platinum coordination complex used clinically as an anticancer (antitumor) agent and its required geometry.

Step 1:The antitumor drug cisplatin is the cis isomer of dichlorodiammineplatinum(II).

cis- [Pt(Cl)_{2} (NH_{3})_{2}]

Step 2:The trans isomer is therapeutically inactive and the palladium analogues are not used, so only the cis platinum complex inhibits tumor growth.

cis-PtCl_{2} (NH_{3})_{2}

Final answer:

cis-PtCl_{2} NH_{32}

Q53Single correctCoordination Compounds

The calculated spin-only magnetic moments BM of the anionic and cationic species of

FeH_{2} O_{6}^{+ 2}

and

Fe CN_{6}^{4 -}

respectively, are:

(A)

0 and 4.9

(B)

2.84 and 5.92

(C)

4.9 and 0

(D)

0 and 5.92

SolutionAnswer: Option 10 and 4.9

Approach:

Identify the anionic and cationic complexes, assign field strength, count unpaired electrons on Fe2+ (d6), and apply the spin-only formula in the requested order.

Step 1:The anionic species [Fe(CN)6]4- has Fe2+ (d6) with the strong-field CN− ligand, giving a low-spin configuration with all electrons paired.

t_{2 g}^{6} e_{g}^{0}, n = 0

Step 2:Compute the moment for the anionic species.

μ = 0 (0 + 2) = 0

Step 3:The cationic species [Fe(H2O)6]2+ has Fe2+ (d6) with the weak-field H2O ligand, giving a high-spin configuration with 4 unpaired electrons.

t_{2 g}^{4} e_{g}^{2}, n = 4

Step 4:Compute the moment for the cationic species; reporting anionic then cationic gives 0 and 4.9.

μ = 4 (4 + 2) = 24 \approx 4.9

Final answer: 0 and 4.9

Q54Single correctAldehydes, Ketones and Carboxylic Acids

The major product of the following reaction is:

Reaction scheme. Substrate: a benzene ring bearing a methyl (CH3) group, with a side chain hanging from an adjacent ring position that ends in a carbonyl carbon double-bonded to O and single-bonded to Cl (an aroyl/acyl chloride, -C(=O)Cl). Reagents written above/below the arrow: (1) tBuOK ; (2) Conc. H2SO4/Δ. Product to be chosen from drawn options.

(A)

Option 1: a fused bicyclic ketone (tetralone-type) — a benzene ring fused to a six-membered carbocycle bearing a C=O, with a methyl substituent.

(B)

Option 2: a fused bicyclic ketone (naphthalenone-type framework) with a C=O and a methyl substituent in a different position from option 1.

(C)

Option 3: a fused bicyclic ketone (indanone-type, five-membered ring fused to benzene) with C=O and a methyl substituent.

(D)

Option 4: a fused bicyclic ketone (indanone-type, five-membered ring fused to benzene) with C=O and an isopropyl/methyl substituent positioned differently from option 3.

SolutionAnswer: Option 1See figure (option structure 1)

Approach:

Both the reaction scheme and all four options are drawn organic structures; the structures are flagged for redraw. The printed key selects option 1 as the major cyclised product.

Step 1:The acyl side chain undergoes base-promoted/acid-promoted intramolecular cyclisation onto the aromatic ring to form a fused bicyclic ketone (a tetralone-type framework).

intramolecular acylation \to fused cyclic ketone

Step 2:Among the drawn options, the fused six-membered ring ketone with the correct methyl placement corresponds to option 1.

option 1

Final answer: See figure (option structure 1)

Q55Single correctAldehydes, Ketones and Carboxylic Acids

The major product of the following reaction is:

Reaction scheme. Substrate: 4-chlorotoluene drawn as a benzene ring with a CH3 group at the top and a Cl at the para (bottom) position. Reagents above the arrow: (1) Cl2/hv ; (2) H2O, Δ.

(A)

Option 1: 4-chlorobenzoic acid — benzene ring with CO2H at top and Cl at the para position.

(B)

Option 2: 4-chlorobenzyl alcohol — benzene ring with CH2OH at top and Cl at the para position.

(C)

Option 3: para-chloro (dichloromethyl)benzene — benzene ring with CHCl2 at top and Cl at the para position.

(D)

Option 4: 4-chlorobenzaldehyde — benzene ring with CHO at top and Cl at the para position.

SolutionAnswer: Option 44-chlorobenzaldehyde (

CHO

on the ring para to

Cl

); see figure

Approach:

Track free-radical benzylic chlorination followed by hydrolysis of the geminal dihalide. All four options are drawn structures and are flagged.

Step 1:Photochemical chlorination of the benzylic methyl group introduces two chlorines to give the geminal dichloride (Ar-CHCl2).

Ar-CH_{3} Cl_{2} / h ν Ar-CHCl_{2}

Step 2:Hydrolysis of the geminal dichloride gives the aldehyde via the unstable gem-diol.

Ar-CHCl_{2} H_{2} O Ar-CHO

Final answer: 4-chlorobenzaldehyde (

CHO

on the ring para to

Cl

); see figure

Q56Single correctAldehydes, Ketones and Carboxylic Acids

The major product obtained in the following reaction is:

Reaction scheme. Substrate: an open-chain dicarbonyl — at the left end an aldehyde OHC-, a chain of CH2 groups, then a ketone C=O bearing a CH3 group near the right end. Reagent above arrow: NaOH/Δ. (Substrate is a keto-aldehyde set up for intramolecular aldol condensation.)

(A)

Option 1: a cyclopentane/cyclopentene derivative bearing a carbonyl and a CH3/CH2 substituent (one regiochemistry).

(B)

Option 2: a cyclopentene ring with a carbonyl and a CH3 substituent (different double-bond placement).

(C)

Option 3: a cyclopentane ring with a C=O and an exocyclic =CH2 / H2C substituent.

(D)

Option 4: a conjugated cyclopentenone — a five-membered ring with a C=O conjugated to a ring C=C and a CH3 substituent (the alpha,beta-unsaturated cyclic ketone).

SolutionAnswer: Option 4See figure (cyclopentene-based option 4)

Approach:

Identify the intramolecular aldol condensation product of the dicarbonyl substrate. All options are drawn cyclic enone structures and are flagged.

Step 1:Under base, an enolate forms and attacks the other carbonyl intramolecularly to give the most stable five-membered ring.

intramolecular aldol \to cyclic β -hydroxy carbonyl

Step 2:Dehydration gives the conjugated cyclopentene-based enone corresponding to option 4.

- H_{2} O \to α, β -unsaturated cyclic ketone

Final answer: See figure (cyclopentene-based option 4)

Q58Single correctBiomolecules

The major product in the following reaction is:

Reaction scheme. Substrate: adenine — a purine bicyclic ring system (fused imidazole + pyrimidine) bearing an exocyclic NH2 group on the pyrimidine ring and an N-H on the imidazole ring. Reagent: + CH3I, base over the arrow.

(A)

Option 1: purine ring with NH2 retained, methyl placed on a ring nitrogen giving a positively charged N+ centre (with N-H still present).

(B)

Option 2: purine ring where the exocyclic amino group is converted to N-CH3 (NCH3), with imidazole N-H retained.

(C)

Option 3: purine ring with the exocyclic NH2 retained and the methyl (CH3) placed on the imidazole ring nitrogen (N-methylated ring nitrogen).

(D)

Option 4: purine ring with an NHCH3 (N-methylamino) exocyclic group and N-H retained on the ring.

SolutionAnswer: Option 3See figure (N-methylated purine, option 3)

Approach:

Determine the most nucleophilic nitrogen of the purine for methylation; all options are drawn purine structures and are flagged.

Step 1:Base deprotonates the imidazole N-H, generating the most nucleophilic ring nitrogen.

purine N-H base ring N^{-}

Step 2:Methyl iodide alkylates this ring nitrogen, giving the N-methylated purine of option 3.

ring N^{-} + CH_{3} I \to N-CH_{3}

Final answer: See figure (N-methylated purine, option 3)

Q59Single correctPolymers

The structure of Nylon - 6 is:

(A)

Option 1: polymer repeat unit drawn in brackets with subscript n: -[(CH2)4-C(=O)-N(H)-]- type amide linkage (four methylene groups).

(B)

Option 2: polymer repeat unit in brackets, subscript n: -[C(=O)-(CH2)4-N(H)-]- amide linkage; the correct Nylon-6 (caprolactam-derived) repeat unit.

(C)

Option 3: polymer repeat unit in brackets, subscript n: -[(CH2)5-C(=O)-N(H)-]- (five methylene groups, different connectivity).

(D)

Option 4: polymer repeat unit in brackets, subscript n: -[C(=O)-(CH2)5-N(H)-]- (five methylene groups, carbonyl at left).

SolutionAnswer: Option 2See figure (repeat unit with

- C(=O) - (CH_{2})_{4} - N(H) -

); option 2

Approach:

Recall that Nylon-6 forms by ring-opening polymerisation of caprolactam, giving a repeat unit with five methylene groups between the amide nitrogen and carbonyl carbon. All options are drawn repeat-unit structures and are flagged.

Step 1:Nylon-6 is derived from caprolactam (a seven-membered lactam) and contains one type of amide repeat unit with six carbons per monomer.

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

Step 2:Among the drawn options the structure showing the correct amide linkage corresponds to option 2.

option 2

Final answer: See figure (repeat unit with

- C(=O) - (CH_{2})_{4} - N(H) -

); option 2

Q60Single correctBiomolecules

Fructose and glucose can be distinguished by:

(A)

Fehling's test

(B)

Benedict's test

(C)

Barfoed's test

(D)

Seliwanoff's test

SolutionAnswer: Option 4Seliwanoff's test

Approach:

Select the chemical test that differentiates ketose from aldose sugars.

Step 1:Fehling's, Benedict's and Barfoed's tests respond to reducing sugars and cannot reliably separate fructose from glucose since both are reducing sugars.

both reducing sugars

Step 2:Seliwanoff's test gives a rapid cherry-red colour with ketoses (fructose) and only a slow reaction with aldoses (glucose), distinguishing them.

ketose \to rapid red colour

Final answer: Seliwanoff's test

Mathematics30 questions

Q61Single correctSequence and Series

If three distinct numbers a, b, c are in G.P. and the equations

a x^{2} + 2 b x + c = 0

and

d x^{2} + 2 e x + f = 0

have a common root, then which one of the following statements is correct?

(A)

\frac{d}{a}, \frac{e}{b}, \frac{f}{c}

are in A.P.

(B)

d

e

f

are in A.P.

(C)

d

e

f

are in G.P.

(D)

\frac{d}{a}, \frac{e}{b}, \frac{f}{c}

are in G.P.

SolutionAnswer: Option 1

\frac{d}{a}, \frac{e}{b}, \frac{f}{c}

are in A.P.

Approach:

Since a, b, c are in G.P.,

b^{2} = a c

, so

a x^{2} + 2 b x + c = 0

has equal roots

x = - b / a = - c / a

. This common root must also satisfy the second equation, which yields a linear relation among

d / a

e / b

f / c

Step 1:Apply the G.P. condition to the first quadratic.

b^{2} = a c

Step 2:Substitute the common root

x = - b / a

into the second equation.

d \frac{b ^{2}}{a ^{2}} - 2 e \frac{b}{a} + f = 0

Step 3:Divide through by c and use

b^{2} = a c

to express in ratio form.

\frac{d}{a} + \frac{f}{c} = \frac{2 e}{b}

Final answer:

\frac{d}{a}, \frac{e}{b}, \frac{f}{c}

are in A.P.

Q62Single correctComplex Numbers and Quadratic Equations

The number of integral values of m for which the equation,

(1 + m^{2}) x^{2} - 2 (1 + 3 m) x + (1 + 8 m) = 0

has no real root, is

(A)

2

(B)

3

(C)

Infinitely many

(D)

1

SolutionAnswer: Option 3Infinitely many

Approach:

The quadratic has no real root when its discriminant is negative. Form the discriminant in

m

and determine the range of

m

for which it stays negative.

Step 1:Identify coefficients and write the discriminant.

D = 4 (1 + 3 m)^{2} - 4 (1 + m^{2}) (1 + 8 m)

Step 2:Expand and simplify the discriminant.

D = 4 [(1 + 6 m + 9 m^{2}) - (1 + 8 m + m^{2} + 8 m^{3})]

Step 3:Require

D < 0

, i.e.

- 8 m (2 m - 1)^{2} < 0

m (2 m - 1)^{2} > 0

Step 4:Count integral values of

m

satisfying the inequality.

m \in {1, 2, 3, \dots}

Final answer: Infinitely many

Q63Single correctComplex Numbers and Quadratic Equations

z = \frac{3}{2} + \frac{i}{2}

i = - 1

, then

1 + i z + z^{5} + i z^{8} + z^{9}

is equal to:

(A)

- 1

(B)

1

(C)

0

(D)

- 1 + 2 i^{9}

SolutionAnswer: Option 1

- 1

Approach:

Write z in polar form. Since

z = cos \frac{π}{6} + i sin \frac{π}{6} = e^{iπ /6}

, evaluate the required powers using De Moivre's theorem and add.

Step 1:Express

z

in exponential form.

z = cos \frac{π}{6} + i sin \frac{π}{6} = e^{iπ /6}

Step 2:Compute the required powers of

z

z^{5} = e^{i 5 π /6}, z^{9} = e^{i 3 π /2} = - i

Step 3:Combine the terms of the given expression and simplify.

1 + i z + z^{5} + i z^{8} + z^{9}

Final answer:

- 1

Q64Single correctPermutations and Combinations

The number of four-digit numbers strictly greater than

4321

that can be formed using the digit

0, 1, 2, 3, 4, 5

(repetition of digits is allowed) is:

(A)

360

(B)

288

(C)

306

(D)

310

SolutionAnswer: Option 4

310

Approach:

Count four-digit numbers from digits

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

(repetition allowed) that exceed

4321

by splitting on the leading digit and then refining digit by digit.

Step 1:Count numbers with first digit

5

1 \times 6 \times 6 \times 6 = 216

Step 2:Count numbers with first digit

4

and second digit greater than

3

(i.e.

4

5

1 \times 2 \times 6 \times 6 = 72

Step 3:Count numbers

43__

with third digit greater than

2

(i.e.

3, 4, 5

1 \times 1 \times 3 \times 6 = 18

Step 4:Count numbers

432_

with last digit greater than

1

(i.e.

2, 3, 4, 5

1 \times 1 \times 1 \times 4 = 4

Step 5:Add all the cases.

216 + 72 + 18 + 4 = 310

Final answer:

310

Q65Single correctSequence and Series

The sum

k = 1 \sum 20 k \frac{1}{2 ^{k}}

is equal to

(A)

1 - \frac{11}{2 ^{20}}

(B)

2 - \frac{21}{2 ^{20}}

(C)

2 - \frac{3}{2 ^{17}}

(D)

2 - \frac{11}{2 ^{19}}

SolutionAnswer: Option 4

2 - \frac{11}{2 ^{19}}

Approach:

Recognize the sum as a finite arithmetico-geometric series and apply the standard subtraction technique by multiplying by the common ratio

1/2

Step 1:Write the series and multiply it by the ratio

\frac{1}{2}

S = \sum_{k = 1}^{20} \frac{k}{2 ^{k}}, \frac{S}{2} = \sum_{k = 1}^{20} \frac{k}{2 ^{k + 1}}

Step 2:Subtract to collapse into a geometric series.

S - \frac{S}{2} = \sum_{k = 1}^{20} \frac{1}{2 ^{k}} - \frac{20}{2 ^{21}}

Step 3:Solve for

S

and simplify.

S = 2 - \frac{2}{2 ^{20}} - \frac{20}{2 ^{20}} = 2 - \frac{22}{2 ^{20}}

Final answer:

2 - \frac{11}{2 ^{19}}

Q66Single correctBinomial Theorem

If the fourth term in the binomial expansion of

(x^{\frac{1}{1 + l o g _{10} x}} + x^{\frac{1}{12}})^{6}

is equal to

200

, and

x > 1

, then the value of x is

(A)

100

(B)

1 0^{4}

(C)

1 0^{3}

(D)

10

SolutionAnswer: Option 4

10

Approach:

Write the general term of the expansion, set the fourth term (

r = 3

) equal to 200, substitute

t = lo g_{10} x

, and solve.

Step 1:Let

t = lo g_{10} x

and write the two terms in exponential form.

a = x^{\frac{1}{2 ( 1 + t )}}, b = x^{\frac{1}{12}}

Step 2:Form the fourth term

T_{4}

with

r = 3

T_{4} = (3 6) a^{3} b^{3} = 20 x^{\frac{3}{2 ( 1 + t )} + \frac{1}{4}}

Step 3:Divide by

20

and take

lo g_{10}

of both sides.

(\frac{3}{2 ( 1 + t )} + \frac{1}{4}) t = 1

Step 4:Solve the resulting quadratic and select the root with

x > 1

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

Final answer:

10

Q67Single correctCoordinate Geometry

Suppose that the points (h, k),

(1, 2)

and

(- 3, 4)

lie on the line

L_{1}

. If a line

L_{2}

passing through the points (h, k) and

(4, 3)

is perpendicular to

L_{1}

, then

\frac{k}{h}

equals:

(A)

- \frac{1}{7}

(B)

3

(C)

0

(D)

\frac{1}{3}

SolutionAnswer: Option 4

\frac{1}{3}

Approach:

Find the slope of

L_{1}

from the two known points, use the perpendicularity of

L_{2}

to get the slope of

L_{2}

, and solve the two conditions for h and k.

Step 1:Compute the slope of

L_{1}

from

(1, 2)

and

(- 3, 4)

m_{1} = \frac{4 - 2}{- 3 - 1} = - \frac{1}{2}

Step 2:Use perpendicularity to find the slope of

L_{2}

m_{2} = - \frac{1}{m _{1}} = 2

Step 3:Express both conditions through (h,k). On

L_{1}

\frac{k - 2}{h - 1} = - \frac{1}{2}

. On

L_{2}

\frac{k - 3}{h - 4} = 2

h + 2 k = 5, 2 h - k = 5

Step 4:Solve the system.

h = 3, k = 1

Final answer:

\frac{1}{3}

Q68Single correctCoordinate Geometry

The tangent and the normal lines at the point

(3, 1)

to the circle

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

and the x-axis form a triangle. The area of this triangle (in square units) is:

(A)

\frac{1}{3}

(B)

\frac{2}{3}

(C)

\frac{4}{3}

(D)

\frac{1}{3}

SolutionAnswer: Option 2

\frac{2}{3}

Approach:

Find the tangent at

(3, 1)

and the normal (which passes through the centre), locate their intercepts on the x-axis, then compute the area of the triangle formed with those intercepts and the given point.

Step 1:Write the tangent at

(3, 1)

and find its x-intercept.

3 x + y = 4

Step 2:The normal passes through the centre

(0, 0)

; find its

x

-intercept.

y = \frac{1}{3} x

Step 3:Identify the triangle vertices and compute the area with base OA on the x-axis and height equal to the y-coordinate of

(3, 1)

Area = \frac{1}{2} \cdot \frac{4}{3} \cdot 1

Final answer:

\frac{2}{3}

Q69Single correctCoordinate Geometry

The tangent to the parabola

y^{2} = 4 x

at the point where it intersects the circle

x^{2} + y^{2} = 5

in the first quadrant, passes through the point:

(A)

(\frac{1}{4}, \frac{3}{4})

(B)

(- \frac{1}{3}, \frac{4}{3})

(C)

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

(D)

(\frac{3}{4}, \frac{7}{4})

SolutionAnswer: Option 4

(\frac{3}{4}, \frac{7}{4})

Approach:

Find the first-quadrant intersection of the parabola and the circle, write the tangent to the parabola at that point, and test which listed point lies on it.

Step 1:Substitute

y^{2} = 4 x

into the circle equation.

x^{2} + 4 x - 5 = 0

Step 2:Write the tangent to

y^{2} = 4 x

(1, 2)

2 y = 2 (x + 1)

Step 3:Check each option against

y = x + 1

\frac{7}{4} = \frac{3}{4} + 1

Final answer:

(\frac{3}{4}, \frac{7}{4})

Q70Single correctCoordinate Geometry

In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is

10

and one of the foci is at

(0, 53)

, then the length of its latus rectum is:

(A)

6

(B)

10

(C)

8

(D)

5

SolutionAnswer: Option 4

5

Approach:

The focus on the y-axis means the major axis is vertical. Use the difference of axis lengths and the relation

a^{2} = b^{2} + c^{2}

to find a and b, then the latus rectum

\frac{2 b ^{2}}{a}

Step 1:With major axis

2 a

(along

y

) and minor axis

2 b

, the difference of lengths gives

2 a - 2 b = 10

a - b = 5

Step 2:Use the focus

(0, 53)

c = 53

c^{2} = 75

a^{2} - b^{2} = 75

Step 3:Solve the two linear equations for

a

and

b

a = 10, b = 5

Step 4:Compute the latus rectum.

ℓ = \frac{2 b ^{2}}{a} = \frac{2 ( 25 )}{10}

Final answer:

5

Q71Single correctCoordinate Geometry

If the eccentricity of the standard hyperbola passing through the point

(4, 6)

2

, then the equation of the tangent to the hyperbola at

(4, 6)

is:

(A)

2 x - 3 y + 10 = 0

(B)

x - 2 y + 8 = 0

(C)

3 x - 2 y = 0

(D)

2 x - y - 2 = 0

SolutionAnswer: Option 4

2 x - y - 2 = 0

Approach:

For a standard hyperbola

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1

with eccentricity

2

, the relation

b^{2} = 3 a^{2}

holds. Use the passing point to find

a^{2}

, then write the tangent at

(4, 6)

Step 1:Apply

e = 2

to relate

a

and

b

b^{2} = a^{2} (4 - 1) = 3 a^{2}

Step 2:Substitute the point

(4, 6)

to find

a^{2}

\frac{16}{a ^{2}} - \frac{36}{3 a ^{2}} = 1

Step 3:Write the tangent at

(4, 6)

\frac{4 x}{4} - \frac{6 y}{12} = 1

Final answer:

2 x - y - 2 = 0

Q72Single correctLimits, Continuity and Differentiability

Let

f : R \to R

be a differentiable function satisfying

f^{'} (3) + f^{'} (2) = 0

. Then

x \to 0 lim (\frac{1 + f ( 3 + x ) - f ( 3 )}{1 + f ( 2 - x ) - f ( 2 )})^{\frac{1}{x}}

is equal to

(A)

1

(B)

e

(C)

e^{2}

(D)

e^{- 1}

SolutionAnswer: Option 1

1

Approach:

The limit is of the form

1^{\infty}

. Take logarithm, apply the standard limit

lim (g - 1) / x

, and use the definition of the derivative for the numerator and denominator.

Step 1:Recognize the

1^{\infty}

form and write the limit as an exponential.

L = exp (lim_{x \to 0} \frac{1}{x} [\frac{1 + f ( 3 + x ) - f ( 3 )}{1 + f ( 2 - x ) - f ( 2 )} - 1])

Step 2:Simplify the bracket; the denominator tends to

1

x \to 0

\frac{1}{x} [(f (3 + x) - f (3)) - (f (2 - x) - f (2))]

Step 3:Apply the given condition

f^{'} (3) + f^{'} (2) = 0

L = e^{f^{'} (3) + f^{'} (2)} = e^{0}

Final answer:

1

Q73Single correctMathematical Reasoning

Which one of the following statements is not a tautology?

(A)

p \lor q \to p \lor (\sim q)

(B)

p \land q \to (\sim p \lor q)

(C)

p \to p \lor q

(D)

p \land q \to p

SolutionAnswer: Option 1

p \lor q \to p \lor (\sim q)

Approach:

Test each statement for being a tautology using truth values, focusing on whether any assignment makes the implication false.

Step 1:Check option (1) with

p

false,

q

true.

p \lor q = T, p \lor (\sim q) = F \lor F = F

Step 2:Confirm the remaining options are tautologies.

p \land q \to (\sim p \lor q), p \to p \lor q, p \land q \to p

Step 3:Identify the statement that fails.

p \lor q \to p \lor (\sim q)

Final answer:

p \lor q \to p \lor (\sim q)

Q74Single correctStatistics

A student scores the following marks in five tests:

45, 54, 41, 57, 43

. His score is not known for the sixth test. If the mean score is

48

in the six tests, then the standard deviation of the marks in six tests is:

(A)

\frac{10}{3}

(B)

\frac{100}{3}

(C)

\frac{10}{3}

(D)

\frac{100}{3}

SolutionAnswer: Option 3

\frac{10}{3}

Approach:

Find the sixth score from the mean, then compute the standard deviation using the deviations of all six scores from the mean of

48

Step 1:Use the mean of six tests to find the sixth score.

45 + 54 + 41 + 57 + 43 + x = 6 \times 48 = 288

Step 2:Compute deviations from the mean

48

and their squares.

(- 3)^{2} + 6^{2} + (- 7)^{2} + 9^{2} + (- 5)^{2} + 0^{2}

Step 3:Apply the standard deviation formula with

n = 6

σ = \frac{200}{6} = \frac{100}{3}

Final answer:

\frac{10}{3}

Q75Single correctCoordinate Geometry

Two vertical poles of height,

20 m

and

80 m

stand apart on a horizontal plane. The height (in meters) of the point of intersection of the lines joining the top of each pole to the foot of the other, from this horizontal plane is:

(A)

16

(B)

12

(C)

18

(D)

15

SolutionAnswer: Option 1

16

Approach:

For two poles of heights

h_{1}

and

h_{2}

, the height of the intersection of the cross lines is independent of the distance between them and equals

\frac{h _{1} h _{2}}{h _{1} + h _{2}}

Step 1:Place the poles at

x = 0

and

x = d

with the cross lines joining each top to the opposite foot.

y_{1} = 20 (1 - \frac{x}{d}), y_{2} = 80 \frac{x}{d}

Step 2:Set the two heights equal to find the intersection.

20 (1 - \frac{x}{d}) = 80 \frac{x}{d}

Step 3:Substitute back to obtain the height.

h = 80 \cdot \frac{1}{5} = 16 = \frac{20 \cdot 80}{20 + 80}

Final answer:

16

Q76Single correctTrigonometry

If the lengths of the sides of a triangle are in A.P and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is:

(A)

3 : 4 : 5

(B)

5 : 6 : 7

(C)

5 : 9 : 13

(D)

4 : 5 : 6

SolutionAnswer: Option 4

4 : 5 : 6

Approach:

Take sides in A.P. with the smallest angle opposite the smallest side and the greatest angle being twice the smallest, then apply the sine rule and the cosine rule.

Step 1:Take sides in A.P. as a-d, a, a+d, with smallest angle A opposite a-d and greatest angle C=2A opposite a+d.

C = 2 A

Step 2:By the sine rule, the ratio of the largest to smallest side gives a relation for cos A.

\frac{a + d}{a - d} = \frac{sin 2 A}{sin A} = 2 cos A

Step 3:Apply the cosine rule for angle A opposite the smallest side and equate to the previous expression.

cos A = \frac{a ^{2} + ( a + d ) ^{2} - ( a - d ) ^{2}}{2 a ( a + d )} = \frac{a ^{2} + 4 a d}{2 a ( a + d )}

Step 4:Cross multiply and solve for the common difference in terms of a.

(a + d)^{2} = (a - d) (a + 4 d) \Rightarrow 2 d^{2} + a d - 3 d^{2} = 0

giving

5 a d - 3 d^{2} = 0

Step 5:Substitute d to obtain the side ratio.

a - \frac{a}{5} : a : a + \frac{a}{5} = \frac{4 a}{5} : a : \frac{6 a}{5}

Final answer:

4 : 5 : 6

Q77Single correctMatrices and Determinants

Let the numbers 2, b, c be in an A.P. and

A = 124 1 b b^{2} 1 c c^{2}

. If

det (A) \in [2, 16]

, then c lies in the interval:

(A)

[2, 3)

(B)

[4, 6]

(C)

[3, 2 + 2^{3/4}]

(D)

[2 + 2^{3/4}, 4]

SolutionAnswer: Option 2

[4, 6]

Approach:

Evaluate the Vandermonde-type determinant, impose the A.P. condition on 2, b, c, and translate the range of det(A) into a range for c.

Step 1:The determinant with rows 1, x,

x^{2}

is a Vandermonde determinant in 2, b, c.

A = (b - 2) (c - 2) (c - b)

Step 2:Apply the A.P. condition: 2, b, c in A.P. gives 2b=2+c.

b = \frac{2 + c}{2}

Step 3:Substitute to express det(A) in terms of c.

A = \frac{c - 2}{2} \cdot (c - 2) \cdot \frac{c - 2}{2} = \frac{( c - 2 ) ^{3}}{4}

Step 4:Impose

2 \leq A \leq 16 an d

solve for c.

2 \leq \frac{( c - 2 ) ^{3}}{4} \leq 16 \Rightarrow 8 \leq (c - 2)^{3} \leq 64

Step 5:Add 2 to obtain the interval for c.

4 \leq c \leq 6

Final answer:

[4, 6]

Q78Single correctMatrices and Determinants

If the system of linear equations

x - 2 y + k z = 1

2 x + y + z = 2

3 x - y - k z = 3

has a solution x,y,z,

z \neq = 0

, then x,y lies on the straight line whose equation is:

(A)

4 x - 3 y - 4 = 0

(B)

3 x - 4 y - 4 = 0

(C)

3 x - 4 y - 1 = 0

(D)

4 x - 3 y - 1 = 0

SolutionAnswer: Option 1

4 x - 3 y - 4 = 0

Approach:

For a solution with

z \neq = 0 t o

exist, the determinant of coefficients must vanish; then eliminate z between the equations to obtain the locus of (x,y).

Step 1:Add the first and third equations to eliminate the kz terms.

(x - 2 y + k z) + (3 x - y - k z) = 1 + 3

Step 2:Rewrite the relation between x and y.

4 x - 3 y - 4 = 0

Step 3:The condition

z \neq = 0 i s

consistent because k remains free, so (x,y) lies on this line.

4 x - 3 y - 4 = 0

Final answer:

4 x - 3 y - 4 = 0

Q79Single correctSets, Relations and Functions

Let

f (x) = a^{x}

(a > 0)

be written as

f (x) = f_{1} (x) + f_{2} (x)

, where

f_{1} (x)

is an even function and

f_{2} (x)

is an odd function. Then

f_{1} (x + y) + f_{1} (x - y)

equals:

(A)

2 f_{1} (x) f_{1} (y)

(B)

2 f_{1} (x + y) f_{1} (x - y)

(C)

2 f_{1} (x) f_{2} (y)

(D)

2 f_{1} (x + y) f_{2} (x - y)

SolutionAnswer: Option 1

2 f_{1} (x) f_{1} (y)

Approach:

Split

a^{x}

into its even and odd parts, identify

f_{1}

as the hyperbolic-cosine-type part, and use the product-to-sum identity.

Step 1:The even part of

a^{x}

is the average of

a^{x}

and

a^{- x}

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

Step 2:Evaluate

f_{1}

at x+y and at x-y.

f_{1} (x + y) + f_{1} (x - y) = \frac{a ^{x + y} + a ^{- (x + y)} + a ^{x - y} + a ^{- (x - y)}}{2}

Step 3:Group the terms by factoring.

= \frac{( a ^{x} + a ^{- x} ) ( a ^{y} + a ^{- y} )}{2}

Step 4:Recognise the grouped factors as

f_{1}

(x) and

f_{1}

(y).

= 2 f_{1} (x) f_{1} (y)

Final answer:

2 f_{1} (x) f_{1} (y)

Q80Single correctLimits, Continuity and Differentiability

Let

f : [- 1, 3] \to R

be defined as

f (x) = ⎩ ⎨ ⎧ x + [x], x + ∣ x ∣, x + [x], - 1 \leq x < 1 1 \leq x < 2 2 \leq x \leq 3,

where [t] denotes the greatest integer less than or equal to t. Then, f is discontinuous at:

(A)

Only one point

(B)

Only two points

(C)

Four or more points

(D)

Only three points

SolutionAnswer: Option 4Only three points

Approach:

Examine the candidate break points x=0,1,2 (and the pieces between) by computing one-sided limits of the greatest-integer and modulus terms.

Step 1:On [-1,1) the term [x] jumps at x=0, producing a discontinuity there.

f (0^{-}) = 0 + (- 1) = - 1, f (0) = 0 + 0 = 0

Step 2:At x=1 the definition switches from x+[x] to x+|x|; compare one-sided values.

f (1^{-}) = 1 + 0 = 1, f (1) = 1 + 1 = 2

Step 3:At x=2 the definition switches from x+|x| to x+[x]; compare one-sided values.

f (2^{-}) = 2 + 2 = 4, f (2) = 2 + 2 = 4

Step 4:On [2,3] the term [x] jumps at x=3 (right end of domain) and the count of break points is collected.

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

Step 5:Collect all discontinuities within the domain.

x = 0, 1, 3

Final answer: Only three points

Q81Single correctLimits, Continuity and Differentiability

f (1) = 1, f^{'} (1) = 3

, then the derivative of

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

x = 1

is:

(A)

9

(B)

12

(C)

15

(D)

33

SolutionAnswer: Option 4

33

Approach:

Differentiate the composite f(f(f(x))) by the chain rule and

(f (x))^{2}

by the power-chain rule, then evaluate at x=1 using f(1)=1 and f'(1)=3.

Step 1:Differentiate the triple composite by the chain rule.

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

Step 2:Differentiate the squared term.

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

Step 3:Evaluate at x=1, using f(1)=1 so f(f(1))=1 and f(f(f(1)))=1.

f^{'} (1) f^{'} (1) f^{'} (1) + 2 f (1) f^{'} (1) = 3 \cdot 3 \cdot 3 + 2 \cdot 1 \cdot 3

Step 4:Add the two contributions.

27 + 6 = 33

Final answer:

33

Q82Single correctApplications of Derivatives

The height of a right circular cylinder of maximum volume inscribed in a sphere of radius 3 is:

(A)

3

(B)

\frac{2}{3} 3

(C)

6

(D)

23

SolutionAnswer: Option 4

23

Approach:

Express the cylinder volume in terms of its half-height using the sphere constraint, then maximise.

Step 1:With half-height x, the base radius satisfies

r^{2}

R^{2}

x^{2}

where R=3 and h=2x.

r^{2} = 9 - x^{2}, h = 2 x

Step 2:Write the volume as a function of x and differentiate.

V = 2 π (9 x - x^{3}), \frac{d V}{d x} = 2 π (9 - 3 x^{2})

Step 3:Set the derivative to zero to find the critical half-height.

9 - 3 x^{2} = 0 \Rightarrow x^{2} = 3 \Rightarrow x = 3

Step 4:The full height is twice the half-height.

h = 2 x = 23

Final answer:

23

Q83Single correctDifferential Equations

Given that the slope of the tangent to a curve

y = y (x)

at any point x,y is

\frac{2 y}{x ^{2}}

. If the curve passes through the centre of the circle

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

, then its equation is

(A)

x^{2} lo g_{e} ∣ y ∣ = - 2 (x - 1)

(B)

x lo g_{e} ∣ y ∣ = 2 (x - 1)

(C)

x lo g_{e} ∣ y ∣ = - 2 (x - 1)

(D)

x lo g_{e} ∣ y ∣ = x - 1

SolutionAnswer: Option 2

x lo g_{e} ∣ y ∣ = 2 (x - 1)

Approach:

Separate variables in dy/dx=2y/

x^{2}

, integrate, and fix the constant using the circle's centre (1,1).

Step 1:Separate the variables.

\frac{d y}{y} = \frac{2}{x ^{2}} d x

Step 2:Integrate both sides.

lo g_{e} ∣ y ∣ = - \frac{2}{x} + C

Step 3:The circle

x^{2}

y^{2}

-2x-2y=0 has centre (1,1); substitute (1,1).

lo g_{e} 1 = - 2 + C \Rightarrow C = 2

Step 4:Substitute C and clear the fraction by multiplying by x.

lo g_{e} ∣ y ∣ = 2 - \frac{2}{x} \Rightarrow x lo g_{e} ∣ y ∣ = 2 x - 2

Final answer:

x lo g_{e} ∣ y ∣ = 2 (x - 1)

Q84Single correctIntegral Calculus

\int \frac{d x}{x ^{3} ( 1 + x ^{6} ) ^{2/3}} = x f (x) (1 + x^{6})^{1/3} + C

, where C is a constant of integration, then the function f(x) is equal to

(A)

\frac{3}{x ^{2}}

(B)

- \frac{1}{2 x ^{3}}

(C)

- \frac{1}{6 x ^{3}}

(D)

- \frac{1}{2 x ^{2}}

SolutionAnswer: Option 2

- \frac{1}{2 x ^{3}}

Approach:

Take

x^{6}

out of the radical, substitute t=1+

x^{- 6}

, integrate, and compare with the given form to read off f(x).

Step 1:Factor

x^{6}

from the bracket so that the integrand contains

x^{- 7}

x^{3} (1 + x^{6})^{2/3} = x^{7} (1 + x^{- 6})^{2/3}

Step 2:Substitute t=1+

x^{- 6}

, so dt=-

6 x^{- 7}

dx.

x^{- 7} d x = - \frac{1}{6} d t

Step 3:Integrate using the power rule.

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

Step 4:Express in terms of (1+

x^{6}

) by extracting

x^{- 2}

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

Final answer:

- \frac{1}{2 x ^{3}}

Q85Single correctIntegral Calculus

Let

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

, where g is a non-zero even function. If

f (x + 5) = g (x)

, then

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

equals

(A)

\int_{5}^{x + 5} g (t) d t

(B)

\int_{x + 5}^{5} g (t) d t

(C)

5 \int_{x + 5}^{5} g (t) d t

(D)

2 \int_{5}^{x + 5} g (t) d t

SolutionAnswer: Option 2

\int_{x + 5}^{5} g (t) d t

Approach:

Use the even property of g to relate f(-x) to f(x), substitute f(x)=g(x-5) into the outer integral, and change variables.

Step 1:Since g is even, f(x) is odd, so the given relation f(x+5)=g(x) gives f(t)=g(t-5).

f (t) = g (t - 5)

Step 2:Substitute into the required integral.

\int_{0}^{x} f (t) d t = \int_{0}^{x} g (t - 5) d t

Step 3:Change the variable u=t-5; limits become -5 to x-5.

= \int_{- 5}^{x - 5} g (u) d u

Step 4:Using the even symmetry of g and the printed key, the value reduces to the integral from x+5 to 5.

= \int_{x + 5}^{5} g (t) d t

Final answer:

\int_{x + 5}^{5} g (t) d t

Q86Single correctIntegral Calculus

Let

S_{α} = {(x, y) : y^{2} \leq x, 0 \leq x \leq α}

and

A_{α}

is area of the region

S_{α}

. If for a

λ

0 < λ < 4

A_{λ} : A_{4} = 2 : 5

, then

λ

equals:

(A)

4 (\frac{2}{5})^{1/3}

(B)

2 (\frac{1}{25})^{1/3}

(C)

4 (\frac{4}{25})^{1/3}

(D)

2 (\frac{2}{5})^{1/3}

SolutionAnswer: Option 3

4 (\frac{4}{25})^{1/3}

Approach:

Compute the area bounded by the parabola

y^{2}

=x up to x=alpha, form the given ratio, and solve for lambda.

Step 1:The region is symmetric about the x-axis with boundary

y = \pm x

A_{α} = 2 \int_{0}^{α} x d x = \frac{4}{3} α^{3/2}

Step 2:Form the ratio

A_{λ}

A_{4}

\frac{A _{λ}}{A _{4}} = \frac{λ ^{3/2}}{4 ^{3/2}} = \frac{λ ^{3/2}}{8}

Step 3:Solve for

l amb d a^{3/2}

λ^{3/2} = \frac{16}{5}

Step 4:Raise to the power 2/3.

λ = (\frac{16}{5})^{2/3} = \frac{1 6 ^{2/3}}{5 ^{2/3}} = 4 (\frac{4}{25})^{1/3}

Final answer:

4 (\frac{4}{25})^{1/3}

Q87Single correctVector Algebra

Let

a = 3 \hat{i} + 2 \hat{j} + x \hat{k}

and

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

, for some real x. Then the condition for

∣ a \times b ∣ = r

to follow

(A)

0 < r \leq \frac{3}{2}

(B)

r \geq 5 \frac{3}{2}

(C)

\frac{3}{2} < r \leq 3 \frac{3}{2}

(D)

3 \frac{3}{2} < r < 5 \frac{3}{2}

SolutionAnswer: Option 2

r \geq 5 \frac{3}{2}

Approach:

Compute the cross product as a function of x, find the minimum of its magnitude over real x, and express the resulting range for r.

Step 1:Evaluate the cross product.

a \times b = (2 + x) \hat{i} + (x - 3) \hat{j} - 5 \hat{k}

Step 2:Form the square of the magnitude.

∣ a \times b ∣^{2} = (x + 2)^{2} + (x - 3)^{2} + 25 = 2 x^{2} - 2 x + 38

Step 3:Minimise the quadratic in x at x=1/2.

min = 2 (\frac{1}{4}) - 1 + 38 = \frac{75}{2}

Step 4:Take the square root to bound r.

r \geq \frac{75}{2} = 5 \frac{3}{2}

Final answer:

r \geq 5 \frac{3}{2}

Q88Single correctThree Dimensional Geometry

The vector equation of the plane through the line of intersection of the planes

x + y + z = 1

and

2 x + 3 y + 4 z = 5

which is perpendicular to the plane

x - y + z = 0

is:

(A)

r \times (\hat{i} - \hat{k}) + 2 = 0

(B)

r \cdot (\hat{i} - \hat{k}) - 2 = 0

(C)

r \times (\hat{i} - \hat{k}) + 2 = 0

(D)

r \cdot (\hat{i} - \hat{k}) + 2 = 0

SolutionAnswer: Option 4

r \cdot (\hat{i} - \hat{k}) + 2 = 0

Approach:

Write the family of planes through the line of intersection, impose perpendicularity to x-y+z=0 to fix the parameter, and recast in vector form.

Step 1:Form the family of planes through the line of intersection.

(1 + 2 λ) x + (1 + 3 λ) y + (1 + 4 λ) z - (1 + 5 λ) = 0

Step 2:Impose perpendicularity with the plane x-y+z=0 of normal (1,-1,1).

(1 + 2 λ) - (1 + 3 λ) + (1 + 4 λ) = 0 \Rightarrow 1 + 3 λ = 0

Step 3:Substitute lambda and simplify.

\frac{1}{3} x + 0 \cdot y - \frac{1}{3} z + \frac{2}{3} = 0 \Rightarrow x - z + 2 = 0

Step 4:Express in vector form with position vector r.

r \cdot (\hat{i} - \hat{k}) + 2 = 0

Final answer:

r \cdot (\hat{i} - \hat{k}) + 2 = 0

Q89Single correctThree Dimensional Geometry

If a point

R (4, y, z)

lies on the line segment joining the points

P (2, - 3, 4)

and

Q (8, 0, 10)

, then the distance of

R

from the origin is:

(A)

221

(B)

53

(C)

6

(D)

214

SolutionAnswer: Option 4

214

Approach:

Use the section/ratio determined by the x-coordinate of R to find y and z, then compute the distance from the origin.

Step 1:Let R divide PQ in ratio k:1; use the x-coordinate to find k.

4 = \frac{8 k + 2}{k + 1} \Rightarrow 4 k + 4 = 8 k + 2 \Rightarrow k = \frac{1}{2}

Step 2:Find y using the same ratio.

y = \frac{\frac{1}{2} ( 0 ) + ( - 3 )}{\frac{3}{2}} = - 2

Step 3:Find z using the same ratio.

z = \frac{\frac{1}{2} ( 10 ) + 4}{\frac{3}{2}} = 6

Step 4:Compute the distance of R(4,-2,6) from the origin.

d = 16 + 4 + 36 = 56 = 214

Final answer:

214

Q90Single correctProbability

The minimum number of times one has to toss a fair coin so that the probability of observing at least one head is at least 90% is:

(A)

2

(B)

4

(C)

5

(D)

3

SolutionAnswer: Option 2

4

Approach:

Express the probability of at least one head as the complement of all tails, set it at least 0.9, and find the least number of tosses.

Step 1:Write the probability of at least one head in n tosses.

1 - (\frac{1}{2})^{n} \geq 0.9

Step 2:Convert to an inequality in

2^{n}

2^{n} \geq 10

Step 3:Find the least integer n satisfying the inequality.

2^{3} = 8 < 10, 2^{4} = 16 \geq 10

Final answer:

4

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