NEET UG 2023 — Physics Electromagnetism

2023mcqhard

$10$ resistors, each of resistance $R$ are connected in series to a battery of emf $E$ and negligible internal resistance. Then those are connected in parallel to the same battery, the current is increased $n$ times. The value of $n$ is:

Official previous-year question

Held on 30 Apr 2023 · Verified 9 Jul 2026.

Options

A
$1000$
B
$10$
C
$100$
D
$1$

Did you get this right?

Solution

When the resistors are connected in series with the battery, the equivalent resistance $({R}_{s})$ can be written as

$\begin{matrix}{R}_{s} & = & R+R+....10\mathrm{times} \\ & = & 10R...(1)\end{matrix}$

Hence, the current $({i}_{s})$ is given by

$\begin{matrix}{i}_{s} & = & \frac{E}{{R}_{s}} \\ & = & \frac{E}{10R}...(2)\end{matrix}$

When the resistors are connected in parallel, the equivalent resistance $({R}_{p})$ is given by

$\begin{matrix}{R}_{p} & = & \frac{1}{\frac{1}{R}+\frac{1}{R}+....10\mathrm{times}} \\ & = & \frac{R}{10}...(3)\end{matrix}$

Hence, the current $({i}_{p})$ is given by

$\begin{matrix}{i}_{p} & = & \frac{E}{{R}_{p}} \\ & = & \frac{E}{\frac{R}{10}} \\ & = & \frac{10E}{R}...(4)\end{matrix}$

Divide equation (4) by equation (2) to obtain the required value of the multiplier $n$ .

$\begin{matrix}\frac{{i}_{p}}{{i}_{s}} & = & \frac{\frac{10E}{R}}{\frac{E}{10R}} \\ & = & 100\end{matrix}$

Your note

NEET UG Electromagnetism in other years

Electromagnetism 2026 Electromagnetism 2025 Electromagnetism 2024 Electromagnetism 2022 Electromagnetism 2021 Electromagnetism 2020 Electromagnetism 2019 Electromagnetism 2018 Electromagnetism 2017 Electromagnetism 2016 Electromagnetism 2015 Electromagnetism 2014

More NEET UG Electromagnetism questions

Two point charges +2μC and -2μC are placed 10 cm apart. The electric field at the midpoint is:

2026medium

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

2025easy

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 :

2025medium

The current passing through the battery in the given circuit, is : ![NEET 2025 Physics, Current Electricity — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/616059730283de43c87e3e22/Physics/images/Current_Electricity/682c9a94c2281abe1696a0a6/question_1__q_682c9a94c2281abe1696a0a6__cdn-question-pool.getmarks.app__yKwOmrdZMSd9Ml6L9jUq5lCIlM2f2z32bgW-sqY2ViU.original.fullsiz__b2faee45bc_final_ppt_sync.png)

2025hard

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 $\mathrm{c}=3 \times 10^8 \mathrm{~ms}^{-1}$ )

2025medium

View all Electromagnetism questions

Keep practicing Electromagnetism

Work through every NEET UG Electromagnetism PYQ, year by year.

Browse all Physics topics

Solution

When the resistors are connected in series with the battery, the equivalent resistance $({R}_{s})$ can be written as

$\begin{matrix}{R}_{s} & = & R+R+....10\mathrm{times} \\ & = & 10R...(1)\end{matrix}$

Hence, the current $({i}_{s})$ is given by

$\begin{matrix}{i}_{s} & = & \frac{E}{{R}_{s}} \\ & = & \frac{E}{10R}...(2)\end{matrix}$

When the resistors are connected in parallel, the equivalent resistance $({R}_{p})$ is given by

$\begin{matrix}{R}_{p} & = & \frac{1}{\frac{1}{R}+\frac{1}{R}+....10\mathrm{times}} \\ & = & \frac{R}{10}...(3)\end{matrix}$

Hence, the current $({i}_{p})$ is given by

$\begin{matrix}{i}_{p} & = & \frac{E}{{R}_{p}} \\ & = & \frac{E}{\frac{R}{10}} \\ & = & \frac{10E}{R}...(4)\end{matrix}$

Divide equation (4) by equation (2) to obtain the required value of the multiplier $n$ .

$\begin{matrix}\frac{{i}_{p}}{{i}_{s}} & = & \frac{\frac{10E}{R}}{\frac{E}{10R}} \\ & = & 100\end{matrix}$

NEET UG 2023 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

NEET UG Electromagnetism in other years

More NEET UG Electromagnetism questions

Other Physics topics for NEET UG

Mechanics

Modern Physics

Thermodynamics

Optics

Waves & Oscillations

Keep practicing Electromagnetism

NEET UG 2023 — Physics Electromagnetism

Options

Did you get this right?

Solution

Your note

NEET UG Electromagnetism in other years

More NEET UG Electromagnetism questions

Other Physics topics for NEET UG

Mechanics

Modern Physics

Thermodynamics

Optics

Waves & Oscillations

Keep practicing Electromagnetism