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JEE Advanced Physics Flashcards
54 question-and-answer cards covering Physics as it is examined in JEE Advanced. 24 of them are printed below, taken from across the deck — no signup, no paywall on the preview.
24 sample cards from the Physics deck
Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.
State the mirror formula and its sign convention basis.
$$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$ Using the Cartesian sign convention, distances measured against the incident light are negative; for a concave mirror $f$ is negative.
State the thin lens formula used in the $u$-$v$ method.
$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$ For a convex lens, real object and real image give $u$ negative and $v$ positive, yielding a positive focal length $f$.
In the $u$-$v$ method, how is focal length found from a graph of $\frac{1}{v}$ versus $\frac{1}{u}$?
The plot is a straight line whose intercepts on both axes have magnitude $\dfrac{1}{f}$; thus $f$ is the reciprocal of the magnitude of either intercept.
In the $u$-$v$ method, how does a graph of $v$ versus $u$ give the focal length?
The $v$–$u$ curve is a rectangular hyperbola; the line $v=u$ meets it at the point $(2f,2f)$, so the focal length is half that coordinate: $f=\dfrac{\text{coordinate}}{2}$.
State the resonance condition (first and second resonance) for a closed (resonance column) air pipe.
For a closed pipe, resonance lengths are $l_1=\dfrac{\lambda}{4}-e$ and $l_2=\dfrac{3\lambda}{4}-e$, where $e$ is the end correction. Subtracting: $l_2-l_1=\dfrac{\lambda}{2}$.
How is the speed of sound found in the resonance column experiment?
$$v=f\lambda=2f(l_2-l_1)$$ since $\lambda=2(l_2-l_1)$; here $f$ is the tuning-fork frequency and $l_1,l_2$ the first two resonance lengths.
What is the end correction in the resonance column experiment, and how is it obtained?
The end correction $e$ accounts for the antinode forming slightly above the open tube end. $e=\dfrac{l_2-3l_1}{2}$; it is approximately $0.3\,d$, where $d$ is the inner diameter.
State Ohm's law and the condition under which it holds.
At constant temperature, the current through a conductor is directly proportional to the potential difference across it: $V\propto I$, i.e. $V=IR$, where $R$ is constant (the resistance).
In verifying Ohm's law, how are the voltmeter and ammeter connected, and why?
The voltmeter is connected in parallel across the resistor (it has high resistance, drawing negligible current), and the ammeter in series with it (low resistance, minimal voltage drop).
What does the $V$-$I$ graph for an ohmic conductor look like, and how is resistance obtained?
It is a straight line through the origin; the resistance equals the slope: $R=\dfrac{V}{I}=\text{slope of }V\text{ vs }I$.
Define specific resistance (resistivity) and give its relation to resistance.
Resistivity $\rho$ is the resistance of a conductor of unit length and unit cross-sectional area: $R=\dfrac{\rho L}{A}$, so $\rho=\dfrac{RA}{L}$. SI unit: $\Omega\,\text{m}$.
State the balance condition of a metre (Wheatstone) bridge for finding an unknown resistance.
At balance (null deflection), with balancing length $l$ from one end, $$\frac{R}{S}=\frac{l}{100-l}\quad\Rightarrow\quad S=R\,\frac{100-l}{l}.$$
How is the resistivity of a wire computed after finding its resistance $X$ on a metre bridge?
$$\rho=\frac{X\,\pi r^{2}}{L}=\frac{X\pi d^{2}}{4L}$$ where $r$ (or diameter $d$, from screw gauge) is the wire radius and $L$ its length.
What is a post office box and what does it measure?
A post office box is a Wheatstone-bridge-based instrument with ratio arms ($P$, $Q$) and a variable resistance arm ($R$); at balance the unknown $X=\dfrac{Q}{P}R$, used to measure resistance/resistivity.
Distinguish between distance and displacement.
Distance is the total path length travelled (a scalar, always $\geq 0$). Displacement is the shortest straight-line vector from initial to final position; $|\text{displacement}|\leq\text{distance}$.
Define average velocity and instantaneous velocity.
Average velocity $=\dfrac{\Delta \vec{x}}{\Delta t}$. Instantaneous velocity $=\lim\limits_{\Delta t\to 0}\dfrac{\Delta \vec{x}}{\Delta t}=\dfrac{d\vec{x}}{dt}$.
Write the three equations of motion for constant acceleration in one dimension.
$$v=u+at,\qquad s=ut+\tfrac{1}{2}at^{2},\qquad v^{2}=u^{2}+2as.$$
What is the displacement of a uniformly accelerated body in the $n$th second of its motion?
$$s_{n}=u+\frac{a}{2}(2n-1).$$ This gives the distance covered during the $n$th second alone.
For a projectile launched with speed $u$ at angle $\theta$, write the time of flight, maximum height, and horizontal range.
$$T=\frac{2u\sin\theta}{g},\quad H=\frac{u^{2}\sin^{2}\theta}{2g},\quad R=\frac{u^{2}\sin 2\theta}{g}.$$
At what angle is the horizontal range of a projectile maximum, and what is that range?
Range is maximum at $\theta=45^{\circ}$ (since $\sin 2\theta=1$), giving $R_{\max}=\dfrac{u^{2}}{g}$.
What is the shape of a projectile's trajectory, and give its equation.
The path is a parabola: $$y=x\tan\theta-\frac{g x^{2}}{2u^{2}\cos^{2}\theta}.$$
Define angular velocity and relate it to linear speed in circular motion.
Angular velocity $\omega=\dfrac{d\theta}{dt}$ (rad s$^{-1}$). Linear speed $v=r\omega$, where $r$ is the radius of the circular path.
Write the expression for centripetal acceleration in uniform circular motion.
$$a_c=\frac{v^{2}}{r}=\omega^{2}r$$ directed radially inward (toward the centre of the circle).
Distinguish between centripetal and tangential acceleration in non-uniform circular motion.
Centripetal acceleration $a_c=\dfrac{v^{2}}{r}$ points toward the centre and changes the direction of velocity; tangential acceleration $a_t=\dfrac{dv}{dt}$ is along the path and changes the speed. Net $a=\sqrt{a_c^{2}+a_t^{2}}$.
What this deck covers
The Physics deck follows the JEE Advanced Physics syllabus — 7 chapters and 23 topics — so questions land on material that is genuinely examinable rather than trivia around it. That works out to roughly 7.7 cards per chapter.
Answers are written to be recallable, not just readable — averaging about 148 characters, which is long enough to carry the reasoning and short enough to say out loud.
A deck like this earns its keep on the second and third pass. Read the syllabus first so you know the shape of the subject, then use the cards to find the specific facts that have not stuck.
Physics flashcards FAQ
How many Physics flashcards are in this JEE Advanced deck?
54 cards. This page previews 24 of them, sampled evenly across the deck so you can judge the difficulty before installing anything.
Are these JEE Advanced flashcards free?
Yes. The preview here is free to read with no signup, and the full 54-card deck is free inside the Examius app.
What do the Physics cards cover?
They follow the JEE Advanced Physics syllabus — 7 chapters and 23 topics — so the questions track what is actually examinable.
How should I use these flashcards?
Read the syllabus first so you know the shape of the subject, then drill the deck. Examius schedules each card with spaced repetition, so cards you keep missing come back sooner and ones you know drift further apart.