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Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index \(n_{1}\) surrounded by a medium of lower refractive index \(n_{2}\). The light guidance in the structure takes place due to successive total internal reflections at the interface of the media \(n_{1}\) and \(n_{2}\) as shown in the figure. All rays with the angle of incidence \(i\) less than a particular value \(i_{m}\) are confined in the medium of refractive index \(n_{1}\). The numerical aperture (NA) of the structure is defined as \(\sin i_{m}\).



Question:

If two structures of same cross-sectional area, but different numerical apertures \(N A_{1}\) and \(N A_{2}\left(N A_{2} < N A_{1}\right)\) are joined longitudinally, the numerical aperture of the combined structure is

The work done on a particle of mass m by a force $\text{K}\left[\frac{\text{x}}{{\left({\text{x}}^2+{\text{y}}^2\right)}^{3/2}}\widehat{\text{i}}+\frac{\text{y}}{{\left({\text{x}}^2+{\text{y}}^2\right)}^{3/2}}\widehat{\text{j}}\right]$ (K being a constant of appropriate dimensions, when the particle is taken from the point (a,0) to the point (0,a) along a circular path of radius a about the origin in the x-y plane is

The isotope ${}_5{}^{12}B$ having a mass 12.014 u undergoes $\beta$ - decay to ${}_6{}^{12}C . {}_6{}^{12}C$ has an excited state of the nucleus $\left({}_6{}^{12}{C}^{*}\right)$ at 4.041 MeV above its ground state. If ${}_5{}^{12}B$ decays to ${}_6{}^{12}{C}^{*},$ the maximum kinetic energy of the $\beta$ - particle in units of MeV is ( $1\mu =931.5 Me\frac{V}{c^2},$ where c is the speed of light in vacuum).

Look at the drawing given in the figure, which has been drawn with ink of uniform line-thickness.

The mass of ink used to draw each of the two inner circles, and each of the two line segments is \(m\). The mass of the ink used to draw the outer circle is \(6 \mathrm{~m}\). The coordinates of the centres of the different parts are, outer circle 10 , \(0)\), left inner circle \((0,0)\), right inner circle \((a, a)\), vertical line \((0,0)\) and horizontal line \((0,-a)\). The \(y\)-coordinate of the centre of mass of the ink in this drawing is

Let us consider a system of units in which mass and angular momentum are dimensionless. If length has dimension of $\mathrm{L},$ which of the following statement (s) is/are correct?

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Consider an evacuated cylindrical chamber of height \(h\) having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius \(r \ll h\). Now a high voltage source \((HV)\) is connected across the conducting plates such that the bottom plate is at \(+V_{0}\) and the top plate at \(-V_{0} .\) Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity)



Question :

Which one of the following statements is correct?

A police car with a siren of frequency \(8 \mathrm{kHz}\) is moving with uniform velocity \(36 \mathrm{~km} / \mathrm{h}\) towards a tall building which reflects the sound waves. The speed of sound in air is \(320 \mathrm{~m} / \mathrm{s}\). The frequency of the siren heard by the car driver is

A solution curve of differential equation $\left(x^2+xy+4x+2y+4\right)\frac{dy}{dx}-y^2=0, x>\mathrm{0 }$ passes through the point (1, 3). Then the solution curve -

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Consider a simple \(R C\) circuit as shown in Figure \(1\).

Process 1: In the circuit the switch \(S\) is closed at \(t=0\) and the capacitor is fully charged to voltage \(V_{0}\) (i.e., charging continues for time \(T>>R C\) ). In the process some dissipation \(\left(E_{D}\right)\) occurs across the resistance \(R\). The amount of energy finally stored in the fully charged capacitor is \(E_{C}\).

Process 2: In a different process the voltage is first set to \(\frac{V_{0}}{3}\) and maintained for a charging time \(T>>R C\). Then the voltage is raised to \(\frac{2 V_{0}}{3}\) without discharging the capacitor and again maintained for a time \(T>>R C\). The process is repeated one more time by raising the voltage to \(V_{0}\) and the capacitor is charged to the same final voltage \(V_{0}\) as in Process 1.

These two processes are depicted in Figure \(2 .\)







Question:

In Process 2, total energy dissipated across the resistance \(E_{D}\) is:

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The amount of energy required to break a bond is same as the amount of energy released when the same bond is formed. In gaseous state, the energy required for homolytic cleavage of a bond is called Bond Dissociation Energy (BDE) or Bond Strength. BDE is affected by \(s\)-character of the bond and the stability of the radicals formed. Shorter bonds are typically stronger bonds. BDEs for some bonds are given below:












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Correct match of the \(\mathbf{C}-\mathbf{H}\) bonds (shown in bold) in Column \(\mathbf{J}\) with their BDE in Column \(\mathbf{K}\) is

The total number of lone pairs of electrons in ${\mathrm{N}}_2{\mathrm{O}}_3$ is

The tangent \(P T\) and the normal \(P N\) to the parabola \(y^2=4 a x\) at a point \(P\) on

${ }^{131}I is$ an isotope of Iodine that $\beta$ decays to an isotope of Xenon with a half-life of $8 days$ . A small amount of a serum labelled with ${ }^{131}I$ is injected into the blood of a person. The activity of the amount of ${ }^{131}I$ injected was $2.4\times {10}^5$ Becquerel (Bq). It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After 11.5 hours, 2.5 ml of blood is drawn from the person's body, and gives an activity of $115 Bq$ . The total volume of blood in the person's body, in litres is approximately (you may use \(\mathrm{e}^{\mathrm{x}} \approx 1+\mathrm{x}\) for \(|\mathrm{x}|<<1\) and \(\ln 2 \approx 0.7\)).