Physics Test - 1

The electric and magnetic fields in electromagnetic waves are perpendicular to the wave's direction of travel, and also perpendicular to each other.

We know that,

The electric field is always perpendicular to the magnetic field, and both fields are directed at right-angles to the direction of propagation of the wave.

The wave propagates in the direction \({E} \times {B}\). Electromagnetic waves are a type of transverse wave.

In electromagnetic waves, the electric field is always perpendicular to the magnetic field and both fields are directed at right –angles to the direction of propagation of the wave.

Consider the diagram for reflection at general point A shown below.

As the Incident ray is horizontal 0 deg, and reacted ray is vertical 90 deg.

So the slope at point of reaction should be \(\frac{90-0} { 2}=45^{\circ}\)

The derivative of equation of mirror surface, we get:

\(y=\frac{2 L}{\pi} \sin \left(\frac{\pi}{L} x\right)\)

\(\Rightarrow \frac{d y}{d x}=\frac{2 L}{\pi} \cos \left(\frac{\pi}{L} x\right) \times \frac{\pi}{L}\)

And,

\(I=2 \cos \left(\frac{\pi}{L} x\right)\)

\(\Rightarrow\frac{\pi x}{L}=\frac{\pi}{3}\)

Therefore,

\(y=\frac{2 L}{\pi} \sin \left(\frac{\pi}{L} \times \frac{L}{3}\right)\)

\(\Rightarrow\frac{2 L}{\pi} \times \frac{\sqrt{3}}{2}\)

\(\Rightarrow\frac{\sqrt{3} L}{\pi}\)

Therefore, the point is \(\left(\frac{L}{3}, \frac{\sqrt{3} L}{\pi}\right) .\)

Electric and magnetic field both will be perpendicular to each other and both will also be perpendicular to direction of propagation of wave.

Let

\(\overrightarrow{\mathbf{E}}=\mathbf{E}_{0} \cos (\mathrm{kx}-\omega \mathrm{t}) \hat{\mathbf{j}}\)

\(\vec{B}=\mathrm{B}_{0} \cos (\mathrm{kx}-\omega \mathrm{t}) \hat{\mathrm{k}}\)

Now,

\(\overrightarrow{\mathrm{E}}=\overrightarrow{\mathrm{B}} \times \overline{\mathrm{C}} \quad\left(\begin{array}{l}\overrightarrow{\mathrm{C}} \rightarrow \text { direction of }^{\prime} \mathrm{Cis} \\ \text { propagation direction }\end{array}\right)\)

and \(C=\frac{1}{\sqrt{\mu_{0} \varepsilon_{0}}}\)

So \(, \overline{\mathrm{E}}=\mathrm{E}_{0} \cos \left(\frac{2 \pi}{\lambda} \mathrm{x}-2 \pi \mathrm{v} \mathrm{t}\right) \hat{\mathrm{j}}\)

\(v=\frac{6}{\pi} \times 10^{10} \Rightarrow 2 \pi v=12 \times 10^{10}\)

\(\Rightarrow \lambda=\frac{C}{v}=\frac{3 \times 10^{8}}{6 / \pi \times 10^{10}}\)

\(\frac{2 \pi}{\lambda}=\frac{2 \pi \times 6 \times 10^{10}}{\pi \times 3 \times 10^{8}}=4 \times 10^{2}\)

\(\vec{E}=\mathrm{E}_{0} \cos \left(\left(4 \times 10^{2}\right) \mathrm{x}-\left(12 \times 10^{10}\right) \mathrm{t}\right) \hat{\mathrm{j}}\)

\(\vec{B}=\mathrm{E}_{0} \sqrt{\mu_{0} \varepsilon_{0}} \cos \left(\left(4 \times 10^{2}\right) \mathrm{x}-\left(12 \times 10^{10}\right) \mathrm{t}\right) \hat{\mathrm{k}}\)

Hence, The correct answer is A.

Let us examine the four alternatives separately.

(1) for 1 < A < 50, on fusion mass number of the resulting nucleus will be less than 100.

(2) For 51 < A < 100, on fusion mass number the resulting nucleus is between 100 and 200. B/A increases, energy will be released.

(3) On fission for 100 < A < 200, the mass number for fission nuclei will be between 50 to 100. B/A decreases, no energy will be released.

(4) On fission for 200 < A < 260, the mass number for fission nuclei will be between 100 to 130, B/A will increases, energy will be released.

Hence options D is correct answers.

Force corresponding to potential energy is, \(F=-\frac{d U}{d x}=-k\left[1-\frac{x^{2}}{a^{2}}\right]\)

For equilibrium, \(\mathrm{F}=0 .\) Therefore

\(1-\frac{x^{2}}{a^{2}}=0\)

\(x=\pm a\)

For stable equilibrium,

\(\frac{d^{2} U}{d x^{2}}>0\)

\(\frac{d^{2} U}{d x^{2}}=-\frac{2 x}{a^{2}}\)

\(\left.\frac{d^{2} U}{d x^{2}}\right|_{a+x=+a}=-\frac{2}{a}<0,\) so unstable

\(\left.\frac{d^{2} U}{d x^{2}}\right|_{\text {at } x=-a}=\frac{2}{a}>0,\) so stable

Potential energy at \(\mathrm{x}=-\mathrm{a}\) is

\(U=k\left[-a+\frac{a^{3}}{3 a^{2}}\right]=-\frac{2 k a}{3}\)

Hence, this is the required solution.

Hence options D is correct answers.

At resonance RLC circuit become pure resistive circuit and current in the circuit is given by\(\mathrm{i}=\frac{\mathrm{v}_{\mathrm{R}}}{\mathrm{R}}\left\{\begin{array}{c}\mathrm{V}_{\mathrm{R}}=\text { potential difference } \\ \text { across resistor }\end{array}\right.\)

\(\mathrm{i}=\frac{30}{60}=0.5 \mathrm{Ampere}\)

Also

Potential drop across inductor

\(V_{L}=i X_{L}\)

\(V_{L}=i \omega L\)

\(\Rightarrow L=\frac{V_{L}}{i \omega}=\frac{20}{0.5 \times 2000}=20 \mathrm{mH}\)

At resonance,

\(\omega L=\frac{1}{\omega C}\)

\(\Rightarrow C=\frac{1}{\omega^{2} L}\)

\(=\frac{1}{(2000)^{2} \times\left(20 \times 10^{-3}\right)}\)

\(=12.5 \mu \mathrm{F}\)

Hence, The correct answer is D.

We know that,

Resonance frequency: The frequency of an AC circuit at which the impedance of the circuit becomes minimum or current in the circuit becomes maximum is called resonance frequency.

The resonance frequency is given by:

\(f=\frac{1}{2 \pi \sqrt{L C}}\)

Where \(L\) is inductance and \(C\) is the capacitance of the circuit.

Given that:

Capacitance \((C)=18 \mu {F}=18 \times 10^{-6} {~F}\)

Inductance \(({L})=8 {H}\)

\(f=\frac{1}{2 \pi \sqrt{L C}}=\frac{1}{2 \pi \sqrt{8 \times 18 \times 10^{-6}}}=\frac{1}{2 \pi \sqrt{144} \times 10^{-3}}=\frac{1000}{24 \pi}=\frac{125}{3 \pi}\)

This resonance frequency \((f)=\frac{125}{3 \pi} {Hz}\)

When another battery is connected with reverse polarity, steady state change.

Now, charge on capacitor \(q^{\prime}=2 C V=2 q\)

Change in charge \(\Delta q=-2 q-q=-3 q\)

Heat generation \(=\frac{(\Delta q)^{2}}{2 C}\)

\(=\frac{9 \mathrm{q}^{2}}{2 \mathrm{C}}=\frac{9 \mathrm{C}^{2} \mathrm{V}^{2}}{2 \mathrm{C}}\)

\(=\frac{9}{2} \mathrm{CV}^{2}\)

Hence, The correct answer is D.

For semi circle

\(\ell=\pi r \ldots(i)\)

\(B_{A}=\left(\frac{\mu_{0} i}{4 R}\right)\)

\(=\frac{\mu_{0} i}{4 \cdot C / \pi}\)

\(=\frac{\mu_{0} \pi}{4 !}\)

For square

\(4 a=\ell\)

\(a=\ell / 4\)

\(B_{B}=4\left(\frac{\mu_{0} i}{4 \pi(a / 2)}\right)\)

\(\left[\sin 45^{\circ}+\sin 45^{\circ}\right]\)

\(B_{B}=\frac{2 \mu_{0} i}{\pi(C / 4)}(\sqrt{2})\)

\(B_{B}=\frac{8 \sqrt{2} \mu_{0} i}{\pi \ell}\)

\(\Rightarrow \frac{B_{1}}{B_{B}}=\frac{\pi^{2}}{32 \sqrt{2}}\)

Hence, The correct answer is A.

The current carrying wire is placed at O. It carries a current I₁, which flows out from the plane of paper. Field lines are circular around the wire, due to I₁.

Force on AB = zero, as the magnetic lines are parallel to the arm AB of the loop.

dF = IdIB sinθ .

or dF = IdlB sinθ = zero.

Force on CD = zero, as the magnetic lines of I₁ are antiparallel to the arm CD of the loop.

dF = IdIBsinθ = IdIB sin 180° = zero.

Force on BC is perpendicular to plane of paper outwards, according to Fleming's left hand rule.

Force on CD is perpendicular to plane of paper inwards, according to Fleming's left hand rule.

The force on BC and CD are equal and opposite.

The net force on ABCD is zero.

Torque on loop ABCD:

The equal and opposite forces, on BC and AD, constitute a torque which, as seen from O, will rotate the loop in clockwise direction, along OO' as axis.

Hence option D is the correct answer.

The potential of the grounded sphere has to be zero, as the positive charge approaches the sphere, negative charge will be induced on the surface near to the charge ( bounded charge under the influence of the approaching positive charge so it cannot move). The other face will be positively charged. So, negative charge must flow from earth to the sphere to neutralize the positive charge. This negative charge is acquired by the sphere from the ground, hence current flows into the ground. As the charged particle comes nearer to the sphere, the potential of the sphere due to the positive charge increases and hence more electrons per unit time will flow from the ground to the sphere and thus magnitude of current increases. Hence, it is the correct answer.

Hence option D is correct answer.

Shift in fringes \(=\frac{D}{d}(\mu-1) t\)

\(x=\frac{D}{d}\left(\mu_{1}-1\right) t \ldots \ldots .\) (i)

\(\frac{2} { 3} x=\frac{D}{d}\left(\mu_{2}-1\right) t \ldots \ldots .\) (ii)

Dividing (i) by (ii)

\(\frac{3} { 2}=\left(\frac{\mu_{1}-1}{\mu_{2}-1}\right)\)

\(\Rightarrow \mu_{2}-1=\frac{2( \mu_{1}-1)} { 3}\)

\(\Rightarrow \mu_{2}-1=\frac{2( 1.75-1)} { 3}\)

\(\Rightarrow \mu_{2}=\frac{1.5 }{ 3}+1\)

\(\Rightarrow \mu_{2}=1.5\)

\(V=\sqrt{V_{R}^{2}+V_{L}^{2}}\)

\(100^{2}=10^{2}+V_{L}^{2} V_{L}^{2}=99.5 V\)

\(i=\frac{V}{P}=\frac{10}{60}=6 A \Rightarrow V_{L}=i X_{L} \Rightarrow\)

\(=i(2 \pi \mathrm{V} \mathrm{L})\)

\(99.5=6 \times 2 \times 3.14 \times 50 \times L\)

\(L=0.052 \mathrm{H}\)

Hence, The correct answer is A.

Resistance of \(1^{\mathrm{st}}\) bulb \(\mathrm{R}_{1}=\frac{\mathrm{V}^{2}}{\mathrm{P}_{1}}=\frac{(200)^{2}}{200}\)

\(R_{1}=200\)

Resistance of II'nd bulb \(\mathrm{R}_{2}=\frac{(200)^{2}}{\mathrm{P}_{2}}=\frac{(200)^{2}}{400}\)

\(R_{2}=100\)

When bulbs are in series with 100 V battery.

Current, \(i=\frac{\left(V_{\text {batlery }}\right)}{R_{1}+R_{2}}\)

\(i=\frac{(100)}{100+200}=\frac{1}{3} A\)

\(=0.33 \mathrm{Amp}\)

Hence, The correct answer is C.

Image distance "v' from lens (After refraction from lens only) \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \quad(f=20 \mathrm{cm}, u=-40 \mathrm{cm})\)

\(\Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u}\)

\(\frac{1}{v}=\frac{1}{20}+\left(\frac{-1}{40}\right)\)

\(\mathbf{v}=40 \mathrm{cm}\)

But there is a glass plate so shift in image (I I')

\(=t\left(1-\frac{1}{\mu}\right)\)

\(=3\left(1-\frac{1}{3 / 2}\right)\)

\(=1 \mathrm{cm}\)

So, distance of image from lens

\(=40+1\)

\(=41 \mathrm{cm}\)

Distance from glass plate \(=41-10=31 \mathrm{cm}\)

Hence, The correct answer is C.