Class 11th

Acceleration due to gravity at r distance above the surface = $\frac{GM}{{\left(R+r\right)}^2}$  

Acceleration due to gravity at r distance below the surface = $\frac{GM}{R^3}\left(R-r\right)$

So, ratio = $\frac{\left(R-r\right){\left(R+r\right)}^2}{R^3}=\frac{\left(R-r\right)\left(R^2+r^2+2Rr\right)}{R^3}=1+\frac{r}{R}-\frac{r^2}{R^2}-\frac{r^3}{R^3}$

Equation of plane is 3x – 2y + 4z – 7 + $\lambda$ (x + 5y – 2z + 9) = 0. (i)

It passes through (1, 4, -3) and we get $\lambda =\frac{2}{3}$

$\therefore$ from (i) we get 11x + 4y + 8z – 3 = 0 Þ -11x – 4y – 8z + 3 = 0

$\therefore \alpha +\beta +\lambda =-11-4-8=-23$  

Using parallel axis theorem:

$l=2*\left[l_{cm}+m{d}^2\right]$  

$l=2*\left[\frac{2}{5}\left(1.5\right){\left(0.5\right)}^2+\left(1.5\right){\left(2.5\right)}^2\right]$

l = 19.05 kgm²

Kindly consider following image

PV = nRT             PV = constant (at constant T)

Pressure increases & volume decreases, PV remains constant at constant T.

Since A (sec θ, 2 tanθ) & B $\left(sec?,2tan?\right)$  lies on 2x² – y² = 2 then

2sec² θ - 4tan^2_θ = 2 or sec² θ - 2 tan² θ = 1

-> $ta{n}^2\theta =0so\theta =0$

Similarly    $?=0but\theta +?=\frac{\pi }{2}$ (given) so not possible

Hence question is not correct

Let $l={\displaystyle \underset{6}{\overset{16}{\int }}\frac{lo{g}_e{x}^2}{lo{g}_e{x}^2+lo{g}_e\left(x^2-44x+484\right)}dx..........\left(i\right)}$

By property ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}$

$\left(i\right)\Rightarrow l={\displaystyle \underset{6}{\overset{16}{\int }}\frac{lo{g}_e{\left(22-x\right)}^2}{lo{g}_e{\left(22-x\right)}^2+lo{g}_e{x}^2}dx}..........\left(ii\right)$      

(i) + (ii) 2l = ${\displaystyle \underset{6}{\overset{16}{\int }}1dx=10}$

l = 5

Kindly consider the following figure

$3*7^{22}+2*10^{22}-44=3*{\left(1+6\right)}^{22}+2{\left(1+9\right)}^{22}-44$

$=3\left[1{+}^{22}{C}_1*6{+}^{22}{C}_2*6^2{+}^{22}{C}_3*6^3+....{.}^{22}{C}_{22}{6}^{22}\right]$                

= -39 on division by 18

= (-54 + 15) on division by 18 = 15

 Slope of C₁C₂ = $\frac{3}{4}=tan\theta$  

By parametric form C₁ (1 + 5 cosθ, 2 + 5sinθ)

& C₂ (1 – 5 cos θ, 2 – 5 sinθ)

C_{1 $\left(1+5*\frac{4}{5}.2+5*\frac{3}{5}\right)\&C_2\left(1-5*\frac{4}{5}.2-5*\frac{3}{5}\right)$}

So, $\left|\left(\alpha +\beta \right)\left(r+\delta \right)\right|=40$