Physics

$Current in circuit i=\frac{10}{4+1}=2 A$

$Terminal voltage =E-iR$

$=10-2*1=8 V$

de-Broglie wavelength $\lambda =\frac{h}{P}=\frac{h}{mv}=\frac{h}{\sqrt[]{2mE}}$ where $E=\frac{1}{2}m{v}^2$

Squaring both sides,

${\lambda }^2=\frac{h^2}{4m^{2}E}$

$\Rightarrow \frac{1}{{\lambda }^2}=\left( constant \right)E$

Graph passes through origin with constant slope.

Solar cell characteristics

B: In reverse biased $pn$ junction diode, the current measured in $\left(?A\right)$ , is due to minority charge carrier.

Resonance occurs when the frequency of an external periodic force matches the natural frequency of a system. From that, physicists know that resonance causes the amplitude of oscillations to increase significantly. This can be beneficial in devices, such as musical instruments, but dangerous in structures like bridges.

The phase in SHM tells us the position and direction of motion of the particle at a specific instant. It determines the state of oscillation and includes both displacement and time information.

The restoring force in SHM is the force that always acts towards the mean position and is directly proportional to the displacement from it. It follows F=? kx. Here, the negative sign indicates the force is in the opposite direction to the displacement.

Time period of Oscillation, $T=2\pi \sqrt[]{\frac{I}{MB}}$

$\Rightarrow \frac{1}{4}=2\pi \sqrt[]{\frac{9.8*10^{-6}}{M*0.049}}$

$\Rightarrow \frac{1}{16}=4{\pi }^2*\frac{9.8*10^{-6}}{M*49*10^{-3}}$

$\Rightarrow M=\frac{4{\pi }^2*9.8*10^{-6}}{49*10^{-3}}*16$

$=\frac{4{\pi }^2*9.8*16*10^{-3}}{49}$

$=12.8{\pi }^2*10^{-3}*10^{-2}*10^2$

$=1280{\pi }^2*10^{-5}A{m}^2$

Sol. Before collision $\Rightarrow$

It undergoes completely inelastic collision

Using conservation of linear momentum

Initial momentum = Final momentum

$\Rightarrow m{v}_1=m{v}_2+m{v}_2$

$\Rightarrow m{v}_1=2m{v}_2$

$\Rightarrow \frac{v_1}{v_2}=\frac{2}{1}$

The potential $V$ at any point, at distance $r$ from centre of dipole $=\frac{KPcos\theta }{r^2}$

At axial point where  $\theta =0^{?},V=\frac{KP}{r^2}=\frac{9*10^9*4*10^{-6}}{2^2}=9*10^3 V$

At axial point where  $\theta =180^{?},V=\frac{-KP}{r^2}=-9*10^3 V$