Physics Test - 24

Given:

Resistance \(R =1 \Omega\), maximum voltage \(\left( v _{ m }\right)=1 V\) and \(\omega=2 \pi\) rad/s

The oscillating potential of an AC circuit is given by,

\(v=v_{m} \sin \omega t\)

Where '\(v_{ m }\)' is the amplitude of the oscillating potential difference and '\(\omega\)' is the angular frequency

The oscillating current in an AC circuit is given by,

\(i=\frac{v_{m}}{R} \sin \omega t\)

Where 'i' is the amplitude of the oscillating current.

Since '\(v_{m}\)' and 'R' are constants.

\(i = i _{ m } \sin \omega t\)

Where \(i_{m}=\frac{v_{m}}{R}\)

The average value of power over a cycle is,

\(\bar{p}=\frac{1}{2} i_{m}^{2} R\)

The instantaneous power dissipated in the resistor is,

\(p=i^{2} R=i_{m}^{2} R \sin ^{2} \omega t\)

\(=\frac{v_{m}^{2}}{R} \sin ^{2} \omega t\)

\(=\frac{1^{2}}{1} \sin ^{2}(2 \pi \times 0.125)\)

\(=\sin ^{2}(0.25 \pi)\)

\(=0.5 W\)

Bulk modulus \(k\) is the reciprocal of compressibility \(fi\).

\(k=\frac{1}{f i}\)

Ideal fluids are incompressible which means \(fi = 0\). Thus, \(k\) will be infinity.

Newton's law of cooling states that

\(\frac{\left(T_1-T_2\right)}{t}=K\left(\left(\frac{T_1+T_2}{2}\right)-T_0\right) \quad \ldots(i)\)

where \(T_0\) is the temperature of the surrounding. From equation (i),

\(\begin{aligned}& \frac{60-40}{7 \times 60}=K\left(\frac{60+40}{2}-10\right) \\& \Rightarrow \frac{20}{7 \times 60}=40 K \\& \Rightarrow K=\frac{30}{7 \times 60 \times 40}\end{aligned}\)

Let the temperature of the body after the given time be \(x^{\circ} C\). By Newton's law of cooling,

\(\begin{aligned}& \frac{40-z}{7 \times 600}=K\left(\frac{40+x}{2}-10\right) \\& \Rightarrow \frac{40-x}{7 \times 60}=\frac{1}{14 \times 60}\left(\frac{20+x}{2}\right) \\& \Rightarrow 160-4 x=20+x \\& \Rightarrow x=28^{\circ} \mathrm{C}\end{aligned}\)

Both Dyne and Newton are units of force in the CGS and S.I. system.

Newton and dyne can also be written as,

\(\left(1 \mathrm{~N}=\mathrm{kg}\mathrm{m} / \mathrm{s}^{2}\right)\) and \(\left(1\right.\) dyne \(=\mathrm{gm}\mathrm{cm} / \mathrm{s}^{2}\) )

\(\Rightarrow 1\) dyne \(=1 \mathrm{gm}\mathrm{cm} / \mathrm{s}^{2}\) and 1 newton \(=1 \mathrm{~kg}\mathrm{m} / \mathrm{s}^{2}\)

\(\Rightarrow 1 \mathrm{~kg}=1000 \mathrm{gm}\)

\(\Rightarrow 1 \mathrm{~m}=100 \mathrm{~cm}\)

\(\Rightarrow 1 \mathrm{~kg}\mathrm{m} / \mathrm{s}^{2}=1000 \mathrm{gm} \times 100 \mathrm{~cm} / \mathrm{s}^{2}\)

\( \Rightarrow 10^{5} \mathrm{gm}\mathrm{cm} / \mathrm{s}^{2}=\) \(10^{5}\) dyne

\(\Rightarrow 1\) dyne \(=10^{-5} \mathrm{~kg}\mathrm{m} / \mathrm{s}^{2} \)

\(\Rightarrow 10^{-5} \mathrm{~N}\)

\(\Rightarrow 1 \mathrm{~N}=10^{5}\) dyne

We know that:

Permittivity of free space is given by:

\(\epsilon_{0}=\frac{ q _{1} q _{2}}{4 \pi Fr ^{2}}\)

Dimensions of \([ F ]=\left[ M L T ^{-2}\right]\)

where, M means mass, L means length, T means time.

Here, Dimension of \(q_{1}\) = Dimension of \(q_{2}\)

Dimensions of \((q_{1})=A T\)

Dimensions of \((q_{2})=A T\)

Dimensions of \(( r )=[ L ]\)

Thus, dimensional formula of \(\left[\epsilon_{0}\right]=\frac{[ AT ][ A T ]}{\left[ MLT ^{-2}\right]\left[ L ^{2}\right]}\)

\(\Rightarrow\left[\epsilon_{0}\right]=\left[ M ^{-1} L ^{-3} T ^{4} A ^{2}\right]\)

\(\mathrm{I}=10 \mathrm{~A}\)

\(\mathrm{f}=50 \mathrm{~Hz}\)

\(\mathrm{I}_{\mathrm{rms}}=\left(\frac{\mathrm{I}_{\mathrm{m}}}{\sqrt{2}}\right)\)

where \(I_{m}\) is peak value \(I_{m}=\sqrt{2} \cdot I_{\text {rms }}\)

\(\therefore I_{m}=10 \sqrt{2} A=10 \times 1.41=14.1 \mathrm{~A}\)

Time taken to reach form \(\mathrm{0}\) to \(\mathrm{I}_{\mathrm{m}}\) is \([(\frac{\mathrm{T}} { 4})-0]=(\frac{\mathrm{T}}{4})\)

\(\mathrm{t=(\frac{T}{ 4})=(\frac{1} { 4 f})}\)

\(\therefore \mathrm{t}=[\frac{1}{ \{4 \times 50}\}]=[\frac{1} { 200}] \mathrm{sec}\)

\(\therefore \mathrm{t}=0.5 \times 10^{-2} \mathrm{sec}\)

\(\mathrm{t}=5 \times 10^{-3} \mathrm{sec}\)

\(\begin{aligned} & T \cos \theta=m g \\ & T \sin \theta=\frac{k q^2}{x^2} \\ & \tan \theta=\frac{k q^2}{x^2 m g} \\ & \text { as } \tan \theta \approx \sin \theta \approx \frac{x}{2 L} \\ & \frac{x}{2 L}=\frac{K q^2}{x^2 m g} \\ & x=\left(\frac{q^2 L}{2 \pi \varepsilon_0 m g}\right)^{1 / 3}\end{aligned}\)

The equation of motion,

\(s=ut+\frac{1}{2} at^{2}\)

\(\Rightarrow s=0+\frac{1}{2} at^{2}\)

\(\Rightarrow s=\frac{1}{2} a t^{2}\)

The above equation is similar to the fundamental equation of the parabola about the y-axis. As shown in the graph:

When heat radiation falls on a surface, then itgives energy and exerts pressure.

• Heat radiation is thermal energy carried in the form of electromagnetic waves.
• Since it is a form of electromagnetic radiation, they also exert radiation pressure on the surface on which it falls.
• Therefore, heat radiation gives energy and exerts pressure as well.