datavorous/entrance-exam-dataset huggingface dataset:
A uniform cylinder of mass M and radius R is to be pulled over a step of height a a <R by applying a force F at its centre 'O' perpendicular to the plane through the axes of the cylinder on the edge of the step (see figure). The minimum value of F required is :

Options:
A Mg1-R-aR2
B MgRR-a2-1
C MgaR
D Mg1-a2R2

The correct answer is:
Mg1-R-aR2

Solution:
Mg1-R-aR2

FR>Mgcosθ R

F>Mgcosθ

F>MgR2-(R-a)2R.

⇒Mg1-R-aR2

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2020 (02 Sep Shift 1)
A uniform cylinder of the radius R=3 m is spin about its axis at an angular velocity Ï0=40Ï rad s-1 and placed between two perpendicular walls. The coefficient of friction between the walls and cylinder is Î¼=2. Then, how many rotations will the cylinder make before it comes to rest?

Options:

The correct answer is:
50

Solution:
Î¼N1+N2=mg â¦â¦..(i)
N1=Î¼N2 â¦â¦..(ii)
N2=mg1+Î¼2
N1=Î¼mg1+Î¼2

The torque about the center of mass of the cylinder is

Ï=-Î¼N1+Î¼N2R

âÏ=-Î¼mgR1+Î¼2Î¼+1

-Î¼mgRÎ¼+11+Î¼2ÃÎ¸ = 0 - 12mR22Ï2

âÎ¸ = 1+Î¼2RÏ24Î¼gÎ¼+1

ân = Î¸2Ï = 1+Î¼2RÏ28ÏÎ¼gÎ¼+1

n=50

Topic: Physics, Rotational Motion, JEE Main
A uniform cylindrical rod of length L and radius r, is made from a material whose Young’s modulus of Elasticity equals Y. When this rod is heated by temperature T and simultaneously subjected to a net longutudinal compressional force F, its length remains unchanged. The coefficient of volume expansion, of the material of the rod, is (nearly) equal to:

Options:
A F/3πr2YT
B 9F/πr2YT
C 6F/πr2YT
D 3F/(πr2YT)

The correct answer is:
3F/(πr2YT)

Solution:
Stress =FA
YαT=FA
α=FAYT
γ=3α=3FAYT
γ=3Fπr2YT

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A uniform disc is rotating at a constant speed in a vertical plane about a fixed horizontal axis passing through the centre of the disc. A piece of the disc from its rim detaches itself from the disc at the instant when it is at a horizontal level with the centre of the disc and moving upward. Then about the fixed axis, the angular speed of the remaining disc is

Options:
A remains unchanged
B decreases
C increases
D initially increases and later decreases

The correct answer is:
remains unchanged

Solution:
Since the broken away piece of mass m (say) has the same linear speed as it had before detaching from the disc, the angular speed of the remaining disc will remain unchanged.

Topic: Physics, Rotational Motion, NEET
A uniform disc of mass 300kg is rotating freely about a vertical axis through its centre at a constant angular velocity ÏÂ°. A boy of mass 30kg starts from the centre and moves along a radius to the edge of the disc. Angular velocity of the disc will now become

Options:
A ω06
B ω05
C 4ω05
D 5Ï06

The correct answer is:
5ω06

Solution:
As Στ=0, angular momentum remains conseved:
∴ L=0+300 R22ω0=300 R22+30 R2ω
⇒150 ω0=180 ω⇒ ω=56 ω0

Topic: Physics, Rotational Motion, NEET
A uniform disc of mass M and radius R is mounted on a fixed horizontal axis. A block of mass m hangs from a massless string that is wrapped around the rim of the disc. The magnitude of the acceleration of the falling block (m) is

Options:
A 2MM+2m g
B 2mM+2m g
C M+2m2 M g
D 2M+m2M g

The correct answer is:
2mM+2m g

Solution:
When pulley has a finite mass M and radius R, then tension in two segments of the string are different.
Here, ma=mg-T
a=mm+M2g=2m2m+Mg

Topic: Physics, Rotational Motion, JEE Main
A uniform disc of mass M and radius R is supported vertically by a pivot at its periphery as shown. A particle of mass M is fixed to the rim and raised to the highest point above the centre. The system is released from rest and it can rotate about pivot freely. The angular speed of the system when it attached object is directly beneath the pivot, is

Options:
A 24g11R
B 8g11R
C 8g3R
D 14g3R

The correct answer is:
24g11R

Solution:
Mg2R+Mg4R=12 32MR2+M(2R)2ω2

⇒6MgR=114MR2ω2
⇒ ω=24g11R

Topic: Physics, Rotational Motion, JEE Main
A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is:

Options:
A 2 Mg2m+M
B 2 Mg2M+m
C 2Â mg2M+m
D 2 mg2m+M

The correct answer is:
2 mg2m+M

Solution:
mg-T=ma
RT=Iα
RT=MR22 .aR
T=Ma2
mg-Ma2= ma
mg=aM2+m
mg=a M+2m2
a=2mgM+2m

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2017 (08 Apr Online)
A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity then

Options:
A
its velocity will decrease.

B
its velocity will increase.

C
it will turn towards right of direction of motion.

D
it will turn towards left of direction of motion

The correct answer is:
its velocity will decrease.

Solution:
$\vec{F}=-e[\vec{E}+\vec{v} \times \vec{B}]=-e \vec{E}$
$\vec{a}=-\frac{e \vec{E}}{m}$
$v(t)=v_{\circ}-\frac{e E}{m} t$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2005
A uniform electric field and a uniform magnetic field exist in a region in the same direction. An electron is projected with a velocity pointed in the same direction. Then the electron will be

Options:
A Be deflected to the left without increase in speed
B Be deflected to the right without increase in speed
C Not be deflected but its speed will decrease
D Not be deflected but its speed will increase

The correct answer is:
Not be deflected but its speed will decrease

Solution:
As the electron is moving parallel to both the fields, so magnetic force does not affect the electron's motion. But electric force (qE) acts opposite to the motion of the electron. Hence, the electron will not be deflected but its speed will decrease.

Topic: Physics, Mathematics in Physics, JEE Main
A uniform electric field E=8meV m-1 is created between two parallel plates of length 1 m as shown in figure, (where m= mass of electron and e= charge of electron). An electron enters the field symmetrically between the plates with a speed of 2 m s-1. The angle of the deviation θ of the path of the electron as it comes out of the field will be _____ .

Options:
A tan-14
B tan-12
C tan-113
D tan-13

The correct answer is:
tan-12

Solution:
Time taken by the electron to cross will be, t=1vx=12 s.

Acceleration of the electron along y axis

ag=Fm=Eem=8me×em=8 m s-2

vy of the electron when it has crossed the capacitor will be,

vy=ayt=8×12=4 m s-1

Therefore,

tanθ=vyvx=42=2⇒θ=tan-12

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (28 Jul Shift 2)
A uniform electric field $\vec{E}$ exists between the plates of a charged condenser. A charged particle enters the space between the plates and perpendicular to $\vec{E}$. The path of the particle between the plates is a:

Options:
A
straight line

B
hyperbola

C
parabola

D
circle

The correct answer is:
parabola

Solution:
When charged particle enters perpendicularly in an electric field, it describes a parabolic path
$$
\mathrm{y}=\frac{1}{2}\left(\frac{\mathrm{QE}}{\mathrm{m}}\right)\left(\frac{\mathrm{x}}{4}\right)^2
$$
This is the equation of parabola.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2013 (09 Apr Online)
A uniform electric field of 10 N C-1 is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy 0.5 eV. The length of each plate is 10 cm. The angle (θ) of deviation of the path of electron as it comes out of the field is ______(in degree).

Options:

The correct answer is:
45

Solution:
Given here, kinetic energy 12e=12mux2⇒ux=em

Acceleration of electron due to electric field is a=eEm, where, e is charge on electron.

Using equation of motion, s=uxt+12axt2 along x direction, we get

L=uxt+0∵ax=0⇒10×10-2=uxt⇒t=10-1ux ....1

Using equation of motion, vy=uy+ayt along y direction, we get

vy=0+eEmt=eEmt ...2

From equation 1 and 2, we have

vy=eEm10-1ux

Now, tanθ=vyvx=eEm10-1ux2=e×10×10-1m×em=1

The angle of deviation of the path of electron as it comes out of the field is θ=45°.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (25 Jan Shift 1)
A uniform heating wire of resistance 36 Ω is connected across a potential difference of 240 V. The wire is then cut into half and a potential difference of 240 V is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be 1: x, where x is

Options:

The correct answer is:
4

Solution:
P=V2/R

P1=V2R2 ⇒ P2=V2R2 ⇒ P1+P2=4V2R

Now
P1+P2P=4V2RV2R=4

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (01 Sep Shift 2)
A uniform heavy rod of mass 20 kg. Cross sectional area 0.4 m2 and length 20 m is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is x×10-9 m. The value of x is _____ .
(Given. Young's modulus Y=2×1011Nm-2 and g=10 m s-2 )

Options:

The correct answer is:
25

Solution:
Tension at a distance x from lower end =mglx

If the elongation in the element is taken as d∆l then, using Hooke's law,

Y=mgxdxAld∆l⇒d∆l=maxdxAlY⇒∫0∆ld∆l=∫0lmgxdxAlY⇒∆l=mgl2AY

⇒Δl=20×10×202×0.4×2×1011

⇒Δl=25×10-9

⇒x=25

Topic: Physics, Mechanical Properties of Solids, JEE Main
A uniform heavy rod of mass 20 kg. Cross sectional area 0.4 m2 and length 20 m is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is x×10-9 m. The value of x is _____ .
(Given. Young's modulus Y=2×1011Nm-2 and g=10 m s-2 )

Options:

The correct answer is:
25

Solution:
Tension at a distance x from lower end =mglx

If the elongation in the element is taken as d∆l then, using Hooke's law,

Y=mgxdxAld∆l⇒d∆l=maxdxAlY⇒∫0∆ld∆l=∫0lmgxdxAlY⇒∆l=mgl2AY

⇒Δl=20×10×202×0.4×2×1011

⇒Δl=25×10-9

⇒x=25

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2022 (26 Jul Shift 2)
A uniform heavy rod of weight 10 kg m s-2, cross-sectional area 100 cm2 and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2×1011 N m-2. Neglecting the lateral contraction, find the elongation of rod due to its own weight:

Options:
A 5×10-10 m
B 4×10-8 m
C 5×10-8 m
D 2×10-9 m

The correct answer is:
5×10-10 m

Solution:
Elongation due to self weight,
Wl2AY=10×0.22×2×1011×100×10-4=5×10-10 m

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2021 (31 Aug Shift 1)
A uniform heavy rod of weight $10 \mathrm{~kg} \mathrm{~ms}^{-2}$, cross-sectional area $100 \mathrm{~cm}^2$ and length $20 \mathrm{~cm}$ is hanging from a fixed support. Young modulus of the material of the rod is $2 \times 10^{11}$ $\mathrm{Nm}^{-2}$. Neglecting the lateral contraction, find the elongation of rod due to its own weight.

Options:
A $2 \times 10^{-9} \mathrm{~m}$
B $5 \times 10^{-8} \mathrm{~m}$
C $4 \times 10^{-8} \mathrm{~m}$
D $5 \times 10^{-10} \mathrm{~m}$

The correct answer is:
$5 \times 10^{-10} \mathrm{~m}$

Solution:
We know,
$\begin{aligned} & \Delta \ell=\frac{W L}{2 \mathrm{AY}} \\ & \Delta \ell=\frac{10 \times 1}{2 \times 5} \times 100 \times 10^{-4} \times 2 \times 10^{11} \\ & \Delta \ell=\frac{1}{2} \times 10^{-9}=5 \times 10^{-10} \mathrm{~m}\end{aligned}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A uniform L shaped rod each of side a is held, as shown in the figure. The angle θ such that rod remains stable will be.

Options:
A tan-112
B tan-113
C tan-12
D tan-13

The correct answer is:
tan-113

Solution:
At equilibrium, the centre of mass will be below the point of suspension.
Taking torque about point O,
mgl2sinθ+mglsinθ-l2cosθ=0
⇒3l2sinθ=l2cosθ
⇒tanθ=13
⇒θ=tan-113

Topic: Physics, Laws of Motion, NEET
A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field changes with time at a steady rate dB/dt=0.032Ts-1. The induced current in the loopis close to (Resistivity of the metal wire is 1.23×10-8 Ωm )

Options:
A 0.43 A
B 0.61 A
C 0.34 A
D 0.53 A

The correct answer is:
0.61 A

Solution:
Radius =d/2=2 mm

Rwire=ρlA

ϕ=BA ; e=dϕdt=dBdtA

i=eR=dBdtA2ρl=0.032×π×2×10-321.23×10-8×4×0.3

=0.61 A

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2020 (03 Sep Shift 2)
A uniform magnetic field B of 0.3T is along the positive Z -direction. A rectangular loop (abcd) of sides 10 cm×5 cm carries a current I of 12 A. Out of the following different orientations which one corresponds to stable equilibrium?

Options:
A
B
C
D

The correct answer is:

Solution:
U=-M→·B→=-MBcosθ
For stable equilibrium potential energy should be minimum and τ=0
Umin=-MB when θ=0°

⇒cos⁡θ = 1
⇒M→ and B→ are on same direction.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2017 (09 Apr Online)
A uniform magnetic field is restricted within a region of radius, r. The magnetic field changes with time at a rate, dB→dt. Loop one of radius R>r encloses the region, r and loop two of radius, R is outside the region of magnetic field as shown in the figure below. Then the emf generated is

Options:
A 0 in loop one and zero in loop two.
B -dB→dtπr2 in loop one and -dB→dtπr2 in loop two.
C -dB→dtπr2 in loop one and infinite in loop two.
D -dB→dtπr2 in loop one and zero in loop two.

The correct answer is:
-dB→dtπr2 in loop one and zero in loop two.

Solution:
Magnetic flux linked with area of loop one is Bπr2.

So, emf in loop one is, -dB→dtπr2 .

Magnetic flux linked with area of loop two is zero.

So, emf in loop two is equal to zero.

Topic: Physics, Electromagnetic Induction, JEE Main
A uniform magnetic field of 2×10-3 T acts along positive Y-direction. A rectangular loop of sides 20 cm and 10 cm with current of 5 A is in Y-Z plane. The current is in anticlockwise sense with reference to negative X axis. Magnitude and direction of the torque is :

Options:
A 2×10-4 N m along positive Z –direction
B 2×10-4 N m along negative Z-direction
C 2×10-4 N m along positive X-direction
D 2×10-4 N m along positive Y-direction

The correct answer is:
2×10-4 N m along negative Z-direction

Solution:
Magnetic moment is given by, M→=iA→

=5×0.2×0.1-i^

=0.1-i^

τ→=M→×B→=0.1−i^×2×10−3j^

=2×10−4−k^ N m

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2024 (31 Jan Shift 2)
A uniform magnetic field of $1.5 \mathrm{~T}$ exists in a cylindrical region of radius $10.0 \mathrm{~cm}$, its direction parallel to the axis along east to west. A wire carrying current of $7.0 \mathrm{~A}$ in the north to south direction passes through this region. What is the magnitude and direction of the force on the wire if,
(a) the wire intersects the axis,
(b) the wire is turned from $\mathrm{N}-\mathrm{S}$ to northeast-northwest direction.
(c) the wire in the $\mathrm{N}-\mathrm{S}$ direction is lowered from the axis by a distance of $6.0 \mathrm{~cm}$ ?

Options:

None

Solution:
The magnetic field is in the direction east to west and in the cylindrical region of radius $10 \mathrm{~cm}$.
(a) A current carrying wire, intersects the axis.
Force on wire
$$
\begin{aligned}
F &=I B l=I B(2 r)=7 \times 1.5 \times 2 \times 1 \times 10^{-2} \\
&=2.1 \mathrm{~N} \text { upwards }
\end{aligned}
$$
(b) By turning the wire by an angle $45^{\circ}$ in NE and SW direction the force remain the same $f=2.1 \mathrm{~N}$.
(c) Now the wire is lowered from the axis by a distance of $6.0 \mathrm{~cm}$.
Length of wire inside cylindrical region is $16 \mathrm{~cm}$ now.
So, the force is $F=I B l=7 \times 1.5 \times 16 \times 10^{-2}$
$$
=1.68 \mathrm{~N}
$$

Topic: Physics, Magnetic Effects of Current
A uniform metal plate shaped like a triangle $\mathrm{ABC}$ has a mass of $540 \mathrm{gm}$. the length of the sides $\mathrm{AB}, \mathrm{BC}$ and CA are $3 \mathrm{~cm}, 5 \mathrm{~cm}$ and $4 \mathrm{~cm}$, respectively. The plate is pivoted freely about the point $\mathrm{A}$. What mass must be added to a vertex, so that the plate can hang with the long edge horizontal ?

Options:
A $140 \mathrm{gm}$ at $\mathrm{C}$
B $540 \mathrm{gm}$ at $\mathrm{C}$
C $140 \mathrm{gm}$ at $\mathrm{B}$
D $540 \mathrm{gm}$ at $\mathrm{B}$

The correct answer is:
$140 \mathrm{gm}$ at $\mathrm{C}$

Solution:
$\mathrm{CM}$ of triangular plate is on the median. If we put a mass say $\mathrm{m}_{1}$ on $\mathrm{C}$ it will produce torque about A which balance the torque produce $m g$ about $A$. Thus plate will can be in equilibrium position
$\begin{array}{l}
\mathrm{m}_{1} \mathrm{~g} \times 4 \cos 37=\mathrm{mg} \times \mathrm{y} \\
\mathrm{m}_{1} \mathrm{~g} \times 4 \times \frac{4}{5}=\mathrm{mg} \times \mathrm{y} \\
\mathrm{m}_{1}=\mathrm{m} \times \mathrm{y} \times \frac{5}{16} \\
\frac{\mathrm{m}_{1}}{\mathrm{~m}}=\mathrm{y} \times \frac{5}{16} \\
\mathrm{y} < 3 \quad \therefore \frac{\mathrm{m}_{1}}{\mathrm{~m}} < 1 \\
\mathrm{~m}_{1} < \mathrm{m} \\
\mathrm{m}_{1} < 540 \mathrm{~g}
\end{array}$
from given option Ans. (A)

Topic: Physics, Rotational Motion, JEE Main
A uniform metal rod is moving with a uniform velocity v parallel to a long straight wire carrying a current I. The rod is perpendicular to the wire with its ends at distances r1 and r2 (with r2>r1) from it. The E.M.F. induced in the rod at that instant is

Options:
A zero
B μ0Iv2πlnr2r1
C μ0Iv2πlnr1r2
D μ0Iv4π1-r1r2

The correct answer is:
μ0Iv2πlnr2r1

Solution:
B=μ0i2πx
dε= Bvdx
ε=μ0iv2π∫r2r1dxx=μ0iv2πlnr2r1

Topic: Physics, Electromagnetic Induction, JEE Main
A uniform metal rod is used as a bar pendulum. If the room temperature rises by $10^{\circ} \mathrm{C}$, and the coefficient of linear expansion of the metal of the rod is $2 \times 10^{-6}$ per ${ }^{\circ} \mathrm{C}$, the period of the pendulum will have percentage increase of

Options:
A $-2 \times 10^{-3}$
B $-1 \times 10^{-3}$
C $2 \times 10^{-3}$
D $1 \times 10^{-3}$

The correct answer is:
$1 \times 10^{-3}$

Solution:
Fractional change in period
$\frac{\Delta T}{T}=\frac{1}{2} \alpha \Delta \theta=\frac{1}{2} \times 2 \times 10^{-6} \times 10=10^{-5}$
$\%$ change $=\frac{\Delta T}{T} \times 100=10^{-5} \times 100=10^{-3} \%$

Topic: Physics, Thermal Properties of Matter, JEE Main
A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly

Options:
A
its speed of rotation increases

B
its speed of rotation decreases

C
its speed of rotation remains same

D
its speed increases because its moment of inertia increases

The correct answer is:
its speed of rotation decreases

Solution:
On heating a uniform metallic rod its length will increase so moment of inertia of rod increased from $I_1$ to $I_2$.
No external torque is acting on the system so angular momentum should be conserved.
$L=$ Angular momentum $=I \omega=$ constant
$I_1 \omega_1=I_2 \omega_2$
So, $\frac{\omega_2}{\omega_1}=\frac{I_1}{I_2} < 1$
$\omega_2 < \omega_1$
$\left(\because\right.$ Due to expansion of the $\left.\operatorname{rod} I_2>I_1\right)$
So, angular veloctiy decreases.

Topic: Physics, Thermal Properties of Matter
A uniform metallic wire carries a current 2 A, when 3.4 V battery is connected across it. The mass of uniform metallic wire is 8.92×10-3 kg, density is 8.92×103 kg m-3 and resistivity is 1.7×10-8 Ω-m The length of wire is :

Options:
A

l=6.8 m

B

l=10 m

C

l=5 m

D

l=100 m

The correct answer is:

l=10 m

Solution:
Given here, I=2 A, V=3.4 V, m=8.92×10-3 kg,

d=8.92×103 kg m-3 and ρ=1.7×10-8 Ω-m

From Ohm's law, resistance is R=VI=3.42=1.7 Ω

As resistance of wire is given by R=ρlA, where, A is area and l is length of wire.

⇒1.7=ρlA ⇒l=1.7Aρ

Now, volume=md=8.92×10-38.92×103=10-6 m3

Thus,

l2=1.7Aρ×l=1.7volumelρ×l=1.7ρ10-6=1.71.7×10-8×10-6

Hence, l=10 m

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (25 Jan Shift 1)
A uniform metallic wire has a resistance of 18 Ω and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is:

Options:
A 2 Ω
B 12 Ω
C 8 Ω
D 4 Ω

The correct answer is:
4 Ω

Solution:
Resistance,

R=ρlAR α lsay Reff=l=18as l1=l2=l3=l/3⇒R1=R2=R3=R=18/3=6

So from diagram

2R=12 and R=6also we know RAB=R2RRRR2R+RRRAB=6×126+12=4 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2019 (10 Jan Shift 1)
A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be _________ cm.

Options:

The correct answer is:
2

Solution:
F=Y.A·Δll⇒Δl=FY.A.·l⇒Δl=F.lY.πr2⇒Δl∝lr2

Δl2Δl1=l2l1r1r22=2122⇒Δl2Δl1=12⇒Δl2=Δl12

=0.042=0.02 m⇒Δl2=2 cm.

Topic: Physics, Mechanical Properties of Solids, JEE Main
A uniform metallic wire is elongated by 0.04 m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be _________ cm.

Options:

The correct answer is:
2

Solution:
F=Y.A·Δll⇒Δl=FY.A.·l⇒Δl=F.lY.πr2⇒Δl∝lr2

Δl2Δl1=l2l1r1r22=2122⇒Δl2Δl1=12⇒Δl2=Δl12

=0.042=0.02 m⇒Δl2=2 cm.

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2021 (24 Feb Shift 2)
A uniform metallic wire is elongated by $0.04 \mathrm{~m}$ when subjected to a linear force $\mathrm{F}$. The elongation, if its length and diameter is doubled and subjected to the same force will be $\qquad$ $\mathrm{cm}$.

Options:
A $1 \mathrm{~cm}$
B $2 \mathrm{~cm}$
C $4 \mathrm{~cm}$
D $8 \mathrm{~cm}$

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

Solution:
$\begin{aligned} & \mathrm{F}=\mathrm{Y} . \mathrm{A} \cdot \frac{\Delta \ell}{\ell} \\ & \Delta \ell=\frac{\mathrm{F}}{\text { Y.A. }} . \ell \\ & \Delta \ell=\frac{\mathrm{F} \cdot \ell}{\mathrm{Y} . \pi \mathrm{r}^2} \\ & \Delta \ell \propto \frac{\ell}{\mathrm{r}^2} \\ & \frac{\Delta \ell_2}{\Delta \ell_1}=\left(\frac{\ell_2}{\ell_1}\right)\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^2=\text { (2) }\left(\frac{1}{2}\right)^2 \\ & \frac{\Delta \ell_2}{\Delta \ell_1}=\frac{1}{2} \\ & \Delta \ell_2=\frac{\Delta \ell_1}{2} \\ & =\frac{0.04}{2}=0.02 \mathrm{~m} \\ & \end{aligned}$
$\Delta \ell_2=2 \mathrm{~cm}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A uniform metallic wire of lenght $L$ is mounted in two configurations. In configuration I (triangle), it is an equilateral triangle and a voltage $V$ is applied to corners A and B. In configuration 2 (circle), it is bent in the form of a circle, and the potential $v$ is applied at diameterically opposite points $P$ and $Q$. The ratio of the power dissipated in configuration 1 to configuration 2 is.

Options:
A $2 / 3$
B $9 / 8$
C $5 / 4$
D $7 / 8$

The correct answer is:
$9 / 8$

Solution:
$\mathrm{R}=\frac{\rho \ell}{\mathrm{A}}$
$\left(\mathrm{R}_{1}=\frac{\rho \frac{\ell}{3}}{\mathrm{~A}}\right)$
$\frac{1}{\mathrm{R}_{\mathrm{ep}}} \Rightarrow \frac{1}{2 \mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{1}} \Rightarrow 2$
$\mathrm{R}_{\mathrm{eq}}=\frac{2 \mathrm{R}_{1}}{3}=\frac{2 \rho \frac{\ell}{3}}{3 \mathrm{~A}}$
$\mathrm{i}_{1}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}}=\frac{\mathrm{V}}{2 \mathrm{e} \ell} 9 \mathrm{~A}$
$\mathrm{R}_{\mathrm{eq}} \Rightarrow \frac{\mathrm{R}}{2} \Rightarrow \frac{1}{2}\left(\frac{\rho . \ell / 2}{\mathrm{~A}}\right)$
$\mathrm{i}_{2}=\frac{\mathrm{V}}{\mathrm{R}_{\mathrm{eq}}} \Rightarrow \frac{\mathrm{V}}{\rho \ell} 4 \mathrm{~A}$
$\frac{\mathrm{i}_{1}}{\mathrm{i}_{2}}=\frac{9}{8}$

Topic: Physics, Current Electricity, JEE Main
A uniform rectangular thin sheet ABCD of mass M has length a and breadth b, as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be:

Options:
A 5a12,5b12
B 5a3,5b3
C 2a3,2b3
D 3a4,3b4

The correct answer is:
5a12,5b12

Solution:
If the total mass of object mass M then the mass removed part will be M4 .

XCM=M·a2-M43a4M-M4=Ma2-3Ma163M4=512a
YCM=M·b2-M43b4M-M4=Mb2-3Mb163M4=512b

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2019 (08 Apr Shift 2)