datavorous/entrance-exam-dataset huggingface dataset:
In a face centred cubic lattice, atoms of A form the corner points and atoms of B form the face centred points. If two atoms of $\mathrm{A}$ are missing from the corner points, the formula of the ionic compound is :

Options:
A
$\mathrm{AB}_3$

B
$\mathrm{AB}_4$

C
$\mathrm{A}_2 \mathrm{~B}_5$

D
$\mathrm{AB}_2$

The correct answer is:
$\mathrm{AB}_4$

Solution:
A form corner points and two atoms of A are missing from corner
$$
\begin{aligned}
\therefore \quad & \text { Atoms at corner }(\mathrm{A})=6 \times \frac{1}{8}=\frac{3}{4} \\
& \text { Atoms at face centre }(\mathrm{B})=6 \times \frac{1}{2}=3 \\
\therefore & \mathrm{A}_{3 / 4} \mathrm{~B}_3 \text { i.e., } \mathrm{AB}_4
\end{aligned}
$$

Topic: Chemistry, Solid State, JEE Main, JEE Main 2013 (23 Apr Online)
In a factory, workers work in three shifts, say shift 1, shit 2 and shift 3 and they get wages in the ratio 3:4:8 depending on the shift 1, 2 and 3 respectively. Number of workers in the shifts are in the ratio 3:2:5. If the total number of workers working is 1500 and wages per worker in shift 1 is Rs. 300, then the mean wage of a worker is

Options:
A Rs. 460
B Rs. 520
C Rs. 570
D Rs. 420

The correct answer is:
Rs. 570

Solution:
Workers in 1st shift =450
Wages in 1st shift =Rs.300
Workers in 2nd shift =300
Wages in 2nd shift =Rs. 400
Workers in 3rd shift =800
Mean=450×300+300×400+750×8001500
=90+80+400
=570
Mean wage =Rs.570

Topic: Mathematics, Statistics, JEE Main
In a famous experiment, A. Weismann cut off tails of new-born mice, generation after generation. At the end of the experiment, tails of mice neither disappeared nor shortened. This experiment:

Options:
A Supports Darwin's theory of natural selection
B Proves that tails is an essential organ of vertebrates
C Disproves de Vries mutation theory
D Show that Lamarck was wrong in his theory of inheritance of acquired characters

The correct answer is:
Show that Lamarck was wrong in his theory of inheritance of acquired characters

Solution:
Lamarck gave the theory of inheritance of acquired characteristics wherein the organism can pass on the physical characteristics acquired during its lifetime via use and disuse to its offspring. Hence, the giraffe has a long neck due to the regular use of stretching the neck to reach the leaves at the top part of the plant/tree. This theory was rejected by August Weismann where he experimented by cutting the tails of mice at each new generation. But found that, every new generation formed from the tail-less acquired parent had a tail in them. Weisman developed the theory of inheritance where the germ-plasm remained intact without change from the main body and that is passed to the next generation.

Topic: Biology, Evolution, NEET
In a ferromagnetic material, below the curie temperature, a domain is defined as:

Options:
A a macroscopic region with saturation magnetization.
B a macroscopic region with consecutive magnetic dipoles oriented in opposite direction.
C a macroscopic region with zero magnetization.
D a macroscopic region with randomly oriented magnetic dipoles.

The correct answer is:
a macroscopic region with saturation magnetization.

Solution:
In a ferromagnetic material, below the curie temperature, a domain is defined as a macroscopic region with saturation magnetization.

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2021 (25 Feb Shift 2)
In a fertilised embryo sac, the haploid, diploid and triploid structures are:

Options:
A Antipodal, zygote and primary endosperm nucleus
B Synergid, antipodal and polar nuclei
C Antipodal, synergid and primary endosperm nucleus
D Synergid, polar nuclei and zygote

The correct answer is:
Antipodal, zygote and primary endosperm nucleus

Solution:
Embryo sac is made up of 7 cells. Three antipodal cells, one central cell, two synergids and one egg cell. These antipodal cells are haploid and they degenerate after fertilization, central cell bear two haploid nuclei and post second fertilization central cell becomes triploid endosperm. Egg cell after first fertilization becomes diploid zygote whereas synergids always remain haploid.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
In a first order reaction, $10 \%$ of the reactant is consumed in 25 minutes. Calculate :
(1) The half-life period of the reaction.

Options:
A 164 min
B 180 min
C 130 min
D 86 min

The correct answer is:
164 min

Solution:
$t=10 \%$ reactant $\rightarrow 25$ minute
$\begin{aligned} & \text { (1) } \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \\ & \mathrm{~K}=\frac{2.303}{\mathrm{t}} \log \frac{100}{90} \\ & \mathrm{~K}=0.004215 \\ & \mathrm{t}_{1 / 2}=\frac{0.693}{\mathrm{~K}} \\ & \mathrm{t}_{1 / 2}=164.405 \text { minute. }\end{aligned}$
(2) $\mathrm{t}=\frac{2.303}{\mathrm{~K}} \log \frac{100}{12.5}$
$t=493.43$ minute.

Topic: Physics, Nuclear Physics, JEE Main
In a first order reaction, $10 \%$ of the reactant is consumed in 25 minutes. Calculate: The time required for completing $17 \%$ of the reaction.

Options:

The correct answer is:
44.21

Solution:
Again, for the given reaction
$\begin{aligned} & \mathrm{K}=\frac{2.303}{\mathrm{t}} \log \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \\ & \mathrm{t}=\frac{2.303}{0.0042145} \log \frac{100}{100-17} \\ & \mathrm{t}=44.21 \text { minutes. }\end{aligned}$
$\therefore$ time required for completing $17 \%$ of the reaction is 44.21 minutes.

Topic: Physics, Nuclear Physics, JEE Main
In a first order reaction $87.5 \%$ of reactant is converted in to product in 15 minutes. The rate constant for the reaction is given by

Options:
A $\frac{0.693}{5} \mathrm{~min}^{-1}$
B $\frac{0.693}{15} \mathrm{~min}^{-1}$
C $\frac{5}{0.693} \min ^{-1}$
D $0.693 \times 5 \mathrm{~min}^{-1}$

The correct answer is:
$\frac{0.693}{5} \mathrm{~min}^{-1}$

Solution:
$87.5 \%$ of reactant is converted into product in 15 minutes, means $12.5 \%$ remains unreacted
$[\mathrm{A}]_{0}=100, \quad[\mathrm{~A}]_{t}=100-87.5=12.5$
$k=\frac{2.303}{15} \log _{10} \frac{[A]_{0}}{[A]_{t}}$
$k=\frac{2.303}{15} \log _{10} \frac{100}{12.5}$
$k=\frac{2.303}{15} \log _{10} 8=\frac{0.693}{5} \min ^{-1}$

Topic: Chemistry, Chemical Kinetics, MHT CET, MHT CET 2020 (13 Oct Shift 1)
In a first order reaction the concentration of reactant decreases from 800  mol/dm 3 50  mol/dm 3 in 200 sec . The rate constant of reaction in s −1 is

Options:
A 2Ã 10 &minus;4 ââ s â1
B 1.386Ã 10 â2 ââ s â1
C 3.45× 10 5    s −1
D 2× 10 4   s −1

The correct answer is:
1.386× 10 −2    s −1

Solution:
C t  =  C 0 2 n
2 n = 800 50 =16
2 n = 2 4
So, n = 4
4T 1 2  = 200 sec
T 1 2  = 50 sec
K =  0.693 T 1 2  =  0.693 50  = 1.386 ×  10 −2   sec −1

Topic: Chemistry, Chemical Kinetics, JEE Main
In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is -

Options:
A 30 minutes
B 15Â minutes
C 7.5Â minutes
D 60Â minutes

The correct answer is:
30 minutes

Solution:
0.1M ât1/20.05M ât1/20.025M
for the I st order reaction :- t 3/4 =2Ã t 1/2 =2Ã15=30 minute
successive half lives in first order reactions remain constant irrespective of initial concentration.

Topic: Chemistry, Chemical Kinetics, JEE Main
In a first order reaction, the concentration of the reactant, decreases from $0.8 M$ to $0.4 M$ in 15 minutes. The time taken for the concentration to change from $0.1 \mathrm{M}$ to $0.025 \mathrm{M}$ is

Options:
A 7.5 minutes
B 15 minutes
C 30 minutes
D 60 minutes

The correct answer is:
30 minutes

Solution:
The concentration of the reactants decrease from 0.8 to 0.4 in $15 \mathrm{~min}$ i.e., $T_{1 / 2}=15 \mathrm{~min}$, concentration from $0.1 \mathrm{~m}$ to 0.025 will fall in 2 half lives so total time taken $=2 \times T_{1 / 2}=2 \times 15=30 \mathrm{~min}$.

Topic: Chemistry, Chemical Kinetics, JEE Main
In a fluorescent lamp choke (a small transformer) 100V of reverse voltage is produced when the choke current changes uniformly from 0.25A to 0 in a duration of 0.025ms . The self-inductance of the choke (in mH ) is estimated to be ________

Options:

The correct answer is:
10

Solution:
100=L0.250.025×103
∴L=100×10-4H
=10 mH

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2020 (09 Jan Shift 1)
In a football championship, there were played 153 matches. Every team played one match with each other. The number of teams participating in the championship is

Options:
A $17$
B $18$
C $9$
D $13$

The correct answer is:
$18$

Solution:
${ }^n C_2=153 \Rightarrow \frac{n(n-1)}{2}=153 \Rightarrow n=18$.

Topic: Mathematics, Permutation Combination, JEE Main
In a Frank - Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges with an energy 0.7 eV. The minimum wavelength of photons emitted by mercury atoms is close to:

Options:
A 250 nm
B 1700 nm
C 220 nm
D 2020 nm

The correct answer is:
250 nm

Solution:
When electron pass through the mercury vapor, it losses some of its energy. The loss in KE of electron =5⋅6-0.7eV =4.9 eV
∴ energy of radiation emitted = 4.9 eV
∴ wavelength of radiation, λ=1.24 × 1044.9A≈ 250 nm

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2019 (12 Jan Shift 2)
In a Fraunhofer diffraction at a single slit, if yellow light illuminating slit is replaced by blue light, then diffraction bands

Options:
A Remain unchanged
B Become wider
C Disappear
D Become narrower

The correct answer is:
Become narrower

Solution:
Bandwidth ∝λ
λblue<λyellow⇒ Bandwidth becomes narrower when yellow light is replaced by blue light.

Topic: Physics, Wave Optics, NEET
In a Fraunhofer diffraction at a single slit, if yellow light illuminating the slit is replaced by blue
light, then diffraction bands

Options:
A Remains unchanged
B Become wider
C Disappear
D Become narrower

The correct answer is:
Become narrower

Solution:
As we know Wavelength of yellow light \( > \) Wavelength of blue light
Therefore, if yellow light used in illuminating the slit is replaced by blue light then the diffraction bands will become
narrower.

Topic: Physics, Wave Optics, KCET, KCET 2018
In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is $\mathrm{CH}_3 \mathrm{OH}(\ell)+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\ell)$ At $298 \mathrm{~K}$ standard Gibb's energies of formation for $\mathrm{CH}_3 \mathrm{OH}(\ell), \mathrm{H}_2 \mathrm{O}(\ell)$ and $\mathrm{CO}_2(\mathrm{~g})$ are $-166.2,-237.2$ and $-394.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. If standard enthalpy of combustion of methanol is $-726 \mathrm{~kJ} \mathrm{~mol}^{-1}$, efficiency of the fuel cell will be

Options:
A
$80 \%$

B
$87 \%$

C
$90 \%$

D
$97 \%$

The correct answer is:
$97 \%$

Solution:
$$
\mathrm{CH}_3 \mathrm{OH}(\ell)+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\ell) \quad \Delta \mathrm{H}=-726 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$
Also $\Delta \mathrm{G}_{\mathrm{f}}^0 \mathrm{CH}_3 \mathrm{OH}(\ell)=-166.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$$
\begin{aligned}
& \Delta \mathrm{G}_{\mathrm{f}}^0 \mathrm{H}_2 \mathrm{O}(\ell)=-237.2 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{G}_{\mathrm{f}}^0 \mathrm{CO}_2(\ell)=-394.4 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \because \Delta \mathrm{G}=\Sigma \Delta \mathrm{G}_{\mathrm{f}}^0 \text { products }-\Sigma \Delta \mathrm{G}_{\mathrm{f}}^0 \text { reactants. } \\
& =-394.4-2(237.2)+166.2 \\
& =-702.6 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
& \text { now Efficiency of fuel cell }=\frac{\Delta \mathrm{G}}{\Delta \mathrm{H}} \times 100 \\
& =\frac{702.6}{726} \times 100 \\
& =97 \%
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, JEE Main
In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is $\mathrm{CH}_3 \mathrm{OH}(\ell)+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\ell)$ At $298 \mathrm{~K}$ standard Gibb's energies of formation for $\mathrm{CH}_3 \mathrm{OH}(\ell), \mathrm{H}_2 \mathrm{O}(\ell)$ and $\mathrm{CO}_2(\mathrm{~g})$ are $-166.2,-237.2$ and $-394.4 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. If standard enthalpy of combustion of methanol is $-726 \mathrm{~kJ} \mathrm{~mol}^{-1}$, efficiency of the fuel cell will be

Options:
A
$80 \%$

B
$87 \%$

C
$90 \%$

D
$97 \%$

The correct answer is:
$97 \%$

Solution:
$$
\mathrm{CH}_3 \mathrm{OH}(\ell)+\frac{3}{2} \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\ell) \quad \Delta \mathrm{H}=-726 \mathrm{~kJ} \mathrm{~mol}^{-1}
$$
Also $\Delta \mathrm{G}_{\mathrm{f}}^0 \mathrm{CH}_3 \mathrm{OH}(\ell)=-166.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$$
\begin{aligned}
& \Delta \mathrm{G}_{\mathrm{f}}^0 \mathrm{H}_2 \mathrm{O}(\ell)=-237.2 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \Delta \mathrm{G}_{\mathrm{f}}^0 \mathrm{CO}_2(\ell)=-394.4 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
& \because \Delta \mathrm{G}=\Sigma \Delta \mathrm{G}_{\mathrm{f}}^0 \text { products }-\Sigma \Delta \mathrm{G}_{\mathrm{f}}^0 \text { reactants. } \\
& =-394.4-2(237.2)+166.2 \\
& =-702.6 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
& \text { now Efficiency of fuel cell }=\frac{\Delta \mathrm{G}}{\Delta \mathrm{H}} \times 100 \\
& =\frac{702.6}{726} \times 100 \\
& =97 \%
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2009
In a full wave rectifier circuit operating from $50 \mathrm{~Hz}$ mains frequency, the fundamental frequency in the ripple would be

Options:
A
$50 \mathrm{~Hz}$

B
$25 \mathrm{~Hz}$

C
$100 \mathrm{~Hz}$

D
$70.7 \mathrm{~Hz}$

The correct answer is:
$100 \mathrm{~Hz}$

Solution:
frequency $=2$ (frequency of input signal).

Topic: Physics, Semiconductors, JEE Main, JEE Main 2005
In a full wave rectifier, input \( \mathrm{AC} \) current has a frequency \( \nu \). The output frequency of current is,

Options:
A \( \frac{\nu}{2} \)
B \( \nu \)
C \( 2 \nu \)
D \( 320 \mathrm{~Hz} \)

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

Solution:
In a full wave rectifier, for some input,

Input frequency=2×frequency of input

Input frequency=2ν.

Topic: Physics, Semiconductors, JEE Main
In a full wave rectifier with input frequency of $50 \mathrm{~Hz}$, the frequency of the output is:

Options:
A $50 \mathrm{~Hz}$
B $100 \mathrm{~Hz}$
C $200 \mathrm{~Hz}$
D $500 \mathrm{~Hz}$

The correct answer is:
$100 \mathrm{~Hz}$

Solution:
Concept:
For a full-wave rectifier with input frequency fin, the output frequency is double the input frequency, i.e.
$\text { fout }=2 \text { fin }$
Application:
Given fin $=50 \mathrm{~Hz}$
Output frequency out fout will be:
$\text { fout }=100 \mathrm{~Hz}$
In the case of a half-wave rectifier, the output waveform completes one cycle at the same time as that of the input waveform. Therefore, the frequency of the output wave is equal to the frequency of the input wave.

Topic: Physics, Semiconductors, JEE Main
In a full-wave rectifier circuit operating from 50 Hz mains frequency, the fundamental frequency in the rectifier wave would be

Options:
A 100 Hz

B 70.7 Hz

C 50 Hz

D 25 Hz

The correct answer is:
100 Hz

Solution:
Frequency of full wave rectifier = 2 x input frequency = 2 × 50 = 100 Hz.

Topic: Physics, Semiconductors, NEET
In $A g B r$ crystal, the ion size lies in the order $A g^{+}\lt \lt \mathrm{Br}^{-}$. The $A g B r$ crystal should have the following characteristics

Options:
A Defectless (perfect) crystal
B Schottky defect only
C Frenkel defect only
D Both Schottky and Frenkel defects

The correct answer is:
Frenkel defect only

Solution:
$\mathrm{AgBr}$ exhibits Frenkel defect due to large difference in the size of $\mathrm{Ag}^{+}$and $\mathrm{Br}^{-}$ions.

Topic: Chemistry, Solid State, JEE Main
In a galvanic cell, after running the cell for sometime, the concentration of the electrolyte is automatically raised to 3 m HCl. Molar conductivity of the 3 m HCl is about 240 S cm2mol-1 and limiting molar conductivity of HCl is about 420 S cm2mol-1 . If Kb of water is 0.52 K kg mol-1 , calculate the boiling point of the electrolyte at the end of the experiment.

Options:
A 375.6 K
B 376.3 K
C 378.1 K
D 380.3 K

The correct answer is:
375.6 K

Solution:
α=Molar conductivityLimiting molar conductivity=240420=1221
α=i-1n-1=i-12-1=1221
i=1221+1=3321
ΔTb=iKbm=3321×0.52×3=2.45K
Tb=Tbo+ΔTb=373.15+2.45=375.6K

Topic: Chemistry, Solutions, JEE Main
In a game, a man wins Rs. 100 if he gets 5 or 6 on a throw of a fair die and loses Rs. 50 for getting any other number on the die. If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is :

Options:
A 4003 gain
B 4009 gain
C 4003 loss
D 0

The correct answer is:
0

Solution:
Probability of success i.e. getting a 5 or 6=Ps=26=13.

So, probability of failure i.e. getting any number other than 5 or 6=Pf=1-Ps=1-13=23.

Let X be the amount that a man wins in a maximum of three throws.

So, X can take values :

Case I : First two throws are 5 or 6, game stops

⇒X=100⇒PX=100=13

Case II : First throw is other than 5 or 6 and second and third throws are 5 or 6, game stops

⇒X=50⇒PX=50=23×13=29

Case III : First two throws are other than 5 or 6 and last throw is 5 or 6

⇒X=0⇒PX=0=23×23×13=427

Case IV : All three throws are other than 5 or 6

⇒X=-150⇒PX=-150=23×23×23=827

Therefore, expected gain or loss

=EX=100×13+50×29+0×427-150×827=0

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (12 Jan Shift 2)
In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression equals

Options:
A
$\frac{1}{2}(1-\sqrt{5})$

B
$\frac{1}{2} \sqrt{5}$

C
$\sqrt{5}$

D
$\frac{1}{2}(\sqrt{5}-1)$

The correct answer is:
$\frac{1}{2}(\sqrt{5}-1)$

Solution:
Given $a r^{n-1}=a r^n+a r^{n+1}$
$\begin{aligned}
& \Rightarrow 1=r+r^2 \\
& \therefore r=\frac{\sqrt{5}-1}{2} .
\end{aligned}$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2007
In a geometric progression, if the ratio of the sum of first 5 terms to the sum of their reciprocals is 49 , and the sum of the first and the third term is 35. Then the first term of this geometric progression is:

Options:
A
7

B
21

C
28

D
42

The correct answer is:
28

Solution:
According to Question
$$
\Rightarrow \frac{S_5}{S_5^{\prime}}=49
$$
(here, $S_5=$ Sum of first 5 terms and $S_5=$ Sum of their reciprocals)
$$
\begin{aligned}
&\Rightarrow \frac{\frac{a\left(r^5-1\right)}{(r-1)}}{\frac{a^{-1}\left(r^{-5}-1\right)}{\left(r^{-1}-1\right)}}=49 \\
&\Rightarrow \frac{a\left(r^5-1\right) \times\left(r^{-1}-1\right)}{a^{-1}\left(r^{-5}-1\right) \times(r-1)}=49 \\
&\text { or } \frac{a^2\left(1-r^5\right) \times(1-r) \times r^5}{\left(1-r^5\right) \times(1-\mathrm{r}) \times r}=49 \\
&\Rightarrow a^2 r^4=49 \Rightarrow a^2 r^4=7^2 \\
&\Rightarrow a r^2=7
\end{aligned}
$$
Also, given, $S_1+S_3=35$ $a+a r^2=35$
Now substituting the value of eq. (1) in eq. (2)
$a+7=35$
$a=28$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2014 (11 Apr Online)
In a geometric progression, if the ratio of the sum of first \( 5 \) terms to the sum of their reciprocals is \( 49 \), and the sum of the first and the third term is \( 35 \). Then the first term of this geometric progression is

Options:
A \( 28 \)
B \( 21 \)
C \( 7 \)
D \( 42 \)

The correct answer is:
\( 28 \)

Solution:
a+ar+ar2+ar3+ar41a+1ar+1ar2+1ar3+1ar4=49
⇒a1+r+r2+r3+r41ar4r4+r3+r2+r+1=49
⇒a2r4=49
⇒ar2=7
⇒T3=7
T1+T3=35
T1+7=35
T1=28

Topic: Mathematics, Sequences and Series, JEE Main
In a girdled plant, when water is supplied to the leaves above the girdle, leaves may remain green for sometime then wilt and ultimately die. What does it indicate?

Options:

None

Solution:
When water is supplied in a girdle plant to the leaves above the girdle, leaves may remain green for sometime because leaves can synthesise their own carbohydrate food through photosynthesis, they however, gradvally wilt due to non-availability of water.
The system of xylem vessels from root to the leaf vein can supply the needed water during girding there is a possible loss of xylem vessels and the water supply is cut off, resulting in death of the plant.

Topic: Biology, Transport in Plants
In a given circuit diagram, a 5 V zener diode along with a series resistance is connected across a 50 V power supply. The minimum value of the resistance required, if the maximum zener current is 90 mA will be Ω.

Options:

The correct answer is:
500

Solution:
Voltage across, RL=5 V

⇒i2=5RL

Also, voltage across R=50-5=45 volt

v=iR⇒R=vi=45ii+i2

R=4590mA+5RL

Current in zener diode is maximum when RL→∞
i2→0 and ii=i

So R=4590 mA=500 Ω

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (22 Jul Shift 1)
In a given process for an ideal gas, $d W=0$ and $d Q \lt 0$. Then for the gas

Options:
A The temperature will decrease
B The volume will increase
C The pressure will remain constant
D The temperature will increase

The correct answer is:
The temperature will decrease

Solution:
From FLOT
$\begin{aligned} \Rightarrow d U & =d Q-d W \Rightarrow d U=d Q(\lt 0), (d W=0) \\ \Rightarrow d U & \lt 0 \text { So temperature will decrease }\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main
In a given process, for an ideal gas, $\Delta W=0$ and $\Delta Q < 0$. Then, for the gas,

Options:
A the temperature will decrease
B the volume will increase
C the pressure will remain constant
D the temperature will increase

The correct answer is:
the temperature will decrease

Solution:
In the given process, for an ideal gas
$\Delta W=0$ and $\Delta Q < 0$
By lst law of thermodynamics,
$\Delta Q=\Delta W+\Delta U$
where, $\Delta U=$ internal energy of gas.
$\begin{aligned} & \Rightarrow \quad \Delta Q=\Delta U \\ & \text { and also } \Delta U < 0 .\end{aligned}$
As the internal energy depends on temperature.
Therefore, if change in internal energy is negative, it mean temperature of gas is decreased.

Topic: Physics, Thermodynamics, AP EAMCET, AP EAMCET 2021 (23 Aug Shift 2)
In a given process of an ideal gas, dW = 0 and dQ < 0 . Then for the gas

Options:
A The temperature will decrease
B The volume will increase
C The pressure will remain contain
D The temperature will increase

The correct answer is:
The temperature will decrease

Solution:
From first law of thermodynamics,
dQ = dU + dW
dQ = dU , if dW = 0
Since, dQ < 0
Therefore, dU < 0 or U Final < U initial
so, the temperature will decrease.
Note: Internal energy U of an ideal gas depends only on the temperature of the gas. The internal energy of n moles of an ideal gas is given by nCv dT

Topic: Physics, Thermodynamics, JEE Main
In a given process on an ideal gas, dW = 0 and dQ < 0. Then for the gas

Options:
A The temperature will decrease
B The volume will increase
C The pressure will remain constant
D The temperature will increase

The correct answer is:
The temperature will decrease

Solution:
From the first law of thermodynamics
dQ = dU + dW
Hence dW = (given)
∴ dQ = dU
Now since dQ < 0 (given)
∴ dQ is negative ⇒ dU = - ve
⇒ dU decreases ⇒ Temperature decreases
∴ The correct option is (1).

Topic: Physics, Thermodynamics, NEET
In a graph of population, on x-axis time and on y-axis population is plotted. A parallel line to x-axis shows:

Options:
A Natality equal to mortality
B Natality decreases mortality increases
C Natality constant mortality increases
D Natality increases mortality decreases

The correct answer is:
Natality equal to mortality

Solution:
On X-axis showing time and Y-axis showing population. A parallel line to X-axis represents the Population is maintained with time because the birth rate will be equal to the death rate. The birth rate is Natality and death rate is mortality. So Natality will be equal to Mortality and the correct answer is A.

Topic: Biology, Organisms and Populations, NEET