datavorous/entrance-exam-dataset huggingface dataset:
In an atom, total number of electrons having quantum numbers \(\mathrm{n}=4,\left|\mathrm{~m}_l\right|=1\) and \(\mathrm{m}_{\mathrm{s}}=-\frac{1}{2}\) is
______

Options:

The correct answer is:
6

Solution:
$\mathrm{n}=4$
$\ell \quad \mathrm{m}_{\ell}$
$\begin{array}{ll}
0 & 0 \\
1 & -1,0,+1 \\
2 & -2,-1,0,+1,+2,+3
\end{array}$
So number of orbital associated with
$\mathrm{n}=4,\left|\mathrm{~m}_{\ell}\right|=1 \text { are } 6$
Now each orbital contain one $e^{-}$with $m_s=-\frac{1}{2}$

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2024 (05 Apr Shift 2)
In an atom, two electrons revolve around the nucleus in circular orbits of radii r and 4r. The ratio of the time taken by them (T 2  :  T 1 ) to complete one revolution is

Options:
A 1:4
B 4:1
C 1:8
D 8:1

The correct answer is:
1:8

Solution:
rn=n2×η
r2=4η
Tn=n3Z2×1.5×10-16sec
T1=1312×1.5×10-16sec
T2=2312×1.5×10-16sec
On dividing T2 by T1 we get
T2T1=81
or T1T2=18

Topic: Chemistry, Structure of Atom, JEE Main
In an atom, what ensures electrical neutrality?

Options:
A The presence of protons in the nucleus
B The presence of neutrons in the nucleus
C The equal number of electrons and protons
D The presence of alpha particles in the nucleus

The correct answer is:
The equal number of electrons and protons

Solution:
Electrical neutrality in an atom is maintained by having an equal number of electrons and protons.

Topic: Chemistry, Structure of Atom, NEET
In an average menstrual cycle of 28 days, which of these events is correctly matched with the time of its occurrence?

Options:
A Release of ovum: 5th day
B Endometrium degenerates: 5th -10th day
C Endometrium secretes nutrients for implantation:15th -28th day
D Rise in progesterone level: 1st -15th day

The correct answer is:
Endometrium secretes nutrients for implantation:15th -28th day

Solution:
The series of cyclic changes that take place simultaneously in the ovaries and uterine wall (endometrium) of the female reproductive tract stimulated by different concentrations of gonadotropins like FSH and LH is called the menstrual cycle. It is mainly made up of 4 phases:
Menstrual phase or Destructive phase -1st - 4th day - endometrium degenerate.
Proliferative phase or Follicular phase - 5th - 13th day - High F.S.H. level stimulates the primordial follicle to mature into the Graafian follicle.
Ovulatory phase - 14th day - the release of the ovum.
Luteal phase or secretory phase - 15th - 28th day - rise in progesterone levels and the endometrium secretes nutrients for implantation.

Topic: Biology, Human Reproduction, NEET
In an elastic collision of two billiard balls, which of the following quantities remain conserved during the short time of collision of the balls (i.e., when they are in contact)?
(a) Kinetic energy
(b)Total linear momentum.
Give reason for your answer in each case.

Options:

None

Solution:
When two billiard balls collide each other then their linear momentum remain conserved. When balls are in contact there may be deformation which means elastic PE which came from part of KE So, KE may not be conserved.

Topic: Physics, Work Power Energy
In an electric circuit, a call of certain emf provides a potential difference of 1.25 V across a load resistance of 5 Ω. However, it provides a potential difference of 1 V across a load resistance of 2 Ω.The emf of the cell is given by x10 V. Then the value of x is .

Options:

The correct answer is:
15

Solution:
Terminal voltage, v=iR=ERR+r

1st→1.25=E(5)5+r …i
2nd→1=E(2)2+r …ii

By i and ii,
r=1 Ω, E=32 V=1510 volt
⇒x=15

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (22 Jul Shift 1)
In an electrical circuit, a battery is connected to pass 20 C of charge through it in a certain given time. The potential difference between two plates of the battery is maintained at 15 V. The workdone by the battery is _______ J

Options:

The correct answer is:
300

Solution:
Work done by battery =QΔV

⇒20×15=300 J

=300

Topic: Physics, Electrostatics, JEE Main, JEE Main 2021 (26 Feb Shift 1)
In an electrical circuit drawn below the amount of charge stored in the capacitor is _______ μC.

Options:

The correct answer is:
60

Solution:
In steady state there will be no current in branch of capacitor, so no voltage drop across R2=5 Ω

I2=0

I1=I3=104+6=1 A

VR3=Vc+VR2 VR2=0

⇒I3R3=Vc

⇒Vc=1×6=6 volt

Therefore, charge on capacitor qc=CVc=10×6=60 μC

Topic: Physics, Capacitance, JEE Main, JEE Main 2024 (01 Feb Shift 2)
In an electrical circuit $R, L, C$ and an $\mathrm{AC}$ voltage source are all connected in series. When $L$ is removed from the circuit, the phase difference between the voltage and the current in the circuit is $\pi / 3$. If instead, $C$ is removed from the circuit, the phase difference is again $\pi / 3$. The power factor of the circuit is

Options:
A $1 / 2$
B $1 / \sqrt{2}$
C 1
D $\sqrt{3} / 2$

The correct answer is:
1

Solution:
Here, phase difference
$\begin{aligned}
\tan \phi & =\frac{X_L-X_C}{R} \\
\tan \frac{\pi}{3} & =\frac{X_L-X_C}{R}
\end{aligned}$
When, $L$ is removed
$\begin{aligned}
& \sqrt{3}=\frac{X_C}{R} \\
& X_C=\sqrt{3} R
\end{aligned}$
When $C$ is removed
$\begin{gathered}
\tan \frac{\pi}{3}=\sqrt{3}=\frac{X_L}{R} \\
X_L=R \sqrt{3}
\end{gathered}$
Hence, in resonant circuit
$\begin{aligned}
\tan \phi & =\frac{\sqrt{3} R-\sqrt{3} R}{R}=0 \\
\phi & =0
\end{aligned}$
$\therefore$ Power factor $\cos \phi=1$
It is the condition of resonance therefore phase difference between voltage and current is zero and power factor is $\cos \phi=1$.

Topic: Physics, Alternating Current, JEE Main
In an electrochemical reaction of lead, at standard temperature, if EPb2+/Pbo=m Volt and EPb4+/Pbo=n Volt,then the value of E°Pb2+/Pb4+ is given by m – xn. The value of x is _______. (Nearest integer)

Options:

The correct answer is:
2

Solution:
The relation between Gibbs Free energy and emf of the cell can be related as follows,

∆G=-nFEcell

The electrode reactions can be written as follows,

Pb2++2e-→Pb ∆G1=-2FEPb2+/Pbo

Pb4++4e-→Pb ∆G2=-4FEPb4+/Pbo

Pb4++2e-→Pb2+ ∆G3=-2FEPb4+/Pb2+o

Now, ∆G3=∆G2-∆G1

-2FEPb4+/Pb2+o=-4FEPb4+/Pbo+2FEPb2+/Pbo

⇒4EPb+4/Pbo=2EPb+2/Pbo+2EPb+4/Pb+2o

⇒4n=2m+2EPb+4/Pb+2o⇒EPb+4/Pb+2o=2n-m⇒EPb+2/Pb+4o=m-2n

Hence, the value of x=2

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2023 (11 Apr Shift 1)
In an electromagnetic wave, at an instant and at a particular position, the electric field is along the negative z-axis and magnetic field is along the positive x-axis. Then the direction of propagation of electromagnetic wave is:

Options:
A positive z-axis
B positive y-axis
C at 45°angle from positive y -axis
D negative y-axis

The correct answer is:
negative y-axis

Solution:
Direction of propagation of the electromagnetic wave is along E →×B→

Thus,

n→=E-k^×Bi^

Therefore,

n^=-k^×i^=-j^

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2023 (13 Apr Shift 2)
In an electromagnetic wave in free space the root mean square value of the electric field is Erms=6 V m-1. The peak value of the magnetic field is

Options:
A 1.41Ã10-8 T
B 2.83Ã10-8 T
C 0.70Ã10-8Â T
D 4.23Ã10-8Â T

The correct answer is:
2.83Ã10-8 T

Solution:
Speed of electromagnetic wave c is given by, ErmsBrms=c
Brms=Ermsc
=63×108
Brms=2×10-8
Brms=B02
B0= 2×Brms
= 2×2×10-8
=2.83×10-8 T

Topic: Physics, Electromagnetic Waves, JEE Main
In an electromagnetic wave, the electric field vector and magnetic field vector are given as E→=E0i^ and B→=B0k^, respectively. The direction of
propagation of electromagnetic wave is along:

Options:
A (k^)
B J^
C (-k^)
D (-j^)

The correct answer is:
(-j^)

Solution:
Direction of propagation =E→×B→=i^×k^=-j^.

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2021 (20 Jul Shift 2)
In an electromagnetic wave, the maximum value of the electric field is 100 V m-1. The average intensity is [ε0=8.8×10-12 C-2 N-1 m2]

Options:
A 13.2 W m-2
B 36.5 W m-2
C 46.7 W m-2
D 765 W m-2

The correct answer is:
13.2 W m-2

Solution:
Iavg=12cε0E02

⇒Iavg=12×3×108×8.8×10-12×1002

⇒Iavg=13.2 W m-2

Topic: Physics, Electromagnetic Waves, JEE Main
In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5×10-12 m, the minimum electron energy required is close to:

Options:
A 100 keV
B 25 keV
C 1 keV
D 500 keV

The correct answer is:
25 keV

Solution:
λ=150V⇒7.5×10-2=150V
V=1507.5×7.5×10-4=803kV
V≈25 KeV

Topic: Physics, Dual Nature of Matter, JEE Main
In an electron microscope, the resolution that can be achieved is of the order of the wavelength of electrons used. To resolve a width of 7.5×10-12 m, the minimum electron energy required is close to:

Options:
A 100 keV
B 25 keV
C 1 keV
D 500 keV

The correct answer is:
25 keV

Solution:
λ=150V⇒7.5×10-2=150V
V=1507.5×7.5×10-4=803kV
V≈25 KeV

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2019 (10 Jan Shift 1)
In an electrostatic precipitator,

Options:
A The velocity of air between the plates must be very high to purify the air.
B The electrode wires maintained at several thousand volts produces a corona which releases positively charged particles.
C Particulate matter that are very small can be removed by precipitation.
D Collecting plates are grounded and collect negatively charged dust particles.

The correct answer is:
Collecting plates are grounded and collect negatively charged dust particles.

Solution:
Electrostatic precipitator has electrodes maintained at many thousand volts. They produce a corona that releases negatively charged electrons which combine with dust particles to give them a negative charge. Grounded collecting plates collect these negatively charged dust particles if the air velocity is low allowing the dust to fall.

Topic: Biology, Environmental Issues, NEET
In an ellipse $9 x^2+5 y^2=45$, the distance between the foci is

Options:
A $4 \sqrt{5}$
B $3 \sqrt{5}$
C $3$
D $4$

The correct answer is:
$4$

Solution:
Given equation may be written as $\frac{x^2}{5}+\frac{y^2}{9}=1$. Comparing the given equation with standard equation, we get $a^2=5$ and $b^2=9$. We also know that in an ellipse (where $\left.b^2 \gt a^2\right) a^2=b^2\left(1-e^2\right) \lambda$
or $e^2=\frac{b^2-a^2}{b^2}=\frac{9-5}{9}=\frac{4}{9}$ or $e=\frac{2}{3}$.
Therefore distance between foci $=2 b e=2 \times 3 \times \frac{2}{3}=4$.

Topic: Mathematics, Ellipse, JEE Main
In an ellipse, the distance between its foci is 6 and minor axis is 8. Then its eccentricity is

Options:
A
$\frac{3}{5}$

B
$\frac{1}{2}$

C
$\frac{4}{5}$

D
$\frac{1}{\sqrt{5}}$

The correct answer is:
$\frac{3}{5}$

Solution:
$2 a e=6 \Rightarrow a e=3$
$2 b=8 \Rightarrow b=4$
$b^2=a^2\left(1-e^2\right)$
$16=a^2-a^2 e^2$
$a^2=16+9=25$
$a=5$
$\therefore e=\frac{3}{a}=\frac{3}{5}$

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2006
In an ellipse, the distance from one of the foci to its corresponding end of the major axis is $4-\sqrt{7}$ and the distance from same focus to one end of the minor axis is 4 . Then the cosine of the angle subtended by the line segment joining its foci at one end of its minor axis is

Options:
A $\frac{1}{8}$
B $\frac{3}{4}$
C $\frac{\sqrt{7}}{3}$
D $\frac{1}{3 \sqrt{7}}$

The correct answer is:
$\frac{1}{8}$

Solution:
$\begin{aligned} & \text { } S_1 A=4-\sqrt{7} \Rightarrow|a e-a|=4-\sqrt{7} \\ & a^2(1-e)^2=(4-\sqrt{7})^2 \\ & B S_1=4\end{aligned}$
$\begin{aligned} & (a e)^2+b^2=16 \Rightarrow(a e)^2+a^2-(a e)^2=16 \\ & \Rightarrow \quad a^2=16 \Rightarrow a=4 \\ & \therefore \quad 4(1-e)=4-\sqrt{7} \\ & \Rightarrow \quad 1-\left(\frac{4-\sqrt{7}}{4}\right)=e \Rightarrow e=\frac{\sqrt{7}}{4} \\ & \therefore \quad b=a \sqrt{1-e^2}=4 \times \frac{3}{4}=3 \\ & \quad B S_1=B S_2=4 \\ & S_1 S_2=2 a e=2 \sqrt{7} \\ & \cos \theta=\frac{B S_1^2+B S_2^2-\left(S_1 S_2\right)^2}{2\left(B S_1\right)\left(B S_2\right)}=\frac{16+16-28}{2(4)(4)} \\ & \therefore \quad \cos \theta=\frac{1}{8} .\end{aligned}$

Topic: Mathematics, Ellipse, JEE Main
In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at 0,53, then the length of its latus rectum is:

Options:
A 6
B 10
C 8
D 5

The correct answer is:
5

Solution:
Given, difference of the lengths of major and minor axis is 10.
⇒2b-2a=10 … (1)
One of the focii is 0, 53
⇒be=53 … (2)
Also a2=b21-e2 .......(3)
⇒a2=b2-75
⇒b-aa+b=75
⇒a+b=15 … (4)
From (1) and (4)
a=5 and b=10
Hence length of the latus rectum =2a2b=2⋅2510=5

Topic: Mathematics, Ellipse, JEE Main, JEE Main 2019 (08 Apr Shift 2)
In an energy recycling process, Xâg of steam at 100 Â°C becomes water at 100 Â°C which converts Yâg of ice at 0 Â°C into water at 100 Â°C. The ratio of XY will be (specific heat of water =4200 J kg-1 K, specific latent heat of fusion =3.36Ã105 J kg-1, specific latent heat of vaporization =22.68Ã105 J kg-1)

Options:
A 13
B 23
C 3
D 2

The correct answer is:
13

Solution:
Heat loss = Heat gain
m 1 L v = m 2 .L f + m 2 .S.ΔT
XÃ10-3Ã22.68Ã105 =YÃ10-3Ã3.36Ã105+YÃ10-3Ã4200Ã100
â´XY=7.5622.68=13

Topic: Physics, Thermodynamics, JEE Main
In an energy recycling process, \( x \mathrm{~g} \) of steam at \( 100^{\circ} \mathrm{C} \) becomes water at \( 100^{\circ} \mathrm{C} \) which converts \( y \mathrm{~g} \) of ice at \( 0^{\circ} \mathrm{C} \) into
water at \( 100^{\circ} \mathrm{C} \). The ratio of \( \frac{x}{y} \) will be

Options:
A \( \frac{1}{3} \)
B \( \frac{2}{3} \)
C \( 3 . \)
D \( 2 \).

The correct answer is:
\( \frac{1}{3} \)

Solution:
Specific heat of water=4200 J kg-1 K-1
Specific latent heat of fusion=3.36×105 J kg
Specific latent heat of vaporization=22.68×105 J kg-1
=x×10-3×22.68×105
=y×10-3×3.36×105+y×10-3×4200×100
∴xy=7.5622.68=13

Topic: Physics, Thermal Properties of Matter, JEE Main
In an engine the piston undergoes vertical simple harmonic motion with amplitude 7cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston, is close to :

Options:
A 0.7 Hz
B 1.9 Hz
C 1.2 Hz
D 0.1 Hz

The correct answer is:
1.9 Hz

Solution:
No contact ⇒ N=0
⇒ amax=g=w2A
w= gA=100.07
f=w2π=1000712π=1.9 Hz

Topic: Physics, Oscillations, JEE Main, JEE Main 2016 (10 Apr Online)
In an entrance test there are multiple choice questions. There are four possible answers to each question of which one is correct. The probability that a student knows the answer to a question is $90 \%$. If he gets the correct answer to a question, then the probability that he was guessing, is

Options:
A $\frac{37}{40}$
B $\frac{1}{37}$
C $\frac{36}{37}$
D $\frac{1}{9}$

The correct answer is:
$\frac{1}{37}$

Solution:
We define the following events :
$A_1$ : He knows the answer.
$A_2$ : He does not know the answer.
E: He gets the correct answer.
then $P\left(A_1\right)=\frac{9}{10}, P\left(A_2\right)=1-\frac{9}{10}=\frac{1}{10}, P\left(\frac{E}{A_1}\right)=1$,
$P\left(\frac{E}{A_2}\right)=\frac{1}{4}$
$\therefore$ Required probability
$=P\left(\frac{A_2}{E}\right)=\frac{P\left(A_2\right) P\left(E / A_2\right)}{P\left(A_1\right) P\left(E / A_1\right)+P\left(A_2\right) P\left(E / A_2\right)}=\frac{1}{37}$.

Topic: Mathematics, Probability, JEE Main
In an equilateral triangle ABC, equation of the side BC is x+y-2=0 and the centroid of ÎABC is 0,0. If points A, B and C are in anticlockwise order, then the equation of side AC is

Options:
A y+2=2-3x+2
B y+2=2+3x+2
C y+1=2+3x+1
D x+2=0

The correct answer is:
y+2=2+3x+2

Solution:
Foot of 0,0 on line BC is D1,1
Because 0,0 divides AD internally in the ratio 2:1
Hence, by using section formula A≡-2,-2
Slope of AC is tan⁡45°+30°=2+3
Equation of AC is y+2=2+3x+2

Topic: Mathematics, Straight Lines, JEE Main
In an estimation of bromine by Carius method, 1.6g of an organic compound gave 1.88g of AgBr. The mass percentage of bromine in the compound is …..
(Atomic mass, Ag=108, Br=80 gmol-1

Options:

The correct answer is:
50

Solution:
if molar mass of compound is M
1.6 / M=1.88 / 188=0.01
so m=160 gram/mole
Now % of bromine =80/160×100%=50%

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2020 (06 Sep Shift 1)
In an examination, there are 3 multiple choice questions and each question has 4 choices. If a student randomly selects answer for all the 3 questions, what is the probability that the student will not answer all the 3 questions correctly?

Options:
A $1 / 64$
B $63 / 64$
C $1 / 12$
D $11 / 12$

The correct answer is:
$63 / 64$

Solution:
Since, probability of answering all the three questions correctly $=\frac{1}{4} \times \frac{1}{4} \times \frac{1}{4}=\frac{1}{64}$
$\therefore$ Probability of not answering all the three questions correctly $=1-\frac{1}{64}=\frac{63}{64}$

Topic: Mathematics, Probability, JEE Main
In an examination, there are 10 true-false type questions. Out of 10, a student can guess the answer of 4 questions correctly with probability 34 and the remaining 6 questions correctly with probability 14. If the probability that the student guesses the answers of exactly 8 questions correctly out of 10 is 27k410, then k is equal to

Options:

The correct answer is:
479

Solution:
Given,

A=1,2,3,4:PA=34→Correct answer

B=5,6,7,8,9,10;PB=14 Correct answer

Now total ways in which 8 Correct answer is given will be,

⇒Case (i) 4,4:C44344·C46·144·342

⇒Case (ii) 3,5:C34343·141·C56145·34

⇒Case (iii) 2,6:C24342142·C66146

Total probability =141034×15×32+4×33×6×3+6×32

=274102.7×15+72+2

Now comparing with 27k410

We get,K=479

Topic: Mathematics, Probability, JEE Main, JEE Main 2022 (24 Jun Shift 2)
In an examination there are three multiple choice questions and each question has 4 choices. Number of ways in which a student can fail to get all answers correct, is

Options:
A $11$
B $12$
C $27$
D $63$

The correct answer is:
$63$

Solution:
Each question can be answered in 4 ways and all questions can be answered correctly in only one way, so required number of ways
$4^{\wedge} 3-1=63$

Topic: Mathematics, Permutation Combination, JEE Main
In an exothermic reaction, how does ∆H relate to the surroundings?

Options:
A ∆H absorbs heat from the surroundings.
B ∆H releases heat to the surroundings.
C ∆H remains constant with the surroundings.
D ∆H is unrelated to the surroundings.

The correct answer is:
∆H releases heat to the surroundings.

Solution:
In an exothermic reaction, ∆H is negative, indicating that heat is released from the system to the surroundings, making option B the correct choice.

Topic: Chemistry, Thermodynamics (C), NEET
In an exothermic reaction, how does the enthalpy change (∆rH) typically behave?

Options:
A ∆rH is negative
B ∆rH is positive
C ∆rH remains constant
D ∆rH is zero

The correct answer is:
∆rH is negative

Solution:
In an exothermic reaction, heat is evolved, and as a result, ∆rH is negative.

Topic: Chemistry, Thermodynamics (C), NEET
In an experiment, 2.4 g of iron oxide on reduction with hydrogen gave 1.68 g of iron. In another experiment, 2.9 g of iron oxide gave 2.09 g of iron on reduction. Which law is illustrated from the above data?

Options:
A Law of constant proportions
B Law of multiple proportions
C Law of conservation of mass
D Law of reciprocal proportions

The correct answer is:
Law of constant proportions

Solution:
In first experiment,
24 g of iron oxide gave 1.68 g of iron.
Mass of iron oxide =2.4 g
Mass of iron =1.68 g
Mass of oxygen =2.4-1.68=0.72
Ratio of masses of iron and oxygen =1.680.72=7:3
In second experiment,
Mass of iron oxide =2.9 g
Mass of iron =2.09 g
Mass of oxygen =2.9-2.09=0.81
Ratio of masses of iron and oxygen =2.090.81=7:3

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
In an experiment, a boy plots a graph between vmax2 and amax2 for a simple pendulum for different values of (small) amplitudes, where vmax and amax is the maximum velocity and the maximum acceleration respectively. He found the graph to be a straight line with a negative slope, making an angle of 30Â° when the experiment was conducted on the earth surface. When the same experiment was conducted at a height h above the surface, the line was at an angle of 60Â°. The value of h is [radius of the earth = R]

Options:
A 0.5R
B 0.24R
C 0.73R
D R

The correct answer is:
0.73R

Solution:
For a simple pendulum,

ω=gl

vmax=Aω

amax=-Aω2

vmaxamax2=1ω2=tanθ

⇒ω=gl=1tan θ

∴ g=ltan θ

∴ gsurfacegheight=tan60°tan30°=R+h2R2
⇒h=0.73R

Topic: Physics, Gravitation, JEE Main
In an experiment a convex lens of focal length 15 cm is placed coaxially on an optical bench in front of a convex mirror at a distance of 5 cm from it. It is found that an object and its image coincide, if the object is placed at a distance of 20 cm from the lens. The focal length of the convex mirror is-

Options:
A 20.0 cm
B 30.5 cm
C 25.0 cm
D 27.5 cm

The correct answer is:
27.5 cm

Solution:
Image by Lens -
-1-20+1v=1+15
V=+60
Distance of Image from mirror =60-5=55 cm
Path retraces when that image in at center of curvature of the mirror. Hence
R=55 cm
fmirror=R2=27.5 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2017 (09 Apr Online)
In an experiment, a small steel ball falls through a liquid at a constant speed of $10 \mathrm{~cm} / \mathrm{s}$. If the steel ball is pulled upward with a force equal to twice its effective weight, how fast will it move upward ?

Options:
A
$5 \mathrm{~cm} / \mathrm{s}$

B
Zero

C
$10 \mathrm{~cm} / \mathrm{s}$

D
$20 \mathrm{~cm} / \mathrm{s}$

The correct answer is:
$10 \mathrm{~cm} / \mathrm{s}$

Solution:
Weight of the body
$\mathrm{W}=\mathrm{mg}=\frac{4}{3} \pi \mathrm{r}^3 \rho \mathrm{g}$
$\mathrm{T}=\frac{4}{3} \pi \mathrm{r}^3 \sigma \mathrm{g}$
and $\mathrm{F}=6 \pi \eta \mathrm{vr}$
When the body attains terminal velocity net force acting on the body is zero. i.e.,
$\mathrm{W}-\mathrm{T}-\mathrm{F}=0$
And terminal velocity $\mathrm{v}=\frac{2}{9} \frac{\mathrm{r}^2(\rho-\sigma) \mathrm{g}}{\eta}$
As in case of upward motion upward force is twice its effective weight, therefore, it will move with same speed $10 \mathrm{~cm} / \mathrm{s}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2013 (25 Apr Online)
In an experiment, a sphere of aluminium of mass 0.20 kg is heated up to 150°C . Immediately, it is put into water of volume 150 cc at 27oC kept in a calorimeter of water equivalent to 0.025 kg . The final temperature of the system is 40oC . The specific heat of the aluminium is(take 4.2 Joule=1 calorie )

Options:
A 434 J kg-1 oC
B 378 J kg-1°C
C 315 J kg-1 oC
D 476 J kg-1 oC

The correct answer is:
434 J kg-1 oC

Solution:
Let S be the specific heat of aluminium
Q given = Q used
⇒1000×0.2×S×(150−40)=150×1×(40−27)+25×1×(40−27)
⇒200×S×110=150×13+25×13
⇒S=13×175200×110=0.1034 cal g-1 °C

⇒S=103.4 ×4.2 J kg-1 °C=434 J kg-1°C

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2017 (08 Apr Online)
In an experiment according to set up, when E1=12 V and internal resistance zero, e=2 V. The galvanometer reads zero, then X would be-

Options:
A 200
B 500
C 100
D 10

The correct answer is:
100

Solution:
Voltage across X is 2v
So E1500+XX=2
â12500+X X=2
â12 X=1000+2X
â10X=1000
X=100 Î©

Topic: Physics, Current Electricity, JEE Main
In an experiment, brass and steel wires of length 1 m each with areas of cross section 1 mm2 are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2 mm is,
[Given, the Young's Modulus for steel and brass are, respectively, 120×109 N/m2 and 60×109 N/m2 ]

Options:
A 8.0×106 N/m2
B 1.2×106 N/m2
C 0.2×106 N/m2
D 1.8×106 N/m2

The correct answer is:
8.0×106 N/m2

Solution:
Stress will be same in both rods
Stress =Y1Δl1l=Y2Δl2l … (1)
And
Δl1+Δl2=0.2×10-3 m … (2)
From equation (1) and (2)
Stress ×lY1+lY2=0.2×10-3 m
Stress ×1120×109+160×109=0.2×10-3 m
Stress =8×106 N/m2

Topic: Physics, Mechanical Properties of Solids, JEE Main