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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:
$\vec F = I\vec L\times\vec B = IL\,\hat i\times(2\hat i+3\hat j-4\hat k) = IL(3\hat k + 4\hat j)$. $|\vec F| = IL\sqrt{9+16} = 5IL$.
A bullet from a gun is fired on a rectangular wooden block with velocity $u$. When bullet travels $24\ \text{cm}$ through the block along its length horizontally, velocity of bullet becomes $\dfrac{u}{3}$. Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is:
$v^2 - u^2 = 2as$. From $u$ to $u/3$ in 24 cm: $\dfrac{8u^2}{9} = 2a(24)$. From $u/3$ to 0: $\dfrac{u^2}{9} = 2a\,s\Rightarrow s = 3$ cm. Total $= 27$ cm.
Two thin lenses are of same focal lengths ($f$), but one is convex and the other one is concave. When they are placed in contact with each other, the equivalent focal length of the combination will be:
$\dfrac{1}{F} = \dfrac{1}{f} - \dfrac{1}{f} = 0\Rightarrow F\to\infty$.
In the figure shown here, what is the equivalent focal length of the combination of lenses (Assume that all layers are thin)?
Power of combination $= (n_g - n_1)/R_1 + (n_2 - n_g)/(-R_2)$ with $n_g = 1.5$, $n_1 = 1.5$, $n_2 = 1.6$, $R_1 = R_2 = 20$ cm gives $F = -100$ cm.
A satellite is orbiting just above the surface of the earth with period $T$. If $d$ is the density of the earth and $G$ is the universal constant of gravitation, the quantity $\dfrac{3\pi}{Gd}$ represents:
$T^2 = \dfrac{4\pi^2 R^3}{GM} = \dfrac{4\pi^2 R^3}{G\cdot (4/3)\pi R^3 d} = \dfrac{3\pi}{Gd}$.
For the following logic circuit, the truth table is:
Two NOT gates feed a NAND: $Y = \overline{\bar A \cdot \bar B} = A + B$ (OR gate by De Morgan).
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.