Physics Test-6

Let each side of the square lamina is \(d\).

So, \( I_{E F}=I_{G H} \quad\) (due to symmetry)

\(I_{A C}=I_{B D} \quad\) (due to symmetry)

Now, according to the theorem of perpendicular axis

\(I_{A C}+I_{B D}=I_{0}\)

Or\(\space\) \(2 I_{A C}=I_{0}\) .........(i)

And \(I_{E F}+I_{G H}=I_{0}\)

Or\(\space\) \(2 I_{E F}=I_{0}\) ..............(ii)

From Equation (i) and (ii), we get

\(I_{A C}=I_{E F}\)

\(\therefore I_{A D}=I_{E F}+\frac{m d^{2}}{4}\)

\(=\frac{m d^{2}}{12}+\frac{m d^{2}}{4} \left ( I_{E F}=\frac{m d^{2}}{12}\right)\)

So, \(I_{A D}=\frac{m d^{2}}{2}=4 I_{E F}\)

If a body is rotating about an axis, passing through its centre of mass then its angular momentum is directed along its axis of rotation.\(\vec{L}=I \times \vec{\omega}\)Where I = Moment of inertia and \(\omega=\) Angular velocity

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

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

If operated by a spring, it is elastic energy due to putting energy into the spring by mechanical winding. If operated by a battery it is the chemical energy stored in the battery.

A battery sends an electric current to a tiny, tuning-fork-shaped piece of quartz, causing it to oscillate at \(32,768\) vibrations persecond. (This is why mechanical watches, whose springs steadily release tension, have smoother second-hand movement than quartzwatches).

We know that,

\({R D}=\frac{\text { density of mercury }}{\text { density of water }}\)

\(13.6=\frac{\text { density of mercury }}{1 {gcm}^{-3}}\)

So, Density of mercury in \({CGS}=13.6 {gcm}^{-3}\)

So, Density of mercury in \(SI\) unit,

\(=\frac{13.6 \times 100 \times 100 \times 100}{1000} {kgm}^{-3}\)

\(\Rightarrow 13.6 \times 10^{3} {kgm}^{-3}\)

Coulomb's law in Electrostatics: It states that force of interaction between two stationary point charges is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them and acts along the straight line joining the two charges.

\({F} \propto {q}_{1} \times {q}_{2}\)

\(F \propto \frac{1}{r^{2}}\)

\(F=K \frac{q_{1} \times q_{2}}{r^{2}}\)

Where \(K=\frac{1}{4 \pi \epsilon_{o}}\) constant called electrostatic force constant or Coulomb's Constant.

The value of \({K}\) depends on the nature of the medium between the two charges and the system of units chosen.

From above it is clear that the constant in Coulomb's Law equation is \(\frac{1}{4 \pi \epsilon_{o}}=9 \times 10^{9} {Nm}^{2} / C^{2}\), where \(\varepsilon_{0}\) is the permittivity of free space.

The process of applying a protective layer of zinc in steel or iron to prevent it from rusting is known as galvanization. Galvanization protects the iron and steel by forming a coat of corrosion resistant zinc which prevents the corrosive nature substance to reach the delicate part of the steel or iron.

We know that,

The energy density (energy per unit volume) in an electric field E in vacuum is given by

\(\mathrm{U}=\frac{1}{2} \varepsilon_0 \mathrm{E}^2\)

where \(\varepsilon_0=8.86 \times 10^{-12} \mathrm{C}^2 / \mathrm{N}-\mathrm{m}^2\)

Given: \(\mathrm{E}=1 \mathrm{~V} / \mathrm{m}\)

\(U=\frac{1}{2} \times 8.86 \times 10^{-12} \times(1)^2 \)

\(U=4.43 \times 10^{-12}\)

So, the average energy density of magnetic field,

\(\mathrm{U}=\frac{4.43 \times 10^{-12}}{2}\)

\(\mathrm{U}=2.21 \times 10^{-12} \mathrm{~J} / \mathrm{m}^2\)

The minority carriers in a diode are the 'Holes' and 'Free Electrons' for the \(\mathrm{p}\) - side and \(\mathrm{n}\) - side, respectively. The minority charge carrier's movement depends on the thermal generation rate at the junction inside the diode. Which further depends on the rate of thermal generation of electron-hole pair.