Class 11th

Boiling point: $\stackrel{\left(239.7\right)}{N{H}_3}>\stackrel{\left(185.5\right)}{P{H}_3}$  due to hydrogen bonding in NH₃.

CuSO₄ acts as catalyst in Kjeldahl's method of estimation of nitrogen.

Kindly go though the solution 

$v=\sqrt[]{\frac{\gamma RT}{M}}=330m/s$

$I=\left(\frac{2}{5}M{R}^2+M{R}^2\right)*2$

$=\frac{14}{5}*2*\frac{9}{16}$

$=\frac{63}{20}$          

= 3.15 kg m²

[A] = L²

$B=\frac{x^2}{tE}\equiv \frac{L^2}{TM{L}^2{T}^{-2}}=\frac{1}{M{T}^{-1}}$            

[B] = M^–1T

[AB] = [M^–1L²T]

$P=\overrightarrow{F}\cdot \overrightarrow{v}$

= 18t⁴ – 18t²

->P (t = 2) =18 [16 − 4] = 216 W

f_K = μmg cosq

= 0.1 * $\frac{50*\sqrt[]{3}}{2}$

$=2.5\sqrt[]{3}N$

 F₁    = mg sinq + f_K

= 25 + 2.5 $\sqrt[]{3}$  

F₂    = mg sinq – f_K

= 25 – 2.5 

F₁ – F₂ = 5 $\sqrt[]{3}$  N

After giving 2 apples to each child 15 apples left now 15 apples can be distributed in
^15+3–1C₂ = ¹⁷C₂ ways

  $=\frac{17*16}{2}=136$          

${\displaystyle \underset{0}{\overset{\pi }{\int }}\frac{x^{2}sinx\cdot cosx}{si{n}^{4}x+co{s}^{4}x}dx}$

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}\left(x^2-{\left(\pi -x\right)}^2\right)dx}$

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinx\cdot cosx\left(2\pi x-{\pi }^2\right)}{si{n}^{4}x+co{s}^{4}x}x}$

$=2\pi \cdot \frac{\pi }{4}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx}-{\pi }^2{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sinxcosx}{si{n}^{4}x+co{s}^{4}x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\frac{1}{2}sin2x}{1-\frac{1}{2}si{n}^{2}2x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sin2x}{2-si{n}^{2}2x}dx}$

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{sin2x}{1+co{s}^{2}2x}dx}$

Let cot2x = t

$=-\frac{{\pi }^2}{2}{\displaystyle \underset{1}{\overset{-1}{\int }}\frac{-\frac{1}{2}dt}{1+t^2}}$

$=-\frac{{\pi }^2}{4}{\displaystyle \underset{-1}{\overset{1}{\int }}\frac{dt}{1+t^2}}$

$=-\frac{{\pi }^2}{4}\cdot \frac{\pi }{2}=-\frac{{\pi }^2}{8}$

$\frac{120}{{\pi }^3}\left|-\frac{{\pi }^3}{8}\right|=15$