Class 12th

Candidates seeking admission to various programme can enrol for admission with Class 12 marks. The college offers various courses such as Commercial Pilot Training, Cabin Crew Training, etc. Candidates must complete Class 12 as the basic eligibility criteria for all courses. 

Amplitude is proportional to the slit – width, thus,

$\frac{A_1}{A_2}=3$            

$\frac{l_{max}}{l_{min}}=\frac{{\left(A_1+A_2\right)}^2}{{\left(A_1-A_2\right)}^2}=\frac{{\left(\frac{A_1}{A_2}+1\right)}^2}{{\left(\frac{A_1}{A_2}-1\right)}^2}=\frac{{\left(3+1\right)}^2}{{\left(3-1\right)}^2}=4.$              

$\alpha =\frac{\Delta l_C}{\Delta l_E}=\frac{3.5}{4}=\frac{7}{8}$  

$\beta =\frac{\alpha }{1-\alpha }=\frac{7/8}{1-7/8}=7$            

Focus of a spherical convex mirror is in the same side of centre of curvature. Thus, f = + $\frac{1}{2}r.$

If charge (-Q) at origin is replaced by (+Q), then electric field at the centre of the cube is zero. Thus, electric field at the centre of the cube is as if only (-2Q) charge is present at the origin.

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{\left(-2Q\right)}{{\left(\frac{\sqrt[]{3}}{2}a\right)}^2}.\frac{1}{\sqrt[]{3}}\left(\widehat{x}+\widehat{y}+\widehat{z}\right)=\frac{|-2Q}{3\sqrt[]{3}\pi {\epsilon }_0{a}^2}\left(\widehat{x}+\widehat{y}+\widehat{z}\right)$

Moles of SO₂ = $\frac{224*10^{-3}}{22.4}$

=0.01 mole

Moles of NaOH = 0.1 * 0.1

= 0.01 mole

SO₂+NaOHNaHSO₃

0.01 mole0.01 mole-

-0.01 mole

Non-volatile solute is NaHSO3

Moles of water = $\frac{36}{18}=2$

Using ; relative lowering in V.P

$\frac{P^o-P}{P^o}=i{x}_B$

Where; $\Delta P=P^0-P$ is lowering in V.P

$\Delta P=P^{0}i{x}_B$

i for NaHSO₃ = 2

here; $x_B=\frac{n_B}{n_A}$ since solution is dilute

$\Delta P=24*2*\frac{0.01}{2}=0.24$

$\Delta P=24*10^{-2}mmHg$

So; x = 24