datavorous/entrance-exam-dataset huggingface dataset:
$A^{2+}, B^{2+}$ and $C^{-}$form an ionic complex like $A_{x-2}\left[B(C)_x\right]_2$. If the complex is $75 \%$ dissociated in a solvent with $i=4$, the coordination number of $B$ is

Options:
A 3
B 4
C 5
D 6

The correct answer is:
5

Solution:
$\begin{gathered}A_{x-2}\left[B(C)_x\right]_2 \longrightarrow(x-2) A^{+2}+\left[B(C)_x\right]^{-(x-2)} \\ A_{x-2}\left[B(C)_x\right] \cdot A^{+2} \text { and }\left[B(C)_2\right]^{-(x-2)} \\ i=1-\alpha+x \alpha-2 \alpha+2 \alpha \\ (x-1) \alpha=3 \text { or } x-1=\frac{3}{\alpha} \\ \alpha=0.75, i=4, x=5\end{gathered}$

Topic: Chemistry, Coordination Compounds, JEE Main
A+2B→C , the rate equation for the reaction is given as
Rate =k[A][B]
If the concentration of A is kept the same but that of B is doubled what will happen to the rate itself?

Options:
A Doubled
B Halved
C The same
D Quadrupled

The correct answer is:
Doubled

Solution:
The order of the overall reaction is 2.

Rate =k[A][B]=R

R'=k[A][2B]

RR'=k[A][B]k[A][2B]=k[A][B]2k[A][B]

⇒2R=R' i.e, rate become doubles.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2015 (11 Apr Online)
A2B has antifluorite structure ( B forms FCC lattice and A occupies tetrahedral voids). If all ions along any one body diagonal are removed, then new formula of compound will be:

Options:
A A 4 B 5
B A 8 B 5
C A 7 B 6
D A 8 B 4

The correct answer is:
A 8 B 5

Solution:
On one body diagonal, 2 A-atoms(full contribution) 2 B-atoms 18contributions lie. So, new formula = A8-2B4-14=A8B5

Topic: Chemistry, Solid State, JEE Main
A 2μF capacitor C1 is first charged to a potential difference of 10 V using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor C2 of 8μF. The charge in C2 on equilibrium condition is μC. (Round off to the Nearest Integer)

Options:

The correct answer is:
16

Solution:
20=C1+C2 V⇒V=2 volt .

⇒Q2=C2V=16 μC

=16

Topic: Physics, Capacitance, JEE Main, JEE Main 2021 (17 Mar Shift 2)
$A_{2(g)}+B_{2(g)} \rightleftharpoons 2 A B_{(g)} ; \Delta H=+v e$

Options:
A Unaffected by pressure
B It occurs at 1000 pressure
C It occurs at high temperature
D It occurs at high pressure and high temperature

The correct answer is:
It occurs at high temperature

Solution:
It occurs at high temperature

Topic: Chemistry, Chemical Equilibrium, JEE Main
A $2 \mathrm{~kg}$ ball moving at $24 \mathrm{~ms}^{-1}$ undergoes inelastic head-on collision with a $4 \mathrm{~kg}$ ball moving in the opposite direction at $48 \mathrm{~ms}^{-1}$. If the coefficient of restitution is $2 / 3$, their velocities in $\mathrm{ms}^{-1}$ after impact are

Options:
A $-56,-8$
B $-28,-4$
C $-14,-2$
D $-7,-1$

The correct answer is:
$-56,-8$

Solution:
$\begin{aligned} & \begin{aligned} & v_1^{\prime}= {\left[\frac{m_1-e m_2}{m_1+m_2}\right] v_1-\left[\frac{(1+e) m_2}{m_1+m_2}\right] v_2 } \\ &=\left[\frac{2-\left(\frac{2}{3}\right) 4}{2+4}\right] 24-\left[\frac{\left(1+\frac{2}{3}\right) 4}{2+4}\right] 48 \\ &=-56 \mathrm{~m} / \mathrm{s} \\ & \text { and } v_2^{\prime}=\left[\frac{(1+e) m_1}{m_1+m_2}\right] v_1-\left[\frac{m_2-e m_1}{m_1+m_2}\right] v_2 \\ &=\left[\frac{\left(1+\frac{2}{3}\right) 2}{2+4}\right] 24-\left[\frac{4-\left(\frac{2}{3}\right) 2}{2+4}\right] 48 \\ &=-8 \mathrm{~m} / \mathrm{s}\end{aligned}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A $2 \mathrm{~kg}$ block slides on a horizontal floor with a speed of $4 \mathrm{~m} / \mathrm{s}$. It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is $15 \mathrm{~N}$ and spring constant is $10,000 . \mathrm{N} / \mathrm{m}$. The spring compresses by

Options:
A
$5.5 \mathrm{~cm}$

B
$2.5 \mathrm{~cm}$

C
$11.0 \mathrm{~cm}$

D
$8.5 \mathrm{~cm}$

The correct answer is:
$5.5 \mathrm{~cm}$

Solution:
No Solution Available

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2007
A $2 \mathrm{~kg}$ brick begins to slide over a surface which is inclined at an angle of $45^{\circ}$ with respect to horizontal axis. The co-efficient of static friction between their surfaces is:

Options:
A $1.7$
B $\frac{1}{\sqrt{3}}$
C $0.5$
D $1$

The correct answer is:
$1$

Solution:
$\begin{aligned}
& \mathrm{mg} \sin 45=\mathrm{f}_{\mathrm{L}} \\
& \mathrm{mg} \cos 45=\mathrm{N} \\
& \mathrm{f}_{\mathrm{L}}=\mu_{\mathrm{s}} N \\
& \mu_{\mathrm{s}}=\tan 45=1
\end{aligned}$
or
$\tan \theta=\mu_s$ ( $\theta$ is angle of repose)
$\tan 45=\mu_{\mathrm{s}}=1$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (04 Apr Shift 2)
A $2 \mathrm{~kg}$ mass lying on a table is displaced in the horizontal direction through $50 \mathrm{~cm}$. The work done by the normal reaction will be

Options:
A $10 \mathrm{~J}$
B 0
C $100 \mathrm{erg}$
D $100 \mathrm{~J}$

The correct answer is:
0

Solution:
Since, displacement on the body of mass $2 \mathrm{~kg}$ is in the horizontal direction while normal reaction acts on the body in perpendicular to the displacement.
Hence, $\quad \theta=0$
$\therefore$ Work done, $W=F s \cos 90^{\circ}=0$

Topic: Physics, Work Power Energy, JEE Main
A 2 kg steel rod of length 0.6 m is clamped on a table vertically at its lower end and is free to rotate in the vertical plane. The upper end is pushed so that the rod falls under gravity. Ignoring the friction due to clamping at its lower end, the speed of the free end of the rod when it passes through its lowest position is _______m s-1.
(Take g=10 m s-2 )

Options:

The correct answer is:
6

Solution:
In the above case the torque due to weight of the rod, here use the concept of translation work done and rotational kinetic energy,

⇒W=K.E.⇒mgl=12Iω2, now the moment of inertia of the rod when the axis of rotation passes through one end of the rod, I=1×ml23,
⇒mgl=12×ml23ω2
⇒ω=6gl, now use the relation between linear velocity and angular velocity,
⇒v=lω=3.6g=6 m s-1

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2021 (01 Sep Shift 2)
A 2 kg stone tied at the end of a string 1 m long is whirled along a vertical circle at a constant speed of 4 m s-1. What will be the tension in the string (in newton) at the bottom-most point?

Options:

The correct answer is:
52

Solution:
Here, m=2 kg, r=1 m, v=4 m s-1
Tension at the bottom of the circle,
TL=mg+mv2r
=20+2×421=52 N

Topic: Physics, Laws of Motion, JEE Main
A 2 L vessel is filled with air 50° C and a pressure of 3 atm. The temperature is now raised to 200 °C . A value is now opened so that the pressure inside drops to one atm. What will be the fraction of the total number of moles, inside escaped on opening the value? (Assume no change in the volume of the container).

Options:
A 7.7
B 9.9
C 8.9
D 0.77

The correct answer is:
0.77

Solution:
Given that, V=2L,T1=50+273=323K,P1=3atm
On heating V=2L,T2=200+273=473,P2=?
Using P-T law, P1T1=P2T2
P2=3×473323=4.39 atm
and n=PVRT=3×20.0821×323=0.226
Now value is opened till the pressure is maintained at 1 atm.
Thus, at constant V and T, P∝n
So 4.39∝0.226
So 1∝nleft
So n=0.2264.39=0.052
So moles escaped out =0.226-0.052=0.174
So fraction of moles escaped out =0.1740.226=0.77

Topic: Chemistry, States of Matter, JEE Main
A 2m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate 25cm/sec , then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the horizontal ground when the top of the ladder is 1 m above the ground is:

Options:
A 25
B 253
C 253
D 253

The correct answer is:
253

Solution:
dydt=-25 cm/sec , dxdt=?
Now, x2+y2=22=4
differentiating w.r.t. t both side
2xdxdt+2ydydt=0 x2+y2=4 when y=1 ⇒x2=3⇒x=3
⇒dxdt=-yxdydt
⇒dxdt 3,1=-13×-25=253 cm/sec

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (12 Apr Shift 1)
A 2 m long wire of resistance 4 ohm and diameter 0.64 mm is coated with plastic insulation of thickness 0.06 mm. When a current of 5 ampere flows through the wire, find the temperature difference across the insulation in steady-state if K = 0.16 Ã 10-2 cal cm-1 Â°C-1 s-1.

Options:
A 1 Â°C

B 2 Â°C

C 3 °C

D 4 °C

The correct answer is:
2 Â°C

Solution:
Consider a coaxial cylindrical shell of radius r and thickness dr as shown in diagram. The radial rate of flow of heat through this shell in steady state will be

H = d Q dt = - K A d Î¸ dr
[Negative sign is used as with increase in r, Î¸ decreases.] Now as for cylindrical shell A = 2ÏrL,
H = - 2 Ï r L K d Î¸ dr
or â« a b dr r = - 2 Ï LK H â« Î¸ 1 Î¸ 2 d Î¸
which on integration and simplification gives:
H = d Q d t = 2 Ï KL Î¸ 1 - Î¸ 2 log e b / a ....(i)
Here H = I 2 R 4.2 = 5 2 Ã 4 4.2 = 24 cal/s, L = 2 m = 200 cm
r1 = (0.64/2) mm = 0.032 cm
and r2 = r1 + d = 0.032 + 0.006 = 0.038 cm
So Î¸ 1 - Î¸ 2 = 24 Ã log e 38 / 32 2 Ã 3.14 Ã 200 Ã 0.16 Ã 10 - 2
= 24 Ã 2.3026 [ log 10 38 - log 10 32 ] 3.14 Ã 0.64
orÎ¸1-Î¸2=55Ã[1.57-1.50]2â2 Â°C

Topic: Physics, Thermal Properties of Matter, JEE Main
A 2 mW laser operates at a wavelength of 500 nm . The number of photons that will be emitted per second is:
[Given Planck's constant h=6.6×10-34 J s, speed of light c=3.0×108 m/s ]

Options:
A 1.5×1016
B 5×1015
C 2×1016
D 1×1016

The correct answer is:
5×1015

Solution:
P=2 mW
λ=500 nm
Number of photons will be emitted per second is,
n=Phcλ=2×10-3×500×10-96.6×10-34×3×108
=5×1015

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (10 Apr Shift 2)
A 2 m wide truck is moving with a uniform speed v0 = 8 m s-1 along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed v moving in a straight line, when the truck is 4 m away from him. The minimum value of v so that he can cross the road safely is

Options:
A 65 m s-1
B 45 m s-1
C 85 m s-1
D 25 m s-1

The correct answer is:
85 m s-1

Solution:
Let the man starts crossing the road at an angle Î¸ shown in figure. For safe crossing the condition is that the man must cross the road by the time the truck describes the distance 4 + AC or 4 + 2 cot Î¸.
â â´ 4 + 2 cot Î¸ 8 = 2 / sin Î¸ v
or v = 8 2 sin Î¸ + cos Î¸ ...(1)
For minimum v, dv d Î¸ = 0
or - 8 2 cos Î¸ - sin Î¸ ( 2 sin Î¸ + cos Î¸ ) 2 = 0
or 2 cos Î¸ - sin Î¸ = 0
or tan Î¸ = 2
from Equation (1)
vmin=82(25)+15=85 m s-1

Topic: Physics, Motion In One Dimension, JEE Main
A $2 \mathrm{MeV}$ proton is moving perpendicular to a uniform magnetic field of 2.5 tes/a. The force on the proton is

Options:
A $2.5 \times 10^{-10} \mathrm{~N}$
B $7.6 \times 10^{-11} \mathrm{~N}$
C $2.5 \times 10^{-11} \mathrm{~N}$
D $7.6 \times 10^{-12} \mathrm{~N}$

The correct answer is:
$7.6 \times 10^{-12} \mathrm{~N}$

Solution:
\(\begin{aligned} & E_k=\frac{1}{2} m v^2\left\{v=2 \times 10^7 \mathrm{~m} / \mathrm{s}\right) \\ & F=q v B \sin 90=8 \times 10^{-12} N\end{aligned}\)

Topic: Physics, Magnetic Effects of Current, JEE Main
A 2 V battery is connected across AB as shown in the figure. The value of the current supplied by the battery when in first case battery's positive terminal is connected to A and in second case when positive terminal of battery is connected to B will respectively be:

Options:
A 0.2 A and 0.1 A
B 0.1 A and 0.2 A
C 0.2 A and 0.4 A
D 0.4 A and 0.2 A

The correct answer is:
0.4 A and 0.2 A

Solution:
When +ve polarity is connected to A
Then D1 is forward biased and D2 is reversed biased i.e.
∴ I1 current through D1=25=0.4 A
When +ve polarity is connected to B then D2 is forward biased and D1 is reversed biased then
⇒I2=210=0.2 A

Topic: Physics, Semiconductors, JEE Main, JEE Main 2015 (11 Apr Online)
A 2 W carbon resistor is color coded with green, black, red and silver respectively. The maximum current which can be passed through this resistor is:

Options:
A 100Â mA
B 0.4Â mA
C 20 mA
D 63Â mA

The correct answer is:
20 mA

Solution:
Colour band A and B indicate the first two significant figures of resistance in "ohm" , while the C band gives the decimal multiplier i.e., the number of zeros that follows the two significant figures A and B.

The last band D indicates the tolerance in percentage about the indicate value or in another ward it represents the percentage accuracy of the indicated value.

According to given color-coding, the values of colors are

Green →5, Black →0 Red →102 Silver →10% tolerance

so, resistance is R =50×102±10%

The maximum current is
I2R=P
I2=250×102=4×10-4 A
I=2×10-2A=20 mA

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (10 Jan Shift 1)
$\vec{a}=3 \hat{i}-5 \hat{j}$ and $\vec{b}=6 \hat{i}+3 \hat{j}$ are two vectors and $\vec{c}$ is a vector such that $\vec{c}=\vec{a} \times \vec{b}$ then
$$
|\vec{a}|:|\vec{b}|:|\vec{c}|
$$

Options:
A
$\sqrt{34}: \sqrt{45}: \sqrt{39}$

B
$\sqrt{34}: \sqrt{45}: 39$

C
$34: 39: 45$

D
$39: 35: 34$

The correct answer is:
$\sqrt{34}: \sqrt{45}: 39$

Solution:
We have $\vec{a} \times \vec{b}=39 \vec{k}=\vec{c}$
Also $|\vec{a}|=\sqrt{34},|\vec{b}|=\sqrt{45},|\vec{c}|=39 \therefore \quad|\vec{a}|:|\vec{b}|:|\vec{c}|=\sqrt{34}: \sqrt{45}: 39$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2002
A 3 kg object has initial velocity 6i^-2j^ m s-1. What will be the total work done (in joule) on the object if its velocity changes to (8i^+4j^) m s-1?

Options:

The correct answer is:
60

Solution:
The net work done on the object is equal to the change in the kinetic energy of the object
Wnet= Kf-Ki=ÎK
Kinetic energy K=12mv2
Velocity v2=vx2+vy2
=( 36+4 )ââ m 2 /s 2
v 2 =40ââ m 2 /s 2
Kf=12Ã3Ã40=60 J
Again, velocity v=82+42
v=64+16=80 m/s
Kinetic energy Kf=123Ã80 m 2 /s 2 =120 J
From work energy theorem
Wnet= Kf-Ki
=120 J-60 J=60 J

Topic: Physics, Work Power Energy, JEE Main
A $0.3 \mathrm{M~} \mathrm{HCl}$ solution contains the following ions $\mathrm{Hg}^{2+}, \mathrm{Cd}^{2+}, \mathrm{Sr}^{2+}, \mathrm{Fe}^{2+}, \mathrm{Cu}^{2+}$. The addition of $\mathrm{H}_2 \mathrm{~S}$ to above solution will precipitate

Options:
A $\mathrm{Cd}, \mathrm{Cu}$ and $\mathrm{Hg}$
B $\mathrm{Cd}, \mathrm{Fe}$ and $\mathrm{Sr}$
C $\mathrm{Hg}, \mathrm{Cu}$ and $\mathrm{Fe}$
D $\mathrm{Cu}, \mathrm{Sr}$ and $\mathrm{Fe}$

The correct answer is:
$\mathrm{Cd}, \mathrm{Cu}$ and $\mathrm{Hg}$

Solution:
The second group radicals will precipitate because their solubility product is very low so sulphates will be precipitated.

Topic: Chemistry, Practical Chemistry, JEE Main
A 3 volt battery with negligible internal resistance is connected in a circuit as shown in the figure. The current $\mathrm{I}$, in the circuit will be

Options:
A
$1 \mathrm{~A}$

B
$1.5 \mathrm{~A}$

C
$2 \mathrm{~A}$

D
$1 / 3 \mathrm{~A}$

The correct answer is:
$1.5 \mathrm{~A}$

Solution:
$I=\frac{3}{2}=1.5 \mathrm{~A}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2003
$A=\left[\begin{array}{ll}4 & 2 \\ 3 & 4\end{array}\right]$ then $|\operatorname{adj} A|$ is equal to

Options:
A $16$
B $10$
C $6$
D None of these

The correct answer is:
$10$

Solution:
$A=\left[\begin{array}{ll}4 & 2 \\ 3 & 4\end{array}\right]$
$\operatorname{adj} A=\left[\begin{array}{cc}4 & -2 \\ -3 & 4\end{array}\right]$
$|\operatorname{adj} A|=(4 \times 4)-(-3 \times-2)=16-6$
$|\operatorname{adj} A|=10$.

Topic: Mathematics, Matrices, JEE Main
A $4 \mu \mathrm{F}$ capacitor is charged by a $200 \mathrm{~V}$ battery. It is then disconnected from the supply and is connected to another uncharged $2 \mu \mathrm{F}$ capacitor. During the process, loss of energy (in $\mathrm{J}$ ) is

Options:
A $3.43 \times 10^{-2}$
B $2.67 \times 10^{-2}$
C $2.67 \times 10^{-4}$
D $3.43 \times 10^{-4}$

The correct answer is:
$2.67 \times 10^{-2}$

Solution:
Charge stored at the capacitor
$q=C_1 V_1=4 \times 200=800 \mu \mathrm{C}$
When this capacitor is connected with a uncharged capacitor, then common potential on both capacitors
$V=\frac{C_1 V_1+C_2 V_2}{C_1+C_2}=\frac{800+0}{4+2}=\frac{800}{6} \mathrm{~V}$
Loss in energy = Initial energy - Final energy
$\begin{aligned}
= & \frac{1}{2} C_1 V_1^2-\frac{1}{2}\left(C_1+C_2\right) V^2 \\
= & \frac{1}{2} \times 4 \times 10^{-6} \times(200)^2 \\
& \quad-\frac{1}{2}(4+2) \times 10^{-6} \times\left(\frac{800}{6}\right)^2 \\
= & 2 \times 10^{-6} \times 4 \times 10^4-\frac{3 \times 10^{-6} \times 64 \times 10^4}{36} \\
= & 8 \times 10^{-2}-\frac{64}{12} \times 10^{-2} \\
= & 8 \times 10^{-2}-5.33 \times 10^{-2} \\
= & 2.67 \times 10^{-2} \mathrm{~J}
\end{aligned}$

Topic: Physics, Capacitance, JEE Main
$\int_{-a}^{4}\left(x^{8}-x^{4}+x^{2}+1\right) d x=2 \int_{0}^{4}\left(x^{8}-x^{4}+x^{2}+1\right) d x$ then $a=$

Options:
A 3
B 4
C 6
D 8

The correct answer is:
4

Solution:
If $\mathrm{f}(x)$ is an even function then

$\int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x$

Here $\mathrm{f}(x)=x^{8}-x^{4}+x^{2}+1$ is an even function, therefore $a=4$.

Topic: Mathematics, Definite Integration, JEE Main
A 5.0 m mol dm-3 aqueous solution of KCl has a conductance of 0.55 mS when measured in a cell constant 1.3 cm-1. The molar conductivity of this solution is _______ mSm2 mol-1. (Round off to the Nearest Integer)

Options:

The correct answer is:
14

Solution:
Given concn of KCl=m.molL

: Conductance G=0.55 mS

: Cell constant ℓA=1.3 cm-1

To Calculate : Molar conductivity λm of sol.

→ Since Λm=11000×kM ...(1)

→ Molarity =5×10-3molL

→ Conductivity =G×ℓA=0.55 mS×1.31100 m-1

=55×1.3 mSm-1

eqn 1 Λm=11000×55×1.351000mSm2 mol

⇒Λm=14.3mSm2 mol

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2021 (16 Mar Shift 2)
A 5 digit number is selected at random. The probability that the chosen number has distinct digits and digits in the odd places are odd and digits in the even places are even is

Options:
A 5P2×5P39×104
B 5P2×5P3105
C 5C2×5C3×2104×9
D 5C2×5C39×104

The correct answer is:
5P2×5P39×104

Solution:
Total cases =9Ã104
Favourable cases =50=5P3Ã5P2
Hence, required probability =5P3Ã5P29Ã104

Topic: Mathematics, Probability, JEE Main
A $5 \mu \mathrm{F}$ capacitor is charged by being connected to a $3 \mathrm{~V}$ battery. The battery is then disconnected. If the resistance of the dielectric material between the capacitor plates is $10^9 \Omega$, what is the charge remaining on the capacitor $1 \mathrm{~h}$ after it has been disconnected ?

Options:
A $1.5 \times 10^{-5} \mathrm{C}$
B $7.4 \times 10^{-6} \mathrm{C}$
C $5.4 \times 10^{-6} \mathrm{C}$
D $8.4 \times 10^{-5} \mathrm{C}$

The correct answer is:
$7.4 \times 10^{-6} \mathrm{C}$

Solution:
$$
\begin{aligned}
& \text { } \begin{aligned}
T=R C & =\left(10^9 \Omega\right)\left(5 \times 10^{-6} \mathrm{~F}\right) \\
& =5 \times 10^3 \mathrm{~s}
\end{aligned} \\
& =\frac{5 \times 10^3 \mathrm{~s}}{(60 \mathrm{~s} / \mathrm{min})(60 \mathrm{~min} / \mathrm{h})}=\frac{5000 \mathrm{~s}}{3600 \mathrm{~s} / \mathrm{h}}=1.4 \mathrm{~h}
\end{aligned}
$$
The initial charge on the capacitor is
$$
\begin{aligned}
& Q_0=C V=\left(5 \times 10^{-6} \mathrm{~F}\right)(3 \mathrm{~V}) \\
&= 1.5 \times 10^{-5} \mathrm{C} \\
& \text { After } t=1 \mathrm{~h}, t / T=(1 \mathrm{~h}) /(1.4 \mathrm{~h}) \\
&=0.71 \\
& \text { and } \begin{aligned}
Q & =Q_0 e^{-t / T} \\
& =\left(1.5 \times 10^{-5} \mathrm{C}\right)\left(e^{-0.71}\right) \\
& =\left(1.5 \times 10^{-5} \mathrm{C}\right)(0.49) \\
& =7.4 \times 10^{-6} \mathrm{C}
\end{aligned}
\end{aligned}
$$

Topic: Physics, Capacitance, AIIMS, AIIMS 2015
A 5 μF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 μF capacitor. If the energy change during the charge redistribution is X100 J then value of X to the nearest integer is :

Options:

The correct answer is:
4

Solution:
C1=5μF V1=220Volt

C2=2.5μF V2=0

Heat loss;

ΔH=U1-Ut=12C1C2C1+C2v1-v22

=12×5×2.5(5+2.5)(220-0)2μJ

=52×3×22×22×100×10-6J

=5×11×223×10-4J=55×223×10-4J

=12103×10-4J=12103×10-3J=4×10-2

According to questions

x100=4×10-2

so, x=4

Topic: Physics, Capacitance, JEE Main
A 5 μF capacitor is charged fully by a 220 V supply. It is then disconnected from the supply and is connected in series to another uncharged 2.5 μF capacitor. If the energy change during the charge redistribution is X100 J then value of X to the nearest integer is :

Options:

The correct answer is:
4

Solution:
C1=5μF V1=220Volt

C2=2.5μF V2=0

Heat loss;

ΔH=U1-Ut=12C1C2C1+C2v1-v22

=12×5×2.5(5+2.5)(220-0)2μJ

=52×3×22×22×100×10-6J

=5×11×223×10-4J=55×223×10-4J

=12103×10-4J=12103×10-3J=4×10-2

According to questions

x100=4×10-2

so, x=4

Topic: Physics, Capacitance, JEE Main, JEE Main 2020 (02 Sep Shift 1)
A 0.5 kg block slides from A on horizontal track with an initial speed of 3 m sâ1 towards a weightless horizontal spring of length 1 m and force constant 2 N mâ1. The part AB of the track is frictionless and the part BC has the coefficient of static and kinetic friction as 0.22 and 0.20 respectively. If the distance AB and BD are 2 m and 2.14 m respectively, the total distance through which the block moves before it comes to rest completely is -

Options:
A 2.5Â m
B 4.42Â m
C 4.24 m
D 2.44Â m

The correct answer is:
4.24 m

Solution:
At D, KE=12 mv2 – μkmg BD
= 12 × 0.5 × 32 – 0.2 × 0.5 × 10 × 2.14
= 2.25 – 2.14=0.11 J
0.11=12 kx2 + μRmgx⇒x2+x–0.11=0
⇒x=0.1 m
Total distance covered=2 + 2.14 + 0.1 =4.24 m

Topic: Physics, Work Power Energy, JEE Main
A 0.5ââkg ball moving with a speed of 12 m s-1 strikes a hard wall at an angle of 30Â° with the wall. It is reflected with the same speed at the same angle. If the ball is in contact with the wall for 0.25 s, the average force acting on the wall is

Options:
A 96 N
B 48 N
C 24 N
D 12 N

The correct answer is:
24 N

Solution:
Here, FÃt= change in momentum
=2âmvâsinÎ¸

â´ F = 2 mv sin Î¸ t
= 2 Ã 0.5 Ã 12 Ã sin 30 o 0.25 = 48 Ã 1 2 = 24 N

Topic: Physics, Laws of Motion, JEE Main
A $5 \mathrm{~kg}$ rifle fires a $6 \mathrm{~g}$ bullet with $S$ peed of $500 \mathrm{~m} / \mathrm{s}$. Find the ratio of the distance the rifle move backward while the bullet is in the barrel to the distance the bullet moves forward?

Options:
A $1 / 600$
B $1 / 1000$
C $1 / 667$
D $3 / 2500$

The correct answer is:
$1 / 1000$

Solution:
By conservation of momentum, since no external impulses force acts on the system
$\begin{aligned} & \mathrm{p}_{\mathrm{i}}=\mathrm{p}_{\mathrm{f}} \\ & \text { Or } \\ & 0=\mathrm{m}_{\mathrm{r}} \mathrm{v}_{\mathrm{r}}+\mathrm{m}_{\mathrm{b}} \mathrm{v}_{\mathrm{b}} \cdots(1) \\ & \mathrm{v}_{\mathrm{r}}=\frac{-\mathrm{m}_{\mathrm{b}}}{\mathrm{m}_{\mathrm{f}}} \mathrm{v}_{\mathrm{b}}\end{aligned}$
From equation (1)
$\begin{aligned} & 0=m_r \frac{\Delta x_r}{\Delta t}+m_b \frac{\Delta x_b}{\Delta t} \\ & 0=m_r \Delta x_r+m_b \Delta x_b\end{aligned}$
Which shows
$\left|\frac{\Delta x_{\mathrm{f}}}{\Delta x_b}\right|=\frac{m_b}{m_t}=\frac{6}{6000}=\frac{1}{1000}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A $5 \mathrm{~kg}$ rock flying through the air is traveling at a velocity of $12 \frac{\mathrm{m}}{\mathrm{s}}$ when it collides into and sticks to a stationary bean bag, weighing $20 \mathrm{~kg}$. What is the velocity of the two objects?

Options:
A $3 \frac{m}{s}$
B $6 \frac{m}{s}$
C $3.6 \frac{\mathrm{m}}{\mathrm{s}}$
D $2.4 \frac{m}{s}$

The correct answer is:
$2.4 \frac{m}{s}$

Solution:
The equation for momentum is:
$p=m v$
To maintain conservation of momentum, a new state must have the same momentum as a previous state:
$m_{1 i} v_{1 i}+m_{2 i} v_{2 i}=m_{1 f} v_{1 f}+m_{2 f} v_{2 f}$
Since the rock and the bean bag move together after the collision, $v_{1 f}=v_{2 f}$. And since the bean bag is initially stationary, $v_{2 i}=0$ Plug in known values and solve.
$\begin{aligned} & 5 \mathrm{~kg}\left(12 \frac{\mathrm{m}}{\mathrm{s}}\right)+20 \mathrm{~kg}(0)=(5 \mathrm{~kg}+20 \mathrm{~kg})\left(v_f\right) \\ & 60 \frac{\mathrm{kgm}}{\mathrm{s}}=25 \mathrm{~kg}\left(v_f\right) \\ & v_f=2.4 \frac{\mathrm{m}}{\mathrm{s}}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main