datavorous/entrance-exam-dataset huggingface dataset:
Two identical spheres each of mass 2 kg and radius 50 cm are fixed at the ends of a light rod so that the separation between the centers is 150 cm. Then, moment of inertia of the system about an axis perpendicular to the rod and passing through its middle point is x20 kg m2, where the value of x is

Options:

The correct answer is:
53

Solution:
Using parallel axis theorem, we can write total moment of inertia as, I=25mR2+md2×2

⇒I=225×2×122+2×342=5320 kg m2

Therefore, x=53.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Two identical spheres, each of mass m are suspended by vertical strings such that they are in contact with their centres at the same level. A third identical sphere strikes the other two spheres simultaneously with a velocity u such that the centres of the spheres at the instant of impact form an equilateral triangle in a vertical plane. If the collision is perfectly elastic, then the combined impulse due to the strings is

Options:
A 127mu
B 67mu
C 237mu
D 87mu

The correct answer is:
127mu

Solution:
Let us assume that after the collision, the velocity of the incoming ball changes from u to u' and the other two balls move in the opposite directions with the same speed v, then

-2N∆tcos30°= mu'-mu

⇒-N∆t3=mu'-mu

For the ball attached to the string,

N∆tsin30°=mv

⇒N∆t=2mv

Eliminating N∆t, we obtain

-23mv=mu'-mu

Using the coefficient of restitution equation we get

1=vcos60°-u'cos30°ucos30°

⇒3u=v-3u'

⇒23v=6u+6u'

6u+6u'=u-u'

⇒u'=-57u

The total impulse of tension is

-2T∆t=mu'-mu

2T∆t=m5u7+mu=127mu

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two identical spherical balls of mass M and radius R each are stuck on two ends of a rod of length 2R and mass M(see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is

Options:
A 20915MR2.
B 15215MR2.
C 13715MR2.
D 175MR2.

The correct answer is:
13715MR2.

Solution:
Applying parallel axis theorem for the balls,
Ib=25MR2+M2R2

Itotal=2Ib+IR
Itotal=25MR2+M2R2×2+M122R2
Itotal=13715MR2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Two identical springs of constant ' $\mathrm{K}$ ' are connected in series and parallel in shown in figure. A mass ' $\mathrm{M}$ ' is suspended from them. The ratio of their frequencies is series to parallel combination will be

Options:
A 1:2
B 1:4
C 4:1
D $1: \sqrt{2}$

The correct answer is:
1:2

Solution:
In series combination, the effective spring constant
$$
\mathrm{k}_1=\frac{\mathrm{k}}{2}
$$
In the parallel combination, the effective spring constant is
$$
\mathrm{k}_2=2 \mathrm{k}
$$
$$
\begin{aligned}
& \mathrm{f}_1=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}_1}{\mathrm{~m}}}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}}{2 \mathrm{~m}}} \\
& \mathrm{f}_2=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{k}_2}{\mathrm{~m}}}=\frac{1}{2 \pi} \sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}} \\
& \therefore \frac{\mathrm{f}_1}{\mathrm{f}_2}=\sqrt{\frac{1}{4}}=\frac{1}{2}
\end{aligned}
$$

Topic: Physics, Oscillations, JEE Main
Two identical springs of spring constant 2k are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is :

Options:
A πmk
B 2πm2k
C πm2k
D 2πmk

The correct answer is:
πmk

Solution:
For parallel combination keq=k1+k2

keq=4k

T=2πmkeq⇒T=2πm4k⇒T=πmk

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Two identical springs (spring constant = k) are connected to two light rigid rods in parallel combination and then a third spring (identical to the other two) is connected to this combination in series. The combined system is fixed to the ceiling and a block of mass m is attached as shown in the figure. If the mass is slightly displaced and then let go, then the system shall oscillate with a time period of

Options:
A 2πm3k
B 2π3m2k
C 2π2m3k
D 2π3km

The correct answer is:
2π3m2k

Solution:
Consider the system as shown in figure below.

Here, k2 and k3 are in parallel. k1 is in series with parallel combination of k2 and k3.
Effective spring constant of the combination is
keff=k2+k3k1k1+k2+k3
=2k×k3k=23k
∴ Time period of oscillation
T=2πmkeff
=2πm23k
=2π3m2k

Topic: Physics, Oscillations, JEE Main
Two identical strings X and Z made of same material have tension TX and TZ in then if their fundamental frequencies are 450 Hz and 300 Hz, respectively, then the ratio TX/TZ is :

Options:
A 2.25
B 0.44
C 1.25
D 1.5

The correct answer is:
2.25

Solution:
fX=12ℓTXμ

fZ=12ℓTZμ

fXfZ=450300=TXTZ

⇒TX/TZ=9/4=2.25

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2020 (02 Sep Shift 1)
Two identical systems, with heat capacity at a constant volume that varies as Cv=bT3 (where b is a constant) are thermally isolated. Initially, one system is at a temperature 100 K and the other is at 200K . The systems are then brought to thermal contact and the combined system is allowed to reach thermal equilibrium. If T04=n×108 K then what will be the value of n, where T0 is the final temperature in kelvin?

Options:

The correct answer is:
8.5

Solution:
Energy given by a system = Energy taken by another system
dQ1=dQ2
⇒ ∫100TmcvdT=-∫200TmcvdT
⇒ ∫100TbT3dT=-∫200TbT3dT⇒T410T=-T4200T
⇒ T4-1004=-T4-2004
⇒ 2T4=17×108⇒T=171 K

⇒ T4=8.5×108 K

Topic: Physics, Thermodynamics, JEE Main
Two identical tennis balls each having mass m and charge q are suspended from a fixed point by threads of length l. What is the equilibrium separation when each thread makes a small angle θ with the vertical?

Options:
A x=q2l2πε0mg12
B x=q2l2πε0mg13
C x=q2l22πε0 m2 g13
D x=q2l22πε0m2g213

The correct answer is:
x=q2l2πε0mg13

Solution:
Tcosθ=mg
Tsinθ=kq2x2
tanθ=kq2x2mg
as tanθ≈sinθ≈x2L
x2L=Kq2x2mg
x=q2 L2πε0mg13

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Two identical thin biconvex lenses of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is _____ cm.

Options:

The correct answer is:
10

Solution:
For convex lenses, let radius of curvature is R. Then,

115 cm=1.5-11R-1-R

⇒1R=115 ⇒R=15 cm

Now for concave lens,

1fconcave=1.25-11-R-1R

=14×-2R=-130

1feff=115+115-130=-110

⇒feff=-10 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (24 Jun Shift 1)
Two identical thin Plano-convex glass lenses (refractive index 1.5 ) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the center. The intervening space is filled with oil of refractive index 1.7 The focal length of the combination is

Options:
A -20 cm
B -25 cm
C -50 cm
D 50 cm

The correct answer is:
-50 cm

Solution:
Len's maker's formula
1f = μ-11R1-1R2
1f1=μ-11 20 ⇒ 1.5-1 120=140
1f1=140
For lens 1f1+1f2 = 140+140=120
For concave 1f3=1.7-1 (-1R-1R )= -0.7×2R
1feq=120+ -0.7×2R
=120 - 1.420
feq= -200.4 =-50 cm

Topic: Physics, Ray Optics, JEE Main
Two identical uniform rectangular blocks (with longest side $\mathrm{L}$ ) and a solid sphere of radius $\mathrm{R}$ are to be balanced at the edge of a heavy table such that the centre of the sphere remains at the maximum possible horizontal distance from the vertical edge of the table without toppling as indicated in the figure.
If the mass of each block is $\mathrm{M}$ and of the sphere is $\mathrm{M/2}$, then the maximum distance $\mathrm{x}$ that can be achieved is

Options:
A $8 \mathrm{L} / 15$
B $5 \mathrm{L} / 6$
C $(3 \mathrm{L} / 4+\mathrm{R})$
D $(7 \mathrm{L} / 15+\mathrm{R})$

The correct answer is:
$8 \mathrm{L} / 15$

Solution:
$2+\mathrm{S}$ System lie above edge of 1 .
$\begin{array}{l}
\frac{M}{2} y-M\left(\frac{L}{2}-y\right)=0 \\
\frac{y}{2}+y=\frac{L}{2} \\
y=\frac{L}{3}
\end{array}$
Now $1+2+\mathrm{S}$ centre of mass wili lie above the table
$\frac{3 M}{2}\left(x-\frac{L}{3}\right)+M\left(x-\frac{L}{3}-\frac{L}{2}\right)=0$
$\frac{3 x}{2}-\frac{L}{2}+x-\frac{L}{3}-\frac{L}{2}=0$
$\frac{5 x}{2}=\frac{4 L}{3} \quad \quad x=\frac{8 L}{15}$

Topic: Physics, Center of Mass Momentum and Collision, KVPY, KVPY 2013 (SA)
Two identical vessels $A$ and $B$ with frictionless pistons contain the same ideal gas at the same temperature and the same volume $V$. The masses of gas in $A$ and $B$ and $m_A$ and $m_B$, respectively. The gases are allowed to expand isothermally to the same final volume $2 V$. The change in pressures of the gas in $A$ and $B$ are found to be $\Delta P$ and $1.5 \Delta P$, respectively. Then

Options:
A $9 m_A=4 m_B$
B $3 m_A=2 m_B$
C $2 m_A=3 m_B$
D $4 m_A=9 m_B$

The correct answer is:
$3 m_A=2 m_B$

Solution:
Let $p_1$ and $p_2$ are the initial and final pressures of the gas filled in $A$. Then
$$
p_1=\frac{\mu_{\mathrm{A}} R T}{V} \text { and } P_2=\frac{\mu_A R T}{2 V}
$$
Dividing Eq. (ii) by (i), we get
$\begin{aligned} 1.5 & =\frac{m_B}{m_A} \\ \text { or } \quad \frac{3}{2} & =\frac{m_B}{m_A} \\ \text { or } \quad 3 m_A & =2 m_B .\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
Two identical vessels contain two different ideal gases at the same temperature. If the average speed of gas molecules in the first vessel is equal to the most probable speed of molecules in the second vessel, then the ratio of the mass of gas molecules in the first vessel to that in the second vessel is

Options:
A 4π
B 8π
C 2π
D π2

The correct answer is:
4π

Solution:
vavg=8RTπM1
vmp=2RTM2
8M1π=2M2
M1M2=4π

Topic: Physics, Kinetic Theory of Gases, JEE Main
Two identical wires A and B, each of length l, carry the same current I . Wire A is bent into a circle of radius R and wire B is bent to form a square of side a. If BA and BB are the values of magnetic field at the centres of the circle and square respectively, then the ratio BABB is

Options:
A π216
B Ï282
C Ï28
D Ï2162

The correct answer is:
π282

Solution:
As length of both the wires is same, 2πR=4a
⇒aR=2π4

⇒aR=π2

BA=μ0i2R and
BB=μ0iπa22
Therefore, ratio will be BABB=μ0i2R×πa22 μ0i
=πa42R=π42π2=π282

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2016 (03 Apr)
Two identical wires A and B, each of length ' l ', carry the same current I . Wire A is bent into a circle of radius R and wire B is bent to form a square of side 'a'. If BA and BB are the values of magnetic field at the centres of the circle and square respectively, then the ratio BABB is

Options:
A π216
B π282
C π28
D π2162

The correct answer is:
π282

Solution:
2ÏR=4a
aR=2Ï4â aR=Ï2
BA=Î¼0i2R
BB=Î¼0iÏa22
BABB=Î¼0i2RÃÏa22 Î¼0i
=Ïa42R=Ï42Ï2=Ï282

Topic: Physics, Magnetic Effects of Current, JEE Main
Two immiscible liquids of refractive indices 85 and 32 respectively are put in a beaker as shown in the figure. The height of each column is 6 cm. A coin is placed at the bottom of the beaker. For near normal vision, the apparent depth of the coin is α4 cm. The value of α is _______.

Options:

The correct answer is:
31

Solution:
The formula to calculate the apparent height of the coin when seen through the combination of liquids can be written as

happ=h1μ1+h2μ2 ...1

From equation (1), it follows that

happ=63/2+68/5 cm=4+154 cm=314 cm

Hence, α=31.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2024 (27 Jan Shift 1)
Two important ores of metals are given below
Malachite CuCO3.Cu(OH)2.xH2O
Carnallite KCl.MgCl2.yH2O
What is y‐x?

Options:

The correct answer is:
6

Solution:
Malachite CuCO3.Cu(OH)2
Carnallite KCl.MgCl2.6H2O

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
Two inclined planes are placed as shown in figure.
A block is projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10 m. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is t2+1s. The value of t is _____ (use g=10 m s-2)

Options:

The correct answer is:
2

Solution:
From energy conservation at point A and B,

12mv02=mgh

⇒v0=gh=10×10

⇒v0=102 m s-1

For A→B

At B, v=0

Acceleration of a particle moving on a smooth incline is gsinθ. Therefore, along AB, a=-gsin45°=-102 m s-2.

Using equation of motion,

v=u+at1

0=102-102t1⇒t1=2 s

For B→C

Using second equation of motion,

s=ut2+12at22

⇒10sin30°=1210sin30°t22

t2=22 s

So total time T=t1+t2 =22+2

=22+1 s

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (27 Jul Shift 2)
Two indentical magnetic dipoles of magnetic moment $1.0 \mathrm{Am}^{2}$ each,placed at a separation of $2 \mathrm{~m}$ with their axes perpendicular to each other. The resultant magnetic field at a point midway between the dipoles is

Options:
A $\quad \sqrt{5} \times 10^{-7} T$
B $5 \times 10^{-7} T$
C $10^{-7} T$
D $2 \times 10^{-7} T$

The correct answer is:
$\quad \sqrt{5} \times 10^{-7} T$

Solution:
Since axes are perpendicular so mid point lies on axial line of one magnet and on equitorial line of other magnet
$\therefore \mathrm{B}_{1}=\frac{\mu_{0}}{4 \pi} \frac{2 \mathrm{M}}{\mathrm{d}^{3}}=\frac{10^{-7} \times 2 \times 1}{1^{3}}=2 \times 10^{-7} \mathrm{~T}$ and $\mathrm{B}_{2}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{M}}{\mathrm{d}^{3}}=\frac{10^{-7} \times 1}{1^{3}}=10^{-7} \mathrm{~T}$
As $\mathrm{B}_{1} \perp \mathrm{B}_{2}$
$\therefore$ Resultant magnetic field
$=\sqrt{\mathrm{B}_{1}^{2}+\mathrm{B}_{2}^{2}}=\sqrt{5} \times 10^{-7} \mathrm{~T}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
Two inductors of inductance \( L \) each are connected in series with opposite magnetic fluxes. What is the resultant inductance? (Ignore mutual inductance)

Options:
A Zero
B \( L \)
C \( 2 L \)
D \( 3 L \)

The correct answer is:
\( 2 L \)

Solution:
Ignoring mutual induction, resultant, inductance
L′= L1+L2
= L+L=2L

Topic: Physics, Electromagnetic Induction, JEE Main
Two infinite line-charges parallel to each other are moving with a constant velocity $v$ in the same direction as shown in the figure. The separation between two line-charges is $d$. The magnetic attraction balances the electric repulsion when, [c = speed of light in free space]

Options:
A $\quad v=\sqrt{2} c$
B $v=\frac{c}{\sqrt{2}}$
C $v=c$
D $v=\frac{c}{2}$

The correct answer is:
$v=c$

Solution:
$\mathrm{F}_{\mathrm{E}}=\mathrm{q}_2 \mathrm{E}_1=\lambda_2 \ell \frac{2 \mathrm{k} \lambda_1}{\mathrm{~d}}$
$\mathrm{~F}_{\mathrm{B}}=\frac{\mu_0 \mathrm{i}_1 \mathrm{i}_2 \ell}{2 \pi \mathrm{d}}=\frac{\mu_0 \lambda_1 \mathrm{~V} \lambda_2 \mathrm{~V}}{2 \pi \mathrm{d}} \times \ell$
$\mathrm{F}_{\mathrm{E}}=\mathrm{F}_{\mathrm{B}}$
$\Rightarrow \frac{2 \mathrm{k} \lambda_1 \lambda_2 \ell}{\mathrm{d}}=\frac{\mu_0 \lambda_1 \lambda_2 \mathrm{~V}^2 \ell}{2 \pi \mathrm{d}}$
$\Rightarrow \frac{2 \lambda_1 \lambda_2 \ell}{4 \pi \varepsilon_0}=\frac{\mu_0 \lambda_1 \lambda_2 \mathrm{~V}^2 \ell}{2 \pi}$
$\mathrm{V}^2=\frac{1}{\mu_0 \varepsilon_0}=\mathrm{c}^2 \Rightarrow \mathrm{V}=\mathrm{c}$

Topic: Physics, Magnetic Effects of Current, JEE Main
Two infinite planes each with uniform surface charge density +σ are kept in such a way that the angle between them is 30o . The electric field in the region shown between them is given by:

Options:
A σ2∈01+3y^-x^2
B σ∈01+32y^+x^2
C σ2∈01+3y^+x^2
D σ2∈01-32y^-x^2

The correct answer is:
σ2∈01-32y^-x^2

Solution:
E→=σ2ε0cos⁡60o-x^+σ2ε0-σ2ε0sin⁡60oy^
E→=σ2ε01-32y^-12x^

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Two infinite sheets carrying current in same direction (of equal current per unit length K ) are separated by a distance d. A proton is released from a point between the plates with a velocity parallel to the sheets but perpendicular to the direction of current in the sheets. Then the path of the proton is

Options:
A circle
B helix
C straight line
D straight line only if it is released from a point exactly midway between the two plates.

The correct answer is:
straight line

Solution:
As net magnetic field due to sheets will be zero, net force acting on the proton also becomes zero, it will continue to move on a straight line. âµ F â = q V â Ã B â

Topic: Physics, Magnetic Effects of Current, JEE Main
Two infinitely long wires carry linear charge densities + Î» and -Î» respectively, as shown.The potential difference between points A (at a distance a from the first wire) and B (at a distance b from the second wire) is

Options:
A λ2πε0lnd-ad-bab
B Î»2ÏÎµ0lnâ¡d2abÂ
C Î»4ÏÎµ0lnâ¡d-ad-babÂ
D Î»2ÏÎµ0lnâ¡d-bb/d-aa

The correct answer is:
λ2πε0lnd-ad-bab

Solution:
The potential difference between A and B is

VB-VA=-∫ABE→.dr→

VB-VA=-λ2πε0∫AB1r+1d-rdr

VB-VA=-λ2πε0∫ad-bdrr+∫ad-bdrd-r

VA-VB=λ2πε0lnd-ba-lnbd-a

VA-VB=λ2πε0lnd-ad-bab

Topic: Physics, Electrostatics, JEE Main
Two inflated balloons I and II (thin skin) having volume 600 mL and 1500 mL at 300 K are taken as shown in diagram. If maximum volume of inner and outer balloons are 800 mL and 1800 mL respectively then find the balloon which will burst first on gradual heating.

Options:
A inner balloon
B outer balloon
C both simultaneously
D unpredictable

The correct answer is:
outer balloon

Solution:
Using Charle's law
V1T1=V2T2
Sppose inner balloon burst first 600300=800T2
T 2 =400âK
At 400 K the volume of outer balloon is 2000 ml so outer ballon burst first.

Topic: Chemistry, States of Matter, JEE Main
Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V_2-V_1=20 \mathrm{~V}$. (i.e. plate 2 is at a higher potential). The plates are separated by $d=0.1 \mathrm{~m}$ and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1 . What is its speed when it hits plate $2 ?$
$\left(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}, \mathrm{m}_{\mathrm{e}}=9.11 \times 10^{-31} \mathrm{~kg}\right)$

Options:
A
$32 \times 10^{-19} \mathrm{~m} / \mathrm{s}$

B
$2.65 \times 10^6 \mathrm{~m} / \mathrm{s}$

C
$7.02 \times 10^{12} \mathrm{~m} / \mathrm{s}$

D
$1.87 \times 10^6 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$2.65 \times 10^6 \mathrm{~m} / \mathrm{s}$

Solution:
$\frac{1}{2} \mathrm{mv}^2=\mathrm{eV}$
$\mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}=2.65 \times 10^6 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2006
Two integers are selected at random from the set $\{1,2, \ldots, 11\}$. Given that the sum of selected numbers is even, the conditional probability that both the numbers are even is :

Options:
A $\frac{7}{10}$
B $\frac{1}{2}$
C $\frac{2}{5}$
D $\frac{3}{5}$

The correct answer is:
$\frac{2}{5}$

Solution:
Probability of getting sum of selected two numbers is even

$=P\left(E_{1}\right)=\frac{{ }^{5} C_{2}+{ }^{5} C_{2}}{{ }^{11} C_{2}}$

Probability of getting sum is even and selected numbers are also even $P\left(E_{2}\right)=\frac{{ }^{5} C_{2}}{{ }^{11} C_{2}}$

Hence, $P\left(\frac{E_{2}}{E_{1}}\right)=\frac{{ }^{5} C_{2}}{{ }^{6} C_{2}+{ }^{5} C_{2}}=\frac{10}{15+10}=\frac{2}{5}$.

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (11 Jan Shift 1)
Two intersecting lines lying in plane P1 have equations x-11=y-32=z-43 and x-12=y-33=z-41. If the equation of plane P2 is 7x-5y+z-6=0 , then the distance between planes P1 and P2 is

Options:
A 1153
B 23
C 13
D 753

The correct answer is:
23

Solution:
Equation of plane P1 is x-1y-3z-4123231=0
x-1-7-y-3-5+z-4-1=0
7x-5y+z+4=0
So, distance between planes P1 & P2 is =1049+25+1=1053
=23

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Two intervals of time are measured as $\Delta t_1=(2.00 \pm 0.02) \mathrm{s}$ and $\Delta t_2=(4.00 \pm 0.02) \mathrm{s}$. The value of $\sqrt{\left(\Delta t_1\right)\left(\Delta t_2\right)}$ with correct significant figures and error is

Options:
A $(2.828 \pm 0.01) \mathrm{s}$
B $(2.83 \pm 0.01) \mathrm{s}$
C $(2.828 \pm 0.0075) \mathrm{s}$
D $(2.83 \pm 0.0075) \mathrm{s}$

The correct answer is:
$(2.83 \pm 0.01) \mathrm{s}$

Solution:
Here, $\Delta t_1=(2.00 \pm 0.02) \mathrm{s}$ and $\Delta t_2=(4.00 \pm 0.02) \mathrm{s}$
Given, $T=\sqrt{\left(\Delta t_1\right)\left(\Delta t_2\right)}=\sqrt{(2.00)(4.00)}=2.828427 \mathrm{~s}$
Now, according to the relative error of product
$\pm \frac{\Delta T}{T}= \pm\left(\frac{1}{2} \frac{\Delta t_1}{t_1}+\frac{1}{2} \frac{\Delta t_2}{t_2}\right)$
Since, $\Delta t_1$ and $\Delta t_2$ have 3 significant figures, then $T$ also has 3 significant figures. Hence, after rounding off,
$\begin{array}{rlrl}
& \therefore & T & =2.83 \mathrm{~s} \\
& \text {Now, } & \Delta T & = \pm \frac{1}{2}\left(\frac{0.02}{2.00}+\frac{0.02}{4.00}\right) \times 2.8284 \\
\Delta T & =0.02121 \mathrm{~s}
\end{array}$
So, after rounding off,
$\begin{aligned}
\Delta T & =0.02 \\
T & =(2.83 \pm 0.02) \mathrm{s}
\end{aligned}$
Hence,
So, most matched answer is (b).

Topic: Physics, Mathematics in Physics, JEE Main
Two ions having same mass have charges in the ratio 1: 2. They are projected normally in a uniform magnetic field with their speeds in the ratio 2: 3. The ratio of the radii of their circular trajectories is,

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

The correct answer is:
4 : 3

Solution:
R=mvqB⇒R1R2=mv1q1 Bmv2q2 B=v1q1×q2v2=q2q1×v1v2

=21×23=43

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Two ions of masses 4 amu and 16 amu have charges +2 e and +3 e respectively. These ions pass through the region of the constant perpendicular magnetic field. The kinetic energy of both ions is the same. Then:

Options:
A lighter ion will be deflected less than heavier ion
B lighter ion will be deflected more than heavier ion
C both ions will be deflected equally
D no ion will be deflected

The correct answer is:
lighter ion will be deflected more than heavier ion

Solution:
The relation between linear moment and radius in terms of the magnitude of magnetic field and charge of a particle, r=PqB=2mkeqB

⇒r∝mq⇒r1r2=34⇒r2>r1.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Two isolated conducting spheres S1 and S2 of radius 23R and 13R have 12 μC and -3 μC charges, respectively, and are at a large distance from each other, They are now connected by a conducting wire. A long time after this is done the charges on S1 and S2 are respectively:

Options:
A 4.5 μC of both
B +4.5 μC and -4.5 μC
C 3 μC and 6 μC
D 6 μC and 3 μC

The correct answer is:
6 μC and 3 μC

Solution:
Total charge =12-3=9 μC

If final charges are q1 and q2

q1q2=R1R2=21

q1=6 μC

q2=3 μC

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (03 Sep Shift 1)
Two isolated metallic solid spheres of radii R and 2R are charged such that both have same charge density σ . The spheres are then connected by a thin conducting wire. If the new charge density of the bigger sphere is σ'. The ratio σ'σ is :

Options:
A 94
B 43
C 53
D 56

The correct answer is:
56

Solution:
Charge and charge density on both isolated metallic solid spheres are show below.

Let the charges and the charge densities on both the spheres after connected by thin conducting wire be Q1', Q2', σ and σ' as shown below.

Once connected, charge on both the spheres will redistribute such that potentials at the surface of both spheres is the same.

∴kQ1'R=kQ2'2R

⇒Q1'4πε0R=Q2'4πε02R

∴Q2'=2Q1'

Using charge conservation,

Q1'+Q2'=Q1+Q2

∴Q2'2+Q2'=20πR2σ

32Q2'=20πR2σ

∴Q2'4π2R2=23×20πR2σ16πR2

∴σ'σ=56

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Two isomeric halo alkenes A and B have molecular formula C5H9Cl . 'A' gives optically inactive compound while 'B' gives optically active compound in hydrogenation. The two isomers respectively are

Options:
A 3-chloro-2-pentene and 1-chloro-2-pentene
B 4-chloro-1-pentene and 2-chloro-2-pentene
C 3-chloro-1-pentene and 4-chloro-2-pentene
D 1-chloro-1-pentene and 5-chloro-1-pentene

The correct answer is:
3-chloro-1-pentene and 4-chloro-2-pentene

Solution:
Here A and B are 3-chloro-1-pentene and 4-chloro-2-pentene respectively and the reactions are as follows

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main