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

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

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.

The displacement current through the capacitor is given by
\(I_{d}=I_{c}=\frac{d q}{d t}\)
Given, \(q=q_{0} \cos 2 \pi v t\)
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
\(I_{d}=I_{c}=-\left(q_{0} \sin 2 \pi v t\right) \times 2 \pi v=-2 \pi v q_{0} \sin 2 \pi v t\)

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

Since the repulsive force between protons is missing in \(_{ 1 }^{ 3 }{ He }\), so the building energy of \(_{ 1 }^{ 3 }{ He }\) is greater than that of \(_{ 2 }^{ 3 }{ He }\)

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

No, because the TV signal of frequency 60 MHz is greater than 30 MHz (i.e., the upper limit of signal frequency for sky wave propagation). Therefore, T.V. signal will not be reflected by the ionosphere but will penetrate through the ionosphere.