Class 11th

Given sequence is -16, 8, -4, 2, .

are in G.P. with first term a = -16 & common ratio r =   $\frac{-1}{2}$

Now $t_p=a{r}^{P-1}=-16{\left(-\frac{1}{2}\right)}^{p-1}\&t_q=a{r}^{q-1}=-16{\left(-\frac{1}{2}\right)}^{q-1}$  

So A.M. = -8 $\left[{\left(-\frac{1}{2}\right)}^{p-1}+{\left(\frac{-1}{2}\right)}^{q-1}\right]=\alpha \left(let\right)$

Given equation is 4x² – 9x + 5 = 0 gives $x=1,\frac{5}{4}$  

From roots we get possible value of b = 1 so

$16{\left(-\frac{1}{2}\right)}^{\frac{P+q-2}{2}}=1OR{\left(-\frac{1}{2}\right)}^{\frac{p+q-2}{2}}=\frac{1}{16}-{\left(-\frac{1}{2}\right)}^4$

$\Rightarrow \frac{p+q-2}{2}=4\Rightarrow p+q=10$       

Blood – Negatively charged colloid.

According to Hardy – Schulze Rule

FeCl₃ is more efficient for blood clotting

Density = $\frac{mass}{volume}$

Cu > Co > Fe > Cr > Zn

Let v is velocity at the highest position.

$T_{max}=5T_{min}\Rightarrow mg+\frac{m\left(v^2+4g\mathcal{l}\right)}{\mathcal{l}}=5\left(\frac{m{v}^2}{\mathcal{l}}-mg\right)\Rightarrow 4.\frac{v^2}{\mathcal{l}}=10g$

$\Rightarrow v=\sqrt[]{\frac{5}{2}g\mathcal{l}}=\sqrt[]{\frac{5}{2}*10*1}=5m/s$

$\Delta U+\Delta KE=0$

$\Rightarrow \frac{f}{2}nR\Delta T=\frac{1}{2}m{v}^2\Rightarrow \Delta T=\frac{m}{n}\frac{v^2}{fR}=4*10^{-3}*\frac{30^2}{3R}=\frac{3.6}{3R}K.$

For equilibrium,

$\frac{dU}{dr}=0\Rightarrow \frac{-10\alpha }{r^{11}}+\frac{5\beta }{r^6}=0\Rightarrow r={\left(\frac{2\alpha }{\beta }\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$P{A}^2+P{B}^2=co{s}^2\theta +{\left(sin\theta -3\right)}^2+co{s}^2\theta +{\left(sin\theta +6\right)}^2$       

= 2 cos² θ + 2 sin² θ + 6 sin θ + 45

= 6 sin θ + 47

for maximum of PA² + PB², sin q = 1

then P (1, 2)

Hence P, A & B will lie on a straight line.

$T=27°C,P=1atm=10^{5}N/m^2,V_{rms}=200m/s$

As we know

$v_{rms}=\sqrt[]{\frac{3RT}{M}}\Rightarrow \frac{v_{rms}}{v{\text{'}}_{rms}}=\frac{\sqrt[]{\frac{3RT}{M}}}{\sqrt[]{\frac{3RT\text{'}}{M}}}$

$V{\text{'}}_{rms}=200*\frac{2}{\sqrt[]{3}}=\frac{x}{\sqrt[]{3}}$

x = 400

Assume the length of train be I and its acceleration be a.

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

Velocity when middle point crosses the post,

$V_m=\sqrt[]{u^2+2a\frac{\mathcal{l}}{2}}.=\sqrt[]{u^2+\frac{v^2-u^2}{2}}=\sqrt[]{\frac{u^2+v^2}{2}}$

$T=2\pi \sqrt[]{\frac{r^3}{G{M}_e}}$

$T_B-T_A=\frac{2\pi }{\sqrt[]{G{M}_e}}\left(r_B^{3/2}-r_A^{3/2}\right)=\frac{2*3.14}{\sqrt[]{6.67*10^{-11}*6*10^{24}}}\left[{\left(8*10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}-{\left(7*10^6\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$