IIT JEE 1973 Chemistry Question Paper with Answer and Solution

2 QuestionsEnglishWith Solutions

ChemistryQ12 of 2 questions

Page 1 of 1 · English

1
ChemistryMCQIIT JEE · 1973
$0.4\overline{23} = .......$
A
$\frac{419}{990}$
B
$\frac{419}{999}$
C
$\frac{417}{990}$
D
$\frac{417}{999}$

Solution

(A) Let $x = 0.4\overline{23} = 0.4232323...$
Multiply by $10$: $10x = 4.232323...$
Multiply by $1000$: $1000x = 423.232323...$
Subtract the two equations: $1000x - 10x = 423.232323... - 4.232323...$
$990x = 419$
$x = \frac{419}{990}$
2
ChemistryMCQIIT JEE · 1973
The circles $x^2 + y^2 - 2x - 4y = 0$ and $x^2 + y^2 - 8y - 4 = 0$:
A
Touch each other internally.
B
Intersect each other at two points.
C
Touch each other externally.
D
None of these.

Solution

(A) For the circle $x^2 + y^2 + 2gx + 2fy + c = 0$,the center is $(-g, -f)$ and the radius is $r = \sqrt{g^2 + f^2 - c}$.
For the first circle $C_1: x^2 + y^2 - 2x - 4y = 0$:
Center $C_1 = (1, 2)$,Radius $r_1 = \sqrt{1^2 + 2^2 - 0} = \sqrt{5}$.
For the second circle $C_2: x^2 + y^2 - 8y - 4 = 0$:
Center $C_2 = (0, 4)$,Radius $r_2 = \sqrt{0^2 + 4^2 - (-4)} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$.
The distance between the centers $d = \sqrt{(1 - 0)^2 + (2 - 4)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.
We observe that $|r_2 - r_1| = |2\sqrt{5} - \sqrt{5}| = \sqrt{5}$.
Since the distance between the centers $d = |r_2 - r_1|$,the circles touch each other internally.

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