Physics

$\frac{h}{p_1}=\frac{h}{p_2}$

$\sqrt[]{2m_1{k}_1}=\sqrt[]{2m_2{k}_2}$

$\frac{k_2}{k_1}=\frac{m_1}{m_2}=1835$

S₁ = $\frac{1}{2}$ a (p – 1)²

S₂ = $\frac{1}{2}$ ap²

S₁ + S₂ = $\frac{1}{2}$ a [ (p – 1)² + p²] = $\frac{1}{2}$ at²

$t=\sqrt[]{2p^2+1-2p}$

$h=\frac{2Scos\theta }{\rho gR}$

as T­, S¯

The correct answer is Option (2).

  $\frac{GMm}{{\left(R+x\right)}^2}=\frac{GMm\left(R-x\right)}{R^3}$

->R³ = (R + x)² (R – x)

->R³ = (R²– x²) (R + x)

->x² + Rx – R² = 0

$x=\frac{-R\pm \sqrt[]{R+4R^2}}{2}$

$x=\frac{\left(\sqrt[]{5R}-1\right)}{2}R$

  $\frac{v^2}{2a}=s$ (a = mg)

v² = 2 * mgs

v² = 2 * (0.4) * 10 * 10

v² = 80

W_f = Dk

= $-\frac{1}{2}*100*80$  

W_f = – 4000

a = $\frac{3}{9}g$ , For C, 3g – T = 3a = (3) $\frac{3}{g}$ g

T = 2g = 20 N

$\frac{\sigma }{\in }=Y$

$\frac{\sigma }{Y}=\in \Rightarrow \frac{20}{5*10^{-8}*2*10^{11}}$

$=2*10^{-3}$  

$=20*10^{-4}\Rightarrow 20$  

Charge on C₁ = CV

            Charge on C₂ = 4CV

When connected in parallel

$V_C=\frac{5V}{3}$  

$Q{\text{'}}_1=\frac{5}{3}CV$ $Q{\text{'}}_2=\frac{10}{3}CV$

$?$ $E=\frac{1}{2}C{V}^2$

$E\text{'}=\frac{25}{18}C{V}^2+\frac{25}{9}C{V}^2$  

$\frac{25}{6}C{V}^2=\frac{50E}{6}$

x = 50

$-0.85=\frac{-13.6}{n^2}$  

n = 4

Number of transitions = $\frac{4*3}{2}=6$  

Current in circuit (I) = $\frac{8-2}{6}$  = 1 A.

So, V_C – 4 (1) – 2 + 8 – 2 (1) = V_A

V_C – 6 – 2 + 8 = V_A

V_C – V_A = 0 V

COAM gives Iw₀ = 2Iw

$\omega =\frac{{\omega }_0}{2}$

Loss in KE =  $\frac{1}{2}I{\omega }_0^2-\frac{1}{2}\left(2I\right){\left(\frac{{\omega }_0}{2}\right)}^2$

$=\frac{1}{2}I{\omega }_0^2$

$=\frac{1}{4}*5*\frac{2}{2}*100=250J$