In Bohr’s hydrogen atom, the electronic transition emitting light of longest wavelength is:

P1 = 1;
P2 = $1 half$ ;
P3 = $fraction numerator 1 over denominator square root of 7 end fraction$ ;
>>> Therefore, we can say that P1>P2>P3.

If θ is angle between pair of lines $x squared minus 3 x y plus lambda y squared plus 3 x minus 5 y plus 2 equals 0$ , then $cosec squared invisible function application theta equals$

>>> $lambda$ = 2.
>>> tan $theta$ = $1 third$
>>> $cos e c squared theta$ = 10.

If θ is angle between pair of lines $x squared minus 3 x y plus lambda y squared plus 3 x minus 5 y plus 2 equals 0$ , then $cosec squared invisible function application theta equals$

>>> $lambda$ = 2.
>>> tan $theta$ = $1 third$
>>> $cos e c squared theta$ = 10.

If the pair of lines $a x squared plus 2 h x y plus b y squared plus 2 g x plus 2 f y plus c equals 0$ intersect on the x-axis, then 2fgh=

2 f g h = b g 2 + c h 2

If the pair of lines $a x squared plus 2 h x y plus b y squared plus 2 g x plus 2 f y plus c equals 0$ intersect on the x-axis, then 2fgh=

2 f g h = b g 2 + c h 2

If the pair of lines $a x squared plus 2 h x y plus b y squared plus 2 g x plus 2 f y plus c equals 0$ intersect on the x-axis, then ac=

$g squared equals a c$

If the pair of lines $a x squared plus 2 h x y plus b y squared plus 2 g x plus 2 f y plus c equals 0$ intersect on the x-axis, then ac=

$g squared equals a c$

maths-

If the equation $x squared plus p y squared plus q y equals a squared$ represents a pair of perpendicular lines then its point of intersection is

maths-General

If the lines $x squared plus 2 x y minus 35 y squared minus 4 x plus 44 y minus 12 equals 0$ and $5 x plus lambda y minus 8 equals 0$ are concurrent then λ

>>> The value of $lambda$ is 2.

If the lines $x squared plus 2 x y minus 35 y squared minus 4 x plus 44 y minus 12 equals 0$ and $5 x plus lambda y minus 8 equals 0$ are concurrent then λ

>>> The value of $lambda$ is 2.

The equation of the line concurrent with the pair of lines $3 x y minus 2 y squared minus 7 x plus y plus 3 equals 0$ is

Hence, x=y is the the line that is concurrent with the pair of straight lines.

The equation of the line concurrent with the pair of lines $3 x y minus 2 y squared minus 7 x plus y plus 3 equals 0$ is

Hence, x=y is the the line that is concurrent with the pair of straight lines.

If the equation $lambda x squared minus 5 x y plus 6 y squared plus x minus 3 y equals 0$ represents a pair of straight lines then their point of intersection is

>>>The point of intersection of the pair of straight lines x² – 5xy + 6y² + x – 3y = 0 is (-3, -1)

If the equation $lambda x squared minus 5 x y plus 6 y squared plus x minus 3 y equals 0$ represents a pair of straight lines then their point of intersection is

>>>The point of intersection of the pair of straight lines x² – 5xy + 6y² + x – 3y = 0 is (-3, -1)

maths-

The point of intersection of the perpendicular lines $a x squared plus 3 x y minus 2 y squared minus 5 x plus 5 y plus c equals 0$ is

maths-General

chemistry-

The transformed equation of $x squared plus y squared equals r squared$ when the axes are rotated through an angle 36° is

>>> Given equation is $x 2 + y 2 = r 2 .$ After rotation

$>>> x = X cos 3 6^{\circ} - Y sin 3 6^{\circ}$ and $y = X sin 3 6^{o} + Y cos 3 6^{\circ}$

$∴ X 2 (cos^{2} 3 6^{o} + sin^{2} 3 6^{o}) + Y 2 (sin^{2} 3 6^{o} + cos^{2} 3 6^{o}) = r 2$

$>>> \Rightarrow X 2 + Y 2 = r 2$

The transformed equation of $x squared plus y squared equals r squared$ when the axes are rotated through an angle 36° is

>>> Given equation is $x 2 + y 2 = r 2 .$ After rotation

$>>> x = X cos 3 6^{\circ} - Y sin 3 6^{\circ}$ and $y = X sin 3 6^{o} + Y cos 3 6^{\circ}$

$∴ X 2 (cos^{2} 3 6^{o} + sin^{2} 3 6^{o}) + Y 2 (sin^{2} 3 6^{o} + cos^{2} 3 6^{o}) = r 2$

$>>> \Rightarrow X 2 + Y 2 = r 2$