datavorous/entrance-exam-dataset huggingface dataset:
A wire $34 \mathrm{~cm}$ long is to be bent in the form of a quadrilateral of which each angle is $90^{\circ} .$ What is the maximum area which can be enclosed inside the quadrilateral?

Options:
A $68 \mathrm{~cm}^{2}$
B $70 \mathrm{~cm}^{2}$
C $71.25 \mathrm{~cm}^{2}$
D $72.25 \mathrm{~cm}^{2}$

The correct answer is:
$72.25 \mathrm{~cm}^{2}$

Solution:
Let one side of quadrilateral be $\mathrm{x}$ and another side be $y$ so, $2(x+y)=34$

or $(x+y)=17$

We know from the basic principle that for a given perimeter square has the maximum area, $\mathrm{so}, \mathrm{x}=\mathrm{y}$ and putting this value in equation (i)

$\mathrm{x}=\mathrm{y}=\frac{17}{2}$

Area $=x \cdot y=\frac{17}{2} \times \frac{17}{2}=\frac{289}{4}=72.25$

Topic: Mathematics, Area Under Curves, JEE Main
A wire ab of length l , mass m and resistance R slides on a smooth thick pair of metallic rails joined at the bottom as shown in fig. The plane of the rails makes an angle θ with the horizontal. A vertical magnetic field B exist in the region. If the wire slides on the rails at a constant speed v , then the value of B is -

Options:
A mgRvl2cos2⁡θ
B mgRcos⁡θvl2sin2θ
C mgRv2l2sin2θ
D mgRsin⁡θv l2cos2⁡θ

The correct answer is:
mgRsin⁡θv l2cos2⁡θ

Solution:
mgsin⁡θ=IlBcos⁡θ

=l Bcos⁡θ×vBl cos⁡θR
B=mgRsinθvl2cos2⁡θ

Topic: Physics, Magnetic Effects of Current, JEE Main
A wire bent as a parabola y=kx2 is located in a uniform magnetic field of induction B, the vector B being perpendicular to the plane xy. At t=0, the sliding wire starts sliding from the vertex O with a constant acceleration a linearly as shown in Fig. Find the emf induced in the loop -

Options:
A By2ak
B By4ak
C By8ak
D Byak

The correct answer is:
By8ak

Solution:
dÏ=B.dA=2B x dy and y=kx2
â´x=yk
â´ Emf E=dÏdt=2Bykdydt
using v2=2as
dydt=v=2ay
orâââÎµ=dÏdt=2Byk2ay
orâââÎµ= By 8a k

Topic: Physics, Electromagnetic Induction, JEE Main
A wire carrying current I is bent in the shapeABCDEFA as shown, where rectangle ABCDA and ADEFA are perpendicular to each other. If the sides of the rectangles are of lengths a and b, then the magnitude and direction of magnetic moment of the loop ABCDEFA is :

Options:
A 2abI along j^2+k^2
B abI along j^2+k^2
C 2abI along j^5+2k^5
D abI along j^5+k^5

The correct answer is:
2abI along j^2+k^2

Solution:
Loop ABCD

M→1=(abI)(k^)
For Loop DEFA M→2=(abI)(j^)
M→=M→1+M→2;M→=(abI)(j^+k^)
2abI along j^2+k^2.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2020 (02 Sep Shift 2)
A wire carrying current I is tied between points P and Q and is in the shape of a circular arc of radius R due to a uniform magnetic field B (perpendicular to the plane of the paper, as shown in the figure) in the vicinity of the wire. If the wire subtends an angle 2θo at the center of the circle (of which it forms an arch) then the tension in the wire is:

Options:
A IBR
B IBRsinθ0
C IBR2sinθ0
D IBRθ0sinθ0

The correct answer is:
IBR

Solution:
Consider any small element dl of arc of angle dθ . Then magnetic force dF=Idl.B outwards
For equilibrium of element
Idl B=2T sindθ2
For small angle sinθ≃θ
IdlB=2Tdθ2
Idl B=TdlR
T=IBR

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2015 (11 Apr Online)
A wire elongates by $\ell \mathrm{mm}$ when a load $\mathrm{W}$ is hanged from it. If the wire goes over a pulley and two weights $\mathrm{W}$ each are hung at the two ends, the elongation of the wire will be (in $\mathrm{mm}$ )

Options:
A
$\ell / 2$

B
$\ell$

C
$2 \ell$

D
zero

The correct answer is:
$\ell$

Solution:
No Solution Available

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2006
A wire fixed at the upper end stretches by length by applying a force F. The work done in stretching is

Options:
A
$\mathrm{F} / 2$

B
$\mathrm{F} \ell$

C
$2 \mathrm{~F} \ell$

D
$\mathrm{F} \ell / 2$

The correct answer is:
$\mathrm{F} \ell / 2$

Solution:
Work done $=\frac{1}{2} \mathrm{kx}^2=\frac{1}{2} \mathrm{k} \ell^2$ where $\ell$ is the total extensions.
$$
=\frac{1}{2}(\mathrm{k} \ell) \ell=\frac{1}{2} \mathrm{~F} \ell
$$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2004
A wire fixed to the ceiling at one end is being pulled slowly by a force F at the other end. If the change in length of the wire is l, then the work done by the force F is

Options:
A F2l
B Fl
C 2Fl
D Fl2

The correct answer is:
Fl2

Solution:
Young's modulus Y=FLAl ............(i)
â´ F=YAlL
or dW=Fdl=YAldlL
or â«dW=YALâ«0lldl=YAl22L
or Work done =YAl22L ...............(ii)
From (i) and (ii)
Work done = Fl2

Topic: Physics, Mechanical Properties of Solids, JEE Main
A wire has a mass 0.3 Â±0.003 g, radius 0.5 Â±0.005 mm and length 6 Â±0.06 cm. The maximum percentage error in the measurement of its density is

Options:
A 1
B 2
C 3
D 4

The correct answer is:
4

Solution:
Density
ρ=mπr2L
∴∆ρρ×100=∆mm+2∆rr+∆LL×100
After substituting the values we get the maximum percentage error in density = 4%

Topic: Physics, Units and Dimensions, JEE Main
A wire having tension 225 N produces six beats per second when it vibrates with some tuning fork. When the tension changes to 256 N, it vibrates with the same fork with the number of beats remaining unchanged. The frequency of the fork will be

Options:
A 186 Hz
B 225 Hz
C 256 Hz
D 280 Hz

The correct answer is:
186 Hz

Solution:
For wire vibrating under tension, the fundamental frequency is given by
f1=12LTμ , Where T is tension and μ linear mass density of string
It is given that, when it is sounded with a tuning fork of frequency x, 6 beats per second were heard
∴ x-f=6
f1=12L225μ
f2=12L256μ undefined
∴ x+6=1615(x-6)
∴ 15x+90=16x-96
∴ x=186 Hz

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2016
A wire having tension 225 N produces six beats per second when it vibrates with some tuning fork. When the tension changes to 256 N, it vibrates with the same fork with the number of beats remaining unchanged. The frequency of the fork will be

Options:
A 186 Hz
B 225 Hz
C 256 Hz
D 280 Hz

The correct answer is:
186 Hz

Solution:
For wire vibrating under tension, the fundamental frequency is given by
f1=12LTμ , Where T is tension and μ linear mass density of string
It is given that, when it is sounded with a tuning fork of frequency x, 6 beats per second were heard
∴ x-f=6
f1=12L225μ
f2=12L256μ undefined
∴ x+6=1615(x-6)
∴ 15x+90=16x-96
∴ x=186 Hz

Topic: Physics, Waves and Sound, JEE Main
A wire in the form of a square of side $a$ carries a current $i$. Then, the magnetic induction at the centre of the square is
(Magnetic permeability of free space $=\mu_0$ )

Options:
A $\frac{\mu_d i}{2 \pi a}$
B $\frac{\mu j \sqrt{2}}{\pi a}$
C $\frac{2 \sqrt{2} \mu_d}{\pi a}$
D $\frac{\mu_0 j}{\sqrt{2} \pi \alpha}$

The correct answer is:
$\frac{2 \sqrt{2} \mu_d}{\pi a}$

Solution:
Magnetic field produced by side ' $A B^{\prime}$ at the centre.
$\begin{aligned} B_1 & =\frac{\mu_0 j^j}{4 \pi r}\left(\sin \phi_1+\sin \phi_2\right) \\ & =\frac{\mu_{\omega^j}}{4 \pi \frac{a}{2}}\left(\sin 45^{\circ}+\sin 45^{\circ}\right) \\ & =\frac{2 \mu_0 i}{4 \pi a}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right) \Rightarrow=\frac{2 \sqrt{2} \mu_0 i}{4 \pi a}\end{aligned}$
$\therefore$ Total magnetic field at the centre
$\begin{aligned} B & =4 B_1 \\ & =4\left(\frac{2 \sqrt{2} \mu_0 i}{4 \pi a}\right) \\ & =\frac{2 \sqrt{2} \mu_0 i}{\pi a}\end{aligned}$

Topic: Physics, Magnetic Effects of Current, JEE Main
A wire in the form of a square of side a carries a current i . Then, the magnetic induction at the centre of the square wire is (magnetic permeability of free space =Î¼0 ).

Options:
A Î¼0i2Ïa
B Î¼0iÂ 2Ïa
C 22 μ0iπa
D Î¼0i2Â Ïa

The correct answer is:
22 μ0iπa

Solution:
Magnetic field due to one side of the square at centre O.
B1=μ04π .2isinâ¡45oa2
⇒ B1= μ04π .22 ia
Hence, magnetic field at centre due to all sides
Bnet=4B1=μ022 iπa

Topic: Physics, Magnetic Effects of Current, JEE Main
A wire is bent in the form of a circular arc of radius r with a straight portion AB. If the current in the wire is i, then the magnetic induction at point O is

Options:
A μ0i2πrtanâ¡Ï
B μ0i2πrπ-ϕ
C μ0i2πrπ-ϕ+tan⁡ϕ
D μ0i2πrπ-tan⁡ϕ

The correct answer is:
μ0i2πrπ-ϕ+tan⁡ϕ

Solution:
B=B1+B2
=μ0i4πr2π-2ϕ+μ0i4πd2sin⁡ϕ
Where cos⁡ϕ=dr
d=rcos⁡ϕ
B= μ 0 i 4πr [2π−2ϕ+2tanϕ] B= μ 0 i 2πr [π−ϕ+tanϕ]

Topic: Physics, Magnetic Effects of Current, JEE Main
A wire is made by attaching two segments together end to end. One segment is made of aluminium and the other is of steel. The effective coefficient of linear expansion of the two-segment wire is 19 Ã 10-6 Â°C-1. If the fractional length of aluminium is pq where p and q are the least positive integers, then what is the value of p+q? (linear coefficients of thermal expansion of aluminium and steel are 23 Ã 10-6 Â°C-1 and 12 Ã 10-6 Â°C-1, respectively)

Options:

The correct answer is:
18

Solution:
The change in length of wire = l⁡ A l⁡ α A l⁡ Δ θ + l⁡ st α st Δ θ associated with a temperature change of Δ θ , where α A l⁡ and α St are the coefficient of linear expansion of aluminium and steel, respectively
αAl⁡=23×10-6 ∘C-1
αSt=12×10-6 ∘C-1
The effective coefficient of linear expansion of the two segments of wire = 1 9 × 1 0 - 6 / ∘ C
l⁡ 1 α A l⁡ Δ θ + l⁡ 2 α St Δ θ = l⁡ 1 + l⁡ 2 α Δ θ
l⁡ 1 l⁡ 1 + l⁡ 2 = α - l⁡ 2 l⁡ 1 + l⁡ 2 α st α A l⁡
l⁡ 1 l⁡ 1 + l⁡ 2 = x l⁡ 1 + l⁡ 2 = l⁡ l⁡ 2 l⁡ 1 + l⁡ 2 = 1 - x
x = α - 1 - x α st α A l⁡
x=α-αstαAl⁡-αst=19×10-6-12×10-623×10-6-12×10-6=711 =pq

So, p+q=18

Topic: Physics, Thermal Properties of Matter, JEE Main
A wire is stretched such that its volume remains constant. The Poission's ratio of the material
of the wire is

Options:
A \( 0.25 \)
B \( 0.50 \)
C \( -0.25 \)
D \( -0.50 \)

The correct answer is:
\( 0.50 \)

Solution:
Poisson's ratio of a stable, isotropic, linear elastic material lies in the range $-1 < \sigma < 0.5$, theoretically.
For most of the materials $\sigma < 0.5 .$ This is because, moduli of elasticity should have positive values.
Negative Poisson's ratio in DESIGNED materials and in some anisotropic materials has been observed.
$\therefore$ For most of the materials $\sigma < 0.5 .$
$\therefore$ option (B) is correct.

Topic: Physics, Mechanical Properties of Solids, KCET, KCET 2019
A wire is tightly wound to form of a spiral of N turns with the inner and outer radii equal to a and b respectively. When a current I is established in the coil, the magnetic field at the centre is

Options:
A μ0NIb
B μ0NIa
C Î¼0NI2b-alnba
D μ0NIb-alnba

The correct answer is:
μ0NI2b-alnba

Solution:
An element is assumed at distance x from center whose width is dx. No. of turns in width b-a=N
∴ No. of turns is width dx=n=Nb-adx
dB=μ0ni2x=μ0iN2b-adxx   ∴B=μ0iN2b-a∫abdxx=μ0Ni2b-aloge⁡ba

Topic: Physics, Magnetic Effects of Current, JEE Main
A wire length L and radius r is rigidly fixed at one end. On stretching the other end of the wire with a force F, the increase in length is l . If another wire of the same material but double the length and radius is stretched with a force 2F, then increase in length is

Options:
A l4
B 2l
C l2
D l

The correct answer is:
l

Solution:
Let the length of wire = L
Radius = r
When a force F is applied, then increase in length = l
Since, Young’s modulus
Y=Fπr2lL⇒Y=FLπr2l …(i)
For another wire,
Length of wire, L’ = 2L
Radius of wire, r’ = 2r
Applied force, F’ = 2F
∴Y=F'L'πr2l' where, l ’ = elongation in length
=2F.2Lπ2r2.c'=4FL4πr2l'
Y=FLπr2l' ….(ii)
From Eqs. (i) and (ii), we have l ’ = l

Topic: Physics, Mechanical Properties of Solids, JEE Main
A wire loop carrying current i is placed in the x-y plane as shown in the figure. If an external uniform magnetic field Bâ=Bi^ is switched on, then the torque acting on the loop is

Options:
A iBa2π3-34j^
B iBa23π-34j^
C iBa2π3-32j^
D iBa23π-32j^

The correct answer is:
iBa2π3-34j^

Solution:
âµ Magnetic dipole moment of current loop is M â =I A â
here area of loop A = Area of sector of circle - area of â
13Ïa2-12a cos60Â°Ãa sin60Â°Ã2
A = Ï a 2 Ã 1 3 - 2 Ã 1 2 Ã a 2 3 2 a . = Ï a 2 3 - 3 4 a 2 , A â = Ï 3 - 3 4 a 2 k ^
âµ torque on loop placed in uniform magnetic field is
Ï â = M â Ã B â = I A â Ã B â = I Ï 3 - 3 4 a 2 k ^ Ã B i ^ = IB a 2 Ï 3 - 3 4 j ^

Topic: Physics, Magnetic Effects of Current, JEE Main
A wire loop PQR having the shape of a right-angle triangle with PQ=3x , PR=4x and QR=5x carries a steady current I. If the magnitude of the magnetic field at the point P due to this loop is kμ0I48πx, then find the value of k.

Options:

The correct answer is:
7

Solution:
The magnetic field at P due to current elements PQ and PR

is zero as the point P is on the conductor. Therefore, the magnetic field at P due to current element QR is

B=Î¼0I4Ï(PD)sinÏ1+sinÏ2

=Î¼01Ã54ÏÃ12x9x53x+16x54x

=Î¼0IÃ548Ïx35+45

=748Î¼0IÏx

Topic: Physics, Magnetic Effects of Current, JEE Main
A wire of 1 Ω has a length of 1 m. It is stretched till its length increases by 25%. The percentage change in resistance to the nearest integer is :

Options:
A 12.5%
B 76%
C 25%
D 56%

The correct answer is:
56%

Solution:
R0=1 Ω R1=

ℓ0=1 m ℓ1=1.25 m

A0=A

As volume of wire remains constant so

A0ℓ0=A1ℓ1⇒A1=ℓ0 A0ℓ1

Now, Resistance R=ρℓA

R0R1=ℓ0/A0ρℓ1/A1

1R1=ℓ0 A0ℓ0 A0ℓ1×ℓ1R1=ℓ12ℓ02=1.5625 Ω

So % change in resistance

=R1-R0R0×100%

=1.5625-11×100%

=56.25%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (26 Feb Shift 2)
A wire of a certain material is stretched slowly by $10 \%$. Its new resistance and specific resistance becomes respectively

Options:
A 1.2 times, 1.1 times
B 1.21 times, 1.21 times
C both remains the same
D None of these

The correct answer is:
None of these

Solution:
Let initial length of the wire be $l$. After stretching new length of the wire $l^{\prime}$ is given as
$$
l^{\prime}=l+10 \% \text { of } l=l+\frac{10}{100} l=\frac{11 l}{10}=1 . l
$$
i.e. length is increased by 1.1 times
$$
\therefore \quad n=1.1
$$
We know that, when length of a wire is increased by $n$ times, then its new resistance is increased by $n^2$ times i.e. $R^{\prime}=n^2 R=(1.1)^2 R \quad[\because$ Here, $n=1.1]$ $=1.21 \mathrm{R}$
Specific resistance (resistivity) of a wire does not depends on its dimensions (length, width etc.) because it is a characteristic property of a materia and depends only nature of material of wire. Hence, specific resistance remains same.

Topic: Physics, Current Electricity, KCET, KCET 2022
A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively

Options:
A 1.2 times, 1.1 times
B 1.21 times, same
C both remain the same
D 1.1 times, 1.1 times

The correct answer is:
1.21 times, same

Solution:
Key Idea : In stretching, specific resistance remains unchanged.
After stretching, specific resistance $(\rho)$ will remain same.
Original resistance of the wire,
$R=\frac{\rho l}{A}$
$R \propto \frac{l}{A} \text { or } R \propto \frac{l^2}{V}$ (as $V=A l$)
and $R^{\prime} \propto \frac{(l+10 \% l)^2}{V}$
Therefore, $\frac{R^{\prime}}{R}=\frac{\left(l+\frac{10}{100} l\right)^2}{l^2}$
$\frac{R^{\prime}}{R}=\frac{\left(\frac{11 l}{10}\right)^2}{l^2}=\frac{121}{100}$
$R^{\prime}=1.21 R$

Topic: Physics, Current Electricity, NEET, NEET 2008 (Screening)
A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively

Options:
A 1.2 times, 1.1 times
B 1.21 times, same
C both remain the same
D 1.1 times, 1.1 times

The correct answer is:
1.21 times, same

Solution:
Key Idea : In stretching, specific resistance remains unchanged.
After stretching, specific resistance $(\rho)$ will remain same.
Original resistance of the wire,
$R=\frac{\rho l}{A}$
$R \propto \frac{l}{A} \text { or } R \propto \frac{l^2}{V}$ (as $V=A l$)
and $R^{\prime} \propto \frac{(l+10 \% l)^2}{V}$
Therefore, $\frac{R^{\prime}}{R}=\frac{\left(l+\frac{10}{100} l\right)^2}{l^2}$
$\frac{R^{\prime}}{R}=\frac{\left(\frac{11 l}{10}\right)^2}{l^2}=\frac{121}{100}$
$R^{\prime}=1.21 R$

Topic: Physics, Current Electricity, JEE Main
A wire of a certain material is stretched slowly by ten percent new resistance and specific resistance become respectively

Options:
A 1.1 times, 1.1 times
B 1.2 times, 1.1 times
C 1.21 times, same
D Both remain the same

The correct answer is:
1.21 times, same

Solution:
$\begin{aligned}
& \frac{R_2}{R_1}=\left(\frac{l_2}{l_1}\right)^2=\left(\frac{1.1 l_1}{l_1}\right)^2=1.21 \\
& R_2=1.21 R
\end{aligned}$

Topic: Physics, Current Electricity, NEET, NEET 2008 (Mains)
A wire of cross sectional area A, modulus of elasticity $2 \times 10^{11} \mathrm{Nm}^{-2}$ and length $2 \mathrm{~m}$ is stretched between two vertical rigid supports. When a mass of $2 \mathrm{~kg}$ is suspended at the middle it sags lower from its original position making angle $\theta=\frac{1}{100}$ radian on the points of support. The value of $\mathrm{A}$ is _______ $\times 10^{-4} \mathrm{~m}^2$ (consider $x< < \mathrm{L}$ ).
(given : $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:

The correct answer is:
1

Solution:
In vertical direction
$2 \mathrm{~T} \sin \theta=20$
using small angle approximation $\sin \theta=\theta$
$\begin{aligned}
& \theta=\frac{1}{100} \\
& \therefore \mathrm{T}=\frac{10}{\theta} \\
& \mathrm{T}=1000 \mathrm{~N}
\end{aligned}$
$\begin{array}{ll}
\text {Change in length } \Delta \mathrm{L} =2 \sqrt{\mathrm{x}^2+\mathrm{L}^2}-2 \mathrm{~L} \\
=2 \mathrm{~L}\left[1+\frac{\mathrm{x}^2}{2 \mathrm{~L}^2}-1\right] \\
\Delta \mathrm{L} =\frac{\mathrm{x}^2}{\mathrm{~L}}
\end{array}$
$\therefore$ Modulus of elasticity $=\frac{\text { stress }}{\text { strain }}$
$\begin{aligned}
& 2 \times 10^{11}=\frac{10^3}{\mathrm{~A} \times \frac{\mathrm{x}^2}{\mathrm{~L}}} \times 2 \mathrm{~L} \\
\therefore & \mathrm{A}=1 \times 10^{-4} \mathrm{~m}^2
\end{aligned}$

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2024 (06 Apr Shift 2)
A wire of cross-sectional area $3 \mathrm{~mm}^2$ is first stretched between two fixed points at a temperature of $20^{\circ} \mathrm{C}$. Determine the tension when the temperature falls to $10^{\circ} \mathrm{C}$. Coefficient of linear expansion $\alpha=10^{-5}{ }^{\circ} \mathrm{C} \mathrm{C}^{-1}$ and $Y=2 \times 10^{11} \mathrm{~N} / \mathrm{m}^2$

Options:
A $20 \mathrm{~N}$
B $30 \mathrm{~N}$
C $60 \mathrm{~N}$
D $120 \mathrm{~N}$

The correct answer is:
$60 \mathrm{~N}$

Solution:
$F=Y A \alpha \Delta t=2 \times 10^{11} \times 3 \times 10^{-6} \times 10^{-5} \times(20-10)=60 \mathrm{~N}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A wire of cross-sectional area Ais stretched horizontally between two clamps loaded at a distance 2l metres from each other. A weight w kg suspended from the midpoint of the wire. The strain produced in the wire, (if the vertical distance through which the midpoint of the wire moves down x <l) will be

Options:
A x2l2
B 2x2l2
C x22l2
D x2l

The correct answer is:
x22l2

Solution:
From figure the increase in length âl=PR+RQ- PQ
= 2PR â PQ
=2l2+x21/2-2l=2l1+x2l21/2-2l
=2l1+12x2l2-2l ( By Binomial theorem)
=x2/l
â´ Strain= âl/2l=x2/2l2

Topic: Physics, Mechanical Properties of Solids, JEE Main
A wire of density 8×103 kg m-3 is stretched between two clamps 0.5 m apart. The extension developed in the wire is 3.2×10-4 m. If Y=8×1010 N m-2, the fundamental frequency of vibration in the wire will be _____ Hz

Options:

The correct answer is:
80

Solution:
Using the relation of Young's modulus,

TA=YΔLL
⇒T=YΔLL×A

The linear mass density is μ=mL.

So,
Tμ=YΔLALmL=Y(ΔL)×LAL(m)=YΔLL×1ρ

Substituting the values,
Tμ=8×1010×3.2×10-40.5×18×103=6.4×103
⇒Tμ=64×102

The fundamental frequency is given by f=12LTμ.
⇒Tμ=8×10=80 m s-1
Therefore,

f=801=80 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (11 Apr Shift 2)
A wire of density 9×10–3 kg cm–3 is stretched between two clamps 1 m apart. The resulting strain in the wire is 4.9 × 10–4. The lowest frequency of the transverse vibrations in the wire (Young's modulus of wire Y = 9×1010 Nm–2 ), (to the nearest integer),_______

Options:

The correct answer is:
35

Solution:
Fundamental frequency in the string

f=12ℓTμ=12ℓTρA=12ℓYΔℓρℓ
f=12ℓYΔℓρℓ
=129×1010×4.9×10-49×10-3×10-6=35

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2020 (02 Sep Shift 2)
A wire of initial length $L$ and raidus $r$ is stretched by a length $l$. Another wire of same material but with initial length $2 L$ and radius $2 r$ is stretched by a length $2 l$. The ratio of stored elastic energy per unit volume in the first and second wire is

Options:
A $1: 4$
B $1: 2$
C $2: 1$
D $1: 1$

The correct answer is:
$1: 1$

Solution:
Elastic energy per unit volume of wire. $\mu=\frac{1}{2} \times$ Young's modulus $\times(\text { Strain })^{2}$
$\therefore \quad \frac{u_{1}}{u_{2}}=\left\{\frac{(\text { Strain })_{1}}{(\text { Strain })_{2}}\right\}^{2}=\frac{l^{2}}{L^{2}} \times \frac{4 L^{2}}{4 l^{2}}=1: 1$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A wire of length 1 m moving with velocity 8 m s-1 at right angles to a magnetic field of 2 T. The magnitude of induced emf, between the ends of wire will be ___________.

Options:
A 20 V
B 8 V
C 12 V
D 16 V

The correct answer is:
16 V

Solution:
The expression of motional emf is e=BLvsinθ, where, θ is angle between magnetic field and direction of motion.

Here, θ=90°,

So, induced emf across the ends of wire is e= BLvsin90°=BLv

= 2 × 1 × 8 = 16 V

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2023 (25 Jan Shift 2)
A wire of length 2L, is made by joining two wires A and B of same length but different radii r and 2r and made of the same material. It is vibrating at a frequency such that the joint of the two wires forms a node. If the number of antinodes in wire A is p and that in B is q then ratio p:q is:

Options:
A 3 :5
B 4 :9
C 1 :4
D 1 :2

The correct answer is:
1 :2

Solution:
For rod A, f1=n12lTρA1
For rod B, f2=n22lTρA2
Given f1=f2
n12lTρA1=n22lTρA2
n12lTρA1=n22lTρ(4A1) ∵rB=2rAA2=4A1
n1=n22
n1n2=12

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (08 Apr Shift 1)
A wire of length 2 units is cut into two parts which are bent respectively to form a square of side =x units and a circle of radius =r units. If the sum of the areas of the square and the circle so formed is minimum, then

Options:
A x=2r
B 2x=r
C 2x=π+4r
D 4-πx=πr

The correct answer is:
x=2r

Solution:
Given that 4x+2πr=2
i.e., 2x+πr=1
∴ r=1-2xπ .......(i)
Area A=x2+πr2
=x2+1π 2x-12
For min vale of area A
dAdx=0 gives x=2π+4 .......(ii)
From (i) and (ii)
r=1π+4 .......(iii)
∴ x=2r

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2016 (03 Apr)
A wire of length 10 cm and radius 7×10-4 m connected across the right gap of a meter bridge. When a resistance of 4.5 Ω is connected on the left gap by using a resistance box, the balance length is found to be at 60 cm from the left end. If the resistivity of the wire is R×10-7 Ω m, then value of R is :

Options:
A 63
B 70
C 66
D 35

The correct answer is:
66

Solution:
From the null point condition of a Meter Bridge, it follows that,

4.5 Ω60=X100-60⇒X=4060×4.5 Ω=3 Ω

Also,

X=ρlA=ρlπr2 ...1

From equation (1), it follows that

3=ρ×0.1π×7×10-42⇒ρ=3×π×7×10-420.1≈66×10-7 Ω m

Therefore, R=66.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (27 Jan Shift 1)
A wire of length 20 m is to be cut into two pieces. A piece of length ℓ1 is bent to make a square of area A1 and the other piece of length ℓ2 is made into a circle of area A2. If 2 A1+3 A2 is minimum then πℓ1:ℓ2 is equal to:

Options:
A 6: 1
B 3: 1
C 1: 6
D 4: 1

The correct answer is:
6: 1

Solution:
The length of wire is 20 m.

According to the question:

ℓ1+ℓ2=20

⇒dℓ2 dℓ1=-1

Let the side of the square be x, then

4x=ℓ1

⇒x=ℓ14 ...1

Let the radius of the circle be y, then

2πy=ℓ2

⇒y=ℓ22π ...2

Hence,

A1=ℓ142 and A2=πℓ22π2

Let S=2A1+3A2

⇒S=(ℓ1)28+3(ℓ2)24π

⇒dSdℓ1=2ℓ18+6ℓ24π·dℓ2 dℓ1

For S to be minimum,

dsdℓ1=0

⇒2ℓ18+6ℓ24π·dℓ2 dℓ1=0

⇒2ℓ18-6ℓ24π=0 ∵ dℓ2 dℓ1=-1

⇒ℓ14=6ℓ24π

⇒πℓ1ℓ2=6

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2023 (31 Jan Shift 1)