Physics Test - 16

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

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

Also we know that the frequency of a light wave is given by \(\nu=\frac{c}{\lambda} \ldots \ldots(2)\)

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

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

According to the question,

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

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 \(1.42 \mathrm{eV}\).

We know that:

A capacitor consists of two plates of a conductor and a dielectric insulator between them:

\(C=\frac{\epsilon A}{d}\)

Where

C = capacitance in farad

ϵ = Permittivity of dielectric

A = area of plate overlap in square meters

d = distance between plates in meters

If the area of the plates is halved i.e., \(\frac{{A}}{2}\) and the distance between them is halved i.e., \(\frac{d}{2}\), then new capacitance is:

\(C^{\prime}=\frac{\epsilon \frac{A}{2}}{\frac{d}{2}}\)

\(\therefore {C}^{\prime}={C}\)

\(\left[M L T^{-2}\right]=\left[L^{2 a}\right] \times\left[L^{b} T^{-b}\right]\left[M^{c} L^{-3 c}\right]\)

\(=\left[M^{c} L^{2 a+b-3 c} T^{-b}\right]\)

Comparing powers of M, L and T, we get

c = 1, 2a+b−3c = 1

−b = −2 or b = 2

Also, 2a+2−3(1) = 1

⇒ 2a = 2 or a = 1

∴ This is 1, 2, 1

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

We know that: \(c=3 \times 10^{8}\) m/s

Mass of a proton \(=1.67 \times 10^{-27}\) Kg

According to einstein law, \(E=m c^{2}\)

Put the values in above formula.

\(E=\left(2 \times 1.67 \times 10^{-27}\right) \times\left(3 \times 10^{8}\right)^{2} \mathrm{~J} \)

\(=3.006 \times 10^{-10} \mathrm{~J}\)

Also We know that De-brogli equation.

\(2 h \nu=E \text { or } 2 h \frac{c}{\lambda}=E \)

\(\therefore \lambda=\frac{2 h c}{E}\)

Put the values in above formula.

\(=\frac{2 \times 6.62 \times 10^{-34} \times 3 \times 10^{8}}{3.006 \times 10^{-10}} \mathrm{~m} \)

\(=1.323 \times 10^{-15} \mathrm{~m}\)

\(=1.3 \times 10^{-15} \mathrm{~m}\)

We know that the kinetic energy of the satellite is given as,

\( E_{K}=\frac{G M m}{2 r}\) .....(i)

We know that the potential energy of the satellite is given as,

\( E_{P}=\frac{G M m}{r}\) .....(ii)

Where \(r=\) distance of the satellite from the center of the planet. By equation (i) and equation (ii).

\(\frac{E_{K}}{E_{P}}=\frac{1}{2}\)

In conductors average relaxation time decreases with increase in temperature resulting in increase in resistivity. The variation of resistivity of copper with temperature is parabolic in nature.

We know that,\(h =6.63 \times 10^{-34} \mathrm{~J} \mathrm{~s}\)
Each photon has energy,\( E =h v=\left(6.63 \times 10^{-34} \mathrm{~J} \mathrm{s}\right)\left(6.0 \times 10^{14} \mathrm{~Hz}\right) =3.98 \times 10^{-19} \mathrm{~J}\)
If N is the number of photons emitted by the source per second,the power P transmitted in the beam equals N times the energyper photon E, so that P = N E. Then,
\(N=\frac{P}{E}=\frac{2.0 \times 10^{-3} \mathrm{~W}}{3.98 \times 10^{-19} \mathrm{~J}}=5.0 \times 10^{15}\) photons per second.

Given:

Acceleration \((\mathrm{a})=9.8 \mathrm{~m} / \mathrm{s}^{2}\)

Weight of the body \(=9.8 \mathrm{~N}\)

Since, Weight of the body \(=\mathrm{mg}\)

Mass \(=\) Weight \(/ \mathrm{g}\)

\(=\frac{9.8 }{ 9.8}\)

\(=1 \mathrm{~kg}\)

We know that,

\(F=m \times a \)

\(\Rightarrow F=1 \mathrm{~kg} \times 9.8 \mathrm{~m} / \mathrm{s}^{2} \)

\(\Rightarrow F=9.8 \text { Newton }\)

The Magnus effect is not related to electromagnetic waves.

• The force exerted on a rapidly spinning cylinder or sphere moving through air or another fluid in a direction at an angle to the axis of spin is called the Magnus effect.
• This force is responsible for the swerving of balls when hit or thrown with spin.

The Doppler effect is the change in the observed frequency of an (electromagnetic) wave due to relative motion of the source and observer.

Interference occurs when several waves are added together provided that the phase differences between them remain constant over the observation time.

Diffraction takes place with sound; with electromagnetic radiation, such as light, X-rays, and gamma rays; and with very small moving particles such as atoms, neutrons, and electrons, which show wavelike properties.

Therefore, Magnus effect is our required answer.

Given:

Refractive index of air, \(\mu_{1}=1\)

Refractive index of glass, \(\mu_{2}=1.5\)

Radius of curvature, \(\mathrm{R}=20 \mathrm{~cm}\)

Object distance, \(\mathrm{u}=-\mathrm{1 0 0} \mathrm{cm}\)

We have to find the distance of the image from the spherical surface.

We know that,

\(\frac{\mu_{2}}{v}-\frac{\mu_{1}}{u}=\frac{\mu_{2}-\mu_{1}}{R} \)

\(\Rightarrow \frac{1.5}{v}-\frac{1}{(-100)}=\frac{1.5-1}{20} \)

\(\Rightarrow \frac{1.5}{v}=\frac{0.5}{20}-\frac{1}{100} \)

\(\Rightarrow \frac{1.5}{v}=\frac{2.5-1}{100} \)

\(\Rightarrow v=\frac{1.5 \times 100}{1.5} \)

\(=100 \mathrm{~cm}\)