JEE Advanced – JEE First

A solid sphere of radius $R$ and density $\rho$ is attached to one end of a mass-less spring of force constant $k$ . The other end of the spring is connected to another solid sphere of radius $R$ and density $3 \rho$ . The complete arrangement is placed in a liquid of density $2 \rho$ and is allowed to reach equilibrium. The correct statement(s) is (are)

the net elongation of the spring is $\frac{4 \pi R^{3} \rho g}{3 k}$ .
the net elongation of the spring is $\frac{8 \pi R^{3} \rho g}{3 k}$
the light sphere is partially submerged.
the light sphere is completely submerged.

Solution

Recall that the buoyant force experienced by …

Bernoulli’s principle in a spray gun

July 9th, 2022jeefirst0 Comments

JEE Advanced 2014 Paper 2, Questions 13 and 14

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are $20 {\rm~mm}$ and $1 {\rm~mm}$ , respectively. The upper end of the container is open to the atmosphere.

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Q.1…

Power dissipated in resistance networks

July 8th, 2022jeefirst0 Comments

JEE Advanced 2008 Paper 1, Question 25

The figure below shows three resistor configurations R1, R2 and R3 connected to $3 \mathrm{~V}$ battery. If the power dissipated by the configuration R1, R2 and R3 is P1, P2 and P3 respectively, then

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$P1 > P2 > P3$
$P1 > P3 > P2$
$P2 > P1 > P3$
$P3 > P2 > P1$

Solution

The power dissipated by a resistance network is

(1) $\begin{equation*} P = \frac{V^2}{R} . \end{equation*}$

The potential applied to all three networks arre the same, so we just need to find the resistances. Of circuits R2 and R3, the latter has a greater resistance since

(2) $\begin{equation*} % R_3 = 1 \, \Omega + \frac{2 \, \Omega \cdot 2 \, \Omega}{2 \, \Omega + 2 \, \Omega} = 2 \, \Omega , R_3 = 1 \, \Omega + \left( \frac{1}{2 \Omega} + \frac{1}{2 \Omega} \right)^{-1} = 2 \, \Omega , \end{equation*}$

whereas

(3) $\begin{equation*} % R_2 = \frac{1 \, \Omega \cdot 1 \, \Omega}{1 \, \Omega + 1 \, \Omega} = 0.5 \, \Omega . R_2 = \left( \frac{1}{1 \Omega} + \frac{1}{2 \Omega} + \frac{1}{2 \Omega} \right)^{-1} = 0.5 \, \Omega. \end{equation*}$

The first circuit can be redrawn in the following way to make the comparison easier.…

Refracting surface with a coating

June 7th, 2022jeefirst0 Comments

JEE Advanced 2014 Paper 1, Question 9

A transparent thin film of uniform thickness and refractive index $n_{1}=1.4$ is coated on the convex spherical surface of radius $R$ at one end of a long solid glass cylinder of refractive index $n_{2}=1.5$ , as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance $f_{1}$ from the film, while rays of light traversing from glass to air get focused at distance $f_{2}$ from the film. Then

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$\left|f_{1}\right|=3 R$
$\left|f_{1}\right|=2.8 R$
$\left|f_{2}\right|=2 R$
$\left|f_{2}\right|=1.4 R$

Related Problem: Lensing by oil on water

Solution

We can solve this problem using …

Lensing by oil on water

May 23rd, 2022jeefirst0 Comments

JEE Advanced 2011 Paper 2, Question 36

Water (with refractive index $=\frac{4}{3}$ ) in a tank is $18 \mathrm{~cm}$ deep. Oil of refractive index $\frac{7}{4}$ lies on water making a convex surface of radius of curvature $R=6 {\rm ~cm}$ as shown. Consider oil to act as a thin lens. An object ‘ $S$ ’ is placed $24 \mathrm{~cm}$ above water surface. The location of its image is at ‘ $x$ ’ cm above the bottom of the tank. What is ‘ $x$ ’?