datavorous/entrance-exam-dataset huggingface dataset:
A convex lens with lateral magnification 2 is used to image a point at the bottom of a tank. The image of the point is formed $60 \mathrm{~cm}$ above the lens. Now a liquid is filled into the tank to a height of $24 \mathrm{~cm}$. It is found that the distance of the image of the same point is now $120 \mathrm{~cm}$ above the lens. Find the refractive index of the liquid.

Options:
A 1.31
B 1.33
C 1.36
D 1.39

The correct answer is:
1.33

Solution:
The given situation is shown below.
Let $u_1=$ initial object distance from lens $f=$ focal length of lens
$v_1=60 \mathrm{~cm}$ (initial image distance) $m=2$ (lateral magnification)
Using, $m=-\frac{v_1}{u_1}=2$
$$
\frac{-60}{u_1}=2 \Rightarrow u_1=-30 \mathrm{~cm}
$$
After filling liquid into the tank.
Let $u_2=$ object distance $v_2=120 \mathrm{~cm}$ (final image distance) In case 1, (without liquid) By lens formula,
$$
\begin{aligned}
& \frac{1}{f}=\frac{1}{v_1}-\frac{1}{u_1} \\
& \frac{1}{f}=\frac{1}{60}-\frac{1}{(-30)} \Rightarrow f=20 \mathrm{~cm}
\end{aligned}
$$
In case 2 , (with liquid)
By lens formula,
$$
\begin{aligned}
\frac{1}{f} & =\frac{1}{v_2}-\frac{1}{u_2} \Rightarrow \frac{1}{20}=\frac{1}{120}-\frac{1}{u_2} \\
u_2 & =24 \mathrm{~cm}
\end{aligned}
$$
Now, normal shift, $O O^{\prime}=H\left(1-\frac{1}{\mu}\right)$
$$
\begin{array}{cc}
& 30-24=24\left(1-\frac{1}{\mu}\right) \Rightarrow 6=24\left(1-\frac{1}{\mu}\right) \\
& {[\because H=24 \mathrm{~cm}(\text { height of liquid filled })]} \\
& \mu=4 / 3=1.33
\end{array}
$$

Topic: Physics, Ray Optics, JEE Main
A convex lens

Options:
A Converges light rays
B Converges light rays
C Form real images always
D Always forms virtual images

The correct answer is:
Converges light rays

Solution:
A convex lens converges ray of light reflected from an object to form a real and inverted image provided object is NOT located in front of Focal Length $(F)$ of the lens. If object is located in front of $F$ then image formed is virtual and magnified.

Topic: Physics, Ray Optics, JEE Main
A convex mirror has a focal length $f$. A real object is placed at a distance $f$ in front of it from the pole produces an image at

Options:
A Infinity
B $f$
C $f / 2$
D $2 f$=

The correct answer is:
$f$

Solution:
In case of real object placed in front of convex mirror at a distance \(\mathrm{f}\).
According to sign convention.
Position of object, \(u=-f\)
Focal length, \(\mathrm{f}=+\mathrm{f}\)
Applying mirror formula
\(\begin{aligned}
& \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\
& =\frac{1}{v}+\frac{1}{(-f)}=\frac{1}{f} \\
& \frac{1}{v}=\frac{2}{f} \\
& \therefore v=\frac{f}{2}
\end{aligned}\)
The positive sign of \(v\) indicates image produced behind the mirror.

Topic: Physics, Ray Optics, JEE Main
A convex, rearview mirror of focal length 20 cm, is fitted in a car. A second car 2 m broad and 1.6 m high is 6 m away from the first car and overtakes the first car at a relative speed of 15 m s-1, then the speed of the image of the second car with respect to the first car is

Options:
A 0.016 m s-1
B 0.0257 m s-1
C 0.162 m s-1
D 0.0073 m s-1

The correct answer is:
0.016 m s-1

Solution:
Mirror formula 1 v + 1 u = 1 f
Differentiating w.r.t. t,
0 = - 1 v 2 dv dt - 1 u 2 du dt
dv dt = - v 2 u 2 du dt
But du dt = 15 m/s
Now1v=(120+1600)∴v=60031 cm
and v u = 600/31 600 = 1 31
Speed of image = d v dt = - ( 1 31 ) 2 × 15
=0.016 m/s

Topic: Physics, Ray Optics, JEE Main
A conveyor belt is moving at a constant speed of 2 m sâ1. A box is gently dropped on it. The coefficient of friction between them is Î¼=0.5. The distance that the box will move relative to belt before coming to rest on it taking g=10mâsâ2, is

Options:
A 1.2 m
B 0.6 m
C Zero
D 0.4 m

The correct answer is:
0.4 m

Solution:
The frictional force on the box f=μmg

∴ Acceleration in the box

a=μg=5 m  s −2

v2=u2+2as
4=0+2×5×s
s = 0.4 m

Topic: Physics, Laws of Motion, JEE Main
A conveyor belt is moving at a constant speed of 2 m s-1. A box is gently dropped on it. The coefficient of friction between them is Î¼=0.5. The distance that the box will move relative to the belt before coming to rest on it taking g=10 m s-2, is

Options:
A 1.2 m

B 0.6 m

C Zero
D 0.4 m

The correct answer is:
0.4 m

Solution:
Frictional force on the box f=μmg
∴ Acceleration in the box
a=μg=5 m s-2
v2=u2+2as
⇒ 0=22+2×(5)s
⇒ s=-25 m (with respect to the belt)
⇒ distance=0.4 m

Topic: Physics, Laws of Motion, JEE Main
A conveyor belt is moving at a constant speed of $2 \mathrm{~m} / \mathrm{s}$. A box is gently dropped on it. The coefficient of friction between them is $\mu=0.5$. The distance that the box will move relative to belt before coming to rest on it taking $g=10 \mathrm{~ms}^{-2}$, is

Options:
A $1.2 \mathrm{~m}$
B $0.6 \mathrm{~m}$
C zero
D $0.4 \mathrm{~m}$

The correct answer is:
$0.4 \mathrm{~m}$

Solution:
Force, $\quad F=\mu m g$
Retardation of the block on the belt
$$
a=\frac{F}{m}=\frac{\mu m g}{m}=\mu g
$$
From,
$$
\begin{aligned}
v^2 & =u^2+2 a s \\
0 & =(2)^2-2(\mu g) s \\
s & =\frac{4}{2 \times 0.5 \times 10}=0.4 \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Laws of Motion, NEET, NEET 2011 (Mains)
A cooking vessel on a slow burner contains 5 kg of water and an unknown mass of ice in equilibrium at 0oC at time t = 0. The temperature of the mixture is measured at various times and the result is plotted as shown in diagram. During the first 50 min the mixture remains at 0oC. From 50 min to 60 min, the temperature increases to 2oC. Neglecting the heat capacity of the heat capacity of the vessel, the initial mass of the ice is

Options:
A 1 0 7 kg
B 5 7 kg
C 5 4 kg
D 5 8 kg

The correct answer is:
5 7 kg

Solution:
Let m be the mass of ice.
Rate of heat given by the burner is constant.
In the first 50 min
dQ dt = mL t 1 = m kg × 8 0 × 4.2 × 1 0 3 J/kg 5 0 min ...(i)
From 50 min to 60 min
dQ dt = m + 5 S H 2 O Δ θ t 2
= m + 5 4.2 × 1 0 3 J/kg × 2 â C 1 min ...(ii)
From Eqs. (i) and (ii)
8 0 m 5 0 = 2 m + 5 1 0
7 m = 5 ⇒ m = 5 7 kg â 0.7 kg

Topic: Physics, Thermodynamics, JEE Main
A cooling curve is plotted between the temperature of a hot body and time. Which of the following is not true for the cooling curve?

Options:
A Cooling is faster from a calorimeter having a larger surface area than a smaller one.
B Cooling is faster from a calorimeter painted black from outside than that from a polished surfaced calorimeter.
C Cooling is faster from a copper calorimeter than from a steel calorimeter.
D The rate of cooling remains same throughout the experiment. Hence the graph is a straight line.

The correct answer is:
The rate of cooling remains same throughout the experiment. Hence the graph is a straight line.

Solution:
The rate of cooling is faster at the beginning and then decreases and becomes constant.

Topic: Physics, Experimental Physics, NEET
A copper ball of mass 100 g is at a temperature T. It is dropped in a copper calorimeter of mass 100 g, filled with 170 g of water at room temperature. Subsequently, the temperature of the system is found to be 75°C. T is given by:
(Given: room temperature =30°C, specific heat of copper =0.1 cal g-1 °C-1)

Options:
A 825°C
B 800°C
C 885°C
D 1250°C

The correct answer is:
885°C

Solution:
mcb=100 g, ToCmc=100 g, 30oCmw=170 g, 30oC
Net heat loss = Net heat gain
⇒100×0.1×T-75=100×0.1×75-30+170×1×75-30
⇒T-75=45+17×45
⇒T=885°C.

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2017 (02 Apr)
A copper ball of radius $1 \mathrm{~cm}$ and work function $4.47 \mathrm{~eV}$ is irradiated with ultraviolet radiation of wavelength $2500 ~Å$. The effect of irradiation results in the emission of electrons from the ball. Further the ball will acquire charge and due to this there will be a finite value of the potential on the ball. The charge acquired by the ball is :

Options:
A
$5.5 \times 10^{-13} \mathrm{~C}$

B
$7.5 \times 10^{-13} \mathrm{~C}$

C
$4.5 \times 10^{-12} \mathrm{~C}$

D
$2.5 \times 10^{-11} \mathrm{~C}$

The correct answer is:
$5.5 \times 10^{-13} \mathrm{~C}$

Solution:
No Solution Available

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2013 (25 Apr Online)
A copper complex crystallising in a CCP lattice with a cell edge of 0.4518 nm has been revealed by employing X-ray diffraction studies. The density of a copper complex is found to be 7.62 g cm-3. The molar mass of copper complex is gmol-1
(Nearest integer): [ Given :NA=6.022×1023 mol-1]

Options:

The correct answer is:
106

Solution:
dgmcc=4×MNA(a cm)3
7.62=4×M/6.022×10230.4518×10-7 cm3⇒M=105.8 g/mol

Topic: Chemistry, Solid State, JEE Main, JEE Main 2021 (22 Jul Shift 1)
A copper disc of radius \( 0.1 \mathrm{~m} \) is rotated about its centre with \( 10 \) revolutions per second in a uniform magnetic field of \( 0.1 \) tesla with its plane perpendicular to the field. The emf induced across the radius of disc is

Options:
A \( \frac{\pi}{10} \mathrm{~V} \)
B \( \frac{2 \pi}{10} \mathrm{~V} \)
C \( \pi \times 10^{-2} \mathrm{~V} \)
D \( 2 \pi \times 10^{-2} \mathrm{~V} \)

The correct answer is:
\( \pi \times 10^{-2} \mathrm{~V} \)

Solution:
Given,

The magnitude of the magnetic field, B=0.1 T

The radius of disk, R=0.1 m

Rotational speed, ω=2π×revolution per second=20π rad s-1

Consider a small radial segment dr at distance r from the center.

The emf generated in this small elemental rod will be,

dE=vBdr, here v is the velocity of rod.

We know at any radial distance r, linear velocity will be,

v=rω

⇒dE=ωBrdr

Since all these differential rods are connected in series, so net emf will be,

∫dE=∫0RωBrdr⇒E=ωBR22⇒E=20π×0.1×0.122=π×10-2 V

Topic: Physics, Electromagnetic Induction, JEE Main
A copper disc of the radius 0.1 m is rotated about its centre with ω=20 revolutions per second in a uniform magnetic field of 0.1 T with its plane perpendicular to the field. The emf induced across the radius of the disc is

Options:
A π20 V
B π10 V
C 20π mV
D 100π mV

The correct answer is:
20π mV

Solution:
ε = B ω R 2 2
= 0.1 × 2 π × 2 0 0.1 2 2
=20π×10-3 V

Topic: Physics, Electromagnetic Induction, JEE Main
A copper rod and a steel rod maintain a difference in their lengths of 10 cm at all temperatures. If their coefficients of expansion are 1.6Ã10-5 K-1â¡ and 1.2Ã10-5 K-1â¡, then the length of the copper rod is

Options:
A 40Â cm
B 30 cm
C 32Â cm
D 24Â cm

The correct answer is:
30 cm

Solution:
Δl1=Δl2
⇒ l1α1ΔT=l2α2ΔT
⇒ l1α1=l2α2
⇒l2l1=α1α2
⇒ l2l1-1=α1α2-1
l2-l1l1=α1-α2α2
⇒ l1=α2l2-l1α1-α2
l1=1.2×10-5(10)(1.6–1.2)×10-5 cm
⇒ l1=30 cm

Topic: Physics, Thermal Properties of Matter, NEET
A copper rod of 88 cm and an aluminum rod of unknown length have their increase in length independent of increase in temperature. The length of aluminum rod is αCu=1.7×10-5 K-1 and αAl=2.2×10-5 K-1

Options:
A 6.8 cm
B 113.9 cm
C 88 cm
D 68 cm

The correct answer is:
68 cm

Solution:
Increase in length is independent of temperature,
Lcu αcu=LAl αAl
LAl=88 cm1.7×10-52.2×10-5
=68 cm

Topic: Physics, Thermal Properties of Matter, JEE Main
A copper rod of cross-sectional area A carries a uniform current I through it. At temperature T, if the volume charge density of the rod is $\rho$, how long will the charges take to travel a distance $\mathrm{d}$ ?

Options:
A
$\frac{2 \rho \mathrm{dA}}{\mathrm{IT}}$

B
$\frac{2 \rho \mathrm{dA}}{\mathrm{I}}$

C
$\frac{\rho \mathrm{dA}}{\mathrm{I}}$

D
$\frac{\rho \mathrm{dA}}{\mathrm{IT}}$

The correct answer is:
$\frac{\rho \mathrm{dA}}{\mathrm{I}}$

Solution:
Charge density
$$
\begin{aligned}
&\rho=\frac{\text { charge }}{\text { volume }}=\frac{q}{A d} \Rightarrow q=\rho A d \\
&\text { Also, } q=I T \Rightarrow T=\frac{q}{I}=\frac{\rho A d}{I}
\end{aligned}
$$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A copper rod of length $75 \mathrm{~cm}$ and an iron rod of length $125 \mathrm{~cm}$ are joined together end to end. Both are of circular cross-section with diameter $2 \mathrm{~cm}$. The free ends of the copper and iron are maintained at $100^{\circ} \mathrm{C}$ and $0^{\circ} \mathrm{C}$ respectively. The surfaces of the bars are insulated thermally. The temperature of the copper-iron junction is
[Thermal conductivity of copper is $386.4 \mathrm{~W} / \mathrm{m}-\mathrm{K}$ and thermal conductivity of iron is $48.46 \mathrm{~W} / \mathrm{m}-\mathrm{K}$ ]

Options:
A $100^{\circ} \mathrm{C}$
B $0^{\circ} \mathrm{C}$
C $93^{\circ} \mathrm{C}$
D $50^{\circ} \mathrm{C}$

The correct answer is:
$93^{\circ} \mathrm{C}$

Solution:
Let temperature of copper and iron junction is $\theta$
$$
\begin{aligned}
& \therefore \text { Temperature of junction }(\theta)=\frac{K_1 l_2 \theta_1+K_2 l_1 \theta_2}{K_1 l_2+K_2 l_1} \\
&=\frac{386.4 \times 125 \times 100+48.46 \times 75 \times 0}{386.4 \times 125+48.46 \times 75} \\
&=\frac{48,300 \times 100}{48,300+3634.5}=\frac{48,300 \times 100}{51934.5} \\
&=0.93 \times 100=93^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, TS EAMCET, TS EAMCET 2015
A copper rod of length 75 cm and an iron rod of length 125 cm are joined together end to end. Both are of circular cross-section with diameter 2 cm. The free ends of the copper and iron rods are maintained at 100 Â°C and 0 Â°C respectively and the curved surfaces of the bars are insulated thermally. The temperature of the copper-iron junction is [Thermal conductivity of copper is 384 W m-1 K-1 and that of iron is 48 W m-1 K-1]

Options:
A 100 °C
B 0 °C
C 93 °C
D 50 °C

The correct answer is:
93 °C

Solution:
100-θR1=θ-0R2
K1A100-θL1=K2Aθ-0L2
100-θK1L1=θK2L2
100-θ3840.75=θ481.25
⇒ θ=93 °C

Topic: Physics, Thermal Properties of Matter, NEET
A copper rod of mass $\mathrm{m}$ slides under gravity on two smooth parallel rails, with separation $l$ and set at an angle of $\theta$ with the horizontal. At the bottom, rails are joined by a resistance R. There is a uniform magnetic field $\mathrm{B}$ normal to the plane of the rails, as shown in the figure. The terminal speed of the copper rod is:

Options:
A
$\frac{\mathrm{mgR} \cos \theta}{\mathrm{B}^2 l^2}$

B
$\frac{\mathrm{mgR} \sin \theta}{\mathrm{B}^2 l^2}$

C
$\frac{\mathrm{mgR} \tan \theta}{\mathrm{B}^2 l^2}$

D
$\frac{\mathrm{mgR} \cot \theta}{\mathrm{B}^2 l^2}$

The correct answer is:
$\frac{\mathrm{mgR} \sin \theta}{\mathrm{B}^2 l^2}$

Solution:
From Faraday's law of electomagnetic induction,
$$
\begin{aligned}
e &=\frac{d \phi}{d t}=\frac{d(B A)}{l t}=\frac{d(B l l)}{d t} \\
&=\frac{B d l \times l}{d t}=B V l
\end{aligned}
$$
Also, $F=i l B=\left(\frac{B V}{R}\right)\left(l^2 B\right)=\frac{B^2 l^2 V}{R}$
At equilibrium
$$
m g \sin \theta=\frac{B^2 l V}{R} \Rightarrow V=\frac{m g R \sin \theta}{B^2 l^2}
$$

Topic: Physics, Electromagnetic Induction, JEE Main
A copper rod of mass $\mathrm{m}$ slides under gravity on two smooth parallel rails, with separation $l$ and set at an angle of $\theta$ with the horizontal. At the bottom, rails are joined by a resistance R. There is a uniform magnetic field $\mathrm{B}$ normal to the plane of the rails, as shown in the figure. The terminal speed of the copper rod is:

Options:
A
$\frac{\mathrm{mgR} \cos \theta}{\mathrm{B}^2 l^2}$

B
$\frac{\mathrm{mgR} \sin \theta}{\mathrm{B}^2 l^2}$

C
$\frac{\mathrm{mgR} \tan \theta}{\mathrm{B}^2 l^2}$

D
$\frac{\mathrm{mgR} \cot \theta}{\mathrm{B}^2 l^2}$

The correct answer is:
$\frac{\mathrm{mgR} \sin \theta}{\mathrm{B}^2 l^2}$

Solution:
From Faraday's law of electomagnetic induction,
$$
\begin{aligned}
e &=\frac{d \phi}{d t}=\frac{d(B A)}{l t}=\frac{d(B l l)}{d t} \\
&=\frac{B d l \times l}{d t}=B V l
\end{aligned}
$$
Also, $F=i l B=\left(\frac{B V}{R}\right)\left(l^2 B\right)=\frac{B^2 l^2 V}{R}$
At equilibrium
$$
m g \sin \theta=\frac{B^2 l V}{R} \Rightarrow V=\frac{m g R \sin \theta}{B^2 l^2}
$$

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A copper wire and an aluminium wire have lengths in the ratio $5: 2$, diameters in the ratio $4: 3$ and forces applied in the ratio $4: 5$. Find the ratio of increase in length of the copper wire to that aluminium wire. (Let $Y_{\mathrm{Cu}}=1.1 \times 10^{11} \mathrm{Nm}^{-2}, Y_{\mathrm{Al}}=0.7 \times 10^{11} \mathrm{Nm}^{-2}$ )

Options:
A $\frac{176}{63}$
B $\frac{63}{88}$
C $\frac{189}{11}$
D $\frac{33}{89}$

The correct answer is:
$\frac{63}{88}$

Solution:
Given, ratio of length of copper wire and
aluminium wire, $l_1: l_2=5: 2$
and diameter ratio of the wire $=d_1: d_2=4: 3$
$\therefore \quad r_1: r_2=2: \frac{3}{2}$
If $\Delta l_1$ and $\Delta l_2$ are increase in the length of copper and aluminium wire, respectively then for copper wire,
$\Delta l_1=\frac{F_1 l_1}{A_1 y_1}$
For aluminium wire,
$\Delta l_2=\frac{F_2 l_2}{A_2 y_2}$
From Eqs. (i) and (ii), we get
$\frac{\Delta_1}{\Delta l_2}=\frac{l_1}{l_2} \cdot \frac{A_2}{A_1} \cdot \frac{y_2}{y_1} \cdot \frac{F_1}{F_2} \quad\left[\because \frac{y_2}{y_1}=\frac{y_{\mathrm{Al}}}{y_{\mathrm{Cu}}}=\frac{0.7}{1.1}\right]$
$=\frac{5}{2} \cdot \frac{\pi r_2^2}{\pi r_1^2} \cdot \frac{0.7 \times 10^{11}}{1.1 \times 10^{11}} \cdot \frac{4}{5} \quad\left[\because\right.$ Given, $\left.\frac{F_1}{F_2}=\frac{4}{5}\right]$
$=\frac{5}{2} \cdot \frac{\left(\frac{3}{2}\right)^2}{(2)^2} \times \frac{7}{11} \times \frac{4}{5}=\frac{63}{88}$
Hence, the ratio of increase in length of the copper and aluminium wire will be $\frac{63}{88}$.

Topic: Physics, Mechanical Properties of Solids, JEE Main
A copper wire is stretched to make it $0.5 \%$ longer. The percentage change in its electrical resistance if its volume remains unchanged is

Options:
A \(2.0 \%\)
B \(2.5 \%\)
C \(1.0 \%\)
D \(0.5 \%\)

The correct answer is:
\(1.0 \%\)

Solution:
Resistance,$R=\rho l / A$
$R=(\rho l / A) \times(I / I)=\left(\rho l^2 / v\right)=(\because$ volume $(v)=A l)$
Since resistivity and volume remains constant therefore percentage change in resistance
$\Delta R / R=(2 \Delta I / I)=2 \times(0.5)=1 \%$
Answer: (3) 100

Topic: Physics, Mathematics in Physics, JEE Main
A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistance if its volume remains unchanged is:

Options:
A 2.5%
B 1.0%
C 2.0%
D 0.5%

The correct answer is:
1.0%

Solution:
Volume, V=A.l

In differential form,
dVV=dAA+dll0=dAA+dll
⇒dAA=-dll

Also resistance of wire is as follows R=ρlA
diffrentiation formula for resistance is as follows

dRR=dll-dAAdRR=dll-(-dll)
dRR=2·dll

dRR×100=2·dll×100dRR%=2×0.5%dRR%=1%
∴% change in R=2×0.5%=1%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2019 (09 Jan Shift 1)
A copper wire is wound on a wooden frame, whose shape is that of an equilateral triangle. If the linear dimension of each side of the frame is increased by a factor of $3,$ keeping the number of turns of the coil per unit length of the frame the same, then the self inductance of the coil:

Options:
A decreases by a factor of 9
B increases by a factor of 27
C increases by a factor of 3
D decreases by a factor of $9 \sqrt{3}$

The correct answer is:
increases by a factor of 3

Solution:
As total length $L$ of the wire will remain constant

$$

\mathrm{L}=(3 \mathrm{a}) \mathrm{N} \quad(\mathrm{N}=\text { total turns })

$$

and length of winding $=(\mathrm{d}) \mathrm{N}$

(d = diameter of wire)

$$

\begin{array}{l}

\text { self inductance }=\mu_{0} \mathrm{n}^{2} \mathrm{~A} \ell \\

\quad=\mu_{0} \mathrm{n}^{2}\left(\frac{\sqrt{3} \mathrm{a}^{2}}{4}\right) \mathrm{d} \mathrm{N} \\

\quad \propto \mathrm{a}^{2} \mathrm{~N} \propto \mathrm{a}\left[\text { as } \mathrm{N}=\mathrm{L} / 3 \mathrm{a} \Rightarrow \mathrm{N} \propto \frac{1}{\mathrm{a}}\right]

\end{array}

$$

Now 'a' increased to ' $3 \mathrm{a}$ ' So self inductance will become 3 times

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A copper wire of 3 mm2 cross-sectional area carries a current 5 A. The magnitude of the drift velocity for the electrons in the wire is (Assume each copper atom contributes one free electron and density of copper =8920 kg m-3, MCu=63.5 kg mol-1)

Options:
A 0.24 m s-1
B 0.12 m s-1
C 2.4 m s-1
D 0.06 m s-1

The correct answer is:
0.12 m s-1

Solution:
Given, A=3Ã10-6 m2, I=5 A,
MCu=63.5 kg mol-1
ÏCu=8920 kg m-3
Electron density =6.023Ã1023Ã892063.5
=8.46Ã1025 m-3
â´ Drift velocity, vd=IenA
=51.6Ã10-19Ã8.46Ã1025Ã3Ã10-6
=540.608
=0.12 m s-1

Topic: Physics, Current Electricity, JEE Main
A copper wire of length $1.0 \mathrm{~m}$ and a steel wire of length $0.5 \mathrm{~m}$ having equal cross-sectional areas are joined end to end. The composite wire is stretched by a certain load which stretches the copper wire by $1 \mathrm{~mm}$. If the Young's modulii of copper and steel are respectively $1.0 \times 10^{11}$ $\mathrm{Nm}^{-2}$ and $2.0 \times 10^{11} \mathrm{Nm}^{-2}$, the total extension of the composite wire is :

Options:
A
$1.75 \mathrm{~mm}$

B
$2.0 \mathrm{~mm}$

C
$1.50 \mathrm{~mm}$

D
$1.25 \mathrm{~mm}$

The correct answer is:
$1.25 \mathrm{~mm}$

Solution:
$\mathrm{Y}_{\mathrm{c}} \times\left(\Delta \mathrm{L}_{\mathrm{c}} / \mathrm{L}_{\mathrm{c}}\right)=\mathrm{Y}_{\mathrm{s}} \times\left(\Delta \mathrm{L}_{\mathrm{s}} / \mathrm{L}_{\mathrm{s}}\right)$
$$
\Rightarrow 1 \times 10^{11} \times\left(\frac{1 \times 10^{-3}}{1}\right)=2 \times 10^{11} \times\left(\frac{\Delta \mathrm{L}_{\mathrm{s}}}{0.5}\right)
$$
$$
\therefore \Delta \mathrm{L}_{\mathrm{s}}=\frac{0.5 \times 10^{-3}}{2}=0.25 \mathrm{~mm}
$$
Therefore, total extension of the composite
$$
\begin{aligned}
& \text { wire }=\Delta \mathrm{L}_{\mathrm{c}}+\Delta \mathrm{L}_{\mathrm{s}} \\
& =1 \mathrm{~mm}+0.25 \mathrm{~m}=1.25 \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A copper wire of length $1.0 \mathrm{~m}$ and a steel wire of length $0.5 \mathrm{~m}$ having equal cross-sectional areas are joined end to end. The composite wire is stretched by a certain load which stretches the copper wire by $1 \mathrm{~mm}$. If the Young's modulii of copper and steel are respectively $1.0 \times 10^{11}$ $\mathrm{Nm}^{-2}$ and $2.0 \times 10^{11} \mathrm{Nm}^{-2}$, the total extension of the composite wire is :

Options:
A
$1.75 \mathrm{~mm}$

B
$2.0 \mathrm{~mm}$

C
$1.50 \mathrm{~mm}$

D
$1.25 \mathrm{~mm}$

The correct answer is:
$1.25 \mathrm{~mm}$

Solution:
$\mathrm{Y}_{\mathrm{c}} \times\left(\Delta \mathrm{L}_{\mathrm{c}} / \mathrm{L}_{\mathrm{c}}\right)=\mathrm{Y}_{\mathrm{s}} \times\left(\Delta \mathrm{L}_{\mathrm{s}} / \mathrm{L}_{\mathrm{s}}\right)$
$$
\Rightarrow 1 \times 10^{11} \times\left(\frac{1 \times 10^{-3}}{1}\right)=2 \times 10^{11} \times\left(\frac{\Delta \mathrm{L}_{\mathrm{s}}}{0.5}\right)
$$
$$
\therefore \Delta \mathrm{L}_{\mathrm{s}}=\frac{0.5 \times 10^{-3}}{2}=0.25 \mathrm{~mm}
$$
Therefore, total extension of the composite
$$
\begin{aligned}
& \text { wire }=\Delta \mathrm{L}_{\mathrm{c}}+\Delta \mathrm{L}_{\mathrm{s}} \\
& =1 \mathrm{~mm}+0.25 \mathrm{~m}=1.25 \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2013 (23 Apr Online)
A copper wire of resistance $R$ is cut into ten parts of equal length. Two pieces each are joined in series and then five such combinations are joined in parallel. The new combination will have a resistance

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

The correct answer is:
$\frac{R}{25}$

Solution:
Hence every new piece will have a resistance $\frac{R}{10}$. If two pieces are connected in series, then their resistance $=\frac{2 R}{10}=\frac{R}{5}$
If 5 such combinations are joined in parallel, then net resistance $=\frac{R}{5 \times 5}=\frac{R}{25}$

Topic: Physics, Current Electricity, JEE Main
A cord is wound round the circumference of wheel of radius r, The axis of the wheel is horizontal and the moment of inertia about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance h, the square of angular velocity of wheel will be

Options:
A 2mghI+mr2
B 2mghI+2mr2
C 2gh
D 2ghI+mr2

The correct answer is:
2mghI+mr2

Solution:
mgh=12Iω2+12mv2

v=ωr

mgh=12Iω2+12 mω2r2

2mghI+mr2=ω2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (26 Feb Shift 2)