datavorous/entrance-exam-dataset huggingface dataset:
A motor car moving at a speed of $72 \mathrm{~km} / \mathrm{h}$ cannot come to a stop in less than $3.0 \mathrm{~s}$ while for a truck this time interval is $5.0 \mathrm{~s}$. On a highway, the car is behind the truck both moving at $72 \mathrm{~km} / \mathrm{h}$. The truck gives a signal that it is going to stop at emergency. At what distance the car should be from the truck so that it does not bump onto (collide with) the truck. Human response time is $0.5 \mathrm{~s}$.

Options:

None

Solution:
As given that speed of car as well as truck
$$
=72 \mathrm{~km} / \mathrm{h}=72 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}=20 \mathrm{~m} / \mathrm{s}
$$
For retarded motion of truck,
$$
v=u+a_t t \quad\left(\because v=0, a_t=?, t=5 \mathrm{sec} .\right)
$$
$$
0=20+a_t \times 5
$$
or $a_t=-4 \mathrm{~m} / \mathrm{s}^2$
$t=3 \mathrm{sec}, u=20 \mathrm{~m} / \mathrm{sec}, v=0, a_c=$ ?
For retarded motion of the car,
$$
\begin{aligned}
&v=u+a_c t \\
&0=20+a_c \times 3
\end{aligned}
$$
So, $a_c=-\frac{20}{3} \mathrm{~m} / \mathrm{s}^2$
Let car be at a distance $x$ meter from the truck, when truck gives the signal and $t$ be the time taken to cover this distance.
The time of human response is $0.5 \mathrm{~s}$, time of retarded motion of car is $(t-0.5) \mathrm{s}$.
Velocity of car after time $t$,
$$
\begin{aligned}
v_c &=u+a_c t \\
&=20-\left(\frac{20}{3}\right)(t-0.5)
\end{aligned}
$$
Velocity of truck after time $t$
$$
v_t=20-4 t
$$
To avoid the car bump onto the truck,
$$
\begin{aligned}
&v_c=v_t \\
&20-\frac{20}{3}(t-0.5)=20-4 t \\
&\Rightarrow 4 t=\frac{20}{3}(t-0.5) \Rightarrow 3 t=5 t-2.5 \Rightarrow t=1.25 \text { seconds }
\end{aligned}
$$
Distance travelled by car and truck in time $t=\frac{5}{4} \sec$.
$$
\begin{aligned}
&s_t=u_t t+\frac{1}{2} a_t t^2 \\
&=20 \times \frac{5}{4}+\frac{1}{2} \times(-4) \times\left(\frac{5}{4}\right)^2=25-3.125 \\
&\therefore s_t=21.875 \mathrm{~m}
\end{aligned}
$$
Distance travelled by car in time $t$
$=$ Distance travelled by car in $0.5 \mathrm{~s}$ (without retardation) $+$ Distance travelled by car in $(t-0.5) \mathrm{s}$ (with retardation)
$$
\begin{aligned}
&s_c=(20 \times 0.5)+20\left(\frac{5}{4}-0.5\right)-\frac{1}{2}\left(\frac{20}{3}\right)\left(\frac{5}{4}-0.5\right)^2 \\
&s_c=23.125 \mathrm{~m} \\
&s_c-s_t=23.125-21.875=1.250 \mathrm{~m}
\end{aligned}
$$
So avoid the bump onto the truck, the car must maintain a distance from the truck atleast $1.250 \mathrm{~m}$.

Topic: Physics, Motion In One Dimension
A motor car moving with a uniform speed of $20 \mathrm{~m} / \mathrm{sec}$ comes to stop on the application of brakes after travelling a distance of ${ }^{10 m}$ Its acceleration is

Options:
A $20 \mathrm{~m} / \mathrm{sec}^2$
B $-20 \mathrm{~m} / \mathrm{sec}^2$
C $-40 \mathrm{~m} / \mathrm{sec}^2$
D $+2 m / \mathrm{sec}^2$

The correct answer is:
$-20 \mathrm{~m} / \mathrm{sec}^2$

Solution:
From $v^2=u^2+2 a S \Rightarrow 0=u^2+2 a S$
$\Rightarrow a=\frac{-u^2}{2 S}=\frac{-(20)^2}{2 \times 10}=-20 \mathrm{~m} / \mathrm{s}^2$

Topic: Physics, Motion In One Dimension, JEE Main
A motor cycle is going on an overbridge of radius $R$. The driver maintains a constant speed. As the motor cycle is ascending on the overbridge, the normal force on it

Options:
A increases
B decreases
C remains the same
D fluctuates erratically

The correct answer is:
increases

Solution:
Reaction of normal force on the motor cyclist
$M g \cos \theta-N=\frac{m v^2}{R}$
or $\quad N=M g \cos \theta-\frac{M v^2}{R}$
As the motor cyclist moves upward on the bridge, $\theta$ decreases and therefore, $N$ increases and becomes maximum at the highest point.

Topic: Physics, Laws of Motion, JIPMER, JIPMER 2008
A motor cycle starts from rest and accelerates along a straight path at $2 \mathrm{~m} / \mathrm{s}^2$. At the starting point of the motor cycle there is a stationary electric sire. How far has the motor cycle gone when the driver hears the frequency of the siren at $94 \%$ of its value when the motor cycle was at rest? (speed of sound $\left.=330 \mathrm{~ms}^{-1}\right)$

Options:
A
$49 \mathrm{~m}$

B
$98 \mathrm{~m}$

C
$147 \mathrm{~m}$

D
$196 \mathrm{~m}$

The correct answer is:
$98 \mathrm{~m}$

Solution:
Motor cycle, $u=0, a=2 \mathrm{~m} / \mathrm{s}^2$
Observer is in motion and source is at rest.
$$
\begin{aligned}
& \Rightarrow n^{\prime}=n \frac{v-v_O}{v+v_S} \Rightarrow \frac{94}{100} n=n \frac{330-v_O}{330} \Rightarrow 330-v_0=\frac{330 \times 94}{100} \\
& \Rightarrow v_O=330-\frac{94 \times 33}{10}=\frac{33 \times 6}{10} \mathrm{~m} / \mathrm{s} \\
& s=\frac{v^2-u^2}{2 a}=\frac{9 \times 33 \times 33}{100}=\frac{9 \times 1089}{100} \approx 98 \mathrm{~m} .
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2009
A motor generates an output power of 220 W at an angular velocity of 2100 rpm. Calculate the torque (in N m) produced by the motor? [Take Ï=227]

Options:

The correct answer is:
01.00

Solution:
ω=2100×2π60=70π rad s-1
P=τω
τ=Pω=220×770×22=1 N m

Topic: Physics, Rotational Motion, JEE Main
A motorcycle starts from rest and accelerates along a straight path at 2 m s-2. At the starting point, there was a siren which produced sound at a constant frequency. How far has the motorcycle gone from the starting point, If now the rider hears the frequency of the siren at 94% of its value as compared to when the motorcycle was rest? (Speed = 330 m s-1)

Options:
A 49 m

B 98 m

C 147 m

D 196 m

The correct answer is:
98 m

Solution:
For Motorcycle, u=0, a=2 m s-2
Observe is in motion and source is at rest
⟹ n′=nv-v2v+vs
⟹94100n=n330-vo330
⟹330-vo=330×94100
⟹vo=330-94×3310=33×610 m s-1
s=v2-u22a=9×33×33100
=9×1089100=98 m.

Topic: Physics, Waves and Sound, JEE Main
A movable steel plate is placed between fixed steel and brass plates and the stack of plates is
subjected to a weight of $100 \mathrm{~N}$ as shown in the figure. The coefficient of kinetic friction for steel on steel
is 0.57 and for steel on brass is 0.44 . Assuming that the entire weight comes onto the stack and that the weight of the plates is negligible in comparison to the applied weight, the force required to move the middle plate (in $\mathrm{N}$ ) is

Options:
A 13
B 101
C 440
D 570

The correct answer is:
101

Solution:
For given figure,
$\mu_{S S}=$ coefficient of friction between steel and steel $=0.57$
and $\mu_{S B}=$ coefficient of friction between steel and brass $=0.44$.
$$
\text { Free body diagram }
$$
where, $f_{S S}=$ friction between steel plates and $f_{S B}=$ friction between steel and brass plates. $\Rightarrow$ For kinetic-friction, the applied force $(F)$ is equal or greater than friction force for moving the steel plate, so $f_{S S}+f_{S B} \leq F$
Therefore,
$$
\begin{aligned}
& f_{S S}=\mu_{S S} \times R=0.57 \times 100=57 \mathrm{~N} \\
& f_{S B}=\mu_{S B} \times R=0.44 \times 100=44 \mathrm{~N} \\
& 57+44=N \\
& \Rightarrow \quad 101 \text { newton }=N \\
&
\end{aligned}
$$
$$

Topic: Physics, Laws of Motion, JEE Main
A moving coil galvanometer allows a full scale current of 10-4 A . A series resistance of 2 ×104 Ω is required to convert the galvanometer into a voltmeter of range 0-5 V . Therefore, the value of shunt resistance required to convert the above galvanometer into an ammeter of range 0-10 mA is:

Options:
A 100 Ω
B 200 Ω
C 300 Ω
D 10 Ω

The correct answer is:
300 Ω

Solution:
ig=10-4 A

i=10×10-3 A

RV=20 kΩ
igG+ig RV=5
igG=i-igRA
RA=igGi-ig=ig5-igRVi-igig
RA=5-igRVi-ig≈300  Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (10 Apr Shift 1)
A moving coil galvanometer has 100 turns and each turn has an area of 2.0 cm2. The magnetic field produced by the magnet is 0.01 T and the deflection in the coil is 0.05 radian when a current of 10 mA is passed through it. The torsional constant of the suspension wire is x×10-5 N-m/rad. The value of x is ______ .

Options:

The correct answer is:
4

Solution:
Torque produced in a moving coil galvanometer is, τ=BINAsinϕ.

Let torsional constant of the suspension wire be C, then Cθ=BINAsin90°

⇒C= BINA θ=0.01×10×10-3×100×2×10-40.05

=4×10-5 N-m/rad.

Therefore, x=4.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (01 Feb Shift 2)
A moving coil galvanometer has 150 equal divisions. Its current sensitivity is 10 divisions per milliampere and voltage sensitivity is 2 divisions per millivolt. In order that each division reads 1 volt, the resistance in ohms needed to be connected in series with the coil will be

Options:
A
$10^3$

B
$10^5$

C
99995

D
9995

The correct answer is:
9995

Solution:
$I_g=15 \mathrm{~mA} \quad V g=75 \mathrm{mV}$
$\mathrm{R}=\frac{\mathrm{V}}{\mathrm{I}_g}-\frac{\mathrm{V}_g}{I_g}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2005
A moving coil galvanometer has a coil with 175 turns and area 1 cm2. It uses a torsion band of torsion constant 10-6 N m/rad The coil is placed in a magnetic field B parallel to its plane. The coil deflects by 1o for a current of 1 mA The value of B (in tesla) is approximately:

Options:
A 10-2
B 10-4
C 10-3
D 10-1

The correct answer is:
10-3

Solution:
For a moving coil galvanometer.
τ=iANB=cθ
B=cθiAN=10-610-4×175×10-3×π180
B≈10-3 T

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A moving coil galvanometer has a resistance of $900 \Omega$. In order to send only $10 \%$ of the main current through this galvanometer, the resistance of the required shunt is

Options:
A $0.9 \Omega$
B $100 \Omega$
C $405 \Omega$
D $90 \Omega$

The correct answer is:
$100 \Omega$

Solution:
$\begin{aligned} & \text { (b) }: I_g=0.1 I, I_s=0.9 I ; S=I_g G / I_s \\ & =0.1 \times 900 / 0.9=100 \Omega\end{aligned}$

Topic: Physics, Current Electricity, AIIMS, AIIMS 2009
A moving coil galvanometer has resistance 50 Ω and it indicates full deflection at 4 mA current. A voltmeter is made using this galvanometer and a 5 kΩ resistance. The maximum voltage, that can be measured using this voltmeter, will be close to:

Options:
A 40 V
B 15 V
C 20 V
D 10 V

The correct answer is:
20 V

Solution:
Vmax=4×10-3×50+4×10-3×5×103
Vmax=20.2 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (09 Apr Shift 1)
A moving coil galvanometer, having a resistance G produces full scales deflection when a current Ig flows through it. It can be converted into an ammeter of range 0 to I0>Ig by connecting a shunt of resistance RA to it. If it can also be converted into a voltmeter of range 0 to V=I0G by connecting a resistance RV to it in series, then which among the following is correct?

Options:
A RARV=G2I0-IgIg and RARV=IgI0-Ig2
B RARV=G2 and RARv=IgI0-Ig
C RARv=G2 and RARv=IgI0-Ig2
D RARV=G2IgI0-Ig and RARV=I0-IgIg2

The correct answer is:
RARv=G2 and RARv=IgI0-Ig2

Solution:
(I0-Ig)R4=IgG and V0=I0G+Rv

I0=I0G+RARA V0=I0G+RV

V0=I0G( given ) â I0G+RV=IgG+RAGRA

RAG+RVRA=G2+RAG

RVRA=G2

RA=IgGI0-Ig;RV=I0-IgGIg

RARV=Ig2I0-Ig2

Topic: Physics, Magnetic Properties of Matter, JEE Main
A moving coil galvanometer of resistance 100 $\Omega$ shows full scale deflection when a current of 100 micro-amperes passes through it. If it is intended to show full scale deflection when a current of 1 milliampere passes through it, the value of shunt resistance in ohms to be connected to the galvanometer is.

Options:
A $\frac{9}{4}$
B $\frac{10}{3}$
C $\frac{100}{9}$
D $\frac{900}{7}$

The correct answer is:
$\frac{100}{9}$

Solution:
Resistance of galvanometer
$$
\begin{gathered}
G=100 \Omega \\
I_g=100 \mu \mathrm{A}=100 \times 10^{-6} \mathrm{~A}=0.1 \times 10^{-3} \mathrm{~A} \\
I=1 \mathrm{~mA}=1 \times 10^{-3} \mathrm{~A}
\end{gathered}
$$
Shunt resistance
$$
\begin{aligned}
S & =\frac{I_g G}{I-I_g}=\frac{0.1 \times 10^{-3} \times 100}{1 \times 10^{-3}-0.1 \times 10^{-3}} \\
& =\frac{1 \times 10^{-2}}{0.9 \times 10^{-3}}=\frac{10^2}{9}=\frac{100}{9} \Omega
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main
A moving conductor coil in a magnetic field produces an induced e.m.f. This is in accordance with

Options:
A Amperes law
B Coulomb law
C Lenz's law
D Faraday's law

The correct answer is:
Faraday's law

Solution:
Faraday's Law states that when the magnetic flux linking a circuit changes, an electromotive force is induced in the circuit proportional to the rate of change of the flux linkage.
This results in a moving conductor coil producing EMF.

Topic: Physics, Electromagnetic Induction, JEE Main
A moving particle of mass $m$, makes a head on elastic collision with another particle of mass $2 m$, which is initially at rest. The percentage loss in energy of the colliding particle on collision, is close to

Options:
A
$33 \%$

B
$67 \%$

C
$90 \%$

D
$10 \%$

The correct answer is:
$90 \%$

Solution:
Fractional decrease in kinetic energy of mass $m$
$$
\begin{aligned}
& =1-\left(\frac{m_2-m_1}{m_2+m_1}\right)^2=1-\left(\frac{2-1}{2+1}\right)^2 \\
& =1-\left(\frac{1}{3}\right)^2=1-\frac{1}{9}=\frac{8}{9}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2012 (19 May Online)
A moving proton and electron have the same de-Broglie wavelength. If K and P denote the K.E. and momentum respectively. Then choose the correct option :

Options:
A Kp>Ke and Pp=Pe
B Kp<Ke and Pp=Pe
C Kp<Keand Pp<Pe
D Kp=Ke and Pp=Pe

The correct answer is:
Kp<Ke and Pp=Pe

Solution:
λdeb =hp=h2mKE
If λ is same the P will be same
KE∝1/m
mp>me⇒Kp<Ke

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (31 Aug Shift 1)
A multiple choice examination has 5 questions. Each question has three alternative answers out of which exactly one is correct. The probability that a student will get 4 or more correct answers just by guessing is :

Options:
A 1 1 3 5
B 1 0 3 5
C 1 7 3 5
D 1 3 3 5

The correct answer is:
1 1 3 5

Solution:
4 or more correct answers means and 1 or 0 wrong answers.
for 1 wrong questions 5 C 1 × 2 C 1 = 1 0
for 0 wrong questions =1
Hence favourable ways =11
Total ways of answering 35
Hence probability = 1 1 3 5

Topic: Mathematics, Probability, JEE Main, JEE Main 2013 (07 Apr)
A multiple choice test consists of 5 questions, each question having 4 responses. There is only one correct response and the remaining 3 are incorrect responses. If a candidate attempts all the 5 questions then the probability that he answers at least 3 questions incorrectly is

Options:
A $\frac{675}{1024}$
B $\frac{459}{512}$
C $\frac{81}{128}$
D $\frac{135}{512}$

The correct answer is:
$\frac{459}{512}$

Solution:
Given $x=5, P=\frac{1}{4}, q=\frac{3}{4}$
$P$ (at least there incorrect answer)
$$
\begin{aligned}
& ={ }^5 \mathrm{c}_3\left(\frac{3}{4}\right)^3\left(\frac{1}{4}\right)^2+{ }^5 \mathrm{c}_4\left(\frac{3}{4}\right)^4\left(\frac{1}{4}\right)^1+{ }^5 \mathrm{c}_5\left(\frac{3}{4}\right)^5\left(\frac{1}{4}\right) \\
& =\left(\frac{3}{4}\right)^3\left[\frac{5 \times 4}{2} \times \frac{1}{16}+5 \times \frac{3}{16}+1 \times \frac{9}{16}\right] \\
& =\frac{27}{64}\left[\frac{10}{16}+\frac{15}{16}+\frac{9}{16}\right]=\frac{27}{64} \times \frac{34}{16} \\
& =\frac{459}{512}
\end{aligned}
$$

Topic: Mathematics, Probability, AP EAMCET, AP EAMCET 2022 (06 Jul Shift 1)
A multiple-choice question has 5 options of which only one is correct. If a student does home work, then he is sure to identify the correct answer, otherwise the answer is chosen at random. Let A be the event that the student does his home work and B be the event that the student answers correctly. If PA=23, then PAB is equal to

Options:
A 1011
B 45
C 37
D 67

The correct answer is:
1011

Solution:
PBAc=15,PA=23
PAB=PAâ©BPB=PA⋅PBAPAPBA+PAcPBAc
=23Ã123Ã1+13Ã15=1011

Topic: Mathematics, Probability, JEE Main
A musician using an open flute of length 50 cm produces second harmonic sound waves. A person runs towards the musician from another end of a hall at a speed of 10 km h-1. If the wave speed is 330 m s-1, the frequency heard by the running person shall be close to

Options:
A 333 Hz
B 500 Hz
C 666 Hz
D 753 Hz

The correct answer is:
666 Hz

Solution:
Frequency of the observer, fo=2×v2l=3300.5 Hz=660 Hz
Velocity of person=10 km h-1=10×518 m s-1=2.8 m s-1
According to Doppler effect, frequency heard by the person,
f=fov+v0v

f=660×330+2.8330 Hz=666 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (09 Jan Shift 2)
A $~n-p-n$ transistor conducts when:

Options:
A both collector and emitter are positive with respect to the base
B collector is positive and emitter is negative with respect to the base
C collector is positive and emitter is at same potential as the base
D both collector and emitter are negative with respect to the base

The correct answer is:
collector is positive and emitter is negative with respect to the base

Solution:
In active region
Emitter base $p-n$ function is in FB and base collector $p-n$ function is in RB.

Topic: Physics, Semiconductors, JEE Main
A narrow parallel beam of light is incident on a transparent sphere of refractive index n. If the beam finally gets focussed at a point situated at a distance = 2 × (radius of sphere) from the centre of the sphere, then find n?

Options:
A 4 3
B 3 2
C - 4 6
D - 2 3

The correct answer is:
4 3

Solution:
From first refraction
Î¼ 2 v - Î¼ 1 u = Î¼ 2 - Î¼ 1 R
n AI 1 - 1 â = n - 1 R
AI ' = n R n - 1
for 2nd refraction
u = BI â²
= A I â² - 2 R
= n R n - 1 - 2 R
= ( n - 2 n + 2 ) R n - 1
u = ( 2 - n ) ( n - 1 ) R
apply Î¼ 2 v - Î¼ 1 u = Î¼ 2 - Î¼ 1 R
1 R - n ( n - 1 ) ( 2 - n ) R = 1 - n - R
- n ( n - 1 ) ( 2 - n ) R = n - 1 R - 1 R
n ( n - 1 ) ( n - 2 ) = n - 2
n n - 1 = n - 2 2
n 2 - n = n 2 + 4 - 4 n
3 n = 4
n = 4 3

Topic: Physics, Ray Optics, JEE Main
A natural number has prime factorization given by n=2x3y5z, where y and z are such that y+z=5 and y-1+z-1=56,y>z. Then the number of odd divisors of n, including 1, is:

Options:
A 12
B 6
C 11
D 6x

The correct answer is:
12

Solution:
y+z=5

1y+1z=56 y>z

⇒y=3,z=2

⇒n=2x.33.52=2.2.2…3.3.35.5

For calculating the odd divisor ofn=2x3y5z, x must be 0.

Hence, Number of odd divisors =3+1×2+1=4×3=12.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (26 Feb Shift 2)
A nauseating smell in the carbylamine test for primary amines is due to the formation of

Options:
A Isocyanide
B Chloroform
C Cyanide
D DDT

The correct answer is:
Isocyanide

Solution:
A test for primary amines by reaction with an alcoholic solution of potassium hydroxide and trichlormethane (or chloroform) is as follows:
\(\underset{\substack{\text{Primary amine}} \\ (\mathrm{R}= \text{ alkyl or aryl)}}{\mathrm{R}-\mathrm{NH}_2+\mathrm{CHCl}_3+3 \mathrm{KOH(alc)}} \xrightarrow{(\text {warm})} \underset{\text{Alkyl isocyanide}}{\mathrm{R}-\mathrm{NC}+3 \mathrm{KCl}+3 \mathrm{H}_2 \mathrm{O}}\)
The isocyanide RNC is recognized by its unpleasant smell. This reaction of primary amines is called the carbylamine reaction.
Hence, (1) is the correct option

Topic: Chemistry, Amines, JEE Main
A NCC parade is going at a uniform speed of 9 km h-1 under a mango tree on which a monkey is sitting at a height of 19.6 m. At any particular instant, the monkey drops a mango. A cadet will receive the mango whose distance from the tree at time of drop is :
(Given g=9.8 m s-2)

Options:
A 5 m
B 10 m
C 19.8 m
D 24.5 m

The correct answer is:
5 m

Solution:
If a cadet receives the mango, then the time taken by the mango to fall and the cadet to reach there should be equal.

Time taken by mango to fall t=2hg.

⇒t=2×19.69.8=2 s

Therefore, the distance of the cadet from the tree will be, d=vt.

⇒d=9×518×2=5 m

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2022 (28 Jul Shift 1)
A negative catalyst will

Options:
A lower the activation energy of the reaction
B does not alter the activation energy of the reaction
C provide a new and easy path to the reaction
D increases the activation energy of the reaction

The correct answer is:
increases the activation energy of the reaction

Solution:
A negative catalyst will increases the activation energy of the reaction.

Topic: Chemistry, Chemical Kinetics, JEE Main
A negative test charge is moving near a long straight wire carrying a current. The force acting on the test charge is parallel to the direction of the current. The motion of the charge is:

Options:
A Away from the wire
B Towards the wire
C Parallel to the wire along the current
D Parallel to the wire opposite to the current

The correct answer is:
Towards the wire

Solution:
Force on a moving charge, F→=qv→×B→
Here, ∵q=-ve
⇒F→=-qv→×B→
∴motion of-q is towards the wire.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2017 (09 Apr Online)
A network of four capacitors capacity equal $C_1=C, C_2=2 C, C_3=3 C$ and $C_4=$ $4 C$ are conducted to a battery as shown in the figure. The ratio of the change on $C_2$ and $C_4$ is:

Options:
A $\frac{4}{7}$
B $\frac{3}{22}$
C $\frac{7}{4}$
D $\frac{22}{3}$

The correct answer is:
$\frac{3}{22}$

Solution:
Here $C_1, C_2$ and $C_3$ are in series.
$$
\frac{1}{C^{\prime}}=\frac{1}{C}+\frac{1}{2 C}+\frac{1}{3 C}
$$
$$
\begin{aligned}
\frac{1}{C^{\prime}} & =\frac{6+3+2}{6 C}=\frac{11}{6 C} \\
\Rightarrow \quad C & =\frac{6 C}{11}
\end{aligned}
$$
All the capacitors in branch number 1 is in series so the charge on each number capacitor is:
$$
Q^{\prime}=\frac{6}{11} C V
$$
Also charge on capacitor $C_4$ is $Q=4$ V
$$
\therefore \text { Ratio }=\frac{Q^{\prime}}{Q}=\frac{6 \mathrm{CV}}{11 \times 4 C V}=\frac{3}{22}
$$

Topic: Physics, Capacitance, JEE Main
A network of four resistances is connected to 9 V battery, as shown in figure. The magnitude of voltage difference between the points A and B is __________ V.

Options:

The correct answer is:
3

Solution:
The resistance in each branch is 6 Ω. They are in parallel. So, net resistance is R'=62=3 Ω.

The total current flowing through the loop,

I=VR'=9 V3 Ω=3 A

Thus, through each branch the current is 1.5 A.

Hence, the voltage difference across the points is

VA-VB=4×1.5-2×1.5=3 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (15 Apr Shift 1)
A neutral sphere of radius r and density Ï is intended to be suspended with an electric field E existing on the earth's surface directed vertically up. What fraction of the electrons should be removed from the sphere? Given that, the atomic number and atomic mass of the material of the sphere are Z and A respectively. Avogadro number is NA. Assume that the sphere is able to hold the necessary charge without any leakage.

Options:
A 2geEZANA
B 3gAeEZ
C gAeEZNA
D 2gAeEZNA

The correct answer is:
gAeEZNA

Solution:
mg = qE
⇒ mg = neE
⇒ 43πr3 ×ρ × g = neE
n = 4πr3ρg3eE
Number of moles of the metal in sphere= mA
= 43πr3ρA
Number of metal atoms= 43πr3ρ×NAA
Total number of electrons= 43πr3ρNA×ZA = N Say
Required fraction= nN=gAeEZNA

Topic: Physics, Electrostatics, JEE Main
A neutral sphere of radius r and density ρ is placed in a uniform electric field E that exists on the earth's surface in the vertically upward direction. If the atomic number and the mass number of the material of the sphere are Z and A respectively, then the fraction of electrons that should be removed from the sphere for it to remain in equilibrium is [Assume that the sphere is able to hold the necessary charge without any leakage. Here NA -- Avogadro number]

Options:
A nntotal=ρgAeENAZ
B nntotal=4gAπeENAZ
C nntotal=gAeENAZ
D nntotal=πρgA3eENAZ

The correct answer is:
nntotal=gAeENAZ

Solution:
Let us assume that we remove n electrons from the sphere due to which the net charge on the sphere is q=ne

For the sphere to be in equilibrium

mg=qE=neE

43πr3ρg=neE

⇒n=4πr3ρg3eE
The number of moles of the metal in the sphere is

nmol=mA=4πr3ρ3A

So the number of metal atoms will be
nmetal atoms=4πr3ρ3A×NA

This means the total number of electrons in the sphere is

ntotal=4πr3ρNA3A×Z

Hence, the fraction is

nntotal=4πr3ρg3eE×3A4πr3ρNAZ

⇒nntotal=gAeENAZ

Topic: Physics, Electrostatics, JEE Main
A neutron, a proton, an electron and an α particle enter a region of uniform magnetic field with same velocities. The magnetic field is perpendicular and directed into the plane of the paper. The tracks of the particles are labeled in the figure. The electron follows the track

Options:
A D
B C
C B
D A

The correct answer is:
D

Solution:
r=mvBq
r is least when mq is least.
mq is least for electron ,so electron will trace path with minimumm radius i.e. track D

Topic: Physics, Magnetic Effects of Current, JEE Main