Physics Test-11

Given,

\(E=2 \mathrm{~V} \mathrm{~m}\)

\(\varepsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} \cdot \mathrm{m}^{-1}\)

Average,

\(\mathrm{U}=\frac{1}{2} \varepsilon_{0} \bar{E}^{2}\)

\(=\frac{1}{2} \times 8.85 \times 10^{-12} \times(2)^{2}\)

\(=17.71 \times 10^{-12} \mathrm{~J} \mathrm{~m}^{-3}\)

Given,

\(u=20 \mathrm{~m} / \mathrm{sec}\)

\(t=10 \mathrm{sec}\)

\(F=100 \mathrm{~N}\)

\(m=5 \mathrm{~kg}\)

Force,

\(F=m a\)

\(\Rightarrow a=\frac{F}{m}\)

\(\Rightarrow a=\frac{100}{5}\)

\(\Rightarrow a=20 \mathrm{~m} / \mathrm{sec}^{2}\)

By the first law of motion,

\(v=u+a t\)

\(\Rightarrow v=20+20 \times 10\)

\(\Rightarrow v=220 \mathrm{~m} / \mathrm{sec}\)

Given,

In hydrogen atom, the electron makes \(6.6 \times 10^{15}\) revolutions per second around the nucleus in an orbit of radius \(0.5 \times 10^{-10} \mathrm{~m}\).

We know that,

\(I=\frac{Q}{t}=\frac{e}{T}\)

But,\(T=\frac{1}{f}\)

\(\Rightarrow I=e \times f\)

\(\Rightarrow I=1.69 \times 10^{-19} \times 6.6 \times 10^{15}\)

\(\Rightarrow I=1.05 \times 10^{-3} \mathrm{~A}\)

\(\Rightarrow I=1 \mathrm{~mA}\)

According to Einstein's quantum theory light propagates in the form of packets i.e. quanta of energy, which is called a photon.

The rest mass of photons is being zero. It can be shown, according to the relativity theory of light.

According to the relativistic theory equation, the mass of the photon is computed as:

\(m=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\)

\(\Rightarrow m_{0}=m \sqrt{1-\frac{v^{2}}{c^{2}}}\)

When \(v=0\)

So, \(m_{0}=0\)

Where, \(m_{0}\) is the rest mass of the Photon.

Given,

Resistance of coil \(=10 \Omega\)

\(\phi=8 t^{2}-4 t+1\)

We know that,

E.m.f. induced, \(E=-\frac{d \phi}{d t}\)

\(\Rightarrow E=-\frac{d\left(8 t^{2}-4 t+1\right)}{d t}\)

By differentiating, we get

\(E=-16 t+4\)

Here, \(t=0.1 \mathrm{sec}\),

\(E=-1.6+4=2.4 \mathrm{~V}\)

Current induced, \(I=\frac{E}{R}\)

\(\Rightarrow I=\frac{2.4}{10}\)

\(\Rightarrow I=0.24 \mathrm{~A}\)

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

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

\(\Rightarrow m=19.6\)

Given,

The momentum of a body \(=850.00 \mathrm{~kg} \mathrm{~m} / \mathrm{sec}\)

\(m=40.00 \mathrm{~kg}\)

The momentum of a body \(=m v\)

\(m v=850.00\)

\(\Rightarrow 40.00 \times v=850.00\)

\(\Rightarrow v=21.25 \mathrm{~m} / \mathrm{sec}\)

Kinetic Energy \(=\frac{1}{2} m v^{2}\)

\(=\frac{1}{2} \times 40 \times(21.25)^{2}\)

\(=9031.25 \mathrm{~J}\)

Given,

A balloon rises from rest with a constant acceleration \(\frac{g}{8}\).

The velocity of the balloon at height \(h\),

\(v=\sqrt{2\left(\frac{g}{8}\right) h}\)

When the stone released from this balloon, it will go upward with velocity \(=\frac{\sqrt{g h}}{2}\) (same as that of the balloon).

In this condition, time taken by stone to reach the ground,

\(t=\frac{v}{g}\left[1+\sqrt{1+\frac{2 g h}{v^{2}}}\right]\)

\(\Rightarrow t=\frac{\frac{\sqrt{g h}}{2}}{g}\left[1+\frac{2 g h}{\frac{g h}{4}}\right]\)

\(\Rightarrow t=\frac{2 \sqrt{g h}}{g}\)

\(\Rightarrow t=2 \sqrt{\frac{h}{g}}\)

Given,

\(B_{y}=\left(3 \times 10^{-7} \mathrm{~T}\right) \sin \left(10^{3} x+6.28 \times 10^{12} t\right)\)

And the general equation,

\(B_{y}=B_{0} \sin (k x+\omega t)\)

Comparing with the general equation, we get

\(k=10^{3} \mathrm{~m}^{-1}\)

We know that,

Wave number \(\mathrm{k}=\frac{2 \pi}{\lambda}\)

\(\Rightarrow \frac{2 \pi}{\lambda}=10^{3} \mathrm{~m}^{-1}\)

\(\Rightarrow \lambda=\frac{2 \pi}{10^{3}} \mathrm{~m}\)

\(=6.28 \times 10^{-3} \) m

\(=0.63 \) cm

Given,

Wavelength, \(\lambda=2480 \mathrm{~nm}\)

Band gap,

\(E_{g}=\frac{h c}{\lambda}\)

Where,

\(h=\) Planck constant \(=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{Hz}^{-1}\)

\(c=\) The speed of light \(=3 \times 10^{8} \mathrm{~m} / \mathrm{sec}\)

\(1 \mathrm{~V}=\frac{1}{1.6 \times 10^{-19}} \mathrm{eV}\)

\(=\frac{\left(6.63 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{2480 \times 10^{-9} \times 1.6 \times 10^{-19}} \mathrm{eV}\)

\(=0.5 \mathrm{eV}\)

Given,

Electrical field \(E=100 \mathrm{~V} / \mathrm{m}\)

Magnetic field \(B=0.265 \mathrm{~A} / \mathrm{m}\)

The energy flow is given by the Poynting vector,

\(\vec{S}=\vec{E} \times \vec{B}\)

\(\Rightarrow S=E B \sin \phi\)

For electromagnetic waves, E and B are always perpendicular to each other and perpendicular to the direction of propagation.

Then, \(\phi=90^{\circ}\)

\(S=E B\)

\(\Rightarrow S=100 \times 0.265\)

\(\Rightarrow S=26.5 \mathrm{~W} / \mathrm{m}^{2}\)

Given,

\(D=1.5 \mathrm{~m}\)

The separation between the slits, \(d=0.25 \mathrm{~mm}\)

\(\lambda_{1}=480 \mathrm{~nm}=480 \times 10^{-9}\)

\(\lambda_{2}=600 \mathrm{~nm}=600 \times 10^{-9}\)

First Maxima \(=\frac{D \lambda}{d}\)

\(\beta_{1}=\frac{1.5 \times 480 \times 10^{-9}}{0.25 \times 10^{-3}}\)

\(=2.880 \times 10^{-3} \mathrm{~m}\)

\(\beta_{2}=\frac{1.5 \times 600 \times 10^{-9}}{0.25 \times 10^{-3}}\)

\(=3.600 \times 10^{-3} \mathrm{~m}\)

So, \(\beta_{2}-\beta_{1}=0.72 \times 10^{-3} \mathrm{~m}\)

\(=0.72 \mathrm{~mm}\)

Given,

\(A B=r=90 \mathrm{~cm}=0.9 \mathrm{~m}\)

\(q_{A}=10 \mu C=10 \times 10^{-6} C\)

\(q_{B}=40 \mu C=40 \times 10^{-6} C\)

\(A C=?\)

At point C,

\(E_{A}=E_{B}\)

\(\frac{q_{A}}{4 \pi \varepsilon_{0}(A C)^{2}}=\frac{q_{B}}{4 \pi \varepsilon_{0}(B C)^{2}}\)

\(\Rightarrow \frac{q_{A}}{(A C)^{2}}=\frac{q_{B}}{(r-A C)^{2}}\)

\(\Rightarrow \frac{10 \times 10^{-6}}{(A C)^{2}}=\frac{40 \times 10^{-6}}{(0.9-A C)^{2}}\)

\(\Rightarrow \frac{1}{(A C)^{2}}=\frac{4}{(0.9-A C)^{2}}\)

\(\Rightarrow \frac{1}{A C}=\frac{2}{(0.9-A C)}\)

\(\Rightarrow 0.9-A C=2 A C\)

\(\Rightarrow 3 A C=0.9\)

\(\Rightarrow A C=0.3 \mathrm{~m}=30 \mathrm{~cm}\)

Given,

The radius of black coated discs,

\(r_A = 2\) m

\(r_B = 4\) m

\(r_C = 6\) m

Maximum intensity wavelengths,

\(\lambda_A = 300\) nm

\(\lambda_B = 400\) nm

\(\lambda_C = 500\) nm

As we know that from Wein's displacement law,

\(\lambda_{\max } T=\) constant

\(T \propto \frac{1}{\lambda}\)

Also, \(Q=\sigma A T^{4}\)

\(Q \propto \frac{A}{\lambda^{4}}\)

Where, \(A= {\pi}{r^2}\)

Finally,

\(Q \propto \frac{r^2}{\lambda^{4}}\)

Now,

\(Q_{A}: Q_{B}: Q_{C}= \frac{2^{2}}{3^{4}}: \frac{4^{2}}{4^{4}}: \frac{6^{2}}{5^{4}}\)

\(=\frac{4}{81}: \frac{1}{16}: \frac{36}{625}\)

\(\therefore Q_{B}\) is maximum.

From the relation between young's modulus and rigidity modulus, we know that,

\(Y=2 \eta(1+\sigma)\)

Where,

\(Y=\) Young's modulus of the material of the wire

\(\eta=\) modulus of rigidity of the material of the wire

\(\sigma=\) transverse strain of the wire

By the given problem, there is no transverse strain in the wire, so \(\sigma=0\).

So, we have \(Y=2 \eta\) from the above relation.

Here, young's modulus of the material of the wire is \(Y=6 \times 10^{12} \mathrm{Nm}^{-2}\)

Therefore, the value of modulus of rigidity of the material of the wire will be,

\(\Rightarrow \eta=\frac{Y}{2}\)

\(=\frac{6 \times 10^{12}}{2} N m^{-2}\)

\(=3 \times 10^{12} \mathrm{Nm}^{-2}\)