1
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The major product of the following reaction is :
JEE Main 2019 (Online) 9th April Morning Slot Chemistry - Hydrocarbons Question 86 English
A
CH3CD2CH(Cl)(I)
B
CH3CD(Cl)CHD(I)
C
CH3C(I)(Cl)CHD2
D
CH3CD(I)CHD(Cl)
2
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
Match the catalysts (Column I) with products (Column II).

Column I Column II
(A) V2O5 (i) Polyethylene
(B) TiCl4/Al(Me)3 (ii) ethanal
(C) PdCl2 (iii) H2SO4
(D) Iron Oxide (iv) NH3
A
(A)-(iii); (B)-(i); (C)-(ii); (D)-(iv)
B
(A)-(ii); (B)-(iii); (C)-(i); (D)-(iv)
C
(A)-(iii); (B)-(iv); (C)-(i); (D)-(ii)
D
(A)-(iv); (B)-(iii); (C)-(ii); (D)-(i)
3
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Let $$\overrightarrow \alpha = 3\widehat i + \widehat j$$ and $$\overrightarrow \beta = 2\widehat i - \widehat j + 3 \widehat k$$ . If $$\overrightarrow \beta = {\overrightarrow \beta _1} - \overrightarrow {{\beta _2}} $$, where $${\overrightarrow \beta _1}$$ is parallel to $$\overrightarrow \alpha $$ and $$\overrightarrow {{\beta _2}} $$ is perpendicular to $$\overrightarrow \alpha $$ , then $${\overrightarrow \beta _1} \times \overrightarrow {{\beta _2}} $$ is equal to
A
$$ 3\widehat i - 9\widehat j - 5\widehat k$$
B
$${1 \over 2}$$($$ - 3\widehat i + 9\widehat j + 5\widehat k$$)
C
$$ - 3\widehat i + 9\widehat j + 5\widehat k$$
D
$${1 \over 2}$$($$ 3\widehat i - 9\widehat j + 5\widehat k$$)
4
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Out of Syllabus
Change Language
For any two statements p and q, the negation of the expression
p $$ \vee $$ (~p $$ \wedge $$ q) is :
A
p$$ \leftrightarrow $$q
B
~p$$ \wedge $$~q
C
p$$ \wedge $$q
D
~p$$ \vee $$~q
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