Physics Test - 21

Given:

The masses of isotopes of neon are, \(19.99 \mathrm{u}, 20.99 \mathrm{u}\) and \(21.99 \mathrm{u}\). Relative abundance of the isotopes are \(90.51 \%, 0.27 \%\) and \(9.22 \%\).

Therefore, Average atomic mass of neon is,

Average atomic mass \(=\sum_{\mathrm{i}=1}^{\mathrm{n}}\frac{\left(\operatorname{mass}_{(\mathrm{i})}\right)}{\left(\operatorname{ Abundance}_{(\mathrm{i})}\right)}\)

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

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

\(=\frac{18.9 .29+5.67+202.75}{100}\)

\(=\frac{2017.7}{100}\)

\(=20.17 u \)

Refractive index of prism,

\(\mu=\sqrt{3}\)

Angle of prism,

\(A=60^{\circ}\)

Now using the prism formula,

\(\mu = \frac{\frac{\sin \left(A+\delta_{m}\right) }{ 2}}{\frac{\sin A}{ 2}}\)

\(\sqrt{3}=\frac{\sin \frac{\delta_{m}+60^{\circ}}{2}}{\frac{\sin 60^{\circ}}{2}}\)

\(\sqrt{3} \times \sin 30^{\circ}=\sin \left(\frac{\delta_{m}+60^{\circ}}{2}\right)\)

\(\sin \left(\frac{\delta_{m}+60^{\circ}}{2}\right)=\sqrt{3} \times \frac{1}{2}=\sin 60^{\circ}\)

\(\frac{\delta_{m}+60^{\circ}}{2}=60^{\circ}\)

\(\delta_{m}+60^{\circ}=120^{\circ}\)

\(\delta_{m}=60^{\circ}\) is the required minimum angle of deviation.

The initial velocity of the particle, \({v}=5 \hat{{i}}\)

After some time velocity changes to \(5 \hat{j}\)

Change in velocity, \({dv}=5 \hat{{j}}-5 \hat{{i}}\)

Time taken, \(t=10 sec\) .

Average acceleration,

\(=\frac{{dv}}{{t}}=\frac{1}{2}(\hat{{j}}-\hat{{i}})\)

\(=\frac{1}{\sqrt{2}} {m} / {s}^{2}\)

The direction is north-west making \(135\) degrees east as evident from its components.

The gravitational potential energy is maximum at Infinity.

As, \(\mathrm{U}=-\mathrm{G} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\mathrm{r}}\)

it can be used to describe the potential energy in a system of point charges (or radially symmetric spheres) as a function of their separation distance \(r\), then the maximum value is zero at infinite separation.

Given,

The initial speed of the auto rickshaw, \(u=36\) km/h \(=36 \times( \frac 5 { 18})\) m/s \(=10\) m/s

The final speed of the auto rickshaw when it stops, \(v=0\)

Time taken to stop, \(t=4.0 \) sec

From the first equation of motion,

\(v=u+a t\)

\(0=10+a \times 4\)

\(a=-(\frac {10 } 4)\)

\(a=-2.5\) m/s\(^2\)

The mass of the system (auto-rickshaw + driver), \(M=400+65=465\) kg

∴ Average retarding force, \(F=M \times a\)

\(=465  \times (2.5 )\)

\(=1,162.5\)

\(=1.162 \times 10^3\) Newton

The total energy in the H₂ molecule (atom) is given by,

$\Rightarrow E_n=-13.6\frac{Z^2}{n^2}eV$

Where n = principal quantum number and Z = atomic number.

Here for Z = 1, therefore the total energy is proportional to

$\Rightarrow E_n\propto \frac{1}{n^2}$

Wavelength, \(\lambda=\frac{2 \pi}{ k }=\frac{2 \pi}{\frac {2\pi} 3}=3~ m\)

Frequency, \(f =\frac{\omega}{2 \pi}=\frac{120 \pi}{2 \pi}=60~ Hz\)

Speed, \(v = f \lambda=180 ~m / s\)

Speed of a wave in a string is given by

\(v =\sqrt{ \frac T \mu }\)

\(\Rightarrow T = \frac{v ^{2} m} l\)

\(\Rightarrow T =\frac{ 1 8 0 ^{2} \times 3 \times 1 0 ^{-2}}{ 1 . 5 }\)

\(\Rightarrow T= 6 4 8 N\)

The correct sentence is that work has only magnitude but no direction.

• Work is donewhen the body moves with the application of external force or the moving body stop after theexternal forceapplied.
• Work done is thedot productof force and displacement.
• W = F ×d
  1. Where F = force applied
  2. d = displacement
• The dot product of vector quantities is always scalarwhich means work is having only magnitude but no direction.

A gas is compressed to half of its initial volume isothermally. The same gas is compressed again until the volume reduces to half through an adiabatic process. Then work done is more during the adiabatic process.

From the graph we can see that for compression of gas, area under the curve for adiabatic is more than isothermal process. Therefore, compressing the gas through adiabatic process will require more work to be done.

\(W_{e x t}=\) negative of area with volume-axis

\(W(\text { adiabatic })>W(\text { isothermal })\)