Physics Test - 18

Given,

The percentage error in velocity\(=\frac{\Delta \mathrm{v}}{\mathrm{v}} \times 100\)

Kinetic energy\(\mathrm{K} . \mathrm{E.}\) \(=\frac{1}{2} \mathrm{mv}^{2}\)

Percentage errorin the kinetic energy,

\(\frac{\Delta \mathrm{K} \cdot \mathrm{E.}}{\mathrm{K} \cdot \mathrm{E.}} \times 100=\mathrm{m} \times 2 \frac{\Delta \mathrm{v}}{\mathrm{v}} \times 100\)

\(\Rightarrow \frac{\Delta \mathrm{K} . \mathrm{E.}}{\mathrm{K} . \mathrm{E.}} \times 100=2 \times 50 \%\)

\(\Rightarrow \frac{\Delta \mathrm{K.} \mathrm{E.}}{\mathrm{K.}\mathrm{E.}} \times 100=100 \%\)

Thus, the error in the measurement of kinetic energy is \(100 \%\).

The de-Broglie wavelength is given by

\(\lambda=\frac{n}{P}\)

\(\lambda \propto \frac{1}{P}\), 

\(\lambda P=\) constant

Let us study the validity of each of the four statements with explanations:

Statement – A: Base, emitter, and collector regions have similar size and doping concentrations.

The Emitter, Base and Collector regions have different sizes with varying concentrations. The order of size of these regions are: Base < Emitter < Collector The order of the concentration of the regions are: Base < Collector < Emitter

Hence, these regions are doped in a different way and their sizes are also, unsimilar.

Statement – B: The base region must be very thin and lightly doped.

The base region is the thinnest region of the transistor and the electron concentration of this region, as a result, is the lowest among the three regions.

Statement – C: The emitter-base junction is forward biased and base-collector junction is reverse biased.

The emitter region has a high concentration of electrons. So, connecting them across the base will result in their movement to the base and then, the collector. To facilitate the movement of the charge carriers from emitter, the emitter-base junction is forward biased, and the base-collector junction is reverse biased so as to attract these charge carriers towards the collector region.

Statement – D: Both the emitter-base junction as well as the base-collector junction are forward biased.

If both the emitter-base and collector-base junction are forward biased, the charge carriers will not be able to cross the barrier across the base to move to the collector, thereby, decreasing or hindering the functionality of the transistor.

Applying Snell's law at \(B\) and \(C\),

\(\mu \sin i =\) costant or

\(\mu_{1} {\sin i}_{B}=\mu_{4}{\sin i}_{C}\)

But \(A B \| C D\)

\(\therefore  i_{B}=i_{C}\)

or \( \mu_{1}=\mu_{4}\)

or \(\mu_{1} \sin i_{1}=\mu_{2} {\sin i}_{2}\)

Here, \(\mu_{1}=1, \mu_{2}=\mu\)

or \(\sin i =\mu \sin r\) or \(\mu=\frac{\sin i}{\sin r }\)

When a projectile is projected with a velocity, \(u\) and at angle, \(\theta\) to the horizontal, the maximum height reached by the projectile is given by:

\(H =\frac{ u ^{2} \sin ^{2} \theta}{2 g }\)

As both projectiles have same projected velocity, so

\(H \propto \sin ^{2} \theta\)

\(\therefore\) Maximum height at \(30^{\circ}\):

\(\left( H _{30}\right)=\frac{ u ^{2} \sin ^{2} 30}{2 g }\)

Similarly,

Maximum height at \(60^{\circ}\):

\(\left(H_{60}\right)=\frac{ u ^{2} \sin ^{2} 60}{2 g }\)

Thus the ratio of maximum heights

\(\frac{ H _{30}}{ H _{60}}=\frac{\sin ^{2} 30}{\sin ^{2} 60}\)

\(=\frac{\frac{1}{4}}{\frac{3}{4}}\)

\(=\frac{1}{3}\)

So, Thus the ratio of maximum heights is \(1: 3\).

The given equation is,

\(n=\frac{A+B}{\lambda^{2}}\)

Where \(A\) and \(B\) are constants.

\(n=\) Refractive index

\(\lambda=\) Wavelength

By homogeneity principle the dimensions of all the terms on both sides should be same i.e.,

\([B]=[A]=\left[n \lambda^{2}\right]\)

\([B]=\left[\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0}\right]\left[\mathrm{L}^{2}\right]\)

\([B]=\left[\mathrm{M}^{0} \mathrm{L}^{2} \mathrm{T}^{0}\right]\)

Magnetic moment$m=AI={\mathrm{\pi r}}^2\mathrm{I}$where$r$ is the radius of the circular loop. Now, the circumference of the circle = length of the wire, i.e.,

$2\mathrm{\pi r}=\mathrm{L}$

${\mathrm{\pi r}}^2=\frac{{\mathrm{L}}^2}{4{\mathrm{\pi }}^2}$

Therefore,$m={\mathrm{\pi r}}^2\mathrm{I}=\frac{{\mathrm{\pi L}}^2\mathrm{I}}{4{\mathrm{\pi }}^2}=\frac{{\mathrm{L}}^2\mathrm{I}}{4\mathrm{\pi }}$

Thus, the magnitude of the magnetic moment is$\frac{L^{2}I}{4\mathrm{\pi }}$

Hence the correct option is (D).

\(\begin{aligned} & \text { For } A \text { mass number }=34 \\ & \text { Total binding energy }=1.2 \times 34=40.8 \mathrm{MeV} \\ & \text { For } B \text { mass number }=26 \\ & \text { total binding energy }=1.8 \times 26 \mathrm{MeV} \\ & =46.8 \mathrm{MeV} \\ & \text { Difference of } \mathrm{BE}=6 \mathrm{MeV}\end{aligned}\)

Newton's law of cooling states that

\(\frac{\left(T_1-T_2\right)}{t}=K\left(\left(\frac{T_1+T_2}{2}\right)-T_0\right) \quad \ldots(i)\)

where \(T_0\) is the temperature of the surrounding. From equation (i),

\( \frac{60-40}{7 \times 60}=K\left(\frac{60+40}{2}-10\right) \)

\( \Rightarrow \frac{20}{7 \times 60}=40 K \)

\( \Rightarrow K=\frac{30}{7 \times 60 \times 40}\)

Let the temperature of the body after the given time be \(x^{\circ} C\). By Newton's law of cooling,

\( \frac{40-z}{7 \times 600}=K\left(\frac{40+x}{2}-10\right) \)

\( \Rightarrow \frac{40-x}{7 \times 60}=\frac{1}{14 \times 60}\left(\frac{20+x}{2}\right) \)

\( \Rightarrow 160-4 x=20+x \)

\( \Rightarrow x=28^{\circ} C\)