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The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury?
Kepler: $T\propto r^{3/2}$. $T_{Mer} = 687\times\left(\tfrac14\right)^{3/2} = \dfrac{687}{8} \approx 86 \approx 88$ days.
A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:
$W' = W\left(\dfrac{R}{R+R/3}\right)^2 = 48\left(\dfrac{3}{4}\right)^2 = 48\times\dfrac{9}{16} = 27$ N.
A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:
Each piece $= R/8$. Four in parallel $= \dfrac{R/8}{4} = \dfrac{R}{32}$. Two such sets in series $= \dfrac{R}{32}+\dfrac{R}{32} = \dfrac{R}{16}$.
De-Broglie wavelength of an electron orbiting in the $n = 2$ state of hydrogen atom is close to (Given Bohr radius $= 0.052$ nm):
$r_2 = n^2a_0 = 4(0.052) = 0.208$ nm. $2\pi r = n\lambda\Rightarrow\lambda = \dfrac{2\pi r_2}{2} = \pi(0.208) \approx 0.67$ nm.
An electric dipole with dipole moment $5\times10^{-6}$ Cm is aligned with the direction of a uniform electric field of magnitude $4\times10^5$ N/C. The dipole is then rotated through an angle of $60^\circ$ with respect to the electric field. The change in the potential energy of the dipole is:
$\Delta U = pE(1-\cos60^\circ) = (5\times10^{-6})(4\times10^5)(1-0.5) = 2\times0.5 = 1.0$ J.
A constant voltage of 50 V is maintained between the points A and B of the circuit shown in the figure. The current through the branch CD of the circuit is:
With CD a connecting wire, $V_M = 32$ V. Current in $1\,\Omega$ (A→C) $= 18$ A and in $2\,\Omega$ (C→B) $= 16$ A, so $I_{CD} = 18-16 = 2.0$ A.
A photon and an electron (mass $m$) have the same energy $E$. The ratio $(\lambda_{\text{photon}}/\lambda_{\text{electron}})$ of their de Broglie wavelengths is: ($c$ is the speed of light)
$\lambda_{ph} = \dfrac{hc}{E}$, $\lambda_e = \dfrac{h}{\sqrt{2mE}}$. Ratio $= \dfrac{hc/E}{h/\sqrt{2mE}} = c\sqrt{\dfrac{2m}{E}}$.
Which of the following options represent the variation of photoelectric current with property of light shown on the x-axis?
Photoelectric (saturation) current is proportional to intensity (graph A) but is independent of frequency, so the current-vs-frequency graphs C and D are wrong. Only A is correct.
A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:
Taking mass $\propto r^3$: small sphere mass 1 unit, big 8 units. $I_{small} = \tfrac25 mR^2 + mR^2 = \tfrac75$ units; $I_{big} = \tfrac25(8)(2R)^2 = \tfrac{64}{5}$; $I_{rest} = \tfrac{57}{5}$. Ratio $= \dfrac{7}{57}$.
A full wave rectifier circuit with diodes $(D_1)$ and $(D_2)$ is shown in the figure. If input supply voltage $V_{in} = 220\sin(100\pi t)$ volt, then at $t = 15$ msec:
At $t = 15$ ms, $100\pi t = 1.5\pi$, $\sin(1.5\pi) = -1$, so $V_{in}$ is in its negative half cycle. Then $D_2$ conducts (forward biased) and $D_1$ is reverse biased.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.