1
JEE Main 2020 (Online) 6th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A solution of two components containing n1 moles of the 1st component and n2 moles of the 2nd component is prepared. M1 and M2 are the molecular weights of component 1 and 2 respectively. If d is the density of the solution in g mL–1, C2 is the molarity and x2 is the mole fraction of the 2nd component, then C2 can be expressed as :
A
$${C_2} = {{1000{x_2}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}$$
B
$${C_2} = {{1000d{x_2}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}$$
C
$${C_2} = {{d{x_2}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}$$
D
$${C_2} = {{d{x_1}} \over {{M_1} + {x_2}\left( {{M_2} - {M_1}} \right)}}$$
2
JEE Main 2020 (Online) 6th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The increasing order of pKb values of the following compounds is :

JEE Main 2020 (Online) 6th September Morning Slot Chemistry - Compounds Containing Nitrogen Question 164 English
A
I < II < IV < III
B
I < II < III < IV
C
II < I < III < IV
D
II < IV < III < I
3
JEE Main 2020 (Online) 6th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The area (in sq. units) of the region
A = {(x, y) : |x| + |y| $$ \le $$ 1, 2y2 $$ \ge $$ |x|}
A
$${1 \over 6}$$
B
$${5 \over 6}$$
C
$${1 \over 3}$$
D
$${7 \over 6}$$
4
JEE Main 2020 (Online) 6th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The general solution of the differential equation

$$\sqrt {1 + {x^2} + {y^2} + {x^2}{y^2}} $$ + xy$${{dy} \over {dx}}$$ = 0 is :

(where C is a constant of integration)
A
$$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$$
B
$$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} - 1} \over {\sqrt {1 + {x^2}} + 1}}} \right) + C$$
C
$$\sqrt {1 + {y^2}} + \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$
D
$$\sqrt {1 + {y^2}} - \sqrt {1 + {x^2}} = {1 \over 2}{\log _e}\left( {{{\sqrt {1 + {x^2}} + 1} \over {\sqrt {1 + {x^2}} - 1}}} \right) + C$$
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