datavorous/entrance-exam-dataset huggingface dataset:
Fascicular, interfascicular and extra-stelar cambium together constitutes

Options:
A Ground meristem
B Apical meristem
C Intercalary meristem
D Lateral meristem

The correct answer is:
Lateral meristem

Solution:
Lateral meristem is the meristem which occurs on the sides and is helpful in increasing width of stem and root. They divide mainly in one plane (periclinal), increasing the diameter of an organ, e.g., cambium (fascicular and interfascicular cambium), extra stellar cambium, cork cambium and marginal meristem of some leaves.

Topic: Biology, Anatomy of Flowering Plants, NEET
Fastest distribution of some injectible material/medicine and with no risk of any kind can be achieved by injecting it into the

Options:
A muscles
B arteries
C veins
D lymph vessels

The correct answer is:
veins

Solution:
Intravenous therapy or IV therapy is providing the liquid substances (or medicine) through drip, directly into the vein. The word intravenous simply means 'within a vein'. Compared with other routes of administration, the intravenous root is the fastest way to deliver fluids and medications throughout the body. Some medications, as well as blood transfusions and lethal injections, can only be given intravenously.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2010 (Mains)
FeCN63- should be an inner orbital complex. Ignoring the pairing energy, the value of crystal field stabilization energy for this complex is -_____Δ0.

Options:

The correct answer is:
2

Solution:
FeCN63- In this the central metal Fe is in +3 oxidation state. And CN- is strong field ligand so, it will form inner orbital complex.

Electronic configuration in Fe3+: d5

Electronic configuration for inner orbital complex: t52g e0g

CFSE=5-0.4Δ0=-2.0Δ0

So, final answer will be =2

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (29 Jul Shift 1)
FeCN64-

FeCN63-

TiCN63-

NiCN42-

CoCN63-

Among the given complexes, number of paramagnetic complexes is

Options:

The correct answer is:
2

Solution:
Complexes in which central atom has unpaired electrons is said to be paramagnetic in nature and where unpaired electrons are not present are diamagnetic in nature.

FeCN63- , Fe+3 3d5 → t2g2,2 eg1, paramagnetic (one unpaired electron)

FeCN64-, Fe+2 3d6→t2g6eg0 ,diamagnetic

TiCN63-, Ti+3(3d1)→t2g1 , Paramagnetic (1 unpaired electron)

NiCN42-, Ni+2(3d8)→ Diamagnetic

CoCN63-, Co+3(3d6)→t2g6, diamagnetic

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2022 (28 Jun Shift 2)
$\left[\mathrm{Fe}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\right]^{2+}+$ en $\rightarrow$ complex $(\mathrm{X})$. The correct statement about the complex (X) is -

Options:
A it is a low spin complex
B it is diamagnetic
C it shows geometrical isomerism
D (a) and (b) both

The correct answer is:
(a) and (b) both

Solution:
Complex $\mathrm{X}$ is $\left[\mathrm{Fe}(\mathrm{en})_{3} \mathrm{right}\right]^{2+} ;$ as 'en' is a strong field ligand pairing of electrons will take place. $\left[\mathrm{Fe}(\mathrm{en})_{3}\right]^{2+}$ :

$d^{2} s p^{3}$ hybridisation

Hence, hybridisation is $d^{2} s p^{3}$ and complex is diamagnetic. As it has 3 bidentate symmetrical 'en' ligands so it will not show geometrical isomerism.

Topic: Chemistry, Coordination Compounds, JEE Main
$\mathrm{Fe}(\mathrm{OH})_3$ can be separated from $\mathrm{Al}(\mathrm{OH})_3$ by addition of

Options:
A Dil. $\mathrm{HCl}$
B $\mathrm{NaCl}$ solution
C $\mathrm{NaOH}$ solution
D $\mathrm{NH}_4 \mathrm{Cl}$ and $\mathrm{NH}_4 \mathrm{OH}$

The correct answer is:
$\mathrm{NaOH}$ solution

Solution:
As $\mathrm{Fe}(\mathrm{OH})_3$ is soluble in $\mathrm{NaOH}$ whereas $\mathrm{Al}(\mathrm{OH})_3$ is not.

Topic: Chemistry, Practical Chemistry, JEE Main
Fehling solution is

Options:
A Ammoniacal cuprous chloride solution
B Acidified copper sulphate solution
C Copper sulphate and sodium hydroxide + Rochelle salt
D None of these

The correct answer is:
Copper sulphate and sodium hydroxide + Rochelle salt

Solution:
$\mathrm{CuSO}_4+\mathrm{NaOH}+$ Roschelle salt( $\mathrm{Na}-\mathrm{K}$ tartarate $)$

Topic: Chemistry, Aldehydes and Ketones, JEE Main
Female cone of Pinus is a

Options:
A modified needles
B modified long shoot
C modified dwarf shoot
D modified scale

The correct answer is:
modified long shoot

Solution:
Female cone of Pinus is a modified long. It occurs on the main stem singly or in pairs. It is homologous to inflorescence of angiosperms. The female cone consists of numerous megasporophylls or female flowers or carpels, which are arranged around the cone axis in a spiral manner. Each megasporophyll is a modified dwarf shoot since it arises in the axil of scaly leaf which is now termed as bract scale. Hence, option B is correct.

Topic: Biology, Plant Kingdom, JIPMER, JIPMER 2012
Ferric chloride is applied to stop bleeding because

Options:
A Blood absorbs FeCl3 and forms a complex.
B Cl-ions cause coagulation of blood.
C Fe3+ ions coagulate blood which is a negatively charged sol.
D FeCl3 reacts with the constituents of blood which is a positively charged sol.

The correct answer is:
Fe3+ ions coagulate blood which is a negatively charged sol.

Solution:
Ferric chloride is a hemostatic agent, which means that it can be used to stop bleeding by promoting the coagulation of blood. Blood is a colloidal solution of an albuminoid substance and can be coagulated to form clot by applying Fe3+ ion on the open wound. Thus, bleeding caused by a razor during shaving can be treated by rubbing potash alum against the cut which helps in the coagulation of the blood by sealing the blood vessels.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Ferric ion forms a prussian blue precipitate due to the formation of

Options:
A $\mathrm{K}_{4}\left[\mathrm{Fe}\left(\mathrm{CN})_{6}\right]\right.$
B $\mathrm{K}_{3}\left[\mathrm{Fe}\left(\mathrm{CN})_{6}\right)\right.$
C $Fe(CNS)_{3}$
D $Fe_{4}[Fe(CN)_{6}]_{3}$

The correct answer is:
$Fe_{4}[Fe(CN)_{6}]_{3}$

Solution:
Ferric ion forms a prussian blue precipitate due to the formation of $\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}$. This complex is formed during the determine action of presence of nitrogen in the given sample. In this method, to portion of sodium fusion extract. freshly prepared ferrous sulphate, FeSO $_{4}$ solution is added and warmed. Then about 2 to 3 drops of $\mathrm{FeCl}_{3}$ solution are added and acidified with conc. HCl. The appearance of a prussian blue colour indicate the presence of nitrogen.
$$
\mathrm{FeSO}_{4}+2 \mathrm{NaOH} \longrightarrow \mathrm{Fe}(\mathrm{OH})_{2}+\mathrm{Na}_{2} \mathrm{SO}_{4}
$$
$6 \mathrm{NaCN}+\mathrm{Fe}(\mathrm{OH})_{2} \longrightarrow \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+2 \mathrm{NaOH}$
$$
\begin{array}{r}
3 \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]+4 \mathrm{FeCl}_{3} \longrightarrow \\
\mathrm{Fe}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]_{3}+12 \mathrm{NaCl}
\end{array}
$$

Topic: Chemistry, Coordination Compounds, JEE Main
Ferrous ion changes to X ion on reacting with acidified hydrogen peroxide. The number of d-electrons present in X and its magnetic moment (in BM) are respectively

Options:
A 6 and 6.93
B 5 and 5.92
C 5 and 4.9
D 4 and 5.92

The correct answer is:
5 and 5.92

Solution:
Ferrous ion (Fe2+) changes to ferric ion Fe3+ on reacting with acidified H2O2 as.
2Fe2++H2O2+2H+â¶2Fe3++2H2O
Electronic configuration of Fe3+ = 1s22s22p63s23p63d5
Number of d-electrons = 5
Magnetic moment = n(n+2) = 5(5+2) = 5.92 BM

Topic: Chemistry, Structure of Atom, JEE Main
Ferrous ion charges to $X$ ion, on reacting with acidified hydrogen peroxide.
The number of $d$-electrons present in $X$ and its magnetic moment (in BM) are, respectively :

Options:
A 6 and 6.93
B 5 and 5.92
C 5 and 4.9
D 4 and 5.92

The correct answer is:
5 and 5.92

Solution:
Ferrous ion $\left(\mathrm{Fe}^{2+}\right)$ changes to ferric ion $\mathrm{Fe}^{3+}$ on reacting with acidified $\mathrm{H}_2 \mathrm{O}_2$ as
$2 \mathrm{~K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]+\mathrm{H}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O}_2 \longrightarrow$ $2 \mathrm{~K}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]+\mathrm{K}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}$
Electronic configuration of $\mathrm{Fe}^{3+}=$
$1 s^2, 2 s^2, 2 p^6, 3 s^2, 3 p^6, 3 d^5$
Number of $d$ electrons $=5$
Magnetic moment $=\sqrt{n(n+2)}$
$=\sqrt{5(5+2)}=5.92 \mathrm{BM}$

Topic: Chemistry, d and f Block Elements, JEE Main
Ferrous oxide has a cubic structure and each edge of the unit cell is $5.0 \mathrm{~A}$. Assuming density of the oxide as $4.0 \mathrm{~g}-\mathrm{cm}^{-3}$, then the number of $\mathrm{Fe}^{2+}$ and $\mathrm{O}^{2-}$ ions present in each unit cell will be

Options:
A Four $\mathrm{Fe}^{2+}$ and four $\mathrm{O}^{2-}$
B Two $\mathrm{Fe}^{2+}$ and four $\mathrm{O}^{2-}$
C Four $\mathrm{Fe}^{2+}$ and two $\mathrm{O}^{2-}$
D Three $\mathrm{Fe}^{2+}$ and three $\mathrm{O}^{2-}$

The correct answer is:
Four $\mathrm{Fe}^{2+}$ and four $\mathrm{O}^{2-}$

Solution:
Let the units of ferrous oxide in a unit cell $=n$, molecular weight of ferrous oxide $(\mathrm{FeO})=56+16=72 \mathrm{gmol}^{-1}$,
weight of $n$ units $=\frac{72 \times n}{6.023 \times 10^{23}}$
Volume of one unit $=(\text { length of corner })^3$
$=(5 Å)^3=125 \times 10^{-24} \mathrm{~cm}^3$
Density $=\frac{\text { wt. of cell }}{\text { volume }}, \quad 4.09=\frac{72 \times n}{6.023 \times 10^{23} \times 125 \times 10^{-24}}$
$n=\frac{3079.2 \times 10^{-1}}{72}=42.7 \times 10^{-1}=4.27 \approx 4$

Topic: Chemistry, Solid State, JEE Main
Ferrous sulphate heptahydrate is used to fortify foods with iron. The amount (in grams) of the salt required to achieve 10ppm of iron in 100kg of wheat is ____________.
Atomic weight: Fe=55.85;S=32.00;O=16.00

Options:

The correct answer is:
4.95

Solution:
Formula of ppm is as follow-

ppm=weight of soluteweight of solution×106

10=Mass of Fe(ing)100×1000×106
Or mass fe =1g
FeSO4.7H2OM=278
56g in 1 mole
1g-156mole
156×278g=4.96g

PPM is a concentration term and its full form is "parts per million" and it also can be expressed as milligrams per liter (mg/L). This measurement is the mass of a chemical or contaminate per unit volume of water.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Fex2 and Fey3 are known when x and y are :

Options:
A x=F,Cl,Br,I and y=F,Cl,Br
B x=F,Cl,Br and y=F,Cl,Br,I
C x=Cl,Br,I and y=F,Cl,Br,I
D x=F,Cl,Br,I and y=F,Cl,Br,I

The correct answer is:
x=F,Cl,Br,I and y=F,Cl,Br

Solution:
2FeI3Unstable⟶2FeI2Stable+I2

Due to strong reducing nature of I-

2Fe3++2I-⟶2Fe2++I2 remaining halides of Fe2+ and Fe3+ are stable.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2021 (16 Mar Shift 2)
Field inside a solenoid is

Options:
A Directly proportional to its length
B Directly proportional to current
C Inversely proportional to total number of turns
D Inversely proportional to current

The correct answer is:
Directly proportional to current

Solution:
$B=\mu_0 n i$

Topic: Physics, Magnetic Effects of Current, JEE Main
Fifteen coupons are numbered 1, 2, 3,…..,15 respectively. Seven coupons are selected at random one at a time with replacement. The probability, that the largest number appearing on selected coupons is at most 9, is

Options:
A 1 1 5 7
B 3 5 7
C 8 1 5 7
D 2 5 7

The correct answer is:
3 5 7

Solution:
Total coupons =15
1 â¤ selected coupon number â¤ 9
i.e., 1, 2,3, 4,5, 6,7, 8,9
â´ Probability of one selected coupon to have number â¤9 = 9 1 5 = 3 5
Hence, the required probability
= 3 5 Ã 3 5 Ã .... 7 times = 3 5 7 (Using multiplicative principle for independent events)

Topic: Mathematics, Probability, JEE Main
Fifth term of a G.P. is 2 , then the product of its 9 terms is

Options:
A $256$
B $512$
C $1024$
D None of these

The correct answer is:
$512$

Solution:
Let the 9 terms of a G.P. is
$\frac{a}{r^4}, \frac{a}{r^3}, \frac{a}{r^2}, \frac{a}{r}, a, a r, a r^2, a r^3, a r^4$
Given, fifth term $\quad a=2$
Hence product of 9 terms is $a^9=(2)^9=512$

Topic: Mathematics, Sequences and Series, JEE Main
Fifth term of a GP is 2, then the product of its 9 terms is

Options:
A
256

B
512

C
1024

D
none of these

The correct answer is:
512

Solution:
$\operatorname{ar}^4=2$
$\mathrm{a} \times \operatorname{ar} \times \mathrm{ar}^2 \times a r^3 \times \mathrm{ar}^4 \times \mathrm{ar}^5 \times a r^6 \times a r^7 \times a r^8$
$=a^9 r^{36}=\left(\operatorname{ar}^4\right)^9=2^9=512$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2002
Fig. represents vertical sections of four wings moving horizontally in air. In which case the force is upwards

Options:
A

B

C

D

The correct answer is:

Solution:
Correct option is 1.
When air moves faster the pressure of air decreases so the pressure on the top of the wing is less then the pressure of the bottom wing. The difference in pressure creates the force on the wing which lifts the wing upward.
So the only correct option is (1)

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Figure below shows a liquid being pushed out of the tube by a piston having area of cross section 2.0 cm2. The area of cross section at the outlet is 10 mm2. If the piston is pushed at a speed of 4 cm s-1, the speed of outgoing fluid is _____ cm s-1

Options:

The correct answer is:
80

Solution:
The given data is

A1=2×10-4 m2A2=10×10-6 m2v1=4 cm s-1

Using equation of continuity which is a conservation of mass principle,

A1v1=A2v2

⇒v2=A1v1A2=2×10-4×410×10-6

= 80 cm s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (10 Apr Shift 2)
Figure represents a crystal unit of cesium chloride, $\mathrm{CsCl}$ The cesium atoms, represented by open circles are situated at the corners of a cube of side $0.40 \mathrm{~nm}$, whereas a $\mathrm{Cl}$ atom is situated at the centre of the cube. The Cs atoms are deficient in one electron while the $\mathrm{Cl}$ atom carries an excess electron.
(i) What is the net electric field on the $\mathrm{Cl}$ atom due to eight Cs atoms?
(ii) Suppose that the Cs atom at the corner A is missing. What is the net force now on the $\mathrm{Cl}$ atom due to seven remaining Cs atoms?

Options:

None

Solution:
(i) Here, the distance of chlorine atom at the centre of the cube is equal distance from all the eight corners of cube where cesium atoms are placed. So, due to symmetry the forces due to all $\mathrm{Cs}$ ions, on $\mathrm{Cl}$ atom will cancel out.
$\begin{array}{ll}\text { Hence, } & E=\frac{F}{q} \text { where } F=0 \\ \text { So, } & E=0\end{array}$
(ii) The, net force on $\mathrm{Cl}$ atom at $\mathrm{A}$ would be.
$$
\mathrm{F}=\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}^2} \text {, }
$$
where, $r=$ distance between $\mathrm{Cl}$ ion and $\mathrm{Cs}$ ion Applying Pythagorous theorem then,
$$
\begin{aligned}
&r=\sqrt{(0.20)^2+(0.20)^2+(0.20)^2} \times 10^{-9} \mathrm{~m} \\
&r=0.346 \times 10^{-9} \mathrm{~m},
\end{aligned}
$$
We know that,
$$
\mathrm{e}=\left(1.6 \times 10^{-19}\right), \frac{1}{4 \pi \epsilon_0}=4 \times 10^9\left(\frac{\mathrm{N}-\mathrm{n}^2}{\mathrm{e}^2}\right)
$$
So, the force $\quad F=\frac{q^2}{4 \pi \varepsilon_0 r^2}=\frac{\mathrm{e}^2}{4 \pi \varepsilon_0 \mathrm{r}^2}$
$$
=\frac{9 \times 10^9\left(1.6 \times 10^{-19}\right)^2}{\left(0.346 \times 10^{-9}\right)^2}=1.92 \times 10^{-9} \mathrm{~N}
$$

Topic: Physics, Electrostatics
Figure shows a circuit in which three identical diodes are used. Each diode has forward resistance of $20 \Omega$ and infinite backward resistance. Resistors $\mathrm{R}_1=\mathrm{R}_2=\mathrm{R}_3=50 \Omega$. Battery voltage is $6 \mathrm{~V}$. The current through $\mathrm{R}_3$ is :

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

B
$100 \mathrm{~mA}$

C
$60 \mathrm{~mA}$

D
$25 \mathrm{~mA}$

The correct answer is:
$50 \mathrm{~mA}$

Solution:
Here, diodes $D_1$ and $D_2$ are forward biased and $\mathrm{D}_3$ is reverse biased. Therefore current through $\mathrm{R}_3$
$$
\mathrm{i}=\frac{\mathrm{V}}{\mathrm{R}^{\prime}}=\frac{6}{120}=\frac{1}{20} \mathrm{~A}=50 \mathrm{~mA}
$$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2013 (22 Apr Online)
Figure shows a DC voltage regulator circuit, with a Zener diode of breakdown voltage =6 V . If the unregulated input voltage varies between 10 V to 16 V, then what is the maximum Zener current?

Options:
A 1.5Â mA
B 3.5 mA
C 7.5Â mA
D 2.5Â mA

The correct answer is:
3.5 mA

Solution:
Current through 2 KΩ is
i = 102000 = 5 mA
Current through 4 KΩ
iL=i-iz=64000 =1.5 mA
⇒ iz=i-1.5

∴iz=5-1.5=3.5 mA

Topic: Physics, Semiconductors, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Figure shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with $1 \mathrm{~ms}^{-2}$. What is the net force on the man? If the coefficient of static friction between the man's shoes and the belt is $0.2$, up to what acceleration of the belt can the man continue to be stationary relative to the belt? (mass of the man $=65 \mathrm{~kg}$.)

Options:

None

Solution:
Here acceleration of conveyor belt $a=\mathrm{ms}^{-2}$, $\mu_s=0.2$ and mass of man $m=65 \mathrm{~kg}$. As the man is in an accelerating frame, he experiences a pseudo force $F_s=m a$ as shown in Fig. (a). Hence to maintain his equilibrium, he exerts a force $F=-F_s=m a=65 \times 1=65 \mathrm{~N}$ in forward direction i.e., direction of motion of belt.
$\therefore$ Net force acting on man $=65 \mathrm{~N}$ (forward)
As shown in Fig. (b), the man can continue to be stationary with respect to belt, it force of friction
$$
\mu_s \mathrm{~N}=\mu_s m g=m a_{\max }
$$
$a_{\max }=\mu_s . \quad g=0.2 \times 10=2 \mathrm{~ms}^{-2}$

Topic: Physics, Laws of Motion
Figure shows a network of capacitors where the number indicates capacitances in micro Farad. The value of capacitance C if the equivalent capacitance between point A and B is to be 1 μF is :

Options:
A 3223 Î¼F
B 3123Î¼F
C 3323Î¼F
D 3423Î¼F

The correct answer is:
3223 Î¼F

Solution:
Ceq of circuit =329 μF excluding C
Now 329 μFis in series with C so,
1Ceq=1C+932=1
⇒ C=3223 μF

Topic: Physics, Capacitance, JEE Main, JEE Main 2016 (10 Apr Online)
Figure shows a parallel plate capacitor with three dielectric slabs of same area \( \frac{A}{3} \) and thickness \( d . \)
Find \( K_{e q}=? \)

Options:
A \( 12 \)
B \( 36 \)
C \( 18 \)
D \( 24 \)

The correct answer is:
\( 12 \)

Solution:
C=∈0Ad
C1=K1∈0A3d=K1C3
C2=K2C3
C3=K3C3
Ceq=Keq=C3K1+K2+K3
Keq=K1+K2+K33=363=12

Topic: Physics, Capacitance, JEE Main
Figure shows a part of an electric circuit. The potentials at points a, b and c are 30 V, 12 V and 2 V respectively. The current through the 20 Ω resistor will be,

Options:
A 1.0 A
B 0.4 A
C 0.6 A
D 0.2 A

The correct answer is:
0.4 A

Solution:
Let the potential be x V at the junction.

By using V=IR we get,

30-x=10i1x-12=20i2x-2=30i3

By using Kirchhoff's law,

i1=i2+i3

⇒30-x10+12-x20+2-x30=0⇒90-3x+18-1.5x+2-x30=0

⇒x=20 V

The current through the 20 Ω resistor is

i2=30-1220 A

⇒i2=820A=0.4 A

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (06 Apr Shift 2)
Figure shows a potentiometer with a cell of $2.0 \mathrm{~V}$ and internal resistance $0.40 \Omega$ maintaining a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of $1.02 \mathrm{~V}$ (for very moderate currents upto a few $\mathrm{mA}$ ) gives a balance point at $67.3$ cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance
of $600 \mathrm{k} \Omega$ is put in series with it, which is shorted close to the balance point. The standard cell is then replaced by a cell of unknown emf $E$ and the balance point found similarly, turns out to be at $82.3 \mathrm{~cm}$ length of the wire.
(a) What is the value of $E$ ?
(b) What purpose does the high resistance of $600 \mathrm{k} \Omega$ have?
(c) Is the balance point affected by this high resistance?
(d) Is the balance point affected by the internal resistance of the driver cell?
(e) Would the method work in the above situation if the driver cell of the potentiometer had an emf of $1.0 \mathrm{~V}$ instead of $2.0 \mathrm{~V}$ ?
(f) Would the circuit work well for determining an extremely small emf, say of the order of a few mV (such as the typical emf of the thermo-couple)? If not, how will you modify the circuit?

Options:

None

Solution:
(a) By formula, $\frac{\mathrm{E}}{\mathrm{E}_{\text {standard }}}=\frac{l}{l_{\text {standard }}}$ $\Rightarrow \mathrm{E}=\frac{l \times \mathrm{E}_{\text {standard }}}{l_{\text {standard }}}=\frac{82.3 \times 1.02}{67.3}=1.2474 \mathrm{~V}$
(b) To reduce current through the galvanometer when the movable contact is far from the balance point.
(c) No.
(d) No.
(e) No. If E is greater than the emf of the drive cell of the potentiometer, there will be no balance point on the wire $A B$.
(f) The circuit, as it is, would be unsuitable because the balance point (for $\varepsilon$ of the order a few $\mathrm{mV}$ ) will be very close to the end $\mathrm{A}$ and the percentage error in measurement will be very large. The circuit is modified by putting a suitable resistor $\mathrm{R}$ in series with the wire $\mathrm{AB}$ so that potential drop across $\mathrm{AB}$ is only slightly greater than the emf to be measured. Then the balance point will be at larger length of the wire and the percentage error will be much smaller.

Topic: Physics, Current Electricity
Figure shows a ray of light entering and passing through a dense glass slab and emerging from the other side. If the angle $i=60^{\circ}$, slab thickness $b=0.04 \mathrm{~m}$ and the refractive index of glass $=\sqrt{3}$, the parallel shift $d$ between the emerging and entering rays in $\mathrm{mm}$ is

Options:
A $\sqrt{\frac{3}{4}}$
B $\sqrt{\frac{4}{3}}$
C $\frac{40}{\sqrt{3}}$
D $15 \sqrt{3}$

The correct answer is:
$\frac{40}{\sqrt{3}}$

Solution:
We have, $\sin r=\frac{\sin i}{\mu}=\frac{\sin 60^{\circ}}{\sqrt{3}}=\frac{1}{2}$
$\Rightarrow$ Angle of refraction, $r=30^{\circ}$
Now, lateral shift $d$ is given by
$$
\begin{aligned}
& d=t \sec (r) \cdot \sin (i-r) \\
& =0.04 \times \sec \left(30^{\circ}\right) \cdot \sin \left(60^{\circ}-30^{\circ}\right)=0.04 \times \tan 30^{\circ} \\
& =\frac{0.04}{\sqrt{3}} \mathrm{~m}=\frac{0.04 \times 1000}{\sqrt{3}} \mathrm{~mm}=\frac{40}{\sqrt{3}} \mathrm{~mm}
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main
Figure shows a rod AB, which is bent in a 120° circular arc of radius R. A charge (-Q) is uniformly distributed over rod AB. What is the electric filed E→ at the centre of curvature O ?

Options:
A 33Q8π2ε0R2 i^
B 33Q8πε0R2 i^
C -33Q8π2ε0R2 i^
D 33Q16π2ε0R2 i^

The correct answer is:
33Q8π2ε0R2 i^

Solution:
δ=2kλRsinθ2i^

=2kRQ2Rπ3sin60°i^

=3kQπR232i^

=332πkQR2i^

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (27 Aug Shift 2)
Figure shows an amperian path ABCDA. Part ABC is in vertical plane PSTU while part CDA is in horizontal plane PQRS. Direction of circulation along the path is shown by an arrow near point \( \mathrm{B} \) and at \( \mathrm{D} \cdot \oint \vec{B} \cdot \mathrm{d} \vec{l} \) for this path according to Ampere's law will be

Options:
A \( \left(\mathrm{i}_{1}-\mathrm{i}_{2}+\mathrm{i}_{3}\right) \mu_{0} \)
B \( \left(-\mathrm{i}_{1}+\mathrm{i}_{2}\right) \mu_{0} \)
C \( \mathrm{i}_{3} \mu_{0} \)
D \( \left(\mathrm{i}_{1}+\mathrm{i}_{2}\right) \mu_{0} \)

The correct answer is:
\( \left(\mathrm{i}_{1}+\mathrm{i}_{2}\right) \mu_{0} \)

Solution:
If the loop current is coming out it should be taken as positive and if it is going in it is taken as negative.
i3 is coming out and again going in so its net contribution will be zero.
According to Ampere's law,
∮B→·dl→=μ0Σiinside
=μ0i1+i2+i3-i3
=μ0i1+i2

Topic: Physics, Magnetic Effects of Current, JEE Main