Class 11th

If you look closely at the formula of kinetic energy (1/2*m*v^2), the velocity is squared which automatically gives a positive integer. And mass of the body can never be a negative value, which leads to the result being a positive integer.

The 1/2 is a result of mathematical calculation, which occurs when we integrate? vdv in the formula of work done according to Newton's second law of motion. Without this, the final result will turn out to be twice of the actual value.

$ForMedian=L+\frac{\frac{N}{2}-c,}{f}xh$

$14=12+\frac{\frac{a+b}{2}+13-\left(a+b\right)}{12}*6$              

->a + b = 18                                   -(ii)

Solving (i) & (ii) a = 8 & b = 10

->(a – b)² = 4

${\left(2x^r+\frac{1}{x^2}\right)}^{10}$

Let the constant term is (k + 1)^th term.

${\therefore }^{10}{C}_k{\left(2x^r\right)}^{10-k}.x^{-2k}$              

${=}^{10}{C}_k{2}^{10-k}.x^{10r-rk-2k}$              

              For constant term 10r – rk – 2k = 0.

$\therefore k=\frac{10r}{r+2}k\in I$              

$\therefore r=3,k=6andr=8,k=8$               

For k = 6

For k = 8  

$\therefore r=8$

1^st position can be filled in 4 ways as zero cannot appear in 1^st position.

2^nd position can be filled in 4 ways and so on.

Total cases = 4 * 4 * 3 * 2 * 1 = 96

  $11^n-9^n>10^n$             

$\Rightarrow {\left(10+1\right)}^n-{\left(10-1\right)}^n>10^n$           

->For  $n\ge 5$ the inequality Satisfies.

For n= 4.             2 [4 * 10³ + 4 * 10] > 10⁴

->8 * 1010 > 10⁴ which is a contradiction

$\therefore n\ge 5$ all values satisfies. Hence possible values of n is 96

Cross product will represent a vector perpendicular to both the vector.

Dir. Of Motion of P : $=\left(\widehat{i}+\widehat{j}\right)*\left(\widehat{j}*\widehat{k}\right)$ $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$

= $\widehat{k}-\widehat{j}+\widehat{i}$ $\stackrel{}{}\stackrel{}{}$

Dir. Of Motion of Q = $\left(\widehat{i}+\widehat{j}\right)*\left(-\widehat{i}+\widehat{j}\right)$ $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$

$=2\widehat{k}$

$cos\theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{ab}=\frac{2}{2\sqrt[]{3}}=\frac{1}{\sqrt[]{3}}\Rightarrow \theta =co{s}^{-1}\left(\frac{1}{\sqrt[]{3}}\right)$

Energy density (u) = $\frac{1}{2}*\sigma *\epsilon =\frac{1}{2}*Y\epsilon *\epsilon =\frac{1}{2}{\epsilon }^{2}Y$ $\frac{}{}\frac{}{}\frac{}{}{}^{}$

$\Rightarrow energystored/m^2=\frac{1}{2}{\epsilon }^{2}YA$

Strain $\left(\epsilon \right)=\frac{\Delta \mathcal{l}}{{\mathcal{l}}_0}=\alpha \Delta T=10^{-5}*10=10^{-4}$ $\frac{}{{}_{}}{}^{}{}^{}$

$\Rightarrow energystored/m^2=\frac{1}{2}*10^{-8}*10^{11}*10^{-2}=5$