Physics

After switch 'S' is closed

$Q_1+Q_2=C_{1}V$ -    (1)

              Using KVL

              $\frac{Q_1}{C_1}-\frac{Q_2}{C_2}=0$  

              $Q_1=Q_2\frac{C_1}{C_2}$                       - (2)

              from (1) & (2)

              $\Rightarrow Q_2\left[\frac{C_1+C_2}{C_2}\right]=C_{1}V\Rightarrow Q_2=\left(\frac{C_1{C}_2}{C_1+C_2}\right)V$

According to Lorentz's Force, we can write

  $\overrightarrow{F}={\overrightarrow{F}}_{Electric}+{\overrightarrow{F}}_{Magnetic}=q\overrightarrow{E}+q\left(\overrightarrow{v}*\overrightarrow{B}\right),so$

Statement I is correct but Statement II is incorrect

At each point in the orbit, there is a variation in the kinetic and potential energy of the satellite. If one increases, the other one decreases. Similarly, if the other one increases, the first one decreases. But their amount of increment or decrement is such that the total sum i.e. mechanical energy of the satellite remains the same. This in turn allows free movement.

According to question, we can write

Total moles of gas = n = n_Oxygen + n_Oxygen = $\frac{16}{2}+\frac{128}{32}=12moles$  

Volume of gas = 12 * 22.4 litre = 268.8 litre = 2.688 * 10⁵ cm³ $\approx 27*10^{4}c{m}^3$  

$KE=\Delta U$

$\Delta U=\frac{1}{2}m{v}^2$

$n{C}_v\Delta T=\frac{m{v}^2}{2}$

$\Rightarrow \frac{m}{M}\frac{R}{y-1}\Delta T=\frac{m{v}^2}{2}$

$\Delta T=\frac{0.4}{2}\frac{M{v}^2}{R}$

$\Delta T=\frac{M{v}^2}{5R}$

According to question, we can write

$\omega =\pi =\sqrt[]{\frac{g}{\mathcal{l}}}\Rightarrow \mathcal{l}=\frac{g}{{\pi }^2}=\frac{9.8}{{\left(3.14\right)}^2}=0.99395=99.4cm$

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.

$g_{eff}=g+a$

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

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

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

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

According to question, we can write

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

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

According to question, we can write

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

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