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AB is a part of an electrical circuit (see figure). The potential difference $V_A - V_B$, at the instant when current $i = 2$ A and is increasing at a rate of 1 amp/second is:
Traversing A→B in the current direction the drops add: $V_A - V_B = L\dfrac{di}{dt} + \varepsilon + iR = (1)(1) + 5 + (2)(2) = 10\ \text{V}$.
A 2 amp current is flowing through two different small circular copper coils having radii ratio 1:2. The ratio of their respective magnetic moments will be:
$M = I\pi r^2$ with the same $I$, so $\dfrac{M_1}{M_2} = \dfrac{r_1^2}{r_2^2} = \dfrac{1}{4}$.
In a certain camera, a combination of four similar thin convex lenses are arranged axially in contact. Then the power of the combination and the total magnification in comparison to the power $(p)$ and magnification $(m)$ for each lens will be, respectively:
Powers in contact add: $P = 4p$. Magnifications multiply: $M = m^4$.
An oxygen cylinder of volume 30 litre has 18.20 moles of oxygen. After some oxygen is withdrawn from the cylinder, its gauge pressure drops to 11 atmospheric pressure at temperature $27^\circ$C. The mass of the oxygen withdrawn from the cylinder is nearly equal to: [Given, $R = \frac{100}{12}\ \text{J mol}^{-1}\text{K}^{-1}$, molecular mass of $O_2 = 32$, 1 atm pressure $= 1.01\times10^5\ \text{N/m}^2$]
Absolute final pressure $= 11+1 = 12$ atm. $n_2 = \dfrac{PV}{RT} = \dfrac{(12\times1.01\times10^5)(0.03)}{(100/12)(300)} \approx 14.54$ mol. $\Delta n = 18.20-14.54 = 3.66$ mol; mass $= 3.66\times32 \approx 117\ \text{g} \approx 0.116\ \text{kg}$.
In some appropriate units, time ($t$) and position ($x$) relation of a moving particle is given by $t = x^2 + x$. The acceleration of the particle is:
$\dfrac{dt}{dx} = 2x+1\Rightarrow v = \dfrac{1}{2x+1}$. $a = v\dfrac{dv}{dx} = \dfrac{1}{2x+1}\cdot\left(-\dfrac{2}{(2x+1)^2}\right) = -\dfrac{2}{(2x+1)^3}$.
To an ac power supply of 220 V at 50 Hz, a resistor of $20\ \Omega$, a capacitor of reactance $25\ \Omega$ and an inductor of reactance $45\ \Omega$ are connected in series. The corresponding current in the circuit and the phase angle between the current and the voltage is, respectively:
$Z = \sqrt{R^2+(X_L-X_C)^2} = \sqrt{20^2+20^2} = 20\sqrt2 \approx 28.3\ \Omega$. $I = \dfrac{220}{28.3} \approx 7.8$ A; $\tan\phi = \dfrac{20}{20} = 1\Rightarrow\phi = 45^\circ$.
The Sun rotates around its centre once in 27 days. What will be the period of revolution if the Sun were to expand to twice its present radius without any external influence? Assume the Sun to be a sphere of uniform density.
Angular momentum conserved: $I\omega = \text{const}$, $I = \tfrac25 MR^2\propto R^2$, so $T\propto R^2$. $T_2 = 27\times2^2 = 108$ days.
A model for quantized motion of an electron in a uniform magnetic field $B$ states that the flux passing through the orbit of the electron is $n(h/e)$ where $n$ is an integer, $h$ is Planck's constant and $e$ is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be ($m$ is the mass of the electron):
Flux $B\pi r^2 = n\tfrac{h}{e}$; for $n=1$, $Br^2 = \dfrac{h}{\pi e}$. With $r = \dfrac{mv}{eB}$, $\mu = \tfrac12 evr = \dfrac{e^2Br^2}{2m} = \dfrac{e^2}{2m}\cdot\dfrac{h}{\pi e} = \dfrac{he}{2\pi m}$.
Three identical heat conducting rods are connected in series as shown in the figure. The rods on the sides have thermal conductivity $2K$ while that in the middle has thermal conductivity $K$. The left end of the combination is maintained at temperature $3T$ and the right end at $T$. The rods are thermally insulated from outside. In steady state, temperature at the left junction is $T_1$ and that at the right junction is $T_2$. The ratio $T_1/T_2$ is:
Equal heat current: $2(3T-T_1) = (T_1-T_2) = 2(T_2-T)$. Solving gives $T_1 = \tfrac{5T}{2}$, $T_2 = \tfrac{3T}{2}$, so $\dfrac{T_1}{T_2} = \dfrac{5}{3}$.
The plates of a parallel plate capacitor are separated by $d$. Two slabs of different dielectric constant $K_1$ and $K_2$ with thickness $\frac{3}{8}d$ and $\frac{d}{2}$, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If $K_1 = 1.25\,K_2$, the value of $K_1$ is:
Series of three regions (air gap $= d/8$): $\dfrac{d}{2} = \dfrac{3d}{8K_1}+\dfrac{d}{2K_2}+\dfrac{d}{8}$. With $K_2 = 0.8K_1$: $\dfrac{3}{8K_1}+\dfrac{0.625}{K_1} = \dfrac{3}{8}\Rightarrow K_1 = \dfrac{8}{3} \approx 2.66$.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.