Blog Log in Get Started Free

NEET Practice Questions (MCQs) with Answers & Solutions

Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

● ● ● ● ●

Cyclotron cannot be used to accelerate

Electrons

Neutrons

Positive ions

Both (1) and (2)

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(d)

Cyclotron cannot be used to accelerate electrons. Due to the small mass, the speed of electrons increases rapidly. Likewise, due to quick relativistic variation in their mass, the electrons get out of step with the oscillating electric field.

And Cyclotron cannot be used to accelerate neutron because the neutron is an uncharged particle

● ● ● ● ●

Magnetic field due to a ring having n turns at a distance x on its axis is proportional to (if r = radius of ring) :

$\frac{r}{(x^{2} + r^{2})}$

$\frac{r^{2}}{(x^{2} + r^{2})^{3 / 2}}$

$\frac{{nr}^{2}}{(x^{2} + r^{2})^{3 / 2}}$

$\frac{n^{2} r^{2}}{(x^{2} + r^{2})^{3 / 2^{}}}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$B = \frac{μ_{0}}{4 π} \cdot \frac{2 {πnir}^{2}}{(x^{2} + r_{}^{2})^{3 / 2}} \Rightarrow B \propto \frac{{nr}^{2}}{(x^{2} + r^{2})^{3 / 2}}$

● ● ● ● ●

An electric current passes through a long straight wire. At a distance 5 cm from the wire, the magnetic field is B. The field at 20 cm from the wire would be :

$\frac{B}{6}$

$\frac{B}{4}$

$\frac{B}{3}$

$\frac{B}{2}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) $B = 1 \times 10^{- 7} \times \frac{2 i}{r} \Rightarrow \frac{B}{B'} = \frac{20}{5} \Rightarrow B' = B / 4$

● ● ● ● ●

The dimension of the magnetic field intensity B is:

${MLT}^{- 2} A^{- 1}$

${MT}^{- 2} A^{- 1}$

${ML}^{2} {TA}^{- 2}$

$M^{2} {LT}^{- 2} A^{- 1}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(b) F = Bil $\Rightarrow [B] = \frac{[F]}{[i] [L]} = \frac{{MLT}^{- 2}}{AL} = {MT}^{- 2} A^{- 1}$

● ● ● ● ●

An electron moving in a circular orbit of radius r makes n rotation per second. The magnetic field produced at the centre has a magnitude of :

$\frac{μ_{0} ne}{2 r}$

$\frac{μ_{0} n^{2} e}{2 r}$

$\frac{μ_{0} ne}{2 πr}$

Zero

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a) Corresponding current i = en

So $B = \frac{μ_{0}}{4 π} \cdot \frac{2 π (en)}{r} = \frac{μ_{0} ne}{2 r}$

● ● ● ● ●

Two particles X and Y having equal charges, after being accelerated through the same potential difference, enter a region of uniform magnetic field and describes circular path of radius $R_{1}$ and $R_{2}$ respectively. The ratio of mass of X to that of Y is :

${(\frac{R_{1}}{R_{2}})}^{1 / 2}$

$\frac{R_{2}}{R_{1}}$

${(\frac{R_{1}}{R_{2}})}^{2}$

$\frac{R_{1}}{R_{2}}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(c) $r = \frac{\sqrt{2 mk}}{qB} = \frac{1}{B} \sqrt{\frac{2 mV}{q}} \Rightarrow r \propto \sqrt{m} \Rightarrow \frac{m_{1}}{m_{2}} = {(\frac{R_{1}}{R_{2}})}^{2}$

● ● ● ● ●

A beam of ions with velocity $2 \times 10^{5} m / s$ enters normally into a uniform magnetic field of $4 \times 10^{- 2}$ tesla. If the specific charge of the ion is $5 \times 10^{7}$ C/kg , then the radius of the circular path described will be :

0.10 m

0.16 m

0.20 m

0.25 m

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(a) $r = \frac{mv}{Bq} = \frac{v}{(q / m) B} = \frac{2 \times 10^{5}}{5 \times 10^{7} \times 4 \times 10^{- 2}}$ = 0.1 m

● ● ● ● ●

If the direction of the initial velocity of the charged particle is perpendicular to the magnetic field, then the orbit will be
or
The path executed by a charged particle whose motion is perpendicular to magnetic field is :

A straight line

An ellipse

A circle

A helix

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(c)

Since the initial velocity of the particle is perpendicular to the magnetic field, the particle will move in a circular path, because the force vector acts perpendicular to the motion of the body.

● ● ● ● ●

A proton and an $α -$ particle enter a uniform magnetic field perpendicularly with the same speed. If proton takes 25 $μ$ sec to make 5 revolutions, then the periodic time for the $α -$ particle would be :

50 $μ$ sec

25 $μ$ sec

10 $μ$ sec

5 $μ$ sec

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

By using $T = \frac{2 πm}{qB} \Rightarrow \frac{T_{α}}{T_{p}} = \frac{m_{α}}{m_{p}} \times \frac{q_{p}}{q_{α}} = \frac{4 m_{p}}{m_{p}} \times \frac{q_{p}}{2 q_{p}}$

$\Rightarrow T_{α} = 2 T_{p} = 10 μ \sec$

● ● ● ● ●

An $α -$ particle travels in a circular path of radius 0.45 m in a magnetic field $B = 1.2 W b / m^{2}$ with a speed of $2.6 \times 10^{7} m / s e c$ . The period of revolution of the $α -$ particle is :

$1.1 \times 10^{- 5}$ sec

$1.1 \times 10^{- 6}$ sec

$1.1 \times 10^{- 7}$ sec

$1.1 \times 10^{- 8}$ sec

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

(c) $T = \frac{2 πm}{qB} = \frac{2 πr}{V} = \frac{2 \times 3.14 \times 0.45}{2.6 \times 10^{7}} = 1.08 \times 10^{- 7}$ sec

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every NEET question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.