datavorous/entrance-exam-dataset huggingface dataset:
Two wires are made of the same material and have the same volume. However, wire 1 has a cross-sectional area $A$ and wire 2 has cross-sectional area $3 A$. If the length of the wire 1 increases by $\Delta x$ on applying force $F$, how much force is needed to stretch wire 2 by the same amount?

Options:
A $\mathrm{F}$
B $4 \mathrm{~F}$
C $6 F$
D $9 \mathrm{~F}$

The correct answer is:
$9 \mathrm{~F}$

Solution:
For the same material, Young's modulus is the same and it is given that the volume is the same and the area of the cross-section for the wire $L_1$ is and that of $L_2$ is $3 A$
$\begin{aligned} & V=V_1=V_2 \\ & V=A \times L_1=3 A \times L_2 \Rightarrow L_2=L_1 / 3 \\ & Y=(F / A) /(\Delta L / L) \\ & F_1=Y A\left(\Delta L_1 / L_1\right) \\ & F_2=Y 3 A\left(\Delta L_2 / L_2\right)\end{aligned}$
Given $\Delta L_1=\Delta L_2=\Delta x$ (for the same extension)
$F_2=Y 3 A\left(\Delta x /\left(L_1 / 3\right)\right)=9 \cdot\left(Y A \Delta x / L_1\right)=9 F_1=9 F$

Topic: Physics, Mechanical Properties of Solids, JEE Main
Two wires are made of the same material and have the same volume. However wire 1 has crosssectional area $A$ and wire-2 has cross-sectional area $3 A$. If the length of wire 1 increases by $\Delta x$ on applying force $F$, how much force is needed to stretch wire 2 by the same amount?

Options:
A
$\mathrm{F}$

B
$4 \mathrm{~F}$

C
$6 \mathrm{~F}$

D
$9 \mathrm{~F}$

The correct answer is:
$9 \mathrm{~F}$

Solution:
$$
\begin{aligned}
& \mathrm{A}_1 \ell_1=\mathrm{A}_2 \ell_2 \Rightarrow \ell_2=\frac{\mathrm{A}_1 \ell_1}{\mathrm{~A}_2}=\frac{\mathrm{A} \times \ell_1}{3 \mathrm{~A}}=\frac{\ell_1}{3} \Rightarrow \frac{\ell_1}{\ell_2}=3 \\
& \Delta \mathrm{x}_1=\frac{\mathrm{F}_1}{\mathrm{~A} \gamma} \times \ell_1 \\
& \Delta \mathrm{x}_2=\frac{\mathrm{F}_2}{3 \mathrm{~A} \gamma} \ell_2 \\
& \text { Here } \Delta \mathrm{x}_1=\Delta \mathrm{x}_2 \\
& \frac{\mathrm{F}_2}{3 \mathrm{~A} \gamma} \ell_2=\frac{\mathrm{F}_1}{\mathrm{~A} \gamma} \ell_1 \\
& \left.\mathrm{~F}_2=3 \mathrm{~F}_1 \times \frac{\ell_1}{\ell_2}=3 \mathrm{~F}_1 \times 3=\mathrm{i}\right)
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2009
Two wires made of same material are clamped rigidly at one end and pulled by the same force on the other end. The length and the radius of the first wire are three times those of the second wire. If $x$ is the increase in the length of the first wire, then the increase in the length of the second wire is

Options:
A $\frac{1}{3} x$
B $3 x$
C $9 x$
D $\sqrt{3} x$

The correct answer is:
$3 x$

Solution:
As wires are of same material so both are having same young's modulus
$\Rightarrow \quad Y_1=Y_2$
$\Rightarrow \quad \frac{\frac{F_1}{A_1}}{\frac{\Delta L_1}{L_1}}=\frac{\frac{F_2}{A_2}}{\frac{\Delta L_2}{L_2}}$
$\Rightarrow \quad \frac{F_1 L_1}{\Delta L_1 A_1}=\frac{F_2 L_2}{\Delta L_2 A_2}$
Now given,
$\begin{aligned} L_1 & =3 L_2 \text { and } r_1=3 r_2 \\ \Delta L_1 & =x\end{aligned}$
Also, $\quad F_1=F_2$
So from eq. (i), we have
$\frac{3 L_2}{x\left(3 r_2\right)^2}=\frac{L_2}{\Delta L_2 r_2^2} \Rightarrow \Delta L_2=3 x$

Topic: Physics, Mechanical Properties of Solids, JEE Main
Two wires of diameter 0.25 cm, one made of steel and other made of brass are loaded as shown in figure. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Young's modulus of steel is 2.0 Ã 1011 and Young's modulus of Brass 0.91 Ã 1011 Pa. The elongations of steel and brass wires are respectively

Options:
A 1.3 Ã 10-2 m;
1.5 Ã 10-4 m
B 1.5 Ã 10-4 m;
1.3 Ã 10-4 m
C 2.0 Ã 10-5 m;
1.8 Ã 10-3 m
D 1.8 Ã 10-3 m;
2.1 Ã 10-3 m

The correct answer is:
1.5 Ã 10-4 m;
1.3 Ã 10-4 m

Solution:
Given, diameter of wires (2r) = 0.25 cm
r = 0.125 cm = 1.25 Ã 10-3 m
For steel wire,
Load (F1) = (4 + 6) kgf = 10 Ã 9.8 N = 98 N
Length of steel wire (l1) = 1.5 m
Young's modulus (Y1) = 2.0 Ã 1011 Pa
Young's modulus (Y) = F 1 Ã lâ¡ 1 A 1 Ã Î lâ¡ 1
Change in length Î lâ¡ 1 = F 1 Ã lâ¡ 1 A 1 Ã Y 1
= F 1 Ã lâ¡ 1 Ï r 1 2 Ã Y 1
= 98 Ã 1.5 3.14 Ã 1.25 Ã 10 - 3 2 Ã 2.0 Ã 10 11
â= 1.5 Ã 10-4 m
For brass wire,
Load (F2) = 6 kgf = 6 Ã 9.8 N = 58.8 N
Length of brass wire (l2) = 1.0 m
Young's modulus (Y2) = 0.91 Ã 1011 Pa
Change in length lâ¡ 2 = F 2 Ã lâ¡ 2 Ï r 2 2 Ã Y 2
= 58.8 Ã 1.0 3.14 Ã 1.25 Ã 10 - 3 2 Ã 0.91 Ã 10 11
â = 1.3 Ã 10-4 m

Topic: Physics, Mechanical Properties of Solids, JEE Main
Two wires of same diameter of the same material having the length $l$ and $2 l$. If the force $\mathrm{F}$ is applied on each, the ratio of the work done in the two wires will be

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

The correct answer is:
$1: 2$

Solution:
$W=\frac{1}{2} F l \quad \therefore W \propto l$ ( $F$ is constant)
$\therefore \frac{W_1}{W_2}=\frac{l_1}{l_2}=\frac{l}{2 l}=\frac{1}{2}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
Two wires of same length and radius are joined end to end and loaded. The Young's moduli of the materials of the two wires are Y1 and Y2. The combination behaves as a single wire then its Young's modulus is:

Options:
A Y=2Y1Y23Y1+Y2
B Y=2Y1Y2Y1+Y2
C Y=Y1Y22Y1+Y2
D Y=Y1Y2Y1+Y2

The correct answer is:
Y=2Y1Y2Y1+Y2

Solution:
In series combination, Δl=ℓ1+ℓ2

Y=F/AΔℓ/ℓ⇒Δℓ=FℓAY

⇒Δℓ∝ℓY

Equivalent length of rod after joining is =2ℓ

As, lengths are same and force is also same in series

Δℓ=Δℓ1+Δℓ2

ℓeqYeq=ℓY1+ℓY2⇒2ℓY=ℓY1+ℓY2

∴ Y=2Y1Y2Y1+Y2

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Two wires of same length and thickness having specific resistances 6 Ω cm and 3 Ω cm respectively are connected in parallel. The effective resistivity is ρ Ωcm. The value of ρ to the nearest integer, is ___ .

Options:

The correct answer is:
4

Solution:
∵ in parallel, Rnet=R1R2R1+R2

⇒ρl2A=ρ1lA×ρ2lAρ1lA+ρ2lA

⇒ρ2=6×36+3=2⇒ρ=4

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Two wires of same material have length $L$ and $2 L$ and cross-sectional areas $4 A$ and $A$ respectively. The ratio of their specific resistance would be

Options:
A $1: 2$
B $8: 1$
C $1: 8$
D $1: 1$

The correct answer is:
$1: 1$

Solution:
Specific resistance doesn't depend upon length and area.

Topic: Physics, Current Electricity, JEE Main
Two wires of the same dimensions but resistivities $\rho_1$ and $\rho_2$ are connected in series. The equivalent resistivity of the combination is

Options:
A $\rho_1+\rho_2$
B $\frac{\rho_1+\rho_2}{2}$
C $\sqrt{\rho_1 \rho_2}$
D $2\left(\rho_1+\rho_2\right)$

The correct answer is:
$\frac{\rho_1+\rho_2}{2}$

Solution:
$R_{\text {eq }}=R_1+R_2 \Rightarrow \frac{\rho_{\text {eff.2 }}}{A}=\frac{\rho_1 I}{A}+\frac{\rho_2 I}{A} \Rightarrow \rho_{\text {eff. }}=\frac{\rho_1+\rho_2}{2}$.

Topic: Physics, Current Electricity, JEE Main
Two wires of the same material and length but diameters in the ratio $1: 2$ are stretched by the same force. The elastic potential energy per unit volume for the wires, when stretched by the same force will be in the ratio.

Options:
A $16: 1$
B $1: 1$
C $2: 1$
D $4: 1$

The correct answer is:
$16: 1$

Solution:
We know that,
Elastic potential energy in a stretched wire
$$
\begin{aligned}
U & =\frac{1}{2} \text { Young's modulus } \times(\text { strain })^2 \\
U & =\frac{1}{2} \times \frac{F}{A} \times \frac{1}{I} \times \frac{F l}{Y \cdot A} \\
U & =\frac{1}{2} \frac{F^2}{A^2} \cdot \frac{1}{Y} \\
\Rightarrow U & \propto \frac{1}{r^4}
\end{aligned}
$$
So,
$$
\begin{aligned}
\frac{U_1}{U_2} & =\frac{1 / r_1^4}{1 / r_2^4}=\left(\frac{r_2}{r_1}\right)^4 \quad\left[\because \frac{d_1}{d_2}=\frac{r_1}{r_2}=\frac{1}{2}\right] \\
\frac{U_1}{U_2} & =\left(\frac{2}{1}\right)^2=\frac{16}{1} \\
U_1: U_2 & =16: 1
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main
Two wires of the same material (Young's modulus Y) and same length L but radii R and 2R respectively are joined end to end and a weight w is suspended from the combination as shown in the figure. The elastic potential energy in the system is

Options:
A 3w2L4ÏR2Y
B 3w2L8ÏR2Y
C 5w2L8ÏR2Y
D w2LÏR2Y

The correct answer is:
5w2L8πR2Y

Solution:
k1= YÏ2R2L, k2=YÏR2L
Equivalent 1k1+1k2=L4YÏR2+LYÏR2
Since, k1x1=k2x2=w
Elastic potential energy of the system
Uâ¡=12k1x12+12 k2x22
U=12 k1wk12+12k2wk22
=12w2 1k1+1k2=12w25L4YÏR2
Uâ¡=5w2L8ÏYR2

Topic: Physics, Mechanical Properties of Solids, JEE Main
Two wires that are made up of two different materials whose specific resistances are in the ratio 2:3 , length 3:4 and area 4:5. If the ratio of their resistance is ab, then what is the value of a+b, where a and b are the smallest positive integers.?

Options:

The correct answer is:
13

Solution:
Use R=ρ lA
∴ R1R2=ρ1l1A1ρ2l2A2

⇒R1R2= 58

Topic: Physics, Current Electricity, JEE Main
Two wires W1 and W2 have the same radius r and respective, densities ρ1 and ρ2, such that ρ2=4ρ1 . They are joined together at the point O, as shown in the figure. The combination is used as a sonometer wire and kept under tension T. The point O is midway between the two bridges. When a stationary wave is set up in the composite wire, the joint is found to be a node. The ratio of the number of antinodes formed in W1 to W2 is

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

The correct answer is:
1:2

Solution:
Let p1 and p2 be the number of antinodes of both wires.

Fundamental frequency remains the same for both wires.
n1=n2
T→same
r→same
l→same
n=p2l Tπr2ρ

From the above equation, we have
p1ρ1 =p2ρ2

→p1p2= ρ1ρ2
The ratio of antinodes is p1p2=12.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2017 (08 Apr Online)
Two women and some men participated in a chess tournament in which every participant played two games with each of the other participants. If the number of games that the men played between them-selves exceeds the number of games that the men played with the women by 66, then the number of men who participated in the tournament lies in the interval

Options:
A (11,13]
B (14,17)
C [10,12)
D [8,9]

The correct answer is:
[10,12)

Solution:
Let the number of men participating in tournament =m.

Number of women participating in tournament =2.

Each player will play 2 games with every other player and two women will play 2 games with each man.

Total numbers of games played by women =2×2m=4m

Total number of games played by men =2·mC2

From given data

2· mC2-4m=66

⇒2mm-12-4m=66

⇒m2-5m-66=0

⇒m-11m+6=0

⇒m=11 or -6 but m∈N, m≠-6

⇒m∈10,12.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2014 (19 Apr Online)
Two wooden blocks of masses $M$ and $m$ are placed on a smooth horizontal surface as shown in figure. If a force $P$ is applied to the system as shown in figure such that the mass $m$ remains stationary with respect to block of mass $M$, then the magnitude of the force $P$ is

Options:
A $(M+m) g \tan \beta$
B $g \tan \beta$
C $m g \cos \beta$
D $(M+m) g \operatorname{cosec} \beta$

The correct answer is:
$(M+m) g \tan \beta$

Solution:
The free body diagram of the given situation is
$\begin{aligned} & \text { Force }=\text { Mass } \times \text { Acceleration } \\ & \therefore \quad P=(M+m) a \\ & m a \cos \beta=m g \sin \beta \\ & \Rightarrow \quad a=g \frac{\sin \beta}{\cos \beta} \\ & =g \tan \beta \\ & \therefore \quad P=(M+m) g \tan \beta \\ & \end{aligned}$

Topic: Physics, Laws of Motion, AP EAMCET, AP EAMCET 2013
Two wooden blocks of masses $M$ and $m$ are placed on a smooth horizontal surface as shown in figure. If a force $P$ is applied to the system as shown in figure such that the mass $m$ remains stationary with respect to block of mass $M$, then the magnitude of the force $P$ is

Options:
A $(M+m) g \tan \beta$
B $g \tan \beta$
C $m g \cos \beta$
D $(M+m) g \operatorname{cosec} \beta$

The correct answer is:
$(M+m) g \tan \beta$

Solution:
The free body diagram of the given situation is
$\begin{aligned} & \text { Force }=\text { Mass } \times \text { Acceleration } \\ & \therefore \quad P=(M+m) a \\ & m a \cos \beta=m g \sin \beta \\ & \Rightarrow \quad a=g \frac{\sin \beta}{\cos \beta} \\ & =g \tan \beta \\ & \therefore \quad P=(M+m) g \tan \beta \\ & \end{aligned}$

Topic: Physics, Laws of Motion, TS EAMCET, TS EAMCET 2013
Two workers A and B are engaged to do a piece of work. Working alone, A takes 8 hours more to complete the work than if both worked together. On the other hand, working alone, B would need $4 \frac{1}{2}$ hours more to complete the work than if both worked together. How much time would they take to complete the job working together?

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

The correct answer is:
6 Hours

Solution:
If A \& B work together take time t hr. If A work alone takes $t+8 \mathrm{hr}$. to complete the work Work $/ \mathrm{hr}$ by $\mathrm{A}$ is $\frac{1}{\mathrm{t}+8}$. Similarly for B work/hr is $\frac{1}{\mathrm{t}+9 / 2}$
When A \& B work together there work/hr $\frac{1}{t+8}+\frac{1}{t+9 / 2}$
Work will be done in t hr.
$\mathrm{t}\left(\frac{1}{t+8}+\frac{1}{t+9 / 2}\right)=1$
By solving
$\mathrm{t}=6$

Topic: Mathematics, Quadratic Equation, JEE Main
Two Zener diodes (A and B) having breakdown voltages of 6 V and 4 V respectively, are connected as shown in the circuit below. The output voltage V0 variation with input voltage linearly increasing with time, is given by (Vinput = 0V at t = 0)

Options:
A
B
C
D

The correct answer is:

Solution:
At time t=0, voltage across the first zener diode is increases linearly till it reaches it's breakdown voltage of 4 V, after that voltage increases linearly till it reaches the breakdown voltage of second zener diode,

As soon as voltage across the zener diode reaches the breakdown voltage of 6 V of second zener diode, volatge becomes constant and doesn't change with time.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2020 (05 Sep Shift 2)
Two-point charges $+Q$ and $+q$ repel each other with a force of $100 \mathrm{~N}$. Keeping the distance between them unchanged, if $Q$ is increased by $10 \%$ and $q$ is decreased by $q$ the force of repulsion between them will.

Options:
A decrease by $10 \mathrm{~N}$
B decrease by $1 \mathrm{~N}$
C increase by $10 \mathrm{~N}$
D remain same

The correct answer is:
decrease by $1 \mathrm{~N}$

Solution:
Given, $100 \mathrm{~N}=\frac{k Q q}{r^2}$ and $F=\frac{k(1.1 Q)(0.9 q)}{r^2}$
Taking the ratio of above two equations:
$F=99$

Topic: Physics, Electrostatics, MHT CET, MHT CET 2022 (10 Aug Shift 1)
Tyndall effect is more effectively shown by:

Options:
A lyophilic colloid
B suspension
C lyophobic colloid
D true solution

The correct answer is:
lyophobic colloid

Solution:
Tyndall effect is due to Scattering of light.

Property	Lyophilic colloids	Lyophobic colloids

Tyndall effect	They do not show tyndall effect	They show tyndall effect.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2021 (27 Aug Shift 1)
Tyndall effect is observed when :

Options:
A The diameter of dispersed particles is much larger than the wavelength of light used.
B The diameter of dispersed particles is much smaller than the wavelength of light used
C The refractive index od dispersed phase is greater than that of the dispersion medium.
D The diameter of dispersed particles is similar to the wavelength of light used.

The correct answer is:
The diameter of dispersed particles is similar to the wavelength of light used.

Solution:
The diameter of the dispersed particles is not much smaller than the wavelength of the lights used. The intensity of scattered light depends on the difference between the refractive indice of the D.P and D.M., In lyophobic colloids, this difference is appreciable and therefore the tyndal effect is quite we ll defined but in lyophilic sols the difference is very small and the tyndal effect is very weak. So, to show Tyndall effect the refractive indices of the dispersed phase and dispersion medium differ greatly in magnitude.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2020 (03 Sep Shift 1)
Type of amino acids obtained by hydrolysis of proteins is :

Options:
A β
B α
C δ
D γ

The correct answer is:
α

Solution:
Proteins are natural polymers composed of α-amino acids which are connected by peptide linkages.

The smallest unit to which the protein can be broken down is called amino acids. The α-amino acids are the amino acids whose one carbon contains both the amino group NH2 and the carboxyl group COOH. This type of carbon is called α-carbon. Thus, the amino acid that contains an α-carbon is called an α-amino acid.

Proteins upon acidic hydrolysis produce α-amino acids.

Hence, the answer is option B.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2024 (29 Jan Shift 1)
Type of isomerism which exists between
$\left[\mathrm{Pd}\left(\mathrm{C}_6 \mathrm{H}_5\right)_2(\mathrm{SCN})_2\right]$ and $\left[\mathrm{Pd}\left(\mathrm{C}_6 \mathrm{H}_5\right)_2(\mathrm{NCS})_2\right]$ is :

Options:
A
Linkage isomerism

B
Coordination isomerism

C
Ionisation isomerism

D
Solvate isomerism

The correct answer is:
Linkage isomerism

Solution:
The compound shows linkage isomerism because the ligand in the compound is an ambidenate ligand that can bond at more than one atomic site.
i.e., $: \mathrm{NCS}^{-}$and $: \mathrm{SCN}^{-}$

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2013 (09 Apr Online)
Type(s) of isomerism shown by Co en2Cl2]Br :-

Options:
A Geometrical
B Optical
C Ionisation
D All of these

The correct answer is:
All of these

Solution:
[Co(en)2Cl2]Br→ Ionisotion [Co(en)2ClBr]Cl
cis-isomer is optically active.

Topic: Chemistry, Coordination Compounds, JEE Main
Types of isomerism shown by	$\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_5 \mathrm{NO}_2\right] \mathrm{Cl}_2$ is

Options:
A Optical
B Ionisation
C Geometrical
D Linkage

The correct answer is:
Linkage

Solution:
$\mathrm{NO}_2$ is ambident and can be linked either to $\mathrm{N}$-side as $\left(-\mathrm{NO}_2\right)$ or to $\mathrm{O}$-side as $(-\mathrm{ONO})$.

Topic: Chemistry, Coordination Compounds, JEE Main
Typhoid is caused by a

Options:
A Gram positive, flagellated bacterium
B Gram positive, non flagellated bacterium
C Gram negative, flagellated bacterium
D Gram negative, non-flagellated bacterium

The correct answer is:
Gram negative, flagellated bacterium

Solution:
Typhoid is mainly caused by bacterium Salmonella typhimurium or Salmonella typhi. Other species causing typhoid are Salmonella paratyphi A & Salmonella paratyphi B. They are Gram negative bacteria. They cant retain Grams stain in the presence of alcohol. They are rod shaped bacteria measuring 2 - 4 µ in length and 0.5 - 0.8 µ in width. They are motile and possess multiple flagella (Peritrichous flagella). They are facultative anaerobes.

Topic: Biology, Human Health and Diseases, NEET
Typhoid is caused by

Options:
A pneumonia
B filarial worm
C Salmonella typhii
D None of the above

The correct answer is:
Salmonella typhii

Solution:
Typhoid or Enteric fever is a bacteria borne disease caused by bacillus Salmonella typhii. It spreads through consumption of contaminated food and water. Main symptoms are high fever, headache, vomiting, loose stools, etc.
Pneumonia is caused by three different strains of bacterium named Streptococcus pneumoniae, Pneumococcus pneumoniae and Diplococcus pneumoniae. Filarial worm cause filarial disease which is also called elephantiasis. The scientific name of worm is Wuchereria bancroft.

Topic: Biology, Human Health and Diseases, JIPMER, JIPMER 2019
Typical growth curve in plants is:

Options:
A Sigmoid
B Linear
C Stair-steps shaped
D Parabolic

The correct answer is:
Sigmoid

Solution:
The growth pattern of cells, organisms is uniform under favourable conditions & a typical growth curve in plants is sigmoidal. Thus following phases of growth are recognized.
(i) Lag phase: In the lag period the growth is slow. It represents the formative or cell division phase.
(ii) Log phase: Also called as exponential phase. During this phase, growth is maximum & most rapid. It represents the cell elongation phase.
(iii) Steady or stationary phase: It represents the cell maturation phase.
Time taken in growth phases (mainly log phase) is called a grand period of growth.

Topic: Biology, Plant - Growth and Development, NEET, NEET 2015 (Phase 1)
Typical growth curve in plants is:

Options:
A Sigmoid
B Linear
C Stair-steps shaped
D Parabolic

The correct answer is:
Sigmoid

Solution:
Growth pattern of cell, organisms is uniform under favourable conditions & a typical growth curve in plants is sigmoidal. Thus following phases of growth are recognized.
(i) Lag phase: In lag period the growth is slow. It represents formative or cell division phase.
(ii) Log phase: Also called as exponential phase. During this phase growth is maximum & most rapid. It represents cell elongation phase.
(iii) Steady or stationary phase: It represents cell maturation phase.
Time taken in growth phases (mainly log phase) is called as grand period of growth.

Topic: Biology, Plant - Growth and Development, NEET
Tyre of a bicycle has volume $2 \times 10^{-3} \mathrm{~m}^3$. Initially, the tube is filled $75 \%$ of its volume by air at atmospheric pressure $10^5 \mathrm{Nm}^{-2}$. When a rider is on the bicycle, the area of contact of tyre with road is $24 \times 10^{-4} \mathrm{~m}^2$. The mass of rider with bicycle is $120 \mathrm{~kg}$. If a pump delivers a volume $500 \mathrm{~cm}^3$ of air in each stroke, then the number of strokes required to inflate the tyre is $\left(g=10 \mathrm{~ms}^{-2}\right)$

Options:
A 10
B 11
C 21
D 20

The correct answer is:
21

Solution:
Pressure against which pump has to deliver air
$=$ Pressure due to weight of rider and cycle
+ Atmospheric pressure initially inside tyre
$$
=\frac{F}{A}+p_{\mathrm{atm}}=\frac{120 \times 10}{24 \times 10^{-4}}+1 \times 10^5=6 \times 10^5 \frac{\mathrm{N}}{\mathrm{m}^2}
$$
Number of moles of air in the tube
$$
n_1=\frac{\mathrm{pV}}{\mathrm{RT}}=\frac{6 \times 10^5 \times 2 \times 10^{-3}}{\mathrm{RT}}
$$
Volume of these moles at atmospheric pressure is
$$
\begin{aligned}
V_1 & =\frac{n R T}{p_{\text {atm }}}=\frac{6 \times 10^5 \times 2 \times 10^{-3}}{1 \times 10^5} \\
& =12 \times 10^{-3} \mathrm{~m}^3
\end{aligned}
$$
Initial volume of air inside tyre
$$
V_0=\frac{75}{100} \times 2 \times 10^{-3}=1.5 \times 10^{-3} \mathrm{~m}^3
$$
So, to inflate the tube volume to be pumped in is
$$
\begin{aligned}
V_2 & =V_1-V_0=(12-1.5) \times 10^{-3} \\
& =10.5 \times 10^{-3} \mathrm{~m}^3
\end{aligned}
$$
Hence, number of strokes of pump required
$$
=N=\frac{V_2}{V_{\text {pump }}}=\frac{10.5 \times 10^{-3}}{500 \times 10^{-6}}=21
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
ΔU is equal to:

Options:
A Isobaric work
B Adiabatic work
C Isothermal work
D Isochoric work

The correct answer is:
Adiabatic work

Solution:
From the 1st law of Thermodynamics
ÎU=q+w
For adiabatic process
q=0,
So, ÎU=W

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2017 (02 Apr)
Ultraviolet light of wavelength 300 nm and intensity 1.0 W m-2 falls on the surface of a photosensitive material. If one percent of the incident photons produce photoelectrons, then the number of photoelectrons emitted from an area of 1.0 cm2 of the surface is nearly

Options:
A 9.61Ã1014 s-1
B 4.12Ã1013 s-1
C 1.51Ã1012 s-1
D 2.13Ã1011 s-1

The correct answer is:
1.51Ã1012 s-1

Solution:
Energy incident over 1 cm2=1.0×10-4 J;
Energy of photon E= hc λ
Total no of photons falling
= 1.0× 10 −4 hc λ = 10 −4 ×300× 10 −9 6.6× 10 −34 ×3× 10 8
No of photoelectrons emitted per second
= 1 100 × 10 −4 ×300× 10 −9 6.6× 10 −34 ×3× 10 8
=1.51× 10 12

Topic: Physics, Dual Nature of Matter, JEE Main
Uncertainity in position of a minute particle of mass $25 \mathrm{~g}$ in space is $10^{-5} \mathrm{~m}$. What is the uncertainity in its velocity (in $\left.\mathrm{ms}^{-1}\right) ?\left(\mathrm{~h}=6.6 \times 10^{-34} \mathrm{Js}\right)$

Options:
A
$2.1 \times 10^{-34}$

B
$0.5 \times 10^{-34}$

C
$2.1 \times 10^{-28}$

D
$0.5 \times 10^{-23}$

The correct answer is:
$2.1 \times 10^{-28}$

Solution:
$\Delta \mathrm{x} \cdot \Delta \mathrm{v}=\frac{\mathrm{h}}{2 \pi \mathrm{m}}$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2002