datavorous/entrance-exam-dataset huggingface dataset:
An earthen pitcher loses 1 g of water per minute due to evaporation. If the water equivalent of pitcher is 0.5 kg and the pitcher contains 9.5 kg of water, calculate the time required for the water in the pitcher to cool to 28 Â°C from its original temperature of 30 Â°C. Neglect radiation effects. Latent heat of vapourization of water in this range of temperature is 580 cal g-1 and specific heat of water is 1 kcal g-1 Â°C-1.

Options:
A 38.6 min
B 30.5 min
C 34.5 min
D 41.2 min

The correct answer is:
34.5 min

Solution:
As water equivalent of pitcher is 0.5 kg, i..e., pitcher is equivalent to 0.5 kg of water, heat to be extracted from the system of water and pitcher for decreasing its temperature from 30 to 28oC is
Q 1 = m + M c Î T
= 9.5 + 0.5 kg 1 k cal/kg C â 3 0 - 2 8 â C
= 2 0 kcal
And heat extracted from the pitcher through evaporation in t minutes
Q 2 = mL = dm dt Ã t L = 1 g min Ã t 5 8 0 cal g
= 5 8 0 Ã t cal
According to given problem Q 2 = Q 1 , i.e., 5 8 0 Ã t = 2 0 Ã 1 0 3
t = 34.5 min

Topic: Physics, Thermodynamics, JEE Main
An earthquake generates both transverse \(S\) and longitudinal \(P\) waves in the earth with speeds \(4.5 \mathrm{~km} \mathrm{~s}^{-1}\) and \(8.0 \mathrm{~km} \mathrm{~s}^{-1}\), respectively. A seismograph records that the first \(P\)-wave arrives 3.5 minutes earlier than the first S-wave. From the seismograph, the epicentre of the earthquake is located at a distance.

Options:
A \(1080 \mathrm{~km}\)
B \(2468 \mathrm{~km}\)
C \(2160 \mathrm{~km}\)
D \(4320 \mathrm{~km}\)

The correct answer is:
\(2160 \mathrm{~km}\)

Solution:
Let the distance of epicenter of earthquake from point of observation be \(d\).
speed of \(S\)-wave, \(v_S=4.5 \mathrm{~km}^{-1} \mathrm{~s}\)
speed of \(P\)-wave, \(v_P=8 \mathrm{~km}^{-1} \mathrm{~s}\)
then, \(d=V_P t_P=V_S t_S\)
or \(8 t_P=4.5 t_S\)
\(t_P=\frac{4.5}{8} t_S \quad \ldots (i)\)
The first \(P\)-wave arrives \(3.5 \mathrm{~min}\) earlier than the first \(S\)-wave.
Hence,
\(\begin{aligned}
t_S-t_P & =3.5 \times 60 \\
t_S-t_P & =210 \quad \ldots (ii)
\end{aligned}\)
From Eq. (i), we get
\(\begin{gathered}
t_s-\frac{45}{8} t_s=210 \\
\frac{8 t_S-4.5 t_S}{8}=210 \\
3.5 t_s=210 \times 8 \\
t_S=\frac{210 \times 8}{3.5}=480 \mathrm{~s}
\end{gathered}\)
Now, \(d=v_S t_S=4.5 \times 480=2160 \mathrm{~km}\)

Topic: Physics, Waves and Sound, JEE Main
An effective power of a combination of 5 identical convex lenses which are kept in contact along the principal axis is $25 \mathrm{D}$. Focal length of each of the convex lens is :

Options:
A $20 \mathrm{~cm}$
B $50 \mathrm{~cm}$
C $500 \mathrm{~cm}$
D $25 \mathrm{~cm}$

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

Solution:
We know that $P_{\mathrm{eq}}=\Sigma \mathrm{P}_{\mathrm{i}}$
$\therefore$ given all lenses are identical
$\begin{aligned}
& \therefore 5 P=25 D \\
& \therefore P=5 D \\
& \therefore \frac{1}{f}=5 \Rightarrow f=\frac{1}{5} \mathrm{~m}=20 \mathrm{~cm}
\end{aligned}$
Hence option (1)

Topic: Physics, Ray Optics, JEE Main, JEE Main 2024 (04 Apr Shift 1)
An eight digit number divisible by 9 is to be formed using digits from 0 to 9 without repeating the digits. The number of ways in which this can be done is:

Options:
A
$72(7 !)$

B
$18(7 !)$

C
$40(7 !)$

D
$36(7 !)$

The correct answer is:
$36(7 !)$

Solution:
We know that any number is divisible by 9 if sum of the digits of the number is divisible by 9 .
Now sum of the digits from 0 to 9
$$
\begin{aligned}
&=0+1+2+3+4+5+6+7+8+9 \\
&=45
\end{aligned}
$$
Hence to form 8 digits numbers which are divisible by 9 , a pair of digits either 0 and 9,1 and 8,2 and 7,3 and 6 or 4 and 5 are not used.
Hence total number of 8 digits numbers which are divisible by 9
$$
\begin{aligned}
&=8 \times(7 !)+7 \times(7 !)+7 \times(7 !)+7 \times(7 !) \\
&+7 \times(7 !) \\
&=36 \times(7 !)
\end{aligned}
$$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2014 (11 Apr Online)
An elaborate network of filamentous proteinaceous structures present in the cytoplasm which helps in the maintenance of cell shape is called

Options:
A thylakoid
B endoplasmic reticulum
C plasmalemma
D cytoskeleton

The correct answer is:
cytoskeleton

Solution:
The cytoplasm of all eukaryotic cells is criss-crossed by a network of protein fibres that support the shape of the cell and anchor organelles to fixed locations. This network is called cytoskeleton. It consists of three types of cytoskeletal filaments namely act filaments or microfilaments (formed of globular protein actin), microtubules (formed of tubulin protein) and intermediate filaments (formed of variety of proteins like vimentin, keratin, desmin, peripherin, lamin, etc).

Topic: Biology, Cell - The Unit of Life, NEET, NEET 2010 (Mains)
An elastic ball is dropped on a long inclined plane. If bounces, hits the plane again, bounces and so on. let us Label the distance between the point of the first and second hit d12 and the distance between the points of second and the third hit is d23. find the ratio of d12/d23.

Options:
A 21
B 12
C 41
D 14

The correct answer is:
12

Solution:
If we rotate the coordinate system so that the ramp is horizontal, then the free fall acceleration will have two components one downward ay=- g cosâ¡Î¸ and one horizontal ax=g sinâ¡Î¸. In this frame, the initial velocity will have components given by Vy=V0cosâ¡Î¸ & Vx=V0 sinâ¡Î¸ time for each bounce is given by tb=2Vy-ay=2V0g
The horizontal displacement Îx=Vxt + 0.5 axt2
given these equation
d12=V0 sinâ¡Î¸2V0g + 0.5g sinâ¡Î¸ 2V0g2
= 4V02sinâ¡Î¸g
d13=V0sinâ¡Î¸4V0g + 0.5 g sinâ¡Î¸4V0g2
= 12V02sinâ¡Î¸g
Since d23=d13- d12= 8V02sinâ¡Î¸g
d12d23=12

Topic: Physics, Motion In Two Dimensions, JEE Main
An elastic material of Young's modulus $Y$ is subjected to a stress $S$. The elastic energy stored per unit volume of the material is

Options:
A $\frac{2 Y}{S^2}$
B $\frac{S^2}{2 Y}$
C $\frac{S}{2 Y}$
D $\frac{S^2}{Y}$

The correct answer is:
$\frac{S^2}{2 Y}$

Solution:
Given,
Young's modulus $=\mathrm{Y}$,
Streaching stress $=\mathrm{S}$,
Energy stored per unit volume is, $=\frac{1}{2} \times$ stress $\times$ strian
We know that, stress $=$ strain $\times Y$
Therefore energy per unit volume will be $\mathrm{E}=\frac{1}{2} \times \mathrm{S} \times \frac{\mathrm{S}}{\mathrm{Y}}=\frac{\mathrm{S}^2}{2 \mathrm{Y}}$
Therefore option 2 is correct.

Topic: Physics, Mechanical Properties of Solids, JEE Main
An elastic spring under tension of $3 \mathrm{~N}$ has a length $a$. Its length is $b$ under tension $2 \mathrm{~N}$. For its length $(3 a-2 b)$, the value of tension will be______$\mathrm{N}$.

Options:

The correct answer is:
5

Solution:
$\begin{aligned} & 3=\mathrm{K}(\mathrm{a}-\ell) \\ & 2=\mathrm{K}(\mathrm{b}-\ell) \\ & \mathrm{T}=\mathrm{K}(3 \mathrm{a}-2 \mathrm{~b}-\ell) \\ & \mathrm{T}=\mathrm{K}(3(\mathrm{a}-\ell)-2(\mathrm{~b}-\ell) \\ & =\mathrm{K}\left[3\left(\frac{3}{\mathrm{~K}}\right)-2\left(\frac{2}{\mathrm{~K}}\right)\right] \\ & =9-4 \\ & =5 \mathrm{~N}\end{aligned}$

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2024 (04 Apr Shift 1)
An elastic string has a force constant k and mass m. The string hangs vertically and a block of unknown mass is attached to its bottom end. It is known that the mass of the block is much greater than that of the string. The hanging block stretches the string the twice it's the relaxed length. How long t would it take for a low amplitude transverse pulse to travel the length of the string stretched by the hanging block? (Assume that the string follows Hook's law)

Options:
A 2mk
B mk
C m2k
D 2m3k

The correct answer is:
2mk

Solution:
Since the mass of the string is small compared to the mass m , the tension force is approximately constant
∴ T=mg
kx=mg, x=mgk
x=2L given
Where L- Relaxed length.
New length=2L
=2mgk
∴ mass per unit length
µ=mk2mg=k2g
Speed of wave V=Tμ
=mg×2mgmk
V=g2mk
time=2LV
=2mgk×1gk2m
time=2mk

Topic: Physics, Oscillations, JEE Main
An elastic string has a mass $M$ suspended at its lower end,the upper being fixed to a support. When the mass is pulled down over a short distance and let go, then it executes SHM with Time period given by
$T=2 \pi \sqrt{\frac{l_1}{g}}$
Where $I_1$ is the elongation of the string. Now the mass $m$ is added to the mass $M$, then it is found that Time period of oscillation $\left(T_2\right)$ such that
$\frac{T_1}{T_2}=\frac{5}{4}$
Find the ratio $\mathrm{m}: \mathrm{M}$
Given $Y$ is the Young modulus of elasticity. $L$ be the original length of string.

Options:
A $\frac{16}{25}$
B $\frac{9}{16}$
C $\frac{25}{16}$
D $\frac{5}{4}$

The correct answer is:
$\frac{9}{16}$

Solution:
Let $l_2$ be the elongation in second case then
$\begin{aligned} & T_1=2 \pi \sqrt{\frac{l_1}{g}}--(1) \\ & T_2=2 \pi \sqrt{\frac{l_2}{g}}--(2)\end{aligned}$
So that
$\begin{aligned} & \frac{T_1^2}{T_2^2}=\frac{l_1}{l_2} \\ & \text { Or } \\ & \frac{l_1}{l_2}=\frac{16}{25}\end{aligned}$
Now
For first case
$Y=\frac{M g L}{A l_1}--(3)$
Second case
$Y=\frac{(M+m) g L}{A l_2}--(4)$
From 3 and 4
$\frac{l_1}{l_2}=\frac{M}{M+m}$
Or
$\begin{aligned} & \frac{M}{M+m}=\frac{16}{25} \\ & \frac{M+m}{M}=\frac{25}{16}\end{aligned}$
Or
$\begin{aligned} & \frac{M+m-M}{M}=\frac{25-16}{16} \\ & \text { Or } \\ & \frac{m}{M}=\frac{9}{16}\end{aligned}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
An elastic string of unstretched length $\mathrm{L}$ and force constant $\mathrm{k}$ is stretched by a small length $\mathrm{x}$. It is further stretched by another small length y. The work done in the second stretching is

Options:
A $1 / 2 \mathrm{Ky}^{2}$
B $1 / 2 \mathrm{Ky}(2 \mathrm{x}+\mathrm{y})$
C $1 / 2 \mathrm{~K}\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)$
D $1 / 2 \mathrm{k}(\mathrm{x}+\mathrm{y})^{2}$

The correct answer is:
$1 / 2 \mathrm{Ky}(2 \mathrm{x}+\mathrm{y})$

Solution:
In the string elastic force is conservative in nature. $\therefore \mathrm{W}=-\Delta \mathrm{U}$

Work done by elastic force of string,

$$

\begin{aligned}

\mathrm{W} &=-\left(\mathrm{U}_{\mathrm{F}}-\mathrm{U}_{\mathrm{i}}\right)=\mathrm{U}_{\mathrm{i}}-\mathrm{U}_{\mathrm{F}} \\

\mathrm{W} &=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}-\frac{\mathrm{k}}{2}(\mathrm{x}+\mathrm{y})^{2} \\

&=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}-\frac{1}{2} \mathrm{k}\left(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{xy}\right) \\

&=\frac{1}{2} \mathrm{k} \mathrm{x}^{2}-\frac{1}{2} \mathrm{k} \mathrm{y}^{2}-\frac{1}{2} \mathrm{kx}^{2}-\frac{1}{2} \mathrm{k}(2 \mathrm{xy}) \\

&=-\mathrm{k} \mathrm{x} \mathrm{y}-\frac{1}{2} \mathrm{k} \mathrm{y}^{2}

\end{aligned}

$$

Therefore, the work done against elastic force

$\mathrm{W}_{\text {external }}=-\mathrm{W}=\frac{\mathrm{K} y}{2}(2 \mathrm{x}+\mathrm{y})$

Topic: Physics, Work Power Energy, JEE Main
An electric appliance supplies 6000 J min-1, heat to the system. If the system delivers a power of 90 W. How long it would take to increase the internal energy by 2.5×103 J?

Options:
A 2.5×101 s
B 2.5×102 s
C 2.4×103 s
D 4.1×101 s

The correct answer is:
2.5×102 s

Solution:
Recall the formula of the first law of thermodynamics in terms of change in internal energy and work done by the system, ΔU=Q-W, given, ΔU=2500 J, Q=600060×t=100t J and W=P×t=90t,
⇒2500 J=100t J-90t J⇒t=2.5×102 s.

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2021 (26 Aug Shift 1)
An electric bulb is rated 220 volt $-100$ watt. The power consumed by it when operated on 110 volt will be

Options:
A
50 watt

B
75 watt

C
40 watt

D
25 watt

The correct answer is:
25 watt

Solution:
$\frac{V_1^2}{P_1}=\frac{V_2^2}{P_2}=$ Resistance
$\Rightarrow P_2=25 \mathrm{~W}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2006
An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.

Options:
A 1.19×10-8T
B 1.71×10-8T
C 0.84×10-8T
D 3.36×10-8T

The correct answer is:
1.71×10-8T

Solution:
Intensity of light is I=PowerArea=200×3.51004π42=0.034 W m-2

Using the relation between magnetic field and intensity, I=B02c2μ0

We have, magnetic field as B0=I×2μ0c=0.034×2×4π×10-73×108=1.69×10-8 T≈1.71×10-8 T

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2022 (24 Jun Shift 2)
An electric bulb marked as 50W-200V is connected across a $100 \mathrm{V}$ supply. The present power of the bulb is

Options:
A $37.5 \mathrm{W}$
B $25 \mathrm{W}$
C $12.5 \mathrm{W}$
D $10 \mathrm{W}$

The correct answer is:
$12.5 \mathrm{W}$

Solution:
We know that
$$
R=\frac{V^{2}}{P}
$$
Given $\quad V=200 \mathrm{V}, P=40 \mathrm{W}$
and $\quad V=100 \mathrm{V}$
Hence, $\quad R=\frac{V^{2}}{P}=\frac{200 \times 200}{50}$
$$
\begin{array}{l}
P=\frac{V^{2}}{R}=\frac{100 \times 100 \times 50}{200 \times 200} \\
P^{\prime}=12.5 \mathrm{W}
\end{array}
$$

Topic: Physics, Current Electricity, JEE Main
An electric bulb of 500 W at 100 V is used in a circuit having a 200 V supply. Calculate the resistance R to be connected in series with the bulb so that the power delivered by the bulb is 500 W.

Options:
A 30 Ω
B 5 Ω
C 20 Ω
D 10 Ω

The correct answer is:
20 Ω

Solution:
Resistance of bulb

P=V2r

r=(100)2500=100×100500=20

i=PV=500100=5 A

ΔV in resistance R=100 V

R=1005=20 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (26 Aug Shift 2)
An electric bulb of power 30Ï W has an efficiency of 10% and it can be assumed to behave like a point source of light. At a distance of 3 m from the bulb, the peak value of the electric field in the light produced by the bulb is [Take Îµ0â9Ã10-12 C2 N-1 m-2]

Options:
A 1009 V m-1
B 109 V m-1
C 1092 V m-1
D 10092 V m-1

The correct answer is:
10092 V m-1

Solution:
Intensity at distance r from a point source, I=P4πr2

Efficiency, η=OutputInput=PP′
P'=ηP

P'=1010030π=3π W

I=3π4π32=112 W m-2

I=12cε0E02

⇒112=12cε0E02

E0=16×3×108×9×10-12=10092 V m-1

Topic: Physics, Magnetic Effects of Current, JEE Main
An electric bulb rated $50 \mathrm{~W}-200 \mathrm{~V}$ is connected across a $100 \mathrm{~V}$ supply. The power dissipation of the bulb is:

Options:
A $25 \mathrm{~W}$
B $12.5 \mathrm{~W}$
C $50 \mathrm{~W}$
D $100 \mathrm{~W}$

The correct answer is:
$12.5 \mathrm{~W}$

Solution:
Rated power \& voltage gives resistance
$\begin{aligned}
& \mathrm{R}=\frac{\mathrm{V}^2}{\mathrm{P}}=\frac{(200)^2}{50}=\frac{40000}{50} \\
& \mathrm{R}=800 \\
& \mathrm{P}=\frac{\left(\mathrm{V}_{\text {applied }}\right)^2}{\mathrm{R}}=\frac{(100)^2}{800} \\
& \mathrm{P}=12.5 \text { watt }
\end{aligned}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (04 Apr Shift 2)
An electric bulb rated as 200 W at 100 V is used in a circuit having 200 V supply. The resistance R that must be put in series with the bulb so that the bulb delivers the same power is ______ Ω.

Options:

The correct answer is:
50

Solution:
Power, P=V2RB

RB=V2P=100×100200

RB=50 Ω

To produce same power, same voltage (i.e., 100 V) should be across the bulb.

Hence, R=RB

R=50 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (25 Jul Shift 1)
An electric cell of emf $E$ is connected across a copper wire of diameter $d$ and length $l$. The drift velocity of electrons in the wire is $v_{d}$. If the length of the wire is changed to $2 l,$ the new drift velocity of electrons in the copper wire will be

Options:
A $v_{d}$
B $2 v_{d}$
C $v_{d} / 2$
D $v_{d} / 4$

The correct answer is:
$v_{d} / 2$

Solution:
The drift velocity is given as
$$
\begin{aligned}
v_{d} &=\frac{i}{n e A}=\frac{E}{R \times n e A} \\
&=\frac{E \times A}{\rho l \times n e A}=\frac{E}{ln e}
\end{aligned}
$$
When length of wire changed to 2l , the new drift velocity.
$$
\begin{aligned}
v_{d}^{\prime} &=\frac{E}{\rho \times 2 l \times n e} \\
\therefore \quad & \frac{v_{d}^{\prime}}{v_{d}}=\frac{E / \rho 2 \text { lne }}{E / \rho \text { lne }}=\frac{1}{2} \\
\Rightarrow \quad v_{d}^{\prime} &=\frac{v_{d}}{2}
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main
An electric charge $10^{-3} \mu \mathrm{C}$ is placed at the origin $(0,0)$ of $X-Y$ co-ordinate system. Two points $A$ and $\mathrm{B}$ are situated at $(\sqrt{2}, \sqrt{2})$ and $(2,0)$ respectively. The potential difference between the points $A$ and $B$ will be

Options:
A
$9 \mathrm{~volt}$

B
zero

C
$2 \mathrm{~volt}$

D
$4.5 \mathrm{~volt}$

The correct answer is:
zero

Solution:
Both points are at same distance from the charge.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2007
An electric charge 10-6 μC is placed at origin (0, 0) m of X-Y co-ordinate system. Two points P and Q are situated at (3, 3) m and (6, 0) m respectively. The potential difference between the points P and Q will be :

Options:
A 3 V
B 6 V
C 0 V
D 3 V

The correct answer is:
0 V

Solution:
The formula to calculate the potential difference between P and Q is given by

∆V=kq1r1-1r2 ...1

Now,

r1=(3)2+(3)2 m=6 m

And,

r2=(6)2+0 m=6 m

As r1=r2=6 m and the value of the charge remains constant, the potential difference between the points is zero.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2024 (27 Jan Shift 1)
An electric charge of 5 Faradays is passed through three electrolytes AgNO3,CuSO4 and FeCl3 solution. The grams of each metal liberated at cathode will be
[Atomic weight; Fe - 56 g/mol, Cu - 63.5 g/mol, Ag - 108 g/mol]

Options:
A Ag = 10.8 g, Cu = 12.7 g, Fe = 1.11 g
B Ag = 540 g, Cu = 367.5 g, Fe = 325 g
C Ag = 108 g, Cu = 63.5 g, Fe = 56 g
D Ag = 540 g, Cu = 158.8 g, Fe = 93.3 g

The correct answer is:
Ag = 540 g, Cu = 158.8 g, Fe = 93.3 g

Solution:
Ag++e-1 mol ≡ 1F→Ag
Cu2++2e-2 mol ≡ 2F→Cu
Fe3++3e-3 mol ≡ 3F→Fe
Grams of Ag liberated =1081×5=540 g
Grams of Cu liberated =63.52×5=158.75 g
Grams of Fe liberated =563×5=93.3 g

Topic: Chemistry, Electrochemistry, JEE Main
An electric charge +q +q moves with velocity vâ=3i^+4j^+k^, in an electromagnetic field given by : Eâ=3i^+j^+2k^ and Bâ=i^+j^+3k^. The y-component of the force experienced by +q is :

Options:
A -7q
B 5q
C -3q
D 2q

The correct answer is:
-7q

Solution:
F → = q E → + v → × B →
= q 3 i ^ + j ^ + k ^ + i ^ j ^ k ^ 3 4 1 1 1 + 3
= q 3 i ^ + j ^ + 2 k ^ + i ^ 1 2 - 1 - j ^ 9 - 1 + k ^ 3 - 4
=q3i^+j^+2k^+12i^-i^-9j^+j^+3k^-4k^
q=14i^-7j^+k^
= F y = -7 q j ^

Topic: Physics, Magnetic Effects of Current, JEE Main
An electric current is flowing through a circular coil of radius $\mathrm{R}$. The ratio of the magnetic field at the centre of the coil and that at a distance $2 \sqrt{2} R$ from the centre of the coil and on its axis is :

Options:
A
$2 \sqrt{2}$

B
27

C
36

D
8

The correct answer is:
27

Solution:
Given $:$ Radius $=R$
Distance $\mathrm{x}=2 \sqrt{2} \mathrm{R}$
$$
\begin{aligned}
\frac{B_{\text {centre }}}{B_{\text {axis }}} & =\left(1+\frac{x^2}{R^2}\right)^{3 / 2}=\left(1+\frac{(2 \sqrt{2} R)^2}{R^2}\right)^{3 / 2} \\
& =(9)^{3 / 2}=27
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2013 (09 Apr Online)
An electric current is passed through a circuit containing two wires of the same material, connected in parallel. If the length and radii of the wires are in the ratio of $4 / 3$ and $2 / 3$, then the ratio of the currents passing through the wire will be

Options:
A
3

B
$1 / 3$

C
$8 / 9$

D
2

The correct answer is:
$1 / 3$

Solution:
$$
\frac{I_1}{I_2}=\frac{R_2}{R_1}
$$
[current divider rule since voltage is same in parallel]
$$
\begin{aligned}
& \frac{I_1}{I_2}=\frac{L_2}{L_1} \times \frac{r_1^2}{r_2^2} \\
& \therefore \frac{I_1}{I_2}=\frac{3}{4} \times\left(\frac{2}{3}\right)^2=\frac{1}{3} .
\end{aligned}
$$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2004
An electric current is passed through an aqueous solution of a mixture of alanine (isoelectric point 6.0) glutamic acid (3.2) and arginine (10.7) buffered at \(\mathrm{pH}\) 6. What is the fate of the three acids?

Options:
A Glutamic acid migrates to anode at \(\mathrm{pH}\) 6. Arginine is present as a cation and migrates to the cathode. Alanine in a dipolar ion remains uniformly distributed in solution.
B Glutamic acid migrates to cathode and others remain uniformly distributed in solution.
C All three remain uniformly distributed in solution.
D All three move to cathode

The correct answer is:
Glutamic acid migrates to anode at \(\mathrm{pH}\) 6. Arginine is present as a cation and migrates to the cathode. Alanine in a dipolar ion remains uniformly distributed in solution.

Solution:
Hints : At \(\mathrm{pH}=6\), glutamic acid exists as a dianionic species & migrates to anode while arginine exists as cationic species & moves to cathode. Alanine does not migrate to any electrode at its isoelectric point .

Topic: Chemistry, Biomolecules, WBJEE, WBJEE 2011
An electric current passes through a circuit containing two wires of the same material connected in parallel. If the lengths of the wires are in the ratio 43 and radius of the wires are in the ratio 23, then the ratio of the current passing through the wires is mn, where m and n are smallest integers. What is m+n?

Options:

The correct answer is:
4

Solution:
Given that,
l1l2=43andr1r2=23
Here, l1and l2are the length of the wires while r1and r2 are the radii of the wires.
Now, we know that
V=IR⟹IR=constant
⟹I1R1=I2R2
or I1I2=R2R1 ….i
But we know that the resistance of the wire is
R=ρlA
Hence, from Eq (i)
I⁡ 1 I⁡ 2 = ρ l⁡ 2 / A⁡ 2 ρ l⁡ 1 / A⁡ 1
Here,
R1=ρl1A1,R2=ρl2A2and ρ1=ρ2=ρ
Because both wires are of same material.
∴ I1I2=l2A1l1A2
⟹I1I2=l2πr12l1πr22
⟹I1I2=l2r12l1r22
Here, l2l1=34 andr1r2=23
⟹ I1I2=34×232
Or I1I2=13=mn⇒m+n=4

Topic: Physics, Current Electricity, JEE Main
An electric dipole consisting of two opposite charges of $2 \times 10^{-6} \mathrm{C}$ each separated by a distance of $3 \mathrm{~cm}$ is placed in an electric field of $2 \times 10^5 \mathrm{~N} / \mathrm{C}$. The maximum torque on the dipole will be

Options:
A $12 \times 10^{-1} \mathrm{Nm}$
B $12 \times 10^{-3} \mathrm{Nm}$
C $24 \times 10^{-1} \mathrm{Nm}$
D $24 \times 10^{-3} \mathrm{~N} \mathrm{~m}$

The correct answer is:
$12 \times 10^{-3} \mathrm{Nm}$

Solution:
Maximum torque $=p E$
$=2 \times 10^{-6} \times 3 \times 10^{-2} \times 2 \times 10^5=12 \times 10^{-3}$ $N-m$.

Topic: Physics, Electrostatics, JEE Main
An electric dipole has fixed dipole moment p→, which makes angle θ with respect to x-axis. When subjected to an electric field E→1=Ei^, it experiences a torque T→1=τk^. When subjected to another electric field E→2= 3 E1j^ it experiences a torque T→2=-T→1 . The angle θ is:

Options:
A 90o
B 30o
C 45o
D 60o

The correct answer is:
60o

Solution:
P→=Pcos⁡θi^+Psin⁡θj^
T→=Pcos⁡θi^+Psin⁡θj^×Ei^
=-PEsin⁡θk^
T→2=Pcos⁡θi^+Psin⁡θj^×3Ej^
= 3PEcos⁡θk^
T→2= -T→1
∴ sin⁡θ= 3cos⁡θ
∴ tan⁡θ= 3
∴ θ=60o

Topic: Physics, Electrostatics, JEE Main, JEE Main 2017 (02 Apr)
An electric dipole has the magnitude of its charge as \( \mathrm{q} \) and its dipole moment is \( \mathrm{p} \). It is placed in a uniform electric field \( \mathrm{E} \). If its dipole moment is along the direction of the field, the force on it and its potential energy are respectively

Options:
A \( 2 \mathrm{q} . \mathrm{E} \) and minimum
B q.E and p.E
C Zero and minimum
D q.E and maximum

The correct answer is:
Zero and minimum

Solution:
The total force on dipole is zero because F = qE is applied on each charge but in opposite direction. The potential energy is U = - p → · E → , which is minimum when p → and E → are parallel.

Topic: Physics, Electrostatics, JEE Main
An electric dipole is formed by two equal and opposite charges q with separation d . The charges have same mass m. It is kept in a uniform electric field E. If it is slightly rotated from its equilibrium orientation, then its angular frequency ω is:

Options:
A 2qEmd
B 2qEmd
C qE2md
D qEmd

The correct answer is:
2qEmd

Solution:
τ=pEsin⁡θ
For small angle,
τ=pE θ
α=τI=pEI θ
∵ α =ω2θ Angular frequency =pEI
=qd Emd24+md24
=2qEmd

Topic: Physics, Electrostatics, JEE Main