IIT-JEE 2006 | JEE Advanced Year Wise Previous Years Questions

IIT-JEE 2006

MCQ (Single Correct Answer)

-0

$$ \text { Match Column I with Column II : } $$

Column I		Column II
(A)	$\mathrm{CH}_3-\mathrm{CHBr}-\mathrm{CD}_3$ on treatment with alc. KOH gives $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CD}_3$ as a major product.	(P)	E1 reaction
(B)	$$ \begin{aligned} &\mathrm{Ph}-\mathrm{CHBr}-\mathrm{CH}_3\\ &\text { reacts faster than }\\ &\mathrm{Ph}-\mathrm{CHBr}-\mathrm{CD}_3 . \end{aligned} $$	(Q)	E2 reaction
(C)	$$ \mathrm{Ph}-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br} $$ on treatment with $$ \mathrm{C}_2 \mathrm{H}_5 \mathrm{OD} / \mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-} $$ gives $\mathrm{Ph}-\mathrm{CD}=\mathrm{CH}_2$ as the major product.	(R)	E1 cb reaction
(D)	$\mathrm{PhCH}_2, \mathrm{CH}_2 \mathrm{Br}$ and $\mathrm{PhCD}_2 \mathrm{CH}_2 \mathrm{Br}$ react with same rate.	(S)	First order reaction

$$ [\mathrm{A} \rightarrow(\mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{Q}) ; \mathrm{C} \rightarrow( \mathrm{~S}) ; \mathrm{D} \rightarrow(\mathrm{P}, \mathrm{~S})] .$$

$$ [\mathrm{A} \rightarrow(\mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{Q}) ; \mathrm{C} \rightarrow(\mathrm{R}, \mathrm{~S}) ; \mathrm{D} \rightarrow(\mathrm{P})] .$$

$$ [\mathrm{A} \rightarrow(\mathrm{Q}) ; \mathrm{B} \rightarrow(\mathrm{Q}) ; \mathrm{C} \rightarrow(\mathrm{R}, \mathrm{~S}) ; \mathrm{D} \rightarrow(\mathrm{P}, \mathrm{~S})] .$$

$$ [\mathrm{A} \rightarrow(\mathrm{Q}, P) ; \mathrm{B} \rightarrow(\mathrm{Q}) ; \mathrm{C} \rightarrow(\mathrm{R}, \mathrm{~S}) ; \mathrm{D} \rightarrow(\mathrm{P}, \mathrm{~S})] .$$

IIT-JEE 2006

MCQ (More than One Correct Answer)

-1.25

The equations of the common tangents to the parabola $$y = {x^2}$$ and $$y = - {\left( {x - 2} \right)^2}$$ is/are

$$y = 4\left( {x - 1} \right)$$

$$y=0$$

$$y = - 4\left( {x - 1} \right)$$

$$y = - 30x - 50$$

IIT-JEE 2006

MCQ (Single Correct Answer)

-0.75

The axis of a parabola is along the line $$y = x$$ and the distances of its vertex and focus from origin are $$\sqrt 2 $$ and $$2\sqrt 2 $$ respectively. If vertex and focus both lie in the first quadrant, then the equation of the parabola is

$${\left( {x + y} \right)^2} = \left( {x - y - 2} \right)$$

$${\left( {x - y} \right)^2} = \left( {x + y - 2} \right)$$

$${\left( {x - y} \right)^2} = 4\left( {x + y - 2} \right)$$

$${\left( {x - y} \right)^2} = 8\left( {x + y - 2} \right)$$

IIT-JEE 2006

MCQ (More than One Correct Answer)

-1.25

Let $${\overrightarrow A }$$ be vector parallel to line of intersection of planes $${P_1}$$ and $${P_2}.$$ Planes $${P_1}$$ is parallel to the vectors $$2\widehat j + 3\widehat k$$ and $$4\widehat j - 3\widehat k$$ and that $${P_2}$$ is parallel to $$\widehat j - \widehat k$$ and $$3\widehat i + 3\widehat j,$$ then the angle between vector $${\overrightarrow A }$$ and a given vector $$2\widehat i + \widehat j - 2\widehat k$$ is

$${\pi \over 2}$$

$${\pi \over 4}$$

$${\pi \over 6}$$

$${3\pi \over 4}$$

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