Physics Test - 22

Given,

\(\mathrm{I}=1.388 \times 10^{3} \mathrm{Wm}^{-2}\)

\(\lambda=550 \times 10^{-9} \mathrm{~m}\)

\(\mathrm{~h}=6.63 \times 10^{-34} \mathrm{Js}\)

Number of photos incident on earth's surface per second per square metre is,

\(\mathrm{n}=\frac{\mathrm{I}}{\mathrm{E}}=\frac{\mathrm{I} \lambda}{\mathrm{hc}}\)

As we know,

\(\mathrm{E}=\frac{h \mathrm{c}}{\lambda}\)

\(=\frac{1.388 \times 10^{3} \times 550 \times 10^{-9}}{6.63 \times 10^{-34} \times 3 \times 10^{-8}}=4 \times 10^{21}\)

The angle of incidence at which a beam of unpolarized light falling on a transparent surface is reflected as a beam of completely plane polarised light is called polarising or Brewster angle. It is denoted by \(\mathrm{i}_{P}\).

Brewster's law: It states that when a ray is passed through some transparent medium having refractive index \(\mu\) at any particular angle of incidence, the reflected ray is completely polarized; and the angle between reflected and refracted ray is \(90^{\circ}\).

\(\mu=\tan \theta_{\mathrm{B}}\)

Where \(\mu=\) refractive index and \(\theta_{\mathrm{B}}\) is Brewster's angle or polarizing angle \(\left(\mathrm{i}_{\mathrm{p}}\right)\).

Given that:

The light is incident on the glass.

The refractive index of glass \((\mu)=\frac{3 }{2}\)

The angle of incidence \(\left(\mathrm{i}_{\mathrm{p}}\right)=?\)

\(\frac{3}{2}=\tan \theta \)

\(\Rightarrow \theta=\tan ^{-1}\left(\frac{3}{2}\right)\) \(\Rightarrow \theta=56.30^{\circ} \approx 57^{\circ}\)

We know that,

Power \((P)=\) Voltage \(({V}) \times\) Current \(({I})\)

From ohms law, \(V=I R\)

So, \(P=\frac{V^2}R\)

Now,

\(P=500 {~W}\)

\(V=100\) volt

\(P=\frac{V^2}R\)

\(\Rightarrow R=\frac{(100 \times 100)}{500}=20\) ohm

Now, When \({V}=150 {~v}\)

Power \(=\frac{(150 \times 150)}{20}\)

\(=\frac{2250}2\)

\(=1125\) Watt

• Ground wave propagation is mode of wave propogation in which the ground has a strong influence on the propogation of signal waves from the transmitting antenna to receiving antenna. In this propogation, the signal wave glides over the surface of earth.
• Skywave propagation is a mode of wave propogation in which the radiowaves emitted from the transmitter antenna reach the receiving antenna after reflection by the ionosphere.
• Space wave propagation is mode of wave propogation in which the radio waves emitted from transmitter antenna reach the receiving antenna directly through space.These radiowaves are called space waves or tropospheric waves.

By the law of conservation of energy,

\(\frac{1}{2} \mathrm{mV}^{2}-\frac{\mathrm{GMm}}{\mathrm{R}}=\frac{-\mathrm{GMm}}{\mathrm{R}+h}\)

\(\Rightarrow \frac{1}{2} \mathrm{~V}^{2}-\frac{\mathrm{GM}}{\mathrm{R}^{2}} \cdot \mathrm{R}=\frac{-\mathrm{Gm}}{\mathrm{R}^{2}} \frac{\mathrm{R}^{2}}{(\mathrm{R}+h)}\)

\(\Rightarrow \frac{1}{2} \mathrm{~V}^{2}-\mathrm{g} \mathrm{R}=\frac{-\mathrm{gR}^{2}}{(\mathrm{R}+h)}\)

\(\Rightarrow \mathrm{R}+h=\frac{-2 \mathrm{gR}^{2}}{\mathrm{~V}^{2}-2 \mathrm{gR}}\)

\(\Rightarrow h=\frac{-2 \mathrm{gR}^{2}}{\mathrm{~V}^{2}-2 \mathrm{gR}}-\mathrm{R}\)

\(\Rightarrow h=\frac{-\mathrm{RV}^{2}}{\mathrm{~V}^{2}-2 \mathrm{gR}}\)

\(\Rightarrow h=\frac{\mathrm{RV}^{2}}{2 \mathrm{gR}-\mathrm{V}^{2}}\)

In the case of eye-piece lense,

\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

\(\Rightarrow \frac{1}{25}-\frac{1}{u}=\frac{1}{5}\)

\(\Rightarrow \frac{1}{u}=\frac{1}{25}-\frac{1}{5}\)

\(\Rightarrow u=\frac{-5}{4}\)

\(\Rightarrow u=-1.25\) cm

For objective the image distance,

\(v=20-1.25\)

\(=18.75\) cm

Again, by the above formula:

\(\frac{1}{18.75}-\frac{1}{u}=\frac{1}{9.5}\)

\(\Rightarrow \frac{1}{u}=\frac{4}{75}-\frac{2}{19}\)

\(\Rightarrow \frac{1}{u}=\frac{76-150}{75 \times 19}\)

\(\Rightarrow u=\frac{75 \times 19}{-74}\)

\(\Rightarrow u=-19.25 \) cm

Total magnification, 

\(m=m_{1} m_{2}\)

Here, \(m = \frac{v}{u}\)

\(\Rightarrow m=\frac{18.75}{19.25} \times \frac{25}{1.25}\)

\(\Rightarrow m=19.6\)

For a body thrown vertically upwards acceleration remains constant (a = – g) and velocity at anytime t is given by V = u – gt

During rise velocity decreases linearly and during fall velocity increases linearly and direction is opposite to each other.

Hence graph (a) correctly depicts velocity versus time.

\(\begin{aligned} & \mathrm{F}=\eta \mathrm{A} \frac{\mathrm{dv}}{d \mathrm{y}} \\ & {\left[\mathrm{MLT}^{-2}\right]=\eta\left[\mathrm{L}^2\right]\left[\mathrm{T}^{-1}\right]} \\ & \eta=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right] \\ & \mathrm{S} \cdot \mathrm{T}=\frac{\mathrm{F}}{\ell}=\frac{\left[\mathrm{MLT}^{-2}\right]}{[\mathrm{L}]}=\left[\mathrm{ML}^0 \mathrm{~T}^{-2}\right] \\ & \mathrm{L}=\mathrm{mvr}=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right] \\ & \mathrm{K} \cdot \mathrm{E}=\frac{1}{2} \mathrm{I} \omega^2=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\end{aligned}\)