Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40K?

Two solid spheres \(A\) and \(B\) of equal volumes but of different densities \(d_A\) and \(d_B\) are connected by a string. They are fully immersed in a fluid of density \(d_F\). They get arranged into an equilibrium state as shown in the figure with a tension in the string. The arrangement is possible only if

Figure 1 shows the configuration of main scale and Vernier scale before measurement. Fig. 2 shows the configuration corresponding to the measurement of diameter \(D\) of a tube. The measured value of \(D\) is:

A projectile is thrown from a point \(\mathrm{O}\) on the ground at an angle \(45^{\circ}\) from the vertical and with a speed \(5 \sqrt{2} \mathrm{~m} / \mathrm{s}\). The projectile at the highest point of its trajectory splits into two equal parts. One part falls vertically down to the ground, \(0.5 \mathrm{~s}\) after the splitting. The other part, \(t\) seconds after the splitting, falls to the ground at a distance \(x\) meters from the point O. The acceleration due to gravity \(g=10 \mathrm{~m} / \mathrm{s}^{2}\).
The value of $x$ is ________.

A particle of mass $m$ is attached to one end of a massless spring of force constant $k$, lying on a frictionless horizontal plane. The other end of the spring is fixed. The particle starts moving horizontally from its equilibrium position at time $t=0$ with an initial velocity $u_0$. When the speed of the particle is $0.5u_0$, it collides elastically with a rigid wall. After this collision :

Most materials have the refractive index, \(n>1\). So, when a light ray from air enters a naturally occurring material, then by Snell's

law, \(\frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{n_{2}}{n_{1}}\), it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, \(\mathrm{n}=c / v=\pm \sqrt{\varepsilon_{r} \mu_{r}}\), where \(c\) is the speed of electromagnetic waves in vacuum, \(v\) its speed in the medium, \(\varepsilon_{r}\) and \(\mu_{r}\) are the relative permittivity and permeability of the medium respectively. In normal materials, both \(\varepsilon_{r}\) and \(\mu_{r}\), are positive, implying positive \(n\) for the medium. When both \(\varepsilon_{r}\) and \(\mu_{r}\) are negative, one must choose the negative root of \(n\). Such negative refractive index materials can now be artificially prepared and are called metamaterials. They exhibit significantly different optical behavior, without violating any physical laws. Since \(n\) is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.
Question : Choose the correct statement.

A spherical bubble inside water has radius $R$. Take the pressure inside the bubble and the water pressure to be $p_0.$ The bubble now gets compressed radially in an adiabatic manner so that its radius becomes $(R-a)$. For $a\ll R$ the magnitude of the work done in the process is given by $\left(4\pi P_{0}R{a}^2\right)X,$ where $X$ is a constant and $\gamma =C_p/C_V=41/30$. The value of $X$ is______

Let $\mathrm{g}\mathrm{ }:\mathbb{R}\to \mathbb{R}$ be a differentiable function with $\mathrm{g}\left(0\right)=0,\mathrm{ }\mathrm{ }\mathrm{ }\mathrm{ }{\mathrm{g}}^{\mathrm{\prime }}\left(0\right)=0$ and ${\mathrm{g}}^{\mathrm{\prime }}\left(1\right)\mathrm{ }\ne 0.\mathrm{ }$ Let $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{ }\left\{\begin{array}{cc}\frac{\mathrm{x}}{\left|\mathrm{x}\right|}\mathrm{ }\mathrm{g}\left(\mathrm{x}\right),& \mathrm{x}\mathrm{ }\ne 0\\ 0,& \mathrm{x}=0\end{array}\right.$ and $\mathrm{h}\left(\mathrm{x}\right)={\mathrm{e}}^{\left|\mathrm{x}\right|}$ for all $\mathrm{x}\in \mathbb{R}.$ Let $\left(\mathrm{foh}\right)\mathrm{ }\left(\mathrm{x}\right)$ denote $\mathrm{f}\left(\mathrm{h}\left(\mathrm{x}\right)\right)\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\left(\mathrm{hof}\right)\mathrm{ }\left(\mathrm{x}\right)\mathrm{ }$denote $\mathrm{h}\left(\mathrm{f}\left(\mathrm{x}\right)\right)$ .Then which of the following is (are) true?

List-I includes starting materials and reagents of selected chemical reactions. List-II gives structures of compounds that may be formed as intermediate products and/or final products from the reactions of List-I.

List-I	List-II
(I)	$\left(\mathrm{P}\right)$
(II)	$\left(\mathrm{Q}\right)$
(III)	$\left(\mathrm{R}\right)$
(IV)	$\left(\mathrm{S}\right)$
	$\left(\mathrm{T}\right)$
	$\left(\mathrm{U}\right)$

Which of the following options has correct combination considering List-I and List-II?

The reaction of ${\mathrm{HClO}}_3$ with $\mathrm{HCl}$ gives a paramagnetic gas, which upon reaction with ${\mathrm{O}}_3$ produces

Two spherical stars $A$ and $B$ have densities ${\rho }_A$ and ${\rho }_B$, respectively. $A$ and $B$ have the same radius, and their masses $M_A$ and $M_B$ are related by $M_B=2M_A$. Due to an interaction process, star $A$ loses some of its mass, so that its radius is halved, while its spherical shape is retained, and its density remains ${\rho }_A$. The entire mass lost by $A$ is deposited as a thick spherical shell on $B$ with the density of the shell being ${\rho }_A$. If $v_A$ and $v_B$ are the escape velocities from $A$ and $B$ after the interaction process, the ratio $\frac{v_B}{v_A}=\sqrt{\frac{10n}{{15}^{{\displaystyle \frac{1}{3}}}}}$. The value of $n$ is

A line L : $y=\mathit{\text{m}}x+3$ meets $y-\mathrm{a}\mathrm{x}\mathrm{\text{is}}$ at E(0,3) and the arc of the parabola $y^2=16x\text{,}0\le y\le 6$ at the point $F\left(x_0,y_0\right)$.The tangent to the parabola at $F\left(x_0,y_0\right)$ intersects the y-axis at $G\left(0,y_1\right)$.The slope m of the line L is chosen such that the area of the triangle EFG has a local maximum.
Match List I with List II and select the correct answer using the code given below the lists :
 

	List I		List II
A.	m =	P.	$\frac{1}{2}$
B.	Maximum area of ΔEFG is	Q.	4
C.	y0​ =	R.	2
D.	y1 =	S.	1

Among the following, the coloured compound is