datavorous/entrance-exam-dataset huggingface dataset:
A positively charged particle is released from rest in an uniform electric field. The electric potential energy of the charge

Options:
A
remains a constant because the electric field is uniform

B
increases because the charge moves along the electric field

C
decreases because the charge moves along the electric field

D
decreases because the charge moves opposite to the electric field

The correct answer is:
decreases because the charge moves along the electric field

Solution:
The direction of electric field is always perpendicular to the direction of electric field and equipotential surface maintained at high electrostatic potential to other equipotential surface maintained at low electrostatic potential.
The positively charged particle experiences the electrostatic force in the direction of electric field i.e., from high electrostatic potential to low electrostatic potential. Thus, the work done by the electric field on the positive charge, so electrostatic potential energy of the positive charge decreases because speed of charged particle moves in the direction of field due to force $\mathrm{q} \overrightarrow{\mathrm{E}}$.

Topic: Physics, Electrostatic Potential and Capacitance
A positron is emitted from ${ }_{11}^{23} \mathrm{Na}$. The ratio of the atomic mass and atomic number of the resulting nuclide is

Options:
A
22/10

B
22/11

C
23/10

D
$23 / 12$

The correct answer is:
23/10

Solution:
$$
{ }_{11}^{23} \mathrm{Na} \longrightarrow{ }_{10}^{23} \mathrm{Na}+{ }_{-1}^0 e
$$

Topic: Chemistry, Chemical Kinetics, JEE Main
A possible value of tan14sin-1638 is:

Options:
A 22-1
B 122
C 7-1
D 17

The correct answer is:
17

Solution:
Let 14sin-1638=θ

sin4θ=638

cos4θ=18

2cos22θ-1=18

cos22θ=916

cos2θ=34

2cos2θ-1=34

cos2θ=78

cosθ=722

tanθ=17

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2021 (24 Feb Shift 2)
A possible value of x, for which the ninth term in the expansion of 3log325x-1+7+3-18log35x-1+110 in the increasing powers of 3-18log35x-1+1 is equal to 180, is :

Options:
A 0
B -1
C 2
D 1

The correct answer is:
1

Solution:
3log325x-1+7+3-18log35x-1+110

=25x-1+7+5x-1+1-1810, using anlogax=xn

Using the formula of the general term in the Binomial theorem for non negative integral index, i.e., Tr+1=Crnan-rbr in a+bn, we get, T9=T8+1=C81025x-1+7×5x-1+1-1=180

⇒25x-1+75x-1+1=4

⇒t2+7t+1=4, taking t=5x-1

⇒t2+7=4t+1

⇒t2-4t+3=0

⇒t-3t-1=0

⇒t=1,3⇒5x-1=1⇒5x-1=50

⇒x-1=0 (one of the possible value).

⇒x=1

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (27 Jul Shift 2)
A potential barrier of $0.50 \mathrm{~V}$ exists across a $P-N$ junction. If the depletion region is $5.0 \times 10^{-7} \mathrm{~m}$ wide, the intensity of the electric field in this region is

Options:
A $1.0 \times 10^6 \mathrm{~V} / \mathrm{m}$
B $1.0 \times 10^5 \mathrm{~V} / \mathrm{m}$
C $2.0 \times 10^5 \mathrm{~V} / \mathrm{m}$
D $2.0 \times 10^6 \mathrm{~V} / \mathrm{m}$

The correct answer is:
$1.0 \times 10^6 \mathrm{~V} / \mathrm{m}$

Solution:
$E=\frac{V}{d}=\frac{0.5}{5 \times 10^{-7}}=10^6 \mathrm{~V} / \mathrm{m}$.

Topic: Physics, Semiconductors, JEE Main
A potential difference of 20 kV is applied across an X-ray tube. The minimum wavelength of X-rays generated is

Options:
A 0.84 Å
B 0.31 Å
C 0.62 Å
D 0.96 Å

The correct answer is:
0.62 Å

Solution:
E = hâ¡ câ¡ Î»
â´ Î» min = hâ¡ câ¡ eâ¡ Vâ¡ = 6 . 6 3 Ã 1 0 - 3 4 Ã 3 Ã 1 0 8 1 . 6 Ã 1 0 - 1 9 Ã 2 0 Ã 1 0 3 = 0 . 6 2 A o

Topic: Physics, Atomic Physics, JEE Main
A potential difference of 220 V is applied across a rheostat of 12000 Î©. The voltmeter V has a resistance of 6000 Î© and distance BC is one-fourth of the distance from A to B . The error in the reading of voltmeter is approximately -

Options:
A 27%
B 10%
C 40%
D 15%

The correct answer is:
27%

Solution:
It total voltage across AB is 220 then voltage across CB should be 2204= 55 V (voltage to be measured)
Vâ² =220Ã200011000= 40 V (voltage being measured)
â´ % error=V-Vâ²VÃ 100=55-4055Ã 100=27%

Topic: Physics, Current Electricity, JEE Main
A potential difference of $300 \mathrm{volts}$ is applied to a combination of $2.0 \mu \mathrm{F}$ and $8.0 \mu \mathrm{F}$ capacitors connected in series. The charge on the $2.0 \mu \mathrm{F}$ capacitor is

Options:
A $2.4 \times 10^{-4} \mathrm{C}$
B $4.8 \times 10^{-4} \mathrm{C}$
C $7.2 \times 10^{-4} \mathrm{C}$
D $9.6 \times 10^{-4} \mathrm{C}$

The correct answer is:
$4.8 \times 10^{-4} \mathrm{C}$

Solution:
In series combination charge $Q$ is same. So charge on $2 \mu \mathrm{F}$ capacitor is
$Q=C_{\text {eq }} V=\left(\frac{2 \times 8}{2+8}\right) \times 300 \times 10^{-6}=4.8 \times 10^{-4} \mathrm{C}$

Topic: Physics, Capacitance, JEE Main
A potential divider circuit is connected with a dc source of $20 \mathrm{~V}$, a light emitting diode of glow in voltage $1.8 \mathrm{~V}$ and a zener diode of breakdown voltage of $3.2 \mathrm{~V}$. The length (PR) of the resistive wire is $20 \mathrm{~cm}$. The minimum length of PQ to just glow the LED is _____$\mathrm{cm}$

Options:

The correct answer is:
5

Solution:
$\begin{aligned} & \mathrm{PR}=20 \mathrm{~cm} \\ & \mathrm{~V}_{\mathrm{PQ}}=\frac{1}{4} \times \mathrm{R}_{\mathrm{PR}} \\ & \ell_{\min }(\mathrm{PQ})=\frac{1}{4} \times 20 \\ & =5 \mathrm{~cm}\end{aligned}$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2024 (08 Apr Shift 2)
A potential divider circuit is shown in figure. The output voltage V0 is

Options:
A 4 V
B 2 mV
C 0.5 V
D 12 mV

The correct answer is:
0.5 V

Solution:
Equivalent resistance, Req=3300+7100=4000 Ω

Now, current through circuit i=44000=11000 A

Potential drop V0=iR=11000×500=0.5 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (30 Jan Shift 1)
A potential V0 is applied across a uniform wire of resistance R. The power dissipation is P1. The wire is then cut into two equal halves and a potential of V0 is applied across the length of each half. The total power dissipation across two wires is P2. The ratio of P2:P1 is x:1. The value of x is _____.

Options:

The correct answer is:
16

Solution:
The formula to calculate the power dissipated in the first case can be written as

P1=V02R ...1

When the wire is halved, its resistance will also become half the previous value.

Thus, the formula to calculate the power dissipated in the second case can be written as

P2=V02R2×2= 4V02R ...2

Divide equation (2) by equation (1) to calculate the required ratio.

P2P1=4V02RV02R=4=16

Hence, x=16.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (13 Apr Shift 1)
A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of 2.0 V and a negligible internal resistance. The potentiometer wire itself is 4 m long. When the resistance, R, connected across the given cell, has values of.
i infinity ii 9.5 Ω
The 'balancing lengths, on the potentiometer wire are found to be 3m and 2.85 m, respectively. The value of internal resistance of the cell is:

Options:
A 0.25Â Î©
B 9.5 Ω
C 0.5 Î©
D 0.75Â Î©

The correct answer is:
0.5 Î©

Solution:
Here, we have to calculate the internal resistence of the unknown cell.

So, we have to use the following formulae:

Internal resistance of the unknown cell is
r=l1l2-1R=32.85-19.5Ω=0.5Ω

Hence the internal resistance will be 0.5 Ohm

Topic: Physics, Current Electricity, NEET, NEET 2014
A potentiometer is an ideal device of measuring potential difference because

Options:
A It uses a sensitive galvanometer
B It does not disturb the potential difference it measures
C It is an elaborate arrangement
D It has a long wire hence heat developed is quickly radiated

The correct answer is:
It does not disturb the potential difference it measures

Solution:
In balance condition, potentiometer doesn't take the current from secondary circuit.

Topic: Physics, Current Electricity, JEE Main
A potentiometer PQ is set up to compare two resistances, as shown in the figure. The ammeter A in the circuit reads 1.0 A when the two-way key K3 is open. The balance point is at a length l1 cm from P when the two-way key K3 is plugged in between 2 and 1, while the balance point is at a length l2 cm from P when the key K3 is plugged in between 3 and 1 . The ratio of two resistances R1R2, is found to be

Options:
A l1l1-l2
B l2l2-l1
C l1l1+l2
D l1l2-l1

The correct answer is:
l1l2-l1

Solution:
When the key is between 2 and 1
Then, emf V1= IR1=xl1.
When the key is between 3 and 1
Emf, V2=IR1+R2=xl2.
The ratio of both emf is R1R1+R2=l1l2
On simplifying the above relation, we get the ratio of resitances, R1R2=l1l2-l1.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2017 (08 Apr Online)
A potentiometer wire AB having length L and resistance 12r is joined to a cell D of emf ε and internal resistance r. A cell C having EMF ε/2 and internal resistance 3r is connected. The length AJ, at which the galvanometer, as shown in the figure, shows no deflection is

Options:
A 512L
B 1124L
C 1112L
D 1324L

The correct answer is:
1324L

Solution:
Potential difference across AJ=I RAJ,
=ε13r·12rL·x
For no current in galvanometer,
ε2= potential difference across AJ
ε2=12ε13Lx
x=13L24

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2019 (10 Jan Shift 1)
A potentiometer wire is $4 \mathrm{~m}$ long and potential difference of $3 \mathrm{~V}$ is maintained between the ends. The e.m.f. of the cell which balances against a length of $100 \mathrm{~cm}$ of the potentiometer wire is

Options:
A $0.50\mathrm{V}$
B $0.60 \mathrm{~V}$
C $3.0 .75 \mathrm{~V}$
D $4.0 .25 \mathrm{~V}$

The correct answer is:
$3.0 .75 \mathrm{~V}$

Solution:
Potential gradient $=\frac{3}{4} \frac{\mathrm{V}}{\mathrm{m}}$
So, emf against $100 \mathrm{~cm}=\frac{3}{4} \times 1 \mathrm{~V}=0.75 \mathrm{~V}$

Topic: Physics, Current Electricity, MHT CET, MHT CET 2020 (16 Oct Shift 1)
A potentiometer wire is 100 cm long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at 50 cm and 10 cm from the positive end of the wire in the two cases. The ratio of EMF's is :

Options:
A 5 : 1
B 5 : 4
C 3 : 4
D 3 : 2

The correct answer is:
3 : 2

Solution:
as E∝L

E 1 + E 2 E 1 − E 2 = λ50 λ10
E1+E2=5E1-5E2
6E2=4E1
32=E1E2

Topic: Physics, Current Electricity, NEET, NEET 2016 (Phase 1)
A potentiometer wire of length 10 cm and resistance 32 Î© is connected in series with a battery of EMF 2.5 V, internal resistance 3 Î© and external resistance R. If a secondary cell of 0.4 V is balanced at 8 cm length. Then value of R is

Options:
A 100 Ω
B 110 Ω
C 105 Ω
D 115 Î©

The correct answer is:
115 Î©

Solution:
Potential gradient of potentiometer=2.5(3+R+32)Ã3010
Balancing length for cell of emf e=0.4 V is 8 cm.
â´2.5(35+R)Ã3Ã8=0.4
â5Ã3Ã4=0.04Ã(35+R)10
â15Ã10=35+R
âR=115 Î©

Topic: Physics, Current Electricity, JEE Main
A potentiometer wire of length 10 m and resistance 20 Ω is connected in series with a 25 V battery and an external resistance 30 Ω. A cell of emf E in secondary circuit is balanced by 250 cm long potentiometer wire. The value of E (in volt) is x10. The value of x is _____ .

Options:

The correct answer is:
25

Solution:
Since the given resistance is in series, so, total resistance R=20+30=50 Ω

Given,

Potential difference=25 V

Current I=VR=2550=0.5 A

Also, Potential gradient =VpdL=IRL

Where, Vpd=potential drop at wire

L=Length of wire

Potential gradient=20×0.510=1 V m-1
So, Emf of the cell is

E=Potential gradient×Balancing Length

E= 1× 250100
E= 2.5 V=2510 V

Thus, x=25.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (24 Jun Shift 2)
A potentiometer wire of length 300 cm is connected in series with a resistance 780 Ω and a standard cell of emf 4 V. A constant current flows through potentiometer wire. The length of the null point for cell of emf 20mV is found to be 60 cm. The resistance of the potentiometer wire is _____ Ω.

Options:

The correct answer is:
20

Solution:
Let resistance of potentiometers wire is R, then the current through AB

iAB=4R+780

Potential difference across AB

VAB=4RR+780

Potential difference across AC

VAC=VABlL=4RR+780×60300=4R5R+780

This should be equal to 20 mV

⇒4R5R+780=20×10-3=2×10-2

⇒4R=10-1R+780

⇒4R-R10=78

⇒39R10=78

⇒R=20 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (26 Jul Shift 2)
A potentiometer wire PQ of 1m length is connected to a standard cell E1. Another cell E2 of emf 1.02 V is connected with a resistance 'r' and switch S (as shown in figure). With switch S open, the null position is obtained at a distance of 49 cm fromQ. The potential gradient in the potentiometer wire is :

Options:
A 0.02 V/cm
B 0.01 V/cm
C 0.03 V/cm
D 0.04 V/cm

The correct answer is:
0.02 V/cm

Solution:
PQ=1 m

QJ=49 cm

∴ PJ=51 cm

vℓ=1.0251=0.02 v/cm

Topic: Physics, Current Electricity, JEE Main, JEE Main 2020 (02 Sep Shift 2)
A power transmission line feeds input power at 2.3 kV to a step down transformer with its primary winding having 3000 turns. The output power is delivered at 230 V by the transformer. The current in the primary of the transformer is 5 A and its efficiency is 90%. The winding of transformer is made of copper. The output current of transformer is ____A.

Options:

The correct answer is:
45

Solution:
Given:

Input power, Pi=2300×5 watt

Output power, P0=2300×5×0.9

Now, we can write η=PoutPin=VSISVPIP
⇒0.9=230 IS2300×5⇒IS=45 A.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (30 Jan Shift 2)
A power transmission line feeds input power at 2300 V to a step down transformer with its primary windings having 4000 turns. The output power is delivered at 230 V by the transformer. If the current in the primary of the transformer is 5 A and its efficiency is 90 %, the output current would be:

Options:
A 35 A
B 25 A
C 50 A
D 45 A

The correct answer is:
45 A

Solution:
Efficiency n=90%
⇒0.9×power in=Power out
0.9×2300×5=230×Iout
Iout=45 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2019 (09 Jan Shift 2)
A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns giving the output power at 230 V. If the current in the primary coil of the transformer is 5 A and its efficiency is 90 %, the output current would be:

Options:
A 45 A
B 50 A
C 20 A
D 25 A

The correct answer is:
45 A

Solution:
Efficiency, η=0.9=PsPp
VsIs=0.9 × VpIp
Is=0.9 ×2300 ×5230=45 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2018 (16 Apr Online)
A power transmission line feeds input power at 2400 V to a step-down transformer and which delivers power at 240 V with its primary windings having 5000 turns. If the current in the primary coil of the transformer is 5 A and its efficiency is 80%, then what is the output current (in A)?

Options:

The correct answer is:
40

Solution:
Îµp=2400V,Np=5000

Îµs=240V,Ip=5A

efficiency, Î·=80%=0.8,Is=?

Î·=P0Pi=ÎµsIsÎµpIp

âIs=Î·ÎµpIpÎµs=0.8Ã2400Ã5240=40A

Topic: Physics, Alternating Current, JEE Main
A power transmission line feeds input power at 2400 V to a step-down transformer which delivers power at 240 V with its primary windings having 5000 turns. If the current in the primary coil of the transformer is 5 A and its efficiency is 80%, then what is the output current (in A)?

Options:

The correct answer is:
40

Solution:
εp=2400 V, Np=5000

εs=240 V, Ip=5 A

η=80%=0.8

η=PoutPin=εsIsεpIp

⇒Is=ηεpIpεs=0.8×2400×5240=40 A

Topic: Physics, Alternating Current, JEE Main
A precipitate of the following would be obtained when $\mathrm{HCl}$ is added to a solution of stannous sulphide (SnS) in yellow ammonium sulphide

Options:
A $\operatorname{SnS}$
B $\mathrm{SnS}_2$
C $\mathrm{Sn}_2 S_2$
D $\left(\mathrm{NH}_4\right)_2 \mathrm{SnS}_3$

The correct answer is:
$\mathrm{SnS}_2$

Solution:
$\mathrm{SnS}+\left(\mathrm{NH}_4\right)_2 \mathrm{~S}_2 \xrightarrow{\mathrm{HCl}}\left(\mathrm{NH}_4\right)_2 \mathrm{~S}+\mathrm{SnS}_2$

Topic: Chemistry, Practical Chemistry, JEE Main
A (predominantly) is

Options:
A

B

C

D

The correct answer is:

Solution:
Reaction is electrophilic addition which proceeds through carbocation reaction intermediate which undergo hydride shift.

Topic: Chemistry, Hydrocarbons, NEET, NEET 2008 (Mains)
A pressure cooker reduces cooking time for food because

Options:
A
boiling point of water involved in cooking is increased

B
the higher pressure inside the cooker crushes the food material

C
cooking involves chemical changes helped by a rise in temperature

D
heat is more evenly distributed in the cooking space

The correct answer is:
boiling point of water involved in cooking is increased

Solution:
On increasing pressure, the temperature is also increased. Thus in pressure cooker due to increase in pressure the b.p. of water increases.

Topic: Chemistry, Solutions, JEE Main
A pressure cooker reduces cooking time for food because

Options:
A
boiling point of water involved in cooking is increased

B
the higher pressure inside the cooker crushes the food material

C
cooking involves chemical changes helped by a rise in temperature

D
heat is more evenly distributed in the cooking space

The correct answer is:
boiling point of water involved in cooking is increased

Solution:
On increasing pressure, the temperature is also increased. Thus in pressure cooker due to increase in pressure the b.p. of water increases.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2003
(a) Pressure decreases as one ascends the atmosphere. If the density of air is $\rho$, what is the change in pressure $d p$ over a differential height $d h$ ?
(b) Considering the pressure $p$ to be proportional to the density, find the pressure $p$ at a height $h$ if the pressure on the surface of the earth is $p_0$.
(c) If $p_0=1.03 \times 10^5 \mathrm{Nm}^{-2}, \rho_0=1.29 \mathrm{~kg} \mathrm{~m}^{-3}$ and $g=9.8$ $\mathrm{ms}^{-2}$, at what height will be pressure drop to $(1 / 10)$ the value at the surface of the earth?
(d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.

Options:

None

Solution:
Let us consider a horizontal parcel of air with crosssection $A$ and thickness $(d h)$. As the pressure at a point in fluid is equal in all direction.
So the pressure on the upper surface $(p)$ is downward and on bottom surface $(p+d p)$ upward if the parcel is in equilibrium, then the net upward force must be balanced by the weight and Buoyant force by air.
i.e. $(p+d p) A-p A=-\rho g A d h$
$$
\begin{aligned}
&(\because \text { Weight }=\text { Density } \times \text { Volume } \times g) \\
&=-\rho \times A d h \times g \\
&\Rightarrow d p=-\rho g d h \\
&
\end{aligned}
$$
Negative sign shows that pressure decreases as height increases.
(b) Let $\rho_0$ be the density of air on the surface of the earth. According to question pressure $p$ at a point is directly proportional to density, pressure $\propto$ density.
$$
\begin{aligned}
&(p \propto \rho) \text { or } \frac{p}{p_0}=\frac{\rho}{\rho_0} \text { or } \rho=\frac{\rho_0}{p_0} p \\
&d p=-\rho g d h \quad \text { from (i) } \\
&\therefore d p=-\frac{\rho_0 g}{p_0} p d h \Rightarrow \frac{d p}{p}=-\frac{\rho_0 g}{p_0} d h
\end{aligned}
$$
Integrating both side,
$$
\begin{gathered}
\int_{\rho_0}^\rho \frac{d p}{p}=-\frac{\rho_0 g}{p_0} \int_0^h d h \\
\quad\left[\because \text { at } h=0, r=p_0 \text { and at } h=h, p=p\right] \\
\Rightarrow \log \left(\frac{p}{p_0}\right)=-\frac{\rho_0 g}{p_0} h \Rightarrow\left(\frac{p}{p_0}\right)=e\left(\frac{-\rho_0 g h}{p_0}\right)
\end{gathered}
$$
Take antilog,
$$
p=p_0 e\left(-\frac{\rho_0 g h}{p_0}\right)
$$
(c) From(ii), $p=p_0 e^{-\frac{\rho_0 g h}{P_0}}$
$$
\log \left(\frac{p}{p_0}\right)=-\frac{\rho_0 g h}{p_0}
$$
As given that $p=\frac{1}{10} p_0$
$$
\begin{aligned}
&\Rightarrow \log \left(\frac{\frac{1}{10} p_0}{p_0}\right)=-\frac{\rho_0 g}{p_0} h \\
&\Rightarrow \log \frac{1}{10}=-\frac{\rho_0 g}{p_0} h \\
&\therefore h=-\frac{p_0}{\rho_0 g} \log \frac{1}{10}=-\frac{p_0}{\rho_0 g} \ln (10)^{-1}=\frac{p_0}{\rho_0 g} \log 10 \\
&=\frac{p_0}{\rho_0 g} \times 2.303\left(\log _{10} 10\right) \quad\left[\because \log (x)=2.303 \log _{10}(x)\right] \\
&=\frac{1.013 \times 10^5}{1.29 \times 9.8} \times 2.303=0.184 \times 10^5 \mathrm{~m} \\
&=18.4 \times 10^3 \mathrm{~m}=18.4 \mathrm{~km} . \quad\left(\because \log _{10} 10=1\right)
\end{aligned}
$$
(d) As from (b) part, $p \propto \rho$
Temperature $(T)$ remains constant only near the surface of the earth, not at greater heights.

Topic: Physics, Mechanical Properties of Fluids
A primary aliphatic amine on reaction with nitrous acid in cold 273 K and there after raising temperature of reaction mixture to room temperature 298 K, gives a/an

Options:
A Alcohol
B nitrile
C diazonium salt
D secondary amine

The correct answer is:
Alcohol

Solution:
A primary aliphatic amine on reaction with nitrous acid in cold gives a diazonium salt, but diazonium salt of aliphatic amines are not much stable, so, they dissociate on increasing temperature slightly and gives alcohol with water.

R-NH2→273°CHNO2aqRN2+→H2O283°CR-OH+N2

Topic: Chemistry, Amines, JEE Main, JEE Main 2022 (28 Jun Shift 1)
A primitive of $\frac{x}{x^2+1}$ is

Options:
A $\log _e\left(x^2+1\right)$
B $x \tan ^{-1} x$
C $\frac{\log _e\left(x^2+1\right)}{2}$
D $\frac{1}{2} x \tan ^{-1} x$

The correct answer is:
$\frac{\log _e\left(x^2+1\right)}{2}$

Solution:
\(\begin{aligned}
& \rightarrow \int \frac{x}{x^2+1} \cdot d x \\
& \text {So, } x^2+1=t \Rightarrow 2 x \cdot d x=d t \\
& \Rightarrow \int \frac{1}{t} \cdot \frac{d t}{2} \Rightarrow \frac{1}{2} \log _e t+c \\
& \Rightarrow \frac{1}{2} \log _e\left(x^2+1\right)+c
\end{aligned}\)

Topic: Mathematics, Indefinite Integration, JEE Main
A printed page is kept pressed by a transparent cube of edge $t$. The refractive index of the cube varies as $\mu(z)=1+\frac{z}{t}$, where $z$ is the vertical distance from bottom of the cube. If viewed from top, then the printed letters appear to be shifted by an amount

Options:
A $(1-\ln 2) t$
B $(2 \ln 2-1) t$
C $\frac{t}{2 \ln 2}$
D $\frac{2 t}{3 \ln 2}$

The correct answer is:
$(1-\ln 2) t$

Solution:
Consider an elemental strip at a height $z$ of thickness $d z$.
$$
\begin{aligned}
\text { Apparent height } & =\frac{\text { Real height }}{\text { RI }}=\frac{d z}{\mu(z)} \\
d h & =\frac{d z}{1+\frac{z}{t}} \\
d h & =\left(\frac{t}{t+z}\right) d z
\end{aligned}
$$
$$
\begin{aligned}
h & =\int_0^z \frac{t}{t+z} \cdot d z \\
h=[t \ln (t+z)]_0^z & =t[\ln (1+t)-\operatorname{In}(t+0)] \\
h & =t \operatorname{In} \frac{2 t}{t}=t \operatorname{In} 2 \\
\text { Shift, } \Delta x & =t-t \ln 2 \\
\Delta x & =(1-\ln 2) t
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main
A printed page is pressed by a glass of water. The refractive index of the glass and water is $1.5$ and $1.33$, respectively. If the thickness of the bottom of glass is $1 \mathrm{~cm}$ and depth of water is $5 \mathrm{~cm}$, how much the page will appear to be shifted if viewed from the top ?

Options:
A
$1.033 \mathrm{~cm}$

B
$3.581 \mathrm{~cm}$

C
$1.3533 \mathrm{~cm}$

D
$1.90 \mathrm{~cm}$

The correct answer is:
$1.3533 \mathrm{~cm}$

Solution:
Real depth $=5 \mathrm{~cm}+1 \mathrm{~cm}=6 \mathrm{~cm}$
Apparent depth $=\frac{\mathrm{d}_1}{\mu_1}+\frac{\mathrm{d}_2}{\mu_2}+\ldots$
$=\frac{5}{1.33}+\frac{1}{1.5}$
$\simeq 3.8+0.7 \simeq 4.5 \mathrm{~cm}$
$\therefore$ Shift $=6 \mathrm{~cm}-4.5 \mathrm{~cm} \cong 1.5 \mathrm{~cm}$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2013 (25 Apr Online)
A prism $A B C$ of angle $30^{\circ}$ has its face $A C$ silvered. A ray of light incident at an angle of $45^{\circ}$ at the face $A B$ retraces its path after refraction at face $A B$ and reflection at face $A C$. The refractive index of the material of the prism is

Options:
A 1.5
B $\frac{3}{\sqrt{2}}$
C $\sqrt{2}$
D $\frac{4}{3}$

The correct answer is:
$\sqrt{2}$

Solution:
$A=r+0 \Rightarrow r=30^{\circ}$
$\therefore \mu=\frac{\sin i}{\sin r}=\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\sqrt{2}$

Topic: Physics, Ray Optics, JEE Main