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IIT-JEE 2006

MCQ (Single Correct Answer)

-0

Match the extraction processes listed in Column I with metals listed in Column II.

Column I		Column II
(A)	Self-reduction	(P)	Lead
(B)	Carbon reduction	(Q)	Silver
(C)	Complex formation and displacement by metal	(R)	Copper
(D)	Decomposition of iodide	(S)	Boron

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

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

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

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

IIT-JEE 2006

MCQ (Single Correct Answer)

-0

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

Column I		Column II
(A)	$$ \mathrm{Bi}^{3+} \rightarrow(\mathrm{BiO})^{+} $$	(P)	Heat
(B)	$$ \left[\mathrm{AlO}_2\right]^{-} \rightarrow \mathrm{Al}(\mathrm{OH})_3 $$	(Q)	Hydrolysis
(C)	$$ \mathrm{SiO}_4^{4-} \rightarrow \mathrm{Si}_2 \mathrm{O}_7^{6-} $$	(R)	Acidification
(D)	$$ \left(\mathrm{B}_4 \mathrm{O}_7^{2-}\right) \rightarrow\left[\mathrm{B}(\mathrm{OH})_3\right] $$	(S)	Dilution by water

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

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

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

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

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 (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

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

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

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

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

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