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The output ($Y$) of the given logic gate is similar to the output of an/a:
Boolean simplification of the combinational circuit gives $Y = A \cdot B$, i.e. AND.
In a vernier calipers, $(N + 1)$ divisions of vernier scale coincide with $N$ divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is:
LC $= 1\,\text{MSD}/(N+1) = 0.1/(N+1)$ mm $= 1/[100(N+1)]$ cm.
At any instant of time $t$, the displacement of any particle is given by $2t - 1$ (SI unit) under the influence of force of $5N$. The value of instantaneous power is (in SI unit):
$v = dx/dt = 2$ m/s; $P = Fv = 10$ W.
A thin spherical shell is charged by some source. The potential difference between the two points $C$ and $P$ (in $V$) shown in the figure is:
(Take $\dfrac{1}{4\pi\epsilon_0} = 9\times 10^9$ SI units)
Inside/on a charged shell, $V$ is constant — $V_C = V_P$.
A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If surface tension of water is $0.07\ \text{Nm}^{-1}$, then the excess force required to take it away from the surface is:
$F = T \cdot 2\pi r = 0.07 \cdot 2\pi \cdot 0.045 \approx 19.8$ mN.
A logic circuit provides the output $Y$ as per the following truth table:
| A | B | Y |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
The expression for the output $Y$ is:
$Y = 1$ exactly when $B = 0$, so $Y = \overline{B}$.
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: The potential ($V$) at any axial point, at 2 m distance($r$) from the centre of the dipole of dipole moment vector $\vec{P}$ of magnitude, $4 \times 10^{-6}\ \text{C m}$, is $\pm 9\times 10^3\ V$.
(Take $\dfrac{1}{4\pi\epsilon_0} = 9\times 10^9$ SI units)
Reason R: $V = \pm \dfrac{2P}{4\pi\epsilon_0 r^2}$, where $r$ is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below:
Correct axial potential: $V = p/(4\pi\epsilon_0 r^2) = 9\times 10^3$ V (A correct). R's formula has extra factor 2 (that is for field, not potential) — R false.
The terminal voltage of the battery, whose emf is $10V$ and internal resistance $1\ \Omega$, when connected through an external resistance of $4\ \Omega$ as shown in the figure is:
$I = 10/5 = 2$ A; $V_\text{term} = 2 \times 4 = 8$ V.
In the above diagram, a strong bar magnet is moving towards solenoid-2 from solenoid-1. The direction of induced current in solenoid-1 and that in solenoid-2, respectively, are through the directions:
Lenz's law: solenoid-1 sees N-flux decreasing (current $A \to B$); solenoid-2 sees S-flux increasing (current $D \to C$).
In an ideal transformer, the turns ratio is $\dfrac{N_p}{N_s} = \dfrac{1}{2}$. The ratio $V_s : V_p$ is equal to (the symbols carry their usual meaning):
$V_s/V_p = N_s/N_p = 2:1$.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.