All Exams >
JEE >
National Level Test Series for JEE Advanced 2025 >
All Questions
All questions of JEE Advanced 2012 for JEE Exam
Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q. The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the given
concentration cell is (take 2.303 x R x 298/F = 0.059 V)- a)1 x 10–15
- b)4 x 10–15
- c)1 x 10–12
- d)4 x 10–12
Correct answer is option 'B'. Can you explain this answer?
Paragraph for Questions 31 and 32
The electrochemical cell shown below is a concentration cell.
M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | M
The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.
Q. The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available for the given
concentration cell is (take 2.303 x R x 298/F = 0.059 V)
concentration cell is (take 2.303 x R x 298/F = 0.059 V)
a)
1 x 10–15
b)
4 x 10–15
c)
1 x 10–12
d)
4 x 10–12
|
Cstoppers Instructors answered |
The shape of XeO2F2 molecule is- a)trigonal bipyramidal
- b)square planar
- c)tetrahedral
- d)see–saw
Correct answer is option 'D'. Can you explain this answer?
The shape of XeO2F2 molecule is
a)
trigonal bipyramidal
b)
square planar
c)
tetrahedral
d)
see–saw
|
Krishna Iyer answered |
Hybridization = sp3d
Shape = see – saw
Shape = see – saw
Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is(are) correct?- a)Both cylinders P and Q reach the ground at the same time.
- b)Cylinders P has larger linear acceleration than cylinder Q.
- c)Both cylinders reach the ground with same translational kinetic energy.
- d)Cylinder Q reaches the ground with larger angular speed.
Correct answer is option 'D'. Can you explain this answer?
Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is(are) correct?
a)
Both cylinders P and Q reach the ground at the same time.
b)
Cylinders P has larger linear acceleration than cylinder Q.
c)
Both cylinders reach the ground with same translational kinetic energy.
d)
Cylinder Q reaches the ground with larger angular speed.
|
|
Dev Kumar answered |
A=g sintheta × MR^2 / MR^2 + I ....
so a is inversely proportional to moment of inertia I ....
mass of P is concentrated at the surface so it's moment of inertia is more and mass of Q is near the axis so it's moment of inertia is less... so the body with less moment of inertia will have more acceleration as the relationship is inverse... so the body Q will have more acceleration and will reach the bottom with greater angular speed.. so the option D is correct..
With respect to graphite and diamond, which of the statement(s) given below is (are) correct?- a)Graphite is harder than diamond.
- b)Graphite has higher electrical conductivity than diamond.
- c)Graphite has higher thermal conductivity than diamond.
- d)Graphite has higher C–C bond order than diamond.
Correct answer is option 'B,D'. Can you explain this answer?
With respect to graphite and diamond, which of the statement(s) given below is (are) correct?
a)
Graphite is harder than diamond.
b)
Graphite has higher electrical conductivity than diamond.
c)
Graphite has higher thermal conductivity than diamond.
d)
Graphite has higher C–C bond order than diamond.
|
Nandini Patel answered |
Graphite can conduct electricity because of the delocalised (free) electrons in its structure. These arise because each carbon atom is only bonded to 3 other carbon atoms. ... However, in diamond, all 4 outer electrons on each carbon atom are used in covalent bonding, so there are no delocalised electrons.
graphite and graphene, diamond is the best thermal conductor around room temperature, having thermal conductivity of more than 2,000 watts per meter per Kelvin, which is five times higher than the best metals such as copper.
The reaction of white phosphorous with aqueous NaOH gives phosphine along with another phosphorous
containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the other
product are respectively- a)redox reaction; -3 and -5
- b)redox reaction; +3 and +5
- c)disproportionation reaction; -3 and +5
- d)disproportionation reaction; -3 and +3
Correct answer is option 'C'. Can you explain this answer?
The reaction of white phosphorous with aqueous NaOH gives phosphine along with another phosphorous
containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the other
product are respectively
containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the other
product are respectively
a)
redox reaction; -3 and -5
b)
redox reaction; +3 and +5
c)
disproportionation reaction; -3 and +5
d)
disproportionation reaction; -3 and +3
|
Anjali Sharma answered |
The balanced disproportionation reaction involving white phosphorus with aq. NaOH is
Two moles of ideal helium gas are in a rubber balloon at 30°C. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35°C. The amount of heat required in raising the temperature is nearly (take R = 8.31 J/mol.K)- a)62 J
- b)104 J
- c)124 J
- d)208 J
Correct answer is option 'D'. Can you explain this answer?
Two moles of ideal helium gas are in a rubber balloon at 30°C. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35°C. The amount of heat required in raising the temperature is nearly (take R = 8.31 J/mol.K)
a)
62 J
b)
104 J
c)
124 J
d)
208 J
|
|
Pallabi Basak answered |
Paragraph for Questions 53 and 54
Let an denote the number of all n–digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let bn = the number of such n–digit integers ending with digit 1 and cn = the number of such n–digit integers ending with digit 0.Q. Which of the following is correct?- a)a17 = a16 + a15
- b)c17 ≌ c16 + c15
- c)b17 ≌ b16 + c16
- d)a17 = c17 + b16
Correct answer is option 'A'. Can you explain this answer?
Paragraph for Questions 53 and 54
Let an denote the number of all n–digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let bn = the number of such n–digit integers ending with digit 1 and cn = the number of such n–digit integers ending with digit 0.
Let an denote the number of all n–digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let bn = the number of such n–digit integers ending with digit 1 and cn = the number of such n–digit integers ending with digit 0.
Q. Which of the following is correct?
a)
a17 = a16 + a15
b)
c17 ≌ c16 + c15
c)
b17 ≌ b16 + c16
d)
a17 = c17 + b16
|
Shounak Shah answered |
from the point (3, 1, –1) is
For a dilute solution containing 2.5 g of a non–volatile non–electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol-1)- a)724
- b)740
- c)736
- d)718
Correct answer is option 'A'. Can you explain this answer?
For a dilute solution containing 2.5 g of a non–volatile non–electrolyte solute in 100 g of water, the elevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76 K kg mol-1)
a)
724
b)
740
c)
736
d)
718
|
|
Athira Shah answered |
B → Solute; A → Solvent
WB = 2.5 g, WA = 100 g
NiCl2{P(C2H5)2(C6H5)}2 exhibits temperature dependent magnetic behaviour (paramagnetic/diamagnetic).
The coordination geometries of Ni2+ in the paramagnetic and diamagnetic states are respectively- a)tetrahedral and tetrahedral
- b)square planar and square planar
- c)tetrahedral and square planar
- d)square planar and tetrahedral
Correct answer is option 'C'. Can you explain this answer?
NiCl2{P(C2H5)2(C6H5)}2 exhibits temperature dependent magnetic behaviour (paramagnetic/diamagnetic).
The coordination geometries of Ni2+ in the paramagnetic and diamagnetic states are respectively
The coordination geometries of Ni2+ in the paramagnetic and diamagnetic states are respectively
a)
tetrahedral and tetrahedral
b)
square planar and square planar
c)
tetrahedral and square planar
d)
square planar and tetrahedral
|
Mahesh Unni answered |
In both complexes Ni exists as Ni2+.
In sp3 (tetrahedral)
In dsp2 (square planar geometry)
SECTION I : Single Correct Answer TypeThis section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONLY ONE is correct.Q. Let a1, a2, a3, ….. be in harmonic progression with a1 = 5 and a20 = 25. The least positive integer n for
which an < 0- a)22
- b)23
- c)24
- d)25
Correct answer is option 'D'. Can you explain this answer?
SECTION I : Single Correct Answer Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONLY ONE is correct.
which ONLY ONE is correct.
Q. Let a1, a2, a3, ….. be in harmonic progression with a1 = 5 and a20 = 25. The least positive integer n for
which an < 0
which an < 0
a)
22
b)
23
c)
24
d)
25
|
Vaishnavi Basak answered |
Understanding Harmonic Progression
Harmonic progression (HP) means that the reciprocals of the terms form an arithmetic progression (AP). Given that a1 = 5 and a20 = 25, we can derive the necessary values.
Finding the Reciprocals
- The terms in HP are:
a1 = 5, a2, a3, ...
The reciprocals are:
1/a1 = 1/5, 1/a2, 1/a3, ..., 1/a20 = 1/25
- Since these reciprocals are in AP, we can denote:
d = common difference of the AP.
Setting Up the Equations
1. The first term of the AP is 1/5.
2. The 20th term can be expressed as:
1/a20 = 1/5 + 19d = 1/25
- From this, we can solve for d:
1/5 + 19d = 1/25
19d = 1/25 - 1/5
19d = 1/25 - 5/25
19d = -4/25
d = -4/(25*19) = -4/475
Finding the General Term
- The n-th term of the reciprocals in AP is given by:
1/an = 1/5 + (n-1)(-4/475)
= 1/5 - 4(n-1)/475
- Therefore, the n-th term can be expressed as:
an = 1 / (1/5 - 4(n-1)/475)
Finding When an < />
- Setting an < 0:="" />
1/5 - 4(n-1)/475 < 0="" />
-4(n-1)/475 < -1/5="" />
4(n-1)/475 > 1/5
n - 1 > (475/4) * (1/5)
n - 1 > 23.75
n > 24.75
- Therefore, the least positive integer n is 25.
Thus, the answer is option D: 25.
Harmonic progression (HP) means that the reciprocals of the terms form an arithmetic progression (AP). Given that a1 = 5 and a20 = 25, we can derive the necessary values.
Finding the Reciprocals
- The terms in HP are:
a1 = 5, a2, a3, ...
The reciprocals are:
1/a1 = 1/5, 1/a2, 1/a3, ..., 1/a20 = 1/25
- Since these reciprocals are in AP, we can denote:
d = common difference of the AP.
Setting Up the Equations
1. The first term of the AP is 1/5.
2. The 20th term can be expressed as:
1/a20 = 1/5 + 19d = 1/25
- From this, we can solve for d:
1/5 + 19d = 1/25
19d = 1/25 - 1/5
19d = 1/25 - 5/25
19d = -4/25
d = -4/(25*19) = -4/475
Finding the General Term
- The n-th term of the reciprocals in AP is given by:
1/an = 1/5 + (n-1)(-4/475)
= 1/5 - 4(n-1)/475
- Therefore, the n-th term can be expressed as:
an = 1 / (1/5 - 4(n-1)/475)
Finding When an < />
- Setting an < 0:="" />
1/5 - 4(n-1)/475 < 0="" />
-4(n-1)/475 < -1/5="" />
4(n-1)/475 > 1/5
n - 1 > (475/4) * (1/5)
n - 1 > 23.75
n > 24.75
- Therefore, the least positive integer n is 25.
Thus, the answer is option D: 25.
is coming
A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without
touching it, the correct statement(s) is(are)- a)The emf induced in the loop is zero if the current is constant.
- b)The emf induced in the loop is finite if the current is constant.
- c)The emf induced in the loop is zero if the current decreases at a steady rate.
- d)The emf induced in the loop is infinite if the current decreases at a steady rate.
Correct answer is option 'A,C'. Can you explain this answer?
A current carrying infinitely long wire is kept along the diameter of a circular wire loop, without
touching it, the correct statement(s) is(are)
touching it, the correct statement(s) is(are)
a)
The emf induced in the loop is zero if the current is constant.
b)
The emf induced in the loop is finite if the current is constant.
c)
The emf induced in the loop is zero if the current decreases at a steady rate.
d)
The emf induced in the loop is infinite if the current decreases at a steady rate.
|
|
Samridhi Nair answered |
Paragraph for Questions 33 and 34Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.Q. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is- a)Cl2O
- b)Cl2O7
- c)ClO2
- d)Cl2O6
Correct answer is option 'A'. Can you explain this answer?
Paragraph for Questions 33 and 34
Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.
Q. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is
a)
Cl2O
b)
Cl2O7
c)
ClO2
d)
Cl2O6
|
|
Keerthana Sharma answered |
, then the possible value(s) of the determinant of P is (are)
Paragraph for Questions 51 and 52
Q. Consider the statements:P : There exists some x ε such that f(x) + 2x = 2(1 + x2)Q : There exists some x ε such that 2f(x) + 1 = 2x(1 + x)
Then- a)both P and Q are true
- b)P is true and Q is false
- c)P is false and Q is true
- d)both P and Q are false
Correct answer is option 'C'. Can you explain this answer?
Paragraph for Questions 51 and 52
Q. Consider the statements:
P : There exists some x ε such that f(x) + 2x = 2(1 + x2)
Q : There exists some x ε such that 2f(x) + 1 = 2x(1 + x)
Then
Then
a)
both P and Q are true
b)
P is true and Q is false
c)
P is false and Q is true
d)
both P and Q are false
|
Ruchi Kumar answered |
SECTION III : Multiple Correct Answer(s) TypeThe section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.Q. The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure.
Which of the following statement(s) is (are) correct?- a)T1 = T2
- b)T3 > T1
- c)wisothermal > wadiabatic
- d)ΔUisothermal > ΔUadiabatic
Correct answer is option 'A,C,D'. Can you explain this answer?
SECTION III : Multiple Correct Answer(s) Type
The section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of which ONE or MORE are correct.
Q. The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure.
Which of the following statement(s) is (are) correct?
Which of the following statement(s) is (are) correct?
a)
T1 = T2
b)
T3 > T1
c)
wisothermal > wadiabatic
d)
ΔUisothermal > ΔUadiabatic
|
|
Anjali Sharma answered |
T1 = T2 because process is isothermal.
Work done in adiabatic process is less than in isothermal process because area covered by isothermal curve is more than the area co vered by the adiabatic curve.
In adiabatic process expansion occurs by using internal energy hence it decreases while in isothermal process temperature remains constant that’s why no change in internal energy.
SECTION II : Paragraph TypeThis section contains 6 multiple choice questions relating to three paragraphs with two questions on each
paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.Paragraph for Questions 49 and 50A tangent PT is drawn to the circle x2 + y2 = 4 at the point 
. A straight line L, perpendicular to PT is a tangent to the circle (x -3)2 + y2 =1.Q. A possible equation of L is- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
SECTION II : Paragraph Type
This section contains 6 multiple choice questions relating to three paragraphs with two questions on each
paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
Paragraph for Questions 49 and 50
A tangent PT is drawn to the circle x2 + y2 = 4 at the point . A straight line L, perpendicular to PT is a tangent to the circle (x -3)2 + y2 =1.
. A straight line L, perpendicular to PT is a tangent to the circle (x -3)2 + y2 =1.Q. A possible equation of L is
a)
b)
c)
d)
|
Tanuja Kapoor answered |
SECTION III : Multiple Correct Answer(s) TypeThis section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE or MORE are correct.15. In the given circuit, the AC source has ω = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is (are)
- a)The current through the circuit, I is 0.3 A.
- b)The current through the circuit, I is
- c)The voltage across 100 ? resistor =
- d)The voltage across 50 ? resistor = 10 V.
Correct answer is option 'A,C'. Can you explain this answer?
SECTION III : Multiple Correct Answer(s) Type
This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONE or MORE are correct.
which ONE or MORE are correct.
15. In the given circuit, the AC source has ω = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is (are)
a)
The current through the circuit, I is 0.3 A.
b)
The current through the circuit, I is
c)
The voltage across 100 ? resistor =
d)
The voltage across 50 ? resistor = 10 V.
|
Ritika Sen answered |
With reference to the scheme given, which of the given statement(s) about T, U, V and W is (are) correct?
- a)T is soluble in hot aqueous NaOH
- b)U is optically active
- c)Molecular formula of W is C10H18O4
- d)V gives effervescence on treatment with aqueous NaHCO3
Correct answer is option 'A,C,D'. Can you explain this answer?
With reference to the scheme given, which of the given statement(s) about T, U, V and W is (are) correct?
a)
T is soluble in hot aqueous NaOH
b)
U is optically active
c)
Molecular formula of W is C10H18O4
d)
V gives effervescence on treatment with aqueous NaHCO3
|
|
Anjali Sharma answered |
Consider a disc rotating in the horizontal plane with a constant angular speed w about its centre O. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously projected at an angle towards R. The velocity of projection is in the y-z plane and is same for both pebbles with respect to the disc. Assume that(i) they land back on the disc before the disc has 1/8 completed rotation,(ii) their range is less than half the disc radius, and(iii) ω remains constant throughout. Then
- a)P lands in the shaded region and Q in the unshaded region.
- b)P lands in the unshaded region and Q in the shaded region.
- c)Both P and Q land in the unshaded region.
- d)Both P and Q land in the shaded region.
Correct answer is option 'C'. Can you explain this answer?
Consider a disc rotating in the horizontal plane with a constant angular speed w about its centre O. The disc has a shaded region on one side of the diameter and an unshaded region on the other side as shown in the figure. When the disc is in the orientation as shown, two pebbles P and Q are simultaneously projected at an angle towards R. The velocity of projection is in the y-z plane and is same for both pebbles with respect to the disc. Assume that
(i) they land back on the disc before the disc has 1/8 completed rotation,
(ii) their range is less than half the disc radius, and
(iii) ω remains constant throughout. Then
a)
P lands in the shaded region and Q in the unshaded region.
b)
P lands in the unshaded region and Q in the shaded region.
c)
Both P and Q land in the unshaded region.
d)
Both P and Q land in the shaded region.
|
Juhi Das answered |
In the given circuit, a charge of +80 μC is given to the upper plate of the 4 μF capacitor. Then in the steady state, the charge on the upper plate of the 3 μF capacitor is
- a)+32 μC
- b)+40 μC
- c)+48 μC
- d)+80 μC
Correct answer is option 'C'. Can you explain this answer?
In the given circuit, a charge of +80 μC is given to the upper plate of the 4 μF capacitor. Then in the steady state, the charge on the upper plate of the 3 μF capacitor is
a)
+32 μC
b)
+40 μC
c)
+48 μC
d)
+80 μC
|
Prateek Choudhury answered |
Let ‘q’ be the final charge on 3μF capacitor then
SECTION II : Paragraph TypeThis section contains 6 multiple choice questions relating to three paragraphs with two questions on each
paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.Paragraph for Questions 29 and 30In the following reaction sequence, the compound J is an intermediate.
Q. The compound K is- a)
- b)
- c)
- d)
Correct answer is option 'C'. Can you explain this answer?
SECTION II : Paragraph Type
This section contains 6 multiple choice questions relating to three paragraphs with two questions on each
paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.
Paragraph for Questions 29 and 30
In the following reaction sequence, the compound J is an intermediate.
Q. The compound K is
a)
b)
c)
d)
|
|
Anjali Sharma answered |
29- 30
SECTION I : Single Correct Answer TypeThis section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONLY ONE is correct.Q. The major product H of the given reaction sequence is
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
SECTION I : Single Correct Answer Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONLY ONE is correct.
which ONLY ONE is correct.
Q. The major product H of the given reaction sequence is
a)
b)
c)
d)
|
|
Anjali Sharma answered |
An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a uniform current density along its length. The magnitude of the magnetic field, 
as a function of the radial distance r from the axis is best represented by- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a uniform current density along its length. The magnitude of the magnetic field, as a function of the radial distance r from the axis is best represented by
as a function of the radial distance r from the axis is best represented bya)
b)
c)
d)
|
Vaishnavi Chauhan answered |
The given graphs / data I, II, III and IV represent general trends observed for different physisorption and
chemisorption processes under mild conditions of temperature and pressure. Which of the following
choice(s) about I, II, III and IV is (are) correct?
- a)I is physisorption and II is chemisorption
- b)I is physisorption and III is chemisorption
- c)IV is chemisorption and II is chemisorption
- d)IV is chemisorption and III is chemisorption
Correct answer is option 'A,C'. Can you explain this answer?
The given graphs / data I, II, III and IV represent general trends observed for different physisorption and
chemisorption processes under mild conditions of temperature and pressure. Which of the following
choice(s) about I, II, III and IV is (are) correct?
chemisorption processes under mild conditions of temperature and pressure. Which of the following
choice(s) about I, II, III and IV is (are) correct?
a)
I is physisorption and II is chemisorption
b)
I is physisorption and III is chemisorption
c)
IV is chemisorption and II is chemisorption
d)
IV is chemisorption and III is chemisorption
|
Ashutosh Shah answered |
Graph (I) and (III) represent physiosorption because, in physiosorption, the amount of adsorption decreases with the increase of temperature and increases with the increase of pressure.
Graph (II) represent chemisorption, because in chemisorption amount of adsorption increase with the increase of temperature. Graph (IV) is showing the formation of a chemical bond, hence chemisorption.
The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed ω and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed ω/2. The ring and disc are separated by frictionless ball bearings. The point P on the inner disc is at a distance R from the origin, where OP makes an angle of 300 with the horizontal. Then with respect to the horizontal surface,
- a)
- b)
- c)
- d)
Correct answer is option 'A,B'. Can you explain this answer?
The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed ω and (ii) an inner disc of radius 2R rotating anti-clockwise with angular speed ω/2. The ring and disc are separated by frictionless ball bearings. The point P on the inner disc is at a distance R from the origin, where OP makes an angle of 300 with the horizontal. Then with respect to the horizontal surface,
a)
b)
c)
d)
|
|
Meera Bajaj answered |
such that
Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shown
in the figure. Given that 
, which of the following statement(s) is (are) correct?
- a)The electric field at O is 6K along OD.
- b)The potential at O is zero.
- c)The potential at all points on the line PR is same.
- d)The potential at all points on the line ST is same.
Correct answer is option 'A,B,C'. Can you explain this answer?
Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shown
in the figure. Given that, which of the following statement(s) is (are) correct?
in the figure. Given that
, which of the following statement(s) is (are) correct?a)
The electric field at O is 6K along OD.
b)
The potential at O is zero.
c)
The potential at all points on the line PR is same.
d)
The potential at all points on the line ST is same.
|
Surbhi Bose answered |
Line PR is perpendicular bisector of all
the dipoles.
the dipoles.
Paragraph for Questions 49 and 50A tangent PT is drawn to the circle x2 + y2 = 4 at the point . A straight line L, perpendicular to PT is a tangent to the circle (x -3)2 + y2 =1.Q. A common tangent of the two circles is- a)x = 4
- b)y = 2
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Paragraph for Questions 49 and 50
A tangent PT is drawn to the circle x2 + y2 = 4 at the point . A straight line L, perpendicular to PT is a tangent to the circle (x -3)2 + y2 =1.
. A straight line L, perpendicular to PT is a tangent to the circle (x -3)2 + y2 =1.Q. A common tangent of the two circles is
a)
x = 4
b)
y = 2
c)
d)
|
|
Dipika Patel answered |
SECTION I : Single Correct Answer TypeThis section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONLY ONE is correct.Q. Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed ω. The discs are in the same horizontal plane. At time t = 0, the points P and Q are facing each other as shown in the figure. The relative speed between the two points P and Q is vr. In one time period (T) of rotation of the discs, vr as a function of time is best represented by
- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
SECTION I : Single Correct Answer Type
This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out of
which ONLY ONE is correct.
which ONLY ONE is correct.
Q. Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed ω. The discs are in the same horizontal plane. At time t = 0, the points P and Q are facing each other as shown in the figure. The relative speed between the two points P and Q is vr. In one time period (T) of rotation of the discs, vr as a function of time is best represented by
a)
b)
c)
d)
|
|
Krishna Iyer answered |
In each rotation relative speed becomes zero twice and becomes maximum twice.
Paragraph for Questions 53 and 54
Let an denote the number of all n–digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let bn = the number of such n–digit integers ending with digit 1 and cn = the number of such n–digit integers ending with digit 0.Q. The value of b6 is- a)7
- b)8
- c)9
- d)11
Correct answer is option 'B'. Can you explain this answer?
Paragraph for Questions 53 and 54
Let an denote the number of all n–digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let bn = the number of such n–digit integers ending with digit 1 and cn = the number of such n–digit integers ending with digit 0.
Let an denote the number of all n–digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let bn = the number of such n–digit integers ending with digit 1 and cn = the number of such n–digit integers ending with digit 0.
Q. The value of b6 is
a)
7
b)
8
c)
9
d)
11
|
Raghav Bansal answered |
where a, b, and c are the lengths of the sides of the triangle opposite to the angles at P, Q and R respectively. Then 
equals
then a possible
where
are
In the cyanide extraction process of silver from argentite ore, the oxidising and reducing agents used are- a)O2 and CO respectively
- b)O2 and Zn dust respectively
- c)HNO3 and Zn dust respectively
- d)HNO3 and CO respectively
Correct answer is option 'B'. Can you explain this answer?
In the cyanide extraction process of silver from argentite ore, the oxidising and reducing agents used are
a)
O2 and CO respectively
b)
O2 and Zn dust respectively
c)
HNO3 and Zn dust respectively
d)
HNO3 and CO respectively
|
Geetika Shah answered |
The reactions involved in cyanide extraction process are:
For the given aqueous reactions, which of the statement(s) is (are) true?
- a)The first reaction is a redox reaction.
- b)White precipitate is Zn3[Fe(CN)6]2.
- c)Addition of filtrate to starch solution gives blue colour.
- d)White precipitate is soluble in NaOH solution.
Correct answer is option 'A,C,D'. Can you explain this answer?
For the given aqueous reactions, which of the statement(s) is (are) true?
a)
The first reaction is a redox reaction.
b)
White precipitate is Zn3[Fe(CN)6]2.
c)
Addition of filtrate to starch solution gives blue colour.
d)
White precipitate is soluble in NaOH solution.
|
Diya Roy answered |
Paragraph for Questions 33 and 34Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.Q. 25 mL of household solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4 N acetic acid. In the
titration of the liberated iodine, 48 mL of 0.25 N Na2S2O3 was used to reach the end point. The molarity of
the household bleach solution is- a)0.48 M
- b)0.96 M
- c)0.24 M
- d)0.024 M
Correct answer is option 'C'. Can you explain this answer?
Paragraph for Questions 33 and 34
Bleaching powder and bleach solution are produced on a large scale and used in several household products. The effectiveness of bleach solution is often measured by iodometry.
Q. 25 mL of household solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4 N acetic acid. In the
titration of the liberated iodine, 48 mL of 0.25 N Na2S2O3 was used to reach the end point. The molarity of
the household bleach solution is
titration of the liberated iodine, 48 mL of 0.25 N Na2S2O3 was used to reach the end point. The molarity of
the household bleach solution is
a)
0.48 M
b)
0.96 M
c)
0.24 M
d)
0.024 M
|
|
Anjali Sharma answered |
Paragraph for Questions 51 and 52Q. Which of the following is true?- a)g is increasing on (1, ∞)
- b)g is decreasing on (1, ∞)
- c)g is increasing on (1, 2) and decreasing on (2, ∞)
- d)g is decreasing on (1, 2) and increasing on (2, ∞)
Correct answer is option 'B'. Can you explain this answer?
Paragraph for Questions 51 and 52
Q. Which of the following is true?
a)
g is increasing on (1, ∞)
b)
g is decreasing on (1, ∞)
c)
g is increasing on (1, 2) and decreasing on (2, ∞)
d)
g is decreasing on (1, 2) and increasing on (2, ∞)
|
|
Lavanya Menon answered |
Two spherical planets P and Q have the same uniform density r, masses Mp and MQ and surface
areas A and 4A respectively. A spherical planet R also has uniform density ρ and its mass is (MP
+ MQ). The escape velocities from the planets P, Q and R are VP, VQ and VR, respectively. Then- a)VQ > VR > VP
- b)VR > VQ > VP
- c)VR / VP = 3
- d)VP / VQ = (1/2)
Correct answer is option 'B,D'. Can you explain this answer?
Two spherical planets P and Q have the same uniform density r, masses Mp and MQ and surface
areas A and 4A respectively. A spherical planet R also has uniform density ρ and its mass is (MP
+ MQ). The escape velocities from the planets P, Q and R are VP, VQ and VR, respectively. Then
areas A and 4A respectively. A spherical planet R also has uniform density ρ and its mass is (MP
+ MQ). The escape velocities from the planets P, Q and R are VP, VQ and VR, respectively. Then
a)
VQ > VR > VP
b)
VR > VQ > VP
c)
VR / VP = 3
d)
VP / VQ = (1/2)
|
Hansa Sharma answered |
By calculation, if Mass of P = m
and Radius of P = R
Then Mass of Q = 8M
and radius of Q = 2R
and Mass of R = 9M
and radius of R = 91/3R
are coplanar, then the plane(s) containing these two lines is(are)
Paragraph for Questions 31 and 32The electrochemical cell shown below is a concentration cell.M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | MThe emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.Q. The value of ΔG (kJ mol–1) for the given cell is (take 1F = 96500 C mol–1)- a)–5.7
- b)5.7
- c)11.4
- d)-11.4
Correct answer is option 'D'. Can you explain this answer?
Paragraph for Questions 31 and 32
The electrochemical cell shown below is a concentration cell.
M|M2+ (saturated solution of a sparingly soluble salt, MX2) || M2+ (0.001 mol dm–3) | M
The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrodes. The emf of the cell at 298 K is 0.059 V.
Q. The value of ΔG (kJ mol–1) for the given cell is (take 1F = 96500 C mol–1)
a)
–5.7
b)
5.7
c)
11.4
d)
-11.4
|
|
Akshita Roy answered |
Chapter doubts & questions for JEE Advanced 2012 - National Level Test Series for JEE Advanced 2025 2025 is part of JEE exam preparation. The chapters have been prepared according to the JEE exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for JEE 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of JEE Advanced 2012 - National Level Test Series for JEE Advanced 2025 in English & Hindi are available as part of JEE exam.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
National Level Test Series for JEE Advanced 2025
3 videos|2 docs|40 tests
|
Signup to see your scores go up within 7 days!
Study with 1000+ FREE Docs, Videos & Tests
10M+ students study on EduRev
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup