1
JEE Main 2022 (Online) 29th July Evening Shift
+4
-1

Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R.

Assertion A: Amylose is insoluble in water.

Reason R: Amylose is a long linear molecule with more than 200 glucose units.

In the light of the above statements, choose the correct answer from the options given below.

A
$$\mathbf{A}$$ Both $$\mathbf{A}$$ and $$\mathbf{R}$$ are correct and $$\mathbf{R}$$ is the correct explanation of $$\mathbf{A}$$.
B
Both $$\mathbf{A}$$ and $$\mathbf{R}$$ are correct but $$\mathbf{R}$$ is NOT the correct explanation of $$\mathbf{A}$$.
C
$$\mathbf{A}$$ is correct but $$\mathbf{R}$$ is not correct
D
$$\mathbf{A}$$ is not correct but $$\mathbf{R}$$ is correct.
2
JEE Main 2022 (Online) 28th July Evening Shift
+4
-1

Two statements in respect of drug-enzyme interaction are given below

Statement I : Action of an enzyme can be blocked only when an inhibitor blocks the active site of the enzyme.

Statement II : An inhibitor can form a strong covalent bond with the enzyme.

In the light of the above statements, choose the correct answer from the options given below

A
Both Statement I and Statement II are true
B
Both Statement I and Statement II are false
C
Statement I is true but Statement II is false
D
Statement I is false but Statement II is true
3
JEE Main 2022 (Online) 28th July Evening Shift
+4
-1

The formulas of A and B for the following reaction sequence

are

A
$$\mathrm{A}=\mathrm{C}_{7} \mathrm{H}_{14} \mathrm{O}_{8}, \quad \mathrm{~B}=\mathrm{C}_{6} \mathrm{H}_{14}$$
B
$$\mathrm{A}=\mathrm{C}_{7} \mathrm{H}_{13} \mathrm{O}_{7}, \quad \mathrm{~B}=\mathrm{C}_{7} \mathrm{H}_{14} \mathrm{O}$$
C
$$\mathrm{A}=\mathrm{C}_{7} \mathrm{H}_{12} \mathrm{O}_{8}, \quad \mathrm{~B}=\mathrm{C}_{6} \mathrm{H}_{14}$$
D
$$\mathrm{A}=\mathrm{C}_{7} \mathrm{H}_{14} \mathrm{O}_{8}, \quad \mathrm{~B}=\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}_{6}$$
4
JEE Main 2022 (Online) 28th July Morning Shift
+4
-1

Match List - I with List - II.

List - I
Reaction
List - II
Catalyst
(A) $$4N{H_3}(g) + 5{O_2}(g) \to 4NO(g) + 6{H_2}O(g)$$ (I) $$NO(g)$$
(B) $${N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g)$$ (II) $${H_2}S{O_4}(l)$$
(C) $${C_{12}}{H_{22}}{O_{11}}(aq) + {H_2}O(l) \to \mathop {{C_6}{H_{12}}{O_6}}\limits_{Glu\cos e} + \mathop {{C_6}{H_{12}}{O_6}}\limits_{Fructose}$$ (III) $$Pt(s)$$
(D) $$2S{O_2}(g) + {O_2}(g) \to 2S{O_3}(g)$$ (IV) $$Fe(s)$$

Choose the correct answer from the options given below :

A
$$(\mathrm{A})-(\mathrm{II}),(\mathrm{B})-(\mathrm{III}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{IV})$$
B
$$(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{II}),(\mathrm{C})-(\mathrm{I}),(\mathrm{D})-(\mathrm{IV})$$
C
$$(\mathrm{A})-(\mathrm{III}),(\mathrm{B})-(\mathrm{IV}),(\mathrm{C})-(\mathrm{II}),(\mathrm{D})-(\mathrm{I})$$
D
(A) - (III), (B) - (II), (C) - (IV), (D) - (I)
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