datavorous/entrance-exam-dataset huggingface dataset:
In a simple harmonic motion, let \(\mathrm{f}\) be the acceleration and \(\mathrm{T}\) be the time period. If \(x\) denotes the displacement, then |fT| vs. X graph will look like,

Options:
A

B

C

D

The correct answer is:

Solution:
\(\begin{gathered}\text {Hint: } F \cdot T=\omega^2 x \cdot \frac{2 \pi}{\omega}=2 \pi \omega x \\ \because \text { F.T }=2 \pi \omega x\end{gathered}\)

Topic: Physics, Oscillations, JEE Main
In a simple harmonic oscillation, what fraction of total mechanical energy is in the form of kinetic energy, when the particle is midway between mean and extreme position.

Options:
A 12
B 34
C 13
D 14

The correct answer is:
34

Solution:
K=12 mω2 A2-x2

=12 mω2 A2-A24

=12 mω23 A24

K=3412 mω2 A2

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (25 Jul Shift 2)
In a simple harmonic oscillator, at the mean position

Options:
A
kinetic energy is minimum, potential energy is maximum

B
both kinetic and potential energies are maximum

C
kinetic energy is maximum, potential energy is minimum

D
both kinetic and potential energies are minimum

The correct answer is:
kinetic energy is maximum, potential energy is minimum

Solution:
K.E. is maximum and P.E minimum at mean position

Topic: Physics, Oscillations, JEE Main, JEE Main 2002
In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measures by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of the pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to

Options:
A 0.2%
B 6.8%
C 3.5%
D 0.7%

The correct answer is:
6.8%

Solution:
g=4π2lT2
∆gg=∆ll+2∆TT
=0.155+2×130
=0.06848
Percentage error ≈6.8%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2019 (08 Apr Shift 2)
In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measures by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of the pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to

Options:
A 0.2%
B 6.8%
C 3.5%
D 0.7%

The correct answer is:
6.8%

Solution:
g=4π2lT2
∆gg=∆ll+2∆TT
=0.155+2×130
=0.06848
Percentage error ≈6.8%

Topic: Physics, Units and Dimensions, JEE Main
In a simple slit diffraction pattern intensity and width of fringes are

Options:
A unequal width
B equal width
C equal width and equal intensity
D unequal width and unequal intensity

The correct answer is:
equal width and equal intensity

Solution:
In single slit diffraction pattern intensity and width of fringes are unequal width and unequal intensity.

Topic: Physics, Wave Optics, JEE Main
In a simultaneous throw of three coins, what is the probability of getting at least 2 tails

Options:
A $\frac{1}{8}$
B $\frac{1}{4}$
C $\frac{1}{2}$
D None of these

The correct answer is:
$\frac{1}{2}$

Solution:
Total ways are 8 and fayourable ways are 4
$S=\{H H H, H H T \ldots \ldots . T T T\}$.
Hence probability $=\frac{4}{8}=\frac{1}{2}$.

Topic: Mathematics, Probability, JEE Main
In a single slit diffraction pattern, a light of wavelength 6000 Ao is used. The distance between the first and third minima in the diffraction pattern is found to be 3 mm when the screen is placed 50 cm away from slits. The width of the slit is ____×10-4 m.

Options:

The correct answer is:
2

Solution:
For nth minima:

⇒bsinθ=nλ

(λ is very small, so sinθ is very small, hence sinθ≃tanθ.

⇒btanθ=nλ

⇒byD=nλ

⇒yn=nλDb (Position of nth minima)

B→1st minima and A→3rd :

y3=3λDb,y1=λDb

⇒Δy=y3-y1=2λDb

⇒3×10-3=2×6000×10-10×0.5b

⇒b=2×6000×10-10×0.53×10-3

⇒b=2×10-4 m

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (29 Jan Shift 2)
In a single slit experiment, a parallel beam of green light of wavelength $550 \mathrm{~nm}$ passes through a slit of width $0.20 \mathrm{~mm}$. The transmitted light is collected on a screen $100 \mathrm{~cm}$ away. The distance of first order minima from the central maximum will be $x \times 10^{-5} \mathrm{~m}$. The value of $x$ is :

Options:

The correct answer is:
275

Solution:
$\mathrm{y}=\frac{\lambda \mathrm{D}}{\mathrm{d}}=\frac{550 \times 10^{-9} \times 100 \times 10^{-2}}{0.2 \times 10^{-3}}=275$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (05 Apr Shift 2)
In a single slit experiment, the width of the slit is doubled. Which one of the following statements is correct?

Options:
A The intensity and width of the central maximum are unaffected.
B The intensity remains same and angular width becomes half.
C The intensity and angular width both are doubled.
D The intensity increases by a factor 4 and the angular width decreases by a factor of $\frac{1}{2}$.

The correct answer is:
The intensity increases by a factor 4 and the angular width decreases by a factor of $\frac{1}{2}$.

Solution:
For single slit diffraction, Fringe width $\mathrm{W}=\frac{\lambda \mathrm{D}}{\mathrm{d}}$ Also, $I \propto D^2$

Topic: Physics, Wave Optics, MHT CET, MHT CET 2023 (09 May Shift 1)
In a single slit, Fraunhofer diffraction experiment, with a decrease in the width of the slit, the width of the central maximum

Options:
A Remains the same
B Increase
C Decrease
D Can be any of these depending on the intensity of the source

The correct answer is:
Increase

Solution:
Width of the central maximum is independent of intensity of the source.
Its width is given by β=2λDd, where d = width of slit
i.e., β∝1d ; hence as d decreases, β increases.

Topic: Physics, Wave Optics, NEET
In a skeletal muscle, the muscle bundles or fascicles are held together by

Options:
A Fascia (made up of elastin fibres mainly)
B Fascia (made up of collagen fibres mainly)
C Endomysium (made up of elastin fibres mainly)
D Endomysium (made up of collagen fibres mainly)

The correct answer is:
Fascia (made up of collagen fibres mainly)

Solution:
Structure of a skeletal muscle - Each skeletal muscle in is made of a number of muscle bundles (or fascicles) that are held together by a common fascia (collagenous connective tissue). Each muscle bundle is made up of a number of muscle fibres. Each muscle fibre is covered by a plasma membrane, known as sarcolemma. A muscle fibre is an anatomical unit of muscle. A muscle fibre is a syncytium (multinucleated condition) and is having cytoplasm (sarcoplasm), endoplasmic reticulum (sarcoplasmic reticulum - a storehouse of calcium ions) and a large number of parallelly arranged filaments in the sarcoplasm - myofilaments or myofibrils.

Topic: Biology, Locomotion and Movement, NEET
In a slide caliper, $(m+1)$ number of vernier divisions is equal to $m$ number of smallest main scale divisions. If $d$ unit is the magnitude of the smallest main scale division, then the magnitude of the vernicr constant is

Options:
A $d /(m+1)$ unit
B $d / m$ unit
C md/ $(m+1)$ unit
D $(m+1) d / m$ unit

The correct answer is:
$d /(m+1)$ unit

Solution:
$(m+1)$ vernier division $=m$ main scale division One division on vernier scale $=\left(\frac{m}{m+1}\right)$ division on main scale Vernier constant $=\left(1-\frac{m}{m+1}\right) d=\frac{d}{m+1}$

Topic: Physics, Experimental Physics, WBJEE, WBJEE 2012
In a solid ' $A B$ ' having the $\mathrm{NaCl}$ structure, 'A' atoms occupy the corners of the cubic unit cell. If all the face-centered atoms along one of the axes are removed, then the resultant stoichiometry of the solid is

Options:
A $A B_2$
B $A_2 B$
C $A_4 B_3$
D $A_3 B_4$

The correct answer is:
$A_3 B_4$

Solution:
There were $6 A$ atoms on the face-ceptres removing face-ceptred atoms along one of the axes means removal of $2 A$ atoms. Now, number of $A$ atoms per unit cell
$=\underset{\text { (corners) }}{8 \times \frac{1}{8}}+\underset{\text { (face-centred) }}{4 \times \frac{1}{2}}=3$
Number of $B$ atoms per unit cell
$\begin{aligned} & =12 \times \frac{1}{4} \quad 1=4 \\ & \text { (edgecentred) }+ \text { (body centred) } \\ & \end{aligned}$
Hence the resultant stoichiometry is $A_3 B_4$

Topic: Chemistry, Solid State, JEE Main
In a solid AB,A atoms are in ccp arrangement and B atoms occupy all the octahedral sites. If two atoms from the opposite faces are removed, then the resultant stoichiometry of the compound is AxBy. The value of x is [nearest integer]

Options:

The correct answer is:
3

Solution:
A forms ccp structure that means number of A atoms per unit cell →4

B atoms per unit cell →4 (Octahedral voids), as number of octahedral voids will be equal to number of atoms per unit cell.

A removed =2×12=1

A left =4-1=3

Formula becomes A3B4

x=3.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2022 (26 Jun Shift 2)
In a solid AB having NaCl structure, ‘A’ atoms occupy the corners & face centre of the cubic unit cell. If all the face centered atoms along one of the axes are removed, then the resultant stoichiometry of the solid is

Options:
A AB2
B A2B
C A4B3
D A3B4

The correct answer is:
A3B4

Solution:
Atom A occupies all corner and face center of unit cell and atom B in all octahedral void. As all atoms from face center from one axis is removed, hence number of atom A in unit cell remains 3 and B as 4.
A=8×18+6×12=4
A=4-2×12=3 after removing atoms
B=12×14+1×1=4
Formula of compound is A3B4

Topic: Chemistry, Solid State, NEET
In a solid 'AB' having rock salt arrangement ( NaCl structure), 'A' atoms occupy the corners and face center of the cubic unit cell. 'B' atoms occupies all octahedral voids of the unit cell. If all the face-centered atoms along one of the axes are removed, then the resultant stoichiometry of the solid is

Options:
A AB2
B A2B
C A4B3
D A3B4

The correct answer is:
A3B4

Solution:
There were 6 A atoms on the face-centres removing face-centred atoms along one of the axes means removal of 2 A atoms.
Now, number of A atoms per unit cell
Number of B atoms per unit cell
=12×14(edgecentred)+1=4(bodycentred)
Hence the resultant stoichiometry is A3B4

Topic: Chemistry, Solid State, JEE Main
In a solid lattice the cation has left a lattice site and is located at an interstitial position, the lattice defect is

Options:
A Interstitial defect
B Valency defect
C Frenkel defect
D Schottky defect

The correct answer is:
Frenkel defect

Solution:
When cation shifts from lattice to interstitial site, the defect is called Frenkel defect.

Topic: Chemistry, Solid State, JEE Main
In a solid, oxide ions are arranged in ccp. Cations 'A' occupy one-sixth of the tetrahedral voids and cations 'B' occupy one-third of the octahedral voids. Which of the following is the correct formula of the oxide?

Options:
A ABO3
B A2BO3
C AB2O3
D None of these

The correct answer is:
ABO3

Solution:
Cation A=16×8=86=43
Cation B=13×4=43
O2-=8×18+6×12=4
A : B : O2-=43:43:4=1:1:3

Topic: Chemistry, Solid State, JEE Main
In a solid 'XY' having the NaCl structure, 'X' atoms occupy the corners of the cubic unit cell. If all the face centred atoms along one of the axes are removed, then the resultant stiochiometry of the solid is found to be XmYn . Find the sum of m + n?

Options:

The correct answer is:
7

Solution:
Na in NaCl has eight corner and six face atoms and thus, if we remove face centred atom of one axis, then two faces atoms are removed. Thus, X is at 8 corners and Y is at 4 faces.
Number of X atoms per unit cell =8×18+4×12=3
Number of Y atoms per unit cell =12×14+1=4
Thus, formula is X3Y4 .
Hence, m = 3, n = 4
sum of m + n = 3 + 4 = 7.

Topic: Chemistry, Solid State, JEE Main
In a solution, $0.02 \mathrm{M}$ acetic acid is $4 \%$ dissociated. The $\left[\mathrm{OH}^{-}\right]$ in the solution is

Options:
A $8 \times 10^{-4}$
B $2 \times 10^{-14}$
C $8 \times 10^{10}$
D $1.25 \times 10^{-11}$

The correct answer is:
$1.25 \times 10^{-11}$

Solution:
$\left[\mathrm{H}^{+}\right]=\sqrt{\alpha^{2} C^{2}}=\sqrt{(0.02)^{2} \cdot(0.04)^{2}}=8 \times 10^{-4}$
$\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]=10^{-14}$
$\therefore\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{8 \times 10^{-4}}=1.25 \times 10^{-11} \mathrm{M}$

Topic: Chemistry, Ionic Equilibrium, JEE Main
In a solution of acetic acid, sodium acetate is added, then its $\mathrm{pH}$ value

Options:
A Decreases
B Increases
C Remains unchanged
D (1) and (2) both are correct

The correct answer is:
Increases

Solution:
Solution of $\mathrm{CH}_3 \mathrm{COONa}$ on addition to acid shows a decrease in dissociation of acid due to common ion effect.
To decrease in $\left[H^{+}\right]$ or increase $p H$.

Topic: Chemistry, Ionic Equilibrium, JEE Main
In a solution of $p H=5$, more acid is added in order to reduce the $p H=2$. The increase in hydrogen ion concentration is

Options:
A 100 times
B 1000 times
C 3 times
D 5 times

The correct answer is:
1000 times

Solution:
$\left[H^{+}\right]_{\mathrm{I}}=10^{-5}\left[H^{+}\right]_{\text {II }}=10^{-2}$
Thus increase in $\left[H^{+}\right]=\frac{10^{-2}}{10^{-5}}=1000$ times

Topic: Chemistry, Ionic Equilibrium, JEE Main
In a solvent 50% of an acid HA dimerizes and the rest dissociates. The van't Hoff factor of the acid is _____×10-2 (Round off to the nearest integer)

Options:

The correct answer is:
125

Solution:
Let initial moles of HA is a

2HA→H2 A20.5a 0.25a

HA⇌H++ A-0.5a 0.5a 0.5a

Now, i= final moles initial moles =0.25a+0.5a+0.5a0.5a+0.5a

=1.25=125×10-2

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (27 Jul Shift 2)
In a special arrangement of Young's double-slit experiment, the distance between the slits d is twice the distance between the screen and the slits D, i.e d=2D. For this setup, the value of D such that the first minima on the screen fall at a distance D from the centre O is found to be Î»N.

What is the value of N? [Take 5=2.24]

Options:

The correct answer is:
2.48

Solution:
From diagram as provided in question, OP=x
CO=D
S1C=S2C=DT1P=T1O-OP=D-x
T2P=T2O+OP=D+x
Now, S1p=S1T12+T1 P2=D2+D-x2

S2P=S2T22+T2P2=D2+D+x2

For fist minimum to occur,

Path difference

S2P-S1P=Î»2

âD2+D+x2-D2+D-x2=Î»2

The first minimum falls at a distance D from the centre, i.e., x=D

D2+4D212-D=Î»2

â´D5-1=Î»2

âD=Î»22.24-1=Î»2.48

Topic: Physics, Wave Optics, JEE Main
In a special arrangement of Young's double-slit experiment, the distance between the slits d is twice the distance between the screen and the slits D, i.e d=2D. For this setup, the value of D such that the first minima on the screen fall at a distance D from the centre O is found to be λN, where λ is the wavelength of light used. What is the value of N? [Take 5=2.24]

Options:

The correct answer is:
2.48

Solution:
From diagram as provided in question, OP=x
CO=D
S1C=S2C=DT1P=T1O-OP=D-x
T2P=T2O+OP=D+x
Now, S1p=S1T12+T1 P2=D2+D-x2

S2P=S2T22+T2P2=D2+D+x2

For fist minimum to occur,

Path difference

S2P-S1P=λ2

⇒D2+D+x2-D2+D-x2=λ2

The first minimum falls at a distance D from the centre, i.e., x=D

D2+4D212-D=λ2

∴D5-1=λ2

⇒D=λ22.24-1=λ2.48

Topic: Physics, Wave Optics, JEE Main
In a spherical distribution, the charge density varies as Ïr=Ar for a<r<b (as shown) where A is a constant. A point charge Q lies at the centre of the sphere at r=0. The electric field in the region a<r<b has a constant magnitude for

Options:
A A=0
B A=Q
C A = Q 2 Ï a 2
D A = Q 4 π a 2

The correct answer is:
A = Q 2 π a 2

Solution:
Q in = ∫ a r A r × 4 π r 2 dr + Q = 2 π A r 2 - a 2 + Q
E=Qin4πr2ε0⇒E=A2-a2A2r2+Q4πr2ε0
For E = constant,
- a 2 A 2r 2 + Q 4 π r 2 = 0
⇒ A = Q 2 π a 2

Topic: Physics, Electrostatics, JEE Main
In a spring gun having spring constant 100 N m-1 a small ball B of mass 100 g is put in its barrel (as shown in figure) by compressing the spring through 0.05 m. There should be a box placed at a distance d on the ground so that the ball falls in it. If the ball leaves the gun horizontally at a height of 2 m above the ground. The value of d is m.
g=10 m s-2

Options:

The correct answer is:
1

Solution:
12kx2=12mv2
Kx2=mv2
v=xkm=0.051000.1=0.05×1010
v=0.510
From h=12gt2
t=2 h g=2×210=210
∴ d=vt=0.510×210=1 m

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (20 Jul Shift 1)
In a standard ECG which one of the following alphabets is the correct represention of the respective activity of the human heart?

Options:
A P-depolarisation of the atria
B R-repolarisation of ventricles
C S-start of systole
D T-end of diastole

The correct answer is:
P-depolarisation of the atria

Solution:
No Solution Available

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2009 (Mains)
In a standard Young's double-slit experiment, how many maxima can be obtained on a screen (including central maxima), if the distance between the silts is d=5Î»2 (where Î» is the wavelength of light)?

Options:

The correct answer is:
5

Solution:
In YDSE, path difference, Î x = d sin Î¸
For maxima, Î x = n Î» , where n = 0 , Â± 1 , Â± 2 ....
5 Î» 2 sin Î¸ = n Î» (given that d = 5 Î» 2 )

âsinÎ¸=2n5 ...(i)
Since, â1â¤sinÎ¸â¤1 From (i) ââ1â¤ 2n/5 â¤1

So, - 5 2 â¤ n â¤ 5 2
So, possible values of n are -2, -1, 0, 1, 2 Thus a total of 5 maximas will be obtained.

Topic: Physics, Wave Optics, JEE Main
In a stationary lift, time period of a simple pendulum is ' $T$ '.' The lift starts accelerating downwards with acceleration $\left(\frac{\mathrm{g}}{4}\right)$, then the time period of the pendulum will be

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

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

Solution:
Time period of pendulum: $T=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$
$\therefore \quad$ When lift is accelerated downward with acceleration $\frac{\mathrm{g}}{4}$,
$$
\begin{aligned}
& g=g-\frac{g}{4} \\
& g=\frac{3 g}{4}
\end{aligned}
$$
$\therefore \quad$ New Time period will be
$$
\begin{aligned}
& \mathrm{T}_1=2 \pi \sqrt{\frac{4 \mathrm{~L}}{3 \mathrm{~g}}} \\
& \mathrm{~T}_1=2 \pi \frac{2}{\sqrt{3}} \sqrt{\frac{\mathrm{L}}{\mathrm{g}}} \\
& \mathrm{T}_1=\frac{2}{\sqrt{3}} \mathrm{~T}
\end{aligned}
$$

Topic: Physics, Oscillations, JEE Main
In a steady state, the temperature at the end A and end B of a 20 cm long rod AB are 100°C and 0°C. The temperature of a point, 9 cm from A is

Options:
A 55∘C
B 45∘C
C 65∘C
D 50∘C

The correct answer is:
55∘C

Solution:
At steady state, the temperature at different parts of the solid remains constant and is independent of time. Therefore, the ratio of the difference in temperature to the distance between two points in a heated body under a steady state is called the temperature gradient will be constant.

Let a point C between the ends of rod which is 9 cm from A, since temperature gradient is constant in the given thermal conduction process.

TA-TC9=TA-TB20⇒100-TC9=100-020⇒TC=100-90020=55°C

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 1)
In a streamlined flow, the velocity of a fluid at a given point

Options:
A is always constant
B does not remain constant
C changes from low value to high value
D changes from high value to low value

The correct answer is:
is always constant

Solution:
By definition, in a streamlined flow the velocity of flow at any chosen point in the liquid is always the same, i.e., it doesn't vary with time.

Topic: Physics, Mechanical Properties of Fluids, MHT CET, MHT CET 2022 (08 Aug Shift 1)
In a system two particles of masses $m_1=3 \mathrm{~kg}$ and $m_2=2 \mathrm{~kg}$ are placed at certain distance from each other. The particle of mass $m_1$ is moved towards the center of mass of the system through a distance $2 \mathrm{~cm}$. In order to keep the center of mass of the system at the original position, the particle of mass $m_2$ should move towards the center of mass by the distance _____ $\mathrm{cm}$.

Options:

The correct answer is:
3

Solution:
$\begin{aligned} & \Delta \mathrm{X}_{\text {C.O.M. }}=\frac{\mathrm{m}_1 \Delta \mathrm{x}_1+\mathrm{m}_2 \Delta \mathrm{x}_2}{\mathrm{~m}_1+\mathrm{m}_2} \\ & \Rightarrow 0=\frac{3 \times 2+2(-\mathrm{x})}{3+2} \\ & \Rightarrow \mathrm{x}=3 \mathrm{~cm}\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2024 (04 Apr Shift 2)
In a system used for spraying cars, a car body is positively charged. Neutral droplets of paint are then attracted to the car because the positive car body induces a charge on the droplets of paint. Which diagram best shows the charge pattern?

Options:
A
B

C

D

The correct answer is:

Solution:
Positive charges on car surface attracts negative charges of droplets.

Topic: Physics, Electrostatics, JEE Main
In a TCA cycle, how many NADH2 and FADH2 are produced, respectively?

Options:
A 3 NADH2, 1 FADH2
B 2 NADH2, 1 FADH2
C 4 NADH2, 1 FADH2
D 4 NADH2, 2 FADH2

The correct answer is:
3 NADH2, 1 FADH2

Solution:
The tricarboxylic acid (TCA) cycle, also known as the Kreb's or citric acid cycle, is the main source of energy for cells and an important part of aerobic respiration. The cycle harnesses the available chemical energy of acetyl coenzyme A (acetyl CoA) into the reducing power of NADH2 and FADH2. Each molecule of acetyl CoA produces 3 NADH2 and 1 FADH2.

Topic: Biology, Respiration in Plants, NEET
In a terrestrial ecosystem,

Options:
A A smaller proportion of energy flows through the grazing food chain as compared to the detritus food chain
B A smaller proportion of energy flows through the detritus food chain as compared to grazing food chain
C Equal proportion of energy flows through detritus food chain and grazing food chain
D No energy flows through detritus food chain and grazing food chain

The correct answer is:
A smaller proportion of energy flows through the grazing food chain as compared to the detritus food chain

Solution:
A grazing food chain involves producers and various types of consumers. A detritus food chain involves decomposers, mainly fungi and bacteria, which break down dead organic matter. In the aquatic ecosystem, most of the energy flows through the grazing food chain, whereas in the terrestrial ecosystem, most of the energy flows through the detritus food chain.

Topic: Biology, Ecosystem, NEET
In a test, an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is 13 and the probability that he copies the answer is 16. The probability that his answer is correct given that he copies it is 18. The probability that his answer is correct, given that he guesses it is 14. The probability that he knew the answer to the question given that he correctly answered, is

Options:
A 2431
B 1724
C 2429
D 2931

The correct answer is:
2429

Solution:
Let E1,E2 and E3 are the events that the examinee guesses, copies and knows the answer and E is the event that he answers correctly.
Then, P E 1 = 1 3 , P E 2 = 1 6
and P E 3 = 1 - 1 3 + 1 6 = 1 2
â´ Required probability = P E 3 E
= P E E 3 Â· P E 3 P E E 1 Â· P E 1 + P E E 2 Â· P E 2 + P E E 3 Â· P E 3
= 1 Ã 1 2 1 4 Ã 1 3 + 1 8 Ã 1 6 + 1 Ã 1 2 = 2 4 2 9

Topic: Mathematics, Probability, JEE Main
In a test experiment on a model aeroplane in wind tunnel, the flow speeds on the upper and lower surfaces of the wings are 70 m s-1 and 65 m s-1 respectively. If the wing area is 2 m2, the lift of the wing is _______N. (Given density of air =1.2 kg m-3)

Options:

The correct answer is:
810

Solution:
Applying Bernoulli's principle on upper & lower surfaces, we get

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2

Since, height difference between upper and lower surface is negligible.

⇒P2-P1=12ρv12-12ρv22

Therefore, upthrust acting due to pressure difference:

⇒FA=12ρv12-v22⇒F=12×1.2×702-652×2

=810 N

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (29 Jan Shift 1)