datavorous/entrance-exam-dataset huggingface dataset:
A 16 Ω wire is bent to form a square loop. A 9 V supply having an internal resistance of 1 Ω is connected across one of its sides. The potential drop across the diagonals of the square loop is __________×10-1 V.

Options:

The correct answer is:
45

Solution:
Here assume current as

By KVL in the outer loop

9-12i-4i=0

16i=9

8i=92=4.5

=45×10-1

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (25 Jul Shift 2)
A $20 \mathrm{~cm}$ long capillary tube is dipped in water. The water rises up to $8 \mathrm{~cm}$. If the entire arrangement is put in a freely falling elevator the length of water column in the capillary tube will be

Options:
A
$8 \mathrm{~cm}$

B
$10 \mathrm{~cm}$

C
$4 \mathrm{~cm}$

D
$20 \mathrm{~cm}$

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

Solution:
Water will rise to the full length of capillary tube

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2005
A 20 cm long metallic rod is rotated with 210 rpm about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field 0.2 T parallel to the axis exists everywhere. The emf developed between the centre and the ring is _________mV.

(Take π=227)

Options:

The correct answer is:
88

Solution:
The data given is

ω=210×2π60 rad s-1=7π rad s-1l=0.2 mB=0.2 T

The emf is given by

ε=Bωl22

=0.2×7π×0.222

=88 mV

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (15 Apr Shift 1)
A $20 \mathrm{~g}$ copper block is suspended by a vertical spring causing $1 \mathrm{~cm}$ elongation over the natural length of spring. If a beaker of water is placed below the block so that the copper block is completely immersed in the liquid, the elongation of the spring is
(Density of copper $9000 \mathrm{~kg} \mathrm{~m}^{-3}$, Density of water 1000 $\mathrm{kg} \mathrm{m}^{-3}, \mathrm{~g}=10 \mathrm{~ms}^{-2}$ )

Options:
A $0.25 \mathrm{~cm}$
B $0.15 \mathrm{~cm}$
C $0.78 \mathrm{~cm}$
D $0.89 \mathrm{~cm}$

The correct answer is:
$0.89 \mathrm{~cm}$

Solution:
Mass of Block, $\mathrm{m}=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}$ Elongation, $x=1 \mathrm{~cm}=0.01 \mathrm{~m}$
Spring elongation after immersed in water, $\mathrm{x}^{\prime}=$ ?
Density, $\rho=9000 \mathrm{~kg} / \mathrm{m}^3$
Density of water, $\rho_\omega=1000 \mathrm{~kg} / \mathrm{m}^3$
At equilibrium, $\mathrm{kx}=\mathrm{mg}$
$$
\mathrm{k}=\frac{\mathrm{mg}}{\mathrm{x}}=\frac{0.002 \times 10}{0.01}=2 \mathrm{~N} / \mathrm{m}
$$
when mass is immersed in water
$$
\begin{aligned}
& \mathrm{mg}=\mathrm{kx}^{\prime}+\left(\rho_\omega \mathrm{V}\right) \mathrm{g} \\
& \mathrm{mg}=\mathrm{kx}^{\prime}+\rho_\omega\left[\frac{\mathrm{m}}{\rho_{\mathrm{c}}}\right] \mathrm{g} \\
& 20 \times 10^{-3} \times 10=2 \times \mathrm{x}^{\prime}+\frac{1000}{9000} \times 20 \times 10^{-3} \times 10 \\
& 0.2=2 \mathrm{x}^{\prime}+\frac{1}{9} \times 0.2 \\
& \frac{0.1}{10}=\mathrm{x}^{\prime}+\frac{1}{90} \\
& \mathrm{x}^{\prime}=\frac{1}{10}-\frac{1}{90}=0.89 \\
& \mathrm{x}^{\prime}=0.89
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A 2.0 g sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of CO2 ceases. The volume of CO2 at 750 mm Hg pressure and at 298 K is measured to be 123.9 mL. A 1.5 g of the same sample requires 150 mL of (M/10) HCl for complete neutralization. Calculate the percentage composition of Na 2 SO 4 in the original mixture. If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
63

Solution:
In 2g sample â CO2 Liberated
Moles of CO2 = 7 5 0 7 6 0 Ã 1 2 3 . 9 1 0 0 0 Ã 1 0 . 0 8 2 Ã 1 2 9 8 = 5 Ã 1 0 - 3
moles = 5 mM â¡ 1 0 mM of NaHCO3 in mixture
Moles of NaHCO3 in 2g sample = 2 x 5 x 10 mM
1.5g sample â¡ 150 Ã 1 1 0 = 15 mM of HCl
2.0g sample â¡ 15 Ã 2 1 Â· 5 = 20 mM Â·
20 mM HCl will be utilised by
10 mM NaHCO 3 + 5 mM Na 2 CO 3 present in 2g sample
NaHCO 3 = 10 mM = 10 Ã 1 0 - 3 Ã 8 4 = 0.84g 42%
Na 2 CO 3 = 5 mM = 5 Ã 1 0 - 3 Ã 1 0 6 = 0.53g = 26.5%
Na 2 SO 4 = (By difference) = 0.63g 31.5%

Topic: Chemistry, Redox Reactions, JEE Main
A 2.0 g sample of a mixture containing sodium carbonate, sodium bicarbonate and sodium sulphate is gently heated till the evolution of CO2 ceases. The volume of CO2 at 750 mm Hg pressure and at 298 K is measured to be 123.9 mL. A 1.5 g of the same sample requires 150 mL of (M/10) HCl for complete neutralization. Calculate the percentage composition of Na 2 SO 4 in the original mixture.

Options:

The correct answer is:
31.50

Solution:
In 2g sample â CO2 Liberated
Moles of CO2 = 7 5 0 7 6 0 Ã 1 2 3 . 9 1 0 0 0 Ã 1 0 . 0 8 2 Ã 1 2 9 8 = 5 Ã 1 0 - 3
moles = 5 mM â¡ 1 0 mM of NaHCO3 in mixture
Moles of NaHCO3 in 2g sample = 2 x 5 x 10 mM
1.5g sample â¡ 150 Ã 1 1 0 = 15 mM of HCl
2.0g sample â¡ 15 Ã 2 1 Â· 5 = 20 mM Â·
20 mM HCl will be utilised by
10 mM NaHCO 3 + 5 mM Na 2 CO 3 present in 2g sample
NaHCO 3 = 10 mM = 10 Ã 1 0 - 3 Ã 8 4 = 0.84g 42%
Na 2 CO 3 = 5 mM = 5 Ã 1 0 - 3 Ã 1 0 6 = 0.53g = 26.5%
Na 2 SO 4 = (By difference) = 0.63g 31.5%

Topic: Chemistry, Redox Reactions, JEE Main
A 20 H inductor coil is connected to a 10 Ω resistance in series as shown in figure. The time at which rate of dissipation of energy (Joule's heat) across resistance is equal to the rate at which magnetic energy is stored in the inductor, is:

Options:
A 1 2 ln2
B 2ln2
C 2 ln2
D ln2

The correct answer is:
2ln2

Solution:
Rate of dissipation of energy in resistor =i2R
Rate of energy stored in inductor =ddt12Li2=Lididt
i2R=Lididt
didt=iRL …i
In L-R circuit:

i=i01-e-tτ (∴ τ=LR=2)
didt=i0τe-t/τ
From equation (i),

i0τe-t/τ=i01-e-tτRL
e-t/τ=1-e-t/τ
e-tτ=12 ⇒t=τ ln2
=2 ln2

Topic: Physics, Alternating Current, JEE Main, JEE Main 2019 (08 Apr Shift 1)
A $20 \mathrm{~kg}$ block is initially at rest on a rough horizontal surface. A horizontal force of $75 \mathrm{~N}$ is required to set the block in motion. After it is in motion, a horizontal force of $60 \mathrm{~N}$ is required to keep the block moving with constant speed. The coefficient of static friction is

Options:
A 0.38
B 0.44
C 0.52
D 0.60

The correct answer is:
0.38

Solution:
Coefficient of friction
$\mu_s=\frac{F_l}{R}=\frac{75}{m g}=\frac{75}{20 \times 9.8}=0.38$

Topic: Physics, Laws of Motion, JEE Main
A $0.0020 \mathrm{~m}$ aqueous solution of an ionic compound $\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right) \mathrm{Cl}$ freezes at $-0.00732^{\circ} \mathrm{C}$. Number of moles of ions which $1 \mathrm{~mol}$ of ionic compound produces on being dissolved in water will be $\left(\mathrm{k}_{\mathrm{f}}=-1.86^{\circ} \mathrm{C} / \mathrm{m}\right)$

Options:
A 2
B 3
C 4
D 1

The correct answer is:
2

Solution:
Given,
$$
\begin{aligned}
\text { molality, } \mathrm{m} & =0.0020 \mathrm{~m} \\
\Delta \mathrm{T}_{\mathrm{f}} & =0^{\circ} \mathrm{C}-0.00732^{\circ} \mathrm{C}=0.00732^{\circ} \mathrm{C} \\
\mathrm{k}_{\mathrm{f}} & =1.86^{\circ} \mathrm{C} / \mathrm{m} \\
\Delta \mathrm{T}_{\mathrm{f}} & =\mathrm{i} \cdot \mathrm{k}_{\mathrm{f}} \times \mathrm{m} \\
\mathrm{i} & =\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{f}} \times \mathrm{m}} \\
& =\frac{0.00732}{1.86 \times 0.0020} \\
& =1.96 \approx 2
\end{aligned}
$$
Since, the compound is ionic, so number of moles produced is equal to vant' Hoff factor, $i$. Hence, 2 moles of ions are produced.

Topic: Chemistry, Solutions, JEE Main
A $0.0020 \mathrm{M}$ aqueous solution of an ionic compound $\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right) \mathrm{Cl}$ freezes at $-0.00732^{\circ} \mathrm{C}$. Number of moles of ions which 1 mole of ionic compound produces on being dissolved in water will be : $\left(\mathrm{k}_{\mathrm{f}}=1.86^{\circ} \mathrm{C} / \mathrm{m}\right)$ -

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ik}_{\mathrm{f}} \cdot \mathrm{m}$
$$
\begin{aligned}
\mathrm{i} & =\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{f}} \cdot \mathrm{m}} \\
& =\frac{0.00732}{1.86 \times 0.002}=\frac{0.00732}{0.00372} \\
\mathrm{i} & =2
\end{aligned}
$$
Compound will be $\left.\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\right] \mathrm{NO}_5 \mathrm{NO}_2\right] \mathrm{Cl}$
Total possible ions $=2$

Topic: Chemistry, Chemical Kinetics, JEE Main
A $0.0020 \mathrm{M}$ aqueous solution of an ionic compound $\mathrm{Co}\left(\mathrm{NH}_3\right)_5\left(\mathrm{NO}_2\right) \mathrm{Cl}$ freezes at $-0.00732^{\circ} \mathrm{C}$. Number of moles of ions which 1 mole of ionic compound produces on being dissolved in water will be : $\left(\mathrm{k}_{\mathrm{f}}=1.86^{\circ} \mathrm{C} / \mathrm{m}\right)$ -

Options:
A 1
B 2
C 3
D 4

The correct answer is:
2

Solution:
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{ik}_{\mathrm{f}} \cdot \mathrm{m}$
$$
\begin{aligned}
\mathrm{i} & =\frac{\Delta \mathrm{T}_{\mathrm{f}}}{\mathrm{k}_{\mathrm{f}} \cdot \mathrm{m}} \\
& =\frac{0.00732}{1.86 \times 0.002}=\frac{0.00732}{0.00372} \\
\mathrm{i} & =2
\end{aligned}
$$
Compound will be $\left.\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5\right] \mathrm{NO}_5 \mathrm{NO}_2\right] \mathrm{Cl}$
Total possible ions $=2$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2009 (Mains)
A 24 V battery of internal resistance 4 Î© is connected to a variable resistance. If the rate of heat production in the resistor is maximum, then the current in the circuit is

Options:
A 2Â A
B 3 A
C 4Â A
D 6Â A

The correct answer is:
3 A

Solution:
When R=r (power dissipated is maximum)
∴ R=4 Ω
∴ I=VRnet= 248= 3 A

Topic: Physics, Current Electricity, JEE Main
A 24 V battery of internal resistance 4 Î© is connected to a variable resistance. If the rate of heat production in the resistor is maximum, then the current in the circuit is

Options:
A 2Â A
B 3 A
C 4Â A
D 6Â A

The correct answer is:
3 A

Solution:
When R=r (power dissipated is maximum)
â´ R=4 Î©
â´I=VRnet= 248= 3 A

Topic: Physics, Current Electricity, JEE Main
A 25 cm long solenoid has the radius 2 cm and 500 turns. It carries a current of 15 A. If it is equivalent to a magnet of the same size and magnetization M→ Magnetic momentvolume,then M→ is:

Options:
A 3π A m-1
B 30000 A m-1
C 30000π A m-1
D 300 A m-1

The correct answer is:
30000 A m-1

Solution:
Volume = Al
Magnetization M→=magnetic monent Volume
=No. of turns×Current×AreaVolume
=N I AA l
=NIl
=500×15×10025
=60×500
=30×103
=30000 A m-1

Topic: Physics, Magnetic Properties of Matter, JEE Main
A 25 cm long solenoid has the radius 2 cm and 500 turns. It carries a current of 15 A. If it is equivalent to a magnet of the same size and magnetization M→ Magnetic momentvolume,then M→ is:

Options:
A 3π A m-1
B 30000 A m-1
C 30000π A m-1
D 300 A m-1

The correct answer is:
30000 A m-1

Solution:
Volume = Al
Magnetization M→=magnetic monent Volume
=No. of turns×Current×AreaVolume
=N I AA l
=NIl
=500×15×10025
=60×500
=30×103
=30000 A m-1

Topic: Physics, Magnetic Properties of Matter, JEE Main, JEE Main 2015 (10 Apr Online)
A 25 cm long string is fixed at both ends. It has a mass of 2.5 g and vibrates in its first overtone under tension. Also given is a closed pipe which is 30 cm long which vibrates in its fundamental frequency. When both of these are sounded simultaneously, 8 beats per second are heard. It is also observed that decreasing the tension in the string also decreases the beat frequency. What is the tension (in N) in the string? [Given, speed of sound in air is 360 ms-1]

Options:

The correct answer is:
59.29

Solution:
μ=2.525=0.1g/cm=10-2kg/m
1st overtone,

λs=25 cm=0.25 m
⇒fs=1λsTμ

Pipe in fundamental frequency, λp4=0.3
λp=1.2m
⇒fp=vsλp=3601.2=300 Hz

∵ By decreasing the tension, beat
frequency is decreased.
∴ fs>fp

⇒fs-fp=8

⇒10.25T10-2-3601.2=8
⇒10×10025T-300=8

⇒T=308×251000

⇒T=59.29 N

Topic: Physics, Waves and Sound, JEE Main
A 25 m long antenna is mounted on an antenna tower. The height of the antenna tower is 75 m. The wavelength (in meter) of the signal transmitted by this antenna would be :

Options:
A 300
B 400
C 200
D 100

The correct answer is:
100

Solution:
Length of Antena =25 m=λ4

⇒λ=100 m

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (16 Mar Shift 1)
A 25 W, 220 V bulb and 100 W, 220 V bulb are joined in series and connected to the mains. Which bulb will glow brighter -

Options:
A 25 W bulb
B 100 W bulb
C First 25 W bulb and then 100 W bulb
D Both will glow with same brightness

The correct answer is:
25 W bulb

Solution:
(Brightness) â (Consumed power) .....(i)
For series combination
Pconsumed = i2R
â´Pconsumed âR ......(ii)
Also, Rated Power
PRated = V2R
PRated â 1R .....(iii
From, (i), (ii) and (iii)
Brightnessâ 1Prated
â´ 25 W bulb will glow brighter.

Topic: Physics, Current Electricity, NEET
A $27 \mathrm{~mW}$ laser beam has a cross-sectional area of $10 \mathrm{~mm}^{2}$. The magnitude of the maximum electric field in this electromagnetic wave is given by:

[Given permittivity of space $\epsilon_{0}=9 \times 10^{-12} \mathrm{SI}$ units, Speed of light $\left.\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right]$

Options:
A $2 \mathrm{kV} / \mathrm{m}$
B $0.7 \mathrm{kV} / \mathrm{m}$
C $1 \mathrm{kV} / \mathrm{m}$
D $1.4 \mathrm{kV} / \mathrm{m}$

The correct answer is:
$1.4 \mathrm{kV} / \mathrm{m}$

Solution:
EM wave intensity

$\Rightarrow \mathrm{I}=\frac{\text { Power }}{\text { Area }}=\frac{1}{2} \varepsilon_{0} \mathrm{E}_{0}^{2} \mathrm{c}$

[where $\mathrm{E}_{0}=$ maximum electric field] $]$

$\Rightarrow \frac{27 \times 10^{-3}}{10 \times 10^{-6}}=\frac{1}{2} \times 9 \times 10^{-12} \times \mathrm{E}_{0}^{2} \times 3 \times 10^{8}$

$\Rightarrow \mathrm{E}_{0}=\sqrt{2} \times 10^{3} \mathrm{kV} / \mathrm{m}=1.4 \mathrm{kV} / \mathrm{m}$

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A $30 \mathrm{~kg}$ boy stands at the far edge of a floating plank, whose near edge is against the shore of a river. The plank is $10 \mathrm{~m}$ long and weighs $10 \mathrm{~kg}$.
If the boy walks to the near edge of the plank, how far from the shore does the plank move

Options:
A $7 \mathrm{~m}$
B $8 \mathrm{~m}$
C $7.5 \mathrm{~m}$
D $15 \mathrm{~m}$

The correct answer is:
$7.5 \mathrm{~m}$

Solution:
COM of system remains at same position in absence of any external force.
Position ( $x$-coordinate) of COM initially (taking origin at shore is)
$$
\begin{aligned}
x_{\mathrm{COM}} & =\frac{m_1 x_1+m_2 x_2}{m_1+m_2} \\
& =\frac{10 \times 5+30 \times 10}{10+30}=\frac{35}{4} \mathrm{~m}
\end{aligned}
$$
When boy changes his position, let the plank moves by distance $d$, then
$$
\begin{aligned}
& x_{\mathrm{COM}}=\frac{m_1 x_1^{\prime}+m_2 x_2^{\prime}}{m_1+m_2} \\
& \Rightarrow \frac{35}{4}=\frac{10(5+d)+30 d}{10+30} \Rightarrow 350=50+40 d \\
& \Rightarrow \frac{300}{40}=d \quad \Rightarrow d=7.5 \mathrm{~m} \\
&
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, TS EAMCET, TS EAMCET 2018 (05 May Shift 1)
A 34 m long ladder weighing 10 kg leans on a frictionless wall. Its feet rest on the floor 3 m away from the wall as shown in the figure. If Ff and Fw are the reaction forces of the floor and the wall, then ratio of FwFf will be :
(Use g=10 m s-2.)

Options:
A 6110
B 3113
C 3109
D 2109

The correct answer is:
3109

Solution:
Applying force balance in vertical direction

N1=mg=100 N

Applying torque balance at the bottom contact point,

mg×32=N2×5⇒N2=Fw=30 N

Applying force balance in horizontal direction

fs=N2=30 N

Net reaction at the floor will be,

Ff=N12+fs2=1002+302=10109 N

Therefore, FwFf=3010109=3109

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2022 (28 Jun Shift 2)
A \(40 \%\) by mass sucrose solution is heated till, it becomes \(50 \%\) by mass. Calculate the mass of water lost from \(100 \mathrm{~g}\) of the solution is

Options:
A \(10 \mathrm{~g}\)
B \(15 \mathrm{~g}\)
C \(20 \mathrm{~g}\)
D \(25 \mathrm{~g}\)

The correct answer is:
\(20 \mathrm{~g}\)

Solution:
\(40 \%\) sucrose solution means it contains \(60 \%\) water. After heating, till \(50 \%\) by mass sucrose remains.
Thus, \(\%\) water lost \(=\frac{0.4}{0.5} \times 100=80\)
Water lost \(=100-80=20 \mathrm{~g}\)
Hence, statement (c) is correct.

Topic: Chemistry, Some Basic Concepts of Chemistry, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 2)
A \( 0.50 \mathrm{~g} \) mixture of \( \mathrm{Cu}_{2} \mathrm{O} \) and \( \mathrm{CuO} \) contains \( 0.425 \mathrm{~g} \) of \( \mathrm{Cu} \). What is the mass of \( \mathrm{CuO}( \) in \( \mathrm{g}) \) in the mixture?
\[
(\mathrm{Cu}=63.5 \mathrm{u}, \mathrm{O}=16 \mathrm{u})
\]
Answer correct up to two places of decimals.

Options:

The correct answer is:
0.21

Solution:
Let the mass of CuO=w g

For one mole of Cu2O: mass of Cu=127 g; gram molecular mass=143 g

For one mole of CuO: mass of Cu=63.5 g; gram molecular mass=79.5 g

Thus, for w g CuO, mass of Cu=63.5w79.5 g

And, for 0.5-w g Cu2O, mass of Cu=1270.5-w143 g

Now, 63.5w79.5 +1270.5-w143 =0.425

w=0.21 g

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A 50 mL solution of pH = 1 is mixed with a 50 mL solution of pH = 2 . The pH of the mixture is

Options:
A 0.86
B 1.26
C 1.76
D 2.26

The correct answer is:
1.26

Solution:
Vf=100 ml
Cf=10-1×50+10-2×50100=0.055M
pH=-log⁡0.055=1.26

Topic: Chemistry, Ionic Equilibrium, JEE Main
A 50 Ω resistance is connected to a battery of 5 V. A galvanometer of resistance 100 Ω is to be used as an ammeter to measure current through the resistance, for this a resistance rS is connected to the galvanometer. Which of the following connections should be employed if the measured current is with in 1% of the current without the ammeter in the circuit?

Options:
A rS=0.5Î© in series with galvanometer
B rS=1Â Î© in series with galvanometer
C rS=1Â Î© in parallel with galvanometer
D rS=0.5 Î© in parallel with the galvanometer

The correct answer is:
rS=0.5 Î© in parallel with the galvanometer

Solution:
Initially current in the circuit is

I=550=0.1

and the current in circuit when ammeter is connected,is 1 % of initial current , i.e., I′=1100× 0·1 = 0.099 A

To increase the range of galvanometer, let resistance rs
is connected parallel to it and along with 50 Ω is connected in series to ammeter. So, equivalent resistance of circuit is:

Req=50+100 rS100+rS

Now, From equation V= I' Req

5 = 0.099×50+100 rs100+rs⇒50+100 rs100+rs = 50.099⇒50+100 rs100+rs =50.5050⇒100 rs100+rs = 50.5050⇒rs = 0.5 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2016 (09 Apr Online)
A 50 volt potential difference is suddenly applied to a coil with $L=5 \times 10^{-3}$ henry and $R=180 \mathrm{ohm}$. The rate of increase of current after 0.001 second is

Options:
A $27.3 \mathrm{amp} / \mathrm{sec}$
B $27.8 \mathrm{amp} / \mathrm{sec}$
C $2.73 \mathrm{amp} / \mathrm{sec}$
D None of the above

The correct answer is:
None of the above

Solution:
The rate of increase of current
$=\frac{d i}{d t}=\frac{d}{d t} i_0\left(1-\mathrm{e}^{-R t / L}\right)$ $=\frac{d}{d t} i_0-\frac{d}{d t} i_0 \mathrm{e}^{-R t / L}$
$=0-i_0 \mathrm{e}^{-R t / L} \cdot \frac{d}{d t}\left(-\frac{R t}{L}\right)$ $=i_0 \frac{R}{L} e^{-R t / L}$
$=\frac{50}{180} \times \frac{180}{5 \times 10^{-3}} \times e^{-(180 \times 0.001) /\left(5 \times 10^{-3}\right)}$ $=10^4 \times \mathrm{e}^{-36} \mathrm{~A} / \mathrm{sec}$

Topic: Physics, Electromagnetic Induction, JEE Main
A 50 W m-2 energy density of sunlight is incident normally on the surface of a solar panel. Some part of incident energy (25%) is reflected from the surface and the rest is absorbed. The force exerted on 1 m2 surface area will be close to c=3Ã108 m s-1

Options:
A 15×10-8 N
B 20×10-8 N
C 10×10-8 N
D 35×10-8 N

The correct answer is:
20×10-8 N

Solution:
The pressure exerted by the photons

P=0.75×Ic +0.25×2Ic ( P = pressure, c = speed of light, I = intensity )
P=1.25 ×Ic
We know the force can be computed using
F=P×A ( F = force, P = pressure, A = area )
F=1.25×50×13×108
F=20.83×10-8 N

Topic: Physics, Electromagnetic Waves, JEE Main
A $5.2$ molal aqueous solution of methyl alcohol, $\mathrm{CH}_3 \mathrm{OH}$, is supplied. What is the mole fraction of methyl alcohol in the solution?

Options:
A
$0.190$

B
$0.086$

C
$0.050$

D
$0.100$

The correct answer is:
$0.086$

Solution:
$\begin{aligned} \text { Molefraction of solute }\left(\mathrm{X}_2\right) \text { in aqueous solution } & =\frac{\mathrm{m}}{\mathrm{m}+\frac{1000}{18}} \\ & =\frac{5.2}{5.2+\frac{1000}{18}}=0.09\end{aligned}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2011
A 60 μF parallel plate capacitor whose plates are separated by 6 mm is charged to 250 V, and then the charging source is removed. When a slab of dielectric constant 5 and thickness 3 mm is placed between the plates, find the change in the potential difference across the capacitor?

Options:
A 250 V
B 100 V
C 150 V
D 75 V

The correct answer is:
100 V

Solution:
Capacitance of capacitor after inserting a slab of dielectric constant 5 and of thickness t=3 mm is, C'=C11-td+tkd

⇒C'=12 μF11-3 mm6 mm+3 mm5×6 mm=12 μF×2012⇒C'=20 μF

Now, after removing the charging source from the capacitor the charge remain same, q=CV=C'V'

⇒CV=C'V'⇒12 μF×250 V=20 μF×V'⇒V'=150 V

Thus, the change in the potential difference across the capacitor is250 V-150V=100 V

Topic: Physics, Capacitance, AP EAMCET, AP EAMCET 2021 (20 Aug Shift 1)
A 60HP electric motor lifts an elevator having a maximum total load capacity of 2000 kg. If the frictional force on the elevator is 4000 N, the speed of the elevator at full load is close to : 1 HP=746 W, g=10 m s-2

Options:
A 1.7 m s-1
B 1.9 m s-1
C 1.5 m s-1
D 2.0 m s-1

The correct answer is:
1.9 m s-1

Solution:
4000ÃV+mgÃV=P
60Ã7464000+20000=V
V=1.9 m s-1

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2020 (07 Jan Shift 1)
A 60pF capacitor is fully charged by a 20V supply. It is then disconnected from the supply and is connected to another uncharged 60pF capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in nJ ) ________

Options:

The correct answer is:
6

Solution:
V0=20V
Heat loss =Ui-Ut
=12Cvo2-212CV022
=CVo24
=60×10-122024J
=6×10-9J=6 nJ

Topic: Physics, Capacitance, JEE Main, JEE Main 2020 (07 Jan Shift 2)
A $6.0$ volt battery is connected to two light bulbs as shown in figure. Light bulb 1 has resistance 3 ohm while light bulb 2 has resistance $6 \mathrm{ohm}$. Battery has negligible internal resistance. Which bulb will glow brighter?

Options:
A
Bulb 1 will glow more first and then its brightness will become less than bulb 2

B
Bulb 1

C
Bulb 2

D
Both glow equally

The correct answer is:
Bulb 1

Solution:
Total resistance $=\frac{6 \times 3}{6+3}=2 \Omega$ Current in circuit $=\frac{6}{2}=3 A$
Therefore current through bulb 1 is $2 \mathrm{~A}$ and bulb 2 is $1 A$. So bulb 1 will glow more.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2012 (19 May Online)
A $0.66 \mathrm{~kg}$ ball is moving with a speed of $100 \mathrm{~m} / \mathrm{s}$. The associated wavelength will be (h $\left.=6.6 \times 10^{-34} \mathrm{Js}\right)$

Options:
A $6.6 \times 10^{-32} \mathrm{~m}$
B $6.6 \times 10^{-34} \mathrm{~m}$
C $1.0 \times 10^{-35} \mathrm{~m}$
D $1.0 \times 10^{-32} \mathrm{~m}$

The correct answer is:
$1.0 \times 10^{-35} \mathrm{~m}$

Solution:
From de-Broglie equation,
$$
\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.6 \times 10^{-34}}{0.66 \times 100}=10^{-35} \mathrm{~m}
$$

Topic: Chemistry, Structure of Atom, JEE Main
A $0.66 \mathrm{~kg}$ ball is moving with a speed of $100 \mathrm{~m} / \mathrm{s}$. The associated wavelength will be (h $\left.=6.6 \times 10^{-34} \mathrm{Js}\right)$

Options:
A $6.6 \times 10^{-32} \mathrm{~m}$
B $6.6 \times 10^{-34} \mathrm{~m}$
C $1.0 \times 10^{-35} \mathrm{~m}$
D $1.0 \times 10^{-32} \mathrm{~m}$

The correct answer is:
$1.0 \times 10^{-35} \mathrm{~m}$

Solution:
From de-Broglie equation,
$$
\lambda=\frac{\mathrm{h}}{\mathrm{mv}}=\frac{6.6 \times 10^{-34}}{0.66 \times 100}=10^{-35} \mathrm{~m}
$$

Topic: Chemistry, Structure of Atom, NEET, NEET 2010 (Mains)