Trick – JEE First

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 ‘ $x$ ’?

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Solution

To find the location of the final image we consider the oil and the water in the tank as a combination of two elements: a lens and a …

Electric flux through a hemisphere

March 24th, 2022jeefirst0 Comments

JEE Advanced 2017 Paper 2, Question 10

A point charge $+Q$ is placed just outside an imaginary hemispherical surface of radius ${\rm R}$ as shown in the figure. Which of the following statements is/are correct?

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The electric flux passing through the curved surface of the hemisphere is $-\frac{Q}{2 \varepsilon_{0}}\left(1-\frac{1}{\sqrt{2}}\right)$
Total flux through the curved and the flat surfaces is $\frac{Q}{\varepsilon_{0}}$
The component of the electric field normal to the flat surface is constant over the surface
The circumference of the flat surface is an equipotential

Solution

In problems like these a beginner may …

Resistor network with mirror symmetry

February 26th, 2022jeefirst0 Comments

JEE Advanced 2012 Paper 1, Question 12

For the resistance network shown in the figure, choose the correct option(s).

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The current through $PQ$ is zero.
$I_1 = 3$ A.
The potential at $S$ is less than that at $Q$ .
$I_2 = 2$ A.

Solution

We can speed up the solution by noticing the mirror symmetry of the circuit. Let the current $I_1$ split into two parts $I_a$ and $I_b$ at the node $X$ (see figure). Similarly, let the currents going into node $Y$ be $I_a'$ and $I_b'$ . If we were to reverse the polarity of the $12 \, {\rm V}$ battery connected to the circuit, all currents would simply reverse direction, but their magnitudes would remain the …

Terminal velocity in a magnetic field

February 23rd, 2022jeefirst0 Comments

A copper connector of mass $m$ slides down two smooth copper bars, set at an angle $\alpha$ to the horizontal, due to gravity (see figure). At the top the bars are interconnected through a resistance $R$ . The separation between the bars is $\ell$ . The system is located in a uniform magnetic field of induction $B$ , perpendicular to the plane in which the connector slides. The resistance of the bars, the connector and the sliding contacts, as well as the self-inductance of the loop are assumed to be negilible. If the connector is released from rest at $t=0$ ,

Find the velocty $v(t)$ of

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