Physics Test - 32

Given,

Two waves of intensities\(=I\), \(4I\)

\(I_{\max } =\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}\)

\(=(\sqrt{I}+\sqrt{4I})^{2}\)

\(I_{\max } =(3 \sqrt{I})^{2}\)

\(I_{\max }=9I\)

\(I_{\max }=\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)^{2}\)

\(I_{\min }=\left(\sqrt{I_{1}}+\sqrt{4I}\right)^{2}\)

\(\Rightarrow (-\sqrt{I})^{2}=I\)

International systems of units are known as SI units. The SI units are represented without the help of other quantities known as fundamental quantities such as length, temperature etc.

The SI system has seven base units, and they are:

• Unit of length
• Unit of mass
• Unit of time
• Unit of electric current
• Unit of thermodynamic temperature
• Unit of amount of substance
• Unit of luminous intensity

These units pertain to the seven fundamental scientific quantities: metre, kilogram, second, ampere, kelvin, mole, and candela.

Using the expression:

\(E=-\frac{d V}{d r}\)

Since the potential ′V′ is constant, change in potential dV is zero for any displacement dr.

So, the electric field will be zero in the region.

The parallax method is used to determine large distances, such as the distance from Earth to a planet or a star. Parallax is the projected shift in the position of one object with respect to another when we move the point observation slant. The distance between two points of observation is called the base \((b)\). The distance of the object from the two points of view is \(D\). The angle between the two directions along which the object is viewed is the parallax angle or displacement angle \((θ)\).  

If \(S\) is the position of the object and \(AB\) is the two points of observation,

\(\theta=\frac{b}{D}\)

The triangulation method finds the angles in a triangle formed by three observation points. The other distances in a triangle are calculated using trigonometry and the measured length of just one side. It is used for geographic measurement and not astronomical distance.

The echo method uses the principle of reflection of sound or light. Knowing the speed of the wave and the time of reflection back to the source, the distance between the two points is calculated. This is not a very accurate method and therefore astronomical distances cannot be relied upon.

Melting of ice:

• ​When you heat ice, its temperature rises, but as soon as the ice starts to melt, the temperature stays constant until all the ice has melted.
• This happens because all the heat energy goes into breaking the bonds of the ice's crystal lattice structure.

The ice shrinks (decreases volume) and becomes denser. The ice density will increase from 0.92g/cm³ to that of liquid water (1g/cm^3​).

The molar specific heat capacity of a gas at constant volume is defined as the amount of heat required to raise the temperature of 1 mol of the gas by \(1^{\circ} C\) at the constant volume.

\(C_{v}=\left(\frac{\Delta Q}{n \Delta T}\right)_{\text {constant volume }}\)

The molar specific heat of a gas at constant pressure is defined as the amount of heat required to raise the temperature of \(1 mol\) of the gas by 1 \({ }^{\circ} C\) at the constant pressure.

\(C_{p}=\left(\frac{\Delta Q}{n \Delta T}\right)_{\text {constant pressure }}\)

The ratio of the two principal specific heat is represented by \(\gamma\).

\(\therefore \gamma=\frac{C_{p}}{C_{v}}\)

The value of \(\gamma\) depends on the atomicity of the gas.

The total internal energy of a mole of a rigid diatomic gas is,

\(U=\frac{5}{2} R T\)

Given:

\(L=0.05 H, C=80 \mu F\)

\(V_{\max }=200 \mathrm{~V}\)

\(\because\) Voltage equation \(V(t)=V_{m} \sin \omega t\)

\(\therefore\) Current \((\mathbf{i})=\frac{c d V}{d t}=c \frac{d\left(V_{m} \sin \omega t\right)}{d t}=C V_{m} \omega \cos \omega t\)

\(\because \omega=\frac{1}{\sqrt{L C}}\)

\(\therefore i=V_{m} \sqrt{\frac{C}{L}} \cos \omega t\)

\(\therefore\) Maximum current \(\left(i_{m}\right)=V_{m} \sqrt{\frac{C}{L}}=200 \times \sqrt{\frac{80 \mu}{0.05}}\)

\(\Rightarrow i_{m}=8 A\)

At steady state, capacitor behaves as an open circuit and current flows in circuit as shown in the diagram.

\(\mathrm{R}_{\mathrm{cq}}=9 \Omega \)

\(\mathrm{i}=\frac{9 \mathrm{~V}}{9 \Omega}=1 \mathrm{~A} \)

\(\Delta \mathrm{V}_{6 \Omega}=1 \times 6=6 \mathrm{~V} \)

\(\mathrm{~V}_{\mathrm{A}}=3 \mathrm{~V}\)

So, potential difference across \(6 \mu \mathrm{F}\) is \(6 \mathrm{~V}\).

Hence

\(\mathrm{Q}  =\mathrm{C} \Delta \mathrm{V} \)

\(=6 \times 6 \times 10^{-6} \mathrm{C} \)

\(=36 \mu \mathrm{C}\)

Theloudness of the sound produced by a radio increases byincreasing the amplitude.

• Sound is a wave and ahigh amplitude wave carries a large amount of energy. It meansa low amplitude wave carries a small amount of energy.
• Theloudnessof a sound wave isdirectly proportional to the square of the amplitudeof that wave.