1
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
For any given series of spectral lines of atomic hydrogen, let $$\Delta \mathop v\limits^\_ = $$ $$\Delta {\overline v _{\max }} - \Delta {\overline v _{\min }}$$ be the difference in maximum and minimum frequencies in cm–1. The ratio Lyman Balmer $${{\Delta {{\overline v }_{Lyman}}} \over {\Delta {{\overline v }_{Balmer}}}}$$ is :
A
9 : 4
B
4 : 1
C
27 : 5
D
5 : 4
2
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Aniline dissolved in dilute HCl is reacted with sodium nitrite at 0ºC. This solution was added dropwise to a solution containing equimolar mixture of aniline and phenol in dil. HCl. The structure of the major product is
A
JEE Main 2019 (Online) 9th April Morning Slot Chemistry - Alcohols, Phenols and Ethers Question 121 English Option 1
B
JEE Main 2019 (Online) 9th April Morning Slot Chemistry - Alcohols, Phenols and Ethers Question 121 English Option 2
C
JEE Main 2019 (Online) 9th April Morning Slot Chemistry - Alcohols, Phenols and Ethers Question 121 English Option 3
D
JEE Main 2019 (Online) 9th April Morning Slot Chemistry - Alcohols, Phenols and Ethers Question 121 English Option 4
3
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Liquid 'M' and liquid 'N' form an ideal solution. The vapour pressures of pure liquids 'M' and 'N' are 450 and 700 mmHg, respectively, at the same temperature. Then correct statement is:
(xM = Mole fraction of 'M' in solution ;
xN = Mole fraction of 'N' in solution ;
yM = Mole fraction of 'M' in vapour phase ;
yN = Mole fraction of 'N' in vapour phase)
A
$${{{x_M}} \over {{x_N}}} < {{{y_N}} \over {{y_N}}}$$
B
(xM – yM) < (xN – yN)
C
$${{{x_M}} \over {{x_N}}} = {{{y_N}} \over {{y_N}}}$$
D
$${{{x_M}} \over {{x_N}}} > {{{y_M}} \over {{y_N}}}$$
4
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let p, q $$ \in $$ R. If 2 - $$\sqrt 3$$ is a root of the quadratic equation, x2 + px + q = 0, then :
A
p2 – 4q – 12 = 0
B
q2 – 4p – 16 = 0
C
q2 + 4p + 14 = 0
D
p2 – 4q + 12 = 0
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