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JEE Advanced 2021 Question Paper with Answer Key PDFs is available for download. Physics paper was easy and had questions from all the topics of class XI and class XII. The Chemistry paper had NCERT-based questions. Mathematics was the toughest. In Maths Paper 2, most of the questions were asked from Calculus and Coordinate Geometry.
JEE Advanced 2021 Questions Paper with Answer Key
Question Paper | Question Paper Link |
---|---|
Paper 1 Physics | Check Here |
Paper 1 Chemistry | Check Here |
Paper 1 Mathematics | Check Here |
Paper 1 (Hindi) | Check Here |
Paper 2 Physics | Check Here |
Paper 2 Chemistry | Check Here |
Paper 2 Mathematics | Check Here |
Paper 2 (Hindi) | Check Here |
JEE Advanced 2021 Questions
1. Let the function \(f:[1,\infin)→\R\) be defined by
\(f(t) = \begin{cases} (-1)^{n+1}2, & \text{if } t=2n-1,n\in\N, \\ \frac{(2n+1-t)}{2}f(2n-1)+\frac{(t-(2n-1))}{2}f(2n+1) & \text{if } 2n-1<t<2n+1,n\in\N. \end{cases}\)
Define \(g(x)=\int\limits_{1}^{x}f(t)dt,x\in(1,\infin).\) Let α denote the number of solutions of the equation g(x) = 0 in the interval (1, 8] and \(β=\lim\limits_{x→1+}\frac{g(x)}{x-1}\). Then the value of α + β is equal to _____.
\(f(t) = \begin{cases} (-1)^{n+1}2, & \text{if } t=2n-1,n\in\N, \\ \frac{(2n+1-t)}{2}f(2n-1)+\frac{(t-(2n-1))}{2}f(2n+1) & \text{if } 2n-1<t<2n+1,n\in\N. \end{cases}\)
Define \(g(x)=\int\limits_{1}^{x}f(t)dt,x\in(1,\infin).\) Let α denote the number of solutions of the equation g(x) = 0 in the interval (1, 8] and \(β=\lim\limits_{x→1+}\frac{g(x)}{x-1}\). Then the value of α + β is equal to _____.
2. A dimensionless quantity is constructed in terms of electronic charge \(e\), permittivity of free space \(\epsilon_0\) , Planck’s constant ℎ, and speed of light c. If the dimensionless quantity is written as \(e^\alpha\epsilon_0^\beta h^\gamma c^\delta\)and n is a non-zero integer, then\((\alpha, \beta,\gamma,\delta)\) is given by
- \((2n,-n,-n,-n)\)
- \((n,-n,-2n,-n)\)
- \((n,-n,-n,-2n)\)
- \((2n,-n,-2n,-2n)\)
3. A block of mass \(5 kg\) moves along the \(x-\)direction subject to the force \(F = (−20x + 10) N,\) with the value of \(x \) in metre. At time \(t = 0 s,\) it is at rest at position \(x = 1 m\). The position and momentum of the block at \(t = (\pi/4)\) s are
- \(-0.5m,5kg \ \frac{m}{s}\)
- \(0.5m,0kg \ \frac{m}{s}\)
- \(0.5m,0kg \ \frac{m}{s}\)
- \(0.5m,0kg \ \frac{m}{s}\)
4. A region in the form of an equilateral triangle (in x-y plane) of height L has a uniform magnetic field 𝐵⃗ pointing in the +z-direction. A conducting loop PQR, in the form of an equilateral triangle of the same height 𝐿, is placed in the x-y plane with its vertex P at x = 0 in the orientation shown in the figure. At 𝑡 = 0, the loop starts entering the region of the magnetic field with a uniform velocity 𝑣 along the +x-direction. The plane of the loop and its orientation remain unchanged throughout its motion.
Which of the following graph best depicts the variation of the induced emf (E) in the loop as a function of the distance (𝑥) starting from 𝑥 = 0?
Which of the following graph best depicts the variation of the induced emf (E) in the loop as a function of the distance (𝑥) starting from 𝑥 = 0?
5. Two beads, each with charge q and mass m, are on a horizontal, frictionless, non-conducting, circular hoop of radius R. One of the beads is glued to the hoop at some point, while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by [ \(\epsilon_0 \)is the permittivity of free space.]
- \(\frac{q^2}{4\pi\epsilon_0R^3m}\)
- \(\frac{q^2}{32\pi\epsilon_0R^3m}\)
- \(\frac{q^2}{8\pi\epsilon_0R^3m}\)
- \(\frac{q^2}{16\pi\epsilon_0R^3m}\)
*The article might have information for the previous academic years, which will be updated soon subject to the notification issued by the University/College.
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