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Exam Details

JEE ADVANCED 2021 PAPER 2 ONLINE

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Questions 57
Duration 180 mins
Package IIT-JEE Advance - Previous Year Papers

Paper pattern & analysis

Filter this paper by subject, topic or subtopic. Every graph updates from the selected questions.

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Showing all 57 questions in this paper.

Subject distribution

Physics
19 Qs
Chemistry
19 Qs
Mathematics
19 Qs

Topic distribution

Mechanics
10 Qs
Physical Chemistry
9 Qs
Calculus
9 Qs
Electricity
6 Qs
Organic Chemistry
6 Qs
Coordinate Geometry
6 Qs
Inorganic Chemistry
4 Qs
Algebra
3 Qs
Modern Physics
2 Qs
Optics
1 Qs
Trigonometry
1 Qs

Subtopic distribution

Definite Integration
6 Qs
Heat And Thermodynamics
4 Qs
Compounds Containing Nitrogen
4 Qs
Circle
4 Qs
Impulse And Momentum
3 Qs
Magnetism
3 Qs
Redox Reactions
3 Qs
Electrochemistry
3 Qs
Application Of Integration
3 Qs
Alternating Current
2 Qs
Salt Analysis
2 Qs
Basics Of Organic Chemistry
2 Qs

Difficulty distribution

Medium 40 70.2%
Hard 14 24.6%
Easy 3 5.3%

Question type distribution

Multiple Choices 30 52.6%
Numerical Answer Type (NAT) 27 47.4%

Syllabus

Full Syllabus

Sample questions from this paper

Questions are selected across the paper subjects wherever the paper contains that variety.

1
2021 · Physics · Mechanics · Impulse And Momentum
JEE ADVANCED 2021 PAPER 2 ONLINE
A pendulum consists of a bob of mass m = 0.1 kg and a massless inextensible string of length L = 1.0 m. It is suspended from a fixed point at height H = 0.9 m above a frictionless horizontal floor. Initially, the bob of the pendulum is lying on the floor at rest vertically below the point of suspension. A horizontal impulse P = 0.2 kg-m/s is imparted to the bob at some instant. After the bob slides for some distance, the string becomes taut and the bob lifts off the floor. The magnitude of the angular momentum of the pendulum about the point of suspension just before the bob lifts off is J kg-m2/s. The kinetic energy of the pendulum just after the lift-off is K Joules.

The value of J is ___________.
Enter a numerical response
2
2021 · Chemistry · Physical Chemistry · Redox Reactions
JEE ADVANCED 2021 PAPER 2 ONLINE
A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO4 solution to reach the end point. Number of moles of Fe2+ present in 250 mL solution is x \(\times\) 10\(-\)2 (consider complete dissolution of FeCl2). The amount of iron present in the sample is y% by weight.

(Assume : KMnO4 reacts only with Fe2+ in the solution

Use : Molar mass of iron as 56 g mol\(-\)1)

The value of y is ______.
Enter a numerical response
3
2021 · Mathematics · Algebra · Probability
JEE ADVANCED 2021 PAPER 2 ONLINE
A number of chosen at random from the set {1, 2, 3, ....., 2000}. Let p be the probability that the chosen number is a multiple of 3 or a multiple of 7. Then the value of 500p is __________.
Enter a numerical response
4
2021 · Physics · Mechanics · Heat And Thermodynamics
JEE ADVANCED 2021 PAPER 2 ONLINE
A soft plastic bottle, filled with water of density 1 gm/cc, carries an inverted glass test-tube with some air (ideal gas) trapped as shown in the figure. The test-tube has a mass of 5 gm, and it is made of a thick glass of density 2.5 gm/cc. Initially the bottle is sealed at atmosphere pressure p0 = 105 Pa so that the volume of the trapped air is v0 = 3.3 cc. When the bottle is squeezed from outside at constant temperature, the pressure inside rises and the volume of the trapped air reduces. It is found that the test tube begins to sink at pressure p0 + \(\Delta\)p without changing its orientation. At this pressure, the volume of the trapped air is v0 \(-\) \(\Delta\)v. Let \(\Delta\)v = X cc and \(\Delta\)p = Y \(\times\) 103 Pa.

JEE Advanced 2021 Paper 2 Online Physics - Heat and Thermodynamics Question 45 English

The value of X is _______________.
Enter a numerical response
5
2021 · Chemistry · Physical Chemistry · Redox Reactions
JEE ADVANCED 2021 PAPER 2 ONLINE
A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO4 solution to reach the end point. Number of moles of Fe2+ present in 250 mL solution is x \(\times\) 10\(-\)2 (consider complete dissolution of FeCl2). The amount of iron present in the sample is y% by weight.

(Assume : KMnO4 reacts only with Fe2+ in the solution

Use : Molar mass of iron as 56 g mol\(-\)1)

The value of x is ______.
Enter a numerical response
6
2021 · Mathematics · Coordinate Geometry · Circle
JEE ADVANCED 2021 PAPER 2 ONLINE
Consider the region R = {(x, y) \(\in\) R \(\times\) R : x \(\ge\) 0 and y2 \(\le\) 4 \(-\) x}. Let F be the family of all circles that are contained in R and have centers on the x-axis. Let C be the circle that has largest radius among the circles in F. Let (\(\alpha\), \(\beta\)) be a point where the circle C meets the curve y2 = 4 \(-\) x.

The value of \(\alpha\) is ___________.
Enter a numerical response