Physics Test - 24

When the coil and a magnet are moved in the same direction and at the same speed, then there is no change in the magnetic field across the coil and hence no induced current. Therefore option 4 is correct.

Whenever the number of magnetic lines of force (magnetic flux) passing through a circuit/coil changes an emf is produced in the circuit called induced emf.

The induced emf is given by the rate of change of magnetic flux linked with the circuit i.e.

\(e=-N \frac{d \Phi}{d t}\)

When a bar magnet is pushed toward the coil, the magnetic field linked with the coil increases, hence induced current/ induced e.m.f. is set up in the coil. Hence galvanometer deflects right.

When the bar magnet is pulled away from the coil, the magnetic field linked with the coil decreases, hence induced current/ induced e.m.f. is set up in the coil. Hence galvanometer deflects left.

When the bar magnet held stationary inside the coil there is no change in the magnetic field, hence no induced current. Such that galvanometer doesn't show any deflection.

Since the velocity of block A is in downward direction so the velocity of block B is in left direction.

For the left pulley velocity of rope on right side is \(V_1\).

For constraint equation,

\(V_{1}+V_{2}=2 V_{A}\)

\(0+V_{1}=2 \times 10\)

\(V_{1}=20 \) m/s

Now, for the block B,

Free body diagram of block B,

From free body diagram of block B,

\(V_{B} \cos 37^{\circ}=V_{1}\)

\(V_{B}=\frac{V_{1}}{\cos 37^{\circ}}\)

\(=\frac{20}{\left(\frac{4}{5}\right)}\)

\(=25\) m/s

Given: Focal length of the objective lens, \(f_{0}=1.25 \mathrm{~cm}\)

The focal length of the eyepiece, \(f_{e}=5 \mathrm{~cm}\)

Least distance of distinct vision, \(d=25 \mathrm{~cm}\)

Angular magnification of the compound microscope \(=30 \mathrm{X}\)

Total magnifying power of the compound microscope, \(m=30\)

The angular magnification of the eyepiece is given by the relation:

\(m_{e}=\left(1+\frac{d}{f_{e}}\right)\)

\(=\left(1+\frac{25}{5}\right)\)

\(=6\)

The angular magnification of the objective lens \(\left(m_{0}\right)\) is related to \(m_{e}\) as:

\(m_{0} m_{e}=m\)

\(m_{o}=\frac{m}{m_{e}}\)

\(=\frac{30}{6}\)

\(=5\)

We also have the relation:

\(m_{o}=\frac{\text { Image distance for the objective lens }\left(v_{o}\right)}{\text { Object distance for the objective lens }\left(-u_{o}\right)}\)

\(5=\frac{v_{0}}{-u_{0}}\)

Therefore, \(v_{0}=-5 u_{0} \ldots \ldots . . .(1)\)

Applying the lens formula for the objective lens:

\(\frac{1}{f_{0}}=\frac{1}{v_{0}}-\frac{1}{u_{o}}\)

\(\frac{1}{1.25}= \frac{1}{-5u_{0}}-\frac{1}{u_{o}}\quad\because{[v_{0}=-5 \mathrm{u}_{0}] from}(1)\)

\(\frac{1}{1.25}=\frac{-6}{5 u_{0}}\)

Therefore, \(\mathrm{u}_{0}=\frac{-6}{5} \times 1.25\)

\(\mathrm{u}_{0}=-1.5 \mathrm{~cm} \)

\(\text { And } v_{0}=-5 \mathrm{u}_{0} \)

\(=-5 \times(-1.5) \)

\(=7.5 \mathrm{~cm}\)

The object should be placed \(1.5 \mathrm{~cm}\) away from the objective lens to obtain the desired magnification. Applying the lens formula for the eyepiece:

\(\frac{1}{f_{e}}=\frac{1}{v_{e}}-\frac{1}{u_{e}}\)

Where,

\(v_{e}=\) Image distance for the eyepiece \(=-d=-25 \mathrm{~cm}\)

\(\mathrm{u}_{\mathrm{e}}=\) Object distance for the eyepiece

\(\frac{1}{u_{e}}=\frac{1}{v_{e}}-\frac{1}{f_{c}} \)

\(\frac{1}{u_{e}}=\frac{-1}{25}-\frac{1}{5}\)

\(\frac{1}{u_{e}}=\frac{-6}{25}\)

Therefore, \(\mathrm{u}_{\mathrm{e}}=-4.17 \mathrm{~cm}\)

Separation between the objective lens and the eyepiece \(=\left|u_{e}\right|+\left|v_{o}\right|\)

\(=4.17+7.5 \)

\(=11.67 \mathrm{~cm}\)

Therefore, \(11.67 \mathrm{~cm}\) is the separation between the objective lens and the eyepiece.

Electricity and magnetism are complementary to each other.

Changing magnetic field produces electric current that is it produces electric field. And moving charges that is changing electric field produces magnetic effect like moving charges produces magnetic field.

Let \(\mathrm{d}(\mathrm{w})\) and \(\mathrm{d}(\mathrm{o})\) be densities of water and oil resp, then the pressure at the bottom of the tank will be

\(=\mathrm{h}(\mathrm{w}) \times \mathrm{d}(\mathrm{w}) \times \mathrm{g}+\mathrm{h}(\mathrm{o}) \times \mathrm{d}(\mathrm{o}) \times \mathrm{g}\)

Let this pressure be equivalent to pressure due to water of height h

\(\mathrm{h\times d}(\mathrm{w}) \times \mathrm{g}=\mathrm{h}(\mathrm{w}) \times \mathrm{d}(\mathrm{w}) \times \mathrm{g}+\mathrm{h}(\mathrm{o}) \times \mathrm{d}(\mathrm{o}) \times \mathrm{g}\)

\(h=h(w)+[h(o) \times \frac{d(0)} {d(w)}]\)

\(\mathrm{h}=100+[\frac{400(0.9)} {1}]\)

\(\mathrm{h}=100+360=460 \mathrm{~cm}\)

According to Toricelli's theorem

\(v=\sqrt{2 g h} v\)

\(=\sqrt{2 \times 980 \times 460} v\)

\(=\sqrt{920 \times 980} v\)

\(=\sqrt{901600}\)

Velocity of efflux \(=949.53 \frac{{~cm}}{{s}}\)

For \(\mathrm{A}, \mathrm{n}=2\)

\(B, n=3 \)

\(C, n=4 \)

\(\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) \)

\(\frac{1}{\lambda_2}=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right) \)

\(\frac{1}{\lambda_2}=\frac{7 R}{144} \ldots(1) \)

\(\frac{1}{\lambda_1}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \)

\( \frac{1}{\lambda_1}=\frac{5 R}{36} \ldots(2)\)

(1) and (2)

\( \frac{\lambda_1}{\lambda_2}=\frac{7}{20}=\frac{7}{4 \times 5} \)

\( n=5\)

Applying energy conservation from (1) to (2)

\(\frac{1}{2} m \cdot\left(\frac{2 G M}{R_e}\right)-\frac{G M m}{R_e}=\frac{1}{2} m v^2-\frac{G M m}{R+r} \)

\(\Rightarrow \frac{1}{2} m v^2=\frac{G M m}{R+r} \)

\( v=\sqrt{\frac{2 G M}{R+r}}=\frac{d r}{d t} \)

\(\sqrt{2 G M} \int_0^t d t=\int_{R_e}^{R_e+h}(\sqrt{R+r}) d r \)

\(\sqrt{2 G M} \cdot t=\frac{2}{3}\left[(R+r)^{3 / 2}\right]_{R_e}^{R_e+h} \)

\(t=\frac{2}{3} \sqrt{\frac{R_e^3}{2 G M}}\left[\left(1+\frac{h}{R_e}\right)^{3 / 2}-1\right] \)

\( \frac{G M}{R_e^2}=\mathrm{g} \)

\( t=\frac{1}{3} \sqrt{\frac{2 R_e}{g}}\left[\left(1+\frac{h}{R_e}\right)^{3 / 2}-1\right]\)

The elemental semiconductor is made up of a single element from the fourth column elements such as Germanium. Here recombination takes place takes place via traps. It is called indirect band gap semiconductors.

Germanium \((Ge)\) is a group-\(IV\) indirect band gap semiconductor but the difference between its direct and indirect band gap is only \(140meV\).

Given,

\(\lambda=589 \times 10^{-9} \mathrm{~m}\)

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

\(\mathrm{P}=100 \mathrm{~W}\)

Energy of a photon,

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

\(=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{589 \times 10^{-9}}\)

\(=3.38 \times 10^{-19}\)

Number of photons deliverved per second,

\(\mathrm{n}=\frac{\mathrm{P}}{\mathrm{E}}\)

\(=\frac{100}{3.38 \times 10^{-19}}=3 \times 10^{20}\)