datavorous/entrance-exam-dataset huggingface dataset:
A circle of unit radius touches the positive x and y axes. Two parallel tangents AB and DC having slopes -1 are drawn to the circle cutting x-axis at B and C and y-axis at A and D respectively. The square of the area of the trapezium ABCD is equal to

Options:

The correct answer is:
32

Solution:
Let the tangent line be x+y=c
Now, |1+1-c|2=1⇒c=2±2
∴A0,2+2,B2+2,0
C2-2,0,D0,2-2
So, the area of the trapezium ABCD
=12222+2+2·2-2

=42 sq. units

Hence, the square of the area =422=32

Topic: Mathematics, Circle, JEE Main
A circle passes through $(0,0)$ and $(1,0)$ and touches the circle $x^2+y^2=9$, then the centre of circle is

Options:
A $\left(\frac{3}{2}, \frac{1}{2}\right)$
B $\left(\frac{1}{2}, \frac{3}{2}\right)$
C $\left(\frac{1}{2}, \frac{1}{2}\right)$
D $\left(\frac{1}{2}, \pm \sqrt{2}\right)$

The correct answer is:
$\left(\frac{1}{2}, \pm \sqrt{2}\right)$

Solution:
Radius of the circle $r=\frac{3}{2}$
$\frac{1}{4}+k^2=\frac{9}{4} \Rightarrow k= \pm \sqrt{2}$
$\text { Hence centre is }\left(\frac{1}{2}, \pm \sqrt{2}\right)$

Topic: Mathematics, Circle, JEE Main
A circle passes through -2,4 and touches the y-axis at 0,2. Which one of the following equations can represent a diameter of this circle ?

Options:
A 2x-3y+10=0
B 3x+4y-3=0
C 4x+5y-6=0
D 5x+2y+4=0

The correct answer is:
2x-3y+10=0

Solution:
As y-axis is tangent to the circle therefore the centre of the circle would be x,2 and distance of centre from 0,2 and -2,4 should be equal.
∴ h2+02=h+22+22
⇒h2=h2+4h+4+4
⇒h=-2
Hence, centre is -2,2.
∴ 2x-3y+10=0 is the diameter, as centre lies on it.

Topic: Mathematics, Circle, JEE Main
A circle passes through -2,4 and touches the y-axis at 0,2. Which one of the following equations can represent a diameter of this circle ?

Options:
A 2x-3y+10=0
B 3x+4y-3=0
C 4x+5y-6=0
D 5x+2y+4=0

The correct answer is:
2x-3y+10=0

Solution:
As y-axis is tangent to the circle therefore the centre of the circle would be x,2 and distance of centre from 0,2 and -2,4 should be equal.
∴ h2+02=h+22+22
⇒h2=h2+4h+4+4
⇒h=-2
Hence, centre is -2,2.
∴ 2x-3y+10=0 is the diameter, as centre lies on it.

Topic: Mathematics, Circle, JEE Main, JEE Main 2016 (09 Apr Online)
A circle passes through the points 2,3 and 4,5. If its centre lies on the line y-4x+3=0, then its radius is equal to :

Options:
A 5
B 2
C 2
D 1

The correct answer is:
2

Solution:
Let centre of circle be Cα, β.

It lies in line y-4x+3-0⇒β=4α-3

∴ C α, 4α-3

∴CA=CB

By distance formula,

⇒α-22+4α-62=α-42+4α-82

⇒-4α+4-48α+36=-8α+16-64α+64

⇒64+8-4-48α=80-40

⇒α=4020=2

∴ C2,5

∴ r=2-22+5-32=2

Topic: Mathematics, Circle, JEE Main, JEE Main 2018 (15 Apr)
A circle passes through the points $(2,3)$ and $(4$, 5). If its centre lies on the line, $y-4 x+3=0$, then its radius is equal to

Options:
A
$\sqrt{5}$

B
1

C
$\sqrt{2}$

D
2

The correct answer is:
$\sqrt{2}$

Solution:
Equation of the line passing through the points $(2,3)$ and $(4,5)$ is
$$
y-3=\left(\frac{5-3}{4-2}\right) x-2 \Rightarrow x-y+1=0
$$
Equation of the perpendicular line passing through the midpoint $(3,4)$ is $x+y-7=0$
Lines (1) and (2) intersect at the center of the circle. So, the center of the circle is $(3,4)$ Therefore, the radius is
$$
\begin{aligned}
&\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2} \\
&=\sqrt{(2-3)^2+(3-4)^2}=\sqrt{2} \text { units. }
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A circle passing through (0,0),(2,6) (6,2) cut the $x$ -axis at the point $P \neq(0,0)$ Then, the lenght of $O P,$ where $O$ is the origin, is

Options:
A $\frac{5}{2}$
B $\frac{5}{\sqrt{2}}$
C 5
D 10

The correct answer is:
5

Solution:
Let the equation of circle is
$$
x^{2}+y^{2}+2 g x+2 f y+c=0
$$
When, circle (i) passes through the onigin
Then, $\quad c=0$
When, circle (i) passes through the point (2,6).
Then $4+36+4 g+12 f+0=0$
$\Rightarrow \quad 4 g+12 f+40=0$
$\Rightarrow g+3 f=-10$
When, circle (1) passes through the point (6,2) Then, $36+4+12 g+4 t+0=0$ [from Eq. $\Rightarrow \quad 12 g+4 f+40=0$
$\Rightarrow \quad 3 g+t+10=0$
On solving Eqs. (iii) and (iv), we get $g=\frac{-5}{2}$ and $f=\frac{-5}{2}$
$\therefore$ Equation of circle becomes
$$
x^{2}+y^{2}-5 x-5 y=0
$$
Circle cut the $x$ -axis.
So, put $y=0$ in Eq. $(v)$, we get
$$
x^{2}-5 x=0 \Rightarrow x(x-5)=0
$$
$\Rightarrow \quad x=5$
So, the circle cut the $x$ -axis at point $P(5,0)$
$\therefore$ The length of $O P=5$

Topic: Mathematics, Circle, JEE Main
A circle $S=x^2+y^2+2 g x+2 f y+4=0$ cuts the circle $x^2$ $+y^2-4 x-4 y-4=0$ orthogonally and makes an angle of $60^{\circ}$ with the circle $x^2+y^2+4 x+4 y+4=0$. Then the radius of the circle $\mathrm{S}=0$ is

Options:
A 4
B 3
C 5
D 1

The correct answer is:
4

Solution:
2 circles cut orthogonally then
$$
\begin{aligned}
& 2 g_1 g_2+2 f_1 f_2=C_1+C_2 \\
\Rightarrow & 2 g(-2)+2 f(-2)=4-4 \\
\Rightarrow & g+f=0
\end{aligned}
$$
Also angle between circles,
$$
\begin{aligned}
& \cos \theta=\frac{r_1^2+r_2^2-d^2}{2 r_2} \\
\Rightarrow & \cos 60^{\circ}=\frac{\eta^2+4-(2-g)^2-(2-f)^2}{2 r_1 \cdot(2)} \\
\Rightarrow & 2 \eta_1=\left|\eta_1^2+4-\left(4+g^2-4 g\right)-\left(4+f^2-4 f\right)\right| \\
\Rightarrow & \left.2 r_1=\mid \eta^2+4-4-g^2+4 g-4-f^2+4 f\right) \mid \\
\Rightarrow & 2 r_1=\left|\eta_1^2-g^2-f^2-4+4(g+f)\right| \\
\Rightarrow & 2 r_1=\left|\eta_1^2-\left(g^2+f^2\right)-4\right| \\
\Rightarrow & 2 r_1=\left|\eta_1^2-\left(g^2+f^2-4\right)-8\right| \\
\Rightarrow & 2 r_1=\left|r_1^2-r_1^2-8\right| \Rightarrow 2 \eta_1=8 \Rightarrow \eta=4 .
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Main
A circle touches both the y-axis and the line x+y=0. Then the locus of its center

Options:
A y=2x
B x=2y..
C y2-x2=2xy
D x2−y2=2xy

The correct answer is:
x2−y2=2xy

Solution:
Plotting the diagram of given data we have,

Now to find locus, let centre be h,k. According to question

OM=ON distance

⇒h=h+k2

squaring both sides we get

⇒2h2=h+k2

⇒2h2=h2+k2+2hk

Replacing h with x & k with y we get

2x2=x2+y2+2xy

⇒x2−y2=2xy

Topic: Mathematics, Circle, JEE Main, JEE Main 2022 (25 Jun Shift 2)
A circle touches the hypotenuse of a right-angled triangle at its midpoint and passes through the midpoint of the shorter side. If 3 units and 4 units are the length of the sides other than hypoteuse and r be the radius of the circle, then the value of 3r is

Options:

The correct answer is:
5

Solution:
Using coordinate geometry, assume endpoints of hypotenuse as A(3,0) and B(0,4), the third vertex as origin,

such that the midpoint of the hypotenuse is P32,2 and the midpoint of the shorter side is Q32,0

Let centre be C(h,ââk)

âµCPâ¥AB

â2âk32âh=34

â6hâ8k=â7 â¦ (i)

âµCP=CQ

â(hâ32)2+(kâ2)2=(hâ32)2+k2

âk=1 and

6h=1âh=16

â´ radius (r)=CQ=(16â32)2+1=(1â96)2+1=53

â´3r=5

Topic: Mathematics, Circle, JEE Main
A circle touches the parabola $\mathrm{y}^{2}=4 \mathrm{x}$ at $(1,2)$ and also touches its directrix. The $\mathrm{y}$-coordinates of the point of contact of the circle and the directrix is-

Options:
A $\sqrt{2}$
B 2
C $2 \sqrt{2}$
D 4

The correct answer is:
$2 \sqrt{2}$

Solution:
$y^{2}=4 x$
$2 y \frac{d y}{d x}=4$
$m_{\mathrm{T}}=\frac{2}{y}=\frac{2}{2}=1$
Circle $\rightarrow \mathrm{S}+\lambda \mathrm{L}=0$
$(\mathrm{x}+1)^{2}+(\mathrm{y}-\alpha)^{2}+\lambda(\mathrm{x}+1)=0$...(1)
differentiate
$$
\begin{array}{l}
2(x+1)+2(y-\alpha) \frac{d y}{d x}+\lambda=0 \\
x=1, y=2 \\
4+2(2-\alpha) m_{T}+\lambda=0 \\
\lambda=2 \alpha-8...(2)
\end{array}
$$
$(1,2)$ satisfies cq. $(1)$
$$
\begin{array}{l}
2^{2}+(2-\alpha)^{2}+2 \lambda=0 \\
\alpha^{2}-4 \alpha+8+2(2 \alpha-8)=0 \\
\alpha^{2}=8 \\
\alpha=2 \sqrt{2}
\end{array}
$$

Topic: Mathematics, Parabola, JEE Main
A circle touches the $x$-axis and also touches the circle with centre at $(0,3)$ and radius
2. The locus of the centre of the circle is

Options:
A
an ellipse

B
a circle

C
a hyperbola

D
a parabola

The correct answer is:
a parabola

Solution:
Equation of circle with centre $(0,3)$ and radius 2 is $x^2+(y-3)^2=4$
Let locus of the variable circle is $(\alpha, \beta)$
$\because$ It touches $x$-axis.
$\therefore$ It equation $(x-\alpha)^2+(y-\beta)^2=\beta^2$
Circles touch externally
$$
\begin{aligned}
& \therefore \sqrt{\alpha^2+(\beta-3)^2}=2+\beta \\
& \alpha^2+(\beta-3)^2=\beta^2+4+4 \beta \\
& \alpha^2=10(\beta-1 / 2)
\end{aligned}
$$
$\therefore$ Locus is $x^2=10(y-1 / 2)$ which is parabola.

Topic: Mathematics, Circle, JEE Main, JEE Main 2005
A circle touches the y-axis at the point 0,4 and passes through the point 2,0 . Which of the following lines is not a tangent to this circle?

Options:
A 4x-3y+17=0
B 3x-4y-24=0
C 3x+4y-6=0
D 4x+3y-8=0

The correct answer is:
4x+3y-8=0

Solution:
Equation of family of circle
x-02+y-42+Î»x=0 ⇒passes2,0
4+16+2Î»=0⇒Î»=-10
x2+y2-10x-8y+16=0
Centre 5,4,R=25+16-16=5
Check the options.
Option (4)
4Ã5+3Ã4-85=245â 5

Topic: Mathematics, Circle, JEE Main, JEE Main 2020 (09 Jan Shift 1)
A circle touching the x- axis at (3, 0) and making an intercept of length 8 on the y- axis passes through the point:

Options:
A 3, 10
B 2, 3
C 3, 5
D 1, 5

The correct answer is:
3, 10

Solution:
Since circle touches the x-axis at 3, 0, hence centre can be taken as 3,α and radius =α
Since y-intercept is 8, by applying Pythagorean theorem
32+42=α2⇒α=5,-5
Hence two circles are possible
x-32+y-52=25 and x-32+y+52=25.
Clearly (3, 10) satisfies the given equation of circle.

Topic: Mathematics, Circle, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A circle which passes through origin and cuts intercepts on axes $a$ and $b$, the equation of circle is

Options:
A $x^2+y^2-a x-b y=0$
B $x^2+y^2+a x+b y=0$
C $x^2+y^2-a x+b y=0$
D $x^2+y^2+a x-b y=0$

The correct answer is:
$x^2+y^2-a x-b y=0$

Solution:
Centre is $\left(\frac{a}{2}, \frac{b}{2}\right)$ and radius $=\sqrt{\frac{a^2+b^2}{4}}$

Topic: Mathematics, Circle, JEE Main
A circle with centre (2,3)and radius 4 intersects the line x+y=3 at the points P and Q. If the tangents at P and Q intersect at the point S(α,β), then 4α-7β is equal to

Options:

The correct answer is:
11

Solution:
A Circle having the centre 2,3 and radius 4.

So, the equation of circle will be

x-22+y-32=16

⇒x2+y2-4x-6y-3=0

Now, it is given x+y=3 passes through two points P & Q on circle and tangents are drawn at P & Q intersect at Sα,β,

So, x+y=3 will be chord of contact.

Now equation of chord of contact will be

Tα,β=0

⇒αx+βy-2x+α-3y+β-3=0

⇒xα-2+yβ-3-2α-3β-3=0

Now comparing above equation with x+y-3=0, we get

α-21=β-31=2α+3β+33

Now on solving above relation we get,

α=-6 & β=-5

Hence, the value of

4α-7β=-24+35=11

Topic: Mathematics, Circle, JEE Main, JEE Main 2023 (29 Jan Shift 2)
A circle with centre at $(2,4)$ is such that the line $x+y+2=0$ cuts a chord of length 6 . The radius of the circle is

Options:
A $\sqrt{41} \mathrm{~cm}$
B $\sqrt{11} \mathrm{~cm}$
C $\sqrt{21} \mathrm{~cm}$
D $\sqrt{31} \mathrm{~cm}$

The correct answer is:
$\sqrt{41} \mathrm{~cm}$

Solution:
Let $r$ be the radius of the circle.
Now, perpendicular distance
$$
\begin{aligned}
A C & =\frac{|2+4+2|}{\sqrt{1^2+1^2}}=\frac{8}{\sqrt{2}} \\
& =4 \sqrt{2}
\end{aligned}
$$
In right angled $\triangle C A B$,
$$
\begin{aligned}
r^2 & =(A C)^2+(A B)^2 \\
& =(4 \sqrt{2})^2+(3)^2=32+9 \\
\Rightarrow \quad r^2 & =41 \Rightarrow r=\sqrt{41}
\end{aligned}
$$

Topic: Mathematics, Circle, JEE Main
A circle $x^2+y^2+2 g x+2 f y+c=0$ passing through $(4,-2)$ is concentric to the circle $x^2+y^2-2 x+4 y+20=0$, then the value of $c$ will be

Options:
A $-4$
B $4$
C $0$
D $1$

The correct answer is:
$-4$

Solution:
Circle $x^2+y^2+2 g x+2 f y+c=0$ is concentric with $x^2+y^2-2 x+4 y+20=0$. Therefore centre is $(1,-2)$ and radius $=\sqrt{g^2+f^2-c}=\sqrt{(1-4)^2+(-2+2)^2} \Rightarrow c=-4$.

Topic: Mathematics, Circle, JEE Main
A circuit connected to an ac source of emf e=e0sin⁡100t with t in seconds, gives a phase difference of π4 between the emf e and current i . Which of the following circuits will exhibit this?

Options:
A RC circuit with R=1 kΩ and C=1 μF
B RL circuit with R=1 kΩ and L=1 mH
C RL circuit with R=1 kΩ and L=10 mH
D RC circuit with R=1 kΩ and C=10 μF

The correct answer is:
RC circuit with R=1 kΩ and C=10 μF

Solution:
ω=100 rad/sec

phase difference between e and i is given by

tanθ=XR ∴θ=π4

so tanθ=tanπ4=1

in case of RL circuit X=XL=ωL

And in case of RC circuit X=Xc=1ωc

For option A: R=1000 Ω and

XC=1ωC=11001×10-6=10000 Ω
tan⁡θ=XCR≠1
For option B: R=103 Ω and

XL=Lω=10-3100=0.1 Ω

tan⁡θ=XLR≠1
For option C:R=1000 Ω and

XL=ωL=10010=1000 Ω
tan⁡θ=XLR≠1
For optionD:R=1 kΩ=1000 Ω and

XC=1ωC=110010×10-6=1000 Ω
tan⁡θ=XCR=1;θ=45o

Topic: Physics, Alternating Current, JEE Main, JEE Main 2019 (08 Apr Shift 2)
A circuit consists of a coil with inductance L and an uncharged capacitor of capacitance C. The coil is in a constant uniform magnetic field such that the flux through the coil is ϕ. At time t=0 min, the magnetic field is abruptly switched OFF. Let ω0=1/LC and ignore the resistance of the circuit. Then,

Options:
A current in the circuit is It=ϕ/Lcosω0t
B magnitude of the charge on the capacitor is |Qt|=2Cω0sinω0t
C initial current in the circuit is infinite
D initial charge on the capacitor is Cω0ϕ

The correct answer is:
current in the circuit is It=ϕ/Lcosω0t

Solution:
At t=0 capacitor is uncharged and flux of inductor is ϕ.

Now, using ϕ=LI, at t=0 current in circuit is

I0=ϕL

Instantaneous current in circuit is dqdt.

where, q is solution of Ld2qdt2+qC=0

or d2qdt2+qLC=0

At t=0, current in circuit is non-zero, so instantaneous current is given by

=I0cosωt=ϕL·cosωt

Topic: Physics, Alternating Current, JEE Main
A circuit consists of three batteries of emf $E_{1}=1 \mathrm{V}, E_{2}=2 \mathrm{V}$ and $E_{3}=3 \mathrm{V}$ and internal resistances $1 \Omega, 2 \Omega$ and $1 \Omega$ respectively which are connected in parallel as shown in the figure. The potential difference between points $P$ and $Q$ is

Options:
A $1.0 \mathrm{V}$
B $2.0 \mathrm{V}$
C $2.2 \mathrm{V}$
D $3.0 \mathrm{V}$

The correct answer is:
$2.0 \mathrm{V}$

Solution:
$1 \Omega .2 \Omega$ and $1 \Omega$ are in parallel

So, the required internal resistance
$$
\begin{array}{l}
\frac{1}{r}=\frac{1}{r_{1}}+\frac{1}{r-{2}}+\frac{1}{r-{3}} \\
\frac{1}{r}=\frac{1}{1}+\frac{1}{2}+\frac{1}{1} \\
\frac{1}{r}=\frac{2+1+2}{2} \Rightarrow r=\frac{2}{5} \Omega
\end{array}
$$
The potential difference between points $P$ and $Q$
$\begin{aligned} E_{\text {dif }} &=\frac{\frac{E_{1}}{r_{1}}+\frac{E_{2}}{5_{2}}+\frac{E_{3}}{5_{3}}}{1 / r}=\frac{1}{1}+\frac{2}{2}+\frac{3}{1} \\ &=\frac{2+2+6}{5 / 2} \\=& \frac{2}{5 / 2}=\frac{10 / 2}{5 / 2}=\frac{5}{5} \times 2=2 \mathrm{V} \end{aligned}$

Topic: Physics, Current Electricity, JEE Main
A circuit consists of three identical lamps connected to a battery as shown in the figure. When the switch S is closed then the intensities of lamps A and B

Options:
A will increase by eight times
B will decrease by two times
C will increase by more than two times
D will remain the same

The correct answer is:
will increase by more than two times

Solution:
After the closing of S, C bulb or lamp become short-circuited
the potential across A and B before the closing of S=V3
Potential across A and B after the closing of S=V2
As P∝V2
∴ PiPf= 49
Pf=Pi×94= 2.25 Pi

Topic: Physics, Current Electricity, NEET
A circuit consists of three identical lamps connected to a battery as shown in the figure. When the switch S is closed then the intensities of lamps A and B

Options:
A will increase by eight times
B will decrease by two times
C will increase by more than two times
D will remain the same

The correct answer is:
will increase by more than two times

Solution:
After closing of S C bulb or lamp become short circuited
potential across A and B before closing of S=V3
Potential across A and B after closing of S=V2
As PâV2
â´PiPf= 49
Pf=PiÃ94= 2.25 Pi

Topic: Physics, Current Electricity, JEE Main
A circuit element X when connected to an AC supply of peak voltage 100 V gives a peak current of 5 A which is in phase with the voltage. A second element Y when connected to the same AC supply also gives the same value of peak current which lags behind the voltage by π2. If X and Y are connected in series to the same supply, what will be the rms value of the current in ampere?

Options:
A 102
B 52
C 52
D 52

The correct answer is:
52

Solution:
As current is in phase with the applied voltage, element X should be resistive with R=V0I0=1005=20 Ω.

As current lags behind voltage by 90°, element Y should be inductive with XL=V0I0=1005=20 Ω

When X and Y are connector in series,

Impedance, Z=XL2+R2=202+202=202 Ω

Now, peak current, I0=V0Z=100202=52 A

Thus, the rms value of current is Irms=I02=52 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (29 Jul Shift 2)
A circuit has a resistance of $12 \Omega$ and an impedance of $15 \Omega$. The power factor of the circuit will be

Options:
A
0.8

B
0.4

C
1.25

D
0.125

The correct answer is:
0.8

Solution:
$\cos \phi=\frac{R}{Z}=\frac{12}{15}=\frac{4}{5}=0.8$

Topic: Physics, Alternating Current, JEE Main
A circuit has a resistance of $12 \Omega$ and an impedance of $15 \Omega$. The power factor of the circuit will be

Options:
A
0.8

B
0.4

C
1.25

D
0.125

The correct answer is:
0.8

Solution:
$\cos \phi=\frac{R}{Z}=\frac{12}{15}=\frac{4}{5}=0.8$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2005
A circuit has a section \( A B \), shown in the figure. The emf of the source is equal to \( 10 \mathrm{~V} \) and the capacitance of capacitor are equal to \( C_{1}=1.0 \mu \mathrm{F} \) and \( C_{2}=2.0 \mu \mathrm{F} \). If the potential difference between points \( A \) and \( B, \phi_{\mathrm{A}}-\phi_{\mathrm{B}}=5.0 \mathrm{~V} \), then the voltages \( V_{1} \) and \( V_{2} \) across each capacitors, \( C_{1} \) and \( C_{2} \), respectively, are

Options:
A \( V_{1}=\frac{5}{3} \mathrm{~V}, V_{2}=\frac{10}{3} \mathrm{~V} \)
B \( V_{1}=\frac{10}{3} \mathrm{~V}, V_{2}=\frac{10}{3} \mathrm{~V} \)
C \( V_{1}=\frac{10}{3} \mathrm{~V}, V_{2}=\frac{5}{3} \mathrm{~V} \)
D none of these

The correct answer is:
\( V_{1}=\frac{10}{3} \mathrm{~V}, V_{2}=\frac{5}{3} \mathrm{~V} \)

Solution:
Let the potential at point A be VA and at point B be VB.

From the question, we know that,

VA-VB=5 …i.

Let us write the voltage in the circuit starting from point A. Voltage drop across capacitor is given by V=qC. Therefore,

⇒VA+qC1-10+qC2=VB.

Substituting the values and using equation i,

-5=q1+q2-10

⇒q=103 μC

Hence, the voltage drop across C1,

V1=qC1=103×1=103 V.

And voltage drop across C2,

V2=qC2=103×2=53 V.

Topic: Physics, Capacitance, JEE Main
A circuit is arranged as shown in figure. The output voltage Vo is equal to __________ V.

Options:

The correct answer is:
5

Solution:
Diodes are forward biased, so they act as wire so input current is 0 as input source is short circuited output current is also zero, as input current is zero

Now, output voltage =5 V

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (27 Aug Shift 1)
A circular $A B C D$ carrying a current ' $i$ ' is placed in a uniform magnetic field, if the magnetic force on the segment $A B$ is $\vec{F}$, the force on the remaining segment BCDA is:

Options:
A $-\vec{F}$
B $3 \vec{F}$
C $-3 \vec{F}$
D $\vec{F}$

The correct answer is:
$-\vec{F}$

Solution:
We know that net magnetic force on a current loop in a uniform magnetic field is zero.
$$
\begin{aligned}
& \therefore \quad F_{A B}+F_{C B A D}=0 \\
& \therefore \quad \vec{F}_{C B A D}=-\vec{F}_{A B}=-\vec{F} \\
&
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2013 (Karnataka)
A circular coil, carrying a constant current i is kept in the x-y plane. The magnetic flux through the entire x-y plane excluding the area of the circular coil is given by ϕ and the magnetic flux through the area of the circular coil area is given by ϕ0, then

Options:
A ϕ>-ϕ0
B Ï<Ï0
C ϕ=-ϕ0
D ϕ=ϕ0

The correct answer is:
ϕ=-ϕ0

Solution:
As magnetic field lines always form a closed loop, hence every magnetic field line creating magnetic flux in the inner region must be passing through the outer region. So the magnetic flux for one of the regions will be positive and for the other, it will be negative but both will have the same magnitude. This means ϕ=-ϕ0

Topic: Physics, Electromagnetic Induction, JEE Main
A circular coil carrying current I has a radius R and magnetic field at the centre is B. The distance from the centre along the axis of the same coil where the magnetic field will be B8 is

Options:
A R2
B R3
C 2R
D 3R

The correct answer is:
R3

Solution:
Bcenter=Î¼0nI2R=B
Bx=B8=Î¼0nIR22(R2+x2)3/2
18Î¼0nI2R=Î¼0nIR22(R2+x2)3/2
18R3=1(R2+x2)3/2
12R=1(R2+x2)12
4R2=R2+x2
x2=3R2
x= 3 R

Topic: Physics, Magnetic Effects of Current, JEE Main
A circular coil carrying current I has radius R and magnetic field at the centre is B. The distance from the centre along the axis of the same coil where the magnetic field will be B8 is

Options:
A R2
B R3
C 2R
D 3R

The correct answer is:
R3

Solution:
Bcenter=μ0nI2R=B
Bx=B8=μ0nIR22(R2+x2)3/2
18μ0nI2R=μ0nIR22(R2+x2)3/2
18R3=1(R2+x2)3/2
12R=1(R2+x2)12
4R2=R2+x2
x2=3R2
x= 3 R

Topic: Physics, Magnetic Effects of Current, JEE Main
A circular coil connected to a battery of emf E produced a certain magnetic induction field at its centre. The coil is unwound, stretched to double its length rewound into a coil of 1rd3 of the original radius and connected to a battery of emf E' to produce same field at the centre. Then E' is

Options:
A 2E9
B 3E7
C 9E4
D 7E4

The correct answer is:
2E9

Solution:
The expression for the magnetic field at the centre of a circular
coil of n turns.

B=μ0i2rn

If length of the wire is doubled,

2 2πrn=2πr'n'

2 r n=r3n'

n'=6n

For the same field at the centre,

μ0in2r=μ0i'n'2r'

μ0in2r=μ0i 6 n2 r3

⇒ i'=i18

Since volume will remain same.

∴ l×A=2l×A'

⇒ A'=A2

The resistance after doubling the length,

R'=ρl'A'

⇒ R'=ρ 2lA/2

⇒ R'=4ρlA

⇒ R'=4 R

The EMF is,

E'=i'R'

⇒ E'=i18 4 R=29E

Topic: Physics, Magnetic Effects of Current, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 2)
A circular coil expands radially in a region of magnetic field and no electromotive force is produced in the coil. This can be because

Options:
A
the magnetic field is constant

B
the magnetic field is in the same plane as the circular coil and it may or may not vary

C
the magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably

D
there is a constant magnetic field in the perpendicular (to the plane of the coil) direction

The correct answers are:
the magnetic field is in the same plane as the circular coil and it may or may not vary
,
the magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably

Solution:
When circular coil expands in the region of magnetic field such that when it is in the same plane as the circular coil or the magnetic field has a perpendicular to the plane of the coil component whose magnitude is decreasing suitably so that the cross product of magnetic field and surface area of plane of coil remain constant at every instant.

Topic: Physics, Electromagnetic Induction
A circular coil has moment of inertia 0.8 kg m2 around any diameter and is carrying current to produce a magnetic moment of 20 Am2. The coil is kept initially in a vertical position and it can rotate freely around a horizontal diameter. When a uniform magnetic field of 4 T is applied along the vertical, it starts rotating around its horizontal diameter. The angular speed the coil acquires after rotating by 60o will be :

Options:
A 10 rad s-1
B 10π rad s-1
C 20π rad s-1
D 20 rad s-1

The correct answer is:
10 rad s-1

Solution:
From energy conservation

12Iω2=Un-Ut

=-MBcos60°-(-MB)

MB2=12Iω2

20×42=12(08)m2

100=ω2

ω=10rad

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (04 Sep Shift 2)