datavorous/entrance-exam-dataset huggingface dataset:
To an aqueous solution containing ions such as Al3+,Zn2+,Ca2+,Fe3+,Ni2+,Ba2+ and Cu2+ was added conc. HCl, followed by H2S. The total number of cations precipitated during this reaction is/are :

Options:
A 1
B 3
C 4
D 2

The correct answer is:
1

Solution:
Al3+ and Fe3+ sulphides hydrolyse in water. Ni2+ and Zn2+ require basic medium with H2 S to form precipitate. Ca2+ and Ba2+ sulphides are also soluble. only Cu2+ can form the precipitate with H2S in the acidic medium.

Topic: Chemistry, Practical Chemistry, JEE Main, JEE Main 2021 (27 Jul Shift 2)
To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mol of He and 1.0 mol of an unknown compound (vapour pressure 0.68 atm. at 0°C ) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases at 0°C is close to

Options:

The correct answer is:
7

Solution:
PHe=1-0.68=0.32atm,n=0.1
V=nRTP=0.1×0.0821×2730.32=7

Topic: Chemistry, States of Matter, JEE Main
To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mole of He and 1.0 mole of an unknown compound (vapour pressure 0.68 atm at 0οC ) are introduced. Considering ideal gas behaviour, the total volume (in L) of the gas at 0οC is close to x L at equilibrium. Calculate the value of x.

Options:

The correct answer is:
7

Solution:
If external pressure is 1 atm the pressure inside the cylinder will also be equal to 1 atm.
The pressure generated by unknown compound will be fixed and equal to 0.68 atm, so the pressure generated by He will be 1â0.68=0.32 atm
Now by: V=nÃRÃTP
V= 0.1×0.082×273 0.32 =7 L

Topic: Chemistry, States of Matter, JEE Main
To avoid slipping while walking on ice, one should take smaller steps because of the

Options:
A Friction of ice is large
B Larger normal reaction
C Friction of ice is small
D Smaller normal reaction

The correct answer is:
Friction of ice is small

Solution:
To avoid slipping one should take smaller steps because the frictional force of ice is small.
Smaller steps will give larger normal force and more the normal force more the friction.
Also the advantage of talking small steps is that lateral forces are decreased. When we put down are foot it will be pushing forwards on the concrete and at the end of the step it will be pushing backwards. Hence larger the step these forward and backward forces will be larger increasing the chances of slipping.

Topic: Physics, Laws of Motion, JEE Main
To convert a galvanometer into a voltmeter one should connect a:

Options:
A high resistance in series with galvanometer
B low resistance in series with galvanometer
C high resistance in parallel with galvanometer
D low resistance in parallel with galvanometer

The correct answer is:
high resistance in series with galvanometer

Solution:
To convert a galvanometer into a voltmeter one should connect a high resistance in series with galvanometer

Topic: Physics, Current Electricity, JEE Main
To demonstrate the phenomenon of interference we require two sources which emit radiation of

Options:
A nearly the same frequency
B the same frequency
C different wavelength
D the same frequency and having a definite phase relationship

The correct answer is:
the same frequency and having a definite phase relationship

Solution:
For interference phenomenon, two sources should emit radiation of the same frequency and having a definite phase relationship.

Topic: Physics, Wave Optics, JEE Main
To demonstrate the phenomenon of interference, we require two sources which emit radiation

Options:
A
of nearly the same frequency

B
of the same frequency

C
of different wavelengths

D
of the same frequency and having a definite phase relationship

The correct answer is:
of the same frequency and having a definite phase relationship

Solution:
No Solution Available

Topic: Physics, Wave Optics, JEE Main, JEE Main 2003
To detect the reducing and non reducing sugars, which of the following test is used

Options:
A Molisch test
B Biuret test
C Fehling's test
D Millions test

The correct answer is:
Fehling's test

Solution:
1)Molisch's test is a chemical test which is used to check for the presence of carbohydrates in a given analyte.
2) The biuret test, also known as Piotrowski's test, is a chemical test used for detecting the presence of peptide bonds. In the presence of peptides, a copper(II) ion forms mauve-colored coordination complexes in an alkaline solution.
3)Fehling's test is used for reducing sugars and non-reducing sugars, supplementary to the Tollens' reagent test. The test was developed by German chemist Hermann von Fehling in 1849.
4). Millon's reagent is an analytical reagent used to detect the presence of soluble proteins.

Topic: Chemistry, Biomolecules, JEE Main
To determine refractive index of glass slab using a travelling microscope, minimum number of readings required are :

Options:
A Two
B Four
C Three
D Five

The correct answer is:
Three

Solution:
Reading one ⇒ without slab
Reading two ⇒ with slab
Reading three ⇒ with Lycopodium powder

Topic: Physics, Ray Optics, JEE Main, JEE Main 2016 (10 Apr Online)
To determine the half-life of a radioactive element, a student plots graph of lndN (t)dt versus t. Here dN (t)dt is the rate of radioactive decay at time t. If the number of radioactive nuclei of this element decreases by a factor of p after 4.16 yr, the value of p is
e-2.08=0.125

Options:

The correct answer is:
8

Solution:
dNdt= |Activity of radioactive substance|
= Î»N = Î»N0e-Î»t (âµ N = N0e-Î»t)
Taking log both sides
lndNdt= ln( Î»N0)-Î»t
Hence, lndNdt versus t graph is a straight line with slope-Î».
From the graph, we can see that,
Î»=12=0.5 yr-1
Now applying the equation
N = N0 e-Î»t=N0 e-0.5 Ã 4.16
=N0e-2.08 = N0e-2Ã0.693 = 0.125 N0
= N08
i.e. nuclei decrease by a factor of 8.
Hence the answer is 8.

Topic: Physics, Nuclear Physics, JEE Main
To determine the half-life of a radioactive element, a student plots graph of lndN(t)dt versus t. Here dN(t)dt is the rate of radioactive decay at time t. If the number of radioactive nuclei of this element decreases by a factor of p after 4.16 yr, the value of p is

Options:
A 8
B 7
C 4
D 8.5

The correct answer is:
8

Solution:
dNdt= |Activity of radioactive substance|
= λN = λN0e-λt (∵ N = N0e-λt)
Taking log both sides
lndNdt= ln( λN0)-λt
Hence, lndNdt versus t graph is a straight line with slope-λ.
From the graph, we can see that,
λ=12=0.5 yr-1
Now applying the equation
N = N0 e-λt=N0 e-0.5 × 4.16
=N0e-2.08 = N0e-2×0.693 = 0.125 N0
= N08
i.e. nuclei decrease by a factor of 8.
Hence the answer is 8.

Topic: Physics, Nuclear Physics, JEE Main
To determine the resistance $(R)$ of a wire, a circuit is designed below. The $V-I$ characteristic curve for this circuit is plotted for the voltmeter and the ammeter readings as shown in figure. The value of $R$ is ______ $\Omega$.

Options:

The correct answer is:
2500

Solution:
$\begin{aligned} & \text { Req }=\frac{10^4 R}{10^4+R} \\ & E=4 V, I=2 m A \\ & I=\frac{E}{R e q} \Rightarrow 2 \times 10^{-3}=\frac{4\left(10^4+R\right)}{10^4 R} \\ & \Rightarrow 20 R=40000+4 R\end{aligned}$
$\begin{aligned} & 16 R=40000 \\ & R=2500 \Omega\end{aligned}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (09 Apr Shift 2)
To double the covering range of a TV transmitting tower, its height should be multiplied by:

Options:
A 12
B 4
C 2
D 2

The correct answer is:
4

Solution:
The range of a tower in the communication system is given by

Range=2hR

Here, h= Height of the tower,

R= radius of Earth

Given, The range of tower is to be doubled.

As, Range ∝h, to make the range double, the height h have to be made 4 times.

Topic: Physics, Communication System, JEE Main, JEE Main 2019 (12 Jan Shift 2)
To draw maximum current from a combination of cells, how should the cells be grouped

Options:
A Series
B Parallel
C Mixed
D Depends upon the relative values of external and internal resistance Solution:

The correct answer is:
Depends upon the relative values of external and internal resistance Solution:

Solution:
Let each cell of emf ' \(\mathrm{E}\) ' and internal resistance ' \(\mathrm{r}\) '. There are \(\mathrm{n}\) such cells.The external resistance be ' \(R\) '.
If cells are connected in series :
current through 'R' \(=\frac{\mathrm{nE}}{\mathrm{nr}+\mathrm{R}}\)
If cells are connected in parallel :
current through ' \(\mathrm{R}\) ' \(=\frac{\mathrm{nE}}{\mathrm{r}+\mathrm{nR}}\)
Do draw a maximum current :
if \(\mathrm{nr}+\mathrm{R} > \mathrm{r}+\mathrm{nR}\) cells must be connected in parallel.
if \(\mathrm{r}+\mathrm{nR} > \mathrm{nr}+\mathrm{R}\) cells must be connected in series.
So it depends on the relative values of internal and external resistances.

Topic: Physics, Current Electricity, JEE Main
To establish an instantaneous current of $2 \mathrm{~A}$ through a $1 \mu \mathrm{F}$ capacitor ; the potential difference across the capacitor plates should be changed at the rate of:

Options:
A
$2 \times 10^4 \mathrm{~V} / \mathrm{s}$

B
$4 \times 10^6 \mathrm{~V} / \mathrm{s}$

C
$2 \times 10^6 \mathrm{~V} / \mathrm{s}$

D
$4 \times 10^4 \mathrm{~V} / \mathrm{s}$

The correct answer is:
$2 \times 10^6 \mathrm{~V} / \mathrm{s}$

Solution:
As, C $=\frac{\mathrm{Q}}{\mathrm{V}}=\frac{\mathrm{It}}{\mathrm{V}}$
$$
\begin{aligned}
& \Rightarrow \frac{\mathrm{V}}{\mathrm{t}}=\frac{\mathrm{I}}{\mathrm{C}}=\frac{2}{1 \times 10^{-6}} \\
& =2 \times 10^6 \mathrm{~V} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Capacitance, JEE Main, JEE Main 2013 (22 Apr Online)
To find the focal length of a convex mirror, a student records the following data:

Object pin	Convex Lens	Convex Mirror	Image Pin

22.2 cm	32.2 cm	45.8 cm	71.2 cm

The focal length of the convex lens is f1 and that of mirror is f2 . Then taking index correction to be negligibly small, f1 and f2 are close to:

Options:
A f1=7.8 cm f2=12.7 cm
B f1=12.7Â cm f2=7.8Â cm
C f1=15.6Â cm f2=25.4Â cm
D f1=7.8Â cm f2=25.4Â cm

The correct answer is:
f1=7.8 cm f2=12.7 cm

Solution:
Take f2=12.7
1f=1v+1u
112.7=125.4+1u
112.7=125.4=1u
u=25.4=v′
1f=1v-1u
1f1=125.4+13.6+110
1f1=139+110
1f1=39049=7.96
The closest answers is (1) as option (3) and (4) are not possible.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2016 (09 Apr Online)
To find the numerical value of $\int_{-2}^2\left(p x^2+q x+s\right) d x$ it is necessary to know the values of constants

Options:
A $p$
B $q$
C $s$
D $p$ and $s$

The correct answer is:
$p$ and $s$

Solution:
Given integral $I=\int_{-2}^2\left(p x^2+s\right) d x+q \int_{-2}^2 x d x$
$I=2 \int_0^2\left(p x^7+s\right) d x+0=\frac{4}{3}(4 p+3 s)$
Thus, to find the numerical value of $I$, it is necessary to know the value of $p$ and $s$.

Topic: Mathematics, Definite Integration, JEE Main
To find the resistance of a galvanometer by the half deflection method the following circuit is used with resistances $\mathrm{R}_1=9970 \mathrm{~W}, \mathrm{R}_2=30 \mathrm{~W}$ and $\mathrm{R}_3=0$. The deflection in the galvanometer is d. With $\mathrm{R}_3=107 \mathrm{~W}$ the deflection changed to $\frac{d}{2}$. The galvanometer resistance is approximately:

Options:
A
$107 \Omega$

B
$137 \Omega$

C
$107 / 2 \Omega$

D
$77 \Omega$

The correct answer is:
$77 \Omega$

Solution:
No Solution Available

Topic: Physics, Current Electricity, JEE Main, JEE Main 2013 (22 Apr Online)
To find the standard potential of M3+/M electrode, the following cell is constituted: Pt/M/M3+ 0.001 mol L-1/Ag+ 0.01 mol L-1/Ag
The emf of the cell is found to be 0.421 volt at 298 K. The standard potential of half-reaction M3++3e-→M at 298 K will be:
(Given: EAg+Ag⊝ at 298 K=0.80 volt)

Options:
A 0.38 volt
B 1.28 volt
C 0.32 volt
D 0.66 volt

The correct answer is:
0.32 volt

Solution:
Ecell=Eocell−0.0591nlogQ
∴ Q=M3+Ag+3 =0.001(0.01)3=103
Ecell=Eocell−0.05913log103
Ecell=Eocell−0.05913×3=0.421Eocell−0.0591=0.421Eocell=0.48 V =EoAg+/Ag-EoM3+/MEoM3+/M=0.80-0.48 = 0.32 V

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2017 (09 Apr Online)
To find the value of $\int \frac{1+\log x}{x} d x$, the proper substitution is

Options:
A $\log x=t$
B $1+\log x=t$
C $\frac{1}{x}=t$
D None of these

The correct answer is:
$1+\log x=t$

Solution:
\(\begin{aligned}
& \text {So, Let } 1+\log x=t \\
& \frac{1}{x} \cdot dx = dt\\
& \Rightarrow \int \frac{1+\log x}{x} \cdot d x=\int t \cdot d t \\
& \Rightarrow \frac{t^2}{2}+c \\
& \Rightarrow \frac{(1+\log x)^2}{2}+c
\end{aligned}\)

Topic: Mathematics, Indefinite Integration, JEE Main
To get an output 1. from the circuit shown in the figure, the input must be

Options:
A $A=0, B=1, C=0$
B $A=1, B=0, C=0$
C $A=1, B=0, C=1$
D $A=1, B=1, C=0$

The correct answer is:
$A=1, B=0, C=1$

Solution:
The Boolean expression for the given combination is
output $Y=(A+B) . C$
The truth table is
$\begin{array}{|c|c|c|c|} \hline A & B & C & Y=(A+B) \cdot C \\ \hline 0 & 0 & 0 & 0 \\ \hline 1 & 0 & 0 & 0 \\ \hline 0 & 1 & 0 & 0 \\ \hline 0 & 0 & 1 & 0 \\ \hline 1 & 1 & 0 & 0 \\ \hline 0 & 1 & 1 & 1 \\ \hline 1 & 0 & 1 & 1 \\ \hline 1 & 1 & 1 & 1 \\ \hline \end{array}$
Hence $A=1, B=0, C=1$

Topic: Physics, Semiconductors, JEE Main
To get an output of 1 from the circuit shown in the figure the input must be:

Options:
A a=0,Â b=0,Â c=1
B a=1,Â b=0,Â c=0
C a=1, b=0, c=1
D a=0,Â b=1,Â c=0

The correct answer is:
a=1, b=0, c=1

Solution:
Output of OR gate must be 1 and c=1
So, a=1, b=0
or a=0, b=1

Topic: Physics, Semiconductors, JEE Main, JEE Main 2016 (10 Apr Online)
To get an output $Y=1$ in given circuit which of the following input will be correct

Options:
A A-1, B-0, C-0
B A-1, B-0, C-1
C A-1, B-1, C-0
D A-0, B-1, C-0

The correct answer is:
A-1, B-0, C-1

Solution:
Here $Y=(A+B) \cdot C$
Truth Table
$C$ should be 1 either $A$ and $B$ or both $A$ and $B$ are 1
Therefore, $A=1, B=0, C=1$

Topic: Physics, Semiconductors, JEE Main
To get an output $Y=1$ in given circuit which of the following input will be correct

Options:
A A-1, B-0, C-0
B A-1, B-0, C-1
C A-1, B-1, C-0
D A-0, B-1, C-0

The correct answer is:
A-1, B-0, C-1

Solution:
Here $Y=(A+B) \cdot C$
Truth Table
$C$ should be 1 either $A$ and $B$ or both $A$ and $B$ are 1
Therefore, $A=1, B=0, C=1$

Topic: Physics, Semiconductors, NEET, NEET 2012 (Mains)
To get output '1' at R, for the given logic gate circuit the input values must be:

Options:
A X=1,Â Y=1
B X=0,Â Y=0
C X=0,Â Y=1
D X=1, Y=0

The correct answer is:
X=1, Y=0

Solution:
R= ( X ¯ +Y )+( X⋅ Y ¯ ¯ ) ¯ =( X ¯ +Y ¯ )⋅( X. Y ¯ ¯ ¯ )=( X. Y ¯ )⋅( X. Y ¯ )=( X. Y ¯ )

Topic: Physics, Semiconductors, JEE Main, JEE Main 2019 (10 Jan Shift 1)
To get three images of a single object, one should have two plane mirrors at an angle of

Options:
A
$60^{\circ}$

B
$90^{\circ}$

C
$120^{\circ}$

D
$30^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
According to Image formula
$\mathrm{n}=\frac{360}{\theta}-1 \Rightarrow 3=\frac{360}{\theta}-1$
$\Rightarrow \frac{360}{\theta}=4 \Rightarrow \theta=\frac{360}{4}=90$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2003
To increase sugar production in sugarcanes, they are sprayed with:

Options:
A
IAA

B
cytokinin

C
gibberellin

D
ethylene

The correct answer is:
gibberellin

Solution:
Gibberellin sprayed on sugarcane induces growth in the internodal area of the sugarcane plant. Larger is the area, larger will be the sugar content, as sucrose is the main constituent in the of sap of sugarcane stem.

Topic: Biology, Plant - Growth and Development
To increase the resonant frequency in series LCR circuit,

Options:
A Source frequency should be increased
B Another resistance should be added in series with the first resistance.
C Another capacitor should be added in series with the first capacitor
D The source frequency should be decreased

The correct answer is:
Another capacitor should be added in series with the first capacitor

Solution:
The resonant frequency of an LCR circuit is given by f=12πLC. It does not depend on the source frequency as well as the resistance of the circuit.

Therefore, to increase the resonating frequency product of L and C should decrease.

By joining another capacitor in series, the equivalent capacitor will decrease.

Hence, option C is correct option.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (25 Jul Shift 1)
To inhibit the growth of tumours, identify the compounds used from the following:

(A) EDTA
(B) Coordination Compounds of Pt
(C) D-Penicillamine
(D) Cis - Platin

Choose the correct answer from the option given below:

Options:
A B and D Only
B C and D Only
C A and B Only
D A and C Only

The correct answer is:
B and D Only

Solution:
Cis - Platin PtNH32Cl2 is a chemotherapy medication used to treat a number of cancers. It is given by injection into a vein. Cisplatin binds to the N7 reactive center on purine residues and as such can cause deoxyribonucleic acid (DNA) damage in cancer cells, blocking cell division and resulting in apoptotic cell death.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (30 Jan Shift 1)
To know the resistance G of a galvanometer by half deflection method, a battery of emf VE and resistance R is used to deflect the galvanometer by angle θ . If a shunt of resistance S is needed to get half deflection the G, R and S are related by the equation:

Options:
A SR+G=RG
B 2SR+G=RG
C 2G=S
D 2S=G

The correct answer is:
SR+G=RG

Solution:
Let resistance S is connected with galvanometer of resistance G, than

IG=VR+G
Now, as shown below figure a shunt S is connected in parallel with the galvanometer for half deflection in it, that is IG' = IG2, than

I=VR+GSG+S
As, shunt and galvanometer are in parallel then VG=VS
IG'G=I-IG'S
⇒IG'G+S=IS
⇒IG2=ISG+S ∵IG'=IG2

On, substituting the values of IG and I in above equation
V2R+G=VR+GSG+S×SG+S
⇒12R+G=SRG+S+GS
⇒RG+S+GS=2SR+G
⇒RG+RS+GS=2SR+G
⇒RG=2SR+G-SR+G
⇒RG=SR+G

Topic: Physics, Current Electricity, JEE Main, JEE Main 2016 (09 Apr Online)
To light, a 50 W,100 V lamp is connected, in series with a capacitor of capacitance 50πx μF, with 200 V, 50 Hz AC source. The value of x will be _____ .

Options:

The correct answer is:
3

Solution:
Electrical power is given by P=V2R⇒R=V2P

Resistance of lamp is R=100×10050⇒R=200 Ω

Let VC and VR be the voltage across capacitor and resistor.

Here, VR2+VC2=V2

Current through lamp is i=100200=12 A

⇒1002+VC2=2002

⇒VC2=30000⇒VC=1003 V

We know that, V=i×XC

Then, XC=2003 Ω=1ωC

So, capacitance of capacitor is C=120×50×203=50×10-6x

x=50×10-6×100×2003

Thus, value of x=3.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (27 Jul Shift 1)
To maintain a rotor at a uniform angular speed of 200 $\mathrm{rad} / \mathrm{s}$, an engine needs to transmit a torque of $180 \mathrm{~N}-\mathrm{m}$.
What is the power of the engine required?

Options:

None

Solution:
$\omega=200 \mathrm{rad} / \mathrm{s}, \tau=180 \mathrm{~N}-\mathrm{m}, P=\tau \omega=180 \times 200=36000$
$$
\mathrm{W}=36 \mathrm{~kW}
$$

Topic: Physics, System of Particles and Rotational Motion
To maintain a speed of 80 km h-1 by a bus of mass 500 kg on a plane rough road for 4 km distance, the work done by the engine of the bus will be _____ kJ. [The coefficient of friction between tyre of bus and road is 0.04]

Options:

The correct answer is:
784

Solution:
It is given,

v=80 km h-1

m=500 kg

S=4 km

μ=0.04

The power due to friction is

Pf=Fv where F=μmg

=0.04500×9.8×80×518 W

Hence, the work done by the engine in overcoming the friction is

W=Pft=Pfdv

⇒W=20×9.8×80×518×4×10380×518=784 kJ

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (12 Apr Shift 1)
To measure the internal resistance of a battery, potentiometer is used. For $\mathrm{R}=10 \Omega$, the balance point is observed at $l=500 \mathrm{~cm}$ and for $\mathrm{R}=1 \Omega$ the balance point is observed at $l=400 \mathrm{~cm}$. The internal resistance of the battery is approximately :

Options:
A $0.2 \Omega$
B $0.3 \Omega$
C $0.4 \Omega$
D $0.1 \Omega$

The correct answer is:
$0.3 \Omega$

Solution:
Let potential gradient be $\lambda$.
$\begin{aligned}
& \therefore \mathrm{i} \times 10=\lambda \times 500=\varepsilon-\mathrm{ir}_{\mathrm{s}} \\
& \Rightarrow 500 \lambda=\varepsilon-50 \lambda \mathrm{r}_{\mathrm{s}}
\end{aligned}$
Also,
$\begin{aligned}
& \mathrm{i}^{\prime} \times 1=\lambda \times 400=\varepsilon-\mathrm{i}^{\prime} \mathrm{r}_{\mathrm{s}} \\
& \Rightarrow 400 \lambda=\varepsilon-400 \lambda \mathrm{r}_5 \\
& \therefore 100 \lambda=350 \lambda \mathrm{r}_{\mathrm{s}} \Rightarrow \mathrm{r}_{\mathrm{s}}=\frac{10}{35} \approx 0.3 \Omega
\end{aligned}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (04 Apr Shift 1)
To measure the temperature coefficient of resistivity α of a semiconductor, an electrical arrangement shown in the figure is prepared. The arm BC is made up of the semiconductor. The experiment is being conducted at 25°C and resistance of the semiconductor arm is 3 mΩ. Arm BC is cooled at a constant rate of 2 °C s-1. If the galvanometer G shows no deflection after 10 s, then α is:

Options:
A -2×10-2C-1 ∘
B -1.5×10-2C-1 ∘
C -1×10-2 C-1∘
D -2.5×10-2 C-1∘

The correct answer is:
-1×10-2 C-1∘

Solution:
For no deflection(balanced condition) 0.81=R3

⇒R=2.4 mΩ

Temperature fall in 10 s = 20 °C

ΔR=RαΔt

⇒α=ΔRRΔt=2.4-33×20

=-1×10-2 °C-1

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (01 Feb Shift 2)
To measured a magnetic field between the magnetic poles of a loud speaker, a small coil having 30 turns and \(2.5 \mathrm{~cm}^2\) area is placed perpendicular to the field and removed immediately. If the total charge flown through the coil is \(75 \times 10^{-3} \mathrm{C}\) and the total resistance of wire and galvanometer is \(0.3 \Omega\), then the magnitude of the magnetic field is

Options:
A \(0.03 \mathrm{~T}\)
B \(0.3 \mathrm{~T}\)
C \(3 \mathrm{~T}\)
D \(3 \times 10^2 \mathrm{~T}\)

The correct answer is:
\(0.3 \mathrm{~T}\)

Solution:
Given,
number of turns in the coil, \(N=30\) turns
area of the coil, \(A=2.5 \mathrm{~cm}^2=2.5 \times 10^{-4} \mathrm{~m}^2\)
total charge flowing through the coil,
\(Q_{\text {Net }}=7.5 \times 10^{-3} \mathrm{C}\)
and total resistance of wire and galvanometer,
\(R=0.3 \Omega\)
We know that,
\(\text { net charge, } Q_{n e t}=\frac{\text { flux in the coil }}{\text { net Resistance }}\)
\(\therefore \quad Q_{\text {net }}=\frac{\phi}{R}\) ...(i)
We know that, flux \(\phi=\) magnetic field \(\times\) area
\(\times\) number of turns in the coil
or \(\phi=B N A\) ...(ii)
Now, from Eqs. (i) and (ii), we get
\(\therefore \quad Q_{\text {Net }}=\frac{B N A}{R}\)
Putting the given values, we get
\(\begin{aligned}
7.5 \times 10^{-3} & =\frac{B \times 30 \times\left(2.5 \times 10^{-4}\right)}{0.3} \\
7.5 \times 10^{-3} & =\frac{B \times 7.5 \times 10^{-3}}{0.3} \\
B & =0.3 \mathrm{~T}
\end{aligned}\)
Hence, the magnitude of the magnetic field, \(B=0.3 \mathrm{~T}\).

Topic: Physics, Electromagnetic Induction, JEE Main
To mop-clean a floor, a cleaning machine presses a circular mop of radius R vertically down with a total force F and rotates it with a constant angular speed about its axis. If the force F is distributed uniformly over the mop and if coefficient of friction between the mop and the floor is μ, the torque, applied by the machine on the mop is:

Options:
A 2 μFR/3
B μFR/3
C μFR/6
D μFR/2

The correct answer is:
2 μFR/3

Solution:
Net torque =âTorque on ring
â«dÏ=â«0RÎ¼ÃFÏR2Ã2Ïx dxx
Ï=2Î¼3FR

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (10 Jan Shift 1)
To neutralise $20 \mathrm{~m} /$ of $M / 10$ sodium hydroxide, the volume of $M / 20$ hydrochloric acid required is

Options:
A $10 \mathrm{ml}$
B $15 \mathrm{ml}$
C $20 \mathrm{ml}$
D $40 \mathrm{ml}$

The correct answer is:
$40 \mathrm{ml}$

Solution:
$\mathrm{NaOH} \quad \mathrm{HCl}$
$N_1 V_1=N_2 V_2 ; \quad 20 \times \frac{1}{10}=\frac{1}{20} \times V ; V=40 \mathrm{ml}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main