Physics Test - 23

Given,

Atomic mass of \({ }^{20} \mathrm{Ne}_{10}, \mathrm{~m}_{1}=19.99 \mathrm{~u}\)

Abundance of \({ }^{20} \mathrm{Ne}_{10}, \eta_{1}=90.51 \%\)

Atomic mass of \({ }^{21} \mathrm{Ne}_{10}, \mathrm{~m}_{2}=20.99 \mathrm{~u}\)

Abundance of \({ }^{21} \mathrm{Ne}_{10}, \eta_{2}=0.27 \%\)

Atomic mass of \({ }^{22} \mathrm{Ne}_{10}, \mathrm{~m}_{3}=21.99 \mathrm{~u}\)

Abundance of \({ }^{22} \mathrm{Ne}_{10}, \eta_{3}=9.22 \%\)

The average atomic mass of neon is given as:

\(m=\frac{m_{1} \eta_{1}+m_{2} \eta_{2}+m_{3} \eta_{3}}{\eta_{1}+\eta_{2}+\eta_{3}}\)

\(=\frac{19.99 \times 90.51+20.99 \times 0.27+21.99 \times 9.22}{90.51+0.27+9.22}\)

\(=20.1771 \mathrm{~u}\)

The temperature determines the direction of net change of intermolecular kinetic energy.

On heating, the temperature rises, particles gain energy. As the kinetic energy increases, intermolecular space increases as they get separated from each other and the force of attraction decreases as the particles go far away from each other.

Given:

Displacement equation is given as \(\mathrm{Y}=0.08 \sin \left(3 \pi \mathrm{t}+\frac{\pi}{4}\right)\)

On comparing with \(\mathrm{Y}=\mathrm{A} \sin (\mathrm{\omega t}+\pi)\)

We get, \(\mathrm{\omega}=3 \pi\) and \(\phi=\frac{\pi}{4}\)

Time period \(\mathrm{T}=\frac{2 \pi}{\mathrm{\omega}}\)

\(=\frac{2 \pi}{3 \pi}\)

\(=\frac{2}{3} \mathrm{~s}\)

Initial phase \(\phi=\frac{\pi}{4}\)

\(\text { At } \mathrm{t}=\frac{7}{36} \mathrm{~s} \)

\( \mathrm{Y}=0.08 \sin \left(3 \pi \times \frac{7}{36}+\frac{\pi}{4}\right)\)

\(=0.08 \sin (\frac{5 \pi }{ 6}) \)

\(\Rightarrow \mathrm{Y}=0.08 \times 0.5\)

\(=0.04 \mathrm{~m}\)

For photoelectric effect to be observed, wavelength of the incident light \((\lambda)\) should be less than the threshold wavelength \(\left(\lambda_{0}\right)\) of the metal.

Einstein's photoelectric equation:

\(\mathrm{eV}_{0}=\frac{\mathrm{hc}}{\lambda}-\phi_{0}\)

Here, \(V_{0}=\) stopping potential, \(\lambda_{0}=\) threshold wavelength,

\(h\) = Planck"s constant, \(\phi_{0}=\) work-function of metal.

So, the stopping potential is related to the shortest wavelength.

Given,

Mass of \({ }^{6} \mathrm{Li}_{3}\), lithium isotope, \(\mathrm{m} _{1}=6.01512 \mathrm{u}\)

Mass of \({ }^{7} \mathrm{Li}_{3}\), lithium isotope, \(\mathrm{m_2}=7.01600 \mathrm{u}\)

Abundance of \({ }^{6} \mathrm{Li}_{3}, \eta_{1}=7.5 \%\)

Abundance of \({ }^{7} \mathrm{Li}_{3}, \eta_{2}=92.5 \%\)

The atomic mass of lithium atom is given as:

\(m=\frac{m_{1} \eta_{1}+m_{2} \eta_{2}}{\eta_{1}+\eta_{2}}\)

\(=\frac{6.01512 \times 7.5+7.01600 \times 92.5}{92.5+7.5}\)

\(=6.940934 \mathrm{u}\)

According to Kepler's second law of planetary motion, the planets move faster near the Sun and move slowly away from the Sun to cover equal areas in equal intervals of time.

Therefore, as the distance of the planet increases from the Sun, it will orbit the Sun slower and hence, takes a longer time.

Hence, the farther away a planet is from the Sun, the longer it takes to orbit the Sun once.

An image is a copy of an object that is formed by reflected (or refracted) light. Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.

Diffuse reflection occurs when light reflects off a rough surface and forms a blurry image or no image at all.

​So, Light rays from an object fall on a surface and get reflected in a completely diffused manner; No image will be formed.

Impulse is defined as the change in momentum.

Impulse I \(=\Delta \mathrm{p}=\mathrm{mv}-\mathrm{mu}\)

So, both Impulse and momentum have the same units/dimensions.

The efficiency of the carnot's heat engine is given as

\(\eta=\left(1-\frac{T_2}{T_1}\right) \times 100\)

When efficiency is \(40 \%\),

\(\mathrm{T}_1=500 \mathrm{~K} ; \eta=40 \)

\(40=\left(1-\frac{\mathrm{T}_2}{500}\right) \times 100 \)

\(\Rightarrow \frac{40}{100}=1-\frac{\mathrm{T}_2}{500} \)

\(\Rightarrow \frac{\mathrm{T}_2}{500}=\frac{60}{100} \Rightarrow \mathrm{T}_2=300 \mathrm{~K}\)

When efficiency is \(60 \%\), then

\(\frac{60}{100}=\left(1-\frac{300}{\mathrm{~T}_2}\right) \Rightarrow \frac{300}{\mathrm{~T}_2}=\frac{40}{100} \)

\(\Rightarrow \mathrm{T}_2=\frac{100 \times 300}{40} \Rightarrow \mathrm{T}_2=750 \mathrm{~K}\)

Given,

Acceptor impurity concentration, \(n_a = 10^{17}  \mathrm{~cm}^{-3}\)

Donor impurity concentration, \(n_a = 10^{16}  \mathrm{~cm}^{-3}\)

Intrinsic carrier concentration \(n_{\mathrm{i}}=1.6 \times 10^{10} \mathrm{~cm}^{-3}\)

Contact potential at the junction,

\(V=\frac{k T}{e}\) ln \((\frac{n_{a} n_{d}}{n_{i}^{2}})\)

\(\Rightarrow V=\frac{k T}{e}\) ln \(\frac{10^{17} \times 10^{16}}{(1.6 \times 10^{10})^{2}}\)

\(\Rightarrow V=\frac{k T}{e}\) ln \((4 \times 10^{12})\)