Physics Test-26

Given,

The reversible heat engine \((\Delta H) 3000 K J\)

\(T_{1}=650 K \)

\(T_{2}=295 K\)

To find the availability of heat energy we have to use the formula:

\(\Delta G=\Delta H-T_{2} \Delta S \)

\(\Delta S=\frac{\Delta Q}{\Delta T} \text { Where } \Delta Q=3000 K J,\left(\Delta T=T_{1}+T_{2}\right)=650+295 \)

\(=945 K \)

\(\Delta S=\frac{3000}{945}=3.17 K J / K \)

\(\Delta G=\Delta H-T_{2} \Delta S\)

On putting the value from above we get

\(\Delta G=3000-295 \times 3.17 \)

\(=2064.85 K J\)

Given:\(\mathrm{m}_{\mathrm{A}}=20 \mathrm{~kg}\)\(\mathrm{m}_{\mathrm{B}}=5 \mathrm{~kg}\)

Kinetic Energy of both bodies \(\mathrm{A}\) and \(\mathrm{B}\) are the same.

Thus,\(\frac{1}{2} \mathrm{~m}_{\mathrm{A}} \mathrm{V}_{\mathrm{A}}^{2}=\frac{1}{2} \mathrm{~m}_{\mathrm{B}} \mathrm{V}_{\mathrm{B}}^{2}\)

\(\frac{\mathrm{V}_{\mathrm{A}}^{2}}{\mathrm{~V}_{\mathrm{B}}^{2}}=\frac{\mathrm{m}_{\mathrm{B}}}{\mathrm{m}_{\mathrm{A}}}\)

\(\frac{\mathrm{V}_{\mathrm{A}}^{2}}{\mathrm{~V}_{\mathrm{B}}^{2}}=\frac{5}{20}\)

\(\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\frac{1}{2}\)

Solid sodium chloridedoes electric current not flow.

For the electric current to flow a solution is needed so that there is a formation of positively charge cations and negatively charged anions.

A copper wire is a conductor of electricity because it is a metal and has free electrons in it, so electric current can flow through it.

In solid sodium chloride, no formation of mobile ions take place due to which electric current may flow therefore electric current cannot travel through it

The magnetic field in a plane of the electromagnetic wave is given by,\(\mathrm{B}=2 \times 10^{-7} \sin \left(0.5 \times 10^{3} {x}+1.5 \times 10^{11} \mathrm{t}\right)\)
Comparing this equation with the standard wave equation:\(B_{y}=B_{0} \sin [K x+\omega t]\)
From the above equation, \(\mathrm{K}=\mathbf{0 . 5} \times 10^{3}=\) propagation constant
\(\mathrm{K}=\frac{2 \pi}{\lambda}\) where, \(\lambda=\) wavelength of wave
\(\lambda=\frac{2 \times 3.14}{0.3 \times 10^{3}}=1.256 \times 10^{-2} \mathrm{~m}=1.256 \mathrm{~cm}\)
The wavelength range of microwaves is \(10^{-3} \mathrm{~m}\) to \(0.3 \mathrm{~m}\). The wavelength of this wave lies between \(10^{-3} m\) to \(0.3 m\), so the equation represents microwaves.

The current through 500 ohm resistor is \(0.01\) ampere as voltage across is \(5 \mathrm{~V}\).

The current through 1500 ohm resistor asas voltage across is \(10 \mathrm{~V}\)is \(0.0067\) ampere.

So, the current through zener diode is difference in current through these resistors. The current is \(0.01-0.0067=0.0033\) ampere

As we know 1 ampere=1000 mA

So,Current through zener diode \(=0.0033 \times 1000=3.3 \mathrm{~mA}\)

The ratio of intensity of magnetisation to the magnetisation force is known assusceptibility.

In electromagnetism, the magnetic susceptibility is one measure of the magnetic properties of a material. The susceptibility indicates whether a material is attracted into or repelled out of a magnetic field.

A hydraulic press works on the principle of Pascal’s law.

Pascal law states that when pressure is applied to a confined fluid, the pressure change occurs throughout the entire fluid

• Pascal’s principle: Pascal’s Law is the principle of transmission of fluid-pressure.
• It says that "a pressure exerted anywhere in a point of the confined fluid is transmitted equally in all directions throughout the fluid”.
• Hydraulic Lift: The lift that uses pascal's law and used to lift objects with the help of fluid is called hydraulic lift. It works on Pascal's law.

Given:

Mass of spaceship (m) = 1000 kg

\(g = 10 ~m/s^2\)

R (radius of earth) = 6400 km

The energy needed for the spaceship to out into free space must be equal to gravitational potential energy.

\(E=\frac{G M m}{R} \)

\(=\left(\frac{G M}{R^{2}}\right) m R \)

\(=m g R \) \(~~[\because g=\left(\frac{G M}{R^{2}}\right)\)]

\(=1000 \times 10 \times 6400 \times 10^{3} \)

\(=6.4 \times 10^{10} \mathrm{~J}\)

\(Na\) in \(NaCl\) has eight corner and six face atoms and thus, if we remove face centred atom of one axis, then two face atoms are removed. 

Thus, \(A\) is at \(8\) corners \(\&\) at \(4\) face centers. 

Number of \(A\) atoms per unit cell  \(=8 \times \frac{1}{8}+4 \times \frac{1}{2}\)

\(=3\)

\(B\) present in all octahedral voids, thus number of  \(B\) atoms per unit cell \(=12 \times \frac{1}{4}+1\)

\(=4\)

Thus, formula is \(A_{3} B_{4}\).