datavorous/entrance-exam-dataset huggingface dataset:
A physical quantity Q is found to depend on quantities a, b, c by the relation Q=a4b3c2. The percentage error in a, b and c are 3%, 4% and 5% respectively. Then, the percentage error in Q is:

Options:
A 66%
B 43%
C 34%
D 14%

The correct answer is:
34%

Solution:
Given quantity is Q=a4b3c2.

The relative error in measuring Q is given by

ΔQQ=4Δaa+3Δbb+2Δcc ...1

Hence, the required percentage error is given by

ΔQQ×100=4Δaa×100+3Δbb×100+2Δcc×100=4×3%+3×4%+2×5%=12%+12%+10%=34%

Topic: Physics, Mathematics in Physics, JEE Main
A physical quantity Q is found to depend on quantities a, b, c by the relation Q=a4b3c2. The percentage error in a, b and c are 3%, 4% and 5% respectively. Then, the percentage error in Q is:

Options:
A 66%
B 43%
C 34%
D 14%

The correct answer is:
34%

Solution:
Given quantity is Q=a4b3c2.

The relative error in measuring Q is given by

ΔQQ=4Δaa+3Δbb+2Δcc ...1

Hence, the required percentage error is given by

ΔQQ×100=4Δaa×100+3Δbb×100+2Δcc×100=4×3%+3×4%+2×5%=12%+12%+10%=34%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2024 (29 Jan Shift 2)
A physical quantity y is represented by the formula y=m2r-4 gxl-32 If the percentage errors found in y, m, r, l and g are 18, 1, 0.5, 4 and p respectively, then find the value of x and p.

Options:
A 5 and ±2
B 4 and ±3
C 163 and ±32
D 8 and ±2

The correct answer is:
163 and ±32

Solution:
Δyy=2Δmm+4Δrr+xΔgg+32Δll

18=2(1)+4(0.5)+xp+32(4)

8=xp

By checking from options.

x=163, p=±32

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (27 Jul Shift 2)
A physical quantity z depends on four observables a, b, c and d, as z=a2b2/3cd3. The percentage of error in the measurement of a, b, c and d are 2%,1.5%,4% and 2.5% respectively. The percentage of error in z is :

Options:
A 12.25%
B 16.5%
C 13.5%
D 14.5%

The correct answer is:
14.5%

Solution:
Δtt=2Δaa+23Δbb+12Δcc+3Δdd

=2×2+23×1.5+12×4+3×2.5

=4+1+2+7.5

=14.5%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2020 (05 Sep Shift 1)
A physician wishes to prepare a buffer solution at $p H=3.58$ that efficiently resists changes in $p H$ yet contains only small concentration of the buffering agents. Which of the following weak acids together with its sodium salt would be best to use

Options:
A $m$ - chlorgbenzoic acid $\left(p K_a=3.98\right)$
B $p$-chlorocionamic acid $\left(p K_a=4.41\right)$
C $2,5^{-}$ dihydroxy benzoic acid $\left(p K_a=2.97\right)$
D Acetoacetic acid $\left(p K_a=3.58\right)$

The correct answer is:
$2,5^{-}$ dihydroxy benzoic acid $\left(p K_a=2.97\right)$

Solution:
$p H=p K_a+\log \frac{[\text { Salt }]}{[\text { Acid }]}$. For small concentration of buffering agent and for maximum buffer gapacit $\frac{[\text {Salt }]}{[\text { Acid] }} \approx 1$.

Topic: Chemistry, Ionic Equilibrium, JEE Main
A piece of bone of an animal from a ruin is found to have ${ }^{14} \mathrm{C}$ activity of 12 disintegrations per minute per gm of its carbon content. The ${ }^{14} \mathrm{C}$ activity of a living animal is 16 disintegrations per minute per gm. How long ago nearly did the animal die? (Given half life of ${ }^{14} \mathrm{C}$ is $\mathrm{t}_{1 / 2}=5760$ years)

Options:
A
1672 years

B
2391 years

C
3291 years

D
4453 years

The correct answer is:
2391 years

Solution:
Given, for ${ }^{14} \mathrm{C}$
$$
\begin{aligned}
&A_0=16 \text { dis } \min ^{-1} \mathrm{~g}^{-1} \\
&\mathrm{~A}=12 \text { dis } \min ^{-1} \mathrm{~g}^{-1} \\
&\mathrm{t}_{1 / 2}=5760 \text { years }
\end{aligned}
$$
Now, $\lambda=\frac{0.693}{\mathrm{t}_{1 / 2}}$
$\lambda=\frac{0.693}{5760}$ per year
Then, from, $\mathrm{t}=\frac{2.303}{\lambda} \log _{10} \frac{\mathrm{A}_0}{\mathrm{~A}}$
$$
=\frac{2.303 \times 5760}{0.693} \log _{10} \frac{16}{12}
$$
$$
=\frac{2.303 \times 5760}{0.693} \log _{10} 1.333
$$
$=\frac{2.303 \times 5760 \times 0.1249}{0.693}$
$=2390.81 \approx 2391$ years

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2014 (12 Apr Online)
A piece of burnt wood of mass 20 g is found to have a 14C activity of 4 decay s-1 . How long has the tree that this wood belonged to be dead? Given T12 of 14C=5730 year.

Options:
A 1840
B 1830
C 1820
D 1860

The correct answer is:
1840

Solution:
The decay constant of 14C is
λ=0.693T12=0.6935730 ×3.16 ×107=3.83 ×10-12 s-1
∵ 1 year=3.17 ×107s
To find the number of 14C nuclei in 20 g of burnt wood, we first calculate the number of 12C nuclei in 20 g of carbon (burnt wood).
Thus N12C=6.02 ×102312 ×20 ≃1024
Now, assuming that the ratio of 14C to 12C is 1.3 ×10-12, the number of 14C nuclei in 20g before decay is, N014C=1.3 ×10-12 1024=1.3 ×1012
We thus have for the initial activity of the sample
R0=N0λ=1.3 ×1012 ×3.83 ×10-12=4.979 decay s-1
≈5 decay s-1
The age of the sample can now be calculated from the relation,
R=R0e-λt
Or e-λt=R0R
Or t=1λloge⁡R0R
It is given that R = 4 decay s-1 and we have calculated R0=5 decay s-1 .
Thus t=1λloge⁡54=loge⁡1.253.83 ×10-12
=0.2233.83 ×10-12=0.58 ×1011 s
=1840 year .

Topic: Physics, Nuclear Physics, JEE Main
A piece of cheese is located at 12, 10 in a coordinate plane. A mouse is at 4, −2 and is running up the line y=−5x+18 to get closer to the piece of cheese. At the point a, b , the mouse starts getting farther from the cheese rather than closer to it. Then the value of a+b is

Options:
A 6
B 10
C 18
D 14

The correct answer is:
10

Solution:
a,âb is the foot of the perpendicular from 12,â10 on the line y+5x=18

â a - 1 2 5 = b - 1 0 1 = - 1 0 + 5 1 2 - 1 8 2 6
â a, b = 2, 8

Topic: Mathematics, Straight Lines, JEE Main
A piece of copper and another of germanium are cooled from room temperature to $77 \mathrm{~K}$, the resistance of

Options:
A
each of them increases

B
each of them decreases

C
copper decreases and germanium increases

D
copper increases and germanium decreases

The correct answer is:
copper increases and germanium decreases

Solution:
Copper is metallic conductor and germanium is semiconductor therefore as temperature decreases resistance of good conductor decreases while for semiconductor it increases.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2004
A piece of copper and another of germanium are cooled from room temperature to $77 \mathrm{~K}$, the resistance of

Options:
A each of them increases
B each of them decreases
C copper decreases and germanium increases
D copper increases and germanium decreases

The correct answer is:
copper decreases and germanium increases

Solution:
Copper is a conductor. Germanium is a semiconductor. When cooled, the resistance of copper decreases and that of germanium increases.

Topic: Physics, Semiconductors, JEE Main
A piece of copper and another of germanium are cooled from room temperature to 77 K. The resistance of

Options:
A each of them increases
B each of them decreases
C copper decreases and of germanium increases
D copper increases and of germanium decreases

The correct answer is:
copper decreases and of germanium increases

Solution:
Copper is a conductor.
Germanium is a semiconductor
When cooled, the resistance of copper decreases and that of germanium increases.

Topic: Physics, Semiconductors, JEE Main
A piece of ice falls from a height h so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of h is:
(Latent heat of ice is 3.4Ã105âJ kg-1 and g=10âN kg-1)

Options:
A 34  km
B 544  km
C 136  km
D 68  km

The correct answer is:
136  km

Solution:
14mgh=mL
h=4Lg=4Ã3.4Ã10510=13.6Ã104
=136Ã103 m
=136 km

Topic: Physics, Thermodynamics, JEE Main
A piece of metal floats in mercury. The coefficients of volume expansion of the metal and mercury are Î³ 1 and Î³ 2 , respectively. If the temperatures of both mercury and the metal are increased by Î T , the fraction of the volume of the metal submerged in mercury changes by the factor of

Options:
A 1 + γ 2 Δ T 1 + γ 1 Δ T
B 1 + γ 2 Δ T
C 1 + γ 1 Δ T
D 1 + γ 2 Δ T 1 - γ 1 Δ T

The correct answer is:
1 + γ 2 Δ T 1 + γ 1 Δ T

Solution:
As, γ 1 = Δ V 1 V 1 Δ T and γ 2 = Δ V 2 V 2 Δ T
⇒ Δ V 1 = γ 1 V 1 Δ T
and Δ V 2 = γ 2 V 2 Δ T
Fraction of volume submerged before temperature is raised is given by fâ¡ = ρ 1 / ρ 2 .
Fraction of volume submerged after the temperature is raised is given by fâ¡ ′ = ρ 1 ′ / ρ 2 ′ .
α ′ = ρ 1 1 + γ 1 Δ T 1 + γ 2 Δ T ρ 2 = ρ 1 ρ 2 1 + γ 2 Δ T 1 + γ 1 Δ T
⇒ α ′ = fâ¡ 1 + γ 2 Δ T 1 + γ 1 Δ T
α ′ fâ¡ = 1 + γ 2 Δ T 1 + γ 1 Δ T

Topic: Physics, Thermal Properties of Matter, JEE Main
A piece of solid weighs 120 gm in open air, 90 gm in water and 60 gm in a liquid. The relative density of solid and that of liquid are respectively.

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

The correct answer is:
4, 2

Solution:
According to question
120 = ρVg .....(i)
(where ρ is density of solid)
90=(ρ-ρw)Vg⇒ρwVg=30 .....(ii)
(where ρw is density of water)
60=ρ-ρlVg⇒ρlVg=60 .....(iii) )
(where ρl is density of water)
Form (i), (ii), (iii)
ρρw=12030=41
ρlρw=6030=21

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A piece of wood from a recently cut tree shows 20 decays per minute. A wooden piece of the same size placed in a museum (obtained from a tree cut many years back) shows 2 decays per minute. If the half-life of C14 is 5730 years, then the age of the wooden piece placed in the museum is approximately
[This question was awarded a bonus and proper correction was made to avoid that]

Options:
A 10439 years
B 39049 years
C 19042 years
D 13094 years

The correct answer is:
19042 years

Solution:
We know that,

The rate of decay at the time t is

A=A0e-λt

2=20·e-λt

∴ 110=e-λt

∴ eλt=10

⇒lneλt=ln10

⇒λt=ln10

∵ λ=ln2T1/2

t=ln10ln2T1/2

∵ ln10=2.303 and ln2=0.693

t=2.3030.693×5730

=3.32×5730

≈19042 years

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2014 (19 Apr Online)
A piece of wood from an archaeological sample has 5.0 counts min $^{-1}$ per gram of $\mathrm{C} - 14,$ while a fresh sample of wood has a count of $15.0 \mathrm{min}^{-1} \mathrm{g}^{-1}$. If half-life of $\mathrm{C}-14$ is $5770 \mathrm{yr}$ the age of the archaeological sample is

Options:
A $8.500 \mathrm{yr}$
B $9,200 \mathrm{yr}$
C $10,000 \mathrm{yr}$
D $11.000 \mathrm{yr}$

The correct answer is:
$9,200 \mathrm{yr}$

Solution:
Age of archaeological sample.
$$
t=\frac{2.303}{k} \log \left(\frac{N_{0}}{N}\right)
$$
(Here, $N_{0}=$ initial activity. $\mathrm{N}=$ remaining activity)
$$
\begin{aligned}
k &=\frac{0.693}{t_{1 / 2}} \\
\quad \text { (where } t_{1 / 2} &=\text { half-life) } \\
&=\frac{0.693}{5770} \mathrm{yr}^{-1}
\end{aligned}
$$
$\therefore$
$$
t=\frac{2.303 \times 5770}{0.693} \log \left(\frac{15}{5}\right)
$$
$$
\begin{array}{l}
=19175.0 \log 3 \mathrm{yr} \\
=9148 \mathrm{yr} \approx 9200 \mathrm{yr}
\end{array}
$$

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2014
A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms-1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is: g=10 ms-2

Options:
A 40 m
B 20 m
C 10 m
D 30 m

The correct answer is:
40 m

Solution:
Using relative velocity

arel=0
vrel=100
Time t for collision,
100=vrel×t
t=100100=1 sec
vbullet=100-1×10=90m/s
vparticle=10×1=10m/s
S=100×1-12×10×1=95 m

Pi=Pf
⇒90×0.02-10×0.03=0.05V
V=30m/s⇒h=v22g=90020=45 m
So from top of building
45-5=40 m

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A piece of wood of mass 0.03 kg is dropped from the top of a 100 m height building. At the same time, a bullet of mass 0.02 kg is fired vertically upward, with a velocity 100 ms-1, from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is: g=10 ms-2

Options:
A 40 m
B 20 m
C 10 m
D 30 m

The correct answer is:
40 m

Solution:
Using relative velocity

arel=0
vrel=100
Time t for collision,
100=vrel×t
t=100100=1 sec
vbullet=100-1×10=90m/s
vparticle=10×1=10m/s
S=100×1-12×10×1=95 m

Pi=Pf
⇒90×0.02-10×0.03=0.05V
V=30m/s⇒h=v22g=90020=45 m
So from top of building
45-5=40 m

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2019 (10 Jan Shift 1)
A pipe closed at one end has length 83 cm. The number of possible natural oscillations of air column whose frequencies lie below 1000 Hz are (velocity of sound in air =332 m s-1)

Options:
A 3
B 4
C 5
D 6

The correct answer is:
5

Solution:
Fundamental frequency
n= V 4L = 332 4Ã83Ã 10 â2 =100 .
Hence possible frequencies are n=100 .
n 1 =300.
n 2 =500.
n 3 =700.
n 5 =900.

Topic: Physics, Waves and Sound, JEE Main
A pipe of length 1.5 m closed at one end is filled with gas and resonates in its fundamental mode with a tuning fork. Another open organ pipe of same dimensions filled with air resonates in its fundamental mode with the same tuning fork. If this experiment is performed at 30Â° C, then the speed of sound at 0 Â°C in the same gas is (Speed of sound in air is 360 m/s)

Options:
A 580 m/s
B 683 m/s
C 880 m/s
D 743 m/s

The correct answer is:
683 m/s

Solution:
For open pipe f=v2â=3602Ã15=120 Hz

For the closed at one end pipe, let the velocity of sound in gas at 30 Â°C be v'

f'=v'4â=v'4Ã1.5=v'6

Since same tuning fork is used frequency will be same and thus,

f'=fâv'=720 m/s

Also

vâTâv=720Ã273303=683 m/s.

Topic: Physics, Waves and Sound, JEE Main
A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

Options:
A 12
B 8
C 6
D 4

The correct answer is:
6

Solution:
fundamental frequency

L=λ4v1=v4L

L = λ 2 + λ 4 = 3 λ 4
v2=3v4L

∴ v=oddv4L
Now L = 8 5 cm = 0.85 m
v=340m/s
= odd 3 4 0 4 × 0.85
=odd×100 Hz
Hence possible frequencies below 1250 Hz are
100 Hz, 300 Hz, 500 Hz...........1100 Hz
Hence 6

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2014 (06 Apr)
A pipe open at both ends has a fundamental frequency f in air. The pipe is dipped vertically in water so that half of it is in water. The fundamental frequency of the air column is now:

Options:
A 2f
B f
C f2
D 3f4

The correct answer is:
f

Solution:
f=v2l
f′=v4l′=v4l2=f

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2016 (03 Apr)
A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 3 7 . 0 ∘ C. As it does so, it absorbs 208 J of heat. The values of q and w for the process will be
(R = 8.314 J/mol K) (ln7.5 = 2.01)

Options:
A q = - 2 0 8 J , w = + 2 0 8 J
B q = + 2 0 8 J , wâ¡ = + 2 0 8 J
C q = + 2 0 8 J , wâ¡ = - 2 0 8 J
D q = - 2 0 8 J , wâ¡ = - 2 0 8 J

The correct answer is:
q = + 2 0 8 J , w⁡ = - 2 0 8 J

Solution:
Since, temperature remains constant, the process will be isothermal by nature.
For Isothermal reversible expansion:
Δ E = 0 = q + w
q = - w
q = + 2 0 8 J {heat absorbed so +ve}
w = - 2 0 8 J {expansion so work done is -ve}

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2013 (07 Apr)
A piston filled with 0.04 mol of an ideal gas expands reversibly from 50.0 mL to 375 mL at a constant temperature of 3 7 . 0 â C. As it does so, it absorbs 208 J of heat.The values of q and w for the process will be
(R = 8.314 J/mol K) (ln 7.5 = 2.01)

Options:
A q = - 2 0 8 J , w = + 2 0 8 J
B q = + 2 0 8 J , wâ¡ = + 2 0 8 J
C q = + 2 0 8 J , wâ¡ = - 2 0 8 J
D q = - 2 0 8 J , wâ¡ = - 2 0 8 J

The correct answer is:
q = + 2 0 8 J , w⁡ = - 2 0 8 J

Solution:
Since, temperature remains constant, the process will be isothermal by nature.
For Isothermal reversible expansion:
Î E = 0 = q + w
q = - w
q = + 2 0 8 J {heat absorveed so +ve}
w = - 2 0 8 J {expansion so work done by system -ve}

Topic: Chemistry, Thermodynamics (C), NEET
A piston is performing S.H.M. in the vertical direction with a frequency of \( 0.5 \mathrm{~Hz} \). A block of \( 10 \)
\( \mathrm{kg} \) is placed on the piston. The maximum amplitude of the system such that the block remains
in contact with the piston is

Options:
A \( 1.5 \mathrm{~m} \)
B \( 1 \mathrm{~m} \)
C \( 0.1 \mathrm{~m} \)
D \( 0.5 \mathrm{~m} \)

The correct answer is:
\( 1 \mathrm{~m} \)

Solution:
(B)
$\mathrm{f}=0.5 \mathrm{~Hz}$
$\omega=2 \pi \mathrm{f}=\pi$
for block to remain in contact with the piston at amplitude position
weight A the block = Force due to oscillation.
mg $=$ ma
mg $=\mathrm{m}\left(\omega^{2} \mathrm{~A}\right)$
$\mathrm{A}=\frac{g}{\omega^{2}}=\frac{10}{\Pi^{2}}=1 \mathrm{~m}$

Topic: Physics, Oscillations, KCET, KCET 2019
A planar frame is made of thin uniform rods. The length of section AB and EF is l1 and its coefficient of linear expansion is α1. The length of section CD is l2 and its coefficient of linear expansion is α2. CB and DE are of the same length having coefficient of linear expansion α2. Points A, B, E and F reside on same line, that is, sections AB and EF overlap. Then the ratio l1l2 for which the distance between end A and end F remains the same at all temperatures is α2nα1 then find the value of n.

Options:

The correct answer is:
2

Solution:
For distance between A and F to remain constant, extension in CD = extension in AB + extension in EF
∴ Δl2=2Δl1 ⇒ l2α2Δθ2=2l1α1Δθ
Or l1l2=α22α1

n=2

Topic: Physics, Thermal Properties of Matter, JEE Main
A planar loop of wire rotates in a uniform magnetic field. Initially, at t=0 , the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of 10s about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at:

Options:
A 2.5 s and 7.5 s
B 2.5 s and 5.0 s
C 5.0 s and 7.5 s
D 5.0 s and 10.0 s

The correct answer is:
2.5 s and 5.0 s

Solution:
Φ=B→·A →=BAcosθ =BAcosωt

e=dϕdt =BAωsinωt

When ωt=π2
∵ϕ will be minimum.
∴e will be maximum
t=π2π5=2.5 s

When ωt=π
∵ϕ will be maximum.
∴e will be minimum
t=ππ/5=5 s

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2020 (07 Jan Shift 2)
A plane bisects the line segment joining the points $(1,2,3)$ and $(-3,4,5)$ at right angles. Then this plane also passes through the point.

Options:
A
$(-3,2,1)$

B
$(3,2,1)$

C
$(1,2,-3)$

D
$(-1,2,3)$

The correct answer is:
$(-3,2,1)$

Solution:
Since the plane bisects the line joining the points $(1,2,3)$ and $(-3,4,5)$ then the plane passes through the midpoint of the line which is :
$$
\begin{aligned}
\left(\frac{1-3}{2}, \frac{2+4}{2}, \frac{5+3}{2}\right) \equiv\left(\frac{-2}{2},\right.&\left.\frac{6}{2}, \frac{8}{2}\right) \\
& \equiv(-1,3,4) .
\end{aligned}
$$
As plane cuts the line segment at right angle, so the direction cosines of the normal of the plane are $(-3-1,4-2,5-3)=(-4,2,2)$ So the equation of the plane is :
$$
-4 x+2 y+2 z=\lambda
$$
As plane passes through $(-1,3,4)$ so
$$
-4(-1)+2(3)+2(4)=\lambda \Rightarrow \lambda=18
$$
Therefore, equation of plane is :
$$
-4 x+2 y+2 z=18
$$
Now, only $(-3,2,1)$ satisfies the given plane as $-4(-3)+2(2)+2(1)=18$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A plane containing the point (3,2,0) and the line x-11=y-25=z-34 also contains the point

Options:
A (0,Â 7,Â -10)
B (0, 7, 10)
C (0,Â 3,Â 1)
D (0,Â -3,Â 1)

The correct answer is:
(0, 7, 10)

Solution:
The Direction ratio perpendicular to the plane would be cross product of
i^+5j^+4k^ and 3-1i^+2-2j^+0-3k^
= i^j^k^15420-3
= -15i^+11j^-10k^
â´ Equation of plane would be
-15x+11y-10z=Î»
As plane contain 3,2,0
-45+22-0=Î»
-23=Î»
-15x+11y-10z+23=0
This passes through (0,7,10)

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2015 (11 Apr Online)
A plane convex lens of refractive index $1.5$ and radius of curvature $30 \mathrm{~cm}$ is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of the size of the object?

Options:
A
$20 \mathrm{~cm}$

B
$30 \mathrm{~cm}$

C
$60 \mathrm{~cm}$

D
$80 \mathrm{~cm}$

The correct answer is:
$20 \mathrm{~cm}$

Solution:
$$
\begin{aligned}
& \frac{1}{F}=\frac{2}{f_1}+\frac{1}{f_m} \\
& \text { and } \frac{1}{f_1}=(1.5-1)\left(\frac{1}{\infty}-\frac{1}{-30}\right)=\frac{1}{60} \\
& \text { and } f_m=15 \mathrm{~cm} . \\
& \therefore F=10 \mathrm{~cm} .
\end{aligned}
$$
Object should be placed at $20 \mathrm{~cm}$ from the lens.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2004
A plane E is perpendicular to the two planes 2x-2y+z=0 and x-y+2z=4, and passes through the point P1,-1,1. If the distance of the plane E from the point Qa,a,2 is 32, then PQ2 is equal to

Options:
A 9
B 12
C 21
D 33

The correct answer is:
21

Solution:
Given, plane P1=2x-2y+z=0,

Whose normal vector is ≡n¯1=2,-2,1

Second plane, P2≡x-y+2z=4,

Whose normal vector is ≡n¯2=1,-1,2

Now let plane perpendicular to P1 and P2 will have normal vector n¯3

Where n¯3=n¯1×n¯2

⇒n→3=i^j^k^2-211-12=-3i^-3j^

Hence, n¯3=-3,-3,0

Equation of plane E through P1,-1,1 and n¯3 as normal vector will be,

-3x-1-3y+1+0z-1=0

⇒ x+y=0≡E

Now distance of point Qa,a,2 from E=2a2

Now given, 2a2=32⇒a=±3

Hence, Q≡±3,±3,2

Now distance PQ=21⇒PQ2=21

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2022 (25 Jul Shift 2)
A plane electromagnetic wave, has frequency of 2.0×1010 Hz and its energy density is 1.02×10-8 J m-3 in vacuum. The amplitude of the magnetic field of the wave is close to 14πε0=9×109Nm2C2 and speed of light =3×108 m s-1.

Options:
A 150 nT
B 160 nT
C 180 nT
D 190 nT

The correct answer is:
160 nT

Solution:
Energy Denisty =12B2μ0

B=2×μ0× Energy density

B=2×4π×10-7×1.02×10-8=160×10-9=160nT

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2020 (02 Sep Shift 1)
A plane electromagnetic wave having a frequency f=23.9 GHz propagates along the positive z- direction in free space. The peak value of the Electric Field is 60 V/m. Which among the following is the acceptable magnetic field component in the electromagnetic wave?

Options:
A B→=2×10-7sin⁡0.5×103z-1.5×1011ti^
B B→=2×10-7sin⁡1.5×103x+0.5×1011tj^
C B→=2×107sin⁡0.5×103z+1.5×1011ti^
D B→=60sin⁡0.5×103x+1.5×1011ti^

The correct answer is:
B→=2×10-7sin⁡0.5×103z-1.5×1011ti^

Solution:
B=Ec=603×108=2×10-7
The magnetic field can be written as,
k=2πλ=2πc/f≈500
ω=2πf≈1.5×1011
B→=2×10-7sin⁡(0.5×103z-1.5×1011t)i^

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A plane electromagnetic wave in a non-magnetic dielectric medium is given by $\vec{E}=\vec{E}_0\left(4 \times 10^{-7} x-50 t\right)$ with distance being in meter and time in seconds. The dielectric constant of the medium is :

Options:
A
$2.4$

B
$5.8$

C
$8.2$

D
$4.8$

The correct answer is:
$5.8$

Solution:
No Solution Available

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2013 (22 Apr Online)
A plane electromagnetic wave is incident on a plane surface of area A, normally and is perfectly reflected. If energy E strikes the surface in timet then average pressure exerted on the surface is $(\mathrm{c}=$ speed of light $)$

Options:
A zero
B $\mathrm{E} / \mathrm{Atc}$
C $2 \mathrm{E} / \mathrm{Atc}$
D $\mathrm{E} / \mathrm{c}$

The correct answer is:
$2 \mathrm{E} / \mathrm{Atc}$

Solution:
Incident momentum, $\mathrm{p}=\frac{\mathrm{E}}{\mathrm{c}}$ For perfectly reflecting surface with normal incidence

$\Delta \mathrm{p}=2 \mathrm{p}=\frac{2 \mathrm{E}}{\mathrm{c}}$

$\mathrm{F}=\frac{\Delta \mathrm{p}}{\Delta \mathrm{t}}=\frac{2 \mathrm{E}}{\mathrm{ct}}$

$\mathrm{P}=\frac{\mathrm{F}}{\mathrm{A}}=\frac{2 \mathrm{E}}{\mathrm{ctA}}$

Topic: Physics, Dual Nature of Matter, JEE Main
A plane electromagnetic wave of frequency 20 MHz propagates in free space along x-direction. At a particular space and time E→=6.6j ^ V m-1. What is B→ at this point?

Options:
A 2.2×10-8k^T
B -2.2×10-8i^T
C -2.2×10-8k^T
D 2.2×10-8i ^T

The correct answer is:
2.2×10-8k^T

Solution:
It is given that

E→=6.6j^( V m-1).

The relation between the electric, magnetic field and speed of light is E→×B→=c→.
Clearly, B→=B0(k^)

Using the relation,
cB0=E
⇒B0=Ec=6.63×108=2.2×10-8 T

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2023 (11 Apr Shift 2)