Chemistry-
n, l and m values of an electron in 3pyorbital are:
Chemistry-General
- n=3;l=1andm=–1
- None of these
- n=3;l=1 and m=1
- Bothe1and2arecorrect
Answer:The correct answer is: Bothe1and2arecorrect
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If l=3 then type and number of orbital is:
If l=3 then type and number of orbital is:
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Area of the circle in which a chord of length 
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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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Hence (c) is the correct answer.
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Let the circle touching the circle x2 + y2 = 1, be x2 + y2 – 8x – 6y + k = 0,
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
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Hence (d) are the correct answers.
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 =
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Therefore the circles are x2+y2–8x–6y+9 = 0 and x2+y2–8x–6y–11 = 0
Hence (d) are the correct answers.
Equation of a circle with centre (4, 3) touching the circle is
maths-General
Let the circle touching the circle x2 + y2 = 1, be x2 + y2 – 8x – 6y + k = 0,
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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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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