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A Carnot engine has an efficiency of 50% when its source is at a temperature $327^\circ\text{C}$. The temperature of the sink is:
$\eta = 1 - T_2/T_1\Rightarrow 0.5 = 1 - T_2/600\Rightarrow T_2 = 300\ \text{K} = 27^\circ\text{C}$.
Resistance of a carbon resistor determined from colour codes is $(22000 \pm 5\%)\ \Omega$. The colour of third band must be:
$22000 = 22 \times 10^3$. First two bands red, red (2,2); third band (multiplier $10^3$) is orange.
The minimum wavelength of $X$-rays produced by an electron accelerated through a potential difference of $V$ volts is proportional to:
$\lambda_{\min} = \dfrac{hc}{eV}\Rightarrow \lambda_{\min}\propto \dfrac{1}{V}$.
For Young's double slit experiment, two statements are given below:
Statement I: If screen is moved away from the plane of slits, angular separation of the fringes remains constant.
Statement II: If the monochromatic source is replaced by another monochromatic source of larger wavelength, the angular separation of fringes decreases.
In the light of the above statements, choose the correct answer from the options given below:
Angular separation $\theta = \lambda/d$ is independent of screen distance (Statement I true). For larger $\lambda$, $\theta$ increases, not decreases (Statement II false).
The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are $2.14\ \text{eV}$, $2.30\ \text{eV}$ and $2.75\ \text{eV}$ respectively. If incident electromagnetic radiation has an incident energy of $2.20\ \text{eV}$, which of these photosensitive surfaces may emit photoelectrons?
Photoemission requires photon energy $\ge$ work function. Only $W_{Cs} = 2.14 < 2.20$ eV.
The magnitude and direction of the current in the following circuit is:
Net EMF $= 10 - 5 = 5$ V (cells oppose), total $R = 2+1+7 = 10\ \Omega$, current $= 0.5$ A driven by the 10 V cell, so from $A$ to $B$ through $E$.
In hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$. The shortest wavelength in the Bracket series is:
Series limit: $1/\lambda_{\min} = R/n^2$. Balmer ($n=2$): $1/\lambda = R/4$. Bracket ($n=4$): $1/\lambda' = R/16$. So $\lambda' = 4\lambda$.
An electric dipole is placed at an angle of $30^\circ$ with an electric field of intensity $2\times 10^5\ \text{N C}^{-1}$. It experiences a torque equal to $4\ \text{N m}$. Calculate the magnitude of charge on the dipole, if the dipole length is $2\ \text{cm}$.
$\tau = qLE\sin\theta\Rightarrow q = \dfrac{4}{0.02 \times 2\times 10^5 \times 0.5} = 2\times 10^{-3}\ \text{C} = 2\ \text{mC}$.
The half life of a radioactive substance is 20 minutes. In how much time, the activity of substance drops to $\left(\dfrac{1}{16}\right)^{th}$ of its initial value?
$1/16 = (1/2)^4$, so 4 half-lives $= 4\times 20 = 80$ min.
The venturi-meter works on:
A venturi-meter measures flow rate via pressure difference, derived from Bernoulli's equation.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.