As, we know that the diode only works in forward biased, so the output is obtained only when positive input is given, so the output waveform is

Since the repulsive force between protons is missing in \({ }_1^3 \mathrm{He}\) so the building energy of \({ }_1^3 \mathrm{He}\) is greater than that of \({ }_2^3 \mathrm{He}\)

Angle of prism, \(A={ 5 }^{ 0 },{ \mu }_{ R }=1.641,{ \mu }_{ B }=1.659\)
As, we know that, \(\delta =(\mu -1)A\) 
\(So,{ \mu }_{ B }=({ \mu }_{ B }-1)A \ and \ { \mu }_{ R }=({ \mu }_{ R }-1)A\)
\(So,{ \mu }_{ B }-{ \mu }_{ R }=({ \mu }_{ B }-{ \mu }_{ R })A\)
\(=(1.659-1.641)\times 5\)
\( ={ 0.09 }^{ 0 }\)

The system of magnets will be in stable equilibrium, if the net force on the system is zero and net torque on it is also zero. It will be possible, if the poles of the remaining two magnets is as shown in the figure.

No, it does not experience a torque, when it is placed along the direction of electric field

(i) DE is the region of negative resistance because the slope of curve in this part is negative.
(ii) BC is the region where Ohm's law is obeyed because in this part, the current varies linearly with the voltage.

Ratio of magnetic moments,
\(\frac { M_{ s } }{ M_{ c } } =\frac { 2INA_{ s } }{ INA_{ c } } =\frac { 2\left( \frac { R }{ 2 } \right) ^{ 2 } }{ \left( R \right) ^{ 2 } } =\frac { 1 }{ 2 } \) 

At the centre, since the electric field due to two charges is equal and opposite at this point.

The displacement current through the capacitor is given by
\(\begin{aligned} I_d=I_c=\frac{d q}{d t} \end{aligned}\)        ....(i)
Given, \(\begin{aligned} q=q_0 \cos 2 \pi v t \end{aligned}\)
Differentiating w.r.t, t on both sides, we get
\(\frac{d q}{d t}=q_0(-\sin 2 \pi v t)(2 \pi v)\)
Putting the value of \(\frac{d q}{d t}\) in Eq. (i), we get
\(I_d=I_c=-\left(q_0 \sin 2 \pi v t\right) \times 2 \pi v\)
\(=-2 \pi v q_0 \sin 2 \pi v t\)