Physics Test - 6

Formula used:

Torque, \(\tau=\vec{\mu} \times \overrightarrow{B_{H}}\), where \(\mu\) is the magnets dipole moment and \(B_{H}\) is the horizontal component of Earth's magnetic field.

We have been given that the dipole moment of the magnet, \(\mu=2 A m^{2}\) and the horizontal component of the Earth's magnetic field is \(B_{H}=0.4 \times 10^{-4} T\).

It is also given that the angle between the magnet and the magnetic meridian is \(30^{\circ} .\) 

So, we can illustrate the condition as follow

Now, we can find the torque generated by using the formula \(\tau=\vec{\mu} \times \overrightarrow{B_{H}}=\mu B_{H} \sin \theta\), where \(\mu\) is the magnets dipole moment, \(B_{H}\) is the horizontal component of Earth's magnetic field and \(\theta\) is the angle between the magnet and the magnetic meridian.

Therefore, \(\tau=2 \times 0.4 \times 10^{-4} \sin 30^{\circ}\)

\(=\frac{0.8}{2} \times 10^{-4}\)

\(=0.4 \times 10^{-4} N m\)

Given,

Angular momentum, \(\mathrm{L}=\mathrm{I} \omega\) ............(i)

Where, \(I\) is the moment of inertia

Kinetic energy \(\mathrm{K}=\frac{1}{2} \mathrm{I} \omega^{2}\)

\(\mathrm{K}=\frac{1}{2}(\mathrm{I} \times \mathrm{\omega}) \omega\)

(From eq. (i))

\(\Rightarrow \mathrm{K}=\frac{1}{2} \mathrm{L} \omega\)

\(\mathrm{L}=\frac{2 \mathrm{K}}{\omega}\)

Now, given angular frequency is halved, so, let \(\omega^{\prime}=\frac{\omega}{2}\)

And the kinetic energy is doubled, so, \(\mathrm{K}^{\prime}=\mathrm{2 K}\) If \(L^{\prime}\) is the new angular momentum, then we can write,

\(\mathrm{L}^{\prime}=\frac{\mathrm{2 K}^{\prime}}{\mathrm{\omega}^{\prime}}\)

\(\Rightarrow \mathrm{L}^{\prime}=\frac{\mathrm{2}(\mathrm{2 K})}{\frac{\omega}{\mathrm{2}}}\)

\(\Rightarrow \mathrm{L}^{\prime}=4 \frac{2 \mathrm{K}}{\omega}\)

\(\mathrm{L}^{\prime}=\mathrm{4} \mathrm{L}\)

So, angular momentum becomes four times its original value.

Given:

\(m_{p}=1.6 \times 10^{-27} \mathrm{~kg}, g=10 \mathrm{~ms}^{-2}\)

\(\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-27} \mathrm{Kg}\)

\(\mathrm{h}=1 \mathrm{~mm}=10^{-3} \mathrm{~m}\)

\(V=5 \mathrm{~V}\)

Time of flight of electron \(t_{e}=\sqrt{\frac{2 h}{a}}\) (ignoring gravity)

We know that \(\mathrm{F}=\mathrm{ma}\)

Similarly \(\mathrm{F=e E}\)

\(\therefore \mathrm{a}=\frac{e E}{m}\left[\right.\) also \(\left.\mathrm{E}=\mathrm{V} / \mathrm{d}=\frac{5}{10^{-3}}=5000 \mathrm{Vm}^{-1}\right]\)

\(\therefore \mathrm{t}_{\mathrm{e}}=\sqrt{\frac{2 \mathrm{hm}_{e}}{e E}}=\sqrt{\frac{2 \times 10^{-3} \times 9.1 \times 10^{-31}}{1,6 \times 10^{-19} \times 5000}}\)

\(=\left[\frac{2 \times 9.1 \times 10^{-34}}{1.6 \times 5 \times 10^{-16}}\right]^{1 / 2}=\left[2.275 \times 10^{-18}\right]^{1 / 2}\)

\(t_{e} \simeq 1.5 \times 10^{-9} \mathrm{~S}\) or \(\simeq 1.5 \mathrm{~ns}\)

Time flight for proton is

\(t_{p}=\sqrt{\frac{2 h m_{p}}{e E}}\)

\(=\sqrt{\frac{2 \times 10^{-3} \times 1.6 \times 10^{-27}}{1.6 \times 10^{-19} \times 5000}}=\left[\frac{2 \times 10^{-30}}{5 \times 10^{-16}}\right]^{1 / 2}\)

\(=\left[0.4 \times 10^{-14}\right]^{1 / 2}\)

\(=\left[4000 \times 10^{-18}\right]^{1 / 2}\)

\(=63.25 \times 10^{-9}\)

\(t_{p}=63.25 \mathrm{~ns}\)

As we know that the energy possessed by a body by virtue of its position or configuration is called potential energy.

Therefore as the height of an object increases, its potential energy also increases. Thus, if the stone is thrown up vertically and returns to the ground, its potential energy is maximum at the maximum height.

Given that a body is moving in a circular path with constant speed. So it will have a centripetal force. The velocity of the body comes to the initial direction after one complete circle.So when a body is moving in a circular path with a constant speed, its direction changes.So the velocity of the body is changing.

Two wires \(A\) and \(B\) are stretched by the same load. If the area of crosssection of wire ' \(A\) ' is double that of ' \(B\) ', then the stress on ' \(B\) ' is twice that on \(A\).

Stress \((S)=\frac{\operatorname{Force}(F)}{\operatorname{Area}(A)}\)

Given,

\(\frac{\text{Force}}{\text{Load}}\) (F) on both the wires is the same.

Let the cross-sectional area of wire \(\mathrm{B}\) is equal to \(\mathrm{A}\).

Cross-sectional area of wire \(\mathrm{A}\left(\mathrm{A}_{1}\right)=2 \times\) Cross-sectional area of wire \(\mathrm{B}\left(\mathrm{A}_{2}\right)=2 \mathrm{~A}\)

Stress on wire \(A(S)=\frac{\text { Force }(F)}{\text { Area of wires } A\left(A_{1}\right)}=\frac{F}{2 A} \quad \cdots\) (1)

Stress on wire \(B\left(S^{\prime}\right)=\frac{\operatorname{Force}(F)}{\text { Area of wires } B\left(A_{2}\right)}=\frac{F}{A} \cdots(2)\)

On dividing equations 1 and 2 , we get

\(\frac{\text { Stress on wire } A(S)}{\text { Stress on wire } B\left(S^{\prime}\right)}=\frac{1}{2}\)

Stress on wire \(B\left(S^{\prime}\right)=2 \times\) Stress on wire \(A(S)\)

In a transistor the collector current is always less than the emitter current because collector 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.

Let us consider an elementary ring of radius \(r\) and thickness \(\mathrm{dr}\) in which current \(\mathrm{I}\) is flowing.

Number of turns in this elementary ring \(\mathrm{d} \mathrm{N}=\frac{N}{b-a} d r\)

We know that the magnetic field at the centre of a ring is given by:

\(\mathrm{B}=\frac{\mu_{0} N I}{2 r}\)

Where,

\(\mu_{0}\) is the permeability of free space.

\(N\) is the number of terms.

\(\mathrm{L}\) is currently flowing.

\(r\) is the radius of the ring.

Thus magnetic field at the centre \(O\) due to this ring \(\mathrm{dB}=\frac{\mu_{0} d N I}{2 r}\)

Putting the value of \(\mathrm{dN}\), we get:

\(\mathrm{dB}=\frac{\mu_{0} N I d r}{2(b-a) r}\)

To get the net magnetic field at the centre of the spiral, we will integrate the above equation from 'a' to 'b'. \(\int_{0}^{B} d B=\int_{a}^{b} \frac{\mu_{0} N I d r}{2(b-a) r}=\frac{\mu_{0} N I}{2(b-a)} \int_{a}^{b} \frac{d r}{r}=\frac{\mu_{0} N I}{2(b-a)}[\ln r]_{a}^{b}\)

\(\Rightarrow B=\frac{\mu_{0} N I}{2(b-a)} \ln \frac{b}{a}\)

Therefore, the required magnetic field is 

\(\Rightarrow B=\frac{\mu_{0} N I}{2(b-a)} \ln \frac{b}{a}\)

Given,

Flux linked with a coil \(,\phi\)\(=0.8 \mathrm{W b}\)

Current, \(i\)\(=2 \mathrm A\)

Increase rate of current, \(\frac{d i}{d t}\)\(=400 \mathrm{~A} / \mathrm{s}\)

\(\phi=\mathrm{Li}\)

\(\Rightarrow 0.8=\mathrm{L} \times 2\)

\(\mathrm{~L}=\frac{0.8}{2}\)

\(=0.4\)

\(\mathrm{emf}=\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}\)

\(=0.4 \times 400\)

\(=160 \mathrm{~V}\)

From Lens Maker's formula:-

\(\frac{1}{f}=\left(\frac{\mu_{2}}{\mu_{1}}-1\right)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

where, \(\mu_{2}\) and \(\mu_{1}\) are refractive index of lens and surrounding medium respectively.

And, \(R_{1}\) and \(R_{2}\) are radius of 1st and 2nd surfaces respectively.

Here,

\(\mu_{1}=1, \quad \mu_{2}=n, \quad R_{1}=-\infty, \quad R_{2}=-R, \quad f=2 R\)

\(\frac{1}{f}=(n-1)\left(\frac{1}{-\infty}-\frac{1}{-R}\right)\)

\(\Rightarrow \frac{1}{f}=(n-1) \frac{1}{R}\)

\(\Rightarrow f=\frac{R}{n-1}\)