1
JEE Advanced 2019 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
Change Language
Consider the Bohr's model of a one-electron atom where the electron moves around the nucleus. In the following List-I contains some quantities for the nth orbit of the atom and List-II contains options showing how they depend on n.

JEE Advanced 2019 Paper 2 Offline Chemistry - Structure of Atom Question 6 English
Which of the following options has the correct combination considering List-I and List-II?
A
(II), (R)
B
(I), (P)
C
(I), (T)
D
(II), (Q)
2
JEE Advanced 2019 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
Change Language
List-I includes starting materials and reagents of selected chemical reactions. List-II gives structures of compounds that may be formed as intermediate products and/or final products from the reactions of List-I.

JEE Advanced 2019 Paper 2 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 48 English Comprehension
Which of the following options has correct combination considering List-I and List-II?
A
(III), (S) (R)
B
(IV), (Q), (R)
C
(III), (T), (U)
D
(IV), (Q), (U)
3
JEE Advanced 2019 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
Change Language
List-I includes starting materials and reagents of selected chemical reactions. List-II gives structures of compounds that may be formed as intermediate products and/or final products from the reactions of List-I.

JEE Advanced 2019 Paper 2 Offline Chemistry - Aldehydes, Ketones and Carboxylic Acids Question 46 English Comprehension
Which of the following options has correct combination considering List-I and List-II?
A
(II), (P), (S), (U)
B
(I), (Q), (T), (U)
C
(II), (P), (S), (T)
D
(I), (S), (Q), (R)
4
JEE Advanced 2019 Paper 2 Offline
MCQ (More than One Correct Answer)
+4
-1
Change Language
For non-negative integers n, let

$$f(n) = {{\sum\limits_{k = 0}^n {\sin \left( {{{k + 1} \over {n + 2}}\pi } \right)} \sin \left( {{{k + 2} \over {n + 2}}\pi } \right)} \over {\sum\limits_{k = 0}^n {{{\sin }^2}\left( {{{k + 1} \over {n + 2}}\pi } \right)} }}$$

Assuming cos$$-1$$ x takes values in [0, $$\pi $$], which of the following options is/are correct?
A
If $$\alpha $$ = tan(cos$$-$$1 f(6)), then $$\alpha $$2 + 2$$\alpha $$ $$-$$1 = 0
B
$$f(4) = {{\sqrt 3 } \over 2}$$
C
sin(7 cos$$-$$1 f(5)) = 0
D
$$\mathop {\lim }\limits_{n \to \infty } \,f(n) = {1 \over 2}$$
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