datavorous/entrance-exam-dataset huggingface dataset:
Calculate the standard cell potential (in V) of the cell in which the following reaction takes place:

Fe2+ aq+Ag+aq→Fe3+aq+Ag(s)

Given that
EAg+/Ago=x V
EFe2+/Feo=y V
EFe3+/Feo=z V

Options:
A x+2y-3z
B x-y
C x-z
D x+y-z

The correct answer is:
x+2y-3z

Solution:
Fe+2A+Ag+C→Fe3++Ag

Ecello=SRP of cathode - SRP of anode
=EAg+|Ago-EFe+3|Fe+2o
=x-EFe3+|Fe+2o

Since E° cell is an intensive property, the E° cell values are not additive by nature. But the standard Gibbs free energy change is an extensive property and hence can be added.

(i) Fe+3→Fe+3e-→Z=E1o;  ΔG1=-3×F×z

(ii) Fe+2→Fe+2e- Y=E2o; ΔG2=-2×F×y

Subtracting (ii) from (i)
Fe+3→Fe+2 ΔG3=ΔG1-ΔG2

Eo=-2y+3z

Ecello=x--2y+3z

=x+2y-3z

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2019 (08 Apr Shift 2)
Calculate the standard cell potentials of galvanic cell in which the following reactions take place \( 2 \mathrm{Cr}(\mathrm{s})+3 \mathrm{Cd}^{2+}(\mathrm{aq}) \longrightarrow 2 \mathrm{Cr}^{3+}(\mathrm{aq})+3 \mathrm{Cd} \) Calculate the \( \Delta_{\mathrm{r}} \mathrm{G}^{0} \) and equilibrium constant for the reactions. \(\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^{\mathrm{o}}=-0.74 \mathrm{~V}, \mathrm{E}_{\mathrm{Cd}^{2+} / \mathrm{Cd}}^{\mathrm{o}}=-0.40 \mathrm{~V}\) The values of \( E_{\text {cell }}^{0}, \Delta_{r} G^{0} \) and \( \mathrm{K}_{\mathrm{C}} \) may be

Options:
A \( \begin{array}{ccc}E_{\text {cell }}^{0} & \Delta_{r} G^{0} & K_{c} \\ 0 \cdot 34 \mathrm{~V} & -196 \cdot 86 \mathrm{~kJ} & 3 \cdot 17 \times 10^{34}\end{array} \)
B \( \begin{array}{ccc}E_{\text {cell }}^{0} & \Delta_{r} G^{0} & K_{c} \\ 0 \cdot 34 \mathrm{~V} & -146 \cdot 86 \mathrm{~kJ} & 3 \cdot 17 \times 10^{34}\end{array} \)
C \( \begin{array}{ccc}E_{\text {cell }}^{0} & \Delta_{r} G^{0} & K_{c} \\ 0 \cdot 34 \mathrm{~V} & -106 \cdot 86 \mathrm{~kJ} & 1 \cdot 17 \times 10^{34}\end{array} \)
D None of these

The correct answer is:
\( \begin{array}{ccc}E_{\text {cell }}^{0} & \Delta_{r} G^{0} & K_{c} \\ 0 \cdot 34 \mathrm{~V} & -196 \cdot 86 \mathrm{~kJ} & 3 \cdot 17 \times 10^{34}\end{array} \)

Solution:
Given ; 2Crs+3Cd2+aq⟶2Cr3+aq+3Cd

E Cr 3 + / Cr o = - 0 . 7 4 V; E Cd 2 + / Cd o = - 0 . 4 0 V
(a) Calculation of standard cell potential (Eo)
Ecello=Ecathodeo-EanodeoEcello = (-0.40)-(-0.74) VEcello = -0.40 + 0.74V = 0.34V
Ecello=0.34 V
(b) Calculation of Δ r G o .

We know that ,
ΔrGo=-nFEcello

Here the value of n = 6

thus ,ΔrGo=- 6 × 96500 × 0.34 J mol-1ΔrGo = -196860 J mol-1ΔrGo = -196.86 kJmol-1
(c) Calculation of equilibrium constant (Kc)
ΔrGo=-2.303 RT log Kc
log K c = - Δ r G o 2 . 3 0 3 RT
log Kc=-(-196860)2·303×8·314×298log Kc = 34.50Kc= Antilog (34.501) =3.17 × 1034

Topic: Chemistry, Electrochemistry, JEE Main
Calculate the standard electrode potential Eo for

Fe++++e-→Fe++

From the given standard electrode potentials value of two half cells

Fe+++2e-→Fe ; Eo=-0.440 V

Fe++++3e-→Fe ; Eo=-0.036 V

Options:
A + 0.772 V
B – 0.404 V
C + 0.404 V
D – 0.476 V

The correct answer is:
+ 0.772 V

Solution:
ÎGo=-nEoF
Fe2++2e-âFe .....(i)
ÎGo=-2ÃFÃ(-0.440âV)=0.880âF
Fe3++3e-âFe .....(ii)
ΔG o =−3×F× −0.036 =0.108 F
On subtracting equation (i) from (ii)
Fe3++e-âFe2+
ΔG o =0.108 F−0.880 F=−0.772 F
Eo for the reaction =-ÎGonF=-(-â0.772âF)1ÃF=+â0.772âV

Topic: Chemistry, Electrochemistry, JEE Main
Calculate the time interval between 33% decay and 67% decay if half-life of a substance is 20 min.

Options:
A 60 min
B 20 min
C 40 min
D 13 min

The correct answer is:
20 min

Solution:
N1=N0e-λt1

N1 N0=e-λt1

0.67=e-λt1

ln0.67=-λt1

N2=N0e-λt2

N2 N0=e-λt2

0.33=e-λt2

ln0.33=-λt2

ln0.67-ln0.33=λt1-λt2

λt1-t2=ln0.670.33

λt1-t2≅ln2

t1-t2≃ln2λ=t1/2

Half life =t1/2=20 minutes.

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Calculate the time required for complete decomposition of one mole of water using 2 ampere current. .

Options:
A 9650 sec
B 96500 sec
C 1.93 ×  10 5  sec
D 38600 sec

The correct answer is:
96500 sec

Solution:
2H2Oâ2H2+O2
One mole of H 2 O need 2 moles of e â s for electrolysis.
So, Qâ=âIâÃât
⇒t =  Q I
= 2Ã96500 2
â´t=96500âsec

Topic: Chemistry, Electrochemistry, JEE Main
Calculate the uncertainty in velocity if the uncertainty in the position of a moving bullet of mass 10 gm is 10-5m .

Options:
A 5.2×10-28 m/sec
B 3.0×10-28 m/sec
C 5.2×10-22 m/sec
D 3×10-22 m/sec

The correct answer is:
5.2×10-28 m/sec

Solution:
Uncertainty of moving bullet velocity Δυ=h4π×m×Δυ=6.625×10-344×3.14×.01×10-5
=5.2×10-28 m/sec .

Topic: Chemistry, Structure of Atom, JEE Main
Calculate the value of equilibrium constant of the reaction at 227oC , if the ΔGo for the reaction X+Y⇌Z is - 4.606 kcal . (R=2.0 cal. mol-1K-1)

Options:
A 100
B 10
C 2
D 0.01

The correct answer is:
100

Solution:
ΔGo=- 2.303 RT logK
- 4.606=- 2.303×0.002×500 logK
log K=2, K=100 .

Topic: Chemistry, Chemical Equilibrium, JEE Main
Calculate the value of the mean free path λ for oxygen molecules at temperature 27°C and pressure 1.01×105 Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. k=1.38×10-23 J K-1

Options:
A 58 nm
B 32 nm
C 86 nm
D 102 nm

The correct answer is:
102 nm

Solution:
λ=RT2πd2 NAP

λ=8.314×3002×3.14×0.3×10-92×6.023×1023×1.01×105

λ=102 nm

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Calculate the volume of water(in mL) required to dissolve 0.1 g lead (II) chloride to get a saturated solution (Ksp of PbCl2 = 3.21 × 10-8, atomic mass of Pb = 207 u).Report your answer by rounding it up to nearest whole number.

Options:

The correct answer is:
180

Solution:
Suppose solubility of PbCl2 in water is s mol L-1.
PbCl 2 s â Pb 2 + (aq) + 2Cl - (aq) 1 - s s 2s
Ksp = [Pb2+] Â· [Cl-]2
Ksp = [s] [2s]2 = 4s3
3.2 Ã 10-8 = 4s3
s 3 = 3.2 Ã 10 - 8 4 = 0.8 Ã 10 - 8
s3 = 8.0 Ã 10-9
Solubility of PbCl2, = s = 2 Ã 10-3 mol L-1
Solubility of PbCl2 in gL-1 = 278 Ã 2 Ã 10-3 = 0.556 g L-1
(âµ Molar mass of PbCl2 = 207 + (2 Ã 35.5) = 278)
0.556 g of PbCl2 dissolve in 1 L of water.
â´ 0.1 g of PbCl 2 will dissolve in = 1 Ã 0.1 0.556 L of water
â= 0.1798 L = 179.8 mL

Topic: Chemistry, Ionic Equilibrium, JEE Main
Calculate the wavelength (in nanometer) associated with a proton moving at $1.0 \times 10^3 \mathrm{~ms}^{-1}$ (Mass of proton $=1.67 \times 10^{-27} \mathrm{~kg}$ and $\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}$ ):

Options:
A
$0.032 \mathrm{~nm}$

B
$0.40 \mathrm{~nm}$

C
$2.5 \mathrm{~nm}$

D
$14.0 \mathrm{~nm}$

The correct answer is:
$0.40 \mathrm{~nm}$

Solution:
$$
\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.63 \times 10^{-34}}{1.67 \times 10^{-27} \times 10^3} \equiv 0.40 \mathrm{~nm}
$$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2009
Calculate the wavelength (in nm) of the spectral line of Lyman series for $\mathrm{n}=2$. (Rydberg constant $=1.1 \times 10^7 \mathrm{~m}^{-1}$ )

Options:
A $121$
B $221$
C $321$
D $421$

The correct answer is:
$121$

Solution:
For Lymen $n_1=1$ and $n_2=2$
And wavelength associated with electron jumping from any state to the first state according to Lyman series and it can be expressed as
$\frac{1}{\lambda}=R\left(\frac{1}{1}-\frac{1}{2^2}\right)$
i.e. $\lambda=\frac{1}{R \times\left(\frac{4-1}{4}\right)}=\frac{1}{1.1 \times 10^7 \times \frac{3}{4}}=1.215 \times 10^{-7}=121 \mathrm{~nm}$

Topic: Physics, Atomic Physics, JEE Main
Calculate the wavelength of the $k_a$ line for $z=31$, when $a=5 \times 10^7 \mathrm{~Hz}^{1 / 2}$ for a characteristic X-ray spectrum.

Options:
A $1.33 Å$
B $1.33 \mathrm{~nm}$
C $133 \times 10^{-10} \mathrm{~m}$
D $133 \mathrm{~nm}$

The correct answer is:
$1.33 Å$

Solution:
Given, $z=31$ and $a=5 \times 10^7 \mathrm{~Hz}^{1 / 2}$
$\begin{aligned}
\sqrt{v} & =a(z-1) \\
\text { or } \quad v & =a^2(z-1)^2 \\
& =\left(5 \times 10^7\right)^2(31-1)^2 \\
& =25 \times 10^{14} \times 30 \times 30 \\
& =2.25 \times 10^{18} \mathrm{~Hz} \\
\therefore \quad & =\frac{c}{\lambda} \Rightarrow \lambda=\frac{c}{v} \\
& =\frac{3 \times 10^8 \mathrm{~ms}^{-1}}{2.25 \times 10^{18} \mathrm{~Hz}} \\
& =1.33 \times 10^{-10} \mathrm{~m} \\
& =1.33 Å
\end{aligned}$

Topic: Physics, Electromagnetic Waves, JEE Main
Calculate the work done when 2.5 mol of H2O vaporizes at 1.0 atm and 25°C . Assume the volume of liquid H2O is negligible compared to that of vapour. Given 1 L atm = 101.3 J and R=0.082 L atm mol-1K-1 .

Options:
A 6190 kJ
B 6.19 kJ
C 61.1 kJ
D 5.66 kJ

The correct answer is:
6.19 kJ

Solution:
Volume of water vapour at 1.0 atm and 298 K is given by
V=nRTP
=â 2.5âmolâÃâ0.082âLâatmâ mol â1 â K â1 âÃâ298âK 1âatm
=61.09 L
Now change volume
ÎV=Vfinal-Vinitial
=61.09âLâ0âL=61.09âL
So work done against constant pressure of 1 atm =PÎV
=1 atm × 61.09 L=61.09 L atm
=61.09 L atm× 101.3 J 1 L atm
=6.19 kJ

Topic: Chemistry, Thermodynamics (C), JEE Main
Calculate the work done when 112 g iron reacts with dilute HCl at 300 K. The reaction is carried out in an open container
(At mass of Fe = 56)

Options:
A 600 cal
B 300 cal
C 200 cal
D 1200 cal

The correct answer is:
1200 cal

Solution:
The number of moles of Fe taken
=11256=2
So the chemical reaction is
2Fes+4HClaq→2FeCl2aq+2H2g
New work done =PV2-V1
Since V2-V1=VH2
Hence volume of solids and liquids is taken as ≅0
So work done =P.VH2
also PVH2=nRT or VH2=nRTP
So work done =P.nRTP=nRT [ R=2 Cal]
=2×2×300=1200 Cal.

Topic: Chemistry, Thermodynamics (C), JEE Main
Calculate the work (in \( \mathrm{kJ} \) ) of an isobaric reversible expansion of three moles of an ideal gas while it is heated from \( 298 \mathrm{~K} \) to \( 400 \mathrm{~K} \).

Options:

The correct answer is:
-2.54

Solution:
For an isobaric reversible expansion of an ideal gas,

The work done, W=-Pext(V2 -V1)

But, Pext=P1=P2 for an isobaric process.

From the ideal gas equation,

V =nRTP

So, the work done W=-PnRT2P -nRT1P=-nR(T2 -T1)

Given: n=3; T1=298 K and T2=400 K; R=8.314 J K-1 mol-1

W=-3×8.314×400-298=-2544 J

W=-2.54 kJ

Topic: Chemistry, Thermodynamics (C), JEE Main
Calculate value of E°Ce4+/Ce3+ . If E°cell for the cell reaction, 2Ce4++Co→2Ce3++Co2+ is 1.89 V. If E°Co/Co2+ = -0.28 V

Options:
A – 1.64V
B + 1.64V
C – 2.08V
D + 2.17V

The correct answer is:
+ 1.64V

Solution:
In this cell Co is oxidised and it acts as anode and Ce acts as cathode.
ECell0=ECathode0-EAnode0=1.89=ECell0-(-0.28)
ECell0=1.89-0.28=1.61 Volts.

Topic: Chemistry, Electrochemistry, JEE Main
Callose thickening found in endothecium of anther wall is

Options:
A a polysaccharide made up of alpha glucose monomers
B a polysaccharide made up of beta glucose monomers
C a protein made up of mainly acidic amino acids
D a protein made up of mainly basic amino acids

The correct answer is:
a polysaccharide made up of beta glucose monomers

Solution:
Callose is polysaccharide having repeating units of beta glucose with β 1→3 glycosidic bonds with some β 1→6 branches. Callose is produced during pollen grain formation, cell plate formation and also as a stress response. Callose present in the endothecium helps in dispersion of pollen grains.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
Calomel $\left(\mathrm{Hg}_2 \mathrm{Cl}_2\right)$ on reaction with ammonium hydroxide gives

Options:
A
$\mathrm{HgNH}_2 \mathrm{Cl}$

B
$\mathrm{NH}_2-\mathrm{Hg}-\mathrm{Hg}-\mathrm{Cl}$

C
$\mathrm{Hg}_2 \mathrm{O}$

D
$\mathrm{HgO}$

The correct answer is:
$\mathrm{HgNH}_2 \mathrm{Cl}$

Solution:
$\mathrm{Hg} \mathrm{Cl}+2 \mathrm{NHOH} \longrightarrow \mathrm{Hg}+\mathrm{Hg}\left(\mathrm{NH}_2\right) \mathrm{Cl}+\mathrm{NH}_4 \mathrm{Cl}+2 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Redox Reactions, JEE Main
Calomel $\left(\mathrm{Hg}_2 \mathrm{Cl}_2\right)$ on reaction with ammonium hydroxide gives

Options:
A
$\mathrm{HgNH}_2 \mathrm{Cl}$

B
$\mathrm{NH}_2-\mathrm{Hg}-\mathrm{Hg}-\mathrm{Cl}$

C
$\mathrm{Hg}_2 \mathrm{O}$

D
$\mathrm{HgO}$

The correct answer is:
$\mathrm{HgNH}_2 \mathrm{Cl}$

Solution:
$\mathrm{Hg} \mathrm{Cl}+2 \mathrm{NHOH} \longrightarrow \mathrm{Hg}+\mathrm{Hg}\left(\mathrm{NH}_2\right) \mathrm{Cl}+\mathrm{NH}_4 \mathrm{Cl}+2 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2005
Calorific value of producer gas is low because of

Options:
A high percent of $\mathrm{N}_2$
B low percent of $\mathrm{CO}_2$
C high percent of CO
D low percent of $\mathrm{N}_2$

The correct answer is:
high percent of $\mathrm{N}_2$

Solution:
Producer gas is a mixture of $\mathrm{CO}+\mathrm{N}_2$. Its calorific value is low due to high percentage of nitrogen.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Calvin cycle operates in

Options:
A C3 plants, C4 plants, CAM plants
B C3 plants only
C C4 plants, CAM plants only
D C3 and C4 plants only

The correct answer is:
C3 plants, C4 plants, CAM plants

Solution:
The second phase of the photosynthesis process is the light independent reaction where glucose is produced by fixing CO2 . This process occurs in the stroma of the chloroplast and utilizes the products of the light reaction. CO2 fixation is brought about by Calvin cycle or C3 pathway. This cycle involves three mains steps of carboxylation, reduction and regeneration. Even plants with adaptation for glucose synthesis make use of Calvin cycle. C4 plants and CAM plants also run Calvin cycle for CO2 fixation with little variation. In C4, 4C compound intermediate forms before splitting and entering the Calvin cycle. In CAM, plants gather sunlight in day time and run the Calvin cycle in night time to avoid water loss.

Topic: Biology, Photosynthesis in Higher Plants, NEET
CAM helps the plants in

Options:
A secondary growth
B disease resistance
C reproduction
D conserving water

The correct answer is:
conserving water

Solution:
CAM plants are mostly succulent xerophytes. The stomata in these plants remain closed during the day. This helps to check the transpiration. In this way, water is conserved.

Topic: Biology, Transport in Plants, NEET, NEET 2011 (Screening)
Can a system be heated and its temperature remains constant?

Options:

None

Solution:
As given that, $\Delta T=0$
We know that, $\Delta U=0$
$$
\Delta Q=\Delta U+\Delta W \Rightarrow \Delta Q=\Delta W
$$
Hence, it is possible when the entire heat supplied to the system is utilised in isothermal expansion system i.e., its working against the surroundings.

Topic: Physics, Thermodynamics
Can infected females be asymptomatic in the early stages of an STI?

Options:
A No
B Yes
C It depends on the type of STI
D It depends on the stage of the infection

The correct answer is:
Yes

Solution:
Infected females may often be asymptomatic in the early stages of an STI and hence, may remain undetected for a long time.

Topic: Biology, Reproductive Health, NEET
Candela is the unit of

Options:
A Electric intensity
B Luminous intensity
C Sound intensity
D None of these

The correct answer is:
Luminous intensity

Solution:
The correct option is Luminous intensity.
The Candela is the base unit of luminous intensity in the International System of Units. Luminous intensity emitted by a point light source in a particular direction per unit solid angle is expressed in terms of Candela. It is one of the $7$ $\mathrm{SI}$ units.

Topic: Physics, Units and Dimensions, JEE Main
Cannizaro's reaction is not given by :

Options:
A

B

C
$\mathrm{CH}_3 \mathrm{CHO}$

D
$\mathrm{HCHO}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CHO}$

Solution:
Only those aldehydes which do not have $\alpha-\mathrm{H}$ atom undergo Cannizaro's reaction.
Hence $\mathrm{CH}_3 \mathrm{CHO}$ will not undergo Cannizaro's reaction as it has $3 \alpha$ $\mathrm{H}$ atoms.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
Cannizaro's reaction is not given by :

Options:
A

B

C
$\mathrm{CH}_3 \mathrm{CHO}$

D
$\mathrm{HCHO}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CHO}$

Solution:
Only those aldehydes which do not have $\alpha-\mathrm{H}$ atom undergo Cannizaro's reaction.
Hence $\mathrm{CH}_3 \mathrm{CHO}$ will not undergo Cannizaro's reaction as it has $3 \alpha$ $\mathrm{H}$ atoms.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2013 (25 Apr Online)
cannot be prepared by:

Options:
A HCHO+PhCHCH3CH2MgX
B PhCOCH2CH3+CH3MgX
C CH3CH2COCH3+PhMgX
D PhCOCH3+CH3CH2MgX

The correct answer is:
HCHO+PhCHCH3CH2MgX

Solution:
The given compound is tertiary alcohol and formaldehyde forms secondary alcohol on reaction with Grignard reagent.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2019 (12 Jan Shift 1)
CaO and NaCl have the same crystal structure and nearly the same ionic radii. If 'u' is the lattice energy of NaCl, the lattice energy of CaO is very nearly

Options:
A 4u
B 2u
C u
D u/2

The correct answer is:
4u

Solution:
Lattice energy ∝Product of charges on ionsInterionic distance
In CaO, the charges on cation and anion are twice the magnitude of charges on the cation and anion of NaCl. The interionic distance is the same.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
$\mathrm{CaOCl}_2+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Ca}\left(\mathrm{OH}_2+X\right.$
$\begin{aligned} X+\mathrm{CH}_3 \mathrm{CHO} & \longrightarrow Y \\ Y+\mathrm{Ca}(\mathrm{OH})_2 & \longrightarrow \mathrm{CHCl}_3 .\end{aligned}$
What is ' $Y$ '?

Options:
A $\mathrm{CH}_3 \mathrm{CH}(\mathrm{OH})_2$
B $\mathrm{CH}_2 \mathrm{Cl}_2$
C $\mathrm{CCl}_3 \mathrm{CHO}$
D $\mathrm{CCl}_3 \mathrm{COCH}_3$

The correct answer is:
$\mathrm{CCl}_3 \mathrm{CHO}$

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Capacitance (in $F$ ) of a spherical conductor with radius $1 \mathrm{~m}$ is

Options:
A
$1.1 \times 10^{-10}$

B
$10^{-6}$

C
$9 \times 10^{-9}$

D
$10^{-3}$

The correct answer is:
$1.1 \times 10^{-10}$

Solution:
$4 \pi \in_0 R=1.1 \times 10^{-10}$

Topic: Physics, Capacitance, JEE Main, JEE Main 2002
Capacitance of a capacitor made by a thin metal foil is 2 μF.If the foil is folded with paper of thickness 0.15 mm, dielectric constant of paper is 2.5 and width of paper is 400 mm,the length of foil will be

Options:
A 0.34 m
B 1.33 m
C 13.4 m
D 33.9 m

The correct answer is:
33.9 m

Solution:
If length of the foil is l then
C = K ε 0 · l × b d
⇒ 2 × 1 0 - 6=2 · 5 × 8 · 8 5 × 1 0 - 1 2 × lâ¡ × 4 0 0 × 1 0 - 3 0 · 1 5 × 1 0 - 3
⇒ l = 3 3 · 9 m

Topic: Physics, Electrostatics, JEE Main
Capacitance of an isolated conducting sphere of radius R1 becomes n times when it is enclosed by a concentric conducting sphere of radius R2 connected to earth. The ratio of their radii R2R1 is:

Options:
A nn-1
B 2n2n+1
C n+1n
D 2n+1n

The correct answer is:
nn-1

Solution:
Capacitance of isolated conducting sphere C=4πε0R1

By enclosing inside another sphere of radius R2, new capacitance C'=4πε0R1R2R2-R1

Given: 4πε0R1R2R2-R1=n×4πε0R1

⇒R2R2-R1=n⇒R2R1R2R1-1=n

⇒R2R1=nR2R1-n⇒R2R1=nn-1

Topic: Physics, Capacitance, JEE Main
Capacitance of an isolated conducting sphere of radius R1 becomes n times when it is enclosed by a concentric conducting sphere of radius R2 connected to earth. The ratio of their radii R2R1 is:

Options:
A nn-1
B 2n2n+1
C n+1n
D 2n+1n

The correct answer is:
nn-1

Solution:
Capacitance of isolated conducting sphere C=4πε0R1

By enclosing inside another sphere of radius R2, new capacitance C'=4πε0R1R2R2-R1

Given: 4πε0R1R2R2-R1=n×4πε0R1

⇒R2R2-R1=n⇒R2R1R2R1-1=n

⇒R2R1=nR2R1-n⇒R2R1=nn-1

Topic: Physics, Capacitance, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Caprolactam when heated at high temperature in presence of water, gives

Options:
A Teflon
B Dacron
C Nylon 6, 6
D Nylon 6

The correct answer is:
Nylon 6

Solution:
Caprolactam is a cyclic amide. Caprolactam when heated at high temperature in presence of water, gives Nylon 6.

Topic: Chemistry, Polymers, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Car B overtakes another car A at a relative speed of 40 m s-1. How fast will the image of car B appear to move in the mirror of focal length 10 cm fitted in car A, when the car B is 1.9 m away from the car A?

Options:
A 0.1 m s-1
B 0.2 m s-1
C 40 m s-1
D 4 m s-1

The correct answer is:
0.1 m s-1

Solution:
Object distance, u=−1.9 m=190 cm

Focal length, f=10 cm

Magnification, m=ff−u=1010−(−190)

Therefore, m=10200=120

Image velocity, VI=−m2V0=−1202×40

Hence, VI=−0.1 m s-1

Magnitude of velocity =0.1 m s-1

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (26 Aug Shift 1)
Carbohydrates are commonly found as starch in plant storage organs. Which of the following five properties of starch (a - e) make it useful as a storage material?
a. Easily translocated
b. Chemical non-reactive
c. Easily digested by animals
d. Osmotically inactive
e. Synthesized during photosynthesis
The useful properties are

Options:
A Both a & e
B Both b & c
C Both b & d
D a, b, c

The correct answer is:
Both b & d

Solution:
Starch is nonreducing and water insoluble so is preferred as storage material.

Topic: Biology, Biomolecules (B), NEET, NEET 2008 (Mains)
Carbohydrates are stored in human body as

Options:
A Glucose
B Glycogen
C Starch
D Fructose

The correct answer is:
Glycogen

Solution:
Glycogen is a multibranched polysaccharide of glucose that serves as a form of energy storage in animals and fungi. This polysaccharide structure represents the main storage form of glucose in the body.

Topic: Chemistry, Biomolecules, JEE Main
Carbon and silicon belong to the $14^{\text {th }}$ group. The maximum coordination number of carbon in commonly occurring compounds is 4 , whereas that of silicon is 6 . This is due to

Options:
A Large size of silicon
B More electropositive nature of silicon
C Availability of low lying $d$-orbitals in silicon
D Both (a) and (b)

The correct answer is:
Availability of low lying $d$-orbitals in silicon

Solution:
Silicon has a coordination number of six owing to the presence of low energy d-orbitals which can also participate in bonding.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Carbon and silicon belong to the $14^{\text {th }}$ group.The maximum coordination number of carbon in commonly occurring compounds is 4 , whereas that of silicon is 6 . This is due to

Options:
A Availability of low lying $d$-orbitals in silicon
B Large size of silicon
C More electropositive nature of silicon
D Both (a) and (b)

The correct answer is:
Availability of low lying $d$-orbitals in silicon

Solution:
Silicon has a coordination number of six owing to the presence of low energy d-orbitals which can also participate in bonding.

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
Carbon forms two oxides which have different compositions. The equivalent mass of which remains constant?

Options:
A Carbon
B Oxygen
C Neither carbon nor oxygen
D Both carbon and oxygen

The correct answer is:
Oxygen

Solution:
The equivalent mass of oxygen in $\mathrm{CO}$ and $\mathrm{CO}_{2}$ remains constant. While, that of carbon changes as it have different oxidation state in different eompound.

Topic: Chemistry, Redox Reactions, COMEDK, COMEDK 2020
Carbon monoxide gas is more dangerous than carbon dioxide gas. Why ?

Options:

None

Solution:
CO binds to haemoglobin for which it has 200 times more affinity than oxygen and forms carboxyhaemoglobin. In blood, when the concentration of carboxyhaemoglobin reaches \(3.4 \%\), the oxygen carrying capacity of the blood is greatly reduced. In other words, the body becomes oxygen-starved. This results into headache, nervousness, cardiovascular disorder, weak eye-sight etc. On the other hand, \(\mathrm{CO}_2\) does not combine with haemoglobin and hence is less harmful as pollutant. \(\mathrm{CO}_2\) is the main contributor towards green-house effect and global warming.

Topic: Chemistry, Environmental Chemistry