Class 11th

$b=a\sqrt[]{1-e^2}$

$\frac{dA}{dt}=\frac{L}{2m}$

$\frac{A}{T}=\frac{L}{2m}$

$T=\frac{2mA}{L}=\frac{2m\pi ab}{L}=\frac{2m\pi a^2\sqrt[]{1-e^2}}{L}$

x = t² – t + 1        … (1)

y = t² + t + 1    … (2)

y – x = 2t & x + y = 2 (t² + 1)

__________on eliminating 't' we get

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

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

Axis : x – y = 0

Tangent at vertex : x + y – 2 = 0

Vertex : (1, 1) = (x, y)

Angular retardation,

In disproportionation reaction, one element of a compound will simultaneously get reduced and oxidised. In ClO₄ -, oxidation number of Cl is +7 and it can not increase it further. So, ClO₄- will not get oxidised and so will not undergo disporportionation reaction.

Let S be the distance between Two ends a be the constant acceleration

As we know $v^2-u^2=2aS$

or, $aS=\frac{v^2-u^2}{2}$

Let $v_c$  be velocity at mid point.

Therefore, $v_c^2-u^2=2a\frac{S}{2}$

$v_c^2=u^2+aS$

$v_c^2=u^2+\frac{v^2-u^2}{2}$

$v_c=\sqrt[]{\frac{u^2+v^2}{2}}$

P? V indicator diagram for isobaric

P? V indicator diagram for isochoric process

P? V indicator diagram for isothermal process

$\frac{3z_1+i}{4}=\frac{2z_2+2z_3}{4}$

P = point of intersection of AD & BC

$AD=1\Rightarrow DP=\frac{3}{4}$

$BP=\sqrt[]{1-\frac{9}{16}}=\frac{\sqrt[]{7}}{4}\Rightarrow BC=\frac{\sqrt[]{7}}{2}$

area of Quad. ABCD = $\frac{1}{2}.AD*BC$

= $\frac{1}{2}*1*\frac{\sqrt[]{7}}{2}=\frac{\sqrt[]{7}}{4}$

Correct order is Be

A = {1, 2, 3, 4}. As (4, 4) $\notin$  R so not reflexive

Also not transitive