Physics Test - 5

Work energy theorem.

\(W =\frac{1}{2} I \left(\omega_{ f }^{2}-\omega_{ i }^{2}\right)\)

Here, \(\theta=2 \pi\) revolution

\(=2 \pi \times 2 \pi=4 \pi^{2} rad\)

\(W _{ i }=3 \times \frac{2 \pi}{60} rad / s\)

\(\Rightarrow-\tau \theta=\frac{1}{2} \times \frac{1}{2} mr ^{2}\left(0^{2}-\omega_{1}^{2}\right)\)

\(\Rightarrow-\tau=\frac{\frac{1}{2} \times \frac{1}{2} \times 2 \times\left(4 \times 10^{-2}\right)\left(-3 \times \frac{2 \pi}{60}\right)^{2}}{4 \pi^{2}}\)

\(\Rightarrow \tau=2 \times 10^{-6} Nm\)

\(\vec{\tau}=\left(\vec{r}-\vec{r}_{0}\right) \times \vec{F}\)

\(\vec{r}-\vec{r}_{0}=(2 \hat{i}+0 \hat{j}-3 \hat{k})-(2 \hat{i}-2 \hat{j}-2 \hat{k})\)

\(=0 \hat{i}+2 \hat{j}-\hat{k}\)

\(\vec{\tau}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -1 \\ 4 & 5 & -6\end{array}\right|=-7 \hat{i}-4 \hat{j}-8 \hat{k}\)

The work to be done in the process is given by \(d W =P d V \)

\( =1 \times 10^5 Pa \times(1.671-0.001) m ^3 \)

\( =1.670 \times 10^5 J\)

The change in heat energy during the vaporisation process can be calculated as follows-

\(\Delta Q_{\text {supplied }} =2257 \times 1 \times 10^3 J \)

\( =22.57 \times 10^5 J\)

Hence, the change in internal energy in the process is given by

\(\Delta U =\Delta Q_{\text {supplied }}-\Delta W \)

\( =(22.57-1.67) \times 10^5 J \)

\( =20.9 \times 10^5 J \)

\( =2090 kJ\)

Threshold frequency is given as:

\( \mathrm{v}_0=\frac{W_0}{h}\)

Here \(W_0=3.3 \times 10^{-19} \mathrm{~J}\)

\(h=6.6 \times 10^{-34} \mathrm{Js}\)

\(\mathrm{v}_0=\frac{\left(3.3 \times 10^{-19}\right)}{\left(6.6 \times 10^{-34}\right)}\)

\(\mathrm{v}_0=5 \times 10^{14} \mathrm{~Hz}\)

The velocity of sound in air is given by:

\(\mathrm{C_s}=\sqrt{\frac{\gamma \mathrm{RT} }{ \mathrm{M}}}\).....(1)

where \(\gamma\) is the degree of freedom

\(\mathrm{T}=\text { temperature } \)

\(\mathrm{M}=\text { mass }\)

According to the kinetic theory of gases, The rms velocity of sound is given by:

\(\mathrm{C}=\sqrt{\frac{3 \mathrm{RT} }{ \mathrm{M}}}\).....(2)

Dividing equation (1) and (2) we get:

\(\frac{\mathrm{C_s} }{ \mathrm{C}}=\sqrt{\frac{\gamma}{ 3} }\)

\(\Rightarrow \mathrm{C_s}=\mathrm{C} \sqrt{ \frac{\gamma }{ 3}}\)

Equation is given as

\(\sqrt{x}=2 t+3\)

Taking square both side

\((\sqrt{x})^{2}=(2 t+3)^{2}\)

\(x=(2 t+3)^{2}\)

\(x=(2 t)^{2}+(3)^{2}+2(2 t)(3)\)

\(x=4 t^{2}+12 t+9\)

Taking derivative both side relative to \('t'\)

\(\frac{d x}{d t}=2(4 t)+12\)

\(\frac{d x}{d t}=8 t+12\)

We know that,

\(v=\frac{dx}{dt}\)

Hence, \(v=8 t+12\)

at \(t=0\)

\(v=8(0)+12\)

\(v=12 m/s\)

Then, the initial velocity is \(12 m/s\).

\(\begin{aligned} & R_1=\frac{R_2}{2} \\ & R_0\left(A_1\right)^{1 / 3}=\frac{R_0}{2}\left(A_2\right)^{1 / 3} \\ & A_1=\frac{1}{8} A_2 \\ & A_1=\frac{192}{8}=24\end{aligned}\)

The inverter converts DC into AC with the help of inductors and capacitors. 

The inverter is a static device that can convert one form of electrical power into another form of electrical power. The device can be used for back power supply in homes.