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The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are:
Unpredictable, fluctuating sources of error are by definition random errors.
The ratio of radius of gyration of a solid sphere of mass $M$ and radius $R$ about its own axis to the radius of gyration of the thin hollow sphere of same mass and radius about its axis is:
$k_{solid}^2 = \tfrac{2}{5}R^2$, $k_{hollow}^2 = \tfrac{2}{3}R^2$; ratio $k_s:k_h = \sqrt{3}:\sqrt{5}$ (often shown as $3:5$ for squared values).
Two bodies of mass $m$ and $9m$ are placed at a distance $R$. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be ($G =$ gravitational constant):
Null point: distance $R/4$ from $m$, $3R/4$ from $9m$. $V = -Gm/(R/4) - G(9m)/(3R/4) = -4Gm/R - 12Gm/R = -16Gm/R$.
The equivalent capacitance of the system shown in the following circuit is:
The two 3 μF in the parallel branch combine to 6 μF; in series with the leftmost 3 μF: $\dfrac{3\times 6}{3+6} = 2\ \mu\text{F}$.
The ratio of frequencies of fundamental harmonic produced by an open pipe to that of closed pipe having the same length is:
Open pipe: $f = v/(2L)$. Closed pipe: $f = v/(4L)$. Ratio $= 2:1$.
Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is $0.15$ ($g = 10\ \text{m s}^{-2}$).
$a_{\max} = \mu_s g = 0.15 \times 10 = 1.5\ \text{m s}^{-2}$.
The $x$-$t$ graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at $t = 2\ \text{s}$ is:
Period $T = 8$ s, $\omega = \pi/4$. At $t = 2$ s, $x = -1$ m (minimum). $a = -\omega^2 x = (\pi/4)^2 = \pi^2/16\ \text{m s}^{-2}$.
The net impedance of circuit (as shown in figure) will be:
$X_L = 2\pi(50)(50/\pi)\times 10^{-3} = 5\ \Omega$. $X_C = 1/[2\pi(50)(10^3/\pi)\times 10^{-6}] = 10\ \Omega$. $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{100 + 25} = 5\sqrt{5}\ \Omega$.
An electric dipole is placed as shown in the figure.
The electric potential (in $10^2\ \text{V}$) at point $P$ due to the dipole is ($\epsilon_0 =$ permittivity of free space and $\dfrac{1}{4\pi\epsilon_0} = K$):
$P$ on axial line; $V = Kq\!\left(\dfrac{1}{0.02} - \dfrac{1}{0.08}\right) = 37.5\,Kq\ \text{V} = (3/8)\,Kq\ (\times 10^2\ \text{V})$.
A wire carrying a current $I$ along the positive $x$-axis has length $L$. It is kept in a magnetic field $\vec{B} = (2\hat{i} + 3\hat{j} - 4\hat{k})\ \text{T}$. The magnitude of the magnetic force acting on the wire is:
$\vec F = I\vec L\times\vec B = IL\,\hat i\times(2\hat i+3\hat j-4\hat k) = IL(3\hat k + 4\hat j)$. $|\vec F| = IL\sqrt{9+16} = 5IL$.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.