datavorous/entrance-exam-dataset huggingface dataset:
A cord of length h=64 m is used to connect an astronaut (m=100 kg) to a spaceship (Mâ«m) which is orbiting near the earth's surface as shown in the figure. Assuming that the spaceship and the astronaut fall on a straight line from the earth's centre, then the tension in the cord is close to [R=6400 km]

Options:
A 3 Ã 1 0 - 2 N
B 2 × 1 0 - 2 N
C 4 × 1 0 - 2 N
D 5 Ã 1 0 -2 N

The correct answer is:
3 Ã 1 0 - 2 N

Solution:
According to given problem, the mass of satellite M is much greater than that of astronaut m. So the centre of mass of the system will be close to satellite and as the satellite is orbiting close to the surface of earth, the equation of motion of the system (S+A) will be-
GMe⁡M⁡+m⁡R⁡2=M⁡+m⁡R⁡ω2
i.e., R⁡ω2=GMe⁡R⁡2=g…i
And the equation of motion of the astronaut will be
GMe⁡m⁡r⁡2+T⁡=m⁡r⁡ω2,i.e.,T⁡=m⁡r⁡ω2-GMe⁡r⁡2
In the light of equation (i), above equation reduces to
T⁡=mgr⁡R⁡-R⁡r⁡2=mgR⁡+h⁡R⁡-R⁡2R⁡+h⁡2
asr⁡=R⁡+h⁡
or T⁡=mg1+h⁡R⁡-1+h⁡R⁡-2=3mghR⁡
as1+h⁡R⁡-2⋍1-2h⁡R⁡
So substituting the given data,
T⁡=3×100×10×646400×103=3×10-2N

Topic: Physics, Gravitation, JEE Main
A cork is submerged in water by a spring attached to the bottom of a pail. When the pail is kept in a elevator moving with an acceleration downwards, the spring length

Options:
A Increases
B Decreases
C Remains unchanged
D Data insufficient

The correct answer is:
Decreases

Solution:
Density of cork $=d$, Density of water $=\rho$
Resultant upward force on cork $=V(\rho-d) g$
This causes elongation in the spring. When the lift moves down with acceleration $a$, the resultant upward force
on cork $=V(\rho-d)(g-a)$ which is less than the previous value. So the elongation decreases.

Topic: Physics, Laws of Motion, JEE Main
A country with a high rate of population growth took measures to reduce it. The figure below shows age-sex pyramids of populations $A$ and $B$ twenty years apart. Select the correct interpretation about them :-

Options:
A "A" is the earlier pyramid and no change has occurred in the growth rate
B " $\mathrm{A}$ " is more recent shows slight reduction in the growth rate
C "B" is earlier pyramid and shows stabilized growth rate
D " $\mathrm{B}$ " is more recent showing that population is very young

The correct answer is:
" $\mathrm{A}$ " is more recent shows slight reduction in the growth rate

Solution:
No Solution Available

Topic: Biology, Organisms and Populations, NEET, NEET 2009 (Mains)
A couple is of moment $\overrightarrow{\mathrm{G}}$ and the force forming the couple is $\overrightarrow{\mathrm{P}}$. If $\overrightarrow{\mathrm{P}}$ is turned through a right angle the moment of the couple thus formed is $\overrightarrow{\mathrm{H}}$. If instead, the force $\overrightarrow{\mathrm{P}}$ are turned through an angle $\alpha$, then the moment of couple becomes

Options:
A
$\overrightarrow{\mathrm{H}} \sin \alpha-\overrightarrow{\mathrm{G}} \cos \alpha$

B
$\overrightarrow{\mathrm{G}} \sin \alpha-\overrightarrow{\mathrm{H}} \cos \alpha$

C
$\overrightarrow{\mathrm{H}} \sin \alpha+\overrightarrow{\mathrm{G}} \cos \alpha$

D
$\overrightarrow{\mathrm{G}} \sin \alpha+\overrightarrow{\mathrm{H}} \cos \alpha$

The correct answer is:
$\overrightarrow{\mathrm{G}} \sin \alpha-\overrightarrow{\mathrm{H}} \cos \alpha$

Solution:
$\overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{p}} ;|\overrightarrow{\mathrm{a}}|=\mathrm{rp} \sin \theta$
$|\overrightarrow{\mathrm{H}}|=\operatorname{rpcos} \theta \quad\left[\because \sin \left(90^{\circ}+\theta\right)=\cos \theta\right\rfloor$
$\mathrm{G}=\mathrm{rp} \sin \theta$
$\mathrm{H}=\mathrm{rp} \cos \theta$
$\mathrm{x}=\mathrm{rp} \sin (\theta+\alpha)$
From (1), (2) \& (3),
$\mathrm{x}=\overrightarrow{\mathrm{a}} \cos \alpha+\overrightarrow{\mathrm{H}} \sin \alpha$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2003
A cricket ball of mass 0.15 kg is thrown vertically up by a bowling machine so that it rises to a maximum height of 20 m after leaving the machine. If the part pushing the ball applies a constant force F on the ball applies a constant force F on the ball and moves horizontally a distance of 0.2 m while launching the ball, the value of F(in N) is g=10 m s-2

Options:

The correct answer is:
150

Solution:
From work energy theorem F0.2-mg20=0

F=mg200.2

=mg2002

=0.15×10×2002

=150.00 N

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2020 (03 Sep Shift 1)
A cricket ball of mass $150 \mathrm{~g}$ moving with a speed of $126 \mathrm{~km} /$ $\mathrm{h}$ hits at the middle of the bat, held firmly at its position by the batsman. The ball moves straight back to the bowler after hitting the bat. Assuming that collision between ball and bat is completely elastic and the two remain in contact for $0.001 \mathrm{~s}$, the force that the batsman had to apply to hold the bat firmly at its place would be

Options:
A $10.5 \mathrm{~N}$

B $21 \mathrm{~N}$

C $1.05 \times 10^4 \mathrm{~N}$

D $2.1 \times 10^4 \mathrm{~N}$

The correct answer is:
$1.05 \times 10^4 \mathrm{~N}$

Solution:
As given that,
$$
\begin{aligned}
&m=150 \mathrm{~g}=\frac{150}{1000} \mathrm{~kg}=0.15 \mathrm{~kg} \\
&\Delta t=\text { time of contact }=0.001 \mathrm{~s} \\
&u=126 \mathrm{~km} / \mathrm{h}=\frac{126 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{s} \\
&=126 \times \frac{5}{18}=35 \mathrm{~m} / \mathrm{s} \\
&v=-126 \mathrm{~km} / \mathrm{h}=-126 \times \frac{5}{18}=-35 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
So, final velocity is acc. to initial force applied by batsman.
So, change in momentum of the ball
$$
\Delta p=m(v-u)=\frac{3}{20}(-35-35) \mathrm{kg}-\mathrm{m} / \mathrm{s}
$$
$$
=\frac{3}{20}(-70)=-\frac{21}{2}
$$
As we know that, force
$$
F=\frac{\Delta p}{\Delta t}=\frac{-21 / 2}{0.001} \mathrm{~N}=-1.05 \times 10^4 \mathrm{~N}
$$
Hence negative sign shown that direction of force will be opposite to initial velocity which taken positive direction. Hence verify the option (c).

Topic: Physics, Work Power Energy
A cricket ball thrown across a field is at heights $h_{1}$ and $h_{2}$ from the point of projection at times $t_{1}$ and $t_{2}$ respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is

Options:
A $\frac{h_{1} t_{2}^{2}-h_{2} t_{1}^{2}}{h_{1} t_{2}-h_{2} t_{1}}$
B $\frac{h_{1} t_{1}^{2}+h_{2} t_{2}^{2}}{h_{2} t_{1}+h_{1} t_{2}}$
C $\frac{h_{1} t_{2}^{2}+h_{2} t_{1}^{2}}{h_{1} t_{2}+h_{2} t_{1}}$
D $\frac{h t_{1}^{2}-h t_{2}^{2}}{h_{1}-h_{2} t_{2}}$

The correct answer is:
$\frac{h_{1} t_{2}^{2}-h_{2} t_{1}^{2}}{h_{1} t_{2}-h_{2} t_{1}}$

Solution:
For vertically moment (for $\left.h_{1}\right)$
or
$\begin{array}{l}
h_{1}=u \sin \theta t_{1}-\frac{1}{2} g t_{1}^{2} \\
t_{1}=\frac{h_{1}+\frac{1}{2} g t_{1}^{2}}{u \sin \theta}
\end{array}$
$h_{2}=u \sin \theta t_{2}-\frac{1}{2} g t_{2}^{2}$
or $t_{2}=\frac{h_{2}+\frac{1}{2} g t_{2}^{2}}{u \sin \theta}$
Divide Eq. (i) by Eq. (ia)
$\begin{array}{l}
\frac{t_{1}}{t_{2}}=\frac{h_{1}+\frac{1}{2} g t_{1}^{2} / u \sin \theta}{h_{2}+\frac{1}{2} g t_{2}^{2} / u \sin \theta} \\
\Rightarrow h_{1} t_{2}-h_{2} t_{1}=\frac{g}{2}\left(t_{1} t_{2}^{2}-t_{1}^{2} t_{2}\right)
\end{array}$
The time of flight of the ball $T=\frac{2 u \sin \theta}{g}=\frac{2}{g}(u \sin \theta) \quad$
$=\frac{2}{g}\left[\frac{h_{1}+\frac{1}{2} g t_{1}^{2}}{t_{1}}\right]=\frac{2}{t_{1}}\left[\frac{h_{1}}{g}+\frac{t_{1}^{2}}{2}\right]$
$=\frac{h_{1}}{t_{1}} \times \frac{2}{g}+t_{1}=\frac{h_{1}}{t_{1}} \times\left(\frac{t_{1} t_{2}^{2}-t_{1}^{2} t_{2}}{h_{1} t_{2}-h_{2} t_{1}}\right)+t_{1}$
$=\frac{h_{1} t_{1} t_{2}^{2}-h_{1} t_{1}^{2} t_{2}+h_{1} t_{1}^{2} t_{2}-h_{2} t_{1}^{3}}{t_{1}\left(h_{1} t_{2}-h_{2} t_{1}\right)}$
$=\left(\frac{h_{1} t_{2}^{2}-h_{2} t_{1}^{2}}{h_{1} t_{2}-h_{2} t_{1}}\right)$

Topic: Physics, Motion In Two Dimensions, JEE Main
A cricket bat is cut at the location of its centre of mass as shown. Then

Options:
A the two pieces will have the same mass.
B the bottom piece will have larger mass.
C the handle piece will have larger mass.
D mass of handle piece is double the mass of bottom piece.

The correct answer is:
the bottom piece will have larger mass.

Solution:
We know that the centre of mass is given by,

xcm=m1x1+m2x2m1+m2

or for continuos body we have, xcm=∫xdm∫dm.

For a cricket bat we have the mass distributed over the greater distance on the left side of the bat than to the right side of the bat, so the r is greater for the left side, to compensate this larger value of distance the right side will have greater mass.
The centre of mass lies closest to the heavier mass.

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A cricket bowler releases the ball in two different ways
(a) giving it only horizontal velocity, and
(b) giving it horizontal velocity and a small downward velocity.
The speed $v_s$ at the time of release is the same. Both are released at a height $H$ from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.

Options:

None

Solution:
(a) When ball is given only horizontal velocity. Let us consider horizontal velocity at the time of release, $\left(u_x\right)=v_s$
During projectile motion, horizontal velocity remains unchanged.
So, $v_x=u_x=v_s$
In vertical direction,
$v_y^2=u_y^2+2 g H$
(if $u_y=0$ )
$v_y=\sqrt{2 g H}$
$\therefore$ Resultant speed of the
ball at ground,
$v=\sqrt{v_x^2+v_y^2}$
$=\sqrt{v_s^2+2 g H} \quad \quad \ldots(i)$
(b) When ball is given horizontal velocity and a small downward velocity.
Let us consider the ball be given a small downward velocity $u$.
In horizontal direction speed is
$$
v_x^{\prime}=u_x=v_s
$$
In vertical direction speed is
$$
v_y^{\prime 2}=u^2+2 g H
$$
So, $v_y^{\prime}=\sqrt{u^2+2 g H}$
$\therefore$ Resultatnt speed of the ball at the ground
$$
v^{\prime}=\sqrt{v_x^{\prime 2}+v_y^{\prime 2}}=\sqrt{v_s^2+u^2+2 g H}
$$
From eqs. (i) and (ii), we have $v^{\prime}>v$.

Topic: Physics, Laws of Motion
A cricket player catches a ball of mass 120 g moving with 25 m s-1 speed. If the catching process is completed in 0.1 s then the magnitude of force exerted by the ball on the hand of player will be(in SI unit):

Options:
A 24
B 12
C 25
D 30

The correct answer is:
30

Solution:
Average force exerted will be, Fav=ΔpΔt

=0-0.12×250.1=30 N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (01 Feb Shift 2)
A cross in which an organism showing a dominant phenotype is crossed with the receseive parent in order to know its genotype is called

Options:
A monohybrid cross
B back cross
C test cross
D dihybrid cross

The correct answer is:
test cross

Solution:
In Genetics, a test cross first introduced by Mendel is used to determine, if an individual exhibiting a dominant trait is homozygous or heterozygous for that trait. More simply, test cross determines the genotype of an individual with a dominant phenotype. In some sources, the test cross is defined as being a type of back cross between the recessive homozygote and $\mathrm{F}_1$-generation.

Topic: Biology, Principles of Inheritance and Variation, NEET, NEET 2010 (Mains)
A crystal diode that has an internal resistance of $20 \Omega$ is used tor rectification. It the supply voltage is $50 \sin \omega t$ and the load resistance is $800 \Omega$, then find the rms value of the load current.

Options:
A $61 \mathrm{~mA}$
B $30.5 \mathrm{~mA}$
C $19.4 \mathrm{~mA}$
D $38.8 \mathrm{~mA}$

The correct answer is:
$30.5 \mathrm{~mA}$

Solution:
Concept:
In a half-wave rectifier,
The average value of load current, $I_{a v g}=\frac{V_m}{\pi\left(R+R_L\right)}$
The RMS value of load current, $I_{r m s}=\frac{V_m}{2\left(R+R_L\right)}$
$V_m$ is the peak value of supply voltage
$R$ is the internal resistance
$R_{\mathrm{L}}$ is the load resistance
Calculation:
Given that, supply voltage $=50 \sin \omega t$
Peal voltage $\left(\mathrm{V}_{\mathrm{m}}\right)=50 \mathrm{~V}$
Internal resistance $(\mathrm{R})=20 \Omega$
Load resistance $\left(R_L\right)=800 \Omega$
The RMS value of load current, $I_{r m s}=\frac{50}{2(20+800)}=30.5 \mathrm{~mA}$

Topic: Physics, Semiconductors, JEE Main
A crystal is made up of metal ions 'M1' and 'M2' and oxide ions. Oxide ions form a ccp lattice structure. The cation 'M1' occupies 50% of octahedral voids and the cation 'M2' occupies 12.5 % of tetrahedral voids of oxide lattice. The oxidation numbers of 'M1' and 'M2' are respectively :

Options:
A +2, +4
B +1, +3
C +3, +1
D +4, +2

The correct answer is:
+2, +4

Solution:
In the ccp latice of oxide ions effective number of O-2ions =8×18+6×12=4

In the ccp latice,

No. of octahedral voids =4

No. of tetrahedrla voids =8

Given M1 atoms occupies 50% of octahedral voids and M2 atoms occupies 12.5 of tetrahederal voids

No. of M1 metal atoms =4×50100=2

No. of M2 metal atoms =8×12.5100=1

∴ Formula of the compound =M12M2O4

∴ Oxidation states of metals M1 & M2 respectively are +2 and +4

Topic: Chemistry, Solid State, JEE Main, JEE Main 2020 (06 Sep Shift 2)
A crystal made up of particles X, Y, and Z. X forms fcc packing. Y occupies all octahedral voids of X and Z occupies all tetrahedral voids of X. It all particles along one body diagonal are removed, then the formula of the crystal is

Options:
A XYZ2
B X2YZ2
C X8Y4Z5
D X5Y4Z8

The correct answer is:
X5Y4Z8

Solution:
For fcc, number of X atoms = 4/unit cell
Number of Tetrahedral Voids = Z = 8
Number of Octahedral Voids = Y = 4
Number of atoms removed along one body diagonal = 2X (corner) and 2Z (TVs) and 1 Y (OV at body centre)
â´ Number of X atoms left = 4 - 2 Ã 1 8 = 1 5 4
Number of Y atom left = 4 - (1 Ã 1) = 3
Number of Z atom left = 8 - (2 Ã 1) = 6
The simplest formula = X 1 5 4 Y 3 Z 6 â X 1 5 Y 1 2 Z 2 4
â X 5 Y 4 Z 8

Topic: Chemistry, Solid State, JEE Main
A crystalline hydrated salt on being heated loses 45.6% of its weight. Percentage composition of anhydrous salt is Al - 10.5 %, K - 15.1 %, S - 24.8 %, O - 49.6 %. Find molecular formula of salt (hydrated) (M = 948 g/mol for hydrated salt).

Options:
A K2SO4.Al2(SO4)3.24H2O
B K2SO4.Al2(SO4)3.12H2O
C K2SO4.Al2(SO4)3.6H2O
D K2SO4.Al4(SO4)6.48H2O

The correct answer is:
K2SO4.Al2(SO4)3.24H2O

Solution:
For anhydrous salt
Element Al K S O Atomic ratio 1 0 Â· 5 2 7 1 5 Â· 1 3 9 2 4 Â· 8 3 2 4 9 Â· 6 1 6 0 Â· 3 9 0 Â· 3 9 0 Â· 7 8 3 Â· 1 0
Empirical formula of anhydrous salt = KAlS2O8
E.F mass of Anhydrous salt = 258 g/mol
Mass of Moisture = 2 5 8 Ã 4 5 Â· 6 5 4 Â· 4 = 2 1 6 g
= 2 1 6 1 8 â 1 2 mole
Empirical formula of Anhydrous salt KAlS2O8.12H2O
E F Mass = 2 5 8 + 2 1 6 = 4 7 4 g/Empirical formula mass
n = Mole Mass E.F. Mass = 9 4 8 4 7 4 = 2
Molecular Formula = K2Al2S4O16.24H2O

= K2SO4.Al2(SO4)3.24H2O

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
A crystalline solid

Options:
A Changes abruptly from solid to liquid when heated
B Has no definite melting point
C Undergoes deformation of its geometry easily
D Has an irregular 3-dimensional arrangements

The correct answer is:
Changes abruptly from solid to liquid when heated

Solution:
In crystalline solid there is perfect arrangement of the constituent particles only at $0 K$. As the temperature increases the chance that a lattice site may be unoccupied by an ion increases. As the number of defects increases with temperature solid change in liquid.

Topic: Chemistry, Solid State, JEE Main
A crystalline solid is made of X, Y and Z. Atoms of X are present in ccp lattice; Y occupies all the octahedral voids and Z occupies all the tetrahedral voids. If all atoms along a body diagonal are removed then the formula of solid will be

Options:
A X5Y4Z8
B X8Y4Z5
C X2YZ2
D XYZ2

The correct answer is:
X5Y4Z8

Solution:
Number of X-atoms in unit cell = 4
Number of Y-atoms = 4
Number of Z atoms = 8
If atoms along one body diagonal are removed then 2 from corners X, 1 from body center Y and 2 neat two corners Z atoms will be absent.
Effective number of atoms may be calculated as
X=4-2×18=154
Y=4-1=3
Z=8-2=6
Simplex ratio =X:Y:Z
=154:3:6
=5:4:8
Simplest formula =X5Y4Z8 .

Topic: Chemistry, Solid State, JEE Main
A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is 8 g cm-3 , then the number of atoms present in 128 g of the crystal is

Options:
A 2×1024
B 4×1024
C 1×1024
D 3×1024

The correct answer is:
1×1024

Solution:
d=ZÃMa3ÃNAâM=a3ÃdÃNAZ
M=(400)3Ã10-30Ã8ÃNA4
On solving M=64Ã2Ã10-24 NAg
Now, 64Ã2Ã10-24NA g contains NA atoms
128 g will contain=NA64Ã2Ã10-24NAÃ128
=1Ã1024

Topic: Chemistry, Solid State, JEE Main
A crystalline solid of a pure substance has a face-centred cubic structure with a cell edge of 400 pm. If the density of the substance in the crystal is 8 g cm-3 , then the number of atoms present in 256 g of the crystal is N×1024 . The value of N is

Options:

The correct answer is:
2

Solution:
d=Z×M0NAa3
(d = density)
8=4×M06×1023(4×10-10)3
M0=18×6×1.6
Number of moles in 256 g =2568×6×1.6=103
Number of atoms =103×6×1023=2×1024

Topic: Chemistry, Solid State, JEE Main
A cube of aluminium of side 6 cm is subjected to a tangential force such that the top face is shears through 0.012 cm relative to the bottom face. The tangential force is k×1010 dyne. What is the value of k? [Shear modulus of the material is η=2×1011 dyne cm-2]

Options:

The correct answer is:
1.44

Solution:
Lateral displacement, x=0.012cm

â´ Shearing strain,

Î¸=xL=0.0126=0.002

Shearing stress, Fa=Î·Î¸

=2Ã1011Ã0.002

=4Ã108dynecm-2

Shearing force, F= shearing stress Ã area

=4Ã108Ã(L)2

=4Ã108Ã(6)2

=1.44Ã1010 dyne

âK=1.44

Topic: Physics, Mechanical Properties of Solids, JEE Main
A cube of aluminium of sides $0.1 \mathrm{~m}$ is subjected to a shearing force of $100 \mathrm{~N}$. The top face of the cube is displaced through $0.02 \mathrm{~cm}$ with respect to the bottom face. The shearing strain would be

Options:
A 0.02
B 0.1
C 0.005
D 0.002

The correct answer is:
0.002

Solution:
Shearing strain $\phi=\frac{x}{L}=\frac{0.02 \mathrm{~cm}}{10 \mathrm{~cm}} \quad \therefore \phi=0.002$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A cube of ice floats partly in water and partly in kerosene oil. The ratio of volume of ice immersed in water to that in kerosene oil (specific gravity of Kerosene oil $=0.8$, specific gravity of ice $=0.9$)

Options:
A $1: 1$
B $5: 4$
C $8: 9$
D $9: 10$

The correct answer is:
$1: 1$

Solution:
$\begin{aligned} & \mathrm{v}_1=\text { volume immersed in water. } \\ & \mathrm{v}_2=\text { volume immersed in oil. } \\ & \mathrm{v}_1 \rho_{\mathrm{w}} \mathrm{g}+\mathrm{v}_2 \rho_{\mathrm{o}} \mathrm{g}=\left(\mathrm{v}_1+\mathrm{v}_2\right) \rho_{\mathrm{c}} \mathrm{g} \\ & \mathrm{v}_1+\frac{\mathrm{v}_2 \rho_{\mathrm{o}}}{\rho_{\mathrm{w}}}=\left(\mathrm{v}_1+\mathrm{v}_2\right) \frac{\rho_{\mathrm{c}}}{\rho_{\mathrm{w}}} \\ & =\mathrm{v}_1+0.8 \mathrm{v}_2=0.9 \mathrm{v}_1+0.9 \mathrm{v}_2 \\ & =0.1 \mathrm{v}_1=0.1 \mathrm{v}_2 \\ & \mathrm{v}_1: \mathrm{v}_2=1: 1\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2024 (08 Apr Shift 2)
A cube of mass m is placed on top of a wedge of mass 2 m, as shown in figure. There is no friction between the cube and the wedge. The minimum co-efficient of friction between the wedge and the horizontal surface, so that the wedge does not move is

Options:
A 0.1
B 0.2
C 0.3
D 12

The correct answer is:
0.2

Solution:
The normal reaction between the wedge and the block is

N'=mgcosθ

So the net horizontal force on the wedge is
F=mgcos⁡θsin⁡θ
The normal reaction between the wedge and the ground is

N=2 mg+mgcos2⁡θ
N=2 mg+mg cos2θ
Now, F=μN gives
μ=FN=mgcos⁡θsin⁡θ2 mg+mgcos2⁡θ
=cos⁡θsin⁡θ2+cos2⁡θ
For θ=45o, μ=0.2

Topic: Physics, Laws of Motion, JEE Main
A cube of metal is subjected to a hydrostatic pressure 4 GPa. The percentage change in the length of the side of the cube is close to : (Given bulk modulus of metal, B=8×1010 Pa)

Options:
A 5
B 0.6
C 20
D 1.67

The correct answer is:
1.67

Solution:
B=-∆P∆VV
∆VV=-∆PB
Magnitude is given by,
∆VV=∆PB

Volume in terms of length can be represented as,

3×∆LL=∆PB

Therefore,

∆LL=13×∆PB

Substituting the values we get,
∆LL=13×4×1098×1010=160

Percantage change is,
160×100=53=1.67

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2020 (04 Sep Shift 2)
A cube of side $I$ is placed in a uniform field $\boldsymbol{E}$, where $\boldsymbol{E}=E \hat{\boldsymbol{i}}$. The net electric flux through the cube is

Options:
A Zero
B $I^2 E$
C $4 I^2 E$
D $6 I^2 E$

The correct answer is:
Zero

Solution:
As there is no charge residing inside the cube, hence net flux is zero

Topic: Physics, Electrostatics, JEE Main
A cube of side $L$ has point charges $+q$ located at its seven vertices and $-q$ at remaining one vertex. The electric field at its centre is found to be $|\mathbf{E}|=\alpha\left(\frac{q}{4 \pi \varepsilon_0 L^2}\right)$. The magnitude of constant $\alpha$ is

Options:
A $\frac{4}{3}$
B $\frac{8}{3}$
C 3
D 1

The correct answer is:
$\frac{8}{3}$

Solution:
The given situation is shown in the following figure
Clearly, $5 x$-components of $\mathbf{E}$ will be left to right, while $3 x$-components of $\mathbf{E}$ will be from right to left.
From the figure given below
$r^2=\frac{L^2}{4}+\frac{L^2}{2}=\frac{3 L^2}{4}$
So, $E$ due to one point charge
$|E|=\frac{K q}{r^2}=\frac{4}{3} \frac{K q}{L^2}$
Electric field at the centre of cube due to contribution of all charges is given as
$|E|_{\text {Total }}=\frac{4}{3} \frac{K q}{r^2}(5-3)=\frac{8}{3} \frac{K q}{r^2}=\frac{8}{3}\left(\frac{q}{4 \pi \varepsilon_0 L^2}\right)$
Given,$|E|=\alpha\left[\frac{q}{4 \pi \varepsilon_0 L^2}\right] \quad\left[\because r^2=\frac{3 L^2}{4}\right]$
Clearly,
$\alpha=\frac{8}{3}$

Topic: Physics, Electrostatics, TS EAMCET, TS EAMCET 2020 (11 Sep Shift 1)
A cubic solid is made up of two elements X and Y. Atoms of X are present on every alternate corner and one at the center of cube. Y is at 13rd of the total faces. The empirical formula of the compound is

Options:
A X2Y1.5
B X1.5Y
C XY2.5
D X1.5Y2

The correct answer is:
X1.5Y

Solution:
The share of the atom present at corner of the cube is 1/8. The share of the atom at face centre is 1/2. Hence, the molecule can be represented as follows,

X4×18+1×1Y6×13×12

⇒X12+1 Y1

⇒X23 Y1

⇒X1.5 Y1(Empirical formula)

⇒X3 Y2.

Topic: Chemistry, Solid State, JEE Main, JEE Main 2023 (25 Jan Shift 1)
A cubic vessel (with face horizontal + vertical) contains an ideal gas at NTP. The vessel is being carried by a rocket which is moving at a speed of $500 \mathrm{~m} \mathrm{~s}^{-1}$ in vertical direction. The pressure of the gas inside the vessel as observed by us on the ground

Options:
A
remains the same because $500 \mathrm{~ms}^{-1}$ is very much smaller than $v_{\text {rms }}$ of the gas

B
remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls

C
will increase by a factor equal to $\left(v_{\mathrm{ms}}^2+(500)^2\right) / v_{\mathrm{ms}}^2$ where $v_{\mathrm{rms}}$ was the original mean square velocity of the gas

D
will be different on the top wall and bottom wall of the vessel

The correct answer is:
remains the same because motion of the vessel as a whole does not affect the relative motion of the gas molecules and the walls

Solution:
As the relative velocity of molecule with respect to the walls of container does not change in rocket, due to the mass of a molecule is negligible with respect to the mass of whole system and system of gas moves as a whole and $(g=0)$ on molecule energy where. Hence pressure of the gas inside the vessel, as observed by us, on the ground remain the same.

Topic: Physics, Kinetic Theory of Gases
A cubical block is floating in a liquid with half of its volume immersed in the liquid. When the whole system accelerates upwards with a net acceleration of g3. The fraction of volume immersed in the liquid will be

Options:
A 12
B 38
C 23
D 34

The correct answer is:
12

Solution:
B-W=ma
ÏlV′g+a-ÏsVg=ÏsVa
ÏlV′g+a=ÏsVg+a
V′V=ÏsÏl=12
So it remains same.

Topic: Physics, Mechanical Properties of Fluids, NEET