Physics Test - 11

In a transistor the collector current is always less than the emitter current becausecollector side is reverse - biased and the emitter side is forward biased.

From the above figure, it is clear that in order to operate the emitter-base junction of the transistor it must be forward biased and the collector-base junction is reverse biased.

As the emitter-base junction is forward biased due to the repulsion many electrons repel from the emitter to the base and some electrons may combine with the holes which are the majority charge carriers and the rest of the electrons reach the collector, which in turn means Collector current is less than the emitter current.

\(E=5.5 e V\), \(\phi=4.5 e V\)

Therefore, the kinetic energy will be equal to

\( K . E=E-\phi\)

On substituting the values in above formula, we get

\( K . E=(5.5-4.5) e V \Rightarrow K . E=1 e V\)

Wavelength \((\lambda)=\frac{h}{\sqrt{2 M_{e} \times K \cdot E}}\)

\(\lambda\) is the wavelength.

\(h\) is the Planck's constant  \(=6.6 \times 10^{-34}\).

\(M_{e}\) is the mass of an electron  \(=9.1 \times 10^{-31}\).

\(K . E\), is the kinetic energy  \(=1.16 \times 10^{-19}\).

So on substituting the values in the formula of wavelength, we get 

\( \lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31}  \times 1.16 \times 10^{-19}}}=1.24 \times 10^{-9} m\)

As we know, the bending of light is inversely proportional to the wavelength of the light.

Bending \(\propto \frac{1}{\text { Wavelength }}\)

Different colours bend differently on passing through a prism and the bending of colour depends on wavelength. Higher wavelengths bend less, whereas shorter wavelength bend more.

Out of all violet has the least wavelength and hence it bends the most and red has the highest wavelength, so it bends the least.

For the net gravitational force on a particle.

\(F=\frac{G M^{2}}{a^{2}} \)     .....(1)

\(F_{1}=\frac{G M^{2}}{(a \sqrt{2})^{2}}\)

\(=\frac{G M^{2}}{2 a^{2}}\)     .....(2)

According to the figure:

For net force, \(F=\sqrt{F^{2}+F^{2}}+F_{1}\)

\(F_{net}=F \sqrt{2}+F_{1}\)     .....(3)

Put values from (1) and (2) in (3).

\(F_{net}=\frac{G M^{2}}{a^{2}} \sqrt{2}+\frac{G M^{2}}{2 a^{2}}\)

This force will act as centripetal force. Distance of particle from centre of circle is \(\frac{\mathrm{a}}{\sqrt{2}}\).

\(\mathrm{~F}_{\mathrm{C}}=\frac{\mathrm{Mv}^{2}}{\mathrm{r}} \)

\(\mathrm{r}=\frac{\mathrm{a}}{\sqrt{2}}\)

\(F_{C}=F_{net}\)

\(\frac{\mathrm{Mv}^{2}}{\frac{\mathrm{a}}{\sqrt{2}}}=\frac{\mathrm{GM}^{2}}{\mathrm{a}^{2}}\left(\frac{1}{2}+\sqrt{2}\right) \)

\(\Rightarrow \mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{a}}\left(\frac{1}{2 \sqrt{2}}+1\right) \)

\(\Rightarrow \mathrm{v}^{2}=\frac{\mathrm{GM}}{\mathrm{a}}(1.35) \)

\(\Rightarrow \mathrm{v}=1.16 \sqrt{\frac{\mathrm{GM}}{\mathrm{a}}}\)

Given:

Change in an electric field, \(\frac{d E}{d t}=5 \times 10^{12} {Vm}^{-1} s^{-1}\),

Side of the plate (I) \(=2\) \({cm}=2 \times 10^{-2} {~m}\),

and \(\epsilon_{0}=8.85 \times 10^{-12} C^{2} N^{-1} {~m}^{-2}\)

The area of the plate is:

Area of square\(=(\text{side})^2\)

\( A=2 \times 10^{-2} \times 2 \times 10^{-2}=4 \times 10^{-4} {~m}\)

We know that displacement current is given as:

\( I_{d}=\epsilon_{0} A \times \frac{d E}{d t}\)

\( I_{d}=8.85 \times 10^{-12} \times 4 \times 10^{-4} \times 5 \times 10^{12}\)

\( I_{d}=177 \times 10^{-4} A\)

\(I_{d}=17.7 {~m}\)

Combination of Logic Gates

The above combination of the logic gate is the NAND gate.

Let \(C\) be the capacitance of the infinite ladder.

As the ladder is infinite, the addition of one more element of two capacitors \((1 \mu F\) and \(2 \mu F)\) across the points \(X\) and \(Y\) should not change the total capacitance.

Therefore, the total capacity of the arrangement shown in the figure must remain \(C\) only.

In figure, \(2 \mu F\) capacitor is in series with capacitance \(C\).

\(\therefore\) Their combined capacity \(=\frac{2 \times C}{2+C}\)

This combination is in parallel with \(1 \mu F\) capacitor.

The equivalent capacity of the arrangement is

\(1+\frac{2 C}{2+C}=C \)

\(\text { or, } C^{2}+2 C=2+3 C \)

\(\text { or, } C^{2}-C-2=0 \)

\(\therefore C=2,-1\)

As capacitance cannot be negative.

\(\therefore C=2 \mu F\)

Given:

\(I = \frac{Q}{t} =\frac{96000}{3600}=\frac{80}{3A}\)

\(V =50\) V

\(t =1\) hour \(=3600\) sec

Where, \(I =\) current, \(R =\) resistance, \(t =\) the time taken, \(V =\) electric potential and \(Q =\) quantity of charge flowing

We know that:

The amount of heat produced (H) in joules is:

\(H = Vlt\)

\(\Rightarrow H=50 \times \frac{80}{3} \times 3600\)

\(\Rightarrow H =4.8 \times 10^{6} J\)

If \(N =\) number of turns in the coil, \(I =\) current in the loop, \(A =\) area enclosed by the loop, \(B=\) magnetic field intensity, and \(\theta=\) angle between the normal to the plane of the coil and the direction of a uniform magnetic field.

Then the torque on the current-carrying rectangular loop is given as,

\(T = NIAB \cdot \sin \theta \quad \ldots\)(1)

By equation (1) it is clear that the torque will be maximum when the value of \(\sin \theta\) is maximum.

We know that the maximum value of \(\sin \theta\) is 1.

For \(\sin \theta=1\)

\(\Rightarrow \theta=90^{\circ}\)

So, the torque on the loop will be maximum when the angle between the area vector and the magnetic field is \(90^{\circ}\).

Field emission involves the emission of electrons from a material's surface when subjected to a strong electric field. In this process, electrons are emitted from a material's surface when a strong electric field is applied. It's termed "quick" because it occurs rapidly under the influence of the electric field, without requiring significant heating or other conditions.