given
Q=at−bt2
i=dtdq=a−2bt
i=0
at t=2ba
∴dH=i2Rdt
=(a−2bt)2Rdt
H=∫02ba(a−2bt)2Rdt
=3(−2b)(a−2bt)3R∣02ba
=−6b1[(a−2b2ba)3−(a)3]R
=−6b1[(0)3−a3]R
H=6ba3R
NEET UG 2016 — Physics Electromagnetism
The charge flowing through a resistance R varies with time t as Q=at−bt2 where a and b are positive constants. The total heat produced in R is:
Held on 30 Apr 2016 · Verified 9 Jul 2026.
6ba3R
3ba3R
2ba3R
ba3R
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
Two point charges +2μC and -2μC are placed 10 cm apart. The electric field at the midpoint is:
A model for quantized motion of an electron in a uniform magnetic field $B$ states that the flux passing through the orbit of the electron is $n(h / e)$ where $n$ is an integer, $h$ is Planck's constant and $e$ is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be ( $m$ is the mass of the electron)
A 2 amp current is flowing through two different small circular copper coils having radii ratio $1: 2$. The ratio of their respective magnetic moments will be
Two identical charged conducting spheres $A$ and $B$ have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is $F$. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres $A$ and $B$ (Radii of $A$ and $B$ are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as :
An electron (mass $9 \times 10^{-31} \mathrm{~kg}$ and charge $1.6 \times 10^{-19} \mathrm{C}$ ) moving with speed $\mathrm{c} / 100(\mathrm{c}=$ speed of light) is injected into a magnetic field $\vec{B}$ of magnitude $9 \times 10^{-4} \mathrm{~T}$ perpendicular to its direction of motion. We wish to apply an uniform electric field $\vec{E}$ together with the magnetic field so that the electron does not deflect from its path. Then (speed of light $\mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}$ )
Work through every NEET UG Electromagnetism PYQ, year by year.