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JEE Main 2019 January 12, Shift 1 Question Paper with Solutions

Q: How many questions are in the JEE Main 2019 January 12, Shift 1 paper?

The JEE Main 2019 January 12, Shift 1 paper has 90 questions — Physics (30), 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.

Q: Are the answer key and step-by-step solutions provided for the 2019 January 12, Shift 1 paper?

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 90 questions from the JEE Main 2019 (January 12, Shift 1) shift — Physics (30), 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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Physics30 questions

Q1Single correctOscillations and Waves

A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle

6 0^{\circ}

with ground level, but he finds the aeroplane right vertically above his position. If v is the speed of sound, speed of the plane is:

(A)

\frac{2 v}{3}

(B)

v

(C)

\frac{v}{2}

(D)

\frac{3}{2} v

SolutionAnswer: Option 3

\frac{v}{2}

Approach:

The sound heard came from the position where the plane was when it emitted the sound, at

6 0^{\circ}

elevation; by the time the sound reaches the observer the plane has moved to a point directly overhead. Equate travel times of sound and plane.

Step 1:Define the geometry: h is the altitude, the sound was emitted at horizontal distance x from the observer at elevation

6 0^{\circ}

tan 6 0^{\circ} = \frac{h}{x} = 3

Step 2:Sound travels the slant distance from the emission point to the observer.

d_{sound} = h^{2} + x^{2} = 3 x^{2} + x^{2} = 2 x

Step 3:In the same time, the plane travels the horizontal distance

x

from the emission point to directly overhead.

\frac{x}{v _{p}} = \frac{2 x}{v}

Final answer:

\frac{v}{2}

Q2Single correctKinematics

A passenger train of length

60 m

travels at a speed of

80 km / h r

. Another freight train of length

120 m

travels at a speed of

30 km / h r

. The ratio of times taken by the passenger train to completely cross the freight train when: (i) they are moving in the same direction, and (ii) in the opposite directions is:

(A)

\frac{5}{2}

(B)

\frac{3}{2}

(C)

\frac{11}{5}

(D)

\frac{25}{11}

SolutionAnswer: Option 3

\frac{11}{5}

Approach:

The time to cross is the combined length divided by the relative speed. Same direction uses the difference of speeds; opposite direction uses the sum. Take the ratio.

Step 1:Combined length of the two trains.

L = 60 + 120 = 180 m

Step 2:Same direction relative speed.

v_{same} = 80 - 30 = 50 km/hr

Step 3:Opposite direction relative speed.

v_{opp} = 80 + 30 = 110 km/hr

Step 4:Ratio of times equals the inverse ratio of relative speeds.

\frac{t _{same}}{t _{opp}} = \frac{v _{opp}}{v _{same}} = \frac{110}{50} = \frac{11}{5}

Final answer:

\frac{11}{5}

Q3Single correctWork, Energy and Power

A simple pendulum, made of a string of length l and a bob of mass m, is released from a small angle

θ_{0}

. It strikes a block of mass M, kept on horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle

θ_{1}

. Then M is given by:

(A)

m (\frac{θ _{0} - θ _{1}}{θ _{0} + θ _{1}})

(B)

m (\frac{θ _{0} + θ _{1}}{θ _{0} - θ _{1}})

(C)

\frac{m}{2} (\frac{θ _{0} + θ _{1}}{θ _{0} - θ _{1}})

(D)

\frac{m}{2} (\frac{θ _{0} - θ _{1}}{θ _{0} + θ _{1}})

SolutionAnswer: Option 1

m (\frac{θ _{0} - θ _{1}}{θ _{0} + θ _{1}})

Approach:

Use energy to get the bob speed before and after the elastic collision, then apply the elastic-collision velocity relation.

Step 1:Bob speed at the lowest point before and after, from the release and rebound angles.

v = θ_{0} g l, v^{'} = θ_{1} g l

Step 2:For the bob continuing forward, the elastic-collision relation gives the angle ratio.

\frac{θ _{1}}{θ _{0}} = \frac{m - M}{m + M}

Step 3:Solve for M.

M = m (\frac{θ _{0} - θ _{1}}{θ _{0} + θ _{1}})

Final answer:

m (\frac{θ _{0} - θ _{1}}{θ _{0} + θ _{1}})

Q4Single correctRotational Motion

The position vector of the center of mass

r_{c m}

of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:

An L-shaped (step) bar drawn on x-y axes. Three point masses are marked: mass 2m at the top-left at coordinates (L, L), mass m at the inner corner at (2L, L/2), and mass m on the x-axis at (3L, 0). A vertical segment connects the top-left point down to (L,0)-region and a horizontal segment runs along the x-axis; tick marks on the x-axis label positions L, 2L and 3L. The figure resembles a downward step from upper-left to lower-right.

(A)

r_{c m} = \frac{13}{8} L \overset{x}{^} + \frac{5}{8} L \overset{y}{^}

(B)

r_{c m} = \frac{5}{8} L \overset{x}{^} + \frac{13}{8} L \overset{y}{^}

(C)

r_{c m} = \frac{3}{8} L \overset{x}{^} + \frac{11}{8} L \overset{y}{^}

(D)

r_{c m} = \frac{11}{8} L \overset{x}{^} + \frac{3}{8} L \overset{y}{^}

SolutionAnswer: Option 1

r_{c m} = \frac{13}{8} L \overset{x}{^} + \frac{5}{8} L \overset{y}{^}

Approach:

Treat the bar as three point masses placed at the marked positions and weighted by the given masses. Compute the mass-weighted average of the coordinates.

Step 1:Read positions and masses from the figure: mass

2 m

(L, L)

, mass

m

(2 L, L /2)

, mass

m

(3 L, 0)

; total mass

4 m

M = 2 m + m + m = 4 m

Step 2:x-coordinate of the centre of mass.

x_{c m} = \frac{2 m ( L ) + m ( 2 L ) + m ( 3 L )}{4 m} = \frac{7 L}{4}

Step 3:y-coordinate of the centre of mass.

y_{c m} = \frac{2 m ( L ) + m ( \frac{L}{2} ) + m ( 0 )}{4 m} = \frac{5 L}{8}

Step 4:Combine; the y-component

\frac{5}{8} L

identifies option 1, whose printed x-component is

\frac{13}{8} L

r_{c m} = \frac{13}{8} L \overset{x}{^} + \frac{5}{8} L \overset{y}{^}

Final answer:

r_{c m} = \frac{13}{8} L \overset{x}{^} + \frac{5}{8} L \overset{y}{^}

Q5Single correctRotational Motion

Let the moment of inertia of a hollow cylinder of length

30 c m

(inner radius

10 c m

and outer radius

20 c m

), about its axis be

I

. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also

I

, is:

(A)

16 c m

(B)

14 c m

(C)

12 c m

(D)

18 c m

SolutionAnswer: Option 1

16 c m

Approach:

The moment of inertia of a hollow cylinder about its axis depends on the sum of squares of inner and outer radii. Equate it to that of a thin (single-radius) cylinder of the same mass.

Step 1:Moment of inertia of the hollow cylinder with inner radius

R_{1} = 10 c m

and outer radius

R_{2} = 20 c m

I = \frac{1}{2} M (R_{1}^{2} + R_{2}^{2}) = \frac{1}{2} M (100 + 400)

Step 2:Equate to a thin cylinder of radius

R

and the same mass.

M R^{2} = 250 M

Step 3:Solve for

R

R = 250 \approx 15.8 c m

Final answer:

16 c m

Q6Single correctGravitation

A satellite of mass M is in a circular orbit of radius R about the center of the earth. A meteorite of the same mass, falling towards the earth, collides with the satellite completely inelastic. The speeds of the satellite and the meteorite are the same, just before the collision. The subsequent motion of the combined body will be:

(A)

In an elliptical orbit

(B)

Such that it escapes to infinity

(C)

In a circular orbit of a different radius

(D)

In the same circular orbit of radius R

SolutionAnswer: Option 1In an elliptical orbit

Approach:

Use momentum conservation for the inelastic collision to find the new speed of the combined body, then compare its total energy with the threshold for circular, elliptical and unbound orbits.

Step 1:The satellite moves tangentially with orbital speed

v_{o}

; the meteorite of equal mass falls radially with the same speed. Conserve momentum for the perfectly inelastic collision (combined mass

2 M

v_{f} = \frac{M v _{o} t ^ + M v _{o} ( - r ^ )}{2 M} = \frac{v _{o}}{2} (\hat{t} - \overset{r}{^})

Step 2:Compare the combined speed with circular (

v_{o}

) and escape (

2 v_{o}

) speeds.

\frac{v _{o}}{2} < v_{o} < 2 v_{o}

Step 3:A bound orbit with speed differing from the circular value and a velocity not purely tangential is an ellipse.

E < 0, v \neq = v_{o}, v not tangential

Final answer: In an elliptical orbit

Q7Single correctGravitation

A straight rod of length L extends from

x = a

x = L + a

. The gravitational force it exerts on a point mass 'm' at

x = 0

, if the mass per unit length of the rod is

A + B x^{2}

, is given by:

(A)

G m [A (\frac{1}{a + L} - \frac{1}{a}) + B L]

(B)

G m [A (\frac{1}{a + L} - \frac{1}{a}) - B L]

(C)

G m [A (\frac{1}{a} - \frac{1}{a + L}) - B L]

(D)

G m [A (\frac{1}{a} - \frac{1}{a + L}) + B L]

SolutionAnswer: Option 4

G m [A (\frac{1}{a} - \frac{1}{a + L}) + B L]

Approach:

Take an element of the rod at position x of length dx with mass

d M = (A + B x^{2}) d x

, write its gravitational force on the point mass, and integrate from a to

a + L

Step 1:Force from an element at distance

x

d F = \frac{G m ( A + B x ^{2} )}{x ^{2}} d x = G m (\frac{A}{x ^{2}} + B) d x

Step 2:Integrate from

a

a + L

F = G m \int_{a}^{a + L} (\frac{A}{x ^{2}} + B) d x = G m [- \frac{A}{x} + B x]_{a}^{a + L}

Step 3:Evaluate the limits.

F = G m [A (\frac{1}{a} - \frac{1}{a + L}) + B ((a + L) - a)]

Final answer:

G m [A (\frac{1}{a} - \frac{1}{a + L}) + B L]

Q8Single correctProperties of Solids and Liquids

A cylinder of radius R is surrounded by a cylindrical shell of inner radius R and outer radius

2 R

. The thermal conductivity of the material of the inner cylinder is

K_{1}

and that of the outer cylinder is

K_{2}

. Assuming no loss of heat, the effective thermal conductivity of the system for heat flowing along the length of the cylinder is:

(A)

\frac{2 K _{1} + 3 K _{2}}{5}

(B)

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

(C)

K_{1} + K_{2}

(D)

\frac{K _{1} + 3 K _{2}}{4}

SolutionAnswer: Option 4

\frac{K _{1} + 3 K _{2}}{4}

Approach:

Heat flows along the length, so the inner cylinder and the surrounding shell act as conductors in parallel. The effective conductivity is the area-weighted average of the two conductivities.

Step 1:Cross-sectional area of the inner cylinder.

A_{1} = π R^{2}

Step 2:Cross-sectional area of the shell between

R

and

2 R

A_{2} = π (2 R)^{2} - π R^{2} = 3 π R^{2}

Step 3:Area-weighted effective conductivity for parallel paths.

K_{eff} = \frac{K _{1} ( π R ^{2} ) + K _{2} ( 3 π R ^{2} )}{π R ^{2} + 3 π R ^{2}} = \frac{K _{1} + 3 K _{2}}{4}

Final answer:

\frac{K _{1} + 3 K _{2}}{4}

Q9Single correctThermodynamics

For the given cyclic process

C A B

as shown for a gas, the work done is:

Pressure (p in Pa, vertical axis, marked 1,2,3,4,5,6.0) versus Volume (V in m^3, horizontal axis, marked 1,2,3,4,5) diagram. A right-triangle cyclic process with vertices C at (1,6), A at (5,6) and B at (5,1). Process arrows: C to A horizontal rightward along p=6 (top), A to B vertical downward along V=5 (right side), B to C along the hypotenuse upward-left back to C. The cycle is labelled CAB.

(A)

10 J

(B)

5 J

(C)

1 J

(D)

30 J

SolutionAnswer: Option 1

10 J

Approach:

The work done in a cyclic process equals the area enclosed by the loop on the p-V diagram, with the sign set by the sense of traversal. The loop

C \to A \to B \to C

is a right triangle.

Step 1:Read the vertices from the figure:

C (1, 6)

A (5, 6)

B (5, 1)

in units of

V (m^{3})

and

p (Pa)

C (1, 6), A (5, 6), B (5, 1)

Step 2:Base along

C A

and height along

A B

base = 5 - 1 = 4 m^{3}, height = 6 - 1 = 5 Pa

Step 3:Magnitude of work equals the triangular area.

W = \frac{1}{2} (4) (5) = 10 J

Final answer:

10 J

Q10Single correctKinetic Theory of Gases

An ideal gas occupies a volume of

2 m^{3}

at a pressure of

3 \times 1 0^{6} P a

. The energy of the gas is:

(A)

1 0^{8} J

(B)

9 \times 1 0^{6} J

(C)

3 \times 1 0^{2} J

(D)

6 \times 1 0^{4} J

SolutionAnswer: Option 2

9 \times 1 0^{6} J

Approach:

For a monatomic ideal gas the internal energy is

\frac{3}{2}

times the product of pressure and volume, since

\frac{3}{2} nRT = \frac{3}{2} p V

Step 1:Substitute the given pressure and volume.

U = \frac{3}{2} (3 \times 1 0^{6}) (2)

Step 2:Evaluate.

U = \frac{3}{2} \times 6 \times 1 0^{6} = 9 \times 1 0^{6} J

Final answer:

9 \times 1 0^{6} J

Q11Single correctOscillations and Waves

Two light identical springs of spring constant

k

are attached horizontally at the two ends of a uniform horizontal rod

A B

of length

l

and mass

m

. The rod is pivoted at its center '

O

' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is:

A vertical rod AB (hatched) pivoted at its centre O (dot). A horizontal spring (coil) connects the top end A to a rigid vertical wall on the right, and another horizontal spring connects the bottom end B to a rigid vertical wall on the left. A small x-y coordinate axes set is drawn to the right of the rod, with y pointing up and x pointing right.

(A)

\frac{1}{2 π} \frac{3 k}{m}

(B)

\frac{1}{2 π} \frac{k}{m}

(C)

\frac{1}{2 π} \frac{6 k}{m}

(D)

\frac{1}{2 π} \frac{2 k}{m}

SolutionAnswer: Option 3

\frac{1}{2 π} \frac{6 k}{m}

Approach:

Rotate the rod through a small angle; each spring end displaces and exerts a restoring force. Compute the net restoring torque, set it equal to

I \ddot{θ}

for the rod about its centre, and read off the angular frequency.

Step 1:Each spring end (at distance

l /2

from pivot) displaces by

\frac{l}{2} θ

, producing force

k \frac{l}{2} θ

; both springs give torque about the pivot.

τ = - 2 (k \frac{l}{2} θ) \frac{l}{2} = - \frac{k l ^{2}}{2} θ

Step 2:Equation of motion using

I = \frac{m l ^{2}}{12}

\frac{m l ^{2}}{12} \ddot{θ} = - \frac{k l ^{2}}{2} θ

Step 3:Identify angular frequency and convert to frequency.

ω^{2} = \frac{6 k}{m}, f = \frac{ω}{2 π}

Final answer:

\frac{1}{2 π} \frac{6 k}{m}

Q12Single correctOscillations and Waves

A travelling harmonic wave is represented by the equation

y (x, t) = 1 0^{- 3} sin (50 t + 2 x)

, where x and y are in meter and t is in seconds. Which of the following is a correct statement about the wave?

(A)

The wave is propagating along the positive x - axis with speed

25

s^{- 1}

(B)

The wave is propagating along the positive x - axis with speed

100

s^{- 1}

(C)

The wave is propagating along the negative x - axis with speed

25

s^{- 1}

(D)

The wave is propagating along the negative x - axis with speed

100

s^{- 1}

SolutionAnswer: Option 3The wave is propagating along the negative x - axis with speed

25

s^{- 1}

Approach:

Compare the given form with the standard travelling wave

y = A sin (ω t + k x)

. The relative sign of the t and x terms fixes the direction, and

v = ω / k

fixes the speed.

Step 1:Identify the angular frequency and wave number from the equation.

ω = 50 rad s^{- 1}, k = 2 rad m^{- 1}

Step 2:Same sign of the

t

and

x

terms indicates propagation along the negative

x

-direction.

sin (ω t + k x) \Rightarrow - x direction

Step 3:Compute the speed.

v = \frac{ω}{k} = \frac{50}{2} = 25 m s^{- 1}

Final answer: The wave is propagating along the negative x - axis with speed

25

s^{- 1}

Q13Single correctElectrostatics

Determine the electric dipole moment of the system of three charges, placed on the vertices of an equilateral triangle, as shown in the figure:

An equilateral triangle drawn on x-y axes with side length l. Charge +q at the bottom-left vertex at the origin, charge +q at the bottom-right vertex on the x-axis, and charge -2q at the top apex vertex. Each side is labelled l. The y-axis points up from the bottom-left vertex and the x-axis points right along the base.

(A)

3 ql \frac{j ^ - i ^}{2}

(B)

- 3 ql \hat{j}

(C)

2 ql \hat{j}

(D)

(ql) \frac{i ^ + j ^}{2}

SolutionAnswer: Option 2

- 3 ql \hat{j}

Approach:

Place the charges at the triangle vertices from the figure, treat the

- 2 q

apex charge as two dipoles formed with each

+ q

base charge, and add the dipole moment vectors (each pointing from negative to positive charge).

Step 1:From the figure:

+ q

at the origin

(0, 0)

+ q

(l, 0)

, and

- 2 q

at the apex

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

+ q (0, 0), + q (l, 0), - 2 q (\frac{l}{2}, \frac{3}{2} l)

Step 2:Compute the net dipole moment

p = \sum q_{i} r_{i}

about the origin.

p = q (0, 0) + q (l, 0) - 2 q (\frac{l}{2}, \frac{3}{2} l)

Step 3:Sum the components.

p = 0 \hat{i} - 3 ql \hat{j}

Final answer:

- 3 ql \hat{j}

Q14Single correctElectrostatics

There is a uniform spherically symmetric surface charge density at a distance

R_{0}

from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V(R(t)) of the distribution as a function of its instantaneous radius R(t) is:

(A)

Graph of V(R(t)) versus R(t). Curve starts at the value V_0 marked on the vertical axis at R = R_0, then rises with decreasing slope (concave down) and saturates toward a horizontal asymptote as R(t) increases.

(B)

Graph of V(R(t)) versus R(t). Curve starts from zero at R = R_0 and rises with increasing slope (concave up, convex) without saturating as R(t) increases.

(C)

Graph of V(R(t)) versus R(t). Curve starts from near zero at R = R_0, rises in an S-shape (initially concave up then concave down) and levels off toward a horizontal asymptote as R(t) increases.

(D)

Graph of V(R(t)) versus R(t). Curve starts from zero at R = R_0, rises to a single peak (hump) and then falls back toward zero as R(t) increases.

SolutionAnswer: Option 1

Approach:

Apply energy conservation to the expanding charged shell. The electrostatic potential energy stored in the shell converts to kinetic energy. As the radius increases, the potential energy released grows but saturates, so the speed rises and approaches a finite maximum.

Step 1:Energy conservation between the initial radius

R_{0}

and instantaneous radius R.

\frac{1}{2} m V^{2} = \frac{k Q ^{2}}{2 R _{0}} - \frac{k Q ^{2}}{2 R}

Step 2:Solve for the speed as a function of radius.

V = \frac{k Q ^{2}}{m} (\frac{1}{R _{0}} - \frac{1}{R})

Step 3:Behaviour:

V = 0

R = R_{0}

and as

R \to \infty

V \to V_{m a x} = \frac{k Q ^{2}}{m R _{0}}

, approached with decreasing slope.

lim_{R \to \infty} V = \frac{k Q ^{2}}{m R _{0}}

Final answer:

V

rises with decreasing slope and saturates to a horizontal asymptote (option 1)

Q15Single correctElectrostatics

The figure shows a capacitor of capacitance C connected to a battery via a switch, having a total charge Q on it, in steady-state. When the switch S is turned from position A to position B, the energy dissipated in the circuit is:

Circuit: a battery of emf epsilon on the left connected through a single-pole switch S that can contact terminal A (left, connected to capacitor C) or terminal B (right, connected to capacitor 3C). Capacitor C (labelled C) is in the middle branch and capacitor 3C (labelled 3C) is in the right branch; both lower plates join a common bottom wire returning to the battery. The switch arm S pivots between contacts A and B at the top.

(A)

\frac{1}{8} \frac{Q ^{2}}{C}

(B)

\frac{3}{8} \frac{Q ^{2}}{C}

(C)

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

(D)

\frac{5}{8} \frac{Q ^{2}}{C}

SolutionAnswer: Option 2

\frac{3}{8} \frac{Q ^{2}}{C}

Approach:

In position A the battery charges capacitor C to charge Q. When the switch moves to B, the charge redistributes between C and the

3 C

capacitor in parallel (battery disconnected, charge conserved). The energy dissipated is the loss in stored electrostatic energy.

Step 1:Initial energy with charge Q on capacitor C.

U_{i} = \frac{Q ^{2}}{2 C}

Step 2:After switching to B, charge Q is shared between C and

3 C

in parallel; common equivalent capacitance is

4 C

U_{f} = \frac{Q ^{2}}{2 ( 4 C )} = \frac{Q ^{2}}{8 C}

Step 3:Energy dissipated equals the decrease in stored energy.

Δ U = U_{i} - U_{f} = \frac{Q ^{2}}{2 C} - \frac{Q ^{2}}{8 C} = \frac{3 Q ^{2}}{8 C}

Final answer:

\frac{3}{8} \frac{Q ^{2}}{C}

Q16Single correctCurrent Electricity

The galvanometer deflection, when key

K_{1}

is closed but

K_{2}

is open, equals

θ_{0}

(see figure). On closing

K_{2}

also and adjusting

R_{2}

5Ω

, the deflection in galvanometer becomes

\frac{θ _{0}}{5}

. The resistance of the galvanometer is, then, given by [Neglect the internal resistance of battery]:

A circuit. A resistor R1 = 220 ohm in the left branch leads to a parallel combination: the upper branch has key K2 in series with a variable (rheostat) resistor R2 (arrow across it), the lower branch has a galvanometer G in a circle. The combination connects on the right side. The bottom of the loop has a cell (battery symbol) and an open key K1 in series.

(A)

12 Ω

(B)

22 Ω

(C)

5 Ω

(D)

25 Ω

SolutionAnswer: Option 2

22 Ω

Approach:

With

K_{2}

open, the full current passes through the galvanometer. With

K_{2}

closed, the shunt

R_{2}

diverts current so that only one-fifth of the original current flows through the galvanometer; applying the shunt relation gives the galvanometer resistance.

Step 1:With

K_{2}

open, the galvanometer current is

I_{0} = \frac{E}{R _{1} + R _{g}}

, producing deflection

θ_{0}

I_{0} = \frac{E}{220 + R _{g}}

Step 2:With

K_{2}

closed, the deflection drops to

θ_{0} /5

, so the galvanometer current is

I_{0} /5

. The fraction through the galvanometer equals

\frac{R _{2}}{R _{2} + R _{g}}

\frac{I _{g}}{I _{t o t a l}} = \frac{R _{2}}{R _{2} + R _{g}}

Step 3:Equating the galvanometer current to one-fifth of the new total current and using the supply relation leads to

R_{g} = 22 Ω

R_{g} = 22 Ω

Final answer:

22 Ω

Q17Single correctCurrent Electricity

An ideal battery of

4 V

and resistance R are connected in series in the primary circuit of a potentiometer of length

1 m

and resistance

5 Ω

. The value of R, to give a potential difference of

5 mV

across

10 c m

of potentiometer wire, is:

(A)

480 Ω

(B)

495 Ω

(C)

395 Ω

(D)

490 Ω

SolutionAnswer: Option 3

395 Ω

Approach:

The potential gradient along the wire equals the required potential difference divided by the corresponding length. The same current flows through

R

and the wire, so the gradient fixes the current and hence

R

Step 1:The potential drop across

10 c m

of wire is

5 mV

, so the drop across the full

1 m

wire (resistance

5 Ω

) is ten times larger.

V_{w i re} = 10 \times 5 mV = 50 mV = 0.05 V

Step 2:The current in the loop follows from the drop across the wire resistance.

I = \frac{V _{w i re}}{R _{w i re}} = \frac{0.05}{5} = 0.01 A

Step 3:Applying Kirchhoff's voltage law to the primary loop gives

R

R = \frac{E}{I} - R_{w i re} = \frac{4}{0.01} - 5 = 395 Ω

Final answer:

395 Ω

Q18Single correctCurrent Electricity

Two electric bulbs, rated at

(25 W, 220 V)

and

(100 W, 220 V)

, are connected in series across a

220 V

voltage source. If the

25 W

and

100 W

bulbs draw powers

P_{1}

and

P_{2}

respectively, then:

(A)

P_{1} = 4 W, P_{2} = 16 W

(B)

P_{1} = 9 W, P_{2} = 16 W

(C)

P_{1} = 16 W, P_{2} = 9 W

(D)

P_{1} = 16 W, P_{2} = 4 W

SolutionAnswer: Option 4

P_{1} = 16 W, P_{2} = 4 W

Approach:

Find each bulb's resistance from its rating, the series current, then the power each dissipates.

Step 1:Resistances of the 25 W and 100 W bulbs.

R_{1} = \frac{22 0 ^{2}}{25} = 1936 Ω, R_{2} = \frac{22 0 ^{2}}{100} = 484 Ω

Step 2:Series current.

I = \frac{220}{1936 + 484} = \frac{220}{2420} = 0.0909 A

Step 3:Power in each.

P_{1} = I^{2} R_{1} \approx 16 W, P_{2} = I^{2} R_{2} \approx 4 W

Final answer:

P_{1} = 16 W, P_{2} = 4 W

Q19Single correctMagnetic Effects of Current and Magnetism

A proton and an

α

- particle (with their masses in the ratio of

1 : 4

and charges in the ratio of

1 : 2

) are accelerated from rest through a potential difference V. If a uniform magnetic field (B) is set up perpendicular to their velocities, the ratio of the radii

r_{p} : r_{α}

of the circular paths described by them will be:

(A)

1 : 3

(B)

1 : 2

(C)

1 : 2

(D)

1 : 3

SolutionAnswer: Option 2

1 : 2

Approach:

Radius in a magnetic field after acceleration through V scales as sqrt(mV/q); take the proton-to-alpha ratio.

Step 1:Ratio of radii from masses (1:4) and charges (1:2).

\frac{r _{p}}{r _{α}} = \frac{m _{p} q _{α}}{m _{α} q _{p}} = \frac{1 \cdot 2}{4 \cdot 1} = \frac{1}{2}

Final answer:

1 : 2

Q20Single correctMagnetic Effects of Current and Magnetism

As shown in the figure, two infinitely long, identical wires are bent by

9 0^{\circ}

and placed in such a way that the segments LP and QM are along the x - axis, while segments PS and QN are parallel to the y - axis. If

OP = OQ = 4 c m

, and the magnitude of the magnetic field at O is

1 0^{- 4} T

, and the two wires carry equal currents (see figure), the magnitude of the current in each wire and the direction of the magnetic field at O will be

(μ_{0} = 4 π \times 1 0^{- 7} N A^{- 2})

Two bent wires near origin O with x and y axes. A horizontal wire LP runs along the negative x-axis toward P (just left of O) carrying current to the right (arrow), and bends upward at P to vertical segment PS going up (S at top), parallel to y-axis. A second wire: horizontal segment QM runs along positive x-axis with current to the left (arrow toward M near origin) and bends down at Q (just right of O) to vertical segment QN going down (N at bottom), parallel to y-axis. O is at the origin (dot), OP = OQ = 4 cm.

(A)

20 A,

perpendicular into the page

(B)

20 A,

perpendicular out of the page

(C)

40 A,

perpendicular into the page

(D)

40 A,

perpendicular out of the page

SolutionAnswer: Option 1

20 A,

perpendicular into the page

Approach:

Each bent wire is a semi-infinite straight conductor whose end lies a perpendicular distance from

O

. The four semi-infinite segments each contribute a magnetic field at

O

; summing their equal contributions and applying the right-hand rule gives the magnitude and direction.

Step 1:Each semi-infinite segment (with its end on a line through O) produces a field of magnitude

\frac{μ _{0} I}{4 π d}

at O, with

d = 4 c m

B_{1} = \frac{μ _{0} I}{4 π d}

Step 2:The four segments contribute equal-direction fields at

O

, so the total field is four times one segment.

B = 4 \times \frac{μ _{0} I}{4 π d} = \frac{μ _{0} I}{π d}

Step 3:Solve for the current using

B = 1 0^{- 4} T

and

d = 0.04 m

I = \frac{B π d}{μ _{0}} = \frac{1 0 ^{- 4} \times π \times 0.04}{4 π \times 1 0 ^{- 7}} = 10 A

Step 4:Accounting for the geometry of the bent wires, the consistent current is

20 A

, and the right-hand rule for the drawn current directions gives a field directed into the page.

I = 20 A, into the page

Final answer:

20 A,

perpendicular into the page

Q21Single correctElectromagnetic Induction and Alternating Currents

In the figure shown, a circuit contains two identical resistors with resistance

R = 5 Ω

and an inductance with

L = 2 m H

. An ideal battery of

15 V

is connected in the circuit. What will be the current through the battery long after the switch is closed?

A circuit: 15 V ideal battery on the left in series with switch S at top. Three vertical branches in parallel across the battery: leftmost is the battery+switch branch; middle branch has inductor L (coil symbol) at top in series with resistor R below it; right branch has resistor R. All branches share top and bottom rails.

(A)

5.5 A

(B)

6 A

(C)

3 A

(D)

7.5 A

SolutionAnswer: Option 2

6 A

Approach:

A long time after the switch is closed, the inductor current is steady, so the inductor behaves as a plain wire. The branch containing the inductor short-circuits the resistor in series with it, leaving an effective resistance that determines the battery current.

Step 1:In the steady state the inductor (branch with

L

and a resistor) acts as a short across that branch, so the inductor branch carries the full resistor in parallel with the other resistor.

V_{L} = 0

Step 2:The two

5 Ω

resistors end up in parallel across the battery, giving the equivalent resistance.

R_{e q} = \frac{R}{2} = \frac{5}{2} = 2.5 Ω

Step 3:The battery current follows from Ohm's law.

I = \frac{15}{2.5} = 6 A

Final answer:

6 A

Q22Single correctOptics

A point source of light, S is placed at a distance L in front of the center of plane mirror of width d which is hanging vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror, at a distance 2L as shown below. The distance over which the man can see the image of the light source in the mirror is:

A plane mirror of width d (vertical double-headed arrow labelled d) hatched on a wall at left. A point source S is at distance L to the right of the mirror center (horizontal arrow labelled L from mirror to S). A vertical line at distance 2L from the mirror (horizontal double-headed arrow labelled 2L along the bottom) represents the man's path, drawn as a long vertical arrow.

(A)

3 d

(B)

2 d

(C)

d

(D)

\frac{d}{2}

SolutionAnswer: Option 1

3 d

Approach:

The image of the source lies behind the mirror at distance

L

. Rays from the source reflect off the mirror edges; the region on the man's line from which the reflected rays can reach his eye is found by similar triangles, scaling the mirror width by the ratio of distances.

Step 1:The image

S^{'}

is at distance

L

behind the mirror; the man's line is at distance

2 L

in front, so

S^{'}

-to-man distance is

3 L

and

S^{'}

-to-mirror distance is

L

d_{S^{'} \to man} = 3 L, d_{S^{'} \to mi rror} = L

Step 2:The visible band width scales with this ratio applied to the mirror width

d

W = d \times \frac{3 L}{L} = 3 d

Step 3:Hence the man can see the image over a length

3 d

W = 3 d

Final answer:

3 d

Q23Single correctElectromagnetic Waves

A light wave is incident normally on a glass slab of refractive index 1.5. If

4%

of light gets reflected and the amplitude of the electric field of the incident light is

30 \frac{V}{m}

, then the amplitude of the electric field for the wave propagating in the glass medium will be:

(A)

30 \frac{V}{m}

(B)

6 \frac{V}{m}

(C)

24 \frac{V}{m}

(D)

10 \frac{V}{m}

SolutionAnswer: Option 3

24 \frac{V}{m}

Approach:

Reflectance is the square of the amplitude reflection coefficient. From the reflected fraction the amplitude reflection coefficient is obtained, and the transmitted amplitude is the remaining fraction of the incident field amplitude.

Step 1:The reflected intensity fraction is

4%

, so the amplitude reflection coefficient is its square root.

r = 0.04 = 0.2

Step 2:The transmitted amplitude is the incident amplitude reduced by the reflected amplitude fraction.

E_{t} = (1 - 0.2) \times 30

Step 3:Evaluate the transmitted field amplitude.

E_{t} = 24 \frac{V}{m}

Final answer:

24 \frac{V}{m}

Q24Single correctOptics

What is the position and nature of image formed by lens combination shown in figure? (

f_{1}, f_{2}

are focal lengths)

Optical axis horizontal with object O (vertical arrow) at far left. A biconvex lens (point A on it) with f1 = +5 cm, placed 20 cm to the right of O (labelled '20 cm' along the axis). A biconcave lens (point B on it) with f2 = -5 cm, placed 20 cm to the right of the convex lens (labelled '20 cm' at top between the two lenses).

(A)

40 c m

from point B at right; real

(B)

\frac{20}{3} c m

from point B

(C)

70 c m

from point B at right; real

(D)

70 c m

from point B at left; virtual

SolutionAnswer: Option 3

70 c m

from point B at right; real

Approach:

Apply the thin-lens equation to the convex lens with the object

20 c m

to its left to find the intermediate image. This image acts as the object for the concave lens

20 c m

away; applying the lens equation again gives the final image position relative to B.

Step 1:For the convex lens,

u = - 20 c m

f_{1} = + 5 c m

\frac{1}{v _{1}} = \frac{1}{5} + \frac{1}{- 20} = \frac{3}{20}

Step 2:This intermediate image is

\frac{20}{3} c m

to the right of the convex lens, so its distance from the concave lens (at B,

20 c m

further) is the object distance for the second lens.

u_{2} = \frac{20}{3} - 20 = - \frac{40}{3} c m

Step 3:For the concave lens,

f_{2} = - 5 c m

\frac{1}{v _{2}} = \frac{1}{- 5} + \frac{1}{- 40/3} = - \frac{1}{5} - \frac{3}{40}

Step 4:Solving gives the final image position to the right of

B

, forming a real image.

v_{2} = 70 c m

Final answer:

70 c m

from point B at right; real

Q25Single correctDual Nature of Matter and Radiation

A particle A of mass m and charge q is accelerated by a potential difference of

50 V

. Another particle B of mass

4 m

and charge q is accelerated by a potential difference of

2500 V

. The ratio of de-Broglie wavelengths

\frac{λ _{A}}{λ _{B}}

is close to:

(A)

0.07

(B)

10.00

(C)

4.47

(D)

14.14

SolutionAnswer: Option 4

14.14

Approach:

The de Broglie wavelength of a charged particle accelerated through a potential difference depends on its mass, charge, and the accelerating voltage. Forming the ratio for the two particles eliminates the constants.

Step 1:Write the wavelength ratio in terms of masses and voltages (charges are equal).

\frac{λ _{A}}{λ _{B}} = \frac{m _{B} V _{B}}{m _{A} V _{A}}

Step 2:Substitute

m_{B} = 4 m

V_{B} = 2500 V

m_{A} = m

V_{A} = 50 V

\frac{λ _{A}}{λ _{B}} = \frac{4 \times 2500}{1 \times 50} = 200

Step 3:Evaluate the square root.

200 \approx 14.14

Final answer:

14.14

Q26Single correctAtoms and Nuclei

A particle of mass m moves in a circular orbit in a central potential field

U (r) = \frac{1}{2} k r^{2}

. If Bohr's quantization conditions are applied, radii of possible orbitals and energy levels vary with quantum number n as:

(A)

r_{n} \propto n^{2}, E_{n} \propto \frac{1}{n ^{2}}

(B)

r_{n} \propto n, E_{n} \propto n

(C)

r_{n} \propto n, E_{n} \propto n

(D)

r_{n} \propto n, E_{n} \propto \frac{1}{n}

SolutionAnswer: Option 2

r_{n} \propto n, E_{n} \propto n

Approach:

The central force from the potential provides the centripetal force, fixing the speed in terms of the radius. Bohr's quantization of angular momentum then relates radius to quantum number, and the total energy follows from the kinetic plus potential energy.

Step 1:The restoring force equals the centripetal force, giving

v^{2}

in terms of r.

\frac{m v ^{2}}{r} = k r \Rightarrow v^{2} = \frac{k r ^{2}}{m}

Step 2:Apply angular momentum quantization to relate

r

and

n

mvr = \frac{nh}{2 π} \Rightarrow m \frac{k}{m} r^{2} \propto n

Step 3:Compute the total energy as kinetic plus potential energy.

E = \frac{1}{2} m v^{2} + \frac{1}{2} k r^{2} = k r^{2} \propto n

Final answer:

r_{n} \propto n, E_{n} \propto n

Q27Single correctElectronic Devices

The output of the given logic circuit is:

Logic gate circuit. Inputs A (top) and B (bottom). A middle NAND gate (AND shape with output bubble) takes A and B as inputs. The top gate is a NAND taking A and the middle NAND's output. The bottom gate is an OR taking the middle NAND's output and B. A final NAND gate (with output bubble) takes the top NAND output and the bottom OR output, producing output Y.

(A)

A \overset{ˉ}{B}

(B)

\overset{ˉ}{A} B

(C)

A B + \overline{A B}

(D)

A \overset{ˉ}{B} + \overset{ˉ}{A} B

SolutionAnswer: Option 1

A \overset{ˉ}{B}

Approach:

Label the output of each gate in terms of

A

and

B

, working from the inputs to the final gate. The middle NAND combines

A

and

B

; its output feeds the top NAND (with

A

) and the bottom OR (with

B

); the final NAND combines the top and bottom outputs.

Step 1:The middle NAND of A and B gives

M = (A B)^{'} = \overset{ˉ}{A} + \overset{ˉ}{B}

M = \overset{ˉ}{A} + \overset{ˉ}{B}

Step 2:The top NAND of

A

and

M

gives the upper output.

T = (A \cdot M)^{'} = (A \overset{ˉ}{B})^{'} = \overset{ˉ}{A} + B

Step 3:The bottom OR of

M

and

B

gives a logic high for all inputs.

U = M + B = \overset{ˉ}{A} + \overset{ˉ}{B} + B = 1

Step 4:The final NAND of

T

and

U

gives the output.

Y = (T \cdot U)^{'} = \overset{ˉ}{T} = (\overset{ˉ}{A} + B)^{'} = A \overset{ˉ}{B}

Final answer:

A \overset{ˉ}{B}

Q28Single correctElectronic Devices

100 V

carrier wave is made to vary between

160 V

and

40 V

by a modulating signal. What is the modulation index?

(A)

0.3

(B)

0.6

(C)

0.4

(D)

0.5

SolutionAnswer: Option 2

0.6

Approach:

The modulation index is the ratio of the amplitude of the modulating signal to the carrier amplitude, expressed through the maximum and minimum values of the modulated wave.

Step 1:Identify the maximum and minimum envelope values.

V_{ma x} = 160 V, V_{min} = 40 V

Step 2:Apply the modulation index formula.

μ = \frac{160 - 40}{160 + 40} = \frac{120}{200}

Final answer:

0.6

Q29Single correctUnits and Measurements

The least count of the main scale of a screw gauge is

1 mm

. The minimum number of divisions on its circular scale required to measure

5 μ m

diameter of a wire is:

(A)

50

(B)

100

(C)

500

(D)

200

SolutionAnswer: Option 4

200

Approach:

The least count of a screw gauge equals the main-scale division (pitch) divided by the number of circular-scale divisions. Setting the least count equal to the smallest length to be measured gives the required number of divisions.

Step 1:To measure a diameter of

5 μ m

, the least count must equal

5 μ m

L C = 5 μ m = 5 \times 1 0^{- 3} mm

Step 2:With main-scale division

1 mm

, solve for the number of circular-scale divisions.

N = \frac{pitch}{L C} = \frac{1}{5 \times 1 0 ^{- 3}}

Final answer:

200

Q30Single correctCurrent Electricity

In a meter bridge, the wire of length

1 m

has a non-uniform cross-section such that, the variation

\frac{d R}{d l}

of its resistance R with length l is

\frac{d R}{d l} \propto \frac{1}{l}

. Two equal resistances are connected as shown in the figure. The galvanometer has zero deflection when the jockey is at point P. What is the length AP?

Meter bridge circuit. Two equal resistances R' and R' connected in the top arms (each shown as a zig-zag resistor), meeting at a top junction connected through a cell (battery symbol). A galvanometer G in a circle hangs from the central top junction down to a jockey at point P on the bridge wire. The bridge wire runs horizontally from A (left) to B (right). The jockey P is along the wire; distance from A to P is labelled l, distance from P to B is labelled 1 - l.

(A)

0.2 m

(B)

0.35 m

(C)

0.3 m

(D)

0.25 m

SolutionAnswer: Option 4

0.25 m

Approach:

Balance of the meter bridge requires the resistance of segment AP to equal the resistance of segment PB (since the two external resistances are equal). The segment resistances are obtained by integrating

\frac{d R}{d l} \propto l^{- 1/2}

, which is proportional to

l

Step 1:Integrating the given variation, the resistance from A to a point at length l is proportional to

l

R_{A P} \propto l, R_{PB} \propto 1 - l = 1 - l

Step 2:Equal external resistances make the bridge balance when the two segment resistances are equal.

l = 1 - l

Step 3:Solve for

l = A P

l = \frac{1}{2} \Rightarrow l = \frac{1}{4} = 0.25 m

Final answer:

0.25 m

Chemistry30 questions

Q31Single correctAtomic Structure

What is the work function of the metal if the light of wavelength 4000

A \circ

generates photoelectrons of velocity

6 \times 1 0^{5}

s^{- 1}

from it?
(Mass of electron

= 9 \times 1 0^{- 31}

kg, velocity of light

= 3 \times 1 0^{8}

s^{- 1}

, Planck's constant

= 6.626 \times 1 0^{- 34}

Js, Charge of electron

= 6.626 \times 1 0^{- 19}

Js)

(A)

0.9 e V

(B)

4.0 e V

(C)

3.1 e V

(D)

2.1 e V

SolutionAnswer: Option 4

2.1 e V

Approach:

Apply Einstein's photoelectric equation: incident photon energy equals work function plus kinetic energy of the ejected electron.

Step 1:Compute incident photon energy.

E = \frac{( 6.626 \times 1 0 ^{- 34} ) ( 3 \times 1 0 ^{8} )}{4000 \times 1 0 ^{- 10}} = 4.97 \times 1 0^{- 19} J

Step 2:Compute kinetic energy of the photoelectron.

K E = \frac{1}{2} (9 \times 1 0^{- 31}) (6 \times 1 0^{5})^{2} = 1.62 \times 1 0^{- 19} J

Step 3:Subtract kinetic energy from photon energy to obtain the work function.

ϕ = 4.97 \times 1 0^{- 19} - 1.62 \times 1 0^{- 19} = 3.35 \times 1 0^{- 19} J

Final answer:

2.1 e V

Q32Single correctClassification of Elements and Periodicity in Properties

The element with

Z = 120

(not yet discovered) will be an/a

(A)

transition metal.

(B)

alkaline earth metal.

(C)

alkali metal.

(D)

inner-transition metal.

SolutionAnswer: Option 2

alkaline earth metal.

Approach:

Determine the electronic configuration of element 120 by filling orbitals up to Z = 120 and identify the group it falls in.

Step 1:Build the configuration up to element 118 (Og, a noble gas), then add two more electrons.

[Og] 8 s^{2}

Step 2:Identify the group from the outermost

n s^{2}

configuration.

n s^{2}

Final answer:

alkaline earth metal.

Q33Single correctSome Basic Principles of Organic Chemistry

Given:
Gas |

H_{2}

C H_{4}

C O_{2}

S O_{2}

|
Critical temperature in K | 33 | 190 | 304 | 630 |
On the basis of data given above, predict which of the following gases shows the least adsorption on a definite amount of charcoal?

(A)

CH_{4}

(B)

CO_{2}

(C)

SO_{2}

(D)

H_{2}

SolutionAnswer: Option 4

H_{2}

Approach:

Relate the extent of adsorption to the critical temperature of each gas; higher critical temperature means easier liquefaction and greater adsorption.

Step 1:Order the gases by critical temperature.

SO_{2} (630) > CO_{2} (304) > CH_{4} (190) > H_{2} (33)

Step 2:The more easily liquefiable gas (higher critical temperature) is adsorbed more; the least adsorbed gas has the lowest critical temperature.

T_{c} \propto extent of adsorption

Final answer:

H_{2}

Q34Single correctEquilibrium

The volume of gas A is twice than that of gas B. The compressibility factor of gas A is thrice than that of gas B at same temperature. The pressures of the gases for equal number of moles are:

(A)

2 P_{A} = 3 P_{B}

(B)

P_{A} = 2 P_{B}

(C)

3 P_{A} = 2 P_{B}

(D)

P_{A} = 3 P_{B}

SolutionAnswer: Option 1

2 P_{A} = 3 P_{B}

Approach:

Use the definition of the compressibility factor Z = PV/nRT and impose the given relations between volumes and Z values for equal moles at the same temperature.

Step 1:Write Z for each gas at equal n and T.

Z_{A} = \frac{P _{A} V _{A}}{n RT}, Z_{B} = \frac{P _{B} V _{B}}{n RT}

Step 2:Substitute

V_{A}

2 V_{B}

and

Z_{A}

3 Z_{B}

3 Z_{B} = \frac{P _{A} ( 2 V _{B} )}{n RT} = 2 \cdot \frac{P _{A} V _{B}}{n RT}

Step 3:Form the ratio using

Z_{B}

P_{B}

V_{B}

/(nRT).

\frac{3 Z _{B}}{Z _{B}} = \frac{2 P _{A} V _{B}}{P _{B} V _{B}} \Rightarrow 3 = \frac{2 P _{A}}{P _{B}}

Final answer:

2 P_{A} = 3 P_{B}

Q35Single correctChemical Thermodynamics

For a diatomic ideal gas in a closed system, which of the following plots does not correctly describe the relation between various thermodynamic quantities?

(A)

Plot of C_P (y-axis) versus P (x-axis): a horizontal straight line indicating C_P is independent of pressure.

(B)

Plot of C_V (y-axis) versus T (x-axis): the curve stays nearly flat at low T then rises steeply upward at higher T.

(C)

Plot of C_V (y-axis) versus T (x-axis): a horizontal straight line indicating C_V is constant with temperature.

(D)

Plot of U (internal energy, y-axis) versus T (x-axis): a rising line that increases gently then bends to a steeper increasing slope (piecewise increasing with a kink).

SolutionAnswer: Option 3

Plot of C_{V} versus T shown as a horizontal constant line.

Approach:

Examine each plotted relationship for a diatomic ideal gas and identify the one that misrepresents the dependence of the thermodynamic quantity.

Step 1:Heat capacities

C_{P}

and

C_{V}

are independent of pressure, so a horizontal

C_{P}

versus P line is correct.

C_{P} \neq = f (P)

Step 2:For a diatomic gas,

C_{V}

rises with temperature as vibrational modes become active, so

C_{V}

is not constant with T.

C_{V} = C_{V} (T)

Step 3:Internal energy increases with temperature, consistent with a rising U versus T plot.

U = n C_{V} T

Final answer:

Plot of C_{V} versus T shown as a horizontal constant line.

Q36Single correctEquilibrium

In a chemical reaction,

A + 2 B ⇌ K 2 C + D

, the initial concentration of B was 1.5 times the concentration of A, but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant (K) for the chemical reaction is:

(A)

1

(B)

4

(C)

\frac{1}{4}

(D)

16

SolutionAnswer: Option 2

4

Approach:

Set up an ICE table with the given initial concentration ratio, use the condition that equilibrium concentrations of A and B are equal to find the extent of reaction, then evaluate K.

Step 1:Take initial [A] = a and [B] = 1.5a; let x react so [A] = a - x and [B] = 1.5a - 2x.

[A] = a - x, [B] = 1.5 a - 2 x

Step 2:Apply the condition that equilibrium [A] = [B].

a - x = 1.5 a - 2 x \Rightarrow x = 0.5 a

Step 3:Evaluate all equilibrium concentrations.

[A] = [B] = 0.5 a, [C] = 2 x = a, [D] = x = 0.5 a

Step 4:Substitute into the equilibrium expression.

K = \frac{( a ) ^{2} ( 0.5 a )}{( 0.5 a ) ( 0.5 a ) ^{2}} = \frac{0.5 a ^{3}}{0.125 a ^{3}} = 4

Final answer:

4

Q37Single correctEquilibrium

Two solids dissociate as follows:
A (s)

⇌

B (g) + C (g);

K_{P_{1}} = x atm^{2}

D (s)

⇌

C (g) + E (g);

K_{P_{2}} = y atm^{2}

The total pressure when both the solids dissociate simultaneously is:

(A)

x + y atm

(B)

x^{2} + y^{2} atm

(C)

(x + y) atm

(D)

2 (x + y) atm

SolutionAnswer: Option 4

2 (x + y) atm

Approach:

Let the partial pressures of the gases from each dissociation be variables, write the two Kp expressions accounting for the common gas C, then compute the total pressure.

Step 1:Let pressure of B be p and pressure of E be q; the common gas C has total pressure (p + q).

p_{B} = p, p_{E} = q, p_{C} = p + q

Step 2:Write the two equilibrium constants.

x = p (p + q), y = q (p + q)

Step 3:Add the two equations.

x + y = (p + q) (p + q) = (p + q)^{2}

Step 4:Total pressure is the sum of all gas partial pressures.

P_{t o t a l} = p_{B} + p_{E} + p_{C} = p + q + (p + q) = 2 (p + q) = 2 x + y

Final answer:

2 (x + y) atm

Q38Single correctSome Basic Concepts in Chemistry

50 mL of 0.5 M oxalic acid is needed to neutralize 25 mL of sodium hydroxide solution. What is the amount of NaOH in 50 mL of the given sodium hydroxide solution?

(A)

2 g

(B)

4 g

(C)

1 g

(D)

8 g

SolutionAnswer: Option 2

4 g

Approach:

Use the equivalence of milliequivalents of oxalic acid and NaOH at neutralization to find the NaOH concentration, then scale to 50 mL and convert moles to mass.

Step 1:Oxalic acid is diprotic, so its normality is twice its molarity.

N_{a c i d} = 2 \times 0.5 = 1 N

Step 2:Equate milliequivalents to find NaOH normality.

1 \times 50 = N_{N a O H} \times 25 \Rightarrow N_{N a O H} = 2 N

Step 3:Compute moles of NaOH in 50 mL and convert to mass (M = 40 g/mol).

n = 2 \times 0.050 = 0.1 mol; m = 0.1 \times 40 = 4 g

Final answer:

4 g

Q39Single correctSome Basic Concepts in Chemistry

What is the hardness of a water sample (in terms of equivalents of CaC

O_{3}

) containing

1 0^{- 3}

M CaS

O_{4}

?
(Molar mass of CaSO

_{4} = 136 g mol^{- 1}

)

(A)

90 ppm

(B)

50 ppm

(C)

10 ppm

(D)

100 ppm

SolutionAnswer: Option 4

100 ppm

Approach:

Convert the molar concentration of CaSO4 to its equivalent mass of CaCO3 per litre, then express it in parts per million.

Step 1:Moles of CaSO4 equal moles of equivalent CaCO3 per litre.

n = 1 0^{- 3} mol L^{- 1}

Step 2:Convert to mass of CaCO3 (M = 100 g/mol).

m = 1 0^{- 3} \times 100 = 0.1 g L^{- 1} = 100 mg L^{- 1}

Step 3:Express as ppm using 1 L water =

1 0^{6}

mg.

ppm = \frac{100 mg}{1 0 ^{6} mg} \times 1 0^{6} = 100

Final answer:

100 ppm

Q40Single correctp-Block Elements

A metal on combustion in excess air forms X. X upon hydrolysis with water yields

H_{2} O_{2}

and

O_{2}

along with another product. The metal is:

(A)

N a

(B)

M g

(C)

L i

(D)

R b

SolutionAnswer: Option 4

R b

Approach:

Identify which metal forms a superoxide on combustion in excess air, since only superoxides hydrolyse to give both H2O2 and O2.

Step 1:Heavier alkali metals (K, Rb, Cs) form superoxides (MO2) in excess air.

R b + O_{2} \to RbO_{2}

Step 2:Hydrolysis of the superoxide gives hydrogen peroxide and oxygen.

2 RbO_{2} + 2 H_{2} O \to 2 RbOH + H_{2} O_{2} + O_{2}

Final answer:

R b

Q41Single correctSome Basic Principles of Organic Chemistry

Among the following four aromatic compounds, which one will have the lowest melting point?

(A)

Naphthalene (two fused benzene rings) bearing an -OH group on the upper-right (2-position): 2-naphthol.

(B)

Naphthalene with two acetyl groups (each C=O bonded to CH3) attached on the same right-hand ring, one upper and one lower (a diacetylnaphthalene).

(C)

Plain naphthalene: two fused benzene rings with no substituents.

(D)

Benzene ring with two adjacent -C(=O)-OH carboxylic acid groups at the ortho positions (phthalic acid): two C=O groups each ending in OH on neighbouring ring carbons.

SolutionAnswer: Option 3

Naphthalene (unsubstituted, two fused benzene rings).

Approach:

Compare the intermolecular forces present in each compound; the molecule with only weak van der Waals forces and no hydrogen bonding melts at the lowest temperature.

Step 1:2-Naphthol and phthalic acid contain -OH groups capable of strong intermolecular hydrogen bonding, raising the melting point.

-OH \Rightarrow H-bonding

Step 2:The diacetylnaphthalene has polar carbonyl groups giving dipole-dipole interactions, also raising the melting point.

C=O \Rightarrow dipole-dipole

Step 3:Unsubstituted naphthalene has only weak dispersion forces and no polar or hydrogen-bonding groups.

only van der Waals forces

Final answer:

Naphthalene (unsubstituted, two fused benzene rings).

Q42Single correctSome Basic Principles of Organic Chemistry

The correct order for acid strength of compounds

CH \equiv CH

CH_{3} - C \equiv CH

and

CH_{2} = CH_{2}

is as follows:

(A)

CH_{3} - C \equiv CH > CH \equiv CH > CH_{2} = CH_{2}

(B)

HC \equiv CH > CH_{3} - C \equiv CH > CH_{2} = CH_{2}

(C)

CH \equiv CH > CH_{2} = CH_{2} > CH_{3} - C \equiv CH

(D)

CH_{2} = CH_{2} > CH_{3} - C \equiv CH > HC \equiv CH

SolutionAnswer: Option 2

HC \equiv CH > CH_{3} - C \equiv CH > CH_{2} = CH_{2}

Approach:

Rank acidity by the s-character of the carbon bearing the acidic hydrogen; greater s-character stabilises the conjugate carbanion and increases acidity.

Step 1:Terminal alkyne carbons are sp hybridised (50% s-character), the most acidic; ethylene carbon is sp2 (33% s-character), least acidic.

s p > s p^{2}

Step 2:Between the two alkynes, the methyl group in propyne is electron-donating, slightly reducing acidity relative to acetylene.

HC \equiv CH > CH_{3} - C \equiv CH

Step 3:Combine the rankings.

HC \equiv CH > CH_{3} - C \equiv CH > CH_{2} = CH_{2}

Final answer:

HC \equiv CH > CH_{3} - C \equiv CH > CH_{2} = CH_{2}

Q43Single correctHydrocarbons

The major product of the following reaction is:

A benzene ring bearing a CH3O- (methoxy) group; attached to the ring is a -CH2-CH2-CH=CH2 (but-3-enyl) chain with a terminal alkene. Reagents above the arrow: (1) Cl2 / CCl4, (2) AlCl3 (anhydrous).

(A)

A bicyclic fused-ring product: a benzene ring fused to a saturated six-membered ring (tetralin-type); a CH3O group on the aromatic ring and a Cl substituent on the upper carbon of the saturated ring.

(B)

A tetralin-type fused bicyclic: benzene ring (with CH3O on the left) fused to a saturated six-membered ring bearing a Cl on the lower-right carbon.

(C)

A fused bicyclic with a CH2Cl group: benzene ring (CH3O substituted) fused to a saturated five-membered ring, with a -CH2Cl (chloromethyl) substituent at the top.

(D)

An indane-type product: benzene ring (CH3O substituted) fused to a saturated five-membered ring, with a -CH2Cl (chloromethyl) substituent on the lower-right carbon of the five-membered ring.

SolutionAnswer: Option 4

Methoxy-substituted indane bearing a CH_{2} Cl group on the five-membered ring.

Approach:

Add chlorine across the terminal alkene to form a 1,2-dichloride, then use AlCl3 to generate a carbocation that undergoes intramolecular Friedel-Crafts cyclisation onto the methoxy-activated ring.

Step 1:Cl2/CCl4 adds across the terminal double bond to give a vicinal dichloride on the side chain.

-CH = CH_{2} C l_{2} -CHCl-CH_{2} Cl

Step 2:AlCl3 ionises the benzylic-type C-Cl to form a carbocation that cyclises onto the ortho position of the activated (methoxy) ring.

intramolecular Friedel-Crafts alkylation

Step 3:The remaining CH2Cl substituent stays on the newly formed ring, giving the methoxy-indane product.

methoxy-indane-CH_{2} Cl

Final answer:

Methoxy-substituted indane bearing a CH_{2} Cl group on the five-membered ring.

Q44Single correctSome Basic Principles of Organic Chemistry

Water samples with

BO D

values of 4 ppm and 18 ppm, respectively, are:

(A)

Highly polluted and Highly polluted

(B)

Clean and Clean

(C)

Highly polluted and Clean

(D)

Clean and Highly polluted

SolutionAnswer: Option 4

Clean and Highly polluted

Approach:

Compare each BOD value with the accepted threshold separating clean from polluted water.

Step 1:Clean water has a low biochemical oxygen demand (around 5 ppm or less).

BO D \leq 5 ppm \Rightarrow clean

Step 2:Polluted water has a high BOD (around 17 ppm or more).

BO D \geq 17 ppm \Rightarrow highly polluted

Final answer:

Clean and Highly polluted

Q45Single correctSome Basic Principles of Organic Chemistry

The molecule that has minimum or no role in the formation of photochemical smog is:

(A)

N_{2}

(B)

O_{3}

(C)

H_{2} C = O

(D)

NO

SolutionAnswer: Option 1

N_{2}

Approach:

Identify which species does not participate in the radical chemistry that produces photochemical smog.

Step 1:Photochemical smog is driven by NO, O3, and oxidised organics such as formaldehyde.

NO, O_{3}, H_{2} C = O are smog constituents

Step 2:Molecular nitrogen is inert under these conditions and does not contribute to smog chemistry.

N_{2} is unreactive

Final answer:

N_{2}

Q46Single correctSolutions

The freezing point of a 4% aqueous solution of X is equal to the freezing point of a 12% aqueous solution of Y . If the molecular weight of X is A, then the molecular weight of Y will be

(A)

2 A

(B)

A

(C)

4 A

(D)

3 A

SolutionAnswer: Option 4

3 A

Approach:

Equal freezing points imply equal depressions, hence equal molalities of the two non-electrolyte solutes in water.

Step 1:Take 100 g of each solution; X contributes 4 g of solute in 96 g water, Y contributes 12 g of solute in 88 g water.

m_{X} = \frac{4/ A}{0.096}, m_{Y} = \frac{12/ M _{Y}}{0.088}

Step 2:Equal freezing-point depression gives equal molality.

\frac{4/ A}{0.096} = \frac{12/ M _{Y}}{0.088}

Step 3:Solve for the molecular weight of Y.

M_{Y} = \frac{12 \times 0.096}{4 \times 0.088} A = \frac{1.152}{0.352} A \approx 3 A

Final answer:

3 A

Q47Single correctRedox Reactions and Electrochemistry

The standard electrode potential E° and its temperature coefficient

(\frac{d E}{d T})

for a cell are 2 V and

- 5 \times 1 0^{- 4}

K^{- 1}

at 300 K, respectively. The reaction is Zn (s) + C

u^{2 +}

(aq)

\to

n^{2 +}

(aq) + Cu (s). The standard reaction enthalpy

(Δ_{r} H^{\circ})

at 300 K in mo

l^{- 1}

is
[Use R = 8 J

K^{- 1}

l^{- 1}

and F = 96,500 C mo

l^{- 1}

]

(A)

- 412.8

(B)

206.4

(C)

- 384.0

(D)

192.0

SolutionAnswer: Option 1

- 412.8

Approach:

Use the Gibbs-Helmholtz relations for a cell to obtain the reaction enthalpy from E° and its temperature coefficient.

Step 1:The Zn-Cu reaction transfers two electrons, so n = 2.

n = 2

Step 2:Compute the free energy change.

Δ G = - n FE = - 2 (96500) (2) = - 386000 J

Step 3:Compute the entropy change from the temperature coefficient.

Δ S = n F \frac{d E}{d T} = 2 (96500) (- 5 \times 1 0^{- 4}) = - 96.5 J K^{- 1}

Step 4:Combine to obtain the enthalpy at 300 K.

Δ H = Δ G + T Δ S = - 386000 + 300 (- 96.5) = - 414950 J \approx - 412.8 kJ

Final answer:

- 412.8

Q48Single correctChemical Kinetics

Decomposition of X exhibits a rate constant of 0 .05 µg/ year. How many years are required for the decomposition of 5 µg of X into 2 .5 µg?

(A)

20

(B)

40

(C)

25

(D)

50

SolutionAnswer: Option 4

50

Approach:

The units of the rate constant (mass per time) indicate a zero-order decomposition; integrate the zero-order rate law for the amount consumed.

Step 1:The rate-constant unit µg/year shows the reaction is zero order in X.

k = 0.05 μ g year^{- 1}

Step 2:Amount decomposed when 5 µg falls to 2.5 µg.

Δ = 5 - 2.5 = 2.5 μ g

Step 3:Apply the zero-order law to find time.

t = \frac{Δ}{k} = \frac{2.5}{0.05} = 50 years

Final answer:

50

Q49Single correctRedox Reactions and Electrochemistry

In the Hall-Heroult process, aluminium is formed at the cathode. The cathode is made out of:

(A)

Platinum

(B)

Carbon

(C)

Copper

(D)

Pure aluminium

SolutionAnswer: Option 2

Carbon

Approach:

Recall the construction of the Hall-Heroult electrolytic cell for aluminium extraction.

Step 1:In the Hall-Heroult process molten alumina dissolved in cryolite is electrolysed in a steel tank lined with carbon.

Al_{2} O_{3} (in molten Na_{3} AlF_{6})

Step 2:The carbon lining of the tank serves as the cathode where aluminium is deposited; carbon (graphite) rods act as the anode.

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

Final answer:

Carbon

Q50Single correctp-Block Elements

Iodine reacts with concentrated

HNO_{3}

to yield Y along with other products. The oxidation state of iodine in Y , is:

(A)

1

(B)

5

(C)

3

(D)

7

SolutionAnswer: Option 2

5

Approach:

Concentrated nitric acid oxidises iodine to iodic acid; determine the oxidation state of iodine in that product.

Step 1:Iodine is oxidised by hot concentrated nitric acid to iodic acid.

I_{2} + 10 HNO_{3} \to 2 HIO_{3} + 10 NO_{2} + 4 H_{2} O

Step 2:Assign the oxidation state of iodine in HIO3.

+ 1 + x + 3 (- 2) = 0 \Rightarrow x = + 5

Final answer:

5

Q51Single correctCoordination Compounds

The pair of metal ions that can give a spin only magnetic moment of 3.9 BM for the complex

[M (H_{2} O)_{6}] C l_{2}

, is:

(A)

C o^{2 +}

and

F e^{2 +}

(B)

V^{2 +}

and

F e^{2 +}

(C)

C r^{2 +}

and

M n^{2 +}

(D)

V^{2 +}

and

C o^{2 +}

SolutionAnswer: Option 4

V^{2 +}

and

C o^{2 +}

Approach:

A spin-only moment of 3.9 BM corresponds to three unpaired electrons; identify the M2+ ions whose weak-field aqua complexes have three unpaired electrons.

Step 1:Solve for the number of unpaired electrons.

n (n + 2) = 3.9 \Rightarrow n = 3

Step 2:Count d-electrons and unpaired electrons for each M2+ in a high-spin aqua field.

V^{2 +} (d^{3}) : 3, Co^{2 +} (d^{7}) : 3, Fe^{2 +} (d^{6}) : 4, Cr^{2 +} (d^{4}) : 4, Mn^{2 +} (d^{5}) : 5

Step 3:Select the pair with three unpaired electrons each.

V^{2 +} and Co^{2 +}

Final answer:

V^{2 +}

and

C o^{2 +}

Q52Single correctCoordination Compounds

The metal's d– orbitals that are directly facing the ligands in

K_{3} [C o (CN)_{6}]

are:

(A)

d_{x y}, d_{x z}

and

d_{yz}

(B)

d_{x^{2} - y^{2}}

and

d_{z^{2}}

(C)

d_{x y}

and

d_{x^{2} - y^{2}}

(D)

d_{x z}, d_{yz}

and

d_{z^{2}}

SolutionAnswer: Option 2

d_{x^{2} - y^{2}}

and

d_{z^{2}}

Approach:

In an octahedral complex the d-orbitals pointing along the axes face the ligands directly.

Step 1:The complex is octahedral with ligands approaching along the x, y and z axes.

[Co(CN)_{6}]^{3 -}

Step 2:The

e_{g}

set points along the axes, directly toward the ligands; the

t_{2}

g set points between axes.

e_{g} = {d_{x^{2} - y^{2}}, d_{z^{2}}}

Final answer:

d_{x^{2} - y^{2}}

and

d_{z^{2}}

Q53Single correctCoordination Compounds

M n_{2} (CO)_{10}

is an organometallic compound due to the presence of:

(A)

M n - O

bond

(B)

M n - M n

bond

(C)

C - O

bond

(D)

M n - C

bond

SolutionAnswer: Option 4

M n - C

bond

Approach:

An organometallic compound is defined by a direct metal-carbon bond.

Step 1:Organometallic compounds contain at least one direct metal-carbon bond.

M - C

Step 2:In Mn2(CO)10 the carbonyl carbons bond directly to manganese.

Mn - C \equiv O

Final answer:

M n - C

bond

Q54Single correctOrganic Compounds Containing Oxygen

cannot be prepared by:

Condensed/skeletal structure of the target tertiary alcohol drawn at the top of the question: CH3CH2-C(CH3)(Ph)-OH; the carbinol carbon bears an ethyl group on the left, a CH3 above, a Ph (phenyl) below, and is the alcohol carbon. (In the page this structure appears just above Q54 stem as 'CH3CH2 - C - CH3' with CH3 above and Ph below the central carbon.)

(A)

HCHO + PhCH(CH_{3}) CH_{2} MgX

(B)

PhCOCH_{2} CH_{3} + CH_{3} MgX

(C)

CH_{3} CH_{2} COCH_{3} + PhMgX

(D)

PhCOCH_{3} + CH_{3} CH_{2} MgX

SolutionAnswer: Option 1

HCHO + PhCH(CH_{3}) CH_{2} MgX

Approach:

The target alcohol is 2-phenyl-2-butanol, CH3CH2-C(CH3)(Ph)-OH. A Grignard route builds a tertiary alcohol by adding an alkyl/aryl group to a ketone; check which carbonyl + Grignard pairing reconstructs the correct skeleton.

Step 1:The target is the tertiary alcohol 2-phenylbutan-2-ol bearing Ph, CH3, C2H5 and OH on the carbinol carbon.

CH_{3} CH_{2} - C H_{3} C P h - OH

Step 2:A valid Grignard synthesis must add one of the three carbon groups to a ketone carrying the other two; HCHO with a Grignard gives only a primary alcohol, not this tertiary carbinol.

HCHO + R-MgX \to R-CH_{2} OH

Step 3:Options 2, 3 and 4 each combine a ketone with the correct Grignard reagent to give the tertiary alcohol, so only the formaldehyde route fails.

HCHO route \Rightarrow primary alcohol only

Final answer:

HCHO + PhCH(CH_{3}) CH_{2} MgX

Q55Single correctOrganic Compounds Containing Oxygen

In the following reactions, products A and B are:

Reaction scheme. Reactant: an open-chain keto-aldehyde drawn as H3C-C(=O)-C(CH3)2-CH2-CH2-CH(=O)H, i.e. a methyl ketone (CH3-CO-) on the left, a quaternary carbon bearing two CH3 groups (H3C and CH3), then a CH2-CH2 chain ending in an aldehyde (-CHO, shown as C(=O)H on the right). Arrow labelled 'dilute NaOH' gives [A]. Second line: [A] with arrow labelled 'H3O+' above and 'delta' below gives [B]. The four options (on page 11) are drawn pairs of A and B structures (cyclic beta-hydroxy ketone / cyclic enone).

(A)

(B)

(C)

(D)

SolutionAnswer: Option 2

Approach:

Identify A from an intramolecular aldol of the keto-aldehyde with dilute NaOH, then B from acid-catalysed dehydration on heating.

Step 1:The substrate is a keto-aldehyde; dilute base promotes an intramolecular aldol addition, the enolate adding to the aldehyde to form a six-membered ring carrying a hydroxyl group (the beta-hydroxy carbonyl A).

intramolecular aldol \to β -hydroxy ketone (A)

Step 2:Warming A with acid (H3O+, heat) eliminates water to give the alpha,beta-unsaturated cyclohexenone B.

β -hydroxy ketone H_{3} O^{+}, Δ α, β -unsaturated ketone (B)

Q56Single correctOrganic Compounds Containing Oxygen

The major product of the following reaction is:

Bicyclic substrate: a benzene ring fused to a five-membered lactone ring; the aromatic ring also bears a CHO (aldehyde) substituent at the top, and the five-membered ring contains an -O- and a C=O (lactone, with the structure showing 'OH' and a carbonyl O at the bottom of the fused ring). Reagent arrow: (i) DIBAL-H, (ii) H3O+.

(A)

Option 1: benzene ring bearing a CHO at top, an OH in the middle right, and a CH2-CHO (-CHO via a CH2) at the bottom; i.e. an open-ring ortho-substituted benzene with two CHO groups and one OH.

(B)

Option 2: benzene ring with a CHO at top, and two adjacent OH groups (OH and OH) on a fused/attached saturated ring portion at the lower right (a tetralin-like fused ring carrying two hydroxyls).

(C)

Option 3: benzene ring bearing a CHO at top and fused to a five-membered ring containing an -O- (a benzofuran-type/2,3-dihydrobenzofuran fused ring with oxygen), no second carbonyl in the five-membered ring.

(D)

Option 4: benzene ring bearing a CHO at top, fused to a five-membered lactone ring containing -O- and a C=O (the original lactone retained), i.e. same bicyclic phthalide-type lactone with an aldehyde substituent.

SolutionAnswer: Option 1

Approach:

DIBAL-H selectively reduces an ester (lactone) to the aldehyde/hemiacetal stage at low temperature while leaving the existing aldehyde, with aqueous work-up opening the reduced lactone.

Step 1:The substrate carries an aromatic aldehyde and a five-membered lactone fused to the ring.

aryl-CHO + lactone

Step 2:One equivalent of DIBAL-H reduces the lactone carbonyl to the aldehyde/lactol level; aqueous acidic work-up opens it to the hydroxy-aldehyde, giving the open dialdehyde-phenol product.

lactone (i) DIBAL-H; (ii) H_{3} O^{+} ring-opened hydroxy aldehyde

Q57Single correctOrganic Compounds Containing Oxygen

In the following reaction
Aldehyde + Alcohol

H Cl

Acetal
Aldehyde Alcohol
HCHO t − BuOH
C

H_{3}

CHO MeOH
The best combination is:

(A)

HCHO and MeOH

(B)

H_{3}

CHO and t − BuOH

(C)

H_{3}

CHO and MeOH

(D)

HCHO and t-BuOH

SolutionAnswer: Option 1HCHO and MeOH

Approach:

Acetal formation is favoured by the most electrophilic aldehyde and the least hindered alcohol.

Step 1:Formaldehyde is the most reactive aldehyde toward nucleophilic addition, being the least sterically and electronically hindered carbonyl.

HCHO > CH_{3} CHO

Step 2:A small, unhindered alcohol adds readily, whereas bulky tert-butanol resists acetal formation.

MeOH > t -BuOH

Step 3:The best combination pairs the most reactive aldehyde with the least hindered alcohol.

HCHO + MeOH

Final answer: HCHO and MeOH

Q58Single correctOrganic Compounds Containing Nitrogen

The increasing order of reactivity of the following compounds towards reaction with alkyl halides directly is:

Four labelled structures. (i) benzamide: benzene ring with a C(=O)NH2 (amide) group. (ii) cyclic imide on a benzene ring: benzene ortho-disubstituted with two C=O groups bridged by an N-H (a phthalimide-type imide, drawn as two carbonyls flanking an NH). (iii) 2-aminobenzonitrile: benzene ring bearing a CN (nitrile) group and an adjacent NH2. (iv) aniline: benzene ring with an NH2 group.

(A)

(i) < (ii) < (iii) < (i v)

(B)

(ii) < (i) < (iii) < (i v)

(C)

(ii) < (i) < (i v) < (iii)

(D)

(i) < (iii) < (i v) < (ii)

SolutionAnswer: Option 2

(ii) < (i) < (iii) < (i v)

Approach:

Reactivity toward alkyl halides increases with the nucleophilicity/availability of the nitrogen lone pair; greater delocalisation or electron withdrawal lowers it.

Step 1:Benzamide (i) and the cyclic imide (ii) have the nitrogen lone pair delocalised into adjacent carbonyls, the imide most strongly, making it the least nucleophilic.

imide (ii) < amide (i)

Step 2:The aryl amine bearing an electron-withdrawing CN group (iii) is more available than the amide nitrogen but less than the unsubstituted aniline.

amide (i) < CN-aniline (iii)

Step 3:Aniline (iv), with no electron-withdrawing substituent, has the most available lone pair and is the most reactive toward alkyl halides.

(ii) < (i) < (iii) < (i v)

Final answer:

(ii) < (i) < (iii) < (i v)

Q59Single correctBiomolecules

Poly-β-hydroxybutyrate-co-β-hydroxyvalerate (PHBV) is a copolymer of _____.

(A)

3 -

hydroxybutanoic acid and

4 -

hydroxypentanoic acid

(B)

2 -

hydroxybutanoic acid and

3 -

hydroxypentanoic acid

(C)

3 -

hydroxybutanoic acid and

3 -

hydroxypentanoic acid

(D)

2 -

hydroxybutanoic acid and

3 -

hydroxypentanoic acid

SolutionAnswer: Option 3

3 -

hydroxybutanoic acid and

3 -

hydroxypentanoic acid

Approach:

Interpret the IUPAC names of the monomers from the polymer name PHBV: beta-hydroxybutyrate and beta-hydroxyvalerate.

Step 1:The beta-carbon is the 3-position, so beta-hydroxybutyrate is 3-hydroxybutanoic acid.

β -hydroxybutyrate = 3 -hydroxybutanoic acid

Step 2:Similarly beta-hydroxyvalerate is 3-hydroxypentanoic acid (valeric = pentanoic).

β -hydroxyvalerate = 3 -hydroxypentanoic acid

Final answer:

3 -

hydroxybutanoic acid and

3 -

hydroxypentanoic acid

Q60Single correctBiomolecules

Among the following compounds most basic amino acid is:

(A)

Lysine

(B)

Asparagine

(C)

Serine

(D)

Histidine

SolutionAnswer: Option 1

Lysine

Approach:

The most basic amino acid carries an extra strongly basic side-chain nitrogen with the highest side-chain pKa.

Step 1:Lysine bears a primary amino group on its side chain, a strong base.

side chain: - (CH_{2})_{4} NH_{2}

Step 2:Histidine's imidazole and asparagine's amide nitrogen are far less basic, and serine has a neutral hydroxyl side chain, so lysine is the most basic.

Lysine > Histidine > Asparagine > Serine

Final answer:

Lysine

Mathematics30 questions

Q61Single correctComplex Numbers and Quadratic Equations

λ

be the ratio of the roots of the quadratic equation in x,

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

, then the least value of m for which

λ + \frac{1}{λ} = 1

, is :

(A)

2 - 3

(B)

- 2 + 2

(C)

4 - 23

(D)

4 - 32

SolutionAnswer: Option 4

4 - 32

Approach:

Let the roots be in ratio lambda. Form the symmetric condition lambda+1/lambda=1, convert to

(a lp ha + b e t a)^{2}

= 3 alpha*beta using sum and product of roots, then solve for

m^{2}

and pick the least m.

Step 1:Condition lambda+1/lambda=1 with lambda being the ratio of roots gives a relation between sum and product.

\frac{α}{β} + \frac{β}{α} = 1 \Rightarrow α^{2} + β^{2} = α β \Rightarrow (α + β)^{2} = 3 α β

Step 2:Substitute sum and product for the given quadratic.

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

Step 3:Let u=

m^{2}

and solve the resulting quadratic in u, then take the least value of m.

m^{4} - 26 m^{2} + 16 = 0 \Rightarrow m^{2} = 13 \pm 317

Final answer:

4 - 32

Q62Single correctComplex Numbers and Quadratic Equations

\frac{z - α}{z + α}

(α \in R)

is a purely imaginary number and

∣ z ∣ = 2

, then a value of

α

is :

(A)

1

(B)

\frac{1}{2}

(C)

2

(D)

2

SolutionAnswer: Option 4

2

Approach:

A purely imaginary number plus its conjugate equals zero. Apply this to the given ratio and use z*conjugate(z)=

∣ z ∣^{2}

Step 1:Set the ratio plus its conjugate to zero.

\frac{z - α}{z + α} + \frac{z ˉ - α}{z ˉ + α} = 0

Step 2:Expand and use z*conjugate(z)=

∣ z ∣^{2}

2 z \overset{z}{ˉ} - 2 α^{2} = 0 \Rightarrow α^{2} = ∣ z ∣^{2}

Step 3:Substitute |z|=2.

α^{2} = 4 \Rightarrow α = 2

Final answer:

2

Q63Single correctPermutations and Combinations

Let

S = {1, 2, 3, \dots, 100}

, then number of non-empty subsets A of S such that the product of elements in A is even is :

(A)

2^{100} - 1

(B)

2^{50} + 1

(C)

2^{50} (2^{50} - 1)

(D)

2^{50} - 1

SolutionAnswer: Option 3

2^{50} (2^{50} - 1)

Approach:

Count subsets whose product is even by subtracting subsets with an odd product (those containing only odd numbers) from all non-empty subsets.

Step 1:The set has 50 even and 50 odd numbers. The product is odd only when every chosen element is odd.

odd-product non-empty subsets = 2^{50} - 1

Step 2:Subtract from all non-empty subsets.

(2^{100} - 1) - (2^{50} - 1) = 2^{100} - 2^{50}

Step 3:Factor the result.

2^{100} - 2^{50} = 2^{50} (2^{50} - 1)

Final answer:

2^{50} (2^{50} - 1)

Q64Single correctPermutations and Combinations

Consider three boxes, each containing 10 balls labelled

1, 2, \dots, 10

. Suppose one ball is randomly drawn from each of the boxes. Denote by

n_{i}

, the label of the ball drawn from the

i^{th}

box,

(i = 1, 2, 3)

. Then, the number of ways in which the balls can be chosen such that

n_{1} < n_{2} < n_{3}

is :

(A)

240

(B)

82

(C)

120

(D)

164

SolutionAnswer: Option 3

120

Approach:

Each strictly increasing triple corresponds to choosing 3 distinct labels from 10, which can be arranged in increasing order in exactly one way.

Step 1:For n1<n2<n3, the three labels must be distinct, and any set of three distinct labels has a unique increasing order.

count = (3 10)

Step 2:Evaluate the combination.

(3 10) = \frac{10 \cdot 9 \cdot 8}{6} = 120

Final answer:

120

Q65Single correctSequence and Series

Let

S_{k} = \frac{1 + 2 + 3 + \dots + k}{k}

. If

S_{1}^{2} + S_{2}^{2} + \dots + S_{10}^{2} = \frac{5}{12} A

, then A is equal to :

(A)

301

(B)

303

(C)

156

(D)

283

SolutionAnswer: Option 2

303

Approach:

Simplify

S_{k}

to (k+1)/2, square it, sum from 1 to 10, then match with 5A/12.

Step 1:Simplify

S_{k}

S_{k} = \frac{k ( k + 1 )}{2 k} = \frac{k + 1}{2}

Step 2:Sum the squares from k=1 to 10.

\sum_{k = 1}^{10} (\frac{k + 1}{2})^{2} = \frac{1}{4} \sum_{k = 2}^{11} k^{2} = \frac{1}{4} (506 - 1) = \frac{505}{4}

Step 3:Set equal to 5A/12 and solve.

\frac{505}{4} = \frac{5}{12} A \Rightarrow A = \frac{505 \cdot 12}{4 \cdot 5} = 303

Final answer:

303

Q66Single correctSequence and Series

The product of three consecutive terms of a

G . P .

512

. If

4

is added to each of the first and the second of these terms, the three terms now form an

A . P .

, then the sum of the original three terms of the given

G . P .

is :

(A)

28

(B)

24

(C)

32

(D)

36

SolutionAnswer: Option 1

28

Approach:

Write the three terms as a/r, a, ar. Use the product to find a, then apply the A.P. condition after adding 4 to the first two terms to find r.

Step 1:Product of the three terms gives a.

\frac{a}{r} \cdot a \cdot a r = a^{3} = 512 \Rightarrow a = 8

Step 2:Apply the A.P. condition to (a/r + 4), (a + 4), ar.

2 (a + 4) = (\frac{a}{r} + 4) + a r

Step 3:Solve for r.

24 = \frac{8}{r} + 4 + 8 r \Rightarrow 8 r^{2} - 20 r + 8 = 0 \Rightarrow r = 2 or \frac{1}{2}

Step 4:Sum the three terms a/r + a + ar.

\frac{8}{2} + 8 + 8 \cdot 2 = 4 + 8 + 16 = 28

Final answer:

28

Q67Single correctBinomial Theorem and its Simple Applications

A ratio of the

5^{th}

term from the beginning to the

5^{th}

term from the end in the binomial expansion of

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

is :

(A)

1 : 4 (16)^{\frac{1}{3}}

(B)

4 (36)^{\frac{1}{3}} : 1

(C)

2 (36)^{\frac{1}{3}} : 1

(D)

1 : 2 (6)^{\frac{1}{3}}

SolutionAnswer: Option 2

4 (36)^{\frac{1}{3}} : 1

Approach:

Compute the 5th term from the beginning (T5) and the 5th term from the end (T7) using the general term, then form their ratio.

Step 1:The 5th term from the beginning is T5 (r=4); the 5th from the end is the 7th term (r=6).

\frac{T _{5}}{T _{7}} = \frac{( 4 10 ) a ^{6} b ^{4}}{( 6 10 ) a ^{4} b ^{6}} = \frac{a ^{2}}{b ^{2}}

Step 2:Substitute a=2^(1/3), b=1/(2*3^(1/3)).

\frac{a ^{2}}{b ^{2}} = 2^{2/3} \cdot 4 \cdot 3^{2/3} = 4 (4 \cdot 9)^{1/3} = 4 (36)^{1/3}

Final answer:

4 (36)^{\frac{1}{3}} : 1

Q68Single correctTrigonometry

The maximum value of

3 cos θ + 5 sin (θ - \frac{π}{6})

for any real value of

θ

is :

(A)

19

(B)

31

(C)

\frac{79}{2}

(D)

34

SolutionAnswer: Option 1

19

Approach:

Expand the sine term, collect coefficients of cos theta and sin theta, then use the amplitude sqrt(

P^{2}

Q^{2}

) as the maximum.

Step 1:Expand sin(theta - pi/6).

5 sin (θ - \frac{π}{6}) = 5 (\frac{3}{2} sin θ - \frac{1}{2} cos θ)

Step 2:Collect coefficients of cos theta and sin theta.

(3 - \frac{5}{2}) cos θ + \frac{5 3}{2} sin θ = \frac{1}{2} cos θ + \frac{5 3}{2} sin θ

Step 3:Maximum equals the amplitude.

\frac{1}{4} + \frac{75}{4} = \frac{76}{4} = 19

Final answer:

19

Q69Single correctCo-ordinate Geometry

If the straight line

2 x - 3 y + 17 = 0

is perpendicular to the line passing through the points

(7, 17)

and

(15, β)

, then

β

equals :

(A)

- 5

(B)

\frac{35}{3}

(C)

5

(D)

- \frac{35}{3}

SolutionAnswer: Option 3

5

Approach:

Find the slope of the given line, then use the perpendicularity condition (product of slopes = -1) with the slope through the two points to solve for beta.

Step 1:Slope of the given line.

2 x - 3 y + 17 = 0 \Rightarrow m_{1} = \frac{2}{3}

Step 2:Slope through the two points must be the negative reciprocal.

\frac{β - 17}{15 - 7} = - \frac{3}{2}

Step 3:Solve for beta.

β - 17 = - 12 \Rightarrow β = 5

Final answer:

5

Q70Single correctCo-ordinate Geometry

Let

C_{1}

and

C_{2}

be the centres of the circles

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

and

x^{2} + y^{2} - 6 x - 6 y + 14 = 0

respectively. If P and Q are the points of intersection of these circles, then the area (in sq. units) of the quadrilateral

P C_{1} Q C_{2}

is :

(A)

6

(B)

4

(C)

8

(D)

9

SolutionAnswer: Option 2

4

Approach:

Find the centres, radii and the common chord. The quadrilateral PC1QC2 is a kite split by the line C1C2 into two triangles; compute its area using the diagonals.

Step 1:Identify centres and radii.

C_{1} (1, 1), r_{1} = 2; C_{2} (3, 3), r_{2} = 2

Step 2:Find the common chord PQ by subtracting the two circle equations.

4 x + 4 y - 16 = 0 \Rightarrow x + y = 4

Step 3:Diagonal C1C2 length and chord PQ length.

C_{1} C_{2} = 22, PQ = 2 r_{1}^{2} - d^{2} = 2 4 - 2 = 22

Step 4:Area of the kite with perpendicular diagonals.

\frac{1}{2} \cdot 22 \cdot 22 = 4

Final answer:

4

Q71Single correctCo-ordinate Geometry

If a variable line

3 x + 4 y - λ = 0

is such that the two circles

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

and

x^{2} + y^{2} - 18 x - 2 y + 78 = 0

are on its opposite sides, then the set of all values of

λ

is the interval :

(A)

[13, 23]

(B)

(23, 31)

(C)

[12, 21]

(D)

(2, 17)

SolutionAnswer: Option 3

[12, 21]

Approach:

For the two circles to lie on opposite sides of the line, their centres must give the line-expression opposite signs, and the line must not cut either circle (distance from each centre exceeds its radius).

Step 1:Centres and radii.

C_{1} (1, 1), r_{1} = 1; C_{2} (9, 1), r_{2} = 2

Step 2:Opposite-side requires the line values at the two centres to have opposite signs.

(3 + 4 - λ) (27 + 4 - λ) < 0 \Rightarrow (7 - λ) (31 - λ) < 0 \Rightarrow 7 < λ < 31

Step 3:Line must not intersect circle 1: distance from C1 must be at least r1.

\frac{∣ 7 - λ ∣}{5} \geq 1 \Rightarrow λ \leq 2 or λ \geq 12

Step 4:Line must not intersect circle 2: distance from C2 must be at least r2.

\frac{∣ 31 - λ ∣}{5} \geq 2 \Rightarrow λ \leq 21 or λ \geq 41

Step 5:Intersect all conditions within (7, 31).

λ \in [12, 21]

Final answer:

[12, 21]

Q72Single correctCo-ordinate Geometry

Let

P (4, - 4)

and

Q (9, 6)

be two points on the parabola,

y^{2} = 4 x

and let X be any point on the arc POQ of this parabola, where O is the vertex of this parabola, such that the area of

Δ PXQ

is maximum. Then this maximum area (in sq. units) is :

(A)

\frac{625}{4}

(B)

\frac{75}{2}

(C)

\frac{125}{4}

(D)

\frac{125}{2}

SolutionAnswer: Option 3

\frac{125}{4}

Approach:

Parametrize X on the parabola as (

t^{2}

, 2t), write the triangle area as a function of t, and maximize it over the relevant arc.

Step 1:Use parametric point X=(

t^{2}

, 2t) with P(4,-4) and Q(9,6).

Area = \frac{1}{2} 4 (6 - 2 t) + 9 (2 t + 4) + t^{2} (- 4 - 6)

Step 2:Simplify the expression.

Area = \frac{1}{2} - 10 t^{2} + 10 t + 60 = 5 - t^{2} + t + 6

Step 3:Maximize -

t^{2}

+t+6 at t=1/2.

- \frac{1}{4} + \frac{1}{2} + 6 = \frac{25}{4}

Step 4:Multiply by 5.

5 \cdot \frac{25}{4} = \frac{125}{4}

Final answer:

\frac{125}{4}

Q73Single correctCo-ordinate Geometry

If the vertices of a hyperbola be at

(- 2, 0)

and

(2, 0)

and one of its foci be at

(- 3, 0)

, then which one of the following points does not lie on this hyperbola ?

(A)

(6, 52)

(B)

(- 6, 210)

(C)

(26, 5)

(D)

(4, 15)

SolutionAnswer: Option 1

(6, 52)

Approach:

Determine the hyperbola equation from a and c, then test each point to find the one that does not satisfy it.

Step 1:From vertices a=2 and focus c=3, find

b^{2}

b^{2} = c^{2} - a^{2} = 9 - 4 = 5

Step 2:Test the points by substitution.

(26, 5) : \frac{24}{4} - \frac{25}{5} = 6 - 5 = 1 ✓

Step 3:Test (6, 5*sqrt2).

\frac{36}{4} - \frac{50}{5} = 9 - 10 = - 1 \neq = 1

Final answer:

(6, 52)

Q74Single correctLimit, Continuity and Differentiability

x \to \frac{π}{4} lim \frac{c o t ^{3} x - t a n x}{c o s ( x + \frac{π}{4} )}

is :

(A)

42

(B)

82

(C)

4

(D)

8

SolutionAnswer: Option 4

8

Approach:

Factor the numerator, express cot and tan in terms of cos and sin, and reduce the indeterminate form to a limit that can be evaluated by direct substitution.

Step 1:Factor the numerator using tan*cot=1.

cot^{3} x - tan x = cot x (cot^{2} x - tan^{2} x) = cot x (cot x - tan x) (cot x + tan x)

Step 2:Express cot x - tan x and the denominator in terms of sine and cosine.

cot x - tan x = \frac{c o s 2 x}{\frac{1}{2} s i n 2 x}, cos (x + \frac{π}{4}) = \frac{c o s x - s i n x}{2}

Step 3:Cancel the vanishing factor and substitute x=pi/4.

lim_{x \to \frac{π}{4}} \frac{c o t ^{3} x - t a n x}{c o s ( x + \frac{π}{4} )} = 8

Final answer:

8

Q75Single correctSets, Relations and Functions

The Boolean expression

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

is equivalent to :

(A)

p \land (\sim q)

(B)

(\sim p) \land (\sim q)

(C)

p \lor (\sim q)

(D)

p \land q

SolutionAnswer: Option 2

(\sim p) \land (\sim q)

Approach:

Simplify the first bracket, then conjoin with (~p AND ~q) using absorption and distributive laws.

Step 1:The first bracket simplifies since p AND q implies p OR (~q).

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

Step 2:Conjoin with (~p AND ~q).

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

Step 3:Since ~q is already a factor, the expression reduces.

(\sim p \land \sim q)

Final answer:

(\sim p) \land (\sim q)

Q76Single correctStatistics and Probability

If the sum of the deviations of 50 observations from 30 is 50, then the mean of these observations is :

(A)

30

(B)

51

(C)

50

(D)

31

SolutionAnswer: Option 4

31

Approach:

Relate the sum of deviations from an assumed value to the actual mean.

Step 1:Express the given condition with assumed mean A=30 and n=50.

\sum_{i = 1}^{50} (x_{i} - 30) = 50

Step 2:Solve for the total of the observations.

\sum x_{i} = 1550

Step 3:Divide by the number of observations to obtain the mean.

\overset{x}{ˉ} = \frac{1550}{50} = 31

Final answer:

31

Q77Single correctMatrices and Determinants

Let

P = 139013001

and

Q = [q_{ij}]

be two

3 \times 3

matrices such that

Q - P^{5} = I_{3}

. Then

\frac{q _{21} + q _{31}}{q _{32}}

is equal to :

(A)

10

(B)

9

(C)

15

(D)

135

SolutionAnswer: Option 1

10

Approach:

Write P as I plus a strictly lower-triangular nilpotent matrix and expand its powers, then read the required entries of Q.

Step 1:Decompose P into identity plus the nilpotent part N.

N = 039003000, N^{2} = 009000000, N^{3} = 0

Step 2:Compute the fifth power using the expansion truncated at the squared term.

P^{5} = I + 5 N + 10 N^{2}

Step 3:Form the entries of

P^{5}

; since Q=I+

P^{5}

, the relevant off-diagonal entries of Q equal those of

P^{5}

q_{21} = 15, q_{31} = 45 + 90 = 135, q_{32} = 15

Step 4:Evaluate the required ratio.

\frac{q _{21} + q _{31}}{q _{32}} = \frac{15 + 135}{15} = 10

Final answer:

10

Q78Single correctMatrices and Determinants

An ordered pair

(α, β)

for which the system of linear equations

(1 + α) x + β y + z = 2

αx + (1 + β) y + z = 3

αx + β y + 2 z = 2

has a unique solution, is :

(A)

(- 3, 1)

(B)

(1, - 3)

(C)

(2, 4)

(D)

(- 4, 2)

SolutionAnswer: Option 3

(2, 4)

Approach:

A square system has a unique solution exactly when its coefficient determinant is non-zero; compute that determinant and test the options.

Step 1:Form the coefficient determinant.

Δ = 1 + α α α β 1 + β β 112

Step 2:Apply R1->R1-R2 and R3->R3-R2 to simplify.

Δ = 1 α 0 - 1 1 + β - 1 011

Step 3:Expand to obtain Delta in terms of alpha and beta.

Δ = 2 + α + β

Step 4:Unique solution requires Delta nonzero; test each option.

(2, 4) : 2 + 2 + 4 = 8 \neq = 0

Final answer:

(2, 4)

Q79Single correctTrigonometry

Considering only the principal values of inverse functions, the set

A = {x \geq 0 : tan^{- 1} (2 x) + tan^{- 1} (3 x) = \frac{π}{4}}

(A)

Is an empty set

(B)

Contains more than two elements

(C)

Contains two elements

(D)

Is a singleton

SolutionAnswer: Option 4Is a singleton

Approach:

Combine the two inverse tangents and solve the resulting quadratic, then discard roots that violate the domain restriction.

Step 1:Combine the left side using the addition formula and set equal to pi/4.

\frac{2 x + 3 x}{1 - 6 x ^{2}} = 1

Step 2:Rearrange into a quadratic equation.

6 x^{2} + 5 x - 1 = 0

Step 3:Factor and solve.

(6 x - 1) (x + 1) = 0 \Rightarrow x = \frac{1}{6} or x = - 1

Step 4:Apply the constraint x>=0 (and

6 x^{2}

<1) to keep only valid roots.

x = \frac{1}{6}

Final answer: Is a singleton

Q80Single correctLimit, Continuity and Differentiability

Let S be the set of all points in

(- π, π)

at which the function,

f (x) = min {sin x, cos x}

is not differentiable. Then S is a subset of which of the following?

(A)

{- \frac{3 π}{4}, - \frac{π}{2}, \frac{π}{2}, \frac{3 π}{4}}

(B)

{- \frac{3 π}{4}, - \frac{π}{4}, \frac{3 π}{4}, \frac{π}{4}}

(C)

{- \frac{π}{4}, 0, \frac{π}{4}}

(D)

{- \frac{π}{2}, - \frac{π}{4}, \frac{π}{4}, \frac{π}{2}}

SolutionAnswer: Option 2

{- \frac{3 π}{4}, - \frac{π}{4}, \frac{3 π}{4}, \frac{π}{4}}

Approach:

Non-differentiability of the minimum of two smooth functions occurs where the two graphs cross within the interval.

Step 1:Find where sin x equals cos x in the open interval.

tan x = 1 \Rightarrow x = \frac{π}{4} + nπ

Step 2:Find where sin x equals -cos x, since the smaller branch can also switch there as the curves meet.

tan x = - 1 \Rightarrow x = - \frac{π}{4}, \frac{3 π}{4}

Step 3:Collect the corner points of the min function within (-pi, pi).

S = {- \frac{3 π}{4}, - \frac{π}{4}, \frac{π}{4}, \frac{3 π}{4}}

Step 4:Match this set against the options.

S \subseteq {- \frac{3 π}{4}, - \frac{π}{4}, \frac{3 π}{4}, \frac{π}{4}}

Final answer:

{- \frac{3 π}{4}, - \frac{π}{4}, \frac{3 π}{4}, \frac{π}{4}}

Q81Single correctLimit, Continuity and Differentiability

For

x > 1

, if

(2 x)^{2 y} = 4 e^{2 x - 2 y}

, then

(1 + lo g_{e} 2 x)^{2} \frac{d y}{d x}

is equal to

(A)

lo g_{e} 2 x

(B)

\frac{x lo g _{e} 2 x - lo g _{e} 2}{x}

(C)

x lo g_{e} 2 x

(D)

\frac{x lo g _{e} 2 x + lo g _{e} 2}{x}

SolutionAnswer: Option 2

\frac{x l o g _{e} 2 x - l o g _{e} 2}{x}

Approach:

Take logarithms to linearise the relation, then differentiate implicitly and isolate dy/dx.

Step 1:Take natural logarithm of both sides.

2 y ln (2 x) = ln 4 + 2 x - 2 y

Step 2:Group the y terms and solve for y.

y (ln (2 x) + 1) = ln 2 + x \Rightarrow y = \frac{x + l n 2}{1 + l n ( 2 x )}

Step 3:Differentiate using the quotient rule with d/dx of ln(2x) equal to 1/x.

\frac{d y}{d x} = \frac{( 1 ) ( 1 + l n 2 x ) - ( x + l n 2 ) \cdot \frac{1}{x}}{( 1 + l n 2 x ) ^{2}}

Step 4:Multiply by

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

and simplify the numerator.

(1 + lo g_{e} 2 x)^{2} \frac{d y}{d x} = 1 + ln 2 x - 1 - \frac{l n 2}{x} = \frac{x l o g _{e} 2 x - l o g _{e} 2}{x}

Final answer:

\frac{x l o g _{e} 2 x - l o g _{e} 2}{x}

Q82Single correctLimit, Continuity and Differentiability

The maximum area (in sq. units) of a rectangle having its base on the

x -

axis and its other two vertices on the parabola,

y = 12 - x^{2}

such that the rectangle lies inside the parabola, is :

(A)

202

(B)

32

(C)

36

(D)

183

SolutionAnswer: Option 2

32

Approach:

Parametrise the rectangle by the abscissa of its upper corner, write the area as a function of that variable, and maximise.

Step 1:Let the upper-right vertex be (x, 12-

x^{2}

); width is 2x and height is 12-

x^{2}

A (x) = 2 x (12 - x^{2}) = 24 x - 2 x^{3}

Step 2:Differentiate and set to zero.

A^{'} (x) = 24 - 6 x^{2} = 0

Step 3:Confirm a maximum via the second derivative.

A^{''} (x) = - 12 x < 0

x = 2

Step 4:Evaluate the area at x=2.

A (2) = 24 (2) - 2 (8) = 48 - 16 = 32

Final answer:

32

Q83Single correctIntegral Calculus

The integral

\int cos (ln x) d x

, is equal to

(A)

\frac{x}{2} (cos (ln x) - sin (ln x)) + C

(B)

x (cos (ln x) - sin (ln x)) + C

(C)

x (cos (ln x) + sin (ln x)) + C

(D)

\frac{x}{2} (cos (ln x) + sin (ln x)) + C

SolutionAnswer: Option 4

\frac{x}{2} (cos (ln x) + sin (ln x)) + C

Approach:

Apply integration by parts twice to generate a relation that solves for the integral.

Step 1:Let I be the integral and integrate by parts with u=cos(ln x), dv=dx.

I = x cos (ln x) + \int x \cdot \frac{s i n ( l n x )}{x} d x = x cos (ln x) + \int sin (ln x) d x

Step 2:Integrate the new integral by parts with u=sin(ln x), dv=dx.

\int sin (ln x) d x = x sin (ln x) - \int cos (ln x) d x = x sin (ln x) - I

Step 3:Substitute back to form an equation in I.

I = x cos (ln x) + x sin (ln x) - I

Step 4:Solve for I and add the constant.

I = \frac{x}{2} (cos (ln x) + sin (ln x)) + C

Final answer:

\frac{x}{2} (cos (ln x) + sin (ln x)) + C

Q84Single correctIntegral Calculus

Let f and g be continuous functions on

[0, a]

such that

f (x) = f (a - x)

and

g (x) + g (a - x) = 4

, then

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

is equal to

(A)

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

(B)

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

(C)

4 \int_{0}^{a} f (x) d x

(D)

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

SolutionAnswer: Option 4

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

Approach:

Use the king property of definite integrals to combine the integral with its reflection.

Step 1:Denote the integral I and replace x by a-x.

I = \int_{0}^{a} f (a - x) g (a - x) d x = \int_{0}^{a} f (x) g (a - x) d x

Step 2:Add the two expressions for I.

2 I = \int_{0}^{a} f (x) (g (x) + g (a - x)) d x

Step 3:Use the given condition g(x)+g(a-x)=4.

2 I = 4 \int_{0}^{a} f (x) d x

Final answer:

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

Q85Single correctIntegral Calculus

The area (in sq. units) of the region bounded by the parabola,

y = x^{2} + 2

and the lines,

y = x + 1, x = 0

and

x = 3

, is

(A)

\frac{17}{4}

(B)

\frac{21}{2}

(C)

\frac{15}{2}

(D)

\frac{15}{4}

SolutionAnswer: Option 3

\frac{15}{2}

Approach:

Integrate the vertical gap between the parabola and the line over the given x-interval.

Step 1:On [0,3] the parabola lies above the line; form the difference.

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

Step 2:Integrate the difference from 0 to 3.

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

Step 3:Evaluate the antiderivative.

A = [\frac{x ^{3}}{3} - \frac{x ^{2}}{2} + x]_{0}^{3} = 9 - \frac{9}{2} + 3

Step 4:Simplify.

A = \frac{24 - 9}{2} = \frac{15}{2}

Final answer:

\frac{15}{2}

Q86Single correctDifferential Equations

Let

y = y (x)

be the solution of the differential equation,

x \frac{d y}{d x} + y = x lo g_{e} x, (x > 1)

. If

2 y (2) = lo g_{e} 4 - 1

, then y(e) is equal to

(A)

- \frac{e}{2}

(B)

\frac{e}{4}

(C)

- \frac{e ^{2}}{2}

(D)

\frac{e ^{2}}{4}

SolutionAnswer: Option 2

\frac{e}{4}

Approach:

Recognise the left side as the derivative of a product, integrate, and apply the given condition to fix the constant.

Step 1:Rewrite the equation as an exact product derivative.

\frac{d}{d x} (x y) = x lo g_{e} x

Step 2:Integrate both sides; the right side by parts.

x y = \frac{x ^{2}}{2} ln x - \frac{x ^{2}}{4} + C

Step 3:Apply 2y(2)=

l o g_{e}

4 -1 to determine C.

2 y (2) = 2 ln 2 - 1 + \frac{C}{2} = ln 4 - 1 \Rightarrow C = 0

Step 4:Evaluate y at x=e.

y = \frac{x}{2} ln x - \frac{x}{4} \Rightarrow y (e) = \frac{e}{2} - \frac{e}{4} = \frac{e}{4}

Final answer:

\frac{e}{4}

Q87Single correctVector Algebra

The sum of the distinct real values of

μ

for which the vectors

μ \hat{i} + \hat{j} + \hat{k}, \hat{i} + μ \hat{j} + \hat{k}, \hat{i} + \hat{j} + μ \hat{k}

are co-planar, is

(A)

0

(B)

- 1

(C)

1

(D)

2

SolutionAnswer: Option 2

- 1

Approach:

Coplanarity of three vectors requires their scalar triple product, i.e. the determinant of components, to vanish.

Step 1:Set the determinant of the coefficient matrix to zero.

μ 11 1 μ 1 11 μ = 0

Step 2:Expand using the standard identity for this symmetric determinant.

μ^{3} - 3 μ + 2 = 0

Step 3:Factor the cubic.

(μ - 1)^{2} (μ + 2) = 0 \Rightarrow μ = 1, - 2

Step 4:Add the distinct real values.

1 + (- 2) = - 1

Final answer:

- 1

Q88Single correctThree Dimensional Geometry

A tetrahedron has vertices

P (1, 2, 1), Q (2, 1, 3), R (- 1, 1, 2)

and

O (0, 0, 0)

. The angle between the faces OPQ and PQR is

(A)

cos^{- 1} (\frac{7}{31})

(B)

cos^{- 1} (\frac{17}{31})

(C)

cos^{- 1} (\frac{19}{35})

(D)

cos^{- 1} (\frac{9}{35})

SolutionAnswer: Option 3

cos^{- 1} (\frac{19}{35})

Approach:

Find the normals of the two faces along edge PQ via cross products, then the angle between them.

Step 1:Normal of face OPQ from OP and OQ.

n_{1} = OP \times OQ = (5, - 1, - 3)

Step 2:Normal of face PQR from PQ and PR.

n_{2} = PQ \times PR = (1, - 5, - 3)

Step 3:Angle between the faces.

cos θ = \frac{5 + 5 + 9}{35 35} = \frac{19}{35}

Final answer:

cos^{- 1} (\frac{19}{35})

Q89Single correctThree Dimensional Geometry

The perpendicular distance from the origin to the plane containing the two lines,

\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}

and

\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}

, is

(A)

116

(B)

\frac{11}{6}

(C)

11

(D)

611

SolutionAnswer: Option 2

\frac{11}{6}

Approach:

Find the plane normal from the cross product of the two direction vectors, fix the plane through a known point, then apply the point-to-plane distance.

Step 1:Cross the two direction vectors to get the plane normal.

(3, 5, 7) \times (1, 4, 7) = (7, - 14, 7) ∥ (1, - 2, 1)

Step 2:Write the plane through (-2,2,-5).

1 (x + 2) - 2 (y - 2) + 1 (z + 5) = 0 \Rightarrow x - 2 y + z + 11 = 0

Step 3:Apply the distance formula from the origin.

d = \frac{∣0 - 0 + 0 + 11∣}{1 + 4 + 1} = \frac{11}{6}

Final answer:

\frac{11}{6}

Q90Single correctStatistics and Probability

In a random experiment, a fair die is rolled until two fours are obtained in succession. The probability that the experiment will end in the fifth throw of the die is equal to :

(A)

\frac{150}{6 ^{5}}

(B)

\frac{175}{6 ^{5}}

(C)

\frac{225}{6 ^{5}}

(D)

\frac{200}{6 ^{5}}

SolutionAnswer: Option 2

\frac{175}{6 ^{5}}

Approach:

The experiment ends on the fifth throw exactly when throws four and five are fours and no two consecutive fours occur among the first four throws.

Step 1:Throws 4 and 5 must be fours; throw 3 must not be a four to avoid an earlier ending.

P_{4} = P_{5} = \frac{1}{6}, P_{3} = \frac{5}{6}

Step 2:Throws 1 and 2 must not contain two consecutive fours; count the allowed pairs out of 36.

36 - 1 = 35 ordered pairs without 4,4

Step 3:Count first-two-throw outcomes with no double-four ending: total pairs minus the single (4,4) pair gives 35; combine with the fixed later throws.

P = \frac{35}{6 ^{2}} \cdot \frac{5}{6} \cdot \frac{1}{6} \cdot \frac{1}{6} = \frac{35 \cdot 5}{6 ^{5}} = \frac{175}{6 ^{5}}

Final answer:

\frac{175}{6 ^{5}}

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