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Light travels a distance $x$ in time $t_1$ in air and $10x$ in time $t_2$ in another denser medium. What is the critical angle for this medium?
$v_{air} = x/t_1$, $v_{med} = 10x/t_2$; $n = v_{air}/v_{med} = t_2/(10t_1)$. $\sin\theta_c = 1/n = 10t_1/t_2$.
The amount of energy required to form a soap bubble of radius $2\ \text{cm}$ from a soap solution is nearly: (surface tension of soap solution $= 0.03\ \text{N m}^{-1}$)
Bubble has two surfaces: $W = T\times 2(4\pi r^2) = 0.03\times 8\pi\times (0.02)^2 \approx 3.01\times 10^{-4}\ \text{J}$.
A $12\ \text{V}$, $60\ \text{W}$ lamp is connected to the secondary of a step down transformer, whose primary is connected to ac mains of $220\ \text{V}$. Assuming the transformer to be ideal, what is the current in the primary winding?
Ideal transformer: $V_p I_p = V_s I_s = 60\ \text{W}\Rightarrow I_p = 60/220 \approx 0.27$ A.
The angular acceleration of a body, moving along the circumference of a circle, is:
Angular vectors (velocity, acceleration) lie along the axis of rotation by definition.
A vehicle travels half the distance with speed $\vartheta$ and the remaining distance with speed $2\vartheta$. Its average speed is:
$\bar v = \dfrac{d}{d/(2\vartheta) + d/(4\vartheta)} = \dfrac{d}{3d/(4\vartheta)} = \dfrac{4\vartheta}{3}$.
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 $= 2:1$.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.