datavorous/entrance-exam-dataset huggingface dataset:
A stone tied to the end of a string of $1 \mathrm{~m}$ long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolution in 44 seconds, what would be the magnitude and direction of acceleration of the stone?

Options:
A $\pi^2 \mathrm{~ms}^{-2}$ and direction along the tangent to the circle
B $\pi^2 \mathrm{~ms}^{-2}$ and direction along the radius towards the centre.
C $\frac{\pi^2}{4} \mathrm{~ms}^{-2}$ and direction along the radius towards the centre.
D $\pi^2 \mathrm{~ms}^{-2}$ and direction along the radius away from the centre.

The correct answer is:
$\pi^2 \mathrm{~ms}^{-2}$ and direction along the radius towards the centre.

Solution:
Acceleration will be $\pi 2 \mathrm{~m} / \mathrm{s} 2$ radially inwards. Radial acceleration or centripetal accceleration is given by:
$\mathrm{ac}=\mathrm{v} 2 \mathrm{r}=\omega 2 \mathrm{r}$
Frequency of revolution is,
$\begin{aligned}
& \mathrm{f}=2244=12 \mathrm{~s}-1 \\
& \Rightarrow \omega=2 \pi \mathrm{f}=\pi \mathrm{rad} / \mathrm{s} \\
& \therefore \mathrm{ac}=\omega 2 \mathrm{r}=\pi 2 \times 1=\pi 2 \mathrm{~m} / \mathrm{s} 2
\end{aligned}$
Tangential acceleration is the rate of change of speed w.r.t time:
$\Rightarrow a_t=\frac{d v}{d t}=0$
$\because$ speed is constant
Hence net acceleration of stone $\vec{a}$ given by:
$\begin{aligned}
& \vec{a}=\vec{a}_c+\vec{a}_t=\vec{a}_c=\pi^2 \mathrm{~m} / \mathrm{s}^2 \\
& \therefore|\vec{a}|=\left|\vec{a}_c\right|=\pi^2 \mathrm{~m} / \mathrm{s}^2
\end{aligned}$
Direction is along the radius towards the centre i.e. radially inwards.

Topic: Physics, Rotational Motion, JEE Main
A stone tied to the end of a string of 1 $\mathrm{m}$ long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of acceleration of the stone?

Options:
A $\pi^2 \mathrm{~ms}^{-2}$ and direction along the radius towards the centre
B $\pi^2 \mathrm{~ms}^{-2}$ and direction along the radius away from the centre
C $\pi^2 \mathrm{~ms}^{-2}$ and direction along the tangent to the circle
D $\pi^2 / 4 \mathrm{~ms}^{-2}$ and direction along the radius towards the centre

The correct answer is:
$\pi^2 \mathrm{~ms}^{-2}$ and direction along the radius towards the centre

Solution:
\(a=\frac{v^2}{r}=\omega^2 r=4 \pi^2 n^2 r=4 \pi^2\left(\frac{22}{44}\right)^2 \times 1=\pi^2 \mathrm{~m} / \mathrm{s}^2\)
and its direction is always along the radius and towards the centre.

Topic: Physics, Motion In Two Dimensions, JEE Main
A stone weight $100 \mathrm{~N}$ on the surface of the earth. The ratio of its weight at a height of half the radius of the earth to a depth of half the radius of the earth will be approximately

Options:
A $3.6$
B $2.2$
C $1.8$
D $0.9$

The correct answer is:
$0.9$

Solution:
Weight of stone at the surface of the earth, $w=100 \mathrm{~N}$ At height $h$ from the surface of earth, gravitational acceleration,
$$
g_{h}=\frac{g}{\left(1+\frac{h}{R}\right)^{2}}
$$
At depth $d$ from the surface of the earth, gravitational acceleration,
$$
g_{d}=g\left(1-\frac{d}{R}\right)
$$
According to question,
$\frac{\text { Weight at } h=R / 2}{\text { Weight at } d=R / 2}=\frac{m_{h}(\text { at } h=R / 2)}{m g_{d}(\text { at } h=R / 2)}$
$=\frac{\frac{g}{\left(1+\frac{h}{R}\right)^{2}}(\text { at } h=R / 2)}{g\left(1-\frac{d}{R}\right) \text { at }(d=R / 2)}$
$=\frac{\frac{1}{\left(1+\frac{R / 2}{R}\right)^{2}}}{1-\frac{R / 2}{R}}=\frac{4 / 9}{1 / 2}=\frac{8}{9}=0.9$

Topic: Physics, Gravitation, JEE Main
A storage battery has e.m.f. $15 \mathrm{volts}$ and internal resistance $0.05 \mathrm{ohm}$. Its terminal voltage when it is delivering 10 ampere is

Options:
A 30 volts
B 1.00 volts
C 14.5 volts
D 15.5 volts

The correct answer is:
14.5 volts

Solution:
$V=E-I R=15-10.0 .05=14.5 \mathrm{~V}$

Topic: Physics, Current Electricity, JEE Main
A storage tower supplies water, as shown in the figure. If P0 is the atmospheric pressure

h= height of water level, g = acceleration due to gravity, Ï= density and v= velocity of flow in the horizontal pipe at B, then the pressure at B is -

Options:
A P0
B P0+ρgh
C P0+ρgh+12ρv2
D P0+ρgh –12ρv2

The correct answer is:
P0+ρgh –12ρv2

Solution:
Consider the streamline connecting A and B. Then
PA+ρgyA+12ρvA2=PB+ρgyB+12ρvB2
Since PA=P0,yA=h,yB=0,
VB=v and VA is negligible
(in fact VA=d2v and vA is negligible)
in fact VA=d2vBD2, d<<D,VA is negilgible
∴PB=P0+ρgh-12ρv2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A straight conductor $0.1 \mathrm{m}$ long moves in uniform magnetic field $0.1 \mathrm{T}$. The velocity the conductor is $15 \mathrm{m} / \mathrm{s}$ and is directer perpendicular to the field. The emf inducod between the two ends of the conductor is

Options:
A $0.10 \mathrm{V}$
B $0.15 \mathrm{V}$
C $1.50 \mathrm{V}$
D $15.00 \mathrm{v}$

The correct answer is:
$0.15 \mathrm{V}$

Solution:
Given, length of conductor $l=0.1
\mathrm{m}$
Mangetic field. $B=0.1 \mathrm{T}$
Velocity of conductor, $v=15 \mathrm{m} / \mathrm{s}$
The angle between $\mathbf{v}$ and $\mathbf{B}$ is $90^{\circ}$
When $\mathbf{v}$ and $\mathbf{B}$ are mutually perpendicular, then emf (induced) is given by
$$
\varepsilon=v B l=15 \times 0.1 \times 0.1=\frac{15}{100}=0.15 \mathrm{V}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
A straight conductor of length $0.4 \mathrm{~m}$ is moved with a speed of $7 \mathrm{~m} / \mathrm{s}$ perpendicular to the magnetic field of intensity of $0.9 \mathrm{~Wb} / \mathrm{m}^2$. The induced e.m.f. across the conductor will be

Options:
A $7.25 \mathrm{~V}$
B $3.75 \mathrm{~V}$
C $1.25 \mathrm{~V}$
D $2.52 \mathrm{~V}$

The correct answer is:
$2.52 \mathrm{~V}$

Solution:
$e=B v l \Rightarrow e=0.9 \times 7 \times 0.4=2.52 \mathrm{~V}$

Topic: Physics, Electromagnetic Induction, JEE Main
A straight is a five card hand containing consecutive values. If m is equal to the number of ways in which all the five cards, in a straight, are not from the same suit, then the value of m1500 is equal to (Consider the value of J as 11, Q as 12, K as 13 and Ace as 14)

Options:

The correct answer is:
6.12

Solution:
Five consecutive cards can be chosen out of 13 cards in 9 ways. 2,3,4,5,6,3,4,5,6,7,......10,J,Q,K,A. The first card can be chosen in 4 ways (spades, hearts, diamonds and clubs). The second and the subsequent cards, too, can be chosen in 4 ways.
Thus, there are 45 ways to choose the five cards.
The number of ways in which all the five cards are not from the same suit is 45-4=1020.
So, the number of straights in this case is m=9Ã1020âm1500=9Ã10201500=6.12

Topic: Mathematics, Permutation Combination, JEE Main
A straight line cuts off an intercept of 2 units on the positive direction of $\mathrm{x}$ -axis and passes through the point $(-3,5)$. What is the foot of the perpendicular drawn from the point $(3,3)$ on this line?

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

The correct answer is:
$(1,1)$

Solution:
The given line passes through $(-3,5)$ and $(2,0)$. Its equation is
$y-y_{1}=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\left(x-x_{1}\right)$
$\Rightarrow(y-5)=\left(\frac{0-5}{2+3}\right)(x+3)$
$\Rightarrow y=-x+2$ ...(1)
Slope $=m=-1$
and slope of perpendicular line $=-\frac{1}{m}=1$
Equation of this line passing through $(3,3)$ is:
$(y-3)=1(x-3)$
$\Rightarrow \quad y=x .$
From eq. (1) we get, $x=-x+2$
$\Rightarrow x=1$ and $y=1 .$

Topic: Mathematics, Straight Lines, JEE Main
A straight line cuts off the intercepts $\mathrm{OA}=\mathrm{a}$ and $\mathrm{OB}=\mathrm{b}$ on the positive directions of $\mathrm{x}$-axis and $\mathrm{y}-$ axis respectively. If the perpendicular from origin $\mathrm{O}$ to this line makes an angle of $\frac{\pi}{6}$ with positive direction of $y$-axis and the area of $\triangle \mathrm{OAB}$ is $\frac{98}{3} \sqrt{3}$, then $\mathrm{a}^2-\mathrm{b}^2$ is equal to:

Options:
A 3923
B 196
C 1963
D 98

The correct answer is:
3923

Solution:
Plotting the diagram of the given data we get,

Equation of straight line

Here given OA=a & OB=b, so equation of line AB in intercept form will be xa+yb=1 ......1

And given perpendicular line of length p is making π3 angle with the x-axis,

So, the equation of line in normal form will be

xcosπ3+ysinπ3=p

⇒x2p+y32p=1

So, on comparing with the equation 1 we get,

a=2p & b=2p3

Now area of the triangle OAB is

12×a×b=9833

⇒12×2p×2p3=9833

⇒p2=49

Therefore,

a2-b2=4p2-4p23=83p2

⇒a2-b2=8349=3923

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2023 (30 Jan Shift 1)
A straight line joining the points (1,1,1) and (0,0,0) intersects the plane $2 x+2 y+z=10$ at

Options:
A (1,2,5)
B (2,2,2)
C (2,1,5)
D (1,1,6)

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

Solution:
Equation of line joining the points (1,1,1) and (0,0,0) is
$$
\begin{aligned}
\frac{x-0}{1-0} &=\frac{y-0}{1-0}=\frac{z-0}{1-0}=\lambda \\
\Rightarrow & x=y=z=\lambda
\end{aligned}
$$
So, the point is $(\lambda, \lambda, \lambda)$ The point intersects the plane $2 x+2 y+z=10$
$\therefore 2(\lambda)+2(\lambda)+\lambda=10$
$\Rightarrow 5 \lambda=10$
$\Rightarrow \lambda=2$
Hence, the point is (2,2,2) .

Topic: Mathematics, Three Dimensional Geometry, JEE Main
A straight line l1 with equation x-2y+10=0 meets the circle with equation x2+y2=100 at B in the first quadrant. A line through B, perpendicular to l1 cuts the x-axis and y-axis at P and Q respectively. The area (in sq. units) of the triangle OPQ is (where, O is the origin)

Options:
A 120
B 150
C 100
D 125

The correct answer is:
100

Solution:
Slope of l1=12
Slope of l2=-2
Equation of l2 is y=-2x-10

⇒y+2x=20
Hence, P and Q are 10,0 and 0,20
Area of ∆OPQ=12×10×20=100 sq. units

Topic: Mathematics, Circle, JEE Main
A straight line L at a distance of 4 units from the origin makes positive intercepts on the coordinate axes and the perpendicular from the origin to this line makes an angle of 60° with the line x+y=0. Then an equation of the line L is:
Note: In actual JEE Main paper, two options were correct for this question. Hence, we have changed one option.

Options:
A 3+1x+3-1y=82
B x+3y=8
C 3x+y=8
D 3-1x+3y=82

The correct answer is:
3+1x+3-1y=82

Solution:
Hence, equation of required line using normal form is
xcos15°+ysin15°=4 or xcos75°+ysin75°=4
⇒  (3+1)x+(3-1)y=82 or   (3-1)x+(3+1)y=82

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A straight line L cuts the sides AB,AC,AD of a parallelogram ABCD at B1,C1,D1 respectively. If AB→1=λ1AB→, AD→1=λ2AD→ and AC→1=λ3AC→, then 1λ3 is equal to

Options:
A 1λ1+1λ2
B 1λ1-1λ2
C -λ1+λ2
D λ1+λ2

The correct answer is:
1λ1+1λ2

Solution:
Let position vector of A,B,D are o→,b→,d→ respectively
AB→1=λ1AB→=λ1b→
AD→1=λ2AD→=λ2d→
AC→1=λ3AC→=λ3b→+d→
∵B1,C1,D1 are collinear
B1C1→=λ3-λ1b→+λ3d→
B1D1→=-λ1b→+λ2d→
⇒λ3-λ1-λ1=λ3λ2⇒λ3λ2-λ1λ2=-λ1λ3
⇒λ3=λ1λ2λ1+λ2⇒1λ3=1λ1+1λ2

Topic: Mathematics, Vector Algebra, JEE Main
A straight line L through the point (3, -2) is inclined at an angle of 60° to the line 3x+y=1. If L also intersects the X-axis, then the equation of L is:

Options:
A y+3 x+2-33=0
B 3 y-x+3+23=0
C 3 y+x-3+23=0
D y-3x+2+33=0

The correct answer is:
y-3x+2+33=0

Solution:
Let m be the slope of the line,

then tan⁡60o=m+31-m3
⇒3=m+31-3 m
⇒±3=m+31-3 m

∴3-3m=m+3 or -3+3m=m+3
4m=0 or 2m=23
m=0 or m=3
Since L intersects X-axis we take m=3
and get y=3x+c
Since it passes through 3,-2, therefore line will become y-3x+2+33=0

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2015 (11 Apr Online)
A straight line makes an angle of \( 135^{\circ} \) with the \( X \)-axis and cuts \( Y \)-axis at a distance \( -5 \) from the origin. The equation of the line is

Options:
A \( 2 x+y+5=0 \)
B \( x+2 y+3=0 \)
C \( x+y+5=0 \)
D \( x+y+3=0 \)

The correct answer is:
\( x+y+5=0 \)

Solution:
The equation of the straight line in slope intercept form will be y=mx+c.

A straight line makes an angle of 135° with the X-axis and cuts Y-axis at a distance -5 from the origin.

The equation of the line is

y=tan135°x+-5

y=-x-5

⇒x+y+5=0.

Topic: Mathematics, Straight Lines, JEE Main
A straight line meets the co-ordinate axes at $\mathrm{A}$ and $\mathrm{B}$. A circle is circumscribed about the triangle $\mathrm{OAB}, \mathrm{O}$ being the origin. If $\mathrm{m}$ and $\mathrm{n}$ are the distances of the tangent to the circle at the origin from the points $\mathrm{A}$ and $\mathrm{B}$ respectively, the diameter of the circle is

Options:
A $m(m+n)$
B $m+n$
C $n(m+n)$
D $\frac{1}{2}(m+n)$

The correct answer is:
$m+n$

Solution:
Clearly, AB is one of diameter
$\because \mathrm{AM}$ and $\mathrm{BN}$ are parallel and $\angle \mathrm{BNO}=\angle \mathrm{AMO}=\pi / 2$
$\therefore$ Points $\mathrm{N}, \mathrm{O}$ and $\mathrm{M}$ are collinear.
$\therefore \triangle \mathrm{BNMA}$ is a rectangle
$\Rightarrow \mathrm{AB}=\mathrm{MN}=\mathrm{m}+\mathrm{n}$

Topic: Mathematics, Circle, JEE Main
A straight line passing through a fixed point 3, 5 intersects the coordinate axes at two points A and B. If the locus of Cx, y which forms a rectangle with the points A, O (origin) and B is ax+2hxy+by=0 then a+b+h=

Options:
A 172
B 7
C 152
D 10

The correct answer is:
152

Solution:
A Straight line passing through a fixed point 3,5 and intersects coordinate axis at point A and B.

Let coordinates of Ap,0 and B0,q

∴ Intercept form of line is xp+yq=1

Here, point 3,5 will satisfy the line.

⇒ 3p+5q=1

Here, coordinate of point are C(x,y)=(p,q)

Hence, to find locus of C substitute p=x & q=y in the above equation.

⇒ 3x+5y=1

⇒ 5x+3y-xy=0

Now compare ax+by+2hxy=0 to 5x+3y-xy=0

a=5 b=3, h=-12

a+b+h ⇒ 5+3-12=152.

Topic: Mathematics, Straight Lines, JEE Main
A straight line through a fixed point 2,3 intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is:

Options:
A 3x+2y=6xy
B 3x+2y=6
C 2x+3y=xy
D 3x+2y=xy

The correct answer is:
3x+2y=xy

Solution:
Let the equation of line be xa+yb=1 .......1

Clearly Pa,0 and Q0,b. Also h=a, k=b

Now 1 passes through 2,3

⇒2a+3b=1

⇒2h+3k=1

Hence, locus is 3x+2y=xy

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2018 (08 Apr)
A straight line through origin bisect the line passing through the given points $(a \cos \alpha, a \sin \alpha)$ and $(a \cos \beta, a \sin \beta)$, then the lines are

Options:
A Perpendicular
B Parallel
C Angle between them is $\frac{\pi}{4}$
D None of these

The correct answer is:
Perpendicular

Solution:
Mid point of $(a \cos \alpha, a \sin \alpha)$ and $(a \cos \beta, a \sin \beta)$ is $P\left(\frac{a(\cos \alpha+\cos \beta)}{2}, \frac{a(\sin \alpha+\sin \beta)}{2}\right)$
Slope of line $A B$ is $\frac{a \sin \beta-a \sin \alpha}{a \cos \beta-a \cos \alpha}=\frac{\sin \beta-\sin \alpha}{\cos \beta-\cos \alpha}=m_1$
and slope of $O P$ is $\frac{\sin \alpha+\sin \beta}{\cos \alpha+\cos \beta}=m_2$
Now
$m_1 \times m_2=\frac{\sin ^2 \beta-\sin ^2 \alpha}{\cos ^2 \beta-\cos ^2 \alpha}=-1$
Hence the lines are perpendicular.

Topic: Mathematics, Straight Lines, JEE Main
A straight line through origin O meets the lines 3y=10-4x and 8x+6y+5=0 at points A and B respectively. Then, O divides the segment AB in the ratio

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

The correct answer is:
4:1

Solution:
Let Ar1cosθ, r1sinθ
A Lies on straight line 4x+3y=10

⇒4r1cosθ+3r1sinθ=10

⇒r14cosθ+3sinθ=10
⇒r1=104cosθ+3sinθ
Let B-r2cosθ, -r2sinθ
B lies on line 8x+6y+5=0

⇒8-r2cosθ+6-r2sinθ+5=0

⇒r28cosθ+6sinθ-5=0⇒r2=58cosθ+6sinθ
∴r1r2=41

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2016 (10 Apr Online)
A straight line through the point $(1,1)$ meets the $x$-axis at ' $A$ ' and the $y$-axis at ' $B$ '. The locus of the mid-point of $A B$ is

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

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

Solution:
Equation of line passing through point $(1,1)$ is,
$y-1=m(x-1)$
Line (i) meets $x$-axis, so $y=0$
$\therefore \frac{-1}{m}=x-1 \Rightarrow x=1-\frac{1}{m}$
Line (i) meets $y$-axis, so $x=0$
$\therefore \quad y-1=-m \Rightarrow y=1-m$
Let mid point of $A B$ be $(h, k)$,
Then $h=\frac{0+(1-(1 / m))}{2} ; k=\frac{0+(1-m)}{2}$
$\begin{aligned}
& m=\frac{1}{1-2 h} ; m=1-2 k \\
& \quad 1-2 k=\frac{1}{1-2 h} \\
& \Rightarrow 1-2 k-2 h+4 h k=1 \Rightarrow-2 h-2 k+4 h k=0
\end{aligned}$
Hence the Locus of mid point is, $x+y-2 x y=0$

Topic: Mathematics, Straight Lines, JEE Main
A straight line through the point A(3, 4) is such that its intercept between the axes is bisected at A. Its equation is

Options:
A
x + y = 7

B
3x − 4y + 7 = 0

C
4x + 3y = 24

D
3x + 4y = 25

The correct answer is:
4x + 3y = 24

Solution:
The equation of axes is $x y=0$
$\Rightarrow$ the equation of the line is $\frac{x \cdot 4+y \cdot 3}{2}=12 \Rightarrow 4 x+3 y=24$

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2006
A straight magnetic strip has a magnetic moment of $44 \mathrm{Am}^2$. If the strip is bent in a semicircular shape, its magnetic moment will be _______ $\mathrm{Am}^2$.
(given $\pi=\frac{22}{7}$ )

Options:

The correct answer is:
28

Solution:
Magnetic moment of straight wire $=\mathrm{mx} \ell=44$
Magnetic moment of arc
$\begin{aligned}& =\mathrm{m} \times 2 \mathrm{r} \\& =\mathrm{m} \times \frac{2 \ell}{\pi} \\& =\frac{44 \times 2}{\pi}=\frac{88}{\pi}=28\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (09 Apr Shift 2)
A straight rod of length L extends from x=a to x=L+a. The gravitational force it exerts on a point mass 'm' at x=0, if the mass per unit length of the rod is A+Bx2, is given by:

Options:
A GmA1a+L-1a+BL
B GmA1a+L-1a-BL
C GmA1a-1a+L-BL
D GmA1a-1a+L+BL

The correct answer is:
GmA1a-1a+L+BL

Solution:
F=∫aa+LGm dMx2=Gm∫aa+L​(A+Bx2)dxx2
=Gm∫aa+LAx2dx+∫aa+LB  dx
=Gm[A[-1x]aa+L+BL]
=GmA1a-1a+L+BL

Topic: Physics, Gravitation, JEE Main, JEE Main 2019 (12 Jan Shift 1)
A straight wire carrying a current $i_1$ amp runs along the axis of a circular current ${ }^{i_2}$ amp . Then the force of interaction between the two current carrying conductors is

Options:
A $\infty$
B Zero
C $\frac{\mu_0}{4 \pi} \frac{2 i_1 i_2}{r} N / m$
D $\frac{2 i_1 i_2}{r} N / m$

The correct answer is:
Zero

Solution:
Direction of magnetic field due to circular current will be perpendicular to the plane of circle i.e. along the axis as shown in the diagram below.

Force due to magnetic field is given as
\(\mathrm{dF}=I(\overrightarrow{\mathrm{d}} \times \overrightarrow{\mathrm{B}})\)
Since both vector \(\overrightarrow{\mathrm{dl}}\) and \(\vec{B}\) are in same direction, the force of intercation between both the wires will be zero.

Topic: Physics, Magnetic Effects of Current, JEE Main
A straight wire carrying a current of 14 A is bent into a semicircular arc of radius 2.2 cm as shown in the figure. The magnetic field produced by the current at the centre O of the arc is ____ × 10–4 T

Options:

The correct answer is:
2

Solution:
It is given that

i=14 AR=2.2×10-2 m

The magnetic field at the centre O is

B=μ0i4R=4π×10-7×144×2.2×10-2=2×10-4 T

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (10 Apr Shift 2)
A straight wire is placed in a magnetic field that varies with distance $x$ from origin as $\vec{B}=B_0\left(2-\frac{x}{a}\right) \hat{k}$. Ends of wire are at $(a, 0)$ and $(2 a, 0)$ and it carries a current $\mathrm{I}$. If force on wire is $\vec{F}=\mathrm{IB}_0\left(\frac{k a}{2}\right) \hat{\mathrm{j}}$, then value of $k$ is

Options:
A 1
B 5
C $-1$
D $\frac{1}{2}$

The correct answer is:
$-1$

Solution:
$\overrightarrow{\mathrm{dF}}=\mathrm{Idx} \times \overrightarrow{\mathrm{B}}=\operatorname{Idx} \hat{\mathrm{i}} \times \mathrm{B} \hat{\mathrm{k}}=\operatorname{IdxB}(-\hat{\mathrm{j}})$
$\int \mathrm{dF}=\int_{\mathrm{a}}^{2 a} \mathrm{Id} x \mathrm{~B}_0\left(2-\frac{\mathrm{x}}{\mathrm{a}}\right)(-\hat{\mathrm{j}})$
$$
\begin{aligned}
&\mathrm{F}=\mathrm{IB}_0 \int_{\mathrm{a}}^{2 \mathrm{a}}\left(2-\frac{\mathrm{x}}{\mathrm{a}}\right) \mathrm{dx}(-\hat{\mathrm{j}}) \\
&=\operatorname{IB}_0\left(2 \mathrm{x}-\frac{\mathrm{x}^2}{2 \mathrm{a}}\right)_{\mathrm{a}}^{2 \mathrm{a}}(-\hat{\mathrm{j}}) \\
&=\mathrm{IB}_0\left[\left(2 \times 2 \mathrm{a}-\frac{(2 \mathrm{a})^2}{2 \mathrm{a}}\right)-\left(2 \mathrm{a}-\frac{\mathrm{a}^2}{2 \mathrm{a}}\right)\right](-\hat{\mathrm{j}}) \\
&=\mathrm{IB}_0\left[(4 \mathrm{a}-2 \mathrm{a})-\left(\frac{3 \mathrm{a}}{2}\right)\right](-\hat{\mathrm{j}}) \\
&=\mathrm{IB}_0\left[2 \mathrm{a}-\frac{3 \mathrm{a}}{2}\right](-\hat{\mathrm{j}}) \\
&=\operatorname{IB}_0\left[\frac{\mathrm{a}}{2}\right](-\hat{\mathrm{j}})
\end{aligned}
$$
Comparing $\mathrm{F}$ with $\mathrm{IB}_0\left[\frac{\mathrm{ka}}{2}\right] \hat{\mathrm{j}}$, we get $\mathrm{k}=-1$

Topic: Physics, Magnetic Effects of Current, JEE Main
A straight wire of length $2 \mathrm{m}$ carries a current of $10 \mathrm{A}$. If this wire is placed in a uniform magnetic field of 0.15 T making an angle of $45^{\circ}$ with the magnetic field, the applied force on the wire will be

Options:
A $1.5 \mathrm{N}$
B $3 \mathrm{N}$
C $3 \sqrt{2} \mathrm{N}$
D $\frac{3}{\sqrt{2}} \mathrm{N}$

The correct answer is:
$\frac{3}{\sqrt{2}} \mathrm{N}$

Solution:
Given $i=10 \mathrm{A}, B=0.15 \mathrm{T}, \theta=45^{\circ}$ and $l=2 \mathrm{m}$
Here,
$$
\begin{aligned}
F &=i L B \sin \theta \\
&=10 \times 2 \times 0.15 ~ x ~ \sin 45^{\circ} \\
&=\frac{3}{\sqrt{2}} \mathrm{N}
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main
A straight wire of length 0.5 metre and carrying a current of 1.2 ampere placed in a uniform magnetic field of induction 2 Tesla. The magnetic field is perpendicular to the length of the wire. The force on the wire is

Options:
A $2.4 \mathrm{~N}$
B $1.2 \mathrm{~N}$
C $3.0 \mathrm{~N}$
D $2.0 \mathrm{~N}$

The correct answer is:
$1.2 \mathrm{~N}$

Solution:
\(\begin{aligned} & \mathrm{F}=\text {Bilsin } \theta, \text { Here } \theta=90^{\circ} \\ & \mathrm{F}=2 \times 1 \cdot 2 \times \mathrm{o} \cdot 5 \\ & \mathrm{~F}=1 \cdot 2 \mathrm{~N}\end{aligned}\)

Topic: Physics, Magnetic Effects of Current, JEE Main
A stream of a positively charged particles having qm=2×1011 C kg-1 and velocity v→0=3×107 i^ m s-1 is deflected by an electric field 1.8j^ kV m-1. The electric field exists in a region of 10 cm along x direction. Due to the electric field, the deflection of the charge particles in the y direction is _____ mm.

Options:

The correct answer is:
2

Solution:
Acceleration of the particles due to electric field, a=Fm=qEm=2×10111.8×103

=3.6×1014 m s-2

Time to cross plates =dv

⇒t=0.103×107

Now deflection,

y=12at2=123.6×10140.103×1072

=0.002 m

=2 mm

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (24 Jan Shift 1)
A stream of electrons from a heat filament was passed between two charge plates kept at a potential difference V esu. If e and m are charge and mass of an electron, respectively, then the value of hλ (where λ is wavelength associated with the electron wave) is given by:

Options:
A meV
B 2 meV
C meÂ V
D 2Â meÂ V

The correct answer is:
2 meV

Solution:
Given, stream of electron from heated filament was passed between two charge plates at potential difference V.
e and m are charge and mass of electron, respectively.
V=Energye
eV=12×m×v2
λ=hmv v=hmλ
eV=12×m×hmλ2
hλ=2 meV

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2016 (03 Apr)
A stream of non-viscous liquid emerges from a very short outlet tube at the base of a large open tank, in which the depth of liquid is h. The tube is at a fixed angle θ to the ground as shown in the figure. The maximum height of the stream y is

Options:
A h sin2â¡Î¸
B h sin2θ
C 12 hsinâ¡Î¸
D h tan2â¡Î¸

The correct answer is:
h sin2â¡Î¸

Solution:
Let the speed of fluid at outlet is v .
Applying Bernoulli's theorem,
p1+Ïgh+12Ïv2=constant
p0+Ïgh+0=p0+0+12Ïv2
â´v=2gh ...(i)
At outlet,
vsinâ¡Î¸ vertical velocity of the fluid at outlet,
v2=u2+2gh1
For maximum height,
0=vsinâ¡Î¸2-2gh1
â´h1=v2sin2â¡Î¸2g
=2ghsin2â¡Î¸2g [from Equation (i)]
=hsin2â¡Î¸
h1=hsin2â¡Î¸

Topic: Physics, Mechanical Properties of Fluids, JEE Main