Physics Test - 7

As due to mirrors there is reflection of light from one mirror to other and finally fall on the eye of the observer.

When we replace it with the lenses then there must be a refraction of light so that then can go to other lens and finally on observer. The phenomenon of reflection of light by a lens is called as total internal reflection. Total internal reflection is a complete reflection of a ray of light within a medium such as water or glass from the surrounding surfaces back into the medium. 

Given:

Two photons of same frequency are produced due to the annhiliation of a proton and antiproton.

We know that:

In the case of a steady electric field, the electric field does not change with time. Therefore, the electric flux density through a closed-loop will be constant.

This results in zero change in electric flux with respect to time i.e.,

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

The displacement current will be:

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

When a soft iron core is inserted inside the solenoid then the strength of the magnetic field becomes very large because the iron core gets magnetized by induction.

The soft iron core helps in concentrating the magnetic lines of forces through the solenoid so that the magnetic field is almost uniform at the end face of the core.

Thus, the strength of the magnetic field increases when a soft iron core is inserted inside a solenoid.

The optical element used to split a light wave into its component waves is called a diffraction grating.

Diffraction grating: It is an optical element that splits or divides a light of lots of different wavelengths into individual components by wavelengths.

The diffraction grating is a more powerful tool to split the different components of a poly-chromatic light than simple dispersion of light using a prism.

We know that:

Permittivity of free space is given by:

\(\epsilon_{0}=\frac{\mathrm{q}_{1} \mathrm{q}_{2}}{4 \pi \mathrm{Fr}^{2}}\)

Dimensions of \([\mathrm{F}]=\left[\mathrm{MLT}^{-2}\right]\)

Dimensions of \([\mathrm{q}]=[\mathrm{AT}]\)

Dimensions of \([\mathrm{r}]=[\mathrm{L}]\)

Thus, dimensional formula of \(\left[\epsilon_{0}\right]=\frac{[\mathrm{AT}][\mathrm{AT}]}{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{L}^{2}\right]}\)

\(\Rightarrow\left[\epsilon_{0}\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]\)

The output of a NOR gate is high if all inputs are low.

The NAND and NOR gates are universal gates. A universal gate is a gate which can implement any Boolean function without need to use any other gate type. That means we can implement any logic function using NAND and NOR gates without need of AND, OR or NOT gates.

• NOR gate is formed by connecting a NOT gate at the output of OR gate.
• NAND gate is formed by connecting NOT gate the output of AND gate.

1 unit of electric energy: When one-kilowatt load works for 1 hour then the energy consumed is called 1 unit of electricity.

As we know that the Einstein's photo-electric equation is given by,

\( K_{\max }=h \nu-\phi_{0}.....(1)\)

Also we know that the frequency of a light wave is given by

\( \nu=\frac{c}{\lambda}.....(2)\)

Substituting the value from (2) to (1), we get:

\( K_{\max }=\frac{h c}{\lambda}-\phi_{0}\)

So the work function of the metal is given by 

\( \phi_{0}=\frac{h c}{\lambda}-K_{\max }.....(3)\)

According to the question,

\(\lambda=400 \mathrm{~nm}, K_{\max }=1.68 \mathrm{eV}\), \(h c=1240 \mathrm{eV}\mathrm{~nm}\)

Putting these values in \((3)\) we get,

\( \phi_{0}=\frac{1240}{400}-1.68\)

\(\Rightarrow \phi_{0}=1.42 \mathrm{eV}\)

Thus the work function of the metal comes out to be equal to \(1.42 \mathrm{eV}\).