Physics Test - 8

The wavelength of the incident light \(= 4000\) Å \(=4000 \times 10^{-10}m\)

The value of the work function \(=2 \mathrm{eV}\)

As we know,

Work function \(\phi=h f-K E\)

Where \(h f=\) incident energy of the light, \(\mathrm{KE}=\) kinetic energy of the ejected photoelectron.

Therefore, \(\mathrm{KE}=h f-\phi\)

We know that \(f=\frac{c}{\lambda}\)

Thus, \(K E=\frac{h c}{\lambda}-\phi\)........(i)

We know that:

\(c=3 \times 10^{8}\)m/s, \(h=606 \times 10^{-34}\), \(\phi=2\)

Put the given values in equation (i),

\(=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{4000 \times 10^{-10} \times 1.6 \times 10^{-19}}-2\quad\) ( \(1.6 \times 10^{-19}\) is used to get the value in terms of eV)

\(\Rightarrow K E=10.37 \mathrm{eV}\)

Therefore, the value of the kinetic energy of the ejected photoelectron is \(10.37 \mathrm{eV}\).

The process of electroplating, using the phenomenon of electrolysis, is primarily based on chemical effect of electric current.

• The phenomenon of producing heat by the electric current is called the heating effect of electric current.
• The amount of heat produced (H) in joules = I²Rt

Where, I is current, R is resistance and t is the time taken

• Due to the heating effect of the current, the filament of the bulb gets heated to a high temperature and it starts glowing.
• Similarly, the metals are melted and coated on another material to prevent corrosion. It is a chemical effect of electric current.

The combination of AND gate and NOT gate gives the NAND gate.

AND Gate: If both the inputs are high, it produces a high output.

Since the particle carrying positive charge is revolving around another charge,

Electrostatic force = Centrifugal force

\(\Rightarrow\)\(\frac{1}{4 \pi \epsilon_{0}} \frac{{q}_{1} {q}_{2}}{{r}^{2}}={mr} \omega^{2}\)

\(\Rightarrow\)\(\frac{1}{4 \pi \epsilon_{0}} \frac{{q}_{1} {q}_{2}}{{r}^{2}}=\frac{4 \pi^{2} {mr}}{{T}^{2}}\)

\(\Rightarrow\) \({T}^{2}=\frac{\left(4 \pi \epsilon_{0}\right) {r}^{2}\left(4 \pi^{2} {mr}\right)}{{q}_{1} {q}_{2}}\)

\(\Rightarrow\) \({T}=4 \pi {r} \sqrt{\frac{\pi \epsilon_{0} {mr}}{{q}_{1} {q}_{2}}}\)

Distance of the star from the solar system \(=4.29\)light year
1 light year \(=\) Speed of light \(\times 1\) year\(=3 \times 10^{8} \times 365 \times 24 \times 60 \times 60=94608 \times 10^{11} \mathrm{~m}\)
\(\therefore 4.29 \mathrm{ly}= 405868.32 \times 10^{11} \mathrm{~m} \)\(\quad\quad \left[\because 1 \text { parsec }=3.08 \times 10^{16} \mathrm{~m}\right]\)
\( \therefore 4.29\)light year \(= \frac{405868.32 \times 10^{11} \mathrm{~m}}{3.08 \times 10^{16} \mathrm{~m}}=1.32~\mathrm{parsec}\)
Now, using the Young’s Double Slit Experiment,\(\theta=\frac{d}{D}\)
where,\(\mathrm{d}=\) diameter of Earth's orbit \(=3 \times 10^{11} \mathrm{~m}\),\(\mathrm{D}=\) Distance of star from the Earth \(=405868.32 \times 10^{11} \mathrm{~m}\)
\(\therefore \theta=\frac{3 \times 10^{11} \mathrm{~m}}{405868.32 \times 10^{11} \mathrm{~m}}=7.39 \times 10^{-6} \mathrm{radian}\)
But, \(1 \mathrm{sec}=4.85 \times 10^{-6} \mathrm{radian}\)
\(\therefore 7.39 \times 10^{-6} \mathrm{radian}=\frac{7.39 \times 10^{-6}}{4.85 \times 10^{-6}}=1.52 ~\mathrm{radian}\)

We know that,

\(\text { Escape velocity } (v _{ e })=\sqrt{\frac{2 G M}{R}} \ldots \ldots . .(1)\)

Where \(G\) is the Universal Gravitational Constant and \(M\) is the mass of the Earth.

Velocity of the object.

\(v=\frac{v_{e} }{ 3}=\sqrt{\frac{2 G M}{9 R}}\)..... (2)

The initial kinetic energy of the object.

\(K E_{1}=\frac{1}{2} m v^{2}=\frac{1}{2} m \frac{2 G M}{9 R} \ldots\)... (3)

The initial potential energy of the object.

\(P E_{1}=-\frac{G M m}{R} \ldots \ldots .\) (4)

Here \(m\) is the mass of the object.

Let \(h\) be the maximum height reached by the object. when the object reaches the maximum height, its velocity becomes zero which is why its final kinetic energy \(= KE _{2}=0 \ldots \ldots \ldots\) (5)

At the maximum height, the potential energy of the object \(\left( PE _{2}\right)=-\frac{G M m}{R+h} \ldots \ldots .(6)\)

According to the law of conservation of energy,

\(KE _{1}+ PE _{1}= KE _{2}+ PE _{2}\)

(From \(3,4,5\), and 6)

\(\frac{1}{2} m \frac{2 G M}{9 R}-\frac{G M m}{R}=0-\frac{G M m}{R+h}\)

\(\Rightarrow \frac{1}{9 R}-\frac{1}{R}=-\frac{1}{R+h}\)

\(\Rightarrow -\frac{8}{9 R}=-\frac{1}{R+h} \)

\(\Rightarrow 8 R+8 h=9 R\)

\(\Rightarrow h=\frac{R}{8}\)

Thus, the maximum height reached will be \(\frac{R }{ 8}\).

Given:

Net focal length \(f =10\) cm \(=0.1\) m

The focal length of one of the lens \(f_{1}=-20\) cm \(=-0.2\) m (Lens is converging so, negative sign is used)

Focal length of another lens \(f_{2}\)

Power of another lens \(=\frac{1}{f_{2}}\)

So, in order to get combined focal length:

\(\frac{1}{f} =\frac{1}{f_{1}}+\frac{1}{f _{2}}\)

\(\Rightarrow \frac{1}{f_{2}}=\frac{1}{f} \frac{-1}{f _{1}}\)

\(\Rightarrow \frac{1}{f_{2}}= \frac{1}{0.1}-\left(\frac{-1}{0.2}\right)\)

\(=\frac{1}{f_{2}}=15\)

So, power of lens is \(15\) D.

Given:

\(P _{1}=200 W\)

\(P _{2}=100 W\)

\(V =100 V\)

We know that:

The rate at which the electric energy is dissipated or consumed is termed as electric power.

The electric power is given as,

\(P=I V=I^{2} R=\frac{V^{2}}{R}\)

Where, \(P =\) electric power, \(V =\) voltage, \(I =\) current and \(R =\) resistance

\(R=\frac{V^{2}}{P}\)

\(R \propto \frac{1}{P} \quad \ldots(1)\)

For an electric bulb, the resistance of the bulb is inversely proportional to the power of the bulb.

So, the bulb which has more power will have low resistance, therefore the \(100 W\) bulb will have more resistance compared to the \(200 W\) bulb.

The heat dissipated by the bulb is given as,

\(H = I ^{2} R\)

Both the bulbs are connected in series so the current in both the bulbs will be equal. So the heat dissipated will be more in the bulb which has more resistance.

Since the resistance of the \(100 W\) bulb is more, so the heat dissipation of the \(100 W\) bulb will be more.

The brightness of the bulb depends on the heat dissipation by the bulb, so the \(100 W\) bulb will have more brightness.

Given:

\(1^{\text {st }}\) orbital radius \(\left(R_{1}\right)=R\), and \(2^{\text {nd }}\) orbital radius \(\left(R_{2}\right)=4 R\)

According to the law of periods: The square of the period of revolution (T) of any planet around the sun is directly proportional to the cube of the semimajor axis of the orbit i.e., \(T^{2} \propto R^{3}\)

\( \mathrm{T}^{2} \propto \mathrm{R}^{3} \)

\(\Rightarrow\left(\frac{T_{1}}{T_{2}}\right)^{2}=\left(\frac{R_{1}}{R_{2}}\right)^{3} \)

\(\Rightarrow\left(\frac{T_{1}}{T_{2}}\right)^{2}=\left(\frac{R}{4 R}\right)^{3}\)

\(=\frac{1}{64} \)

\(\Rightarrow \frac{T_{1}}{T_{2}}=\left(\frac{1}{64}\right)^{\frac{1}{2}}\)

\(=\frac{1}{8}\)

The idea of displacement current was introduced to the current for making ampere circuital law consistent.

\(\oint \vec{B} . d \vec{l}=\mu_{0}\left(i_{c}+i_{d}\right)\)\(\quad\) (Modified Ampere circuital law)

Where, \(\mu_{0} \) is the permittivity of free space, \({i}_{d}\) is the displacement current, and \({i}_{c}\) is the conduction current. \(\oint \vec{B} \cdot d \vec{l}\) is line integral of the magnetic field over the closed-loop.

The expression for displacement current is given by,

\( i_{d}=\epsilon_{0} \frac{d \phi_{E}}{d t}\)

Where \(\phi_{{E}}\) is the flux of the electric field through the area bounded by the closed curve, \({i}_{{d}}\) is the displacement current, and \(\epsilon_{0} \) is the permittivity of free space.