datavorous/entrance-exam-dataset huggingface dataset:
There are two radioactive substances A and B. Decay constant of B is two times that of A. Initially, both have an equal number of nuclei. After n half lives of A, rate of disintegration of both are equal. The value of n is

Options:
A 4
B 2
C 1
D 5

The correct answer is:
1

Solution:
Let λA=λ∴λB=2λ
If N0 is total number of atoms in A and B at t=0, then initial rate of disintegration of A=λN0, and initial rate of disintegration of B=2λ N0
As λB=2λA
∴TB=12TA
ie, half-life of B is half the half-life of A.
After one half-life of A
-dNdtA=λN02
Equivalently, after two half lives of B
-dNdtB=2λN04=λN02
Clearly, -dNdtA=-dNdtB
after n=1 ie, one half-life of A

Topic: Physics, Nuclear Physics, JEE Main
There are two red, three blue, four white and K green socks in a drawer. If 2 socks are taken at random from the drawer without replacement, the probability that they are of the same colour is 15, then K is equal to

Options:

The correct answer is:
2

Solution:
Total cases =k+9C2
Favourable cases = both are of same colour
= C 2 2 + C 3 2 + C 4 2 + C k 2
So, 2C2+3C2+4C2+kC2k+9C2=15
â10+kk-12k+9k+82=15
â20+k2-kk2+17k+72=15
â4k2-22k+28=0
â2k2-11k+14=0
âk=2,72 (rejected)

Topic: Mathematics, Probability, JEE Main
There are two solenoids of same length and inductance L but their diameters differ to the extent that one can just fit into the other. They are connected in three different ways in series. (1) They are connected in series but separated by large distance, (2) they are connected in series with one inside the other and senses of the turns coinciding, (3) both are connected in series with one inside the other with senses of the turns opposite as depicted in figures 1, 2 and 3 respectively. The total inductance of the solenoids in each of the case 1, 2 and 3 are respectively

Options:
A 0, 4L0, 2L0
B 4L0, 2L0, 0
C 2L0, 0, 4L0
D 2L0, 4L0, 0

The correct answer is:
2L0, 4L0, 0

Solution:
When two solenoids of inductance L0 are connected in series at large distance and current i is passed through them, the total flux linkage ϕtotalis the sum of the flux linkages L0i and L0i, ie,
ϕtotal=L0i+L0i
If L be the equivalent inductance of the system, then
ϕtotal=Li
∴ Li=L0i+L0i
or L=2L0
When solenoids are connected in series with one inside the other and senses of the turns coinciding, then there will be mutual inductance L between them. In this case the resultant induced emf in the coils is the sum of the emf s e1 and e2 in the respective coils, ie,
e=e1+e2
=-L0didt±L0didt+-L0didt±L0didt
Where (+) sign is for positive coupling and (-) sign for negative coupling.
But, e=-L.didt
∴ -Ldidt=-L0didt-L0didt±2L0didt
ie, L=L0+L0+2L0
=4L0 (for positive coupling)
When solenoids are connected in series with one inside the other with senses of the turns opposite, then their is negative coupling.
So, L=L0+L2-2L0=0

Topic: Physics, Electromagnetic Induction, JEE Main
There are two stations \( A, B \) due north, due south of a tower of height \( 15 \mathrm{~m} \). The angle of depression of \( A \) and \( B \) as seen from top of the tower are \( \cot ^{-1} \frac{12}{5}, \sin ^{-1} \frac{3}{5} \), then the distance between \( A \) and \( B \) is

Options:
A \( 48 \)
B \( 56 \)
C \( 25 \)
D None of these

The correct answer is:
\( 56 \)

Solution:
Given cotα=125
and sinβ=35
Now, d=15cotα+cotβ
=15125+45
=56

Topic: Mathematics, Heights and Distances, JEE Main
There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urn two balls are taken out at random and then transferred to the other. The number of ways in which this can be done is

Options:
A
36

B
66

C
108

D
3

The correct answer is:
108

Solution:
Total number of ways $={ }^3 \mathrm{C}_2 \times{ }^9 \mathrm{C}_2$
$$
=3 \times \frac{9 \times 8}{2}=3 \times 36=108
$$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2010
There are two waves having wavelength 100ââcm and 101ââcm and same velocity 303ââm s-1. The beat frequency is

Options:
A 4  Hz
B 1  Hz
C 3ââHz
D 2  Hz

The correct answer is:
3  Hz

Solution:
If f1 and f2 are the frequencies (f1>f2) of the two waves, then the beat frequency fb=f1-f2
Frequency of 1st wave,
f1=V1Î»1=303100Ã102=3.03Ã102 Hz
For 2nd wave, f2=V1Î»2=303101Ã102=3Ã102 Hz
So, beat frequency =f1-f2
=3.03Ã102-3Ã102=.03Ã102=3 Hz

Topic: Physics, Waves and Sound, JEE Main
There are two wire of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be

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

The correct answer is:
4: 1

Solution:
$\because$ Both wires are same materials so both will have same Young's modulus, and let it be Y. $Y=\frac{\text { stress }}{\text { strain }}=\frac{F}{\text { A. }(\Delta L / L)}, F=$ applied force

$\mathrm{A}=$ area of cross-section of wire

Now,

$\mathrm{Y}_{1}=\mathrm{Y}_{2} \Rightarrow \frac{\mathrm{FL}}{\left(\mathrm{A}_{1}\right)\left(\Delta \mathrm{L}_{1}\right)}=\frac{\mathrm{FL}}{\left(\mathrm{A}_{2}\right)\left(\Delta \mathrm{L}_{2}\right)}$

Since load and length are same for both

$\begin{array}{l}

\Rightarrow \mathrm{r}_{1}^{2} \Delta \mathrm{L}_{1}=\mathrm{r}_{2}^{2} \Delta \mathrm{L}_{2},\left(\frac{\Delta \mathrm{L}_{1}}{\Delta \mathrm{L}_{2}}\right)=\left(\frac{\mathrm{r}_{2}}{\mathrm{r}_{1}}\right)^{2}=4 \\

\Delta \mathrm{L}_{1}: \Delta \mathrm{L}_{2}=4: 1

\end{array}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
There is a circular tube in a vertical plane. Two liquids which do not mix and of densities d1 and d2 are filled in the tube. Each liquid subtends 90o angle at centre. Radius joining their interface makes an angle α with vertical. Ratio d 1 d 2 is :

Options:
A 1 + sin α 1 - sin α
B 1 + cos α 1 - cos α
C 1 + tan Î± 1 - tan Î±
D 1 + sin α 1 - cos α

The correct answer is:
1 + tan α 1 - tan α

Solution:
Pressure at common interface of two liquid columns is same.
d 1 gR 1 - sin α = d 1 gR 1 - cos α + d 2 gR sin α + cos α
d 1 cos α - sin α = d 2 sin α + cos α
∴ d 1 d 2 = sin α + cos α cos α - sin α = tan α + 1 1 - tan α

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2014 (06 Apr)
There is a crater of depth R 100 on the surface of the moon, where R is the radius of the moon. A ball is projected vertically upwards from the crater with a velocity which is equal to the escape velocity from the surface of the moon. If the maximum height attained by the particle is close to nR, where n is an integer, then find the value of n?

Options:

The correct answer is:
99

Solution:
Speed of particle at A,

vA=v=2GMR

At point B,vB=0

By conservation of energy,

decrease in KE= increase in GPE

â12MvÎ2=UB-UA

â´ 12MÂ·2GMR=MVB-VA

â´GMR=-GMR+h--GM2R3-R-R1002R2

â´ 1R=-1R+h+32R-12991002Â·1R

â´ 1R=-1R+h+32R-120.98Râ´ 1R+h=12R[1-0.98]

â´ 2R=0.02(R+h)

â´ h=1.980.02

â´ h=99R

Topic: Physics, Gravitation, JEE Main
There is a current of 20 amperes in a copper wire of $10^{-6}$ square metre area of cross-section. If the number of free electrons per cubic metre is $10^{29}$, then the drift velocity is

Options:
A $125 \times 10^{-3} \mathrm{~m} / \mathrm{sec}$
B $12.5 \times 10^{-3} \mathrm{~m} / \mathrm{sec}$
C $1.25 \times 10^{-3} \mathrm{~m} / \mathrm{sec}$
D $1.25 \times 10^{-4} \mathrm{~m} / \mathrm{sec}$

The correct answer is:
$1.25 \times 10^{-3} \mathrm{~m} / \mathrm{sec}$

Solution:
$v_d=\frac{i}{n A e}=\frac{20}{10^{29} \times 10^{-6} \times 1.6 \times 10^{-19}}=1.25 \times 10^{-3} \mathrm{~m} / \mathrm{s}$

Topic: Physics, Current Electricity, JEE Main
There is a cylindrical bottle of negligible mass and radius 2.5 cm floating in the water of density 103 kg m-3 filled with 310 ml of water inside it. Now the bottle is slightly dipped into water and released. The frequency of oscillation is

Options:
A 3.75 s-1â¡
B 1.25 s-1â¡
C 2.75 s-1â¡
D 3.00 s-1â¡

The correct answer is:
1.25 s-1â¡

Solution:
Let at equilibrium, y length of the bottle is immersed in water:-
So, mg=fbuo
mg=ρAyg ..... (1) [Equilibrium Condition]
Now it is further pushed by distance ′x′
Frestoring=ρ[Ax+y)g-mg
mω2x=ρAxg+ρAyg-mg
ω=ρAgm=ρAgρV=πr2gV=2πf
∴f=12ππr2gV=1.25 s-1

Topic: Physics, Oscillations, JEE Main
There is a hole in the bottom of tank having water. If total pressure at bottom is $3 \mathrm{atm}\left(1 \mathrm{atm}=10^5 \mathrm{~N} / \mathrm{m}^2\right)$ then the velocity of water flowing from hole is

Options:
A $\sqrt{400} \mathrm{~m} / \mathrm{s}$
B $\sqrt{600} \mathrm{~m} / \mathrm{s}$
C $\sqrt{60} \mathrm{~m} / \mathrm{s}$
D None of these

The correct answer is:
$\sqrt{400} \mathrm{~m} / \mathrm{s}$

Solution:
Pressure at the bottom of tank
$P=h \rho g=3 \times 10^5 \frac{\mathrm{N}}{\mathrm{m}^2}$ Pressure due to liquid column
$P_l=3 \times 10^5-1 \times 10^5=2 \times 10^5$
and velocity of water $v=\sqrt{2 g h}$
$\therefore v=\sqrt{\frac{2 P_l}{\rho}}=\sqrt{\frac{2 \times 2 \times 10^5}{10^3}}=\sqrt{400} \mathrm{~m} / \mathrm{s}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
There is a pair of fixed, parallel rails of negligible resistance in a horizontal plane, at a distance of L=10 cm. The rails are connected, at one of their ends to an ideal battery of EMF 0.5 V, as shown in the figure. The system is placed in a vertically downward and perpendicular magnetic field of magnetic induction B=1 T. A metal rod of resistance R=10 Î© is placed perpendicularly on the rails. The rod and rails are frictionless. The magnitude of force, which is to be exerted on the rod in order to move it with a constant speed of 1 m sâ1 in the direction shown, is

Options:
A 2Ã10â3Â N
B 4Ã10â3 N
C 8Ã10â3Â N
D 5Ã10â3Â N

The correct answer is:
4Ã10â3 N

Solution:
Îµin=BâLâV=1Ã10Ã10â2Ã1
Îµin=0.1 V
I=0.5-0.110=0.04
Fâ=1Ã0.04Ã10â2Ã10
=0.4Ã10â2
|Fâ| =4Ã10â3 N

Topic: Physics, Electromagnetic Induction, JEE Main
There is a point inside a circle. What is the probability that this point is close to the circumference than to the centre?

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

The correct answer is:
$3 / 4$

Solution:
Let radius of circle be ' $r$ '. Total possible outcomes $=$ Area of circle $=\pi r^{2}$ Observe the figure, we have to find the probability of point, $\mathrm{P}$ in the ring which will be closer to circumference.
Area of ring $=$ Area of outer circle $-$ Area of inner circle
$=\pi r^{2}-\pi\left(\frac{r}{2}\right)^{2}=\pi r^{2}-\frac{\pi r^{2}}{4}=\frac{3 \pi r^{2}}{4}$
So, favourable outcome $=\frac{3 \pi r^{2}}{4}$.
Required probability $=\frac{\frac{3}{4} \pi r^{2}}{\pi r^{2}}=\frac{3}{4}$

Topic: Mathematics, Probability, JEE Main
There is a sample of diatomic gas in a container (whose volume is constant). The temperature of this gas was increased greatly so some molecules fell into atoms (dissociated). The pressure of the gas increased by a factor of 6 and the internal energy of the gas increased to a value of 4.4 times the original internal energy. By what factor did the temperature of the gas (measured in Kelvin) increase?

Options:

The correct answer is:
4

Solution:
PV=nRT ; 6PV=(n2+2n2)RT′⇒6=n2+2n2nT′T ... (i)

U=n(52RT) ; 4.4U=n2(52RT′)+2n2(32RT′)

4.4=n(52RT′)+n23RT′n(52RT) 4.4=n1+65n2nT′T ... (ii)

64.4=n1+2n2n1+65n2=5n1+10n25n1+6n2 ; 30n1+36n2=22n1+44n2 ; 8n1=8n2⇒  n1=n2

Also, n1+n2=n⇒n1=n2=n2 ; 6=(12+1)T′T⇒T′=4T

Topic: Physics, Thermodynamics, JEE Main
There is a small source of light at some depth below the surface of water (refractive index =43 ) in a tank of large cross sectional surface area. Neglecting any reflection from the bottom and absorption by water, percentage of light that emerges out of surface is (nearly):
[Use the fact that surface area of a spherical cap of height h and radius of curvature r is 2πrh ]

Options:
A 21%
B 34%
C 17%
D 50%

The correct answer is:
17%

Solution:
sinβ=34,cos⁡β=74
Solid angle dΩ=2πR21-cos⁡β
Percentage of light =2πR21-cos⁡β4πR2×100
=1-cosβ2×100=4-78×100≈17%

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (09 Jan Shift 2)
There is a source of sound (a tuning fork) moving towards a reflecting wall will a speed of 30 m s-1. The velocity of the sound in air is 330 m s-1 and the frequency of the tuning fork is 600 Hz. The observer is between the source and the wall and is moving with some velocity V1 towards the wall. The beat frequency heard by the observer is 4 Hz. If the tuning fork is waxed, the frequency beats heard by the observer becomes 3 Hz. If the new frequency of tuning fork is k then k50 is

Options:

The correct answer is:
9

Solution:
Apparent frequency of source =n0c-v1c-v2=n′
Apparent frequency of reflected sound =n0c+v1C-V=n′′
Beat frequency =n′′-n′=2n0V1C-V2=2×600×V1330-30=4 Hz
∴ V1= 1 m s-1

When the tuning fork is waxed let its frequency become k
Beat frequency =2kv1C-V2=2×k×1330×30 =3 Hz
∴ k=450 Hz= new frequency of tuning fork.

k50=9

Topic: Physics, Waves and Sound, JEE Main
There is a stream of neutrons with a kinetic energy of 0.0327 eV. If the half-life of the neutron is 700 second, what fraction of neutrons will decay before they travel a distance of 10 m? Given the mass of neutron = 1.675 Ã 10-27 kg.

Options:
A 3.96×10-6

B 3.90×10-6

C 3.85×10-6

D 4.86×10-6

The correct answer is:
3.96×10-6

Solution:
From the given kinetic energy of the neutrons we first calculate their velocity. Thus
12mv2=0.0327×1.6×10-19
∴ v⁡ 2 = 2 × 0 . 0 3 2 7 × 1 . 6 × 1 0 - 1 9 1 . 6 7 5 × 1 0 - 2 7 = 6 2 5 × 1 0 4
or v = 2500 m s-1
With this speed, the time taken by the neutrons to travel a distance of 10 m is,
Δt=102500=4×10-3 s⁡
The fraction of neutrons decayed in time Δt second is,
ΔN⁡N⁡=λΔt
Also, λ = 0 . 6 9 3 T⁡ 1 / 2
∴ Δ N⁡ N⁡ = 0 . 6 9 3 T⁡ 1 / 2 Δ t = 0 . 6 9 3 7 0 0 × 4 × 1 0 - 3 = 3 . 9 6 × 1 0 - 6

Topic: Physics, Nuclear Physics, JEE Main
There is a thin uniform disc of radius R and mass per unit area Ï, in which a hole of radius R/2 has been cut out as shown in the figure. Inside the hole, a square plate of same mass per unit area Ï is inserted so that its corners touch the periphery of the hole. The distance of the centre of mass of the system from the origin is

Options:
A R(2-Ï)23Ï+2
B R(1-Ï)22Ï+1
C 2 R π 2 3 π + 2
D 3 R π 2 2 π + 1

The correct answer is:
R(2-π)23π+2

Solution:
Side of square = R cos 4 5 â = R 2
Area of square = R 2 2
XCOM=ÏÃR2ÃÏÃ0+ÏÃR24-ÏÃR2+R22ÃÏÃR2ÏÃR2ÃÏ+ÏÃR24-Ï+R22ÃÏ
=R(2-Ï)2(3Ï+2)
â´ The centre of mass of the system is at a distance of R(2-Ï)2(3Ï+2) from the centre O towards the plate as shown in the figure

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
There is a trolley in which there is a fixed inclined surface on which a smooth block of mass 5 kg is placed. Two horizontal forces of magnitude F1 and F2 are applied on the trolley as shown to keep the trolley at rest. The value of F1 - F2 is : (Assume there is no friction between the trolley and horizontal ground and g = 10 m s-2)

Options:
A 0 N

B 24 N
C -24 N
D -30 N

The correct answer is:
-24 N

Solution:
There is a trolley for which
F1+Nsin37Â°=F2
F1-F2=-Nsin37Â°. . . (1)
N=mgcos37Â°. . . (2)
F1-F2=-mgcos37Â°sin37Â°
=-5Ã10Ã45Ã35=-24 N

Topic: Physics, Laws of Motion, JEE Main
There is a uniform electric field of strength $10^3 \mathrm{Vm}^{-1}$ along $Y$-axis. A body of mass $1 \mathrm{~g}$ and charge $10^{-6} \mathrm{C}$ is projected into the field from origin along the positive $X$-axis with a velocity $10 \mathrm{~ms}^{-1}$. Its speed in $\mathrm{ms}^{-1}$ after $10 \mathrm{~s}$ is (Neglect gravitation)

Options:
A $10$
B $5 \sqrt{2}$
C $10 \sqrt{2}$
D $20$

The correct answer is:
$10$

Solution:
$\mathrm{E}=10^2 \mathrm{~V} / \mathrm{m}$ (aloeg $Y$-axis)
$\mathrm{m}=1 \mathrm{~g}=1 \times 10^{-2} \mathrm{~kg}$
$\mathrm{q}=10^{-6} \mathrm{C}, v=10 \mathrm{~m} / \mathrm{s}$
The charged body is moving perpendicularly to the direction of electric field, so no forte acts on ir, so the velocity of the body remains same.

Topic: Physics, Electrostatics, AP EAMCET, AP EAMCET 2001
There is a uniform electric field of strength $10^3 \mathrm{Vm}^{-1}$ along $Y$-axis. A body of mass $1 \mathrm{~g}$ and charge $10^{-6} \mathrm{C}$ is projected into the field from origin along the positive $X$-axis with a velocity $10 \mathrm{~ms}^{-1}$. Its speed in $\mathrm{ms}^{-1}$ after $10 \mathrm{~s}$ is (Neglect gravitation)

Options:
A $10$
B $5 \sqrt{2}$
C $10 \sqrt{2}$
D $20$

The correct answer is:
$10$

Solution:
$\mathrm{E}=10^2 \mathrm{~V} / \mathrm{m}$ (aloeg $Y$-axis)
$\mathrm{m}=1 \mathrm{~g}=1 \times 10^{-2} \mathrm{~kg}$
$\mathrm{q}=10^{-6} \mathrm{C}, v=10 \mathrm{~m} / \mathrm{s}$
The charged body is moving perpendicularly to the direction of electric field, so no forte acts on ir, so the velocity of the body remains same.

Topic: Physics, Electrostatics, TS EAMCET, TS EAMCET 2001
There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at P, in the region, is found to vary between the limits 589.0 V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of 60° with the direction of the field?

Options:
A 589.4 V
B 589.5 V
C 589.2 V
D 589.6 V

The correct answer is:
589.6 V

Solution:
ΔV=Ed
Along diameter ∆V=589.8-589.0 V
=0.8 V
0.8=Ed
Now
ΔV=E d 2 cosθ=0.4×cos 60 o
=0.2
∴589.8−V=0.2
V=589.6 V

Topic: Physics, Electrostatics, JEE Main, JEE Main 2017 (08 Apr Online)
There is an air bubble of radius 1.0 mm in a liquid of surface tension 0.075 N m–1 and density 1000 kg m–3 at a depth of 10 cm below the free surface. The amount by which the pressure inside the bubble is greater than the atmospheric pressure is _______ Pag=10 m s-2.

Options:

The correct answer is:
1150

Solution:
The pressure inside the bubble is given by

ΔP=P-P0=ρgh+2Tr

The data given is

r=10-3 mT=0.075 N m-1ρ=1000 kg m-3h=10 cm

Therefore, the pressure in the bubble is

ΔP=1000×10×10×10-2+2×0.07510-3=1000+150=1150 Pa

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (15 Apr Shift 1)
There is an air filled $1 p F$ parallel plate capacitor. When the plate separation is doubled and the space is filled with wax, the capacitance increases to $2 p F$. The dielectric constant of wax is

Options:
A 2
B 4
C 6
D 8

The correct answer is:
4

Solution:
$C=\frac{E_0 A}{d}=1 p F \quad$ and $C^{\prime}=\frac{K_{c_0} A}{2 d}=2 p F \quad \therefore K=4$.

Topic: Physics, Capacitance, JEE Main
There is an unlimited number of identical balls of four different colours. How many arrangements of atmost \( 8 \) balls in row can be made by using them?

Options:
A \( 97380 \)
B \( 87380 \)
C \( 87370 \)
D \( 87480 \)

The correct answer is:
\( 87380 \)

Solution:
The number of arrangements of one ball=4, because there are only four different balls.
The number of arrangements of two balls =4×4=42
∴ The required number of arrangements =4+42+43+....+48

=448-14-1.
=4348-1

=43×65535

=87380.

Topic: Mathematics, Permutation Combination, JEE Main
There is greater biodiversity in tropical/subtropical regions than in temperate region. Explain.

Options:

None

Solution:
The tropical/subtropical region remain undisturbed due to less variable climatic conditions and thus, exhibit maximum biological diversity. Tropics had a comparatively longer evolutionary time for diversification of species. The temperate environment is more seasonal, less constant and unpredictable, and hence niche specialisation and lesser species diversity are observed in the temperate regions.

Topic: Biology, Biodiversity and its Conservation
There is hole of area 'A' at the bottom of a cylindrical vessel. Water is filled to a height ' $h$ ' and water flows out in ' $t$ ' second. If water is filled to a height ' $4 \mathrm{~h}$ ', it will flow out in time (in second)

Options:
A $\mathrm{t}$
B $4 \mathrm{t}$
C $2 \mathrm{t}$
D $\frac{\mathrm{t}}{4}$

The correct answer is:
$2 \mathrm{t}$

Solution:
From equation of continuity,
$A v_1=a_2$
$\mathrm{A}\left(-\frac{\mathrm{dh}}{\mathrm{dt}}\right)=\mathrm{a} \sqrt{2 \mathrm{gh}}$
$\int_{\mathrm{h}_1}^{\mathrm{h}_2} \frac{-\mathrm{dh}}{\sqrt{\mathrm{h}}}=\int_0^{\mathrm{t}} \frac{\mathrm{a}}{\mathrm{A}} \sqrt{2 \mathrm{~g}} \mathrm{dt}$
$-[2 \sqrt{\mathrm{h}}]_{\mathrm{h}_1}^{\mathrm{h}_2}=\frac{\mathrm{a}}{\mathrm{A}} \sqrt{2 \mathrm{~g}} \mathrm{t}$
$2\left(\sqrt{\mathrm{h}_2}-\sqrt{\mathrm{h}_1}\right)=\frac{\mathrm{a}}{\mathrm{A}} \sqrt{2 \mathrm{gt}}$
$\therefore \quad t=\frac{2 A}{a} \sqrt{\frac{H}{2 g}}=\frac{A}{a} \frac{\sqrt{2 H}}{\sqrt{g}}$
For height $h$, the time is $t_1=\frac{A}{A_0} \sqrt{\frac{2 h}{g}}$
For height $4 \mathrm{~h}$, the time is $t_2=\frac{A}{A_0} \sqrt{\frac{8 h}{g}}$
$\therefore \quad \frac{t_2}{t_1}=2$
$\therefore \quad \mathrm{t}_2=2 \mathrm{t}_1$
$\therefore \quad \mathrm{t}_2=2 \mathrm{t}$
$\ldots\left(\because \mathrm{t}_1=\mathrm{t}\right)$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
There is no change in the volume of a wire due to change in its length on stretching. The Poisson's ratio of the material of wire is

Options:
A \( +0.50 \)
B \( -0.50 \)
C \( +0.25 \)
D \( -0.25 \)

The correct answer is:
\( +0.50 \)

Solution:
The Poisson's ratio of the material of the wire is
σ=∆DD∆ll
The relation for volume of wire is
V=πr2l But, r=D2
V=πD22 l=πD2l4 …(i)
Differentiating both sides of Eq. (i)
dV=πl4⋅2D⋅dD+πD2×14dl
As volume remains constant hence, we get
0=πl2DdD+πD24dl
or -πl2D dD= πD24dl
or -dDDldl=24=0.5
Therefore, Poisson's ratio, σ=0.5

Topic: Physics, Mechanical Properties of Solids, JEE Main
There is no change in the volume of a wire due to the change in its length on stretching. The Poisson's ratio of the material of the wire is

Options:
A +12
B -12
C +14
D -14

The correct answer is:
-12

Solution:
Volume of cylindrical wire, V=πr2L4, where x is the diameter of wire.
Differentiating both sides,
dVdx=π42xL+x2⋅dLdx
Also, volume remains constant
∴ dVdx=0
∴ 2xL+x2dLdx=0
⟹ 2xL=-x2dLdx
⟹ dxxdLL=-12
Poisson's ratio =-12

Topic: Physics, Mechanical Properties of Solids, JEE Main
There is no DNA in :

Options:
A Hair root
B An enucleated ovum
C Mature RBCs
D A mature spermatozoan

The correct answer is:
Mature RBCs

Solution:
No Solution Available

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2009 (Mains)
There is no \( \mathrm{d}-\mathrm{d} \) transition in \( \mathrm{Cu}^{+} \)but \( \mathrm{Cu}_{2} \mathrm{O} \) is coloured due to

Options:
A The presence of unpaired electron
B The presence of coloured \( \mathrm{O}^{2-} \) ion
C Charge transfer from oxygen to metal
D Charge transfer from metal to oxygen

The correct answer is:
Charge transfer from oxygen to metal

Solution:
There is no d-d transition in Cu+ because d- orbital is completely filled, but Cu2O is coloured due to charge transfer from oxygen to metal which can undergo an electronic transition into an excited electronic state.

Topic: Chemistry, d and f Block Elements, JEE Main
There is S-S bond in

Options:
A S2O42-
B S2O52-
C S2O62-
D All of these

The correct answer is:
All of these

Solution:
Ion
Structure

(A)
S2O42-
O--S||O-S||O-O-

(B)
S2O52-
O--S||O||O-S||O-O-

(C)
S2O62-
O--S||O||O-S||O||O-O-

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
There is some change in length when a 33000 N tensile force is applied on a steel rod of the area of cross-section10-3 m2. The change of temperature required to produce the same elongation, if the steel rod is heated, is (The modulus of elasticity is 3Ã1011N m-2 and the coefficient of linear expansion of steel is 1.1Ã10-5 â-1)

Options:
A 20 °C
B 15 °C
C 10 °C
D 0 °C

The correct answer is:
10 °C

Solution:
Modulus of elasticity = ForceAreaÃlâl
3Ã1011 =3300010-3Ãlâl
âll =3300010-3Ã13Ã1011
=11Ã10-5
Change in length, âll=Î±âT
11Ã10-5 = 1.1Ã10-5ÃâT
â âT=10 K or 10 Â°C

Topic: Physics, Mechanical Properties of Solids, JEE Main