datavorous/entrance-exam-dataset huggingface dataset:
Two thin wires rings each having a radius $R$ are placed at a distance $d$ apart with their axes coinciding. The charges on the two rings are $+q$ and $-q$. The potential difference between the centres of the two rings is

Options:
A
$\mathrm{QR} / 4 \pi \varepsilon_0 \mathrm{~d}^2$

B
$\frac{Q}{2 \pi \varepsilon_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right]$

C
zero

D
$\frac{Q}{4 \pi \varepsilon_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right]$

The correct answer is:
$\frac{Q}{2 \pi \varepsilon_0}\left[\frac{1}{R}-\frac{1}{\sqrt{R^2+d^2}}\right]$

Solution:
$v_1=\frac{k q}{R}-\frac{k q}{\sqrt{R^2+d^2}}$
$v_2=\frac{-k q}{R}+\frac{k q}{\sqrt{R^2+d^2}}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2005
Two touching blocks 1 and 2 are placed on an inclined plane forming an angle $60^{\circ}$ with the horizontal. The masses are $m_1$ and $m_2$ and the coefficient of friction between the inclined plane and the two blocks are $1.5 \mu$ and $1.0 \mu$, respectively. The force of reaction between the blocks during the motion is ( $g=$ acceleration due to gravity)

Options:
A $\left(m_2-m_1\right) \mu g$
B $\left(m_2+m_1\right) \mu g$
C $\frac{1}{2} \frac{m_1 m_2}{m_1+m_2} \mu g$
D $\frac{1}{4} \frac{m_1 m_2}{m_1+m_2} \mu \mathrm{g}$

The correct answer is:
$\frac{1}{4} \frac{m_1 m_2}{m_1+m_2} \mu \mathrm{g}$

Solution:
According to the question, there are two blocks of masses $m_1$ and $m_2$ which are kept in contact on inclined plane of inclination $60^{\circ}$ as shown in the figure,
Coefficient of friction between the surface of plane and surface of first block, $\mu_1=1.5 \mu$
Coefficient of friction between the surface of plane and surface of second block, $\mu_2=1 \mu$
Now, free body diagram for the first block,
If $a$ be the common acceleration, then
$$
\begin{aligned}
& m_1 g \sin 60^{\circ}-\mu_1 \cdot R_1+R=m_1 a \\
& m_1 g \frac{\sqrt{3}}{2}-1.5 \mu m_1 g \cos 60^{\circ}+R=m_1 a \\
& m_1 g \frac{\sqrt{3}}{2}-\frac{1.5 \mu m_1 g}{2}+R=m_1 a
\end{aligned}
$$
Similarly, for second body,
$$
\begin{gathered}
m_2 g \sin 60^{\circ}-\mu_2 R_2-R=m_2 a \\
m_2 g \frac{\sqrt{3}}{2}-\mu m_2 g \cos 60^{\circ}-R=m_2 a \\
m_2 g \frac{\sqrt{3}}{2}-\frac{\mu m_2 g}{2}-R=m_2 a
\end{gathered}
$$
From Eqs. (i) and (ii), we get
$$
\frac{m_1 g \frac{\sqrt{3}}{2}-\frac{1.5 \mu m_1 g}{2}+R}{m_2 g \frac{\sqrt{3}}{2}-\frac{\mu m_2 g}{2}-R}=\frac{m_1 a}{m_2 a}
$$
$$
\begin{aligned}
& \Rightarrow \quad \frac{\sqrt{3}}{2} m_1 m_2 g-0.75 \mu m_1 m_2 g+m_2 R \\
& =\frac{\sqrt{3}}{2} m_1 m_2 g-0.5 \mu m_1 m_2 g-m_1 R \\
& \Rightarrow\left(m_1+m_2\right) R=0.25 \mu m_1 m_2 g \\
& \Rightarrow \quad R=\frac{\mu m_1 m_2 g}{4\left(m_1+m_2\right)} \\
&
\end{aligned}
$$
Hence, the force of reaction between the blocks doing the motion is, $R=\frac{1}{4} \frac{m_1 m_2}{\left(m_1+m_2\right)} \mu g$.

Topic: Physics, Laws of Motion, JEE Main
Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 kmh in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 ms2. If after 50 s, the guard of B just brushes past the driver of A and the original distance between them is x, then calculate value of x10 ?

Options:

The correct answer is:
45.00

Solution:
Length of each train,
l⁡ A = l⁡ B = 4 0 0 m
u⁡A⁡=72×518m/s
= 20 m/s
Distance travelled by train A in 50 s
s A = u A × t
(As for unaccelerated motion, distance = Speed x Time)
s A = 2 0 × 5 0
= 1000 m
Distance travelled by train B in 50 s,
s B = u B t + 1 2 a B t 2
(As motion of train B is an accelerated motion)
s B = 2 0 × 5 0 + 1 2 × 1 × 5 0 2
= 1000 + 1250
= 2250 m
relative distance between the two trains =S B − S A
= 2250 - 1000
= 1250 m
we should subtract 2l from it where l is length of train so final answer will be 450 m
Alternate Method
Initially, both the trains are moving with same speed, the train B first comes that of the train A in 50 s with an acceleration of 50 s .
∴ Original distance between the trains
= Distance covered by train B with an acceleration a
= ut + 1 2 at 2
= 0 × 5 0 + 1 2 × 1 × 5 0 2
(As both train starts with same velocity, so velocity, so velocity of train when it starts accelerating can be taken as zero).
= 1250 m .
we should subtract 2l from it where l is length of train so final answer will be 450 m

Topic: Physics, Motion In One Dimension, JEE Main
Two trains A and B of length l and 4l are travelling into a tunnel of length L in parallel tracks from opposite directions with velocities 108 km h-1 and 72 km h-1, respectively. If train A take 35 s less time than train B to cross the tunnel then, length L of tunnel is:

(Given L=60 l)

Options:
A 1200 m
B 900 m
C 1800 m
D 2700 m

The correct answer is:
1800 m

Solution:
The time taken for A

ΔtA=l+LVA

The time taken for B

ΔtB=4l+LVB

It is given that B takes more time than A. So,

⇒4l+L20-l+L30=35

⇒64l20-61l30=35

⇒l=35×600700=30 m

Therefore,

L=1800 m.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Two trains are moving in the same direction along parallel tracks. One of them is 200 m long travelling with a velocity of 20 m s-1. The second one is 800 m long travelling with a velocity of 7.5 m s-1. The time taken for the first train to overtake the second is

Options:
A 20 s
B 40 s
C 60 s
D 80 s

The correct answer is:
80 s

Solution:
From the figure, the relative displacement is
srel=800+200m=1000 m
Ïrel=Ï1-Ï2=20-7.5 m s-1
=12.5 m s-1
â´t=srelÏrel=100012.5=80 s

Topic: Physics, Motion In Two Dimensions, JEE Main
Two trains, each $30 \mathrm{~m}$ long are travelling in opposite directions with velocities $5 \mathrm{~m} / \mathrm{s}$ and $10 \mathrm{~m} / \mathrm{s}$. They will cross after

Options:
A $4 \mathrm{~s}$
B $3 \mathrm{~s}$
C $2 \mathrm{~s}$
D $1 \mathrm{~s}$

The correct answer is:
$4 \mathrm{~s}$

Solution:
Relative velocity of one train w. r. t other $=5+0$
$=15 \mathrm{~m} / \mathrm{s}$
Total length to cross $(\mathrm{L})=30+30=60 \mathrm{~m}$
$\therefore \quad t=\frac{L}{V}=\frac{60}{15}=4 s$

Topic: Physics, Motion In One Dimension, JEE Main
Two trains, each $50 \mathrm{~m}$ long are travelling in opposite direction with velocity $10 \mathrm{~m} / \mathrm{s}$ and $15 \mathrm{~m} / \mathrm{s}$. The time of crossing is

Options:
A 2s
B 4s
C $2 \sqrt{3} \mathrm{~s}$
D $4 \sqrt{3} s$

The correct answer is:
4s

Solution:
Time$=\frac{\text { Total length }}{\text { Relative velocity }}=\frac{50+50}{10+15}=\frac{100}{25}=4 \mathrm{sec}$

Topic: Physics, Motion In Two Dimensions, JEE Main
Two trains, which are moving along different tracks in opposite directions towards each other, are put on the same track due to a mistake. Their drivers, on noticing the mistake, start slowing down the trains when the trains are 300 m apart. Graphs given below show their velocities as a function of time as the trains slow down. What is the separation between the trains (in metre) when both have stopped?

Options:

The correct answer is:
20

Solution:
Using the concept that the area under the v-t graph gives the change in displacement.
s1=12×10×40=200 m
s2=-12×8×20=-80 m
Relative displacement
s relative =s1-s2=200--80=280
Final separation=300-280=20 m

Topic: Physics, Motion In Two Dimensions, JEE Main
Two transparent media having refractive indices 1.0 and 1.5 are separated by a spherical refracting surface of radius of curvature 30 cm. The centre of curvature of surface is towards denser medium and a point object is placed on the principal axis in rarer medium at a distance of 15 cm from the pole of the surface. The distance of image from the pole of the surface is cm.

Options:

The correct answer is:
30

Solution:
Let's consider the following diagram-

The relation between the object distance, the image distance and the radius of curvature of the spherical surface can be written as

μ2v-μ1u=μ2-μ1R ...1

Substitute the values of the known parameters into equation (1) and solve to calculate the required image distance.

1.5v-1-15=1.5-1.030⇒1.5v=160-115=-120⇒v=-30

Hence, the image is formed at a distance of 30 cm from the pole on the same direction as the object is placed.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (08 Apr Shift 2)
Two trolleys of mass $m$ and $3 m$ are connected by a spring. They were compressed and released once, they move off in opposite direction and comes to rest after covering distances $S_1$ and $S_2$ respectively. Assuming the coefficient of friction to be uniform, the ratio of distances $S_1: S_2$ is

Options:
A $1: 9$
B $1: 3$
C $3: 1$
D $9: 1$

The correct answer is:
$9: 1$

Solution:
When trolley are released then they posses same linear momentum but in opposite direction. Kinetic energy acquired by any trolley will dissipate against friction.
$\therefore \mu m g s=\frac{p^2}{2 m} \Rightarrow s \propto \frac{1}{m^2} \quad[$ As $P$ and $u$ are constants]

Topic: Physics, Laws of Motion, JEE Main
Two tubes of radii r1 and r2 and lengths l1 and l2, respectively, are connected in series and a liquid flows through each of them in stream line conditions. P1 and P2 are pressure differences across the two tubes. If P2 is 4P1 and l2 is l 1 4 then the radius r2 will be equal to :

Options:
A 4r1
B r1
C 2r1
D r12

The correct answer is:
r12

Solution:
dVdt=π8pr4ηL
p1r14L1=p2r24L2
p1r14l1=4p1r24l14
⇒r24=r1416
⇒r2=r12

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2017 (09 Apr Online)
Two tuning forks $A$ and $B$ sounded together give 6 beats per second. With an air resonance tube closed at one end, the two forks give resonance when the two air columns are $24 \mathrm{~cm}$ and 25 $\mathrm{cm}$ respectively. Calculate the frequencies of forks.

Options:
A $120 \mathrm{~Hz}, 124 \mathrm{~Hz}$
B $110 \mathrm{~Hz}, 114 \mathrm{~Hz}$
C $150 \mathrm{~Hz}, 144 \mathrm{~Hz}$
D $170 \mathrm{~Hz}, 118 \mathrm{~Hz}$

The correct answer is:
$150 \mathrm{~Hz}, 144 \mathrm{~Hz}$

Solution:
Let the frequency of the first fork be $\mathrm{f}_{1}$ and that of second be $\mathrm{f}_{2}$
We then have, $\mathrm{f}_{1}=\frac{\mathrm{v}}{4 \times 24}$ and $\mathrm{f}_{2}=\frac{\mathrm{v}}{4 \times 25}$
We also see that $\mathrm{f}_{1}>\mathrm{f}_{2}$
$$
\therefore \mathrm{f}_{1}-\mathrm{f}_{2}=6
$$
and $\frac{\mathrm{f}_{1}}{\mathrm{f}_{2}}=\frac{24}{25}$
Solving (i) and (ii), we get $\mathrm{f}_{1}=150 \mathrm{~Hz}$ and $\quad \mathrm{f}_{2}=144 \mathrm{~Hz}$

Topic: Physics, Waves and Sound, JEE Main
Two tuning forks \(X\) and \(Y\) are of frequencies \(280 \mathrm{~Hz}\) and \(284 \mathrm{~Hz}\). A third tuning fork \(Z\) is of unknown frequency. When \(X\) and \(Z\) are sounded together certain beats are heard per second. When \(Y\) and \(Z\) are sounded together beat frequency is found to be thrice as great. The frequency of \(Z\) is

Options:
A \(282 \mathrm{~Hz}\)
B \(286 \mathrm{~Hz}\)
C \(280 \mathrm{~Hz}\)
D \(278 \mathrm{~Hz}\)

The correct answer is:
\(278 \mathrm{~Hz}\)

Solution:
Given, frequency of tuning fork \(X\) and \(Y\) are,
\(\begin{aligned}
& n_X=280 \mathrm{~Hz} \\
& n_Y=284 \mathrm{~Hz} \\
& n_Z=?
\end{aligned}\)
When tuning forks \(X\) and \(Z\) are sounded together, then beats produced is \(b\) (assume).
\(\therefore \quad n_Z-n_X=b\) ...(i)
Again, when \(Y\) and \(Z\) are sounded together, then
\(n_Z-n_Y=3 b\) ...(ii)
Dividing Eq. (i) by Eq. (ii), we have
\(\begin{array}{ll}
& \frac{n_Z-n_X}{n_Z-n_Y}=\frac{b}{3 b} \Rightarrow \frac{n_Z-280}{n_Z-284}=\frac{1}{3} \\
\Rightarrow \quad & 3 n_Z-840=n_z-284 \\
\Rightarrow \quad & 3 n_z-n_Z=840-284 \\
\Rightarrow \quad & 2 n_Z=556 \Rightarrow n_Z=278 \mathrm{~Hz}
\end{array}\)

Topic: Physics, Waves and Sound, JEE Main
Two uniform brass body \( A \) and \( B \) of lengths \( l \) and \( 2 l \) and radii \( 2 r \) and \( r \) respectively are heated to the same temperature. The ratio of the increase in the length of \( A \) to that of \( B \) is

Options:
A \( 1: 1 \)
B \( 1: 2 \)
C \( 1: 4 \)
D \( 2: 1 \)

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

Solution:
∵ by thermal expansion,
Δl= ∝lΔθ
∴ for small ∝ and Δθ
Δl∝l
∴ ΔlAΔlB=lAlB=l2l=12

Topic: Physics, Thermal Properties of Matter, JEE Main
Two uniform brass rods A and B of lengths l and 2l and radii 2r and r respectively are heated to the same temperature. The ratio of the increase in the volumes of A to that of B is

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

The correct answer is:
2:1

Solution:
For brass rod A
Volume V1= π2r2 ×l ...(i)
For volume expansion
V 1 ′ = V1(1+γât)
⇒ V 1 ′ - V1 ∝V1
Or âV1 ∝V1 ...(ii)
Similarly, for brass rod B
Volume V2=π(r)2×2l ...(iii)
and âV2 ∝V2 ...(iv)
Dividing Eq. (i) by Eq. (ii), we get
V1V2=π4r2lπr22l=21
From eqs. (ii) and (iv),
âV1âV2=21

Topic: Physics, Thermal Properties of Matter, JEE Main
Two uniform circular discs are rotating independently in the same direction around their common axis passing through their centres. The moment of inertia and angular velocity of the first disc are 0.1 kg-m2 and 10 rad s-1 respectively while those for the second one are 0.2 kg-m2 and 5 rad s-1 respectively. At some instant they get stuck together and start rotating as a single system about their common axis with some angular speed. The kinetic energy of the combined system is :

Options:
A 103J
B 203J
C 53J
D 23J

The correct answer is:
203J

Solution:
Common angular velocity from angular momentum conversation

ω=l1ω1+I2ω2I1+I2=0.1×10+0.2×50.1+0.2=203
Final K.E.
Kf=12I1ω2+12I2ω2
=120.1+0.2×2032⇒kf=203

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Two uniform discs A and B of equal radii but having different masses 1 kg and 2 kg respectively, are connected (at centres) by a massless spring of spring constant k=1000 N m-1 and then placed on a sufficiently rough horizontal surface. If initially, the system is released from rest with the spring being compressed by 10 cm, then the speed of A when the spring comes back to its natural length, is

Options:
A vA=409 m s-1
B vA=209 m s-1
C vA=203 m s-1
D vA=403 m s-1

The correct answer is:
vA=409 m s-1

Solution:
kx-f=Ma

fR=12MR2aR

a=2kx3M

f=kx3

So the friction forces acting on A and B are always equal in magnitude and opposite in direction. Hence, the net force acting on the system is always zero.
By energy and momentum conservation. (When spring in its natural length)

mAvA=mBvB

vB=vA2

34mAvA2+34mBv2B=12kx2

34mAvA2+34mBvA24=12kx2

9vA28=12×1000×0.12

vA=409 m s-1

Topic: Physics, Rotational Motion, JEE Main
Two uniform rods of equal length but different masses are rigidly joint to form an L-shaped body, which is then pivoted as shown. If in equilibrium, the body is in the shown configuration, ratio Mm will be

Options:
A 2
B 3
C 2
D 3

The correct answer is:
3

Solution:
Net torque about O should be zero. Hence
mgl2sin 60Â°=Mgl2sin 30Â°
â´ Mm=sin 60Â°sin 30Â°=3

Topic: Physics, Rotational Motion, JEE Main
Two uniform solid spheres made of copper have radii 15 cm and 20 cm respectively. Both of them are heated to a temperature of 70°C and then placed in a region of ambient temperature equal to 45°C. What will be the ratio of the initial rates of cooling of both the spheres?

Options:

The correct answer is:
01.33

Solution:
For two bodies made up of same material, rate of loss of heat will depend only on their surface area for same temperature change.

∴ dQdt1dQdt2=A1A2=r12r22 ……i

Ratio of initial rates of cooling=dθdt1dθdt2

As, dQdt=mcdθdt

∴ m1c1(dθ/dt)1m2c2(dθ/dt)2=r12r22 ...[from (i)]

since, both the spheres are made up of copper,

c1=c2=c.

∴ dθdt1dθdt2=r12×m2r22×m1

As, spheres have same densities,

m1m2=V1V2=r13r23

∴ (dθ/dt)1(dθ/dt)2=r12r22×r23r13=r2r1=2015=43=1.33

Topic: Physics, Thermal Properties of Matter, JEE Main
Two uniform strings $A$ and $B$ made of steel are made to vibrate under the same tension. If the first overtone of $A$ is equal to the second overtone of $B$ and if the radius of $A$ is twice that of $B$, the ratio of the lengths of the strings is :

Options:
A $1: 2$
B $1: 3$
C $1: 4$
D $1: 5$

The correct answer is:
$1: 3$

Solution:
Fundamental frequency of string,
$n=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
where, $m=$ mass per unit length of the string
$=\pi r^2 \rho$
$\therefore \quad n=\frac{1}{2 l} \sqrt{\frac{T}{\pi r^2 \rho}}$
For string $A$ first overtone,
$n_A=\frac{1}{l_1} \sqrt{\frac{T_1}{m_1}}=\frac{1}{l_1} \sqrt{\frac{T_1}{\pi r_1^2 \rho}}$
For string $B$ second overtone,
$n_B=\frac{3}{2 l_2} \sqrt{\frac{T_2}{\pi r_2^2 \rho_2}}$
According to question,
$n_A=n_B$
$\frac{1}{l_1} \sqrt{\frac{T_1}{\pi r_1^2 l_1}}=\frac{3}{2 l_2} \sqrt{\frac{T_2}{\pi r_2^2 l_2}}$
but $\quad T_1=T_2$
$\therefore \quad \frac{2 l_2}{3 l_1}=\sqrt{\frac{\pi r_1^2 \rho}{\pi r_2^2 \rho}}$
$\frac{4 l_2^2}{9 l_1^2}=\frac{r_1^2}{r_2^2}$
$\left(\frac{l_1}{l_2}\right)^2=\frac{4 r_2^2}{9 r_1^2}$
$\frac{2 l_2}{3 l_1}=\frac{r_1}{r_2}$
$\frac{l_2}{l_1}=\frac{3 r_1}{2 r_2}$
$\frac{l_1}{l_2}=\frac{2 r_2}{3 r_1}$
but $\quad r_1=2 r_2$
$\therefore \quad \frac{l_1}{l_2}=\frac{2 r_2}{3\left(2 r_2\right)}$
$l_1: l_2=1: 3$

Topic: Physics, Waves and Sound, JEE Main
Two uniform wires of the same material are vibrating under the same tension. If the first overtone of the first wire is equal to the second overtone of the second wire and radius of the first wire is twice the radius of the second wire then the ratio of the lengths of the first wire to second wire is

Options:
A 13
B 14
C 15
D 16

The correct answer is:
13

Solution:
Fundamental frequency of the first wire
f=12L1 Tm =12L1TÏr1 2Ï =12L1r1TÏÏ
The first overtone f1=2f=22L1r1TÏÏ =1L1r1TÏÏ
The second overtone of the second wire
f2=32L2r2TÏÏ
f1=f2
1L1r1TÏÏ =32L2 r2 TÏÏ
â´ 3L1r1=2L2r2
L1L2=23.r2r1
=23.r22r2 [â´ r1=2r2]
=13

Topic: Physics, Waves and Sound, JEE Main
Two uniformly charged nonconducting spheres, each of radius R, are fixed in a gravity free space as shown in the figure. If an electron is released at rest from the point A, then its speed just before striking the other sphere is [mass of electron = me]

Options:
A 2Qe9πε0meR
B 8Qe9πε0meR
C 16Qe9πε0meR
D 4Qe9πε0meR

The correct answer is:
8Qe9πε0meR

Solution:
Using the conservation of mechanical energy, we get
KQeR-KQe9R=-KQeR+KQe9R+12mv2
-2KQe9R+2KQeR=12mv2
164πε0eQ9R=12mv2 ⇒ v=8Qe9πε0meR

Topic: Physics, Electrostatics, JEE Main
Two uniformly charged spherical conductors A and B of radii 5 mm and 10 mm are separated by a distance of 2 cm. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitudes of the electric fields at the surface of the sphere A and B will be

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

The correct answer is:
2:1

Solution:
When two conductors are connected by a conducting wire, their potentials become equal. Therefore, VA=VB

The potential of a spherical conductor is given by, V=kQR

⇒kQARA=kQBRB

⇒Q∝R

Now the electric field near the surface of the conductor will be, E=kQR2

⇒EAEB=kQARA2kQBRB2=RBRA

⇒EAEB=105=21

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Two uniformly long charged wires with linear densities λ and 3λ are placed along X and Y axis respectively. Determined the slope of electric field at any point on the line y=3 x .

Options:
A 33
B 332
C 133
D 3

The correct answer is:
133

Solution:
Ex=3λ2πϵox
Ey=λ2πϵox3
E→=3λ2πε0x  i^+λ2πε0x3  j^
Slope=EyEx=13×3=133

Topic: Physics, Electrostatics, JEE Main
Two unknown resistance are connected in the two gaps of a meter-bridge. The null point is obtained at 40 cm from the left end. A 30 Î© resistance is connected in series with the smaller of the two resistances, the null point shifts by 20 cm to the right end. What is the value of the smaller resistance (in Î©)?

Options:

The correct answer is:
24

Solution:
lx=40 cm, lR=60 cm
xR=lxlR=4060=23 ....... (i)
∴ x+30 R=6040=32
∴ x+30x=94

9x=4x+120
5x=120 ⇒ x=24 Ω

Topic: Physics, Current Electricity, JEE Main
Two unknown resistances are connected in two gaps of a meter-bridge. The null point is obtained at 40 cm from the left end. A 30 Î© resistance is connected in series with the smaller of the two resistances, the null point shifts by 20 cm to the right end. The value of smaller resistance in Î© is

Options:

The correct answer is:
24

Solution:
lx=40 cm, lR=60 cm
xR=lxlR=4060=23 ....... (i)
â´ x+30 R=6040=32
â´ x+30R=32
â´ 2x+30=3R
R= 2(x+30)3 ........ (ii)
From (i) and (ii),
x2x+30 3=23
3x2(x+30)=23
9x=4x+120
5x=120 â x=24 Î©

Topic: Physics, Current Electricity, JEE Main
Two van der waal's gases have same value of a but different value of b which of the following statement is correct?

Options:
A The smaller the value of b larger will be compressibility
B The larger the value of b, larger will be compressibility
C Both have same compressibility
D All the with smaller value of b will occupy larger volume

The correct answer is:
The smaller the value of b larger will be compressibility

Solution:
Smaller the value of 'b', smaller the volume occupied by molecules so high compressibility.

Topic: Chemistry, States of Matter, JEE Main
Two vectors A→ and B→ have equal magnitudes. The magnitude of A→+B→ is ' n ' times the magnitude of A→-B→ . The angle between A→ and B→ is:

Options:
A cos-1n2-1n2+1
B sin-1n-1n+1
C cos-1n-1n+1
D sin-1n2-1n2+1

The correct answer is:
cos-1n2-1n2+1

Solution:
Given: A→+B→=nA→-B→, so, the magnitudes of this will be given as,

A2+B2+2ABcosθ=nA2+B2-2ABcosθ;where θ is the angle between A→ and B→.

Also, A→=B→,

2A2+2A2cosθ=n2A2-2A2cosθ

Squaring both sides:
2A21+cosθ=n22A21-cosθ

cosθ=n2-1n2+1

i.e., θ=cos-1n2-1n2+1

Topic: Physics, Mathematics in Physics, JEE Main
Two vectors A→ and B→ have equal magnitudes. The magnitude of A→+B→ is ' n ' times the magnitude of A→-B→ . The angle between A→ and B→ is:

Options:
A cos-1n2-1n2+1
B sin-1n-1n+1
C cos-1n-1n+1
D sin-1n2-1n2+1

The correct answer is:
cos-1n2-1n2+1

Solution:
Given: A→+B→=nA→-B→, so, the magnitudes of this will be given as,

A2+B2+2ABcosθ=nA2+B2-2ABcosθ;where θ is the angle between A→ and B→.

Also, A→=B→,

2A2+2A2cosθ=n2A2-2A2cosθ

Squaring both sides:
2A21+cosθ=n22A21-cosθ

cosθ=n2-1n2+1

i.e., θ=cos-1n2-1n2+1

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Two vectors \( \overrightarrow{\mathrm{a}} \) and \( \overrightarrow{\mathrm{b}} \) are at an angle of \( 60^{\circ} \) with each other. Their resultant makes an angle of \( 45^{\circ} \) with \( \overrightarrow{\mathrm{a}} \). If \( |\overrightarrow{\mathrm{b}}|=2 \) units, then \( |\overrightarrow{\mathrm{a}}| \) is

Options:
A \( \sqrt{3} \)
B \( \sqrt{3}-1 \)
C \( \sqrt{3}+1 \)
D \( \frac{\sqrt{3}}{2} \)

The correct answer is:
\( \sqrt{3}-1 \)

Solution:
Given,
Magnitude of b→ is, b→=2 units.
Angle between the vectors is, θ=60°.
Angle between resultant and a→ is, ϕ=45°.
The above is shown in the below figure,

From parallelogram law, ϕ & θ is related as,
tanϕ=b→sinθa→+b→cosθ.
Substituting all the corresponding given values in above relation, we have,
⇒tan45°=2sin60°a→+2cos60°=3a+1, ∵a→=a
⇒1=3a+1
⇒a+1=3
⇒a=3-1.

Topic: Physics, Mathematics in Physics, JEE Main
Two vectors P→ and Q→ have equal magnitudes. If the magnitude of P→+Q→ is n times the magnitude of P→-Q→, then angle between P→ and Q→ is

Options:
A

sin-1n-1n+1

B cos-1n-1n+1
C sin-1n2-1n2+1
D cos-1n2-1n2+1

The correct answer is:
cos-1n2-1n2+1

Solution:
|P→|=|Q→|=x …i

|P→+Q→|=n|P→-Q→|

P2+Q2+2PQcosθ=n2P2+Q2-2PQcosθ

Using (i) in above equation,

cosθ=n2-11+n2

θ=cos-1n2-1n2+1

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Two vectors X→ and Y→ have equal magnitude. The magnitude of (X→-Y→) is n times the magnitude of (X→+Y→). The angle between X→ and Y→ is :

Options:
A cos-1-n2-1n2-1
B cos-1n2-1-n2-1
C cos-1n2+1-n2-1
D cos-1n2+1n2-1

The correct answer is:
cos-1n2-1-n2-1

Solution:
Given X=Y

X2+Y2-2×Ycosθ

=nX2+Y2+2×Ycosθ

Square both sides

2X2(1-cosθ)=n2·2X2(1+cosθ)

1-cosθ=n2+n2cosθ

cosθ=1-n21+n2

θ=cos-1n2-1-n2-1

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (25 Jul Shift 2)