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The current passing through the battery in the given circuit, is:
The 6 Ω arm bridges the balanced Wheatstone network, so it carries no current. The remaining resistors reduce to an effective resistance giving $I = \dfrac{5\ \text{V}}{2.5\ \Omega} = 2.0\ \text{A}$ through the battery.
A bob of heavy mass $m$ is suspended by a light string of length $l$. The bob is given a horizontal velocity $v_0$ as shown in figure. If the string gets slack at some point $P$ making an angle $\theta$ from the horizontal, the ratio of the speed $v$ of the bob at point $P$ to its initial speed $v_0$ is:
String slack at $P$: $mg\sin\theta = \dfrac{mv^2}{l}\Rightarrow v^2 = gl\sin\theta$. Energy from bottom to $P$ (height $l(1+\sin\theta)$): $v^2 = v_0^2 - 2gl(1+\sin\theta)$. Hence $v_0^2 = gl(2+3\sin\theta)$ and $\dfrac{v}{v_0} = \left(\dfrac{\sin\theta}{2+3\sin\theta}\right)^{1/2}$.
The output (Y) of the given logic implementation is similar to the output of an/a ________ gate.
Each NOR gate gives $(A+B)'$. Feeding both identical outputs to a NAND: $Y = \overline{(A+B)'\cdot(A+B)'} = \overline{(A+B)'} = A+B$, i.e. an OR gate.
The electric field in a plane electromagnetic wave is given by $E_z = 60\cos(5x + 1.5\times10^9\,t)\ \text{V/m}$. Then expression for the corresponding magnetic field is (here subscripts denote the direction of the field):
$B_0 = \dfrac{E_0}{c} = \dfrac{60}{3\times10^8} = 2\times10^{-7}\ \text{T}$. With $\vec E$ along $z$ and propagation along $-x$, $\vec B$ is along $y$: $B_y = 2\times10^{-7}\cos(5x+1.5\times10^9 t)$.
A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take $g = 9.8\ \text{m/s}^2$):
$v_{down} = \sqrt{2g(40)} = 28\ \text{m/s}$, $v_{up} = \sqrt{2g(10)} = 14\ \text{m/s}$. Impulse $= m(v_{up}+v_{down}) = 0.5(14+28) = 21\ \text{N·s}$.
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:
$\dfrac{dt}{dx} = 2x+1\Rightarrow v = \dfrac{1}{2x+1}$. $a = v\dfrac{dv}{dx} = \dfrac{1}{2x+1}\cdot\left(-\dfrac{2}{(2x+1)^2}\right) = -\dfrac{2}{(2x+1)^3}$.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.