datavorous/entrance-exam-dataset huggingface dataset:
A heavy iron bar of weight 12 kg is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle 60° with the horizontal, the normal force applied by the man on bar is:

Options:
A 6 kg-wt
B 12 kg-wt
C 3 kg-wt
D 63 kg-wt

The correct answer is:
3 kg-wt

Solution:
Let's consider the following diagram:

For the equilibrium of the system, the net torque, according to the above figure, on the system about point A should be zero.

With reference to the above figure, the net anti-clockwise torque about A is

τa=F2L ...1

And, the net clockwise torque about the same point is

τc=Wcosθ×L2 ...2

Equations (1) and (2) imply that

Wcosθ×L2=F2L⇒12×10×cos60°×L2=F2 L⇒F2=120×10×cos60°2=30

Hence, the required weight can be expressed as

F2=3 kg-wt

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2024 (27 Jan Shift 2)
A heavy iron bar, of weight $W$ is having its one end on the ground and the other on the shoulder of a person. The bar makes an angle $\theta$ with the horizontal. The weight experienced by the person is :

Options:
A $\mathrm{W} \cos \theta$
B $\frac{W}{2}$
C W
D $W \sin \theta$

The correct answer is:
$\frac{W}{2}$

Solution:
$R=$ net reaction force by shoulder
Balancing torque about pt of contact on ground:
$\begin{aligned} & \mathrm{W}\left(\frac{\mathrm{L}}{2} \cos \theta\right)=\mathrm{R}(\mathrm{L} \cos \theta) \\ & \Rightarrow \mathrm{R}=\frac{\mathrm{W}}{2}\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2024 (09 Apr Shift 1)
A heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio $3: 1$. The ratio of radii of the fragments is

Options:
A $1: 3^{1 / 3}$
B $3^{1 / 3}: 4$
C $4: 1$
D $2: 1$

The correct answer is:
$1: 3^{1 / 3}$

Solution:
From law of conservation of momentum
$\begin{aligned} m v=\text { constant } & \\ V \rho v & =k \\ \frac{4}{3} \pi R^3 \rho v & =k\end{aligned}$
$\left(\frac{R_1}{R_2}\right)^3=\frac{v_2}{v_1} \Rightarrow \frac{R_1}{R_2}=\left(\frac{v_2}{v_1}\right)^{1 / 3}=\left(\frac{1}{3}\right)^{1 / 3}=1: 3^{1 / 3}$

Topic: Physics, Nuclear Physics, JEE Main
A heavy nucleus having mass number 200 gets disintegrated into smaller fragments of mass number 80 and 120. If binding energy per nucleon for parent atom is 6.5 eV and for daughter nuclei are 7 eV and 8 eV respectively and the energy released in the decay is given by XÃ105 eV, then X will be

Options:

The correct answer is:
2200

Solution:
200X â 80Y + 120Z
6.5 7 8
Energy released = 80 Ã 7 + 120 Ã 8 â 200 Ã 6.5
= 220 MeV
= 2200 Ã 105 eV

Topic: Physics, Nuclear Physics, JEE Main
A heavy plank of mass 100 kg hangs on three vertical wires of equal length arranged symmetrically (see figure). All the wires have the same cross-section. The middle wire is of steel and the other two are of copper. The modulus of elasticity of steel is assumed to be double that of copper. Determine the tensions in the wire. ( use g=9.8 m s-2 )

Options:
A Tcu = 245 N, Tsteel = 490 N
B Tcu = 235 N, Tsteel = 490 N
C Tcu = 205 N, Tsteel = 490 N
D Tcu = 295 N, Tsteel = 420 N

The correct answer is:
Tcu = 245 N, Tsteel = 490 N

Solution:
Let tension in copper and steel wires are Tcu and Tst. Assuming same strain (due to heavy plank) we note that
Ysteel=Tâ¡stLâ¡Aâ¡ÎLâ¡ and
or Tâ¡stTâ¡cu=Yâ¡steelYâ¡cu=2givenor Tâ¡st=2Tâ¡cu
Further, Tst + 2Tcu = 100 kg (given)

or 2Tcu + 2Tcu = 100 kg or Tcu = 25 kg
And therefore, Tst = 2Tcu = 50 kg
Thus Tcu is Tcu=25 kgf=245 N

and Tst=50 kgf=490 N

Topic: Physics, Mechanical Properties of Solids, JEE Main
A helicopter is flying along the curve given by y-x32=7, x≥0. A soldier positioned at the point 12, 7, who wants to shoot down the helicopter when it is nearest to him. Then this nearest distance is:

Options:
A 1673
B 12
C 1373
D 56

The correct answer is:
1673

Solution:
We have

y=7+x32

⇒dydx=32x
Let the point on curve be Px1, 7+x132 and given point be A12, 7.
For nearest point normal at P passes through A.
So, the slope of line AP= Slope of normal at P
⇒x132x1-12=-dxdyx1,y1 =-23x1
⇒3x12=1-2x1
⇒3x12+2x1-1=0
⇒x1+13x1-1=0
⇒x1=13 (x1=-1, is not possible as x1>0)
Hence, point P is 13, 7+133.
So, AP=136+127=1673 units.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2019 (10 Jan Shift 2)
A helicopter is flying horizontally with a speed v at an altitude h has to drop a food packet for a man on the ground. What is the distance of helicopter from the man when the food packet is dropped ?

Options:
A 2ghv2+1 h2
B 2ghv2+h2
C 2ghv2+h2
D 2v2 h g+h2

The correct answer is:
2v2 h g+h2

Solution:
The aeroplane imparts v velocity to the food packet horizontally when dropping. Horizontal range of food packet =v×2hg
Also, Horizontal distance, travelled by aeroplane =v×2hg thus, vertical distance between aeroplane and person =h
distance =2v2 h g+h2

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (31 Aug Shift 1)
A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height h. The time taken by the packet to reach the ground is close to [g is the acceleration due to gravity]:

Options:
A t=23hg
B t=1.8hg
C t=3.4hg
D t=2h3g

The correct answer is:
t=3.4hg

Solution:
⇒For upward motion of helicopter

v2=u2+2as

v2=0+2gh

v=2gh

⇒ Now particle will start moving under gravity.

s=ut+12at2

-h=2ght-12gt2

then t=2gh±2gh+4×g2×h2×g2

t=2ghg(1+2)

t=2hg(1+2)

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2020 (05 Sep Shift 1)
A helicopter takes off along the vertical with an acceleration a=3 m s-2 and zero initial velocity. In a certain time t1, the pilot switches off the engine. At the point of take-off, the sound dies away in time t2=30 s. Determine the velocity v of the helicopter (in m s-1) at the moment when its engine is switched off, assuming that the velocity c of sound is 320 m s-1.

Options:

The correct answer is:
80

Solution:
t2=t1+at122c
t1=ca2+2cat2-ca
v=at1=aca2+2ct2a-ca
=c2+2act2-c=80 m s-1

Topic: Physics, Motion In One Dimension, JEE Main
A Helmholtz coil has a pair of loops, each with N turns and radius R . They are placed coaxially at distance R and the same current I flows through the loops in the same direction. The magnitude of the magnetic field at P, midway between the centres A and C, is given by [Refer to figure given below]:

Options:
A Â 8NÎ¼0I51/2R
B 4NÎ¼0I53/2R
C Â 4NÎ¼0I51/2R
D 8NÎ¼0I53/2R

The correct answer is:
8NÎ¼0I53/2R

Solution:
Magnetic field at point P, due to both coils, Bp=2μ0NIR22R2+R243/2 = μ 0 NI R 2 5 3/2 8 R 3 = 8Nμ0I53/2R.

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2018 (15 Apr)
A Helmholtz coil has pair of loops, each with $N$ turns and radius $R$. They are placed coaxially at distance $R$ and the same current $I$ flows through the loops in the same direction. The magnitude of magnetic field at $P$, midway between the centres $A$ and $C$, is given by (Refer to figure):

Options:
A
$\frac{4 N \mu_0 I}{5^{3 / 2} R}$

B
$\frac{8 N \mu_0 I}{5^{3 / 2} R}$

C
$\frac{4 N \mu_0 I}{5^{1 / 2} R}$

D
$\frac{8 N \mu_0 I}{5^{1 / 2} R}$

The correct answer is:
$\frac{8 N \mu_0 I}{5^{3 / 2} R}$

Solution:
Point $P$ is situated at the mid-point of the line joining the centres of the circular wires which have same radii (R). The magnetic fields $(\vec{B})$ at $\mathrm{P}$ due to the currents in the wires are in same direction.
Magnitude of magnetic field at point, $P$
$B=2\left\{\frac{\mu_0 N I R^2}{2\left(R^2+\frac{R^2}{4}\right)^{3 / 2}}\right\}$
$=\frac{\mu_0 N I R^2}{\frac{5^{3 / 2}}{8}}=\frac{8 \mu_0 N I}{5^{3 / 2} R}$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
A hemispherical bowl just floats without sinking in a liquid of density $1.2 \times 10^3 \mathrm{~kg} / \mathrm{m}^3$. If outer diameter and the density of the bowl are $1 \mathrm{~m}$ and $2 \times 10^4 \mathrm{~kg} / \mathrm{m}^{\mathrm{3}}$ respectively, then the inner diameter of the bowl will be

Options:
A $0.94 \mathrm{~m}$
B $0.97 \mathrm{~m}$
C $0.98 \mathrm{~m}$
D $0.99 \mathrm{~m}$

The correct answer is:
$0.98 \mathrm{~m}$

Solution:
Weight of the bowl $=m g$
$=V \rho g=\frac{4}{3} \pi\left[\left(\frac{D}{2}\right)^3-\left(\frac{d}{2}\right)^3\right] \rho g$
where $D=$ Outer diameter.
$d=$ Inner diameter
$\rho=$ Density of bowl
Weight of the liquid displaced by the bowl
$=V \sigma g=\frac{4}{3} \pi\left(\frac{D}{2}\right)^3 \sigma g$
where $\sigma$ is the density of the liquid.
For the flotation
$\frac{4}{3} \pi\left(\frac{D}{2}\right)^3 \sigma g=\frac{4}{3} \pi\left[\left(\frac{D}{2}\right)^3-\left(\frac{d}{2}\right)^3\right] \rho g$
$\Rightarrow\left(\frac{1}{2}\right)^3 \times 1.2 \times 10^3=\left[\left(\frac{1}{2}\right)^3-\left(\frac{d}{2}\right)^3\right] 2 \times 10^4$
By solving we get $d=0.98 \mathrm{~m}$.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A hemispherical bowl just floats without sinking in a liquid of density 1.2Ã103 kg m-3. If the outer diameter and the density of the bowl are 1 m and 2Ã104 kg m-3 respectively, then the inner diameter of the bowl will be

Options:
A 0.88 m
B 0.78 m
C 0.98 m
D 0.68 m

The correct answer is:
0.98 m

Solution:
Weight of the bowl=mg=Vρg=23π[(D2)3-(d2)3]ρg, where D is the outer diameter, d is the inner diameter and ρ is the density of bowl.
Weight of the liquid displaced by the bowl=V'σg=23πD23σg, where σ is the density of the liquid.
For the flotation,

23π(D2)3σg=23π[(D2)3-(d2)3]ρg
By solving we get d=0.98 m

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A hemispherical cavity of radius R is created in a solid sphere of radius 2R as shown in the figure. The y-coordinate of the centre of mass of the remaining sphere is

Options:
A yCM=-R15
B y CM = - R 4 0
C yCM=-R30
D yCM=-R20

The correct answer is:
y CM = - R 4 0

Solution:
x CM = 4 3 π 2 R 3 0 - 2 3 π R 3 R 4 3 π 2 R 3 - 2 3 π R 3 = - R 1 5
y CM = 4 3 π 2 R 3 0 - 2 3 π R 3 3 R 8 4 3 π 2 R 3 - 2 3 π R 3 = - R 4 0
z CM = 0

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A hemispherical glass body of radius 10 cm and refractive index 1.5 is silvered on its curved surface. A small air bubble is 6 cm below the flat surface inside it along the axis. The position of the image of the air bubble made by the mirror is seen :

Options:
A 14cm below flat surface
B 20cm below flat surface
C 16cm below flat surface
D 30cm below flat surface

The correct answer is:
20cm below flat surface

Solution:
Mirror image
1v+1u=1f⇒1v+1-4=1-5
1v=120
v=20
So, it is 30 cm below flat surface.
Now, After refraction
v'=30μ=301.5=20 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2016 (10 Apr Online)
A high density of elephant population in an area can result in:

Options:
A intra specific competition
B inter specific competition
C predation on one another
D mutualism.

The correct answer is:
intra specific competition

Solution:
The intraspecific competition is competition between individuals of same species. The intraspecific competition may be very severe becauseall the members of a species have similar requirements of food, habitat mate, etc and they also have similar adaptations to gettheir needs. An interaction between individuals of different species wherein both are negatively affected is interspecific competition. An interaction between two species wherein both species involved are benefited is Mutualism. Predation is the act of preying on other animals is termed predation. The animal which hunts and kills other animals for food, is called the predator and the attacked animal is called the prey.
Related Theory
The principle of competitive exclusion was proposed by Gause states that two species competing for the same resources cannot coexist. One population will drive off the other one. If any of the members of the depleted population remains, that would be because they have adapted themselves according to the different niche.

Topic: Biology, Organisms and Populations, NEET, NEET 2007
A highly rigid cubical block A of small mass M and side L is fixed rigidly on to another cubical block B of the same dimensions and of low modulus of rigidity Î· such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side faces of A. After the force is withdrawn, block A executes small oscillations, the time period of which is given by

Options:
A 2πMηL
B 2πMηL
C 2πMLη
D 2πMηL

The correct answer is:
2πMηL

Solution:
Modulus of rigidity, Î·= F A.âÎ¸ Here, A= L 2
And
Î¸ = x L

Therefore, restoring force is F=âÎ·âAÎ¸=âÎ·Lx

Or acceleration, a= F M =â Î·L M x
Since, aâx , oscillations are simaple harmonic in nature, time period of which is given by
T=2Ï | displacement acceleration | =2Ï | x a | =2Ï M Î·L

Topic: Physics, Oscillations, JEE Main
A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m s-1 over the hill. The cannon is located at a distance of 800 m from the foot of the hill and can be moved on the ground at a speed of 2 m s-1 so that its distance from the hill can be adjusted. What is the shortest time (in seconds) in which a packet can reach on the ground across the hill? (Take g=10 m s-2)

Options:

The correct answer is:
45

Solution:
Given, height of the hill (h) = 500 m
u = 125 m/s
To cross the hill, the vertical component of the velocity should be sufficient to cross such height.
∴ u y ≥ 2 gh
≥ 2 × 1 0 × 5 0 0
≥ 1 0 0 m/s
But u 2 = u x 2 + u y 2
∴ Horizontal component of initial velocity
u x = u 2 - u y 2
= 1 2 5 2 - 1 0 0 2
= 7 5 m / s
Time taken to reach the top of the hill
t = 2 h g = 2 × 5 0 0 1 0 = 1 0 s
Time taken to reach the ground from the top of the hill t' = t = 10 s
Horizontal distance travelled in 10 s
x = u x × t
= 7 5 × 1 0
= 7 5 0 m
∴ Distance through which canon has to be moved
= 800 - 750
= 50 m
Speed with which canon can move = 2 m/s
∴ Time taken by canon = 5 0 2
t ″ = 2 5 s
∴ Total time taken by a packet to reach on the ground
= t ″ + t + t ′
= 2 5 + 1 0 + 1 0
= 4 5 s .

Topic: Physics, Motion In Two Dimensions, JEE Main
(A) HOCl+H2O2→H3O++Cl-+O2

(B) I2+H2O2+2OH-→2I-+2H2O+O2

Choose the correct option.

Options:
A H2O2 act as oxidizing and reducing agent respectively in equations (A) and (B).
B H2O2 acts as oxidising agent in equations (A) and (B).
C H2O2 acts as reducing and oxidising agent respectively in equations (A) and (B).
D H2O2 acts as reducing agent in equations (A) and (B).

The correct answer is:
H2O2 acts as reducing agent in equations (A) and (B).

Solution:
In (A) reduction of HOCl occurs, so it will be an oxidising agent hence H2O2 will be a reducing agent.

In (B) reduction of I2 occurs, so it will be an oxidising agent and H2O2 will be a reducing agent.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (24 Feb Shift 1)
A hollow charged metal sphere has radius r. If the potential difference between its surface and a point at distance 3r from the centre is V, then the electric field intensity at distance 3r from the centre is

Options:
A V6r
B V4r
C V3r
D V2r

The correct answer is:
V6r

Solution:
VA-VB=kQr-kQ3r=V=2kQ3r
E =kQ3r2=kQ9r2âE=V6r

Topic: Physics, Electrostatics, JEE Main
A hollow conducting sphere is placed in an electric field produced by a point charge placed at $P$ as shown in figure. Let $V_A, V_B, V_C$ be the potentials at points $A, B$ and $C$ respectively. Then

Options:
A $V_C\gtV_B$
B $V_B\gtV_C$
C $V_A\gtV_B$
D $V_A=V_C$

The correct answer is:
$V_A=V_C$

Solution:
At each point on the surface of a conducting sphere, the potential is equal
So, \(V_A=V_B=V_C\)

Topic: Physics, Electrostatics, JEE Main
A hollow cylinder has a charge q placed at its centre. If Ï is the electric flux (in V m) associated with the curved surface B, the flux linked with the plane surface A is

Options:
A 12qε0-ϕ
B q2ε0
C ϕ3
D qε0-ϕ

The correct answer is:
12qε0-ϕ

Solution:
According to Gauss law, ϕtotal=qε0
Let electric flux linked with surfaces A, B and C are ϕA,ϕB and ϕC respectively. That is
ϕtotal=ϕA+ϕB+ϕC
Since ϕC=ϕA
∴2ϕA+ϕB=ϕtotal=qε0
or ϕA=12qε0-ϕB
But ϕB=ϕ given
Hence,
ϕA=12qε0-ϕ

Topic: Physics, Electrostatics, JEE Main
A hollow cylinder of mass m made heavy at its bottom is floating vertically in water. It is tilted from its vertical position through an angle θ and is left. The restoring force acting on it is

Options:
A mgcos⁡θ
B mgsin⁡θ
C mg1cos⁡θ-1
D mg1cos⁡θ+1

The correct answer is:
mg1cos⁡θ-1

Solution:
Let l be the length of the cylinder in water it is in the vertical position and A be the cross-sectional area of the cylinder. As cylinder is floating so
Weight of cylinder = upward thrustmg=A l ρ g or m=Alρ
When the cylinder is tilled through an angle θ, length of cylinder in water =lcos⁡θ
Weight of water displaced =lcos⁡θA ρ g
Restoring force =l A ρ gcos⁡θ=l A ρ g
=l A ρ g1cos⁡θ-1=mg1cos⁡θ-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A hollow cylindrical conductor has length of 3.14 m, while its inner and outer diameters are 4 mm and 8 mm respectively. The resistance of the conductor is n×10-3 Ω.
If the resistivity of the material is 2.4×10-8 Ω m. The value of n is _____.

Options:

The correct answer is:
2

Solution:
Resistance of a conductor with uniform cross-section is given by,

R=ρlA

In this case the cross-sectional area is πb2-a2.

Therefore,

R=ρlπb2-a2=2.4×10-8×3.143.14×42-22×10-6=2×10-3 Ω

Hence, n=2.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2023 (24 Jan Shift 1)
A hollow metal sphere of radius $5 \mathrm{~cm}$ is charged so that the potential on its surface is $10 \mathrm{~V}$. The potential at the centre of the sphere is

Options:
A $0 \mathrm{~V}$
B $10 \mathrm{~V}$
C Same as at point $5 \mathrm{~cm}$ away from the surface
D Same as at point $25 \mathrm{~cm}$ away from the surface

The correct answer is:
$10 \mathrm{~V}$

Solution:
Since potential inside the hollow sphere is same as that on the surface.

Topic: Physics, Electrostatics, JEE Main
A hollow pipe of length $0.8 \mathrm{~m}$ is closed at one end. At its open end a $0.5 \mathrm{~m}$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of the pipe. If the tension in the wire is $50 \mathrm{~N}$ and the speed of sound is $320 \mathrm{~ms}^{-1}$, the mass of the string is

Options:
A
$5 \mathrm{~g}$

B
$10 \mathrm{~g}$

C
$20 \mathrm{~g}$

D
$40 \mathrm{~g}$

The correct answer is:
$10 \mathrm{~g}$

Solution:
$2\left(\frac{v_1}{2 l_1}\right)=\frac{v_2}{4 l_2}$ $\therefore \quad \frac{\sqrt{T / \mu}}{l_1}=\frac{320}{4 l_2}$
( $\mu=$ mass per unit length of wire)
or $\frac{\sqrt{50 / \mu}}{0.5}=\frac{320}{4 \times 0.8}$
Solving we get $\mu=0.02 \frac{\mathrm{kg}}{\mathrm{m}}=20 \frac{\mathrm{g}}{\mathrm{m}}$
$\therefore$ Mass of string
$$
=\left(20 \frac{\mathrm{g}}{\mathrm{m}}\right)(0.5 \mathrm{~m})=10 \mathrm{~g}
$$
$\therefore$ The correct option is (b).

Topic: Physics, Waves and Sound, JEE Main
A hollow sphere has a radius of $6.4 \mathrm{~m}$. Minimum velocity required by a motorcyclist at the bottom to complete the circle will be

Options:
A $17.7 \mathrm{~m} / \mathrm{s}$
B $10.2 \mathrm{~m} / \mathrm{s}$
C $12.4 \mathrm{~m} / \mathrm{s}$
D $16.0 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$17.7 \mathrm{~m} / \mathrm{s}$

Solution:
Minimum velocity required by a
motorcyclist at the bottom to complete
the circle, $\mathrm{v}_{\min }=\sqrt{5 r g}$
$=\sqrt{5 \times 6.4 \times 9.8}$
$=17.7 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Laws of Motion, JEE Main
A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is $\frac{x}{5}$. The value of $x$ is _____ .

Options:

The correct answer is:
2

Solution:
$\begin{aligned} & \frac{\frac{1}{2} \mathrm{I} \omega^2}{\frac{1}{2} \mathrm{I} \omega^2+\frac{1}{2} \mathrm{mv}^2}=\frac{\left(\frac{1}{2}\right)\left(\frac{2}{3} \mathrm{mR}^2\right) \omega^2}{\left(\frac{1}{2}\right)\left(\frac{2}{3} \mathrm{mR}^2\right) \omega^2+\frac{1}{2} \mathrm{~m}(\mathrm{R} \omega)^2} \\ & =\frac{\frac{2}{3}}{\frac{2}{3}+1}=\frac{2}{5} \\ & \mathrm{x}=2\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2024 (05 Apr Shift 2)
A hollow sphere of mass $2 \mathrm{~kg}$ is kept on a rough horizontal surface. A force of $10 \mathrm{~N}$ is applied at the centre of the sphere as shown in the figure. Find the minimum value $\mu$ so that the sphere starts pure rolling. (Take $\left.g=10 \mathrm{~m} / \mathrm{s}^2\right)$

Options:
A $\sqrt{3} \times 0.16$
B $\sqrt{3} \times 0.08$
C $\sqrt{3} \times 0.1$
D Data insufficient

The correct answer is:
$\sqrt{3} \times 0.08$

Solution:
From given figure we get
$\begin{aligned}& N=F \sin \theta+m g \rightarrow(1), F \cos \theta-f=m g \rightarrow(2) \\& f R=I \alpha \rightarrow(3)\end{aligned}$
Solve above equations we can easily get
$\mu=0.08 \times \sqrt{3}$

Topic: Physics, Rotational Motion, JEE Main
A hollow sphere of mass M and radius R is immersed in a tank of water (density Ï w ). The sphere would float if it were set free. The sphere is tied to the bottom of the tank by two wires which makes angle 45Â° with the horizontal as shown in the diagram. The tension T1 in the wire is

Options:
A 43ÏR3Ïwg-Mg2
B 23πR3ρwg-Mg
C 43πR3ρwg-Mg2
D 43πR3ρwg+Mg

The correct answer is:
43πR3ρwg-Mg2

Solution:
â´ 2 T1+Mg=43ÏR3Ïwg
â´ T1= 43ÏR3Ïwg-Mg2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A hollow sphere of volume $V$ is floating on water surface with half immersed in it. What should be the minimum volume of water poured inside the sphere so that the sphere now sinks into the water

Options:
A $V / 2$
B $V / 3$
C $V / 4$
D $V$

The correct answer is:
$V / 2$

Solution:
When body (sphere) is half immersed, then upthrust = weight of sphere
$\Rightarrow \frac{V}{2} \times \rho_{\text {liq }} \times g=V \times \rho \times g \quad \therefore=\frac{\rho_{\text {liq }}}{2}$
When body (sphere) is fully immersed then, Upthrust = wt. of sphere + wt. of water poured in sphere
$\Rightarrow V \times \rho_{\text {liq }} \times g=V \times \rho \times g+V \times \rho_{\text {liq }} \times g$
$\Rightarrow V \times \rho_{\text {liq }}=\frac{V \times \rho_{\text {liq }}}{2}+V \times \rho_{\text {liq }} \Rightarrow V^{\prime}=\frac{V}{2}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A hollow spherical ball of uniform density rolls up a curved surface with an initial velocity 3 m s-1 (as shown in figure). Maximum height with respect to the initial position covered by it will be _____ cm
(take, g=10 m s-2)

Options:

The correct answer is:
75

Solution:
The total kinetic energy of the ball at the bottom of the curve can be calculated as follows:

K=12Mv2+12·23MR2ω2= 12Mv2+13MRω2= 12Mv2+13Mv2 ∵v=ωR= 56Mv2 ...1

The formula to calculate the potential energy acquired by the ball when it attains the maximum height is given by

U=Mgh ...2

Equate equation (1) and (2) and simplify to obtain the required height.

Mgh=56Mv2⇒h=5v26g ...3

Substitute the values of the known parameters into equation (3) to calculate the required height attained by the ball.
h=5×3 m s-126×10 m s-2= 0.75 m×100 cm1 m=75 cm

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2023 (08 Apr Shift 2)
A hollow spherical shell at outer radius R floats just submerged under the water surface. The inner radius of the shell is r . If the specific gravity of the shell material is 278 with respect to water, the value of r is:

Options:
A 89R
B 49R
C 23R
D 13R

The correct answer is:
89R

Solution:
In equilibrium mg=FB

43πR3-r3ρ0g=43πR3ρwg

1-rR3278ρw=ρw

1-r3R3=927

1-13=r3R3

23=r3R3

rR=231/3

1-r3R3=827

r3R3=1-827=1927

r=0.89

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2020 (05 Sep Shift 1)