Class 11th

Surfaces that are hard, smooth, and non-porous are sound reflectors. Common examples of such materials include concrete, metal, glass, plaster, and marble. In contrast, soft and porous materials tend to absorb sound waves. In short, both porous and non-porous reflect sound waves, but the effect of reflection is lesser with porous materials.

A reflection of sound that you hear with a delay after the original sound is an echo. But when there are multiple reflections, they build up. They further blend with the source sound. That phenomenon is reverberation.

Reflection of sound occurs when sound waves bounce off a boundary-like surface. It's pretty similar to how we observe a ball that bounces off a wall after it hits. It practically follows the law of reflection. This law states that the angle of the wave's approach equals the angle of its departure. 

Refraction of sound is similar to what you know about refraction of light. It's the bending of sound waves when passing from one medium to another. Variables, such as different temperature or density, leads to the change in the speed of sound.

$\frac{{\left(x+1\right)}^2}{\lambda +5}=\frac{{\left(y+1\right)}^2}{\frac{\lambda +5}{4}}=1$  

length of latus rectum = $\frac{2b^2}{a}=\frac{2\left(\frac{\lambda +5}{4}\right)}{\sqrt[]{5+\lambda }}=4$  

$\sqrt[]{\lambda +5}=8\Rightarrow \lambda =59$                

Major axis = $2\sqrt[]{\lambda +5}=16$  

$e_H=\sqrt[]{1+\frac{64}{49}}=\frac{\sqrt[]{113}}{7}$

$\Rightarrow e_H.e_E=\frac{1}{2}$

$\Rightarrow \frac{113}{49}.\frac{\left(64-a^2\right)}{64}=\frac{1}{4}\Rightarrow a^2-64=\frac{32^2}{113}$

$\mathcal{l}=\frac{2a^2}{b}=2\left(64+\frac{32^2}{113}\right).\frac{1}{8}$

$113\mathcal{l}=1552$

The central atom of nitrogen has 5 valence electrons as per the electronic configuraion. During the formation of ${\mathrm{N}\mathrm{H}}_3$ , 3 valence electrons forms three sigma bonds with hydrogen and one lone electron pair is left.

• The steric number  of the ammonia molecule: SN=3 bonds +1 lone pair $=4$ total electron domains.
• As per the steric number, there is ${\mathrm{s}\mathrm{p}}^3$ hybridisation in ammonia.
• The lone pair causes repulsion, which leads to a trigonal pyramidal geometry with bond angles of ${107}^{?}$ .

$\overline{x}=\frac{{\displaystyle \sum _{i=1}^{10}{x}_i}}{10}=15;\frac{{\displaystyle \sum _{i=1}^{10}{x_i}^2}}{10}-{\left(\overline{x}\right)}^2=15$

$\Rightarrow \Sigma x_i=150;\Sigma {x_i}^2=2400$

Actual mean $\overline{x}=\frac{\Sigma x_i+15-25}{10}=\frac{140}{10}=14$

Actual variance =  $\frac{\Sigma {x_i}^2+15^2-25^2}{10}-{\left(14\right)}^2$

$=\frac{2400-400}{10}-196$

$\begin{array}{l}{\sigma }^2=4\\ \sigma =2\end{array}$

Circle passes through (6, 1)

12 g – 19 c = 43               …. (i)

Centre lies on x – 2xy = 8

->g + 6c = 8                     …. (ii)

From (i) & (ii), c = 1, 9 = 2

Length of x – intercept -  $2\sqrt[]{g^2-C}$

$\frac{{\Delta }_1}{{\Delta }_2}=\frac{\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill x\hfill & \hfill -4x-x\hfill & \hfill 1\hfill \\ \hfill -4\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right|}{\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill -4\hfill & \hfill 3\hfill & \hfill 1\hfill \\ \hfill -2\hfill & \hfill -5\hfill & \hfill 1\hfill \end{array}\right|}=\frac{4}{7}$

->14 x – 35 y = -95        …. (ii)

Solve (i) & (ii), x =    $\frac{-20}{7},y=\frac{-11}{7}$

$ar\Delta AQR$

$=\frac{1}{2}*1*1=\frac{1}{2}$                

Solve tan 2a = $\frac{h}{b}$  

$tan\alpha =\frac{2h}{\sqrt[]{7}h+b}$