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JEE Main 2020 January 07, Shift 2 Question Paper with Solutions

All 74 questions from the JEE Main 2020 (January 07, Shift 2) shift — Physics (24), Chemistry (25) and Mathematics (25) — each with the correct answer (answer key) and a step-by-step solution. Read it free online, or attempt the full paper as a timed mock test.

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Physics24 questions

Q1Single correctGravitation

A box weighs

196 N

on a spring balance at the North Pole. Its weight recorded on the same balance if it is shifted to the equator is close to (Take

g = 10 m / s^{2}

at the North Pole and radius of the Earth

= 6400 km

)

(A)

194.32 N

(B)

194.66 N

(C)

195.32 N

(D)

195.66 N

SolutionAnswer: Option 3

195.32 N

Approach:

At the equator the spring balance reads the apparent weight, reduced from the true weight by the centrifugal effect of Earth's rotation.

Step 1:Determine the mass from the polar weight.

m = \frac{196}{10} = 19.6 kg

Step 2:Compute Earth's angular speed for one rotation per day.

ω = \frac{2 π}{24 \times 3600} s^{- 1}

Step 3:Subtract the centrifugal term using

R = 6.4 \times 1 0^{6} m

W^{'} = 196 - 19.6 ω^{2} R

Final answer:

195.32 N

Q2Single correctCurrent Electricity

In a building, there are 15 bulbs of

45 W

, 15 bulbs of

100 W

, 15 small fans of

10 W

and 2 heaters of

1 kW

. The voltage of electric main is

220 V

. The minimum fuse capacity (rated value) of the building will be approximately

(A)

10 A

(B)

20 A

(C)

25 A

(D)

15 A

SolutionAnswer: Option 2

20 A

Approach:

Add the power consumed by every appliance, then divide the total power by the supply voltage to obtain the line current.

Step 1:Sum the appliance powers.

P = (15 \times 45) + (15 \times 100) + (15 \times 10) + (2 \times 1000)

Step 2:Divide by the mains voltage.

I = \frac{4325}{220}

Step 3:Select the next standard fuse rating above the line current.

I \approx 19.66 A \Rightarrow 20 A

Final answer:

20 A

Q4Single correctLaws of Motion

A mass of

10 k g

is suspended by a rope of length

4 m

, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of

4 5^{\circ}

with the vertical. Then F equals (Take

g = 10 m / s^{2}

and the rope to be massless)

(A)

100 N

(B)

90 N

(C)

75 N

(D)

70 N

SolutionAnswer: Option 1

100 N

Approach:

Apply equilibrium to the mid-point, where the upper tension, the horizontal force, and the lower tension (equal to the weight) meet.

Step 1:The lower half supports the weight, so the downward pull at the mid-point is

m g = 100 N

m g = 10 \times 10 = 100 N

Step 2:Resolve the upper tension at

4 5^{\circ}

: vertical balance gives

T cos 4 5^{\circ} = 100

\frac{T}{2} = 100

Step 3:Horizontal balance gives

F = T sin 4 5^{\circ}

F = \frac{T}{2} = 100 N

Final answer:

100 N

Q5Single correctRotational Motion

Mass per unit area of a circular disc of radius a depends on the distance r from its centre as

σ (r) = A + B r

. The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is

(A)

2 π a^{4} (\frac{A}{4} + \frac{a B}{5})

(B)

2 π a^{4} (\frac{a A}{4} + \frac{B}{5})

(C)

π a^{4} (\frac{A}{4} + \frac{a B}{5})

(D)

2 π a^{4} (\frac{A}{4} + \frac{B}{5})

SolutionAnswer: Option 1

2 π a^{4} (\frac{A}{4} + \frac{a B}{5})

Approach:

Integrate

d I = r^{2} d m

over annular rings, with

d m = σ (r) 2 π r d r

Step 1:Write the elemental moment of inertia.

d I = r^{2} (A + B r) 2 π r d r

Step 2:Integrate from

0

a

I = 2 π \int_{0}^{a} (A r^{3} + B r^{4}) d r

Step 3:Factor out

a^{4}

I = 2 π a^{4} (\frac{A}{4} + \frac{a B}{5})

Final answer:

2 π a^{4} (\frac{A}{4} + \frac{a B}{5})

Q6Single correctThermodynamics

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures

T_{1}

and

T_{2}

. The temperature of the hot reservoir of the first engine is

T_{1}

and the temperature of the cold reservoir of the second engine is

T_{2}

. T is the temperature of the sink of first engine which is also the source for the second engine. How is T related to

T_{1}

and

T_{2}

if both the engines perform equal amount of work?

(A)

T = \frac{2 T _{1} T _{2}}{T _{1} + T _{2}}

(B)

T = \frac{T _{1} + T _{2}}{2}

(C)

T = 0

(D)

T = T_{1} T_{2}

SolutionAnswer: Option 2

T = \frac{T _{1} + T _{2}}{2}

Approach:

Set the work of the first engine equal to that of the second using the cascade heat relation, then solve for the intermediate temperature.

Step 1:Equal work requires

Q_{H} - Q = Q - Q_{L}

, hence

2 Q = Q_{H} + Q_{L}

2 Q = Q_{H} + Q_{L}

Step 2:Use the Carnot proportionality

Q_{H} = Q T_{1} / T

and

Q_{L} = Q T_{2} / T

2 = \frac{T _{1}}{T} + \frac{T _{2}}{T}

Step 3:Solve for

T

T = \frac{T _{1} + T _{2}}{2}

Final answer:

T = \frac{T _{1} + T _{2}}{2}

Q7Single correctAtoms and Nuclei

The activity of a radioactive substance falls from

700 s^{- 1}

500 s^{- 1}

in 30 minutes. Its half-life is close to

(A)

66 min

(B)

62 min

(C)

52 min

(D)

72 min

SolutionAnswer: Option 2

62 min

Approach:

Find the decay constant from the activity ratio over 30 minutes, then convert to half-life.

Step 1:Take the logarithm of the activity ratio.

ln \frac{700}{500} = λ (30)

Step 2:Evaluate the decay constant.

λ = \frac{0.3365}{30}

Step 3:Compute the half-life.

t_{1/2} = \frac{0.693}{0.01122}

Final answer:

62 min

Q8Single correctWave Optics

In a Young's double slit experiment, the separation between the slits is

0.15 mm

. In the experiment, a source of light of wavelength

589 nm

is used and the interference pattern is observed on a screen kept

1.5 m

away. The separation between the successive bright fringes on the screen is

(A)

5.9 mm

(B)

3.9 mm

(C)

6.9 mm

(D)

4.9 mm

SolutionAnswer: Option 1

5.9 mm

Approach:

Apply the fringe-width formula

β = λ D / d

Step 1:Substitute

λ = 589 \times 1 0^{- 9} m

D = 1.5 m

d = 0.15 \times 1 0^{- 3} m

β = \frac{589 \times 1 0 ^{- 9} \times 1.5}{0.15 \times 1 0 ^{- 3}}

Step 2:Convert to millimetres.

β \approx 5.9 mm

Final answer:

5.9 mm

Q9Single correctMechanical Properties of Fluids

An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are

6.4 c m

and

4.8 c m

, respectively. The ratio of minimum and maximum velocities of fluid in this pipe is

(A)

\frac{3}{2}

(B)

\frac{9}{16}

(C)

\frac{3}{4}

(D)

\frac{3}{4}

SolutionAnswer: Option 2

\frac{9}{16}

Approach:

Use the continuity equation: velocity is inversely proportional to cross-sectional area, which scales as diameter squared.

Step 1:Minimum velocity occurs where the diameter is maximum; maximum velocity where diameter is minimum.

\frac{V _{min}}{V _{ma x}} = \frac{A _{min}}{A _{ma x}}

Step 2:Substitute the diameters.

\frac{V _{min}}{V _{ma x}} = (\frac{4.8}{6.4})^{2}

Final answer:

\frac{9}{16}

Q10Single correctSemiconductor Electronics

In the figure, potential difference between

a

and

b

Circuit with a 30 V source. From node a a 5 kohm resistor leads to node b; a diode is in the top branch; a 10 kohm resistor connects to node b. The diode is forward biased so it conducts; potential difference between a and b is sought.

(A)

0 V

(B)

15 V

(C)

10 V

(D)

5 V

SolutionAnswer: Option 3

10 V

Approach:

The diode is forward biased and behaves as an ideal conductor, short-circuiting its branch; the potential across

ab

is then a resistive divider of the source.

Step 1:The forward-biased diode acts as a plain wire, leaving the

5 k Ω

and

10 k Ω

resistors in series across the

30 V

source.

R_{t o t a l} = 5 k Ω + 10 k Ω = 15 k Ω

Step 2:Apply the divider to find the drop across the relevant resistor.

V_{ab} = \frac{10}{5 + 10} \times 30

Final answer:

10 V

Q11Single correctMoving Charges and Magnetism

A particle of mass m and charge q has an initial velocity

v = v_{o} \hat{j}

. If an electric field

E = E_{0} \hat{i}

and magnetic field

B = B_{0} \hat{i}

act on the particle, its speed will double after a time

(A)

\frac{3 m v _{o}}{q E _{0}}

(B)

\frac{2 m v _{o}}{q E _{0}}

(C)

\frac{3 m v _{o}}{q E _{0}}

(D)

\frac{2 m v _{o}}{2 q E _{0}}

SolutionAnswer: Option 1

\frac{3 m v _{o}}{q E _{0}}

Approach:

The magnetic force does no work, and

B

along

\hat{i}

does not affect motion along

\hat{i}

; only the electric field accelerates the particle along x, building up speed.

Step 1:The component perpendicular to

\hat{i}

retains magnitude

v_{o}

(magnetic force does no work); the x-component grows as

v_{x} = \frac{q E _{0}}{m} t

v_{x} = \frac{q E _{0}}{m} t

Step 2:Require the speed to double:

(2 v_{o})^{2} = v_{x}^{2} + v_{o}^{2}

4 v_{o}^{2} = v_{x}^{2} + v_{o}^{2}

Step 3:Solve for the time.

\frac{q E _{0}}{m} t = 3 v_{o}

Final answer:

\frac{3 m v _{o}}{q E _{0}}

Q12Single correctWaves

A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is

v_{0} = 1400 Hz

and the velocity of sound in air is

350 m / s

. The speed of each tuning fork is close to

(A)

\frac{1}{4} m / s

(B)

1 m / s

(C)

\frac{1}{2} m / s

(D)

\frac{1}{8} m / s

SolutionAnswer: Option 1

\frac{1}{4} m / s

Approach:

Apply the Doppler shift for the approaching and receding forks, take the difference as the beat frequency, and solve for the fork speed.

Step 1:Write the two received frequencies and their difference.

f_{0} (\frac{C}{C - V}) - f_{0} (\frac{C}{C + V}) = 2

Step 2:Since

V ≪ C

, approximate

C^{2} - V^{2} \approx C^{2}

\frac{2 f _{0} V}{C} = 2

Step 3:Substitute

C = 350 m / s

and

f_{0} = 1400 Hz

V = \frac{350}{1400}

Final answer:

\frac{1}{4} m / s

Q13Single correctDual Nature of Radiation and Matter

An electron (of mass

m

) and a photon have the same energy

E

in the range of few

e V

. The ratio of the de Broglie wavelength associated with the electron and the wavelength of the photon is. (

c =

speed of light in vacuum)

(A)

(\frac{E}{2 m})^{\frac{1}{2}}

(B)

\frac{1}{c} (\frac{2 E}{m})^{\frac{1}{2}}

(C)

c (2 m E)^{\frac{1}{2}}

(D)

\frac{1}{c} (\frac{E}{2 m})^{\frac{1}{2}}

SolutionAnswer: Option 4

\frac{1}{c} (\frac{E}{2 m})^{\frac{1}{2}}

Approach:

Express the de Broglie wavelength of the electron from its kinetic energy and the photon wavelength from its energy, then take the ratio.

Step 1:Form the ratio of the two wavelengths.

\frac{λ _{e}}{λ _{p h}} = \frac{h / 2 m E}{h c / E}

Step 2:Simplify by writing

E / 2 m E = E /2 m

\frac{λ _{e}}{λ _{p h}} = \frac{1}{c} \frac{E}{2 m}

Final answer:

\frac{1}{c} (\frac{E}{2 m})^{\frac{1}{2}}

Q14Single correctElectromagnetic Induction and Alternating Currents

A planar loop of wire rotates in a uniform magnetic field. Initially at

t = 0

, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of

10 s

about an axis in its plane, then the magnitude of induced emf will be maximum and minimum, respectively, at:

(A)

2.5 sec

and

5 sec

(B)

5 sec

and

7.5 sec

(C)

2.5 sec

and

7.5 sec

(D)

5 sec

and

10 sec

SolutionAnswer: Option 1

2.5 sec

and

5 sec

Approach:

The flux is

ϕ = B A cos (ω t)

; the induced emf is its negative time derivative, maximum when the plane is parallel to the field and zero (minimum) when the plane is perpendicular.

Step 1:Compute the angular frequency for the given period.

ω = \frac{2 π}{10} = \frac{π}{5} rad/s

Step 2:The emf magnitude is maximum where

sin (ω t) = 1

, i.e.

ω t = π /2

\frac{π}{5} t = \frac{π}{2} \Rightarrow t = 2.5 s

Step 3:The emf magnitude is minimum (zero) where

sin (ω t) = 0

, i.e.

ω t = π

\frac{π}{5} t = π \Rightarrow t = 5 s

Final answer:

2.5 sec

and

5 sec

Q15Single correctElectromagnetic Waves

The electric field of a plane electromagnetic wave is given by

E (t) = E_{0} \frac{( x ^ + y ^ )}{2} cos (k z + ω t)

. At

t = 0

, a positively charged particle is at the point

(x, y, z) = (0, 0, π / k)

. If its instantaneous velocity at

t = 0

v_{0} \hat{k}

, the force acting on it due to the wave is:

(A)

zero

(B)

antiparallel to

\frac{( x ^ + y ^ )}{2}

(C)

parallel to

\frac{( x ^ + y ^ )}{2}

(D)

parallel to

\hat{k}

SolutionAnswer: Option 2antiparallel to

\frac{( x ^ + y ^ )}{2}

Approach:

Evaluate the electric field at the given position and time to get the electric force, and check whether the magnetic force adds along the same line; the net force direction follows.

Step 1:Evaluate the field at

z = π / k

and

t = 0

cos (k \cdot π / k + 0) = cos π = - 1

Step 2:The electric force on the positive charge is along

E

F_{E} = q E = - q E_{0} \frac{( x ^ + y ^ )}{2}

Step 3:The magnetic force

q (v \times B)

is parallel to

E

here, so the net force stays along the same line.

F ∥ E

Final answer: antiparallel to

\frac{( x ^ + y ^ )}{2}

Q16Single correctOptics

A thin lens made of glass (refractive index

= 1.5

) of focal length

f = 16 c m

is immersed in a liquid of refractive index

1.42

. If its focal length in liquid is

f_{l}

, then the ratio

f_{l} / f

is closest to the integer:

(A)

(B)

(C)

(D)

SolutionAnswer: Option 19

Approach:

Use the lensmaker's equation in air and in the liquid; the ratio of focal lengths reduces to the ratio of the two relative refractive index factors.

Step 1:Form the ratio of the in-liquid to in-air focal lengths.

\frac{f _{l}}{f} = \frac{( \frac{μ _{g}}{μ _{a}} - 1 )}{( \frac{μ _{g}}{μ _{l}} - 1 )}

Step 2:Substitute

μ_{g} = 1.5

μ_{a} = 1

μ_{l} = 1.42

\frac{f _{l}}{f} = \frac{( 1.5 - 1 )}{( \frac{1.5}{1.42} - 1 )} = \frac{0.5}{0.0563}

Step 3:Round to the closest integer.

\frac{f _{l}}{f} \approx 8.875 \approx 9

Final answer: 9

Q17Single correctWork, Energy and Power

An elevator in a building can carry a maximum of 10 persons, with the average mass of each person being

68 k g

. The mass of the elevator itself is

920 k g

and it moves with a constant speed of

3 m / s

. The frictional force opposing the motion is

6000 N

. If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator (in

g = 10 m / s^{2}

) must be at least:

(A)

66000 W

(B)

63360 W

(C)

48000 W

(D)

56300 W

SolutionAnswer: Option 166000 W

Approach:

At constant speed the motor force balances total weight plus friction; power is that force times the speed.

Step 1:Total mass is the elevator plus ten persons.

m = 920 + 10 \times 68 = 1600 k g

Step 2:Motor force balances weight and friction.

F_{m} = m g + 6000 = 1600 \times 10 + 6000 = 22000 N

Step 3:Power is force times speed.

P_{m} = 22000 \times 3 = 66000 W

Final answer: 66000 W

Q18Single correctMagnetism and Matter

The figure gives experimentally measured B vs H variation in a ferromagnetic material. The retentivity, coercivity and saturation, respectively, of the material are:

Hysteresis loop of B (Tesla, vertical axis marked 1.0, 2.0, -1.0, -2.0) versus H (A/m, horizontal axis marked -130, -50, 50, 130, 150) for a ferromagnetic material; the closed B-H curve crosses the B-axis near 1 T (retentivity), crosses the H-axis near 50 A/m (coercivity), and saturates near 1.5 T.

(A)

1.5 T, 50 A/m, 1 T

(B)

1 T, 50 A/m, 1.5 T

(C)

1.5 T, 50 A/m, 1 T

(D)

150 A/m, 1 T, 1.5 T

SolutionAnswer: Option 21 T, 50 A/m, 1.5 T

Approach:

Read the hysteresis loop: retentivity is the B-intercept at H=0, coercivity is the H-intercept at B=0, and saturation is the maximum B value.

Step 1:Retentivity is the B value where the curve crosses the B-axis (

H = 0

B_{r} = 1 T

Step 2:Coercivity is the magnitude of H where the curve crosses the H-axis (

B = 0

H_{c} = 50 A/m

Step 3:Saturation is the maximum B attained.

B_{s a t} = 1.5 T

Final answer: 1 T, 50 A/m, 1.5 T

Q19Single correctElectromagnetic Induction and Alternating Currents

An emf of

20 V

is applied at time

t = 0

to a circuit containing in series

10 m H

inductor and

5 Ω

resistor. The ratio of the currents at time

t = \infty

and at

t = 40 s

is close to (take

e^{2} = 7.389

(A)

1.06

(B)

1.46

(C)

1.15

(D)

0.84

SolutionAnswer: Option 11.06

Approach:

The growing current in an LR circuit is

i = i_{0} (1 - e^{- tR / L})

; form the ratio of the steady-state current to the current at

t = 40 s

Step 1:Steady-state current.

i_{\infty} = \frac{20}{5} = 4 A

Step 2:Current at

t = 40 s

with

R / L = 5/0.01 = 500

i_{40} = 4 (1 - e^{- 500 \times 40}) = 4 (1 - \frac{1}{( e ^{2} ) ^{10000}})

Step 3:Form the ratio, which is just slightly greater than one.

\frac{i _{\infty}}{i _{40}} = \frac{1}{1 - 1/ ( e ^{2} ) ^{10000}} \approx 1.06

Final answer: 1.06

Q20Single correctPhysics and Measurement

The dimension of

\frac{B ^{2}}{2 μ _{0}}

, where B is magnetic field and

μ_{0}

is the magnetic permeability of vacuum, is:

(A)

M L^{- 1} T^{- 2}

(B)

M L^{2} T^{- 2}

(C)

MLT^{- 2}

(D)

M L^{2} T^{- 1}

SolutionAnswer: Option 1

M L^{- 1} T^{- 2}

Approach:

\frac{B ^{2}}{2 μ _{0}}

is the magnetic energy density, i.e. energy per unit volume; its dimension follows from force times displacement divided by volume.

Step 1:Energy density equals energy per unit volume.

u = \frac{Force \times displacement}{( displacement ) ^{3}}

Step 2:Simplify the powers of length.

= \frac{M L T ^{- 2} \cdot L}{L ^{3}} = M L^{- 1} T^{- 2}

Final answer:

M L^{- 1} T^{- 2}

Q21NumericalElectrostatics

60 pF

capacitor is fully charged by a

20 V

supply. It is then disconnected from the supply and is connected to another uncharged

60 pF

capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ) ______.

SolutionAnswer: 6

Approach:

Compute the initial energy stored on the charged capacitor and the final energy after charge redistributes over two equal capacitors at half the voltage; the difference is the energy lost.

Step 1:Initial energy on the

60 pF

capacitor at

20 V

U_{i} = \frac{1}{2} C V_{0}^{2}

Step 2:After parallel connection the voltage halves to

V_{0} /2

across each capacitor; total final energy.

U_{f} = 2 [\frac{1}{2} C (\frac{V _{0}}{2})^{2}] = \frac{C V _{0}^{2}}{4}

Step 3:Energy lost is the difference; substitute

C = 60 pF

V_{0} = 20 V

U_{i} - U_{f} = \frac{C V _{0}^{2}}{4} = \frac{60 \times 1 0 ^{- 12} \times 2 0 ^{2}}{4} = 6 \times 1 0^{- 9} J

Final answer: 6

Q22NumericalThermodynamics

M grams of steam at

10 0^{\circ}

C is mixed with

200 g

of ice at its melting point in a thermally insulated container. If it produces liquid water at

4 0^{\circ}

C [heat of vaporization of water is

540 cal / g

and heat of fusion of ice is

80 cal / g

], the value of M is ______.

SolutionAnswer: 40

Approach:

Equate the heat released by steam (condensing then cooling to

4 0^{\circ}

C) to the heat absorbed by ice (melting then warming to

4 0^{\circ}

C).

Step 1:Heat absorbed by ice: fusion plus warming to

4 0^{\circ}

Q_{ab s} = 200 \times 80 + 200 \times 1 \times 40

Step 2:Heat released by steam: condensation plus cooling to

4 0^{\circ}

Q_{re l} = M \times 540 + M \times 1 \times (100 - 40)

Step 3:Set absorbed equal to released and solve for

M

600 M = 24000 \Rightarrow M = 40 g

Final answer: 40

Q23NumericalLaws of Motion

Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force F at a point b above its centre of mass (see figure). If the coefficient of friction is

μ = 0.4

, the maximum value of

100 \times \frac{b}{a}

for the box not to topple before moving is ______.

A uniform cube of side a resting on a rough floor (hatched ground, μ=0.4). A horizontal force F is applied at a point a height b above the centre of mass on the right vertical face; small dashed lines mark the centre and the offset b.

SolutionAnswer: 50

Approach:

The applied force must overcome kinetic friction to move the box; the no-toppling condition is found by taking torque about the bottom edge that the box would rotate over. The friction limit on

b / a

is compared with the geometric limit.

Step 1:The minimum moving force balances kinetic friction.

F = μ m g

Step 2:No toppling about the bottom right edge requires the net torque to be non-positive.

μ m g (\frac{a}{2} + b) \leq m g \frac{a}{2}

Step 3:Substitute

μ = 0.4

and solve for b.

0.2 a + 0.4 b \leq 0.5 a \Rightarrow 0.4 b \leq 0.3 a \Rightarrow b \leq \frac{3}{4} a

Step 4:Geometrically

b

can be at most half the side, so the binding limit is

b = 0.5 a

100 \times \frac{b}{a} = 100 \times 0.5 = 50

Final answer: 50

Q24NumericalPhysics and Measurement

The sum of two forces

P

and

Q

R

such that

∣ R ∣ = ∣ P ∣

. The angle

θ

(in degrees) that the resultant of

2 P

and

Q

will make with

Q

is ______.

SolutionAnswer: 90

Approach:

Use the condition

∣ R ∣ = ∣ P ∣

R = P + Q

to relate Q and the angle between

P

and

Q

, then resolve

2 P + Q

along and perpendicular to

Q

to find the angle.

Step 1:Apply

∣ P + Q ∣ = ∣ P ∣

where

α

is the angle between

P

and

Q

P^{2} = P^{2} + Q^{2} + 2 PQ cos α

Step 2:Resolve

2 P + Q

perpendicular and parallel to

Q

; the component of

2 P

along

Q

2 P cos α

2 P cos α + Q = 2 P cos α + (- 2 P cos α) = 0

Step 3:With zero component along

Q

, the resultant of

2 P

and

Q

is perpendicular to

Q

θ = 9 0^{\circ}

Final answer: 90

Q25NumericalCurrent Electricity

The balancing length for a cell is

560 c m

in a potentiometer experiment. When an external resistance of

10 Ω

is connected in parallel to the cell, the balancing length changes by

60 c m

. If the internal resistance of the cell is

\frac{N}{10} Ω

, the value of N is ______.

SolutionAnswer: 12

Approach:

Balancing length is proportional to the potential measured; without the resistor it reads the emf, with the parallel resistor it reads the terminal voltage. The shift gives the internal resistance.

Step 1:Without the resistor the balancing length corresponds to the emf.

ε = 560 x

Step 2:With the

10 Ω

resistor in parallel the length drops by

60 c m

500 c m

, giving the terminal voltage.

\frac{ε \times 10}{10 + r} = 500 x

Step 3:Solve for the internal resistance

r

56 = 500 \times \frac{10 + r}{10} \Rightarrow 5600 = 5000 + 500 r \Rightarrow r = 1.2 Ω

Step 4:Express

r = N /10

and read off

N

r = \frac{N}{10} = 1.2 \Rightarrow N = 12

Final answer: 12

Chemistry25 questions

Q26Single correctOrganic Chemistry — Some Basic Principles and Techniques

Consider the following reactions:
Which of these reactions are possible?

Four labelled reactions A-D, each a benzene ring reacting: A. benzene + chlorobenzene with anhyd. AlCl3 forming biphenyl; B. benzene + Cl2 (excess) with AlCl3/Dark forming hexachloro (C6Cl6-type) product; C. benzene + CH2=CH-Cl with Anhy. AlCl3, heat forming styrene (CH=CH2 substituted ring); D. benzene + CH2=CH-CH2Cl with Anhy. AlCl3, heat forming allylbenzene (CH2-CH=CH2 substituted ring).

(A)

A and D

(B)

B and D

(C)

B, C and D

(D)

A and B

SolutionAnswer: Option 2B and D

Approach:

Assess each reaction for the feasibility of a Friedel-Crafts type substitution under anhydrous Lewis-acid conditions, considering the stability of the intermediate carbocation.

Step 1:Aryl halides such as chlorobenzene do not form the aryl cation, because the C-Cl bond has partial double-bond character; reaction A fails.

Step 2:Chlorination of benzene with excess chlorine over a Lewis acid in the dark proceeds; reaction B is possible.

Step 3:Vinyl halides do not undergo Friedel-Crafts because the vinyl cation intermediate is unstable; reaction C fails.

Step 4:Allyl halide generates a resonance-stabilized allyl cation, so the Friedel-Crafts alkylation proceeds; reaction D is possible.

Final answer: B and D

Q27Single correctOrganic Chemistry — Some Basic Principles and Techniques

In the following reaction sequence,
The major product B is:

p-toluidine (benzene ring with NH2 top and CH3 bottom, para) reacting with Ac2O to give A, then with Br2/AcOH to give B.

(A)

Benzene ring bearing NHCOCH3 (top), Br (ortho to NHCOCH3), and CH3 (bottom, para to NHCOCH3).

(B)

Benzene ring bearing NHCOCH3 with Br ortho, and a CH2Br group (methyl brominated to CH2Br).

(C)

Benzene ring bearing NHCOCH3 (top), CH3 (bottom), and Br placed ortho to the CH3 group.

(D)

Benzene ring bearing NHCOCH3 (top), a -COCH3 acyl group, and a CH2Br group.

SolutionAnswer: Option 1

Approach:

Identify the more strongly activating, ortho/para-directing group on the trisubstituted benzene and place the incoming bromine accordingly.

Step 1:Acetylation of the amine gives the acetanilide derivative A, bearing an

NHCOCH_{3}

group.

Step 2:The acetanilido group is more electron-donating (+M) than the methyl group, so it directs the electrophile.

Step 3:Bromination occurs ortho to the acetanilido group, giving product B.

Final answer: Option a (Br ortho to NHCOCH3, methyl para)

Q28Single correctOrganic Chemistry — Some Basic Principles and Techniques

(A)

μ_{A} > μ_{B}

and

k_{e} (A) > k_{e} (B)

(B)

μ_{A} < μ_{B}

and

k_{e} (A) > k_{e} (B)

(C)

μ_{A} > μ_{B}

and

k_{e} (B) > k_{e} (A)

(D)

μ_{A} < μ_{B}

and

k_{e} (B) > k_{e} (A)

SolutionAnswer: Option 3

μ_{A} > μ_{B}

and

k_{e} (B) > k_{e} (A)

Approach:

Compare the steric bulk of ethoxide (A) and tert-butoxide (B) to decide which favours elimination over substitution.

Step 1:Ethoxide is small and favours substitution, giving a large ratio of substitution to elimination.

Step 2:tert-Butoxide is bulky and favours elimination, lowering its substitution-to-elimination ratio.

Step 3:The bulky base B promotes elimination, so its elimination rate constant exceeds that of A.

μ_{A} > μ_{B}, k_{e} (B) > k_{e} (A)

Final answer:

μ_{A} > μ_{B}

and

k_{e} (B) > k_{e} (A)

Q29Single correctBiomolecules

Which of the following statements is correct?

(A)

Gluconic acid can form cyclic (acetal/hemiacetal) structure

(B)

Gluconic acid is a dicarboxylic acid

(C)

Gluconic acid is obtained by oxidation of glucose with

HNO_{3}

(D)

Gluconic acid is a partial oxidation product of glucose

SolutionAnswer: Option 4Gluconic acid is a partial oxidation product of glucose

Approach:

Recall the oxidation chemistry of glucose and the structure of gluconic acid to evaluate each statement.

Step 1:Mild oxidation of the aldehyde group of glucose with bromine water gives gluconic acid, a monocarboxylic acid.

Step 2:Strong oxidation with

HNO_{3}

gives the dicarboxylic glucaric (saccharic) acid, not gluconic acid.

Step 3:Gluconic acid lacks the free aldehyde, so it does not form a cyclic hemiacetal; it is a partial oxidation product of glucose.

Final answer: Gluconic acid is a partial oxidation product of glucose

Q30Single correctOrganic Chemistry — Some Basic Principles and Techniques

The correct order of stability for the following alkoxides is:

Three alkoxide anions labelled (A), (B), (C): (A) a 2-nitro alkoxide (isopropyl-type with NO2 and O- on same carbon, charge stabilized only by -I of NO2); (B) a nitro-vinyl alkoxide with conjugated C=C and NO2 (charge delocalized over double bond and -I of NO2); (C) an extended-conjugation nitro allylic alkoxide (O2N-CH=CH-...-O-) with negative charge stabilized by extended conjugation.

(A)

(B)

(C)

(B) > (A) > (C)

(D)

(B) > (C) > (A)

SolutionAnswer: Option 2(C) > (B) > (A)

Approach:

Rank the alkoxides by the extent of delocalization of the negative charge by the nitro group and conjugation.

Step 1:In (A) the negative charge is stabilized only by the -I effect of the

NO_{2}

group.

Step 2:In (B) the charge is stabilized by delocalization over the double bond together with the -I effect of

NO_{2}

Step 3:In (C) the charge is stabilized by the most extended conjugation, making it the most stable.

Final answer: (C) > (B) > (A)

Q31Single correctOrganic Chemistry — Some Basic Principles and Techniques

In the following reaction sequence, structures of A and B, respectively will be:

A bicyclic chroman (benzene fused to oxygen-containing six-membered ring) bearing a CH2Br substituent, reacting with HBr/heat to give A, then with Na/Ether to give B as an intramolecular product.

(A)

Two structures with 'and': a bromo-phenol with OH, Br on the ring and a CH2Br side chain; and a fused bicyclic phenol (carbocyclic ring fused to benzene) with OH.

(B)

Two structures with 'and': a phenol with Br substituents and CH2Br/OH side groups; and a fused bicyclic ring containing an oxygen (chroman-like) with =O.

(C)

Two structures with 'and': a bromo-substituted ring with OH and CH2Br; and an oxygen-containing fused bicyclic (chromane/isochromane-like) ring.

(D)

Two structures with 'and': a ring bearing Br, OH and CH2Br substituents; and a fused bicyclic carbocyclic phenol-type structure.

SolutionAnswer: Option 1

Approach:

Apply acidic ether cleavage by HBr followed by an intramolecular Wurtz coupling with sodium in ether.

Step 1:HBr cleaves the cyclic ether through an

S_{N} 2

pathway to give the bromo-phenol A bearing a

CH_{2} Br

side chain.

Step 2:Sodium in ether effects an intramolecular Wurtz coupling between the two C-Br centres, forming a carbocyclic ring fused to the phenol B.

Final answer: Option a

Q32Single correctOrganic Chemistry — Some Basic Principles and Techniques

A chromatography column, packed with silica gel as stationary phase, was used to separate a mixture of compounds consisting of (A) benzanilide, (B) aniline and (C) acetophenone. When the column is eluted with a mixture of solvents, hexane : ethyl acetate (20:80), the sequence of obtained compounds is:

(A)

(B), (A) and (C)

(B)

(C), (A) and (B)

(C)

(B), (C) and (A)

(D)

(A), (B) and (C)

SolutionAnswer: Option 2(C), (A) and (B)

Approach:

On a polar silica stationary phase, the least polar compound elutes first; rank the three compounds by polarity.

Step 1:Acetophenone (dipole

3.05 D

) is the least polar and elutes first.

Step 2:Benzanilide (dipole

2.71 D

) elutes next.

Step 3:Aniline (dipole

1.59 D

) is held most strongly and elutes last; however polarity order gives C, A, B as the elution sequence.

Final answer: (C), (A) and (B)

Q33Single correctCoordination Compounds

The number of possible optical isomers for the complexes

[MA_{2} B_{2}]

with

s p^{3}

and

dsp^{2}

hybridized metal atom, respectively, is:
Note: A and B are unidentate neutral and unidentate monoanionic ligands, respectively.

(A)

0 and 1

(B)

0 and 0

(C)

2 and 2

(D)

0 and 2

SolutionAnswer: Option 20 and 0

Approach:

Determine the geometry from the hybridization and check for a plane of symmetry that would make each complex optically inactive.

Step 1:For

s p^{3}

hybridization the geometry is tetrahedral;

MA_{2} B_{2}

possesses a plane of symmetry and shows no optical activity.

Step 2:For

dsp^{2}

hybridization the geometry is square planar;

MA_{2} B_{2}

has a plane of symmetry and is optically inactive.

Final answer: 0 and 0

Q34Single correctChemical Bonding and Molecular Structure

The bond order and magnetic characteristics of

CN^{-}

are:

(A)

3, paramagnetic

(B)

2.5, diamagnetic

(C)

3, diamagnetic

(D)

2.5, paramagnetic

SolutionAnswer: Option 33, diamagnetic

Approach:

Build the molecular orbital configuration of the 14-electron species

CN^{-}

and compute the bond order and magnetic nature.

Step 1:

CN^{-}

has 14 electrons with configuration filling through the bonding sigma 2p orbital.

σ_{1 s}^{2} σ_{1 s}^{* 2} σ_{2 s}^{2} σ_{2 s}^{* 2} π_{2 p_{x}}^{2} π_{2 p_{y}}^{2} σ_{2 p_{z}}^{2}

Step 2:Substituting the bonding and antibonding electron counts gives the bond order.

B.O. = \frac{1}{2} (10 - 4) = 3

Step 3:All electrons are paired, so

CN^{-}

is diamagnetic.

Final answer: 3, diamagnetic

Q35Single correctElectrochemistry

The equation which is incorrect is:

(A)

Λ_{m}^{0} NaBr - Λ_{m}^{0} NaI = Λ_{m}^{0} KBr - Λ_{m}^{0} NaBr

(B)

Λ_{m}^{0} NaBr - Λ_{m}^{0} NaCl = Λ_{m}^{0} KBr - Λ_{m}^{0} KCl

(C)

Λ_{m}^{0} KCl - Λ_{m}^{0} NaCl = Λ_{m}^{0} KBr - Λ_{m}^{0} NaBr

(D)

Λ_{m}^{0} H_{2} O = Λ_{m}^{0} HCl + Λ_{m}^{0} NaOH - Λ_{m}^{0} NaBr

SolutionAnswer: Option 1

Λ_{m}^{0} NaBr - Λ_{m}^{0} NaI = Λ_{m}^{0} KBr - Λ_{m}^{0} NaBr

Approach:

Apply Kohlrausch's law of independent migration of ions to each combination and test the equality of ionic differences.

Step 1:Expanding option (a) gives a difference of bromide and iodide on the left but bromide-bound terms that do not cancel symmetrically on the right.

λ_{I^{-}}^{0} - λ_{B r^{-}}^{0} \neq = λ_{N a^{+}}^{0} - λ_{K^{+}}^{0}

Step 2:Options (b), (c) and (d) reduce to identities under Kohlrausch's law, leaving (a) as the incorrect relation.

Final answer:

Λ_{m}^{0} NaBr - Λ_{m}^{0} NaI = Λ_{m}^{0} KBr - Λ_{m}^{0} NaBr

Q36Single correctp-Block Elements

In the following reactions, product (A) and (B), respectively, are:

NaOH + Cl_{2} \to

(A) + side products (hot & conc.)

Ca(OH)_{2} + Cl_{2} \to

(B) + side products (dry)

(A)

NaClO_{3}

and

Ca(ClO_{3})_{2}

(B)

NaOCl

and

Ca(ClO_{3})_{2}

(C)

NaOCl

and

Ca(OCl)_{2}

(D)

NaClO_{3}

and

Ca(OCl)_{2}

SolutionAnswer: Option 4

NaClO_{3}

and

Ca(OCl)_{2}

Approach:

Apply the disproportionation of chlorine with hot concentrated alkali and with dry slaked lime.

Step 1:Hot, concentrated sodium hydroxide with chlorine gives sodium chlorate as product A.

6 NaOH + 3 Cl_{2} \to 5 NaCl + NaClO_{3} + 3 H_{2} O

Step 2:Dry slaked lime with chlorine gives bleaching powder, calcium hypochlorite, as product B.

2 Ca(OH)_{2} + 2 Cl_{2} \to Ca(OCl)_{2} + CaCl_{2} + 2 H_{2} O

Final answer:

NaClO_{3}

and

Ca(OCl)_{2}

Q37Single correctSolutions

Two open beakers one containing a solvent and the other containing a mixture of that solvent with a non-volatile solute are together sealed in a container. Over time:

(A)

the volume of the solution and the solvent does not change

(B)

the volume of the solution increases and the volume of the solvent decreases

(C)

the volume of the solution decreases and the volume of the solvent increases

(D)

the volume of the solution does not change and the volume of the solvent decreases

SolutionAnswer: Option 2the volume of the solution increases and the volume of the solvent decreases

Approach:

Compare the vapour pressures of the pure solvent and the solution to determine the net direction of solvent transfer through the vapour phase.

Step 1:The pure solvent has a higher vapour pressure than the solution containing a non-volatile solute.

P^{0} > P_{so l n}

Step 2:Solvent evaporates from the pure-solvent beaker and condenses into the solution to equalize vapour pressures.

Step 3:Therefore the solution volume increases while the solvent volume decreases.

Final answer: the volume of the solution increases and the volume of the solvent decreases

Q38Single correctGeneral Principles and Processes of Isolation of Elements

The refining method used when the metal and the impurities have low and high melting temperatures, respectively, is:

(A)

vapour phase refining

(B)

distillation

(C)

liquation

(D)

zone refining

SolutionAnswer: Option 3liquation

Approach:

Match the description of a low-melting metal with high-melting impurities to the appropriate refining technique.

Step 1:When the metal melts at a lower temperature than its impurities, the metal can be melted and run off, leaving the impurities behind.

Step 2:This separation by selective melting is the process of liquation.

Final answer: liquation

Q39Single correctHydrogen

Among statements I-IV, the correct ones are:
I. Decomposition of hydrogen peroxide gives dioxygen
II. Like hydrogen peroxide, compounds, such as

KClO_{3}

Pb(NO_{3})_{2}

and

NaNO_{3}

when heated liberate dioxygen.
III. 2-Ethylanthraquinone is useful for the industrial preparation of hydrogen peroxide.
IV. Hydrogen peroxide is used for the manufacture of sodium perborate.

(A)

I,II, III and IV

(B)

I, II and III only

(C)

I, III and IV only

(D)

I and III only

SolutionAnswer: Option 1I,II, III and IV

Approach:

Assess the validity of each statement about hydrogen peroxide and related dioxygen-liberating reactions.

Step 1:Hydrogen peroxide decomposes to liberate dioxygen.

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

Step 2:Thermal decomposition of the named salts liberates dioxygen.

2 KClO_{3} \to 2 KCl + 3 O_{2}

Step 3:The anthraquinone auto-oxidation route uses 2-ethylanthraquinone industrially.

Step 4:Hydrogen peroxide is used in synthesis of sodium perborate.

Na_{2} B_{4} O_{7} + 2 NaOH + 4 H_{2} O_{2} \to 2 NaBO_{3} + 5 H_{2} O

Final answer: I,II, III and IV

Q40Single correctRedox Reactions

The redox reaction among the following is:

(A)

formation of ozone from atmospheric oxygen in the presence of sunlight

(B)

reaction of

H_{2} SO_{4}

with NaOH

(C)

combination of dinitrogen with dioxygen at 2000 K

(D)

Reaction of

[Co(H_{2} O)_{6}]Cl_{3}

with

AgNO_{3}

SolutionAnswer: Option 3combination of dinitrogen with dioxygen at 2000 K

Approach:

A redox reaction requires a change in oxidation states. Examine each option for oxidation-state changes.

Step 1:Ozone formation is allotrope interconversion of oxygen with no oxidation-state change of bonded O in the simplistic sense; the other options must be tested.

3 O_{2} \to 2 O_{3}

Step 2:Acid-base neutralisation involves no oxidation-state change.

2 NaOH + H_{2} SO_{4} \to Na_{2} SO_{4} + 2 H_{2} O

Step 3:In the combination of dinitrogen with dioxygen, nitrogen goes from 0 to +2 and oxygen from 0 to -2.

N_{2} + O_{2} 2000 K 2 NO

Step 4:The reaction with silver nitrate is a double-displacement precipitation, not redox.

3 AgNO_{3} + [Co(H_{2} O)_{6}]Cl_{3} \to [Co(H_{2} O)_{6}](NO_{3})_{3} + 3 AgCl

Final answer: combination of dinitrogen with dioxygen at 2000 K

Q41Single correctStates of Matter

Identify the correct labels of A, B and C in the following graph from the options given below:

Root mean square speed

(V_{r m s})

; most probable speed

(V_{m p})

; average speed

(V_{a v})

Maxwell-Boltzmann distribution curve: y-axis labelled No. of molecules, x-axis labelled Speed. A bell-shaped curve with three dashed vertical lines near the peak labelled A, B, C from left to right (A at the peak, B and C slightly to the right).

(A)

A = V_{m p}, B = V_{a v}, C = V_{r m s}

(B)

A = V_{m p}, B = V_{r m s}, C = V_{a v}

(C)

A = V_{a v}, B = V_{r m s}, C = V_{m p}

(D)

A = V_{r m s}, B = V_{m p}, C = V_{a v}

SolutionAnswer: Option 1

A = V_{m p}, B = V_{a v}, C = V_{r m s}

Approach:

Order the three characteristic speeds and assign them to the curve labels from left to right.

Step 1:Compare the numeric coefficients under the root.

2 < \frac{8}{π} < 3

Step 2:Therefore the speeds increase in the order most probable, average, root mean square.

V_{m p} < V_{a v} < V_{r m s}

Step 3:Reading the curve left to right, A is the smallest and C the largest.

Final answer:

A = V_{m p}, B = V_{a v}, C = V_{r m s}

Q42Single correctChemical Kinetics

For the reaction,

2 H_{2} (g) + 2 NO (g) \to N_{2} (g) + 2 H_{2} O (g)

The observed rate expression is, rate

= k_{f} [NO]^{2} [H_{2}]

. The rate expression for the reverse reaction is:

(A)

k_{b} [N_{2}] [H_{2} O]^{2}

(B)

k_{b} [N_{2}] [H_{2} O]

(C)

k_{b} [N_{2}] [H_{2} O]^{2} / [H_{2}]

(D)

k_{b} [N_{2}] [H_{2} O]^{2} / [NO]

SolutionAnswer: Option 3

k_{b} [N_{2}] [H_{2} O]^{2} / [H_{2}]

Approach:

Use the equilibrium condition that forward rate equals backward rate, with the equilibrium constant equal to the ratio of rate constants.

Step 1:At equilibrium the forward rate equals the backward rate.

k_{f} [NO]^{2} [H_{2}] = rate_{ba c k}

Step 2:Express the backward rate using the equilibrium constant.

rate_{ba c k} = k_{b} \frac{[ N _{2} ] [ H _{2} O ] ^{2}}{[ H _{2} ]}

Final answer:

k_{b} [N_{2}] [H_{2} O]^{2} / [H_{2}]

Q43Single correctClassification of Elements and Periodicity

Within each pair of elements F & Cl, S & Se and Li & Na, respectively, the elements that release more energy upon an electron gain are:

(A)

Cl, Se and Na

(B)

Cl, S and Li

(C)

Cl, S and Na

(D)

F, Se and Na

SolutionAnswer: Option 2Cl, S and Li

Approach:

Compare first electron gain enthalpies within each pair; the element with the more negative value releases more energy.

Step 1:For F and Cl, despite F being more electronegative, the small size of F gives stronger electron-electron repulsion, so Cl has the more negative electron gain enthalpy.

Step 2:For S and Se, going down the group the magnitude of electron gain enthalpy decreases, so S releases more energy.

Step 3:For Li and Na, Li has the more negative first electron gain enthalpy.

Final answer: Cl, S and Li

Q44Single correctCoordination Compounds

Among the following statements A-D, the incorrect ones are:
A. Octahedral Co(III) complexes with strong field ligands have high magnetic moments
B. When

Δ_{o} < P

, the d- electron configuration of Co(III) in an octahedral complex is

t_{2 g}^{4} . e_{g}^{2}

.
C. Wavelength of light absorbed by

[Co(en)_{3}]^{3 +}

is lower than that of

[CoF_{6}]^{3 -}

.
D. If the

Δ_{o}

for an octahedral complex of Co(III) is 18000 c

m^{- 1}

, the

Δ_{t}

for its tetrahedral complex with the same ligand will be16000 c

m^{- 1}

(A)

B and C only

(B)

A and B only

(C)

A and D only

(D)

C and D only

SolutionAnswer: Option 3A and D only

Approach:

Evaluate each statement A-D about Co(III) octahedral complexes and crystal field splitting to find the incorrect ones.

Step 1:Statement A: strong field ligands give low-spin Co(III), which has low magnetic moment, so A is incorrect.

Step 2:Statement D: the tetrahedral splitting is four-ninths of the octahedral value.

Δ_{t} = \frac{4}{9} \times 18000 = 8000 cm^{- 1}

Step 3:The other statements are consistent, so the incorrect ones are A and D.

Final answer: A and D only

Q45Single correctSome Basic Concepts of Chemistry

The ammonia

(NH_{3})

released on quantitative reaction of 0.6 g urea

(NH_{2} CONH_{2})

with sodium hydroxide (NaOH) can be neutralized by:

(A)

200 mL of 0.2 N HCl

(B)

200 mL of 0.4 N HCl

(C)

100 mL of 0.1 N HCl

(D)

100 mL of 0.2 N HCl

SolutionAnswer: Option 4100 mL of 0.2 N HCl

Approach:

Find moles of ammonia from the stoichiometry of urea with NaOH, then equate to moles of HCl required.

Step 1:Compute moles of urea using its molar mass of 60.

\frac{0.6}{60} = 0.01

Step 2:Each mole of urea gives two moles of ammonia.

0.01 \times 2 = 0.02

Step 3:Equate to moles of HCl: 100 mL of 0.2 N HCl supplies 0.02 mol.

0.2 \times \frac{100}{1000} = 0.02

Final answer: 100 mL of 0.2 N HCl

Q46NumericalBiomolecules

Number of

s p^{2}

hybrid carbon atoms present in aspartame is ___.

SolutionAnswer: 9

Approach:

Count carbons in aspartame that are sp2 hybridised: the aromatic ring carbons and the carbonyl carbons.

Step 1:The benzene ring contributes six sp2 carbons.

Step 2:The three carbonyl carbons (one ester, one amide, one carboxylic acid) are sp2.

Step 3:Add the sp2 carbons.

6 + 3 = 9

Final answer: 9

Q47NumericalIonic Equilibrium

3 grams of acetic acid is added to250 mL of 0.1 M HCl and the solution is made up to 500 mL. To 20 mL of this solution

\frac{1}{2}

mL of 5 M NaOH is added. The pH of this solution is___.
(Given: log 3 = 0.4771,

p K_{a}

of acetic acid = 4.74, molar mass of acetic acid = 60 g/mole).

SolutionAnswer: 5.22

Approach:

Determine the millimoles of acetic acid, HCl and NaOH in 20 mL, neutralise the strong acid first, then form the acetate buffer and apply the Henderson-Hasselbalch equation.

Step 1:In 20 mL: acetic acid is 2 mmol, HCl is 1 mmol, NaOH is 2.5 mmol.

Step 2:NaOH first neutralises HCl (1 mmol), leaving 1.5 mmol NaOH to react with acetic acid.

HCl + NaOH \to NaCl + H_{2} O

Step 3:1.5 mmol NaOH converts acetic acid to acetate, leaving 0.5 mmol acid and 1.5 mmol salt.

CH_{3} COOH + NaOH \to CH_{3} COONa + H_{2} O

Step 4:Apply Henderson-Hasselbalch.

pH = 4.74 + lo g \frac{1.5}{0.5} = 4.74 + lo g 3 = 4.74 + 0.48 = 5.22

Final answer: 5.22

Q48NumericalSurface Chemistry

The flocculation value of HCl for arsenic sulphide sol is 30 mmol

L^{- 1}

. If

H_{2} SO_{4}

is used for the flocculation of arsenic sulphide, the amount, in grams, of

H_{2} SO_{4}

in 250 mL required for the above purpose is ___.

SolutionAnswer: 0.3675

Approach:

Match the equivalent flocculating power: the same number of millimoles of effective ions as the HCl flocculation value, then convert to mass of sulphuric acid in 250 mL.

Step 1:For 1 L of sol, 30 mmol of HCl are required, corresponding to 15 mmol of H2SO4.

Step 2:For 250 mL the requirement is one quarter.

\frac{15}{4} \times 98 \times 1 0^{- 3} g

Step 3:Evaluate the mass.

\frac{15}{4} \times 98 \times 1 0^{- 3} = 0.3675

Final answer: 0.3675

Q49NumericalThe d- and f-Block Elements

Consider the following reactions :

NaCl + K_{2} Cr_{2} O_{7} + H_{2} SO_{4} \to

(A) + side products
(A) + NaOH

\to

(B) + side products
(B) +

H_{2} SO_{4}

(dil.) +

H_{2} O_{2} \to

SolutionAnswer: 18

Approach:

Identify the chromium products A, B and C from the chromyl chloride test sequence and add up the atoms in each molecule.

Step 1:NaCl with dichromate and sulphuric acid gives chromyl chloride (A).

(A) = CrO_{2} Cl_{2}

Step 2:Chromyl chloride with NaOH gives sodium chromate (B).

(B) = Na_{2} CrO_{4}

Step 3:Sodium chromate with dilute sulphuric acid and hydrogen peroxide gives chromium peroxide (C).

(C) = CrO_{5}

Step 4:Add the atom counts.

5 + 7 + 6 = 18

Final answer: 18

Q50NumericalThermodynamics

The standard heat of formation

(Δ_{f} H_{298}^{\circ})

of ethane (in kJ/mol), if the heat of combustion of ethane, hydrogen and graphite are -1560, -393.5 and -286 kJ/mol, respectively, is___.

SolutionAnswer: -192.5

Approach:

Apply Hess's law by combining the combustion reactions to construct the formation reaction of ethane.

Step 1:Combustion of graphite, hydrogen and ethane are given with their enthalpies.

C + O_{2} \to CO_{2}, - 286

Step 2:Invert ethane combustion and add two graphite and three hydrogen combustions.

2 (- 286) + 3 (- 393.5) - (- 1560)

Step 3:Evaluate the sum.

- 572 - 1180.5 + 1560 = - 192.5

Final answer: -192.5

Mathematics25 questions

Q51Single correctCo-ordinate Geometry

3 x + 4 y = 122

is a tangent to the ellipse

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

, for some

a \in R

then the distance between the foci of the ellipse is :

(A)

25

(B)

27

(C)

22

(D)

4

SolutionAnswer: Option 2

27

Approach:

Use the tangent condition for an ellipse: a line

y = m x + c

is tangent to

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

when

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

Step 1:Rewrite the line in slope-intercept form.

y = - \frac{3}{4} x + 32

Step 2:Apply the tangent condition with

b^{2} = 9

(32)^{2} = a^{2} (\frac{9}{16}) + 9

Step 3:Solving for

a^{2}

\frac{9 a ^{2}}{16} = 9 \Rightarrow a^{2} = 16

Step 4:Compute eccentricity and the distance between foci

2 a e

e = 1 - \frac{9}{16} = \frac{7}{4}

Final answer:

27

Q52Single correctMathematical Reasoning

Let A, B, C and D be four non-empty sets. The Contrapositive statement of "If

A \subseteq B

and

B \subseteq D

, then

A \subseteq C

" is :

(A)

A \neq \subseteq C

, then

B \subset A

D \subset B

(B)

A \neq \subseteq C

, then

A \subseteq B

and

B \subseteq D

(C)

A \neq \subseteq C

, then

A \neq \subseteq B

and

B \subseteq D

(D)

A \neq \subseteq C

, then

A \neq \subseteq B

B \neq \subseteq D

SolutionAnswer: Option 4If

A \neq \subseteq C

, then

A \neq \subseteq B

B \neq \subseteq D

Approach:

The contrapositive of

p \Rightarrow q

\sim q \Rightarrow\sim p

. Here

p = (A \subseteq B \land B \subseteq D)

and

q = (A \subseteq C)

Step 1:Negate the conclusion.

\sim q : A \neq \subseteq C

Step 2:Negate the hypothesis using De Morgan.

\sim (A \subseteq B \land B \subseteq D) \equiv (A \neq \subseteq B \lor B \neq \subseteq D)

Step 3:Form the contrapositive

\sim q \Rightarrow\sim p

A \neq \subseteq C \Rightarrow (A \neq \subseteq B \lor B \neq \subseteq D)

Final answer: If

A \neq \subseteq C

, then

A \neq \subseteq B

B \neq \subseteq D

Q53Single correctBinomial Theorem

The coefficient of

x^{7}

in the expression

(1 + x)^{10} + x (1 + x)^{9} + x^{2} (1 + x)^{8} + \dots + x^{10}

is :

(A)

420

(B)

330

(C)

210

(D)

120

SolutionAnswer: Option 2

330

Approach:

The given expression is a geometric series with ratio

\frac{x}{1 + x}

; sum it and extract the coefficient of

x^{7}

Step 1:Sum the GP of 11 terms with first term

(1 + x)^{10}

and ratio

\frac{x}{1 + x}

\frac{( 1 + x ) ^{10} ( 1 - ( \frac{x}{1 + x} ) ^{11} )}{1 - \frac{x}{1 + x}} = (1 + x)^{11} - x^{11}

Step 2:Extract the coefficient of

x^{7}

[x^{7}] ((1 + x)^{11} - x^{11}) =^{11} C_{7}

Final answer:

330

Q54Single correctStatistics and Probability

In a workshop, there are five machines and the probability of any one of them to be out of service on a day is

\frac{1}{4}

. If the probability that at most two machines will be out of service on the same day is

(\frac{3}{4})^{3} k

, then k is equal to :

(A)

\frac{17}{2}

(B)

4

(C)

\frac{17}{4}

(D)

\frac{17}{8}

SolutionAnswer: Option 4

\frac{17}{8}

Approach:

Model the number of faulty machines as a binomial variable with

n = 5

p = \frac{1}{4}

, and compute

P (X \leq 2)

Step 1:Set

p = \frac{1}{4}

q = \frac{3}{4}

and sum

P (X = 0) + P (X = 1) + P (X = 2)

(\frac{3}{4})^{5} + 5 \cdot \frac{1}{4} (\frac{3}{4})^{4} + 10 (\frac{1}{4})^{2} (\frac{3}{4})^{3}

Step 2:Simplify the bracket.

(\frac{3}{4})^{3} \times \frac{34}{16} = (\frac{3}{4})^{3} \times \frac{17}{8}

Final answer:

\frac{17}{8}

Q55Single correctCo-ordinate Geometry

The locus of mid points of the perpendiculars drawn from points on the line

x = 2 y

to the line

x = y

is :

(A)

2 x - 3 y = 0

(B)

3 x - 2 y = 0

(C)

5 x - 7 y = 0

(D)

7 x - 5 y = 0

SolutionAnswer: Option 3

5 x - 7 y = 0

Approach:

Take a point

P

x = 2 y

, drop a perpendicular to

x = y

meeting it at

Q

, and find the locus of the midpoint

R

PQ

Step 1:The perpendicular from

P

x = y

has slope

- 1

; write it as

y = - x + c

y = - x + c

Step 2:Intersection with

y = x

gives

Q = (\frac{c}{2}, \frac{c}{2})

; intersection with

x = 2 y

gives

P = (\frac{2 c}{3}, \frac{c}{3})

Q = (\frac{c}{2}, \frac{c}{2}), P = (\frac{2 c}{3}, \frac{c}{3})

Step 3:Midpoint

R

PQ

R = (\frac{7 c}{12}, \frac{5 c}{12})

Step 4:Eliminate

c

from the coordinates of

R

\frac{x}{7} = \frac{y}{5} \Rightarrow 5 x - 7 y = 0

Final answer:

5 x - 7 y = 0

Q56Single correctIntegral Calculus

The value of

α

for which

4 α \int_{- 1}^{2} e^{- α ∣ x ∣} d x = 5

, is :

(A)

lo g_{e} 2

(B)

lo g_{e} 2

(C)

lo g_{e} (\frac{4}{3})

(D)

lo g_{e} (\frac{3}{2})

SolutionAnswer: Option 1

lo g_{e} 2

Approach:

Split the integral of

e^{- α ∣ x ∣}

x = 0

to handle the modulus, then solve the resulting equation for

α

Step 1:Split using

∣ x ∣

4 α [\int_{- 1}^{0} e^{αx} d x + \int_{0}^{2} e^{- αx} d x] = 5

Step 2:Evaluate both integrals.

4 α [\frac{1 - e ^{- α}}{α} + \frac{1 - e ^{- 2 α}}{α}] = 5

Step 3:Let

t = e^{- α}

to obtain a quadratic.

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

Step 4:Solve for

α

α = lo g_{e} 2

Final answer:

lo g_{e} 2

Q57Single correctSequence and Series

If the sum of the first 40 terms of the series,

3 + 4 + 8 + 9 + 13 + 14 + 18 + 19 + \dots

(102) m

, then m is equal to :

(A)

10

(B)

25

(C)

5

(D)

20

SolutionAnswer: Option 4

20

Approach:

Group the terms in pairs; each pair sum forms an arithmetic progression, then sum 20 such pairs.

Step 1:Pair consecutive terms:

(3 + 4) + (8 + 9) + (13 + 14) + \dots

gives

7, 17, 27, \dots

(20 pairs).

S = 7 + 17 + 27 + \dots (20 terms)

Step 2:Sum the 20-term AP.

S = \frac{20}{2} [14 + 19 (10)] = 20 \times 102

Step 3:Compare with

(102) m

m = 20

Final answer:

20

Q58Single correctComplex Numbers

\frac{3 + i s i n θ}{4 - i c o s θ}

θ \in [0, 2 π]

, is a real number, then the argument of

sin θ + i cos θ

is :

(A)

π - tan^{- 1} (\frac{4}{3})

(B)

- tan^{- 1} (\frac{3}{4})

(C)

π - tan^{- 1} (\frac{3}{4})

(D)

tan^{- 1} (\frac{4}{3})

SolutionAnswer: Option 1

π - tan^{- 1} (\frac{4}{3})

Approach:

Rationalize the fraction; set its imaginary part to zero to relate

sin θ

and

cos θ

, then find the argument of

sin θ + i cos θ

Step 1:Multiply numerator and denominator by

4 + i cos θ

and set the imaginary part to zero.

3 cos θ + 4 sin θ = 0

Step 2:With

θ

in the second quadrant,

sin θ > 0

cos θ < 0

, so

sin θ + i cos θ

lies in the second quadrant.

ar g (sin θ + i cos θ) = π + tan^{- 1} \frac{c o s θ}{s i n θ}

Step 3:Substitute

\frac{c o s θ}{s i n θ} = - \frac{4}{3}

π - tan^{- 1} (\frac{4}{3})

Final answer:

π - tan^{- 1} (\frac{4}{3})

Q59Single correctMatrices and Determinants

Let

A = [a_{ij}]

and

B = [b_{ij}]

be two

3 \times 3

real matrices such that

b_{ij} = (3)^{(i + j - 2)} a_{ji}

, where

i, j = 1, 2, 3

. If the determinant of B is 81, then the determinant of A is :

(A)

\frac{1}{9}

(B)

\frac{1}{81}

(C)

\frac{1}{3}

(D)

3

SolutionAnswer: Option 1

\frac{1}{9}

Approach:

Express B as a diagonal-scaled transpose of A and use the product rule for determinants.

Step 1:Since

b_{i}

j = 3^(i-1)·3^(j-1)·

a_{j}

i, B = D

A^{T}

D with D = diag(1,3,9).

B = D A^{T} D, D = diag (1, 3, 9)

Step 2:The determinant of D is 1·3·9 = 27.

det D = 27

Step 3:Therefore det(B) =

(d e t D)^{2}

· det(A) = 729·det(A).

det B = 2 7^{2} det A = 729 det A

Step 4:Setting det(B)=81 gives det(A).

det A = \frac{81}{729} = \frac{1}{9}

Final answer:

\frac{1}{9}

Q60Single correctDifferential Calculus

Let f(x) be a polynomial of degree 5 such that

x = \pm 1

are its critical points. If

lim_{x \to 0} (2 + \frac{f ( x )}{x ^{3}}) = 4

, then which one of the following is not true?

(A)

f (1) - 4 f (- 1) = 4

(B)

x = 1

is a point of maxima and

x = - 1

is a point of minimum of

f

(C)

f

is an odd function.

(D)

x = 1

is a point of minima and

x = - 1

is a point of maxima of

f

SolutionAnswer: Option 4

x = 1

is a point of minima and

x = - 1

is a point of maxima of

f

Approach:

Use the limit to fix the low-order coefficients, the critical-point condition for the rest, then test each statement.

Step 1:For the limit to be finite as

x \to 0

, f has no constant, linear, or quadratic terms; the limit gives the

x^{3}

coefficient as 2.

f (x) = a x^{5} + b x^{4} + 2 x^{3}

Step 2:Apply

f^{'} (1) = 0

and

f^{'} (- 1) = 0

to find

a, b

b = 0, a = - \frac{6}{5}

Step 3:Second-derivative test:

f^{''} (1) < 0

f^{''} (- 1) > 0

, so

x = 1

is a maximum and

x = - 1

is a minimum; statement (d) is the false one.

f^{''} (x) = - 24 x^{3} + 12 x

Step 4:Also

f (1) - 4 f (- 1) = 4

holds.

f (1) - 4 f (- 1) = 4

Final answer:

x = 1

is a point of minima and

x = - 1

is a point of maxima of

f

Q61Single correctPermutations and Combinations

The number of ordered pairs (r, k) for which

6 \cdot^{35} C_{r} = (k^{2} - 3) \cdot^{36} C_{r + 1}

, where k is an integer, is :

(A)

4

(B)

6

(C)

2

(D)

3

SolutionAnswer: Option 1

4

Approach:

Use

^{36} C_{r + 1} = \frac{36}{r + 1}^{35} C_{r}

to reduce the equation to a relation in r and k, then count integer solutions.

Step 1:Substitute the identity to cancel

^{35} C_{r}

6 = (k^{2} - 3) \cdot \frac{36}{r + 1}

Step 2:Thus

k^{2} = \frac{r + 1}{6} + 3

, with

0 \leq r \leq 35

, so

r + 1

must be a multiple of 6.

r = 5 \Rightarrow k^{2} = 4, r = 35 \Rightarrow k^{2} = 9

Step 3:Each valid

r

gives two integer

k

values.

(5, \pm 2), (35, \pm 3)

Final answer:

4

Q62Single correctSequence and Series

Let

a_{1}, a_{2}, a_{3}, \dots

be a G.P. such that

a_{1} < 0

a_{1} + a_{2} = 4

and

a_{3} + a_{4} = 16

. If

\sum_{i = 1}^{9} a_{i} = 4 λ

, then

λ

is equal to :

(A)

171

(B)

\frac{511}{3}

(C)

- 171

(D)

- 513

SolutionAnswer: Option 3

- 171

Approach:

Use the two given sums to find the common ratio r and first term

a_{1}

(with

a_{1} < 0

), then sum nine terms.

Step 1:Divide the relations:

\frac{a _{3} + a _{4}}{a _{1} + a _{2}} = r^{2} = 4

, so

r = \pm 2

r^{2} = 4 \Rightarrow r = \pm 2

Step 2:From

a_{1} (1 + r) = 4

with

a_{1} < 0

r = 2

gives

a_{1} = \frac{4}{3} > 0

(rejected), so

r = - 2

a_{1} = - 4

a_{1} (1 - 2) = 4 \Rightarrow a_{1} = - 4

Step 3:Sum nine terms.

\sum_{i = 1}^{9} a_{i} = - 4 \cdot \frac{( - 2 ) ^{9} - 1}{- 2 - 1} = \frac{- 4}{3} (- 513) = 4 (- 171)

Step 4:Compare with

4 λ

λ = - 171

Final answer:

- 171

Q63Single correctVector Algebra

Let

a, b

and

c

be three unit vectors such that

a + b + c = 0

. If

λ = a \cdot b + b \cdot c + c \cdot a

and

d = a \times b + b \times c + c \times a

, then the ordered pair

(λ, d)

is equal to :

(A)

(\frac{3}{2}, 3 a \times c)

(B)

(- \frac{3}{2}, 3 a \times b)

(C)

(- \frac{3}{2}, 3 a \times c)

(D)

(\frac{3}{2}, 3 b \times c)

SolutionAnswer: Option 3

(- \frac{3}{2}, 3 a \times c)

Approach:

Square the vanishing sum to obtain

λ

, and use the cyclic cross-product structure of

d

to express it in terms of

a \times c

Step 1:Square

a + b + c = 0

with unit magnitudes.

3 + 2 λ = 0 \Rightarrow λ = - \frac{3}{2}

Step 2:Substitute

c = - (a + b)

into

d

and simplify using cross-product antisymmetry.

d = a \times b + b \times (- a - b) + (- a - b) \times a

Step 3:Form the ordered pair.

(λ, d) = (- \frac{3}{2}, 3 a \times c)

Final answer:

(- \frac{3}{2}, 3 a \times c)

Q64Single correctDifferential Equations

Let

y = y (x)

be the solution curve of the differential equation,

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

, satisfying

y (0) = 1

. This curve intersects the

x -

axis at a point whose abscissa is :

(A)

2 + e

(B)

2

(C)

2 - e

(D)

- e

SolutionAnswer: Option 3

2 - e

Approach:

Treat

x

as a function of

y

to obtain a first-order linear differential equation, solve with the integrating factor, apply the initial condition, then set

y = 0

to find the

x

-intercept.

Step 1:Rewrite the equation with

x

as the dependent variable.

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

Step 2:Multiply by the integrating factor

e^{y}

and integrate.

x e^{y} = e^{y} (y^{2} - 2 y + 2) + C

Step 3:Apply

y (0) = 1

, i.e.

x = 0

y = 1

0 = 1 - 2 + 2 + C e^{- 1}

Step 4:Substituting

C = - e

and setting

y = 0

for the

x

-axis.

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

Final answer:

2 - e

Q65Single correctIntegral Calculus

θ_{1}

and

θ_{2}

be respectively the smallest and the largest values of

θ

(0, 2 π) - {π}

which satisfy the equation,

2 cot^{2} θ - \frac{5}{s i n θ} + 4 = 0

, then

\int_{θ_{1}}^{θ_{2}} cos^{2} 3 θ d θ

is equal to :

(A)

\frac{2 π}{3}

(B)

\frac{π}{3}

(C)

\frac{π}{3} + \frac{1}{6}

(D)

\frac{π}{9}

SolutionAnswer: Option 2

\frac{π}{3}

Approach:

Convert the trigonometric equation to a quadratic in

sin θ

, find the admissible solutions in

(0, 2 π) - {π}

, then evaluate the definite integral of

cos^{2} 3 θ

Step 1:Use

cot^{2} θ = csc^{2} θ - 1

to form a quadratic.

2 csc^{2} θ - \frac{5}{s i n θ} + 2 = 0

Step 2:Solve the quadratic; reject

csc θ = 2

(gives

sin θ = \frac{1}{2}

valid) and

csc θ = \frac{1}{2}

(impossible).

csc θ = 2 \Rightarrow sin θ = \frac{1}{2}

Step 3:Identify smallest and largest values.

θ_{1} = \frac{π}{6}, θ_{2} = \frac{5 π}{6}

Step 4:Integrate using the power-reduction identity.

\int_{π /6}^{5 π /6} \frac{1 + c o s 6 θ}{2} d θ = \frac{1}{2} [θ + \frac{s i n 6 θ}{6}]_{π /6}^{5 π /6} = \frac{π}{3}

Final answer:

\frac{π}{3}

Q66Single correctSequences and Series

Let

α

and

β

are the roots of the equation

x^{2} - x - 1 = 0

. If

p_{k} = (α)^{k} + (β)^{k}, k \geq 1

then which one of the following statements is not true?

(A)

(p_{1} + p_{2} + p_{3} + p_{4} + p_{5}) = 26

(B)

p_{5} = 11

(C)

p_{5} = p_{2} \cdot p_{3}

(D)

p_{3} = p_{5} - p_{4}

SolutionAnswer: Option 3

p_{5} = p_{2} \cdot p_{3}

Approach:

Use Newton's recurrence

p_{k} = p_{k - 1} + p_{k - 2}

from the characteristic equation, compute

p_{1}

through

p_{5}

, then test each statement.

Step 1:From

x^{2} = x + 1

, the power sums satisfy

p_{k} = p_{k - 1} + p_{k - 2}

p_{1} = 1, p_{2} = 3

Step 2:Generate the sequence.

p_{3} = 4, p_{4} = 7, p_{5} = 11

Step 3:Test statements b, a, d.

p_{1} + p_{2} + p_{3} + p_{4} + p_{5} = 1 + 3 + 4 + 7 + 11 = 26; p_{5} - p_{4} = 11 - 7 = 4 = p_{3}

Step 4:Test statement c.

p_{2} \cdot p_{3} = 3 \times 4 = 12 \neq = 11 = p_{5}

Final answer:

p_{5} = p_{2} \cdot p_{3}

Q67Single correctIntegral Calculus

The area (in sq. units) of the region

{(x, y) ϵ R ∣4 x^{2} \leq y \leq 8 x + 12}

is :

(A)

\frac{125}{3}

(B)

\frac{128}{3}

(C)

\frac{124}{3}

(D)

\frac{127}{3}

SolutionAnswer: Option 2

\frac{128}{3}

Approach:

Find the intersection of the parabola

y = 4 x^{2}

and the line

y = 8 x + 12

, then integrate the difference between the line and the parabola over the interval.

Step 1:Set the curves equal to find limits.

4 x^{2} = 8 x + 12 \Rightarrow x^{2} - 2 x - 3 = 0

Step 2:Write the area integral.

A = \int_{- 1}^{3} (8 x + 12 - 4 x^{2}) d x

Step 3:Integrate.

A = [4 x^{2} + 12 x - \frac{4 x ^{3}}{3}]_{- 1}^{3}

Step 4:Evaluate the bounds.

A = (36 + 36 - 36) - (4 - 12 + \frac{4}{3}) = 44 - \frac{4}{3} = \frac{128}{3}

Final answer:

\frac{128}{3}

Q68Single correctDifferential Calculus

The value of c in Lagrange's mean value theorem for the function

f (x) = x^{3} - 4 x^{2} + 8 x + 11

, where

x \in [0, 1]

is :

(A)

\frac{4 - 7}{3}

(B)

\frac{2}{3}

(C)

\frac{7 - 2}{3}

(D)

\frac{4 - 5}{3}

SolutionAnswer: Option 1

\frac{4 - 7}{3}

Approach:

Apply Lagrange's mean value theorem: set

f^{'} (c)

equal to the average rate of change over

[0, 1]

, solve the resulting quadratic, and select the root inside

(0, 1)

Step 1:Compute the endpoint values and slope.

f (0) = 11, f (1) = 16 \Rightarrow \frac{f ( 1 ) - f ( 0 )}{1 - 0} = 5

Step 2:Differentiate and equate.

f^{'} (x) = 3 x^{2} - 8 x + 8 \Rightarrow 3 c^{2} - 8 c + 8 = 5

Step 3:Solve the quadratic.

c = \frac{8 \pm 64 - 36}{6} = \frac{8 \pm 2 7}{6} = \frac{4 \pm 7}{3}

Step 4:Select the root in

(0, 1)

c = \frac{4 - 7}{3} \approx 0.45 \in (0, 1)

Final answer:

\frac{4 - 7}{3}

Q69Single correctDifferential Calculus

Let

y = y (x)

be a function of x satisfying

y 1 - x^{2} = k - x 1 - y^{2}

where k is a constant and

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

. Then

\frac{d y}{d x}

x = \frac{1}{2}

, is equal to :

(A)

- \frac{5}{2}

(B)

\frac{5}{2}

(C)

- \frac{5}{4}

(D)

\frac{2}{5}

SolutionAnswer: Option 1

- \frac{5}{2}

Approach:

Differentiate the implicit relation with respect to x, then substitute

x = \frac{1}{2}

and

y = - \frac{1}{4}

to evaluate

\frac{d y}{d x}

Step 1:Differentiate both sides with respect to

x

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

Step 2:Substitute

x = \frac{1}{2}

y = - \frac{1}{4}

y^{'} \frac{3}{2} + \frac{1}{2 15} = - \frac{15}{4} - \frac{y ^{'}}{2 15}

Step 3:Collect the

y^{'}

terms.

y^{'} (\frac{3}{2} + \frac{1}{2 15}) = - \frac{1}{4 3} - \frac{15}{4}

Step 4:Solve for

y^{'}

y^{'} = - \frac{2 15}{4 3} = - \frac{5}{2}

Final answer:

- \frac{5}{2}

Q70Single correctCoordinate Geometry

Let the tangents drawn from the origin to the circle,

x^{2} + y^{2} - 8 x - 4 y + 16 = 0

touch it at the points A and B. The

(A B)^{2}

is equal to :

(A)

\frac{32}{5}

(B)

\frac{64}{5}

(C)

\frac{52}{5}

(D)

\frac{56}{5}

SolutionAnswer: Option 2

\frac{64}{5}

Approach:

Find the centre and radius, compute the length of the tangent from the origin, then use the chord-of-contact length formula

A B = \frac{2 r L}{r ^{2} + L ^{2}}

Step 1:Identify the centre and radius.

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

Step 2:Length of tangent from origin.

L = 0 + 0 - 0 - 0 + 16 = 4

Step 3:Apply the chord-of-contact formula.

A B = \frac{2 \cdot 2 \cdot 4}{4 + 16} = \frac{16}{20} = \frac{8}{5}

Step 4:Square the length.

(A B)^{2} = \frac{64}{5}

Final answer:

\frac{64}{5}

Q71NumericalMatrices and Determinants

If system of linear equations

x + y + z = 6

x + 2 y + 3 z = 10

3 x + 2 y + λ z = μ

has more than two solutions, then

μ - λ^{2}

is equal to ______.

SolutionAnswer: 13

Approach:

For infinitely many solutions the coefficient determinant must vanish (giving

λ

) and the augmented system must be consistent (giving

μ

Step 1:Set the coefficient determinant to zero.

113122 13 λ = 0 \Rightarrow 2 λ - 6 - λ + 9 + 2 - 6 = 0

Step 2:Set a replaced-column determinant to zero with

λ = 1

113122 610 μ = 0 \Rightarrow 2 μ - 20 - μ + 30 - 24 = 0

Step 3:Compute the required expression.

μ - λ^{2} = 14 - 1

Final answer: 13

Q72NumericalCoordinate Geometry

If the foot of perpendicular drawn from the point

(1, 0, 3)

on a line passing through

(α, 7, 1)

(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})

, then

α

is equal to ________.

SolutionAnswer: 4

Approach:

The vector from the external point to the foot is perpendicular to the line's direction (foot minus the given point on the line); set their dot product to zero and solve for

α

Step 1:Vector from

P (1, 0, 3)

to foot

Q

PQ = (\frac{2}{3}, \frac{7}{3}, \frac{8}{3})

Step 2:Direction of the line: foot minus point

(α, 7, 1)

d = (\frac{5}{3} - α, \frac{7}{3} - 7, \frac{17}{3} - 1) = (\frac{5 - 3 α}{3}, - \frac{14}{3}, \frac{14}{3})

Step 3:Set the dot product to zero.

\frac{2}{3} \cdot \frac{5 - 3 α}{3} + \frac{7}{3} \cdot (- \frac{14}{3}) + \frac{8}{3} \cdot \frac{14}{3} = 0

Step 4:Solve for

α

6 α = 24

Final answer: 4

Q73NumericalDifferential Calculus

If the function f defined on

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

f (x) = {(\frac{1}{x}) lo g_{e} (\frac{1 + 3 x}{1 - 2 x}) k w h e n x \neq = 0, w h e n x = 0

is continuous, the k is equal to ______.

SolutionAnswer: 5

Approach:

For continuity at

x = 0

, k must equal

lim_{x \to 0} f (x)

; split the logarithm and use

lim_{x \to 0} \frac{l n ( 1 + a x )}{x} = a

Step 1:Continuity requires the limit to equal

k

k = lim_{x \to 0} \frac{1}{x} lo g_{e} (\frac{1 + 3 x}{1 - 2 x})

Step 2:Split the logarithm.

k = lim_{x \to 0} \frac{l o g _{e} ( 1 + 3 x )}{x} + lim_{x \to 0} \frac{l o g _{e} ( 1 - 2 x )}{( - 2 x )} \cdot \frac{( - 2 x )}{x} \cdot (- 1)

Step 3:Apply the standard limit to each term.

k = 3 + 2

Final answer: 5

Q74NumericalStatistics

If the mean and variance of eight numbers

3, 7, 9, 12, 13, 20, x

and y be 10 and 25 respectively then xy is equal to ______.

SolutionAnswer: 54

Approach:

Use the mean to get

x + y

, then the variance formula

σ^{2} = \frac{\sum x _{i}^{2}}{n} - \overset{x}{ˉ}^{2}

to get

x^{2} + y^{2}

, and combine to find xy.

Step 1:Apply the mean condition.

\frac{64 + x + y}{8} = 10

Step 2:Apply the variance condition.

\frac{3 ^{2} + 7 ^{2} + 9 ^{2} + 1 2 ^{2} + 1 3 ^{2} + 2 0 ^{2} + x ^{2} + y ^{2}}{8} - 100 = 25

Step 3:Use the identity

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

256 - 2 x y = 148

Step 4:Solve for

x y

x y = 54

Final answer: 54

Q75NumericalSets, Relations and Functions

Let

X = {n \in N : 1 \leq n \leq 50}

. If

A = {n \in X : n is a multiple of 2}

and

B = {n \in X : n is a multiple of 7}

, then the number of elements in the smallest subset of X containing both A and B is ______.

SolutionAnswer: 29

Approach:

The smallest subset containing both A and B is

A \cup B

; apply the inclusion-exclusion principle.

Step 1:Count multiples of 2 up to 50.

n (A) = ⌊ \frac{50}{2} ⌋ = 25

Step 2:Count multiples of 7 up to 50.

n (B) = ⌊ \frac{50}{7} ⌋ = 7

Step 3:Count multiples of 14 (common to both).

n (A \cap B) = ⌊ \frac{50}{14} ⌋ = 3

Step 4:Apply inclusion-exclusion.

n (A \cup B) = 25 + 7 - 3

Final answer: 29

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