Physics

For sphere 'C' after contacting with 'A'. q_A = q_C = $\frac{q}{2}.$

offer contacting with 'B'. q_B = q_C = $\frac{3q}{4}$

$F_{Net}=\left|F_1-F_2\right|$

F₂ = $\frac{K\cdot 3q^2*4}{r^{2}8}=\frac{3}{2}\frac{k{q}^2}{r^2}$

$F_1=\frac{9k{q}^2*4}{16r^2}=\frac{9k{q}^2}{4r^2}=\frac{9}{4}F$

$\therefore F_{Net}=\left|\frac{9}{4}F-\frac{3}{2}F\right|$

= $\frac{9-6}{4}F=\frac{3}{4}F$

The two main types of potential energy are:

1. Gravitational Potential Energy - Energy stored due to an object's position in a gravitational field (PE = mgh).
2. Elastic Potential Energy - Energy stored in deformed elastic objects like springs or stretched materials (PE = ½kx²).

The best way to find potential energy is to use the relationship 

${}_{}$ $U_f-U_i=-{\int }_i^f\overrightarrow{F}\cdot d\overrightarrow{r}$

and integrating the conservative force over the path.

For common cases, apply the standard potential energy formulas:

• PE = mgh for gravity
• PE = ½kx² for springs. 

Potential energy is best defined as a stored energy in a system. This energy is stored due to the configuration or position of the objects within a conservative force field. Potential energy in physics tells us that it's the capacity of the work done when the system is allowed to move from that position to the reference point.  

Com

m1v + m2 * 0 = m1v1 + m2v2

75v = $45\frac{v}{3}+50*10$ $\frac{}{}$

$\Rightarrow 75*\frac{2v}{3}=50*10$

$\Rightarrow v=\frac{3*50*10}{75*2}=10m/sec$

Bernoulli's equation between (1) & (2)

$\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{v_2^2}{zg}+z_2$

$\frac{P_{atm}+mg/A}{\rho g}+0+40+10^{-2}$

$\frac{P_{atm}}{\rho g}+\frac{v_2^2}{2g}+0\left[v_1\approx 0\right]$

$\frac{mg}{A\rho g}+40*10^{-2}=\frac{v_2^2}{2g}$

$\frac{m}{A\rho }+40*10^{-2}=\frac{v_2^2}{2g}$

$\Rightarrow \frac{25}{0.5*10^3}+40*10^2=\frac{V_2^2}{2*10}$

$\Rightarrow V_2^2=\left(5*10^{-2}+40*10^{-2}\right)20$

$V_2^2=45*10^{-2}*20$

$V_2^2=90*10^{-1}$

$V_2=3m/sec$

$=300cm/sec$

Given, hT = 25m, R = 6400 km

hR = 49m

dm (Maximum distance for satisfactory communication)

dm = $\sqrt[]{2R{h}_T}+\sqrt[]{2R{h}_R}$ $\text{exception 25:}\text{exception 25:}$

dm = $\sqrt[]{2*6400*10^3*25}+\sqrt[]{2*6400*10^3*49}$ $\text{exception 25:}\text{exception 25:}$

= $\sqrt[]{5*10^4*6400}+\sqrt[]{6400*10^3*49*2}$ $\text{exception 25:}\text{exception 25:}$

= $192\sqrt[]{5}*10^2$ $\text{exception 25:}{}^{}$

= $k\sqrt[]{5}*10^2$ $\text{exception 25:}{}^{}$

k= 192

J (current density) = 4 * 10⁶ Am^-2

Area between radial distance  $\frac{R}{2}toR$

A =  $\pi \left[R^2-\frac{R^2}{4}\right]=\frac{3R^2}{4}\pi$

I = AJ

=  $\frac{3R^2}{4}\pi *4*10^6$

= 3R² * 10⁶ πAmp

= 3 (4 * 10^-3)² * 10⁶ πAmp

= 3 * 16 * 10^-6 * 10⁶π Amp

= 48πA

R shunt Resistance

Given

CD = 3m

CF = 2m

Current through potentiometer wire

I = $\frac{v}{R_0}=\frac{VA}{\rho L}$ $\frac{}{{}_{}}\frac{}{}$ (1)

A Area of potentiometer wire

$$  $\rho \to$ Resistivity of the wire

L Length of the potentiometer wire

Case 1 When 8 $\Omega$ $$ shunt is bal aced at 3m length

$V_{CD}=I{R}_{CD}=\frac{V}{\rho L}A*\frac{\rho CF}{A}$

$V_{CD}=\frac{V\mathcal{l}}{L}=\frac{V*3}{L}$

VCD = VAB = E – I1r

= $-\frac{E}{R+r}r=\frac{V}{L}*3$ $\frac{}{}\frac{}{}$

$$ $\Rightarrow E\left[1-\frac{r}{8+r}\right]=\frac{V}{L}*3$  (2)

Case 2 When 4 $$ $\Omega$ shunt is balanced at 2m length

$V_{CF}=I{R}_{CF}=\frac{V}{\rho L}A*\frac{\rho CF}{A}=\frac{V*2}{L}$

$V_{CF}=V_{AB}=E-I_{2}r$

= $E-\frac{E}{R+r}r$ $\frac{}{}$

= $E\left[1-\frac{r}{4+r}\right]$ $\frac{}{}$

$\frac{}{}\frac{}{}$  $\Rightarrow E\left(1-\frac{r}{4+r}\right)=\frac{V*2}{L}$ (3)

$\left(2\right)÷\left(3\right)$

$\frac{1-\frac{r}{8+r}}{1-\frac{r}{4+r}}=\frac{3}{2}$

$\Rightarrow \left(\frac{8}{8+r}\right)*\left(\frac{4+8}{4}\right)=\frac{3}{2}$

$\Rightarrow 4\left(4+r\right)=3\left(8+r\right)$

$\Rightarrow 16+4r=24+3r$

r = 24 – 16

r = 8 $\Omega$