datavorous/entrance-exam-dataset huggingface dataset:
A current of 2 A flows through a 2 Î© resistor when connected across a battery. The same battery supplies a current of 0.5 A when connected across a 9 Î© resistor. The internal resistance of the battery is

Options:
A 13 Ω
B 14 Ω
C 5Â Î©
D 0.5Â Î©

The correct answer is:
13 Ω

Solution:
Current I=ER+r
⇒2=E2+r .... (i)
⇒0.5=E9+r .... (ii)
On dividing equation (i) by equation (ii), we get
20.5=9+r2+r
⇒4=9+r2+r
⇒3r=1
⇒r=13 Ω

Topic: Physics, Current Electricity, JEE Main
A current of 2 A flows through a wire of cross-sectional area 25.0 mm2. The number of free electrons in a cubic meter are 2.0×1028. The drift velocity of the electrons is _____ ×10-6 ms-1

(given, charge on electron =1.6×10-19C).

Options:

The correct answer is:
25

Solution:
The drift velocity is given by

vd=IneA

⇒I=nAevd ...(i)

The given data is

I=2 AA=25×10-6 m2n=2×1028e=1.6×10-19 C

Substituting the given values in equation (i)

2=2×1028×25×10−6×1.6×10−19×vd⇒vd=22×1028×25×10−6×1.6×10−19

⇒vd=125×1.6×1000 m s-1=25×10−6 m s-1

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (08 Apr Shift 1)
A current of 2 A is flowing in the sides of an equilateral triangle of side 9 cm. The magnetic field at the centroid of the triangle is

Options:
A 1.66×10-6 T
B 1.22×10-4 T
C 1.33×10-6 T
D 1.44×10-4 T

The correct answer is:
1.33×10-6 T

Solution:
The magnetic field at the centre O due to the current through side AB is
B1=μ0I4πasin⁡θ1+sin⁡θ2
As the magnetic field due to each of the three sides is the same in magnitude and direction. So, the total magnetic field at O is sum of all the fields.
i.e. B=3B1=3μ0I4πθ sin⁡θ1+sin⁡θ2
Here, tan⁡θ1=ADOD
⇒ tan⁡60o=12a
⇒ a=123=9×10-223
Now B=3×4π×10-7×24π×9×10-223 [sin⁡60o+sin⁡60o]
=439×10-632+32
=43×10-6
∴ B=1.33×10-6 T

Topic: Physics, Magnetic Effects of Current, JEE Main
A current of 2 amp when passed for 5 hour through a molten salt deposits 22.2 g of metal of atomic mass 177. The oxidation state of the metal in the metal salt is
[Report your answer by rounding it upto nearset whole number]

Options:

The correct answer is:
3.00

Solution:
W=ItE96500
E=WÃ96500It=22.2Ã965002Ã5Ã3600=59.5
Equivalent mass =Atomic massOxidation state
59.5=177n
So nâ3 .

Topic: Chemistry, Electrochemistry, JEE Main
A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is:

Options:
A 11×105 W
B 11×10-3 W
C 11×10-5 W
D 11×10-4 W

The correct answer is:
11×10-5 W

Solution:
P=V2R=i2RResistance R=V2P=Pi2

Here, Power P=4.4 W and i=2 mA
so R=4.44 ×10-6 Ω
When 11 V power supply is connected across R
P′=1124.4×4×10-6
=11×10-5 W.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (10 Jan Shift 2)
A current of \( 2 \mathrm{~A} \) is established in a circuit containing an aluminium resistor of length \( 30 \mathrm{~cm} . \) If the area of cross-section of the resistor is \( 0.003 \mathrm{~cm}^{2} \) at a point P. The current density at that point will be:

Options:
A \( 6.67 \times 10^{6} \mathrm{~A} \mathrm{~m}^{2} \)
B \( 6.67 \times 10^{6} \mathrm{~A} \mathrm{~m}^{-2} \)
C \( 6.67 \times 10^{7} \mathrm{~A} \mathrm{~m}^{-2} \)
D \( 6.67 \times 10^{7} \mathrm{Am}^{2} \)

The correct answer is:
\( 6.67 \times 10^{6} \mathrm{~A} \mathrm{~m}^{-2} \)

Solution:
Current density at any point in a conductor is defined by current per unit area of cross section at that point.

J=IA

Given I=2 A and area A=0.003 cm2

J=23×10-7=6.67×106 A m-2

Topic: Physics, Current Electricity, JEE Main
A current of $3 \mathrm{amp}$. flows through the $2 \Omega$ resistor shown in the circuit. The power dissipated in the $5 \Omega$ resistor is

Options:
A 5 watt
B 4 watt
C 2 watt
D 1 watt

The correct answer is:
5 watt

Solution:
Potential difference across $2 \Omega=6 \mathrm{~V}$
Current through $5 \Omega=\frac{6}{6}=1 \mathrm{~A}$
Power dissipated in $5 \Omega=(1)^2 \times 5=5$ Watt

Topic: Physics, Current Electricity, NEET, NEET 2008 (Mains)
A current of 5.0 A flows for 4.0 h through an electrolytic cell containing a molten salt of metal M. This results in deposition of 0.25 mol of the metal M at the cathode. The oxidation state of M in the molten salt is + x . the value of ' x ' is ( 1 Faraday = 96000 C mol-1)

Options:

The correct answer is:
3

Solution:
I=5.0 A, t=4.0 h
Moles of metal deposited =14
Q=5.0Ã4Ã60Ã60 coulombs
Charge in faradays
Q= 5Ã4Ã60Ã60 96000 =0.75âF
14 moles of metal will be deposited by 34 F charge required
So, for 1 mole, 34Ã4=3F
Oxidation =+3

Topic: Chemistry, Electrochemistry, JEE Main
A current of 5 A is flowing at 220 V in the primary coil of a transformer. If the voltage produced in the secondary coil is 2200 V and 50% of power is lost, then the current in the secondary coil is

Options:
A 0.25 A
B 2.5 A
C 0.5 A
D 5 A

The correct answer is:
0.25 A

Solution:
Given, power output
=12× power input (or 50% of input power)
∴IsVs=12Ip⋅Vp ∵P=VI
Here, Ip=5 A, Vp=220 V
and Vs=2200 V
∴Is=Ip⋅Vp2Vs=5×2202×2200=0.25 A

Topic: Physics, Alternating Current, JEE Main
A current of 5 A is passing through a non-linear magnesium wire of cross-section 0.04 m2. At every point the direction of current density is at an angle of 60° with the unit vector of area of cross-section.
The magnitude of electric field at every point of the conductor is: (resistivity of magnesium ρ=44×10-8 Ωm)

Options:
A 11×10-2 V m-1
B 11×10-7 V m-1
C 11×10-5 V m-1
D 11×10-3 V m-1

The correct answer is:
11×10-5 V m-1

Solution:
I=J→·A→=JAcos(θ)
5=J4100×cos(60)
J=5×50=250 A m-2
Now, E→=ρ×J→
=44×10-8×250=11×10-5 V m-1

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (20 Jul Shift 1)
A current of 6 A enters one corner P of an equilateral triangle PQR having 3 wires of resistance 2 Ω each and leaves by the corner R. The currents i1 in ampere is

Options:

The correct answer is:
2

Solution:
For parallel combination, current divides in the inverse ratio of resistance.

iPQ=26×6 A=2 A.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (25 Feb Shift 2)
A current of 10.0 A flows for 2.00 h through an electrolytic cell containing a molten salt of metal X. This results in the decomposition of 0.250 mol of metal X at the cathode. The oxidation state of X in the molten salt is:

F=96,500 C

Options:
A 1+
B 4+
C 3+
D 2+

The correct answer is:
3+

Solution:
Quantity of Electricity = 1 0 × 3 6 0 0 × 2 9 6 5 0 0
=0.746≈0.75F.
If oxidation state is n+ for 1 mol deposition of a metal:
Electricity=nF (Faradays)
For 0.250 mol=0.250 nF
∴ 0.250 nF=0.75 F
n=3

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2014 (09 Apr Online)
A current of \( 10 A \) passes through two very long wires held parallel to each other and separated by a distance of \( 1 \mathrm{~m} \). The magnitude of force per unit length between them is [Use in \( \frac{\mu_{0}}{4 \pi}=10^{-7} \) S.I. unit]

Options:
A \( 2 \times 10^{-5} \mathrm{Nm}^{-1} \)
B \( 3 \times 10^{-5} \mathrm{Nm}^{-1} \)
C \( 4 \times 10^{-5} \mathrm{Nm}^{-1} \)
D \( 5 \times 10^{-6} \mathrm{Nm}^{-1} \)

The correct answer is:
\( 2 \times 10^{-5} \mathrm{Nm}^{-1} \)

Solution:
The magnitude of force per unit length between the two parallel long wires is

Fl=μ0i22πr=2×10-71021=2×10-5 N m-1.

Topic: Physics, Magnetic Effects of Current, JEE Main
A current of 10 ampere is flowing in a wire of length $1.5 \mathrm{~m}$. A force of $15 \mathrm{~N}$ acts on it when it is placed in a uniform magnetic field of 2 tesla. The angle between the magnetic field and the direction of the current is

Options:
A $30^{\circ}$
B $45^{\circ}$
C $60^{\circ}$
D $90^{\circ}$

The correct answer is:
$30^{\circ}$

Solution:
Force on a current carrying wire of length \(l\) and current i, placed in uniform magnetic field \(B\) is given by
\(\begin{aligned}
& \mathrm{F}=\mathrm{i}(\overrightarrow{\mathrm{l}} \times \overrightarrow{\mathrm{B}}) \\
& \mathrm{F}=\mathrm{IlB} \sin \theta \text { or } \sin \theta=\frac{\mathrm{F}}{\mathrm{IlB}}
\end{aligned}\)
\(\sin \theta=\frac{15}{10 \times 1.5 \times 2}=\frac{1}{2} \text { or } \theta=30^{\circ}\)

Topic: Physics, Magnetic Effects of Current, JEE Main
A current of 15 mA flows in the circuit as shown in figure. The value of potential difference between the points A and B will be

Options:
A 50V
B 75V
C 150V
D 275V

The correct answer is:
275V

Solution:
Using current division rule,

Current, i1=510+5×15 mA=5 mA

Now, using Kirchhoff's voltage law- in lower loop,

VA-5i-10i1-10i=VB

Potential difference between A and B is VA-VB=75+50+150=275 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (25 Jul Shift 2)
A current of $2.5 \mathrm{~A}$ flows through a coil of inductance $5 \mathrm{H}$. The magnetic flux linked with the coil is:

Options:
A $0.5 \mathrm{~Wb}$
B $12.5 \mathrm{~Wb}$
C zero
D $2 \mathrm{~Wb}$

The correct answer is:
$12.5 \mathrm{~Wb}$

Solution:
According to the question
Current $I=2.5 \mathrm{~A}$ and Induction $L=5 \mathrm{H}$
$\therefore$ Magnetic flux linked with coil is
$$
\begin{aligned}
\phi & =L I=(5 \mathrm{H}) \\
(2.5 \mathrm{~A}) & =12.5 \mathrm{~Wb}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, JEE Main
A current of 200 μA deflects the coil of a moving coil galvanometer through 60°. The current to cause deflection through π10 radian is

Options:
A 30 μA
B 120 μA
C 60 μA
D 180 μA

The correct answer is:
60 μA

Solution:
In a moving coil galvanometer, the current is related to the angle of deflection by the formula:

i∝θ ...1

From equation (1), it follow sthat
i2i1=θ2θ1⇒i2200 μA=π10π3=310⇒i2=60 μA

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (27 Jan Shift 2)
A current of dry air was passed first through a series of bulbs containing a solution of C6H5–NO2 in ethanol of molality 0.725 and then through a series of bulbs containing pure ethanol. (T = 284 K) loss in weight of the solvent bulbs was 0.0685 g. Calculate the loss in weight of the solution bulbs.

Options:
A 4.60 g
B 5.20 g
C 2.50 g
D 2.05 g

The correct answer is:
2.05 g

Solution:
If n2 and n1 are the number of moles of solute and solvent
Then n2n1=mM1000
Here m = molality
M = molar mass of the solvent.
n 2 n 1 = 0.725Ã46 1000 i. e.,0.0334
Thus
Weight loss in solution bulb âPs
Weight loss in solution bulb âPo-Ps
Now, Po-PsPs=n2n1
â0.0685Ps=0.0334
â´Ps=2.05g

Topic: Chemistry, Solutions, JEE Main
A current strength of 0.965 amperes is passed through excess fused AlCl3 for 5 hours. How many litres of chlorine will be liberated at STP ? F=96500 C

Options:
A 2.016
B 1.008
C 11.2
D 20.16

The correct answer is:
20.16

Solution:
1âFâ11.2 L Cl 2 at STP
â´No.of Faradays=9.65Ã5Ã60Ã6096500=1.8
â´Volumeâ of Cl 2 at STP=1.8Ã11.2âL
= 20.16 L

Topic: Chemistry, Electrochemistry, JEE Main
A current-carrying loop is dropped from the position shown. The loop starts entering into a region of uniform magnetic field, directed into the plane. (Neglect effects of self-induction and g is the acceleration due to gravity.) Which of the following statements is correct?

Options:
A The loop will fall with an acceleration more than g
B The loop will fall with an acceleration less than g
C The magnitude of acceleration of loop will increase as it moves down
D The magnitude of acceleration of loop will remain constant as it moves down

The correct answer is:
The loop will fall with an acceleration less than g

Solution:
Apply the concept of Lenz's law.

Topic: Physics, Electromagnetic Induction, NEET
A current-carrying loop is shown in the figure. The magnitude of the magnetic field produced at a point O is

Options:
A μ0I4R1+2π
B μ0I4R1-2π
C μ0I2R1+2π
D μ0I4R1+22π

The correct answer is:
μ0I4R1+2π

Solution:
B0=μ0I2R×12+4×μ0I4π×2R×12

B0=μ0I4R1+2π

Topic: Physics, Magnetic Effects of Current, JEE Main
A current-carrying uniform square frame is suspended from hinged supports as shown in the diagram such that it can freely rotate about its upper side. The length and mass of each side of the frame are 2 m and 4 kg respectively. A uniform magnetic field B → = 3 i ^ + 4 j ^ is applied. When the wireframe is rotated to 45° from vertical and released it remains in equilibrium. What is the magnitude of current (in A) in the wireframe?

Options:

The correct answer is:
10

Solution:
Î¼ â (Magnetic moment of loop) when it is lifted by 45â=i l2j^+k^2
â´ Ï â due to magnetic field =Î¼âÃBâ=i l22j^+k^Ã3i^+4j^
Ï â due to mg (about top edge) =4mgl2cos45Â°i^
â´ For equilibrium net torque along x-axis = 0
â´ 4mgl22=4i l22âi=mg2l=10 A

Topic: Physics, Magnetic Effects of Current, JEE Main
A current-carrying wire is placed in the grooves of an insulating semicircular disc of radius R, as shown. The current enters at point A and leaves from point B. Determine the magnetic field at point D.

Options:
A μ 0 I 8 π R 3
B Î¼ 0 I 4 Ï R 3
C 3 μ 0 I 4 π R
D none of these

The correct answer is:
μ 0 I 4 π R 3

Solution:
Magnetic field due to wire AC,
BAC=Î¼0I4Ï2Rsin30cos30-cos90=3Î¼0I8ÏR

Magnetic field due to wire BC,
BBC=Î¼0I4Ï2Rsin60cos60-cos90=Î¼0I83ÏR=3Î¼0I24ÏR
B â net = B A - B B
=3Î¼0I8ÏR1-13
= 2 3 Î¼ 0 I 2 4 Ï R = Î¼ 0 I 4 3 Ï R

Topic: Physics, Magnetic Effects of Current, JEE Main
A current-carrying wire is placed in the grooves of an insulating semicircular disc of radius R, as shown. The current enters at point A and leaves from the point B. Determine the magnetic field at point D.

Options:
A μ0I8πR3
B Î¼0I4ÏR3
C 3μ0I4πR
D none of these

The correct answer is:
μ0I4πR3

Solution:
The angle formed by lines from each end of diameter to any point of the semicircle is 90°.

Therefore, CB ⊥DB and CA⊥DA.

Magnetic field due to wire AC,
BAC=μ0I4π2Rsin30°cos30°-cos90°=3μ0I8πR

Magnetic field due to wire BC,
BBC=μ0I4π2Rsin60°cos60°-cos90°=μ0I83πR=3μ0I24πR
B→net=BA-BB
=3μ0I8πR1-13
=23μ0I24πR=μ0I43πR

Topic: Physics, Magnetic Effects of Current, JEE Main
A current-carrying wire is placed in the grooves of an insulating semicircular disc of radius R as shown. The current enters at point A and leaves from point B. Determine the magnetic field at point D.

Options:
A μ 0 I 8 π R 3
B Î¼ 0 I 4 Ï R 3
C 3 μ 0 I 4 π R
D none of these

The correct answer is:
μ 0 I 4 π R 3

Solution:
Magnetic field due to wire AC,
BAC=μ0I4π2Rsin30cos30-cos90=3μ0I8πR

Magnetic field due to wire BC,
BBC=μ0I4π2Rsin60cos60-cos90=μ0I83πR=3μ0I24πR
B → net = B A - B B
=3μ0I8πR1-13
= 2 3 μ 0 I 2 4 π R = μ 0 I 4 3 π R

Topic: Physics, Magnetic Effects of Current, JEE Main
A current-carrying wire of certain length is bent to form an arc that subtends an angle θ at the centre. Let B be the magnetic field at the centre due to this wire. The correct graph between B and θ is

Options:
A
B
C
D

The correct answer is:

Solution:
lR=θ
B=μ0iθ2π2R
B=μ0ilθ24πl2
B∝θ2

Topic: Physics, Magnetic Effects of Current, JEE Main
A curve amongst the family of curves represented by the differential equation, x2-y2 dx+2xy dy=0 which passes through 1,1, is

Options:
A A circle with centre on the x- axis.
B A circle with centre on the y- axis.
C A hyperbola with transverse axis along the x- axis.
D An ellipse with major axis along the y- axis.

The correct answer is:
A circle with centre on the x- axis.

Solution:
x2dx-y2dx+2xy dy=0

x2dx=y2dx-2x ydy

⇒-dx=dy2x (exact differential equation)

⇒ -x=y2x+C

⇒ x2+y2+Cx=0

∵ It passes through 1,1.

Hence, C=-2

⇒ x2+y2-2x=0

Topic: Mathematics, Differential Equations, JEE Main
A curve amongst the family of curves represented by the differential equation, x2-y2 dx+2xy dy=0 which passes through 1,1, is

Options:
A A circle with centre on the x- axis.
B A circle with centre on the y- axis.
C A hyperbola with transverse axis along the x- axis.
D An ellipse with major axis along the y- axis.

The correct answer is:
A circle with centre on the x- axis.

Solution:
x2dx-y2dx+2xy dy=0

x2dx=y2dx-2x ydy

⇒-dx=dy2x (exact differential equation)

⇒ -x=y2x+C

⇒ x2+y2+Cx=0

∵ It passes through 1,1.

Hence, C=-2

⇒ x2+y2-2x=0

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (10 Jan Shift 2)
A curve between saturation magnetization and temperature of a ferromagnetic sample is

Options:
A
B
C
D

The correct answer is:

Solution:
Saturation magnetization of ferromagnetic materials decrease gently upto a point situated just before the curie temperature. As after curie temperature ferromagnetic materials become paramagnetic materials, saturation magnetization decreases steeply near curie temperature as shown in the figure.

Topic: Physics, Magnetic Properties of Matter, JEE Main
A curve in the first quadrant is such that the slope of OP is twice the slope of the tangent drawn at P to the curve, where O is the origin and P is any general point on the curve. If the curve passes through 4,2, then its equation is

Options:
A y=x2-14
B y2=x
C y=x3-62
D y=sin⁡x-4+2

The correct answer is:
y2=x

Solution:
Slope of OP=2dydxP
⇒yx=2dydx
⇒2∫dyy=∫dxx
⇒2ln⁡y=ln⁡x+ln⁡c
⇒y2=cx
As it passes through 4,2⇒c=1
∴ the equation of the curve is y2=x

Topic: Mathematics, Differential Equations, JEE Main
A curve is such that the slope of the tangent to it at any point P is the product of ordinate of P and abscissa increased by 2. If it passes through -4,e, then its equation is

Options:
A y=x2-16+e
B xy+4e=0
C y=ex22+2x+1
D y=ex+5

The correct answer is:
y=ex22+2x+1

Solution:
dydx=yx+2
⇒∫dyy=∫x+2dx
⇒ln⁡y=x22+2x+c
⇒y=ex22+2x+c
As it passes through -4,e⇒c=1
∴ Equation of the curve is y=ex22+2x+1

Topic: Mathematics, Differential Equations, JEE Main
A curve is such that the x-intercept of the tangent drawn to it at the point Px,y is reciprocal of the abscissa of P. Then, the equation of the curve is (where, c is the constant of integration and x>1)

Options:
A y=cx2-1
B y=cx2+1
C y=cx2-1
D y=cx-1

The correct answer is:
y=cx2-1

Solution:
The general equation of the tangent is
Y-y=dydxX-x
â´x-intercept =x-ym
âx-ym=1x
âx-1x=ym
Or m=yx-1x
âdydx=yx-1x
âdyy=xdxx2-1
On integrating, we get
â«dyy=â«xx2-1dx
âlnâ¡y=12â«2xdxx2-1=12lnâ¡x2-1+lnâ¡c
âlnâ¡y=lnâ¡x2-1c
ây=câx2â1

Topic: Mathematics, Differential Equations, JEE Main
A curve passes through the point $\left(1, \frac{\pi}{6}\right)$. Let the slope of the curve at each point $(x, y)$ be $\frac{y}{x}+\sec \left(\frac{y}{x}\right), x>0$, then, the equation of the curve is

Options:
A $\sin \left(\frac{y}{x}\right)=\log (x)+\frac{1}{2}$
B $\operatorname{cosec}\left(\frac{y}{x}\right)=\log (x)+2$
C $\sec \left(\frac{2 y}{x}\right)=\log (x)+2$
D $\cos \left(\frac{2 y}{x}\right)=\log (x)+\frac{1}{2}$

The correct answer is:
$\sin \left(\frac{y}{x}\right)=\log (x)+\frac{1}{2}$

Solution:
$$
\begin{aligned}
& \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{y}{x}+\sec \left(\frac{y}{x}\right) \\
& \text { Put } y=\mathrm{v} x \\
& \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{v}+x \frac{\mathrm{dv}}{\mathrm{d} x} \\
& \therefore \quad \mathrm{v}+x \frac{\mathrm{dv}}{\mathrm{d} x}=\mathrm{v}+\sec \mathrm{v} \\
& \quad \Rightarrow \cos \mathrm{v} \mathrm{dv}=\frac{1}{x} \mathrm{~d} x
\end{aligned}
$$
Integrating both sides, we get
$$
\begin{aligned}
& \sin \mathrm{V}=\log (x)+\mathrm{c} \\
& \Rightarrow \sin \left(\frac{y}{x}\right)=\log (x)+\mathrm{c}
\end{aligned}
$$
The curve passes through $\left(1, \frac{\pi}{6}\right)$.
$$
\sin \left(\frac{\pi}{6}\right)=\log (1)+c \Rightarrow c=\frac{1}{2}
$$
$\therefore \quad$ Equation (ii) becomes
$$
\sin \left(\frac{y}{x}\right)=\log (x)+\frac{1}{2}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
A curve passes through the point \( \left(1, \frac{\pi}{6}\right) \). Let the slope of the curve at each point \( (\mathrm{x}, \mathrm{y}) \) be \( \frac{\mathrm{y}}{\mathrm{x}}+\sec \left(\frac{\mathrm{y}}{\mathrm{x}}\right), \mathrm{x} > 0 \). Then the equation of the curve is

Options:
A \( \sin \left(\frac{y}{x}\right)=\log x+\frac{1}{2} \)
B \( \operatorname{cosec}\left(\frac{y}{x}\right)=\log x+2 \)
C \( \sec \left(\frac{2 y}{x}\right)=\log x+2 \)
D \( \cos \left(\frac{2 y}{x}\right)=\log x+\frac{1}{2} \)

The correct answer is:
\( \sin \left(\frac{y}{x}\right)=\log x+\frac{1}{2} \)

Solution:
Given slope at (x, y) is

dydx=yx+secyx......i
let yx=t⇒y=xt

Differentiating this wrt x

⇒dydx=t+xdtdx......ii

Now using equation i and ii, we have
t + x dt dx = t + sec t

⇒dtsect=dxx

Integrating both sides.
∫costdt=∫1xdx
sint=lnx+c

Now Put value of t in above equation.
sinyx=lnx+c.....iii
Given that curve passes through 1, π6 so put in above equation.
sinπ6=ln1+c⇒c=12

Now, put the value of c in equation iii
sinyx=lnx+12

Topic: Mathematics, Differential Equations, JEE Main
A curve passing through the point 1,2 and satisfying the condition that slope of the normal at any point is equal to the ratio of ordinate and abscissa of that point, then the curve also passes through the point

Options:
A 0,0
B 2,2
C 2,1
D 3,2

The correct answer is:
2,1

Solution:
Given -1dydx=yx
dydx=-xy
ydy+xdx=0
y2+x2=C
Since, it passes through 1,2⇒4+1=C⇒C=5
x2+y2=5
Also, passes through 2,1

Topic: Mathematics, Application of Derivatives, JEE Main
A curve passing through the point \( (1,2) \) and satisfying the condition that slope of the normal at any point is equal to the
ratio of ordinate and abscissa of that point, then the curve also passes through the point

Options:
A \( (0,0) \)
B \( (2,2) \)
C \( (2,1) \)
D \( (3,2) \)

The correct answer is:
\( (2,1) \)

Solution:
Given -1dydx=yx
dydx=-xy
ydy+xdx=0
y2+x2=C
Since, it passes through 1,2⇒4+1=C⇒C=5
x2+y2=5
Also, passes through 2,1

Topic: Mathematics, Application of Derivatives, JEE Main
A curved in a level road has a radius 75 m. The maximum speed of a car turning this curved road can be 30 m s-1 without skidding. If radius of curved road is changed to 48 m and the coefficient of friction between the tyres and the road remains same, then maximum allowed speed would be _____ m s-1.

Options:

The correct answer is:
24

Solution:
The maximum speed is given as v=μgr, where, μ is coefficient of friction and r is radius of curved road.

⇒v1v2=r1r2

⇒30v2=7548=2516=54

⇒v2=24 m s-1

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2022 (25 Jun Shift 2)
A cut diamond sparkles because of its

Options:
A Hardness
B High refractive index
C Emission of light by the diamond
D Absorption of light by the diamond

The correct answer is:
High refractive index

Solution:
Due to high refractive index its critical angle is very small so that most of the light incident on the diamond is total internally reflected repeatedly and diamond sparkles.

Topic: Physics, Ray Optics, JEE Main
A cycle wheel of radius $0.5 \mathrm{~m}$ is rotated with constant angular velocity of $10 \mathrm{rad} / \mathrm{s}$ in a region of magnetic field of $0.1 \mathrm{~T}$ which is perpendicular to the plane of the wheel. The EMF generated between its centre and the rim is

Options:
A $0.25 \mathrm{~V}$
B $0.125 \mathrm{~V}$
C $0.5 \mathrm{~V}$
D zero

The correct answer is:
$0.125 \mathrm{~V}$

Solution:
When a conducting disc or wheel of radius $r$ rotates with constant angular velocity of $\omega$ about its axis in a uniform magnetic field perpendicular to its plane and parallel to its axis of rotation, then,
Induced emf is given by, $\mathrm{e}=\frac{1}{2}{\mathrm{~B} \omega \mathrm{r}^2}^2$
Here, $B=0.1 \mathrm{~T}, \omega=10 \mathrm{rad} / \mathrm{s}, \mathrm{r}=0.5 \mathrm{~m}$
Substituting these values in Eq. (i), we get
$$
\therefore \quad \mathrm{e}=\frac{1}{2} \times 0.1 \times 10 \times(0.5)^2=\frac{1}{8} \mathrm{~V}=0.125 \mathrm{~V}
$$

Topic: Physics, Electromagnetic Induction, JEE Main
A cyclic process ABCA is shown in the V-T diagram. Process on the P-V diagram is

Options:
A
B
C
D

The correct answer is:

Solution:
From the given VT diagram, In process AB, VâT
The pressure is constant (As the quantity of the gas remains the same).
In process. BC, V= Constant
and in process CA, T= constant
So, These processes are correctly represented on the PV diagram by graph C.

Topic: Physics, Thermodynamics, NEET