Chemistry-
General
Easy
Question
The electronic configuration of element when just above the element with atomic number 43 the same periodic group is:
- 1s2 2s2 2p6 3s2 3p6 3d10 4s1 4p6
- 1s2 2s2 2p6 3s2 3p6 3d5 4s1
- 1s2 2s22p63s23p63d54s2
- 1s22s22p63s23p63d104s24p6
The correct answer is: 1s2 2s22p63s23p63d54s2
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Let AB be the chord of length , O be the centre of the circle and let OC be the perpendicular from O on AB. Then AC = BC = 
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Hence (c) is the correct answer.
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Let AB be the chord of length , O be the centre of the circle and let OC be the perpendicular from O on AB. Then AC = BC =
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Hence (c) is the correct answer.
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\ Area of the circle = p (OB)2 = p.
Hence (c) is the correct answer.
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Equation of a circle with centre (4, 3) touching the circle 
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Let the circle touching the circle x2 + y2 = 1, be x2 + y2 – 8x – 6y + k = 0,
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The equation of the common tangent is S1 – S2 = 0
i.e. 8x + 6y–1–k = 0
This is a tangent to the circle x2+y2 = 1.
Hence ± 1 =
Þ k+1 = ± 10 Þ k = –11 or 9
Therefore the circles are x2+y2–8x–6y+9 = 0 and x2+y2–8x–6y–11 = 0
Hence (d) are the correct answers.
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i.e. 8x + 6y–1–k = 0
This is a tangent to the circle x2+y2 = 1.
Hence ± 1 =
Þ k+1 = ± 10 Þ k = –11 or 9
Therefore the circles are x2+y2–8x–6y+9 = 0 and x2+y2–8x–6y–11 = 0
Hence (d) are the correct answers.
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