Physics Test - 37

Referring to the figure, suppose the charge \(\mathrm{q}_{0}=\mathrm{q} / 2 \) is given a small displacement \(\mathrm{OC}=\mathrm{y}\) along the \(\mathrm{y}\) -axis. The force of repulsion on charge \(\bar{q}_0\) due to the charge \(q_{1}=q\) is,

\(F_{1}=k \frac{9 q_{0}}{(A C)^{2}}=k \frac{q^{x} q / 2}{\left(a^{2}+y^{2}\right)}\)

The force of repulsion on charge \(\mathrm{q}_{0}\) due to charge \(\mathrm{q}_{2}=\mathrm{q}\) is,

\(\mathrm{F}_{2}=\mathrm{k} \frac{\mathrm{q} \mathfrak{q}}{(\mathrm{BC})^{2}}=\mathrm{k} \frac{\mathrm{q}^{\mathrm{x}} \mathrm{q} / 2}{\left(\mathrm{a}^{2}+\mathrm{y}^{2}\right)}\)

The directions of \(F_{1}\) and \(F_{2}\) are shown in the figure. Force \(F_{1}\) can be resolved into two components \(\left(F_{1}\right) x=F_{1} \sin \theta\) along positive x -direction and \(\left(\mathrm{F}_{1}\right) \mathrm{y}=\mathrm{F}_{1}\) cos \(\theta\) along positive y direction. Similarly, force \(\mathrm{F}_{2}\) can be resolved into two components \(\left(F_{2}\right)_{\mathrm{x}}=\mathrm{F}_{2} \sin \theta\) along negative \(\mathrm{x}\) -direction and \(\left(\mathrm{F}_{2}\right) \mathrm{y}=\mathrm{F}_{2} \cos \theta\) along positive y -direction. since \(\mathrm{F}_{1}=\mathrm{F}_{2}\) components \(\left(\mathrm{F}_{1}\right) \mathrm{x}\) and \(\left(\mathrm{F}_{2}\right) \mathrm{x}\) are equal and opposite and hence, they cancel each other. The net force on charge \(\mathrm{q}_{0}\) is along positive y -direction and is given by;

We know that:The equation for SHM is:

\(\mathrm{y}=\operatorname{Asin}(\omega+\phi)\)

As the displacement is half of the amplitudes so:\(\left(\mathrm{y}=\frac{\mathrm{A}}{2}\right)\)

or \(\frac{\mathrm{A}}{2}=\mathrm{A} \sin (\omega \mathrm{t}+\)\(\phi) \)

\((\omega t+\phi)=\frac{1}{2} \)

\(\therefore \omega \mathrm{t}+\phi=30^{\circ} \text { or } 150^{\circ}\)

Since the two particles are going in opposite directions, the phase of one is \(30^{\circ}\) and that of the other \(150^{\circ} .\)

So the phase difference between the two particles:

\(=150^{\circ}-30^{\circ}\)

\(=120^{\circ}\)

\(=\frac{2 \pi}{3}\)

The electric field at a point is continuous if there is no charge at that point and discontinuous if there is a charge at that point.

Electric field is a region of space in which an electric charge experiences a force. The direction of the electric field at a point in space is the direction in which a positive test charge moves if placed at that point.

The electric field at a point is continuous only when there is no negative or positive charge in its vicinity that affects its pathway. If there is no other charge in the medium, The electric field because of any charge will be continuous. It will be discontinuous if there is a charge at that viable point.

Charged particles tend to change the emergence and divergence of continuous electric field lines.

Hydraulic pressure exerted on glass slab, \(P\) \(=10\) atm Bulk modulus of glass, \({B}=37 \times 10^{9} ~\mathrm{Nm}^{-2}\)

Bulk modulus, \(\mathrm{B}= \frac{P}{\frac{\Delta {V}} {V}}\)

where, \(\frac{\Delta {V}} {V}=\) Fractional change in volume \(\frac{\Delta {V}} {V}=\frac{P} {B}\)

\(=\frac{10 \times 1.013 \times 10^{5}}{ (37 \times 10^{9})}\)

\(=2.73 \times 10^{-5}\)

Therefore, the fractional change in the volume of the glass slab is \(2.73 \times 10^{-5}\).

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

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

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

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 \([(\frac{\mathrm{T}} { 4})-0]=(\frac{\mathrm{T}}{4})\)

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

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

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

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

Given,

\(m _{ A }=9 \) kg

\(m _{ B }=25\) kg

\(KE _{ A }= KE _{ B }\)

As we know,

The relation between the kinetic energy and the momentum is given as,

\( P ^{2}=2 mKE\)

Momentum for body A, \((P_{A})^{2}=2 m_{A} \times K E_{A}^{2}\)

\(\therefore (P_{A})^{2}=2 \times 9 \times K E\)

\(\Rightarrow (P_{A})^{2}=18 \times K E\)...(1)

Momentum for body B \((P_{B})^{2}=2 m_{B} \times K E_{B}^{2}\)

\(\therefore (P_{B})^{2}=2 \times 25 \times K E\)

\(\Rightarrow (P_{B})^{2}=50 \times K E\)...(2)

On dividing equation (1) and (2), we get

\(\frac{(P_{A})^{2}}{(P_{B})^{2}}=\frac{18 \times K E}{50 \times K E}\)

\(\Rightarrow \frac{(P_{A})^{2}}{(P_{B})^{2}}=\frac{9}{25}\)

\(\Rightarrow \frac{P_{A}}{P_{B}}=\sqrt{\frac{9}{25}}\)

\(\Rightarrow \frac{P_{A}}{P_{B}}=\frac{3}{5}\)

\(\Rightarrow P_{A}: P_{B}=3: 5\)

So, the ratio of linear momentum is \(3:5\).

\(\begin{aligned} & \text { Fringe Width, } \beta=\frac{\lambda \mathrm{D}}{\mathrm{d}} \\ & \beta_{\max } \Rightarrow \mathrm{d}_{\min } \text { and } \beta_{\min } \Rightarrow \mathrm{d}_{\max } \\ & \mathrm{d}=\mathrm{d}_0+\mathrm{a}_0 \sin \omega \mathrm{t} \\ & \mathrm{d}_{\max }=\mathrm{d}_0+\mathrm{a}_0 \text { and } \mathrm{d}_{\min }=\mathrm{d}_0-\mathrm{a}_0 \\ & \therefore \beta_{\min }=\frac{\lambda \mathrm{D}}{\mathrm{d}_0+\mathrm{a}_0} \text { and } \therefore \beta_{\max }=\frac{\lambda \mathrm{D}}{\mathrm{d}_0-\mathrm{a}_0} \\ & \beta_{\max }-\beta_{\min }=\frac{\lambda \mathrm{D}}{\mathrm{d}_0-\mathrm{a}_0}-\frac{\lambda \mathrm{D}}{\mathrm{d}_0+\mathrm{a}_0}=\frac{2 \lambda \mathrm{Da}_0}{\mathrm{~d}_0^2-\mathrm{a}_0^2}\end{aligned}\)

The nucleus of a heavy atom (such as uranium, plutonium or thorium), when bombarded with low-energy neutrons, can be split apart into lighter nuclei. This process is called nuclear fission.

Fission reaction can further be classified into controlled and uncontrolled fission reaction.

In controlled fission, the chain reaction is controlled and only a controlled amount of reaction is allowed, nuclear reactors in nuclear power plants are one of the examples of the controlled fission reaction.

And for an uncontrolled fission chain reaction, is allowed to happen unless fission material is over, atomic bomb is one of the examples of an uncontrolled fission reaction.