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Part Test - 1 (JEE Advanced)

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66 Questions MCQ Test JEE Main & Advanced Mock Test Series | Part Test - 1 (JEE Advanced)

Part Test - 1 (JEE Advanced) for JEE 2022 is part of JEE Main & Advanced Mock Test Series preparation. The Part Test - 1 (JEE Advanced) questions and answers have been prepared according to the JEE exam syllabus.The Part Test - 1 (JEE Advanced) MCQs are made for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Part Test - 1 (JEE Advanced) below.
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Part Test - 1 (JEE Advanced) - Question 1

A simple pendulum is oscillating in a vertical plane. If resultant acceleration of bob of mass m at a point A is in horizontal direction, find the tangential force at this point in terms of tension T and mg.

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 1

When the acceleration of bob is horizontal, net vertical force on the bob will be zero.
T cos θ – mg = 0
The tangential force at that instant is

Part Test - 1 (JEE Advanced) - Question 2

Hailstones falling vertically with a speed of 10 m/s, hit the wind screen (wind screen makes an angle 30° with the horizontal) of a moving car and rebound elastically. The velocity of the car if the driver finds the hailstones rebound vertically after striking is :

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 2

For the driver to observe the rain move vertically upward after the elastic collision, rain should come at angle 30º with the horizontal (as clear from figure).

Let, velocity of rain w.e.t. car be Vm//c

But , since rain fall vertically down.

[Since ; VR/G(y) = – 10 m/s ; VC/G(y) = 0
⇒ V sin 30° = 10 ⇒ V = 20 m/s.
Substituting V = 20 m/s in equation (i)

Part Test - 1 (JEE Advanced) - Question 3

A water tank stands on the roof of a building as shown. Then the value of 'h' for which the distance covered by the water 'x'  is greatest is -  a

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 3

the roots of x are (0,4) and the maximum of x is at h = 2.
The permitted value of h is 0 to 1 clearly h = 1 will give the maximum value of x is this interval.
Aliter : If the column of water itself were from ground upto a height of 4m, h = 2m would give the maximum range x. Farther the hole is from this midpoint ,lower the range.Here the nearest point possible to this midpoint is the base of the container. Hence h = 1m.

Part Test - 1 (JEE Advanced) - Question 4

A U-tube of base length “l” filled with same volume of two liquids of densities ρ and 2ρ is moving with an acceleration “a” on the horizontal plane. If the height difference between the two surfaces (open to atmosphere) becomes zero, then the height h is given by:

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 4

For the given situation, liquid of density 2ρ should be behind that of ρ. From right limb :

.......(1)
But from left limb : Pc = Patm + (2ρ) gh  ....(2)
From (1) and (2) :

Part Test - 1 (JEE Advanced) - Question 5

A particle A of mass 10/7 kg is moving in the positive direction of x. Its initial position is x = 0 & initial velocity is 1 m/s. The velocity at x = 10 is: (use the graph given)

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 5

Area under P-x graph

from graph ; area
v = 4m/s
ALITER : from graph

or

or

Now integrate both sides,

Part Test - 1 (JEE Advanced) - Question 6

The string of a step rolling wheel is pulled by applying force F with different lines of action in two  situations as shown. The wheel starts rolling without slipping due to application of the force :

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 6

(Pure rotation about instantaneous point of contact)
Note : If line action passes through point of contact, it only spins.

Part Test - 1 (JEE Advanced) - Question 7

Two particles start together from a point O and slide down straight smooth wires inclined at 30º & 60º to the vertical & in the same vertical plane as in figure. The relative acceleration of second with respect to first will be (in magnitude & direction) as :

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 7

Resultant Acceleration
α = 30°
i.e. Resulting acceleration is in vertical direction.

Part Test - 1 (JEE Advanced) - Question 8

Two points A & B on a disc have velocities v1 & v2 at some moment. Their directions make angles 60° and 30° respectively with the line of separation as shown in figure. The angular velocity of disc is :

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 8

For rigid body separation between two point remains same.

Part Test - 1 (JEE Advanced) - Question 9

In the figure shown ADB & BEF are two fixed circular paths. A block of mass m enters in the tube ADB through point A with minimum velocity to reach point B. From there it moves on another circular path of radius R'. There it is just able to complete the circle.

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 9

For minimum velocity. at A ;

For looping the loop ;

*Multiple options can be correct
Part Test - 1 (JEE Advanced) - Question 10

The displacement of a body from a reference point is given by, (√x) = 2t-3  where ' x ' is in metres and t in seconds. This shows that the body :

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 10

Vel c = dx/dt = 2(2t - 3) x 2 = 4 (2t - 3)
if v = 0  t = 3/2
acceleration

so correct ans is (A) and (B)

*Multiple options can be correct
Part Test - 1 (JEE Advanced) - Question 11

If the resultant force on a system of particles is non-zero, then :

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 11

The resultant force can be accelerating or decelerating, hence the momentum can increase or decrease. Hence (A) is wrong.
Since Fnet = M acm
∴ acm  ≠ 0 ;
hence vcm must change
Hence (B) is correct.
In case of a circular motion of centre of mass about a point the distance of centre of mass will remain constant. Hence (C) is correct.
Kinetic energy of some particles may increase and of some particles may decrease at the same time.

*Multiple options can be correct
Part Test - 1 (JEE Advanced) - Question 12

A painter is applying force himself to raise him and the box with an  cceleration of 5 m/s2 by a massless rope and pulley arrangement as shown in figure. Mass of painter is 100 kg and that of box is 50 kg. If g = 10 m/s2, then :

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 12

For the whole system, 2T – 1500 = 150 x 5 ⇒ T = 1125 N
For the person, T – 1000 + N = 100 x 5 N = 1500 – 1125 = 375 N

Part Test - 1 (JEE Advanced) - Question 13

Statement-1 : For a disc undergoing fixed axis rotation, the magnitude of angle between velocity and acceleration vector of any moving point on disc at a particular instant of time are same.
Statement-2 : Each moving point on a disc undergoing fixed axis rotation has same angular speed and same angular acceleration at an instant of time. Hence the ratio of magnitude of tangential acceleration and magnitude of centripetal acceleration is same for all moving points at an instant of time.

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 13

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

Part Test - 1 (JEE Advanced) - Question 14

Statement-1 : The equation of distance travelled by a particle moving in a straight line with constant acceleration in nth second is  where letters have usual meaning, is dimensionally incorrect.

Statement-2 : For every equation relating physical quantities to be true, it must have dimensional homogenity.

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 14

The equation of distance travelled in nth second is dimensionally correct because the interval of time 1 second has already been substituted into the equation and its dimension should be taken into account. Therefore statement-1 is false.

Part Test - 1 (JEE Advanced) - Question 15

Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed spring of spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance 'd' from the base of that surface after falling height 5m. (use π2 = 10; g = 10 m/s2). Assume that the spring does not affect the collision.

Mass of the block B is

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 15

Part Test - 1 (JEE Advanced) - Question 16

Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed spring of spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance 'd' from the base of that surface after falling height 5m. (use π2 = 10; g = 10 m/s2) Assume that the spring does not affect the collision.

Amplitude of the SHM as being executed by block B-spring system, is -

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 16

Immediately after the collision, suppose velocities of the blocks are V1 and V1 as shown 1/2 vel. of approach = velocity of separation.

⇒ 5 = V2 - V1 .... (1)
Using principle of conservation of momentum for the collision
2 = 0.2 V1 + V2
or  10 = V1 + 5V2   .....(2)
On solving V2 = 2.5 m/s; V1 = - 2.5 m/s
Hence block A moves leftward after the collision with speed 2.5 m/s. And the block B moves towards right with speed 2.5 m/s.
The maximum velocity of B = 2.5 = ωA
⇒

Part Test - 1 (JEE Advanced) - Question 17

Figure shows block A of mass 0.2 kg sliding to the right over a frictionless elevated surface at a speed of 10 m/s. The block undergoes a collision with stationary block B, which is connected to a nondeformed spring of spring constant 1000 Nm–1. The coefficient of restitution between the blocks is 0.5. After the collision, block B oscillates in SHM with a period of 0.2 s, and block A slides off the left end of the elevated surface, landing a distance 'd' from the base of that surface after falling height 5m. (use π2 = 10; g = 10 m/s2) Assume that the spring does not affect the collision.

The distance 'd' will be equal to -

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 17

Part Test - 1 (JEE Advanced) - Question 18

Matrix Match Type

This section contains 1 questions. Each question has four statements (A, B, C and D) given in Column-I and five statements (p,q,r, s and t) in Column-II. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. The answers to these questions have to be appropriately marked as illustrated in the following example. If the correct matches are A-p, A-r, B-p, B-s, C-r, C-s, D-q and D-t then the answer should be written as : A→ p,r ; B→ p, s ; C → r, s ; D → q, t

Q. A uniform disc of mass M and radius R lies on a fixed rough horizontal surface at time t = 0. Initial angular velocity ωo of each disc (magnitude and sense of rotation) and horizontal velocity v0 of centre of mass is shown for each situation of column-I. Match each situation in column-I with the results given in column-II.

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 18

In all four situation of column-I, angular momentum of the disc about a point on ground is conserved. Take angular momentum out of the paper as positive

(A) Initial angular momentum about its point of contact on ground = 1/2 {mR2ωo–mR(2Rωo )} = negative. Hence final state of the disc is as shown if figure B.

Hence angular velocity shall first decrease and then increase in opposite sense. The velocity of centre shall decrease till the disc starts rolling without slipping.

(B) The initial angular momentum about its point of contact on ground =0.

Hence angular speed and velocity of centre simultaneously reduce to zero without a change in direction.

(C) Because v0 > Rω0, velocity of centre of mass will decrease and angular velocity will increase without a change in direction till disc starts rolling without slipping.

(D) Because v0 < Rω0, velocity of centre of mass will increase and angular velocity will decrease without a change in direction till disc starts rolling without slipping.?

Part Test - 1 (JEE Advanced) - Question 19

Two particles are projected simultaneously with the same speed vv in the same vertical plane with angles of elevation θ, and 2θ, where θ<45. At what time will velocities be parallel?

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 19

*Answer can only contain numeric values
Part Test - 1 (JEE Advanced) - Question 20

A man can swim in still water with a speed of 3 m/s. x and y axis are drawn along and normal to the bank of river flowing to right with a speed of 1 m/s. The man starts swimming from origin O at t = 0 second. Assume size of man to be negligible. Locus of all the possible points where man can reach at t = 1 sec. is (x–a)2 + y2 = c2  Find value of ac2.

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 20

Method -1
If the river is still, the man will be at a distance 3 metres from origin O after 1 second. The locus of all the point where man can reach at t = 1 second is a semicircle of radius 3 and centre at O (dotted semicircle shown in figure).
The river flows to right with a speed 1 m/s. Hence there shall be additional shift in position by 1 m/s x 1 sec = 1m towards right. Hence the locus of all points giving possible position after one second will be the dotted semicircle shifted to right by 1 m as shown in figure.

Hence locus all the points where the man can be at t = 1 sec. is a semicircle of radius 3 and centre at 0' (1 m, 0 m)
∴ Equation of locus of all the points is
(x - 1)2 + (y - 0)2 = 32
or (x - 1)2 + y2 = 9

Method - 2
Let the relative velocity of the man make angle 'θ' with the x-axis.
Then at time 't' :
x = (3 cosθ + 1) t
and y = 3 sinθ t
⇒(x - t)2 + y2 = (3 cosθ)2 t2 + (3 sinθ)2 t2
(x - t)2 + y2 = 9t2
at t = 1 sec. the required equation is
(x - 1)2 + y2 = 9.

*Answer can only contain numeric values
Part Test - 1 (JEE Advanced) - Question 21

In the figure shown a small block ‘B’ of mass ‘m’ is released from the top of a smooth movable wedge ‘A’ of the same mass ‘m’. ‘B’ ascends another movable smooth wedge ‘C’ of the same mass. Neglecting friction any where the maximum height attained by ‘B’ on ‘C’ is h/2x  Find the value of  x .

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 21

Let u and v be the speed of wedge A and block B at just after the block B gets off the wedge A. Applying conservation of momentum in horizontal direction, we get. mu = mv       ..............(1)  Applying conservation of energy between initial and final state as shown in fig (1), we get

solving (1) and (2) we get
.......(3)

At the instant block B reaches maximum height h' on the wedge C (figure 2),the speed of block B and Wedge C are v'. Applying conservation momentum in horizontal direction,we get mv= (m+m)v'................(4)

Applying conservation of energy between initial and final state
......... (4)
Solving equations (3) (4) and (5) we get  h' = h/4.

*Answer can only contain numeric values
Part Test - 1 (JEE Advanced) - Question 22

A cylinder rotating at an angular speed of 50 rev/s is brought in contact with an identical stationary cylinder. Because of the kinetic friction, torques act on the two cylinders, accelerating the stationary one and decelerating the moving one. If the common magnitude of the acceleration and deceleration be one revolution per second square, how long will it take before the two cylinders have equal angular speed ?

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 22

fR = lα1 fR = lα2
α1 = α2 = 2p red/sec2
For A cylinder : ω = ω0 - αt    ω = 100π - 2πt ...(i)
For B cylinder ω = ω0  αt  ω0 = 0
ω = αt   ω = 27πt ....(ii)
From (i) and (ii) ω= 100 π - ω
2ω = 100π
ω = 50π
From (ii) euqation 50 π = 2 πt
t = 25 sec

Part Test - 1 (JEE Advanced) - Question 23

The volume occupied  by 2.0 mole of N2 at 200K and 8.21 atm pressure, if is

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 23

So
So Volume of 2 moles = 3.6 L

Part Test - 1 (JEE Advanced) - Question 24

If 10 gram of V2O5 is dissolved in acid and is reduced to V2+ by zinc metal, how many mole of I2 could be reduced by the resulting solution if it is further oxidised to VO2+ ions ?
[Assume no change in state of Zn2+ions] (V = 51, O = 16, I = 127) :

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 24

________________________________________
....(1)
Now

_____________________________________________
So we have 1 moles of V2O5 will reduce 2 moles of iodine
So  moles of I2 will be reduced by given amount of V2O5 = 0.11 moles of I2

Part Test - 1 (JEE Advanced) - Question 25

0.5 mole each of two ideal gas A(C= 3/2 R) and B(C= 5/2 R) are taken in a container and expanded reversibley and adiabatically from V = 1 litre to V = 4 litre starting from initial temperature t = 300K. ΔH for the process (in cal/mol) is

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 25

Part Test - 1 (JEE Advanced) - Question 26

When a weak acid is titrated against a strong base. The pH of solution keeps on changing with amount of base added. In this titration there is a formation of buffer also. If the buffer capacity (here defined as the volume of base  of a particular concentration  added per unit change in pH), is plotted against volume of base added for titration of 25 ml, 0.1 M HA (weak acid) solution with 0.1 M strong base solution, then the most appropriate curve will be :

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 26

Initially solution will have low buffer capacity,will have maximum buffer capacity at half neutralisation and will be close to zero at equivalence point.After equivalence point the buffer capacity will keep on  increasing with increase in concentration of base

Part Test - 1 (JEE Advanced) - Question 27

One mole of an ideal monoatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of system in the above process is :
[R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1K–1].

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 27

Part Test - 1 (JEE Advanced) - Question 28

An electron in a hydrogen like atom makes transition from a state in which its de-Broglie wavelength is λ1 to a state where its de-Broglie wavelength is λ2 then wavelength of photon (λ) generated will be : where m is mass of the electron, c is speed of light in vaccum.

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 28

⇒
⇒

Part Test - 1 (JEE Advanced) - Question 29

Given

Find out the negative of logarithm of the solubility of solid Zn(OH)2 at 25°C ,at pH=6.Consider Zn(OH)makes saturated solution at 25°C.

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 29

Dissolved [Zn(OH)2] = [Zn+2]aq + [Zn(OH)+]aq + (Zn(OH)2)aq + [Zn(OH)3-] + [Zn(OH)4]2-
Now, [Zn(OH)2]aq = 10-6 M in saturated solution.
so.
Dissolved Zn (OH)2 + 10-6 + 10-3 [OH-] + 10-2 [OH-]2

+ 10-6 + 10-3 x 10-8 = 10-18 = 10-1 + 10-5 + 10-6 + 10-11 = 10-1

Part Test - 1 (JEE Advanced) - Question 30

The curve of pressure volume (PV) against pressure (P) of the gas at a particular temperature is as shown, according to the graph which of the following is / are incorrect (in the low pressure region):

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 30

If Z > 1 positive deviation
Z < 1 negative deviation

*Multiple options can be correct
Part Test - 1 (JEE Advanced) - Question 31

2CaSO4(s)  2CaO(s) + 2SO2(g) + O2(g), Δ H > 0

Above equilibrium is established by taking some amount of CaSO4(s) in a closed container at 1600K.Then which of the following may be correct options?

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 31

- As reaction is endothermic therefore with increase in temperature it will go in the forward direction hence moles of CaO will increase.
- With increase or decrease of volume partial pressure of the gases will remain same.
- Due to the addition of the insert gas at constan pressure reaction will proceed in the direction in which more number of gases moles are formed.

*Multiple options can be correct
Part Test - 1 (JEE Advanced) - Question 32

The normal boiling point of a liquid `X` is 400 K. Which of the following statement is true about the process X (l) ———? X(g)?

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 32

Boiling of a liquid at normal boiling point is a equilibrium process and on decreasing the pressure equilibrium will go forward and ΔG will be negative and vice versa.

*Multiple options can be correct
Part Test - 1 (JEE Advanced) - Question 33

The variation of pH during the titration of 0.5 N Na2CO3 with 0.5 N HCl is shown in the given graph.The following tables indicates the colour and pH ranges of different indicators:

Based on the graph and table ,Which of the following statement are true?

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 33

(A) First equivalence point pH is in between 8-10 and also this point is inbetween 8.2-9.8 (indicator range of cresolphthalein).(B), (C) second equivalence point (complete neutralisation).
(D) 50/2 = 25.

*Multiple options can be correct
Part Test - 1 (JEE Advanced) - Question 34

In a H-like sample electrons make transition from 4th excited state to 2nd state then

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 34

Transition is taking place from 5 → 2
⇒ so Δn = 3
Hence maximum number of spectral line obsrved

(C) Number of lines belonging to the balmer series = 3

(D) Number of lines belonging to paschen series is 2.

Part Test - 1 (JEE Advanced) - Question 35

Statement-1 : A reaction which is spontaneous and accompanied by decrease of randomness must be exothermic.

Statement-2 : All exothermic reaction are accompanied by decrease of randomness.

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 35

A reaction that is spontaneous and accompanied by a decrease of randomness must be exothermic.

A spontaneous reaction has a negative value of ΔG. Also, the decrease in randomness means a negative value of ΔS. Since, ΔG=ΔH−TΔS, the enthalpy change will be negative. Hence, the reaction will be exothermic.

Those exothermic reactions in which gaseous products are formed from a liquid or solid reactants are accompanied by an increase of randomness.

Part Test - 1 (JEE Advanced) - Question 36

Statement-1 : The titration curve for (WA + WB) is as follows :

For this titration no suitable indicator is present.
Statement-2 : Indicator should change its colour sharply for indication of reaction to be complete so its pH-range should lie perfectly with in sharp change to avoid experimental error.

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 36

When there is no sharp increase (change) in pH of titration curve within pH range of indicator,there will be experimental error

Part Test - 1 (JEE Advanced) - Question 37

Amphoteric oxides, such as aluminium oxide, are soluble both in strongly acidic and in strongly basic
solutions :

Dissolution of Al(OH)3 in excess base is just a special case of the effect of complex-ion formation on solubility. Al(OH)3 dissolves because excess OH¯ ions convert it to the soluble complex ion Al(OH)4¯ (aluminate ion). The effect of pH on the solubility of Al(OH)3 is shown in figure.

Other examples of amphoteric hydroxides include Zn(OH)2, Cr(OH)3, Sn(OH)2 and Pb(OH)2, which react with
excess OH¯ ions to form the soluble complex ions Zn(OH)42- (zincate ion), Cr(OH)4¯ (chromite ion), Sn(OH)3¯
(stannite ion), and Pb(OH)3¯ (plumbite ion), respectively. By contrast, basic hydroxides, such as Mn(OH)2,
Fe(OH)2,and Fe(OH)3, dissolve in strong acid but not in strong base.

Which of the following curves best represents the variation of solubility of ferrous hydroxide Fe(OH)2  with the concentration of [H+] ions in the solution :

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 37

On increasing concentration of [H+] ions the solubility of basic hydroxide, Fe(OH)2 will increase.

Part Test - 1 (JEE Advanced) - Question 38

Amphoteric oxides,such as aluminium oxide, are soluble both in strongly acidic and in strongly basic solutions.
Dissolution of Al(OH)3 in excess base is just a special case of the effect of complex-ion formation on solubility.Al(OH)3  dissolves because excess OHions convert it to the soluble complex ion (aluminate ion). The effect of pH on the solubility of Al(OH)3 is shown in figure.

Other examples of amphoteric hydroxides include Zn(OH)2, Cr(OH)2,  Sn(OH)2  and  Pb(OH)2, which react with excess OH- ions to from the soluble complex ions Zn(OH)42- (zincate ion), Cr(OH)4¯ (chromite ion), Sn(OH)3¯
(stannite ion), and Pb(OH3)¯ (plumbite ion),respectively.By contrast, basic hydroxides, such as Mn(OH)2,
Fe(OH)2,and Fe(OH)3, dissolve in strong acid but not in strong base.

Aluminium is mined as bauxite (Al2O3.xH2O) a hydrated oxide that is always contaminated with Fe2O3 and SiO2. In Bayer process, Al2O3 is purified by some reagents, which of the following reagents will be the best for this purification (Note : Strong heating of precipitate of Al(OH)3 produces Al2O3

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 38

First heat bauxite with NaOH so, Al2O3 get dissolved in it (SiO2 will also get dissolved as it is an acidic oxide but not Fe2O3 (basic acidic),then filter lt. To filterate add a weak acid like CO2 so Al(OH)3 forms Al3+ ions). On heating the ppt. of Al(OH)3 we will get pure Al2O3.

Part Test - 1 (JEE Advanced) - Question 39

Amphoteric oxides, such as aluminium oxide, are soluble both in strongly acidic and in strongly basic
solutions :

Dissolution of Al(OH)3 in excess base is just a special case of the effect of complex-ion formation on solubility. Al(OH)3 dissolves because excess OH¯ ions convert it to the soluble complex ion Al(OH)4¯ (aluminate ion).The effect of pH on the solubility of Al(OH)is shown in figure.

Other examples of amphoteric hydroxides include Zn(OH)2, Cr(OH)3, Sn(OH)2 and Pb(OH)2, which react with
excess OH- ions to form the soluble complex ions Zn(OH)42- (zincate ion),Cr(OH)4- (chromite ion),Sn(OH)3¯(stannite ion), and Pb(OH)3¯ (plumbite ion), respectively. By contrast, basic hydroxides, such as Mn(OH)2, Fe(OH)2,and Fe(OH)3, dissolve in strong acid but not in strong base.

Zn(OH)2 is a amphoteric hydroxide and is involved in the following two equilibria in aqueous solutions

At what pH the solubility of Zn(OH)2 be minimum ?

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 39

Let solubility of Zn(OH)2 = s, some of will go in Zn2+ form and some in complex |Zn(OH)4|2-, of = 10-x M,
........(1)
[S1 + S2 total solubility]
So, S1 S2 = 1.2 x 1.2 10-18
Now, we want S = S1 + S2 to be minimum we will have S1 = S2
So, = S1 = S2 = 1.2 x 10-9 M
Hence from 1" equation we get

So,
Hence x = 10

Part Test - 1 (JEE Advanced) - Question 40

Matrix Match Type

This section contains 1 questions. Each question has four statements (A, B, C and D) given in Column-I and five statements (p,q,r, s and t) in Column-II. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. The answers to these questions have to be appropriately marked as illustrated in the following example. If the correct matches are A-p, A-r, B-p, B-s, C-r, C-s, D-q and D-t then the answer should be written as : A→ p,r ; B→ p, s ; C → r, s ; D → q, t

Q. Match the following :

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 40

*Answer can only contain numeric values
Part Test - 1 (JEE Advanced) - Question 41

0.0333 M KMnO4 is used for the following titration. What volume of the solution in ml will be required to react with 0.158 g of Na2S2O3 ? (mol. wt. of Na2 S2 O3 = 158)
S2O32- + MnO4- +H2O →MnO2(s) + SO42- + OH(not balanced)

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 41

V = 80 ml

Change in oxidation num ber of sulphur per molecule of
Change in oxidation number of Mn per molecule of
No. of moles in 0.158 g of
No. of equivalents = 8 x 10-3
Normality of 0.0333 M KMnO4 solution = 0.0333 x 3 = 0.1
Let V mL of volume of KMnO4 be required; then

or V = 80 ml

*Answer can only contain numeric values
Part Test - 1 (JEE Advanced) - Question 42

Heat of hydrogenation of cyclohexene to cyclohexane is –28.6 kcal/mol. The observed heat of hydrogenation of benzene to cyclohexane is –49.8 kcal/mol calculate the resonance energy of benzene in kcal/mole.

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 42

Resonance energy = -49.8 - (-3 x 28.6) = 36 k cal/mol.

*Answer can only contain numeric values
Part Test - 1 (JEE Advanced) - Question 43

The equilibrium constant for the reaction Br2(l) + Cl2(g)  2Br Cl(g) at 27°C is kp = 1 atm. In a closed container of volume 164 L initially 10 moles of Cl2 are present at 27ºC. What minimum mass of Br2(l) (in grams) must be introduced into this container so that above equilibrium is maintained at total pressure of 2.25 atm. Vapour pressure of Br2(l) at 27ºC is 0.25 atm. Assume that volume occupied by liquid is negligible. [R = 0.082 L atm mole-1 K-1, Atomic mass of bromine = 80].

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 43

To maintain the mentioned equilibrium there must be some (negligible) mass of Br2(l) at equilibrium so there will be an equilibrium

so some Br2(ℓ) will be required for this conversion (into vapour) moles of Br2(ℓ) required for above equilibrium

Now for equilibrium

otal gaseous moles = (10 + x) {Let x be moles of Br2 (ℓ) required just to maintain above equilibrium}
we have

So,

So
⇒  ⇒ 8x2 = 100 - x2 ⇒ 9x2 = 100
so x = 10/3
Hence total moles of Br2(ℓ) required to maintain both of above equilibria

so mass of Br2 (ℓ) required = (80 g/mole) (5 mole) = 400 gm

*Answer can only contain numeric values
Part Test - 1 (JEE Advanced) - Question 44

The equilibrium constants for amino acids are given in terms of succesive ionization constants of the protonated form. For example, equilibrium constants for Glycine (NH2CH2COOH) are Ka1 = 5 × 10-3 M and Ka2 = 2 × 10-10 M. What will be the pH at the Isoelectric point for this amino acid and pH of 0.02 M protonated Glycine in pure water respectively ? [Take : log 2 = 0.30]

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 44

Isoelectric point

Part Test - 1 (JEE Advanced) - Question 45

If (1 + 2x) (1 + x + x2)n

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 45

Integratin both sides of

Part Test - 1 (JEE Advanced) - Question 46

If all the distinct roots of the equation x47 + 2x46 + 3x45 + ........+ 24x24 + 23x23 + ...... + 2x2 + x = 0 are z1, z2 ...... zk and imaginary part of zk2 is bk . Then the value of |b1| + |b2| ..... + |bk| is

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 46

x47 + 2x46 + 3x45 +..... + 24x24 + 23x23 +.... + 2x2 + x ≡ x (x23 + x22 + ...... + x + 1)2

∴ roots of the given equation are 0 and roots of x23 + x22 + .... + x + 1 = 0 each repeated once.
Now roots of x23 + x22 + .....+ x + 1 = 0 are cos where k = 1, 2, ......... 23
the distinct roots of the given equation are 0, cos  where k = 1, 2, .... . . . . 23

Part Test - 1 (JEE Advanced) - Question 47

If all the words formed from the letters of the word "HORROR" are arranged in the opposite order as they are in a dictionary, then the rank of the word "HORROR" is

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 47

RRROOH

Part Test - 1 (JEE Advanced) - Question 48

For complex number z, the minimum value of |z| + |z – cos α – i sin α| + |z – 2(cos α + i sin α)| is

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 48

|z| + |z - cos α - i sin α| + |z - 2(cos α + i sin α)|
As shown in figure A and B are the points cos α + i sin α and 2(cos α + i sin α) respectively and let z be the affix of P. Then |z| + |z - cos α - i sin α| + |z - 2 (cos a + i sin α.)| is minimum only when P coincides with the point A
∴ OP + AP + BP = 2

Part Test - 1 (JEE Advanced) - Question 49

If a, b, c are in H.P. and a, b, c ≠ 9/4 , then  are in

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 49

since a,b,c are in H.P

∴

Part Test - 1 (JEE Advanced) - Question 50

If a determinant of order 3 × 3 is formed by using the numbers 1 or – 1, then minimum value of determinant is

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 50

so mimimum value = -4

Part Test - 1 (JEE Advanced) - Question 51

Range of value of 'θ' for which Arg (z + 2√3 – 2i) = θ and |z|  2 have atleast one z common is

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 51

From the figure range of θ is

Part Test - 1 (JEE Advanced) - Question 52

If (1 + x + x2 + x3)n = a0 + a1x + a2x2 + .......... + a3n . x3n and
A = a0 + a4+ a8 + ......., B = a2 + a6 + a10 + ............., C = a3 + a7 + a11 + .........., then

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 52

As A = B = C
A + C = 2B

*Multiple options can be correct
Part Test - 1 (JEE Advanced) - Question 53

If Arg,  then which of the following is (are) correct :

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 53

z lies on major arc of circle with z0 as centre, where

*Multiple options can be correct
Part Test - 1 (JEE Advanced) - Question 54

If P(x) = ax2 + bx + c satisfy a, b > 0 and | P’(0)|max = α, then

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 54

∵
∴ maximum occurs at x = 1 and minimum occurs at x = 0
∴ P(1) < 1 and P(0) > - 1 ⇒ a + b + c < 1 and c > - 1
⇒  a + b < 2    ∴ b < 2
|P'(0)|max = 2
if b = 2, then c = -1

⇒

*Multiple options can be correct
Part Test - 1 (JEE Advanced) - Question 55

Product of 22 integers is equal to 1, then their sum cannot be

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 55

Integers can be either + 1 or - 1
(+1)r (-1)22-r = 1
⇒ r = even

as r is given
2r - 22 can not be , 0, 1 or 4

*Multiple options can be correct
Part Test - 1 (JEE Advanced) - Question 56

If each root of x2 – ax – b = 0, a, b R has absolute value < 1, then

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 56

Part Test - 1 (JEE Advanced) - Question 57

Statement-1 : Let m, n, a, b and c are non-zero real numbers such that a, b, c are in H.P., then   ,  ,   are also in H.P.
Statement-2 : If a, b, c are in G.P., then a – b/2 , b/2, c – b/2  are in H.P.

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 57

Statment -1 is true :
⇔  ⇔
⇔  ⇔ a,b,c are in H.P.
Statment-2 is false.

Part Test - 1 (JEE Advanced) - Question 58

α, β are the real roots of a1x2 + b1x + c1 = 0; γ, δ are the real roots of a2x2 + b2x + c2 = 0; D1 = b12 – 4a1c1, and D2 = b22 – 4a2c2.
Statement-1 : If α, β, γ, δ are in H.P., then D1 : D2 : : c12 : c22
Statement-2 : If α, β, γ, δ  are in A.P., then D1 : D2 : : a12 : a22

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 58

Statment -2   ⇒
⇒
⇒  ⇒
Statment-1
∴ ⇒
⇒
Statement-1 and statement-2 both are correct.but not each other's correct explanations.

Part Test - 1 (JEE Advanced) - Question 59

Let z = e and a = e. The given figure consists of a unit circle and two parallelograms OAHE and OBDC. AOB and OEF are two line segments and G is the mid point of FH. ΔEFH is isosceles where EF = EH.

If OH is perpendicular to OC, then the value of θ is

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 59

∴  [∵ OAHE is a rhombus]

Part Test - 1 (JEE Advanced) - Question 60

Let z = e and a = e. The given figure consists of a unit circle and two parallelograms OAHE and OBDC. AOB and OEF are two line segments and G is the mid point of FH. ΔEFH is isosceles where EF = EH.

The affix of the point G is

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 60

affix of E is az, affix of H is z + az and affix of F is 2az
∴

Part Test - 1 (JEE Advanced) - Question 61

Let z = e and a = e. The given figure consists of a unit circle and two parallelograms OAHE and OBDC. AOB and OEF are two line segments and G is the mid point of FH. ΔEFH is isosceles where EF = EH.

The length of AF is

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 61

Part Test - 1 (JEE Advanced) - Question 62

Matrix Match Type

This section contains 1 questions. Each question has four statements (A, B, C and D) given in Column-I and five statements (p,q,r, s and t) in Column-II. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. The answers to these questions have to be appropriately marked as illustrated in the following example. If the correct matches are A-p, A-r, B-p, B-s, C-r, C-s, D-q and D-t then the answer should be written as : A→ p,r ; B→ p, s ; C → r, s ; D → q, t

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 62

(A)

m is of the from (i + 2)2 - 2, i = 1, 2, 3 ..., n

(B)
∵

∴ required value = 16 - 2(8) = 0
(C)   ⇒
(D)

*Answer can only contain numeric values
Part Test - 1 (JEE Advanced) - Question 63

If |z1 + z2| = |z1| – |z2| = 2 and |2z2 + 2i(z3 – z2)| = |z2| + |2iz3 + (1 – 2i) z2| = 10 where z1 = 3 + 4i, then find the value of |z1|2 + |z2|2 + |z3|2

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 63

∵
⇒ Z1, Z2 and origin are collinear and origin lies between Z1 and Z2 but |Z1| - |Z2| = 2

∵ |2Z2 + 2i (Z3 - Z2)[ = |Z2| + |2iZ3 + (1 - 2i) Z2|
let (1 - 2i) Z2 + 2iZ3 = Z4
∴ |Z2 + Z4| = |Z2| + |Z4|
∴ Z2, Z4 and origin are collinear and Z2, Z4 lies on the same side of origin
∵ Z4 = Z2 + 2i (Z3 - Z2)

∴ By rotating it about x=π/2 in anticlockwise direction we get Z4 and |Z4 - Z2| = 2 |Z3 - Z2|
|Z4| = 10 - 3 = 7     [∵ |Z2| + |Z4| = 10] |Z4 - Z2| = 4
|Z3 - Z2| = 2

∴ |Z1|2 + IZ2I2 + IZ3I2 = 25 + 9 + 13 = 47

*Answer can only contain numeric values
Part Test - 1 (JEE Advanced) - Question 64

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 64

.......(1)
Also (x + 1)n = C0Xn + C1Xn-1 + C2Xn-1 + .............. + Cn ( 2 )
Multiply (1) & (2) and compare coefficient of xn+1

*Answer can only contain numeric values
Part Test - 1 (JEE Advanced) - Question 65

Each digit has to be use only once then 6 digit numbers are there whose 3 digits are even and 3 are odd
is λ, then find exponent of 10 in λ.

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 65

Even : 0, 2, 4, 6, 8 odd : 1, 3, 5, 7, 9
5C3 . 5C3. 6! - 4C2 5C3 5! = 64800 (Total) - (when zero occupies first place)
5C3 x 5C3. 6!: Selection of 3 even from 5 even numbers and selection of 3 odd from 5 odd number.
4C2 x 5C3. 5 ! ; Selection of 2 even from remaining four numbers and selection of 3 odd from 5 odd numbers.
In order to find numbers divisible by 5, Consider two cases.
In order to find numbers divisible by 5, Consider two cases.
(i) fixing 0 at last digit
4C2 x 5C3 x 5!
(i) fixing 5 at last digit
4C2 x 5C3 x 5! - 4C2 . 4C2. 4!
Number divisible by 5 = 4Cx 5C3 x 5! + 4Cx 5C3 x 5! - 4C2. 4C2. 4! = 13536

*Answer can only contain numeric values
Part Test - 1 (JEE Advanced) - Question 66

If n € odd integers such that (1 + x + x2)n = a0 + a1x + a2 x2 + ........+ a2nx2n and (a1)2 – (2 a2)2 + (3 a3)2 – (4 a4)2 + ......–(2na2n)2 = – λn2 an-1 then find the value of λ.

Detailed Solution for Part Test - 1 (JEE Advanced) - Question 66

∵ (1 + x + x2)n = a0 + a1x + a2x2 + ....+ a2nx2n ........... (1)
differentiate w.r.t. x, we get.
⇒ n (1 + x + x2)n-1 (1 + 2x) = a1 + 2a2 x + 3a3x2 + ...... 2na2nx2n-1 ...... (2)

⇒ n(x2 - x + 1)n -1 (x - 2) = a1x2n-1 - 2a2x2n-2 + 3a3x2n-3 - .............. - 2n a2n ........ (3)
multiplying equation (2) and (3) and compare coefficient of x2n-1 of both sides.
⇒ (a1)2 - (2a2)2 + (3a3)2 - .... - (2n a2n)2 = coefficient of x2n-1 in n2 (1 + x2 + x4)n-1 (2x2- 3 x- 2)
= 2n2 {coefficient of x2-3 in (1 + x2 + x4)n- 1} - 3n2 {coefficient of x2n-2 in (1 + x2 + x4)n-1} - 2n2 {coefficient of x2n-1 in (1 + x2 + x4)n-1}

= 2n2 (0) - 3n2 {coefficient of xn-1 in (1 + x+ x2)n-1} - 2n2(0).
= -3n2 an-1

Note : ∵ n is odd ⇒ (2n - 3) and (2n - 1) both will be odd.
∴ in the expansion of (1 + x2 + x4)n-1 all the powers of x will be even
∴ coefficient of x2n-1 and x2n-3 will be zero.

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