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$g_{eff}=g+a$

$=g+\frac{g}{6}$

$=\frac{7g}{6}$

$T\text{'}=2\pi \sqrt[]{\frac{\mathcal{l}}{g_{eff}}}$

$T\text{'}=\sqrt[]{\frac{6}{7}}T$

According to question, we can write

  $d=\sqrt[]{2hR}\Rightarrow \frac{d_2}{d_1}=\sqrt[]{\frac{h_2}{h_1}}\Rightarrow h_2={\left(\frac{d_2}{d_1}\right)}^2{h}_1={\left(\frac{2}{1}\right)}^2\times 125=500m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

Steam point (T₁) = 100°C = 273 + 100 = 373 K

Ice point (T₂) = 0°C = 273 K

$\eta =\left(1-\frac{T_2}{T_1}\right)\times 100=\left(1-\frac{273}{373}\right)\times 100=26.81\mathrm{\%}$  

This is because the satellite doesn't have any other external third force acting upon it. Its original total energy comprises kinetic and potential energy, which remains constant since the gravity has zero influence over the satellite.

Let AB $\equiv x-2y+1=0$  

AC  $\equiv 2x-y+1=0$  

So vertex A = (1, 1)

altitude from B is perpendicular to AC and passing through

orthocentre.

So, BH = x + 2y – 7 = 0

CH = 2x + y – 7 = 0

now solve AB & BH to get B (3, 2) similarly CH and AC to get C (2, 3) so centroid is at (2, 2)

If currents are flowing in same direction, magnetic field will cancel each other, so the currents must flowing in opposite direction

$B_P=\frac{{\mu }_{0}I}{2\pi r}\times 2$

$\Rightarrow 300\times 10^{-6}=\frac{4\pi \times 10^{-7}}{2\pi \times 4\times 10^{-2}}\times 2$                                                                         

=> I = 30 A

$g\propto r0\le r\le R$

$g\propto \frac{1}{r^2}r>R$

Organic chemistry is applied in pharmaceuticals, the food industry, the manufacture of polymers, agriculture, and environmental science. Most of the things that we use in daily life, such as soaps and medicines, are made of organic molecules.

Line $\perp$  to the normal

->3p + 2q – 1 = 0

$\left(2,-1,-3\right)$ lies in the plane 2p + q = 8

From here p = 15, q = -22

Equation of plane 15x – 22y + z – 5 = 0

Distance from origin =  $\left|\frac{5}{\sqrt[]{15^2+{\left(-22\right)}^2+1^2}}\right|=\sqrt[]{\frac{5}{142}}$  

The minus sign in the formula indicates that:

• A gravitational force exists.
• The satellite cannot leave the orbit by itself.
• A minimum threshold of energy will be required to move the satellite from its existing orbit.

The main difference between organic compounds and inorganic compounds is that organic compounds contain carbon atoms from bonds with hydrogen and other elements, whereas inorganic compounds don't have carbon atoms (except carbonates and cyanides).

${\left(x-\frac{1}{\sqrt[]{2}}\right)}^2+{\left(y-\frac{1}{\sqrt[]{2}}\right)}^2=1$                                                    

here AB =  $\sqrt[]{2}$ , BC = 2, AC = $\sqrt[]{2}$

area =  $\frac{1}{2}\times \sqrt[]{2}\times \sqrt[]{2}=1$  

By conservation of Angular momentum

Li = Lf 

$M{R}^2\omega =\left(m{R}^2+2m{R}^2\right)\omega \text{'}$

$\frac{2M}{M+2m}=\omega \text{'}$

$\Rightarrow$ $\omega \text{'}=\frac{2M}{M+2m}$

Process-AB $\Rightarrow$ Isobaric,                                                           

Process-AC $\Rightarrow$ Isothermal, and

Process-ADÞAdiabatic

W₂ < W₁ < W₃

In class 11 organic chemistry, the important chapters are hydrocarbons, classification of organic compounds, isomerism, method of purification of organic compounds, and Qualitative and quantitative analysis of organic compounds.

Tangents making angle $\frac{\pi }{4}$  with y = 3x + 5.

$tan\frac{\pi }{4}=\left|\frac{m-3}{1+3m}\right|\Rightarrow m=-2,\frac{1}{2}$  

So, these tangents are  . So ASB is a focal chord.

According to Law of discharging of capacitor, we can write

$Q=Q_0{e}^{-\frac{t}{\tau }}\Rightarrow \frac{Q_2}{8}=Q_0{e}^{-\frac{t_2}{\tau }}\Rightarrow t_{2}3\tau \mathcal{l}n\left(2\right),and$

^{$U=\frac{Q^2}{2C}=\frac{Q_0^2}{2C}{e}^{-\frac{2t}{\tau }}\Rightarrow \frac{1}{2}\left(\frac{Q_0^2}{2C}\right)=\frac{Q_0^2}{2C}{e}^{-\frac{2t}{\tau }}\Rightarrow t_1=\frac{\tau }{2}\mathcal{l}n\left(2\right)$}

^{$\Rightarrow \frac{t_1}{t_2}=\frac{1}{6}$}

NLM at point Q

N – mg sin a =  $\frac{m{v}^2}{R}$  

$N=\frac{m{v}^2}{R}+mgsin\alpha$           - (1)

COE : between P & Q

$\frac{m{v}^2}{R}=2mgsin\alpha$               - (2)

Centripetal force = 2mg sin a

 N (Normal Reaction) = 2mg sin a + mg sin a

  = 3mg sin a

$\frac{Centripetalforce}{NormalReaction}=\frac{2mgsin\alpha }{3mgsin\alpha }=\frac{2}{3}$             

The light wave contains two lights of different frequencies, so

  $E_1=h{\upsilon }_1=\frac{4.14\times 10^{-15}\times 6\times 10^{15}}{2\pi }=3.96eV,and$

  $E_2=h{\upsilon }_2=\frac{4.14\times 10^{-15}\times 9\times 10^{15}}{2\pi }=5.92eV$

Maximum kinetic energy of the photoelectron = 5.9-2.5 = 3.42 eV

$\Rightarrow {\displaystyle \underset{}{\overset{}{\int }}\frac{dy}{1+y^2}=-2{\displaystyle \underset{}{\overset{}{\int }}\frac{e^{x}dx}{1+{\left(e^x\right)}^2}+C}}$  

$\Rightarrow ta{n}^{-1}y=-2\cdot ta{n}^{-1}{e}^x+C$

x = 0, y = 0

$\Rightarrow 0=2ta{n}^{-1}+C$

-> $C=+\frac{\pi }{2}$

now at x =  $\mathcal{l}n\sqrt[]{3}$  

$ta{n}^{-1}y=-2ta{n}^{-1}\left(e^{\mathcal{l}n\sqrt[]{3}}\right)+\frac{\pi }{2}$

$6\left(y\text{'}\left(0\right)+{\left(y\left(\mathcal{l}n\sqrt[]{3}\right)\right)}^2\right)=6\left(-1+\frac{1}{3}\right)=-4$