A circular wire loop of radius \(R\) is placed in the \(x-y\) plane centered at the origin \(O\). A square loop of side \(a(a< < R)\) having two turns is placed with its centre at \(z=\sqrt{3} R\) along the axis of the circular wire loop, as shown in figure.

The plane of the square loop makes an angle of \(45^{\circ}\) with respect to the \(z\)-axis. If the mutual inductance between the loops is given by \(\frac{\mu_{0} a^{2}}{2^{p / 2} R}\), then the value of \(p\) is

The figure shows a system consisting of (i) a ring of outer radius \(3 R\) rolling clockwise without slipping on a horizontal surface with angular speed \(\omega\) and (ii) an inner disc of radius \(2 R\) rotating anti-clockwise with angular speed \(\omega / 2\). The ring and disc are separated by frictionless ball bearings. The point \(P\) on the inner disc is at a distance \(R\) from the origin, where \(O P\) makes an angle of \(30^{\circ}\) with the horizontal. Then with respect to the horizontal surface,

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If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation $z =\frac{x}{y}$. If the errors in $x, y \mathrm{a}\mathrm{n}\mathrm{d} z \mathrm{a}\mathrm{r}\mathrm{e} ∆x, ∆y \mathrm{a}\mathrm{n}\mathrm{d} ∆z$ , respectively, then

$z\pm ∆z=\frac{x\pm ∆x}{y\pm ∆y}=\frac{x}{y}\left(1\pm \frac{∆x}{x}\right){\left(1\pm \frac{∆y}{y}\right)}^{-1}$

The series expansion for ${\left(1\pm \frac{∆y}{y}\right)}^{-1}$, to first power in $∆y/y, \mathrm{i}\mathrm{s} 1\mp \left(∆y/y\right)$. The relative errors in independent variables are always added. So the error in $z$ will be

$\text{Δ}z=z\left(\frac{\text{Δ}x}{x}+\frac{\text{Δ}y}{y}\right).$
Question :
The above derivation makes the assumption that $\frac{\text{Δ}r}{x}\ll 1,\frac{\text{Δ}y}{y}\ll 1$. Therefore, the higher powers of these quantities are neglected.

A symmetric star shaped conducting wire loop is carrying a steady state current $I$ as shown in the figure. The distance between the diametrically opposite vertices of the star is $4a$ .

A planar structure of length $\mathrm{L}$ and width $\mathrm{W}$ is made of two different optical media of refractive indices ${\mathrm{n}}_1=1.5$ and ${\mathrm{n}}_2=1.44$ as shown in figure. If $\mathrm{L}>>\mathrm{W},$ a ray entering from end $\mathrm{A}\mathrm{B}$ will emerge from end $\mathrm{C}\mathrm{D}$ only if the total internal reflection condition is met inside the structure. For $\mathrm{L}=9.6\mathrm{}\mathrm{m},$ if the incident angle $\mathrm{\theta }$ is varied, the maximum time taken by a ray to exit the plane $\mathrm{C}\mathrm{D}$ is $\mathrm{t}\times {10}^{-9}\mathrm{s},$ where $\mathrm{t}$ is ____.
[Speed of light $\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}]$

Two vectors $\overrightarrow{A} \mathrm{a}\mathrm{n}\mathrm{d} \overrightarrow{B}$ are defined as $\overrightarrow{A}=a\widehat{i} and \overrightarrow{B}=a\left(\cos \omega t \widehat{i}+\sin \omega t\widehat{j}\right)$, where a is a constant and $\omega =\frac{\pi }{6} rad s^{-1}.$ If $\left|\overrightarrow{A}+\overrightarrow{B}\right|=\sqrt{3}\left|\overrightarrow{A}-\overrightarrow{B}\right|$ at time $t=\tau$ for the first time, the value of $\tau$ in seconds, is_______

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\(\mathbf{P}_{17 \text { - } 19}\) : Paragraph for Questions Nos. 17 to 19 Two discs \(A\) and \(B\) are mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I, respectively about the common axis. Disc \(A\) is imparted an initial angular velocity \(2 \omega\) using the entire potential energy of a spring compressed by a distance \(x_1\). Disc \(B\) is imparted an angular velocity \(\omega\) by a spring having the same spring constant and compressed by a distance \(x_2\). Both the discs rotate in the clockwise direction.Question:
The ratio \(\frac{x_1}{x_2}\) is

The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction ${}_7{}^{16}\mathrm{N}+{}_2{}^4\mathrm{He}\to {}_1{}^1\mathrm{H}+{}_8{}^{19}\mathrm{O}$ in a laboratory frame is $n$ (in $\mathrm{MeV}$). Assume that ${}_7{}^{16}\mathrm{N}$ is at rest in the laboratory frame. The masses of ${}_7{}^{16}\mathrm{N}, {}_2{}^4\mathrm{He}, {}_1{}^1\mathrm{H}$ and ${}_8{}^{19}\mathrm{O}$ can be taken to be $16.006 \mathrm{u}, 4.003 \mathrm{u}$, $1.008 \mathrm{u}$ and $19.003 \mathrm{u}$, respectively, where $1 \mathrm{u}=930 \mathrm{MeV} {\mathrm{c}}^{-2}$. The value of $n$is
If the numerical value has more than two decimal places, truncate/round-off the value to TWO decimal places.

The total number of carboxylic acid groups in the product P is

Let $E_1$ and $E_2$ be two ellipse whose centers are at the origin. The major axes of $E_{1}and E_2$ lie along the x - axis and the y - axis, respectively. Let S be the circle $x^2+{\left(y-1\right)}^2=2.$ The straight line $x+y=3$ touches the curves $S, E_1$ and $E_2$ at $P, Q$ and $R,$ respectively. Suppose that $PQ=PR= \frac{2\sqrt{2}}{3}.$ If$e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2,$respectively, then the correct expression(s) is(are)

The value of the integral $\underset{0}{\overset{\frac{1}{2}}{\int }}\frac{1+\sqrt{3}}{{\left({\left(x+1\right)}^2{\left(1-x\right)}^6\right)}^{\frac{1}{4}}} dx$ is

A rod of length $2 \mathrm{cm}$ makes an angle $\frac{2\pi }{3}\mathrm{rad}$ with the principal axis of a thin convex lens. The lens has a focal length of $10 \mathrm{cm}$ and is placed at a distance of $\frac{40}{3} \mathrm{cm}$ from the object as shown in the figure. The height of the image is $\frac{30\sqrt{3}}{13} \mathrm{cm}$ and the angle made by it with respect to the principal axis is \(\alpha\) rad. The value of $\alpha$ is $\frac{\pi }{n}\mathrm{rad}$, where $n$ is

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Carbon-14 is used to determine the age of organic material. The procedure is based on the formation of \({ }^{14} \mathrm{C}\) by neutron capture in the upper atmosphere.
\(
{ }_7^{14} \mathrm{~N}+{ }_0^1 n \longrightarrow{ }_6^{14} \mathrm{C}+{ }_1 n^1
\)
\({ }^{14} \mathrm{C}\) is absorbed by living organisms during photosynthesis. The \({ }^{14} \mathrm{C}\) content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of \({ }^{14} \mathrm{C}\) in the dead being, falls to the decay which \({ }^{14} \mathrm{C}\) undergoes.
\(
{ }_6^{14} \mathrm{C} \longrightarrow{ }_7^{14} \mathrm{~N}+\beta^{-}
\)
The half life period of \({ }^{14} \mathrm{C}\) is 5770 years. The decay constant \((\lambda)\) can be calculated by using the following formula \(\lambda=\frac{0.693}{t_{1 / 2}}\).
The comparison of the \(\beta^{-}\)activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of \({ }^{14} \mathrm{C}\) to \({ }^{12} \mathrm{C}\) in living matter is \(1: 10^{12}\).
Question:
A nuclear explosion has taken place leading to increase in concentration of \(C^{14}\) in nearby areas. \(\mathrm{C}^{14}\) concentration is \(C_1\) in nearby areas and \(C_2\) in areas far away. If the age of the fossil is determined to be \(T_1\) and \(T_2\) at the places respectively then