datavorous/entrance-exam-dataset huggingface dataset:
A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass 'm' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is:

Options:
A 32m v2
B m v2
C 12m v2
D 2 m v2

The correct answer is:
m v2

Solution:
At height r from the centre of the Earth, orbital velocity is given as:

orbital velocity = GMr

Ve=2v
K.E=12m2v2=mv2

Topic: Physics, Gravitation, JEE Main, JEE Main 2019 (10 Jan Shift 1)
A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is the universal constant of gravitation, the quantity 3πGd represents :

Options:
A T
B T
C T2
D T3

The correct answer is:
T2

Solution:
The formula to calculate the mass of the Earth considering it a sphere is given by

M=43πR3d ...1

The formula to calculate the time period of revolution of the satellite can be written as

T=2πR3GM ...2

Substitute the expression for the mass from equation (1) into equation (2) and simplify to obtain the expression for the required quantity.

T=2πR3G43πR3d=2π34πGd=12π24πGd=3πGd⇒3πGd=T2

Topic: Physics, Gravitation, JEE Main
A satellite is revolving in a circular orbit at a height h from the carth surface, such that $\mathrm{h}< < \mathrm{R}$ where $\mathrm{R}$ is the radius of the earth. Assuming that the effect of earth"s atmosphere can be neglected the minimum increase in the speed required so that the satellite could escape from the gravitational field of earth is

Options:
A $\sqrt{2 g R}$
B $\sqrt{g R}$
C $\sqrt{\frac{g R}{2}}$
D $\sqrt{g R}(\sqrt{2}-1)$

The correct answer is:
$\sqrt{g R}(\sqrt{2}-1)$

Solution:
For a satellite orbiting close to the earth, orbital velocity is given by

$v_{0}=\sqrt{g(R+h)}=\sqrt{g R}$

Escape velocity $\left(\mathrm{v}_{\mathrm{e}}\right)$ is

$v_{e}=\sqrt{2 g(R+h)}=\sqrt{2 g R} \quad[\because h < c R$

$\Delta v=v_{e}-v_{0}=(\sqrt{2}-1) \sqrt{g R}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2019 (11 Jan Shift 1)
A satellite is revolving in a circular orbit at a height h from the earth's surface (radius of earth R; h << R ). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is close to (Neglect the effect of atmosphere.)

Options:
A gR2
B gR 2-1
C 2Â gR
D gR

The correct answer is:
gR 2-1

Solution:
Since, h << R ,
Ve=2gR
& V0=gR
∴The minimum increase in velocity required,
Ve-V0=2gR-gR
=2-1gR

Topic: Physics, Gravitation, JEE Main, JEE Main 2016 (03 Apr)
A satellite is revolving in a circular orbit at a height h from the earth's surface, such that hâªR where R is the radius of the earth. Assuming that the effect of the earth's atmosphere can be neglected. What should be the minimum change required in the speed so that the satellite could escape from the gravitational field of earth?

Options:
A 2-1gR
B 9gR2
C 2gR
D gR

The correct answer is:
2-1gR

Solution:
Orbital velocity v0=GM(R+h)

Velocity required to escape the gravitational field, vs=2GM(R+h)

Change in velocity =2GM(R+h)-GM(R+h)

If h<<R,
Δv=2GMR-GMR
Δv=2gR-gR

Topic: Physics, Gravitation, JEE Main
A satellite is revolving in circular orbit of radius \( r \) around the earth of mass \( M \). Time of revolution of satellite is

Options:
A \( T \propto \frac{r^{3}}{G M} \)
B \( T \propto \sqrt{\frac{r^{3}}{G M}} \)
C \( T \propto \sqrt{\frac{r^{3}}{\frac{G M}{3}}} \)
D \( T \propto \sqrt{\frac{r^{3}}{\frac{G M}{4}}} \)

The correct answer is:
\( T \propto \sqrt{\frac{r^{3}}{G M}} \)

Solution:
When the satellite is moving in circular orbit after launching the satellite, then centripetal force should be equal to the gravitational force,

Fc=FG

12mv02=GMmr2

where G, M, m, h are universal gravitational constant, the mass of the planet, the mass of satellite and height where the satellite is projected,

⇒mv02r=GMmR+h2

⇒v02=GMr.

Now the time period of the satellite,

T=2πrv0=2πrGMr12

⇒T2=2πrGMr122

⇒T∝r3GM .

Topic: Physics, Gravitation, JEE Main
A satellite moving with velocity $v$ in a force free space collects stationary interplanetary dust at a rate of $\frac{d M}{d t}=\alpha v$ where $M$ is the mass (of satellite + dust) at that instant. The instantaneous acceleration of the satellite is

Options:
A
$-\frac{\alpha v^2}{2 M}$

B
$-\frac{\alpha v^2}{M}$

C
$-\alpha v^2$

D
$-\frac{2 \alpha v^2}{M}$

The correct answer is:
$-\frac{\alpha v^2}{M}$

Solution:
$a_{\text {inst }}=\frac{\alpha \mathrm{V}^2}{\mathrm{M}}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2012 (26 May Online)
A satellite of mass \( M \) is launched vertically upwards with an initial speed \( u \) from the surface of the earth. After it reaches height \( R \) ( \( R= \) radius of the earth), it ejects a rocket of mass \( \frac{M}{10} \) so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is ( \( G \) is the gravitational constant; \( M_{e} \) is the mass of the earth):

Options:
A \( \frac{M}{20}\left(u^{2}+\frac{113}{200} \frac{G M_{e}}{R}\right) \)
B \( 5 M\left(u^{2}-\frac{119}{200} \frac{G M_{e}}{R}\right) \)
C \( \frac{3 M}{8}\left(u+\sqrt{\frac{5 G M_{e}}{6 R}}\right)^{2} \)
D \( \frac{M}{20}\left(u-\sqrt{\frac{2 G M_{e}}{3 R}}\right)^{2} \)

The correct answer is:
\( 5 M\left(u^{2}-\frac{119}{200} \frac{G M_{e}}{R}\right) \)

Solution:
-GMeMR+12Mu2=-GMeM2R+12Mv2

v=u2-GMeR

Vτ→ Transverse velocity of rocket
VR→ Radial velocity of rocket

V=GMe2R

M10VT=9M10GMe2R
M10Vr=Mu2-GMeR

Kinetic energy =12M10VT2+Vr2=M2081GMe2R+100u2-100GMeR

=M20100u2-119GMe2R
=5Mu2-119GMe200R

Topic: Physics, Gravitation, JEE Main
A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R ( R= radius of the earth), it ejects a rocket of mass m10 so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is ( G is the gravitational constant; M is the mass of the earth):

Options:
A m20u2+113200GMR
B 5mu2-119200GMR
C 3m8u+5GM6R2
D m20u-2GM3R2

The correct answer is:
5mu2-119200GMR

Solution:
-GMemR+12mu2=-GMem2R+12mv2
v=u2-GMeR
M10vr=Mu2-GMeR
Kinetic energy =12M10VT2+Vr2=M2081GM2R+100u2-100GMR
= M 20 100 u 2 − 119GM 2R
=5mu2-119GM200R

Topic: Physics, Gravitation, JEE Main
A satellite of mass m is orbiting the earth in a circular orbit at a height 2R above earth surface where R is the radius of the earth. If it starts losing energy at a constant rate β, then it will fall on the earth surface after the time t=GMmnβR. Assuming that the satellite is approximately in a circular orbit at all times, find the value of n.

Options:

The correct answer is:
3

Solution:
Ei=-GMm23R, Ef=-GMm2R
∴ βt=GMmR12–16
βt=GMm3R ⇒ t=GMm3βR

Topic: Physics, Gravitation, JEE Main
A satellite of mass m orbits around the Earth of mass M in an elliptical orbit of semi-major and semi-minor axes 2a and a respectively. The angular momentum of the satellite about the centre of the Earth is

Options:
A πmGMa2
B πmGMa4
C mGMa8
D mGMa2

The correct answer is:
mGMa2

Solution:
If the satellite was orbiting the Earth in a circular orbit of radius 2a, then its time period would be

T=2π2a3GM

The same would be the time period of the satellite if it orbits in an ellipse having a semi-major axis 2a

The area of the ellipse is

A=π×2a×a=2πa2

dAdt=L2m=2πa2T, where L is the angular momentum of the satellite about the centre of the Earth.

⇒L=4πma2GM2π8a3=mGMa2

Topic: Physics, Gravitation, JEE Main
A satellite of mass $m$ revolves around the earth of radius $R$ at a height $x$ from its surface. If $g$ is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

Options:
A
$g x$

B
$\frac{\mathrm{gR}}{\mathrm{R}-\mathrm{x}}$

C
$\frac{g^2}{R+x}$

D
$\left(\frac{g R^2}{R+x}\right)^{1 / 2}$

The correct answer is:
$\left(\frac{g R^2}{R+x}\right)^{1 / 2}$

Solution:
For the satellite, the gravitational force provides the necessary centripetal force i.e.
$$
\begin{aligned}
& \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{(\mathrm{R}+\mathrm{X})^2}=\frac{\mathrm{Mv}_0^2}{(\mathrm{R}+\mathrm{X})} \text { and } \frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^2}=\mathrm{g} \\
& \therefore \mathrm{v}_0=\left(\frac{g \mathrm{R}^2}{\mathrm{R}+\mathrm{X}}\right)^{1 / 2} \\
&
\end{aligned}
$$

Topic: Physics, Gravitation, JEE Main, JEE Main 2004
A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is :
$\left(\right.$ Given $=$ Radius of geo-stationary orbit for earth is $4.2 \times 10^4 \mathrm{~km}$ )

Options:
A $1.4 \times 10^4 \mathrm{~km}$
B $1.05 \times 10^4 \mathrm{~km}$
C $8.4 \times 10^4 \mathrm{~km}$
D $1.68 \times 10^5 \mathrm{~km}$

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

Solution:
$\begin{aligned} & \mathrm{T}=\frac{2 \pi \mathrm{r}^{3 / 2}}{\sqrt{\mathrm{GM}}} \\ & \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^{3 / 2}\left(\frac{\mathrm{M}_2}{\mathrm{M}_1}\right)^{1 / 2} \\ & \frac{6}{24}=\frac{\left(\mathrm{r}_1\right)^{3 / 2}}{\left(4.2 \times 10^4\right)^{3 / 2}}\left(\frac{\mathrm{M}}{\mathrm{M} / 4}\right)^{1 / 2} \\ & \mathrm{r}_1=1.05 \times 10^4 \mathrm{~km}\end{aligned}$

Topic: Physics, Gravitation, JEE Main, JEE Main 2024 (05 Apr Shift 2)
A satellite revolving around the earth at a certain height experiences acceleration due to gravity equal to $\frac{16}{49} g_0$, where $g_0$ is the acceleration due to gravity on the earth's surface. If $R$ is the radius of earth, then the square of time period of the satellite's revolution is equal to $K\left[\frac{\pi^2 R^3}{G M}\right]$. The value of $K$ is

Options:
A $\frac{27}{36}$
B $\frac{343}{16}$
C $\frac{125}{64}$
D $\frac{675}{81}$

The correct answer is:
$\frac{343}{16}$

Solution:
Acceleration due to gravity at height $h$ from the surface of earth,
$$
\begin{aligned}
g & =g_0\left(\frac{R}{R+h}\right)^2=\frac{16}{49} g_0 \\
\frac{R}{R+h} & =\frac{4}{7}
\end{aligned}
$$
Now, orbital velocity of satellite, $v_0=\sqrt{G M / r}$
Now, time period of revolution,
$$
\begin{aligned}
T & =\frac{2 \pi r}{\sqrt{\frac{G M}{r}}}=2 \pi \sqrt{\frac{r^3}{G M}} \\
T^2 & =\frac{4 \pi^2 r^3}{G M}=\frac{4 \pi^2}{G M}\left(\frac{7 R}{4}\right)^3\left[\because r=R+h=R+\frac{3 R}{4}=\frac{7 R}{4}\right]
\end{aligned}
$$
Comparing Eqs. (ii) and (iii), we get
$$
K=\frac{343}{16}
$$

Topic: Physics, Gravitation, JEE Main
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Then,

Options:
A the total mechanical energy of S varies periodically with time.
B the linear momentum of S remains constant is magnitude.
C the acceleration of S is always directed towards the centre of the earth.
D the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant.

The correct answer is:
the acceleration of S is always directed towards the centre of the earth.

Solution:
The gravitation force on the satellite will be aiming toward the centre of earth so acceleration of the satellite will also be aiming toward the centre of earth.

Topic: Physics, Gravitation, NEET, NEET 2015 (Phase 2)
A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth. Then

Options:
A the acceleration of S is always directed towards the centre of the earth
B The angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant
C the total mechanical energy of S varies periodically with time
D the linear momentum of S remains constant in magnitude

The correct answer is:
the acceleration of S is always directed towards the centre of the earth

Solution:
The acceleration of S is always directed towards the centre of the earth

Topic: Physics, Gravitation, NEET
A saturated solution of $\mathrm{H}_{2} \mathrm{~S}$ in $0.1 \mathrm{M} \mathrm{HCl}$ at $25^{\circ} \mathrm{C}$ contains $\mathrm{S}^{2-}$ ion concentration of $10^{-23} \mathrm{~mol} \mathrm{~L}^{-1}$. The solubility product of some sulphides are $\mathrm{CuS}=10^{-44}, \mathrm{FeS}=10^{-14}, \mathrm{MnS}=10^{-15}, \mathrm{CdS}=$ $10^{-25}$. If $0.01 \mathrm{M}$ solution of these salts in $1 \mathrm{M}$ $\mathrm{HCl}$ are saturated with $\mathrm{H}_{2} \mathrm{~S}$, which of these will be precipitated?

Options:
A All
B All except MnS
C $\mathrm{AU}$ except $\mathrm{MnS}$ and $\mathrm{FeS}$
D Only-CuS

The correct answer is:
$\mathrm{AU}$ except $\mathrm{MnS}$ and $\mathrm{FeS}$

Solution:
$\left[\mathrm{S}^{2-}\right]=10^{-23} \mathrm{~mol} \mathrm{~L}^{-1}$
$\left[\mathrm{M}^{2+}\right]=10^{-2} \mathrm{M}$
Ionic product, $\mathrm{K}_{1 \mathrm{P}}=\left[\mathrm{M}^{2+}\right]\left[\mathrm{S}^{2-}\right]=10^{-25}$
$\therefore \quad$ ionic product is greater than $\mathrm{K}_{\mathrm{sp}}$ of CuS and $\mathrm{CdS}$,
Therefore, all except $\mathrm{MnS}$ and $\mathrm{FeS}$ are precipitated.

Topic: Chemistry, Ionic Equilibrium, JEE Main
A s-block element (M) reacts with oxygen to form an oxide of the formula MO2. The oxide is pale yellow in colour and paramagnetic. The element (M) is:

Options:
A Mg
B Na
C Ca
D K

The correct answer is:
K

Solution:
(A) 2Mg+O2→2MgO (Diamagnetic)
(B) 2Na+O2→Na2O (Diamagnetic)

2Na+O2(excess)→Na2O2 (Diamagnetic)

(C) 2Ca+O2→2CaO (Diamagnetic)
Ca+O2→CaO2 (Diamagnetic)

(D) K+O2 (excess) →KO2 (Paramagnetic)

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2021 (20 Jul Shift 1)
A schematic plot of ln $\mathrm{K}_{\text {eq }}$ versus inverse of temperature for a reaction is shown below
The reaction must be

Options:
A
exothermic

B
endothermic

C
one with negligible enthalpy change

D
highly spontaneous at ordinary temperature

The correct answer is:
exothermic

Solution:
$\mathrm{K}_{\mathrm{eq}}=\mathrm{A} \mathrm{e}^{-\frac{\Delta \mathrm{H}}{\mathrm{RT}}}$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2005
A scientific committee is to be formed from 6 Indians and 8 foreigners, which includes at least 2 Indians and double the number of foreigners as Indians. Then the number of ways, the committee can be formed, is:

Options:
A 1050
B 1625
C 575
D 560

The correct answer is:
1625

Solution:
Indians	Foreigners	Number of ways

2	4	C26×C48=1050

3	6	C36×C68=560

4	8	C46×C88=15

Total number of ways =1625

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (24 Feb Shift 1)
A scientist is observing a bacteria through a compound microscope. For better analysis and to improve its resolving power he should. (Select the best option)

Options:
A Increase the wave length of the light
B Increase the refractive index of the medium between the object and objective lens
C Decrease the focal length of the eye piece
D Decrease the diameter of the objective lens

The correct answer is:
Increase the refractive index of the medium between the object and objective lens

Solution:
As we know resolving power P∝μsinθλ.

Therefore, as μ increases, P will increase.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (29 Jan Shift 2)
A scion is grafted to a stock. The quality of fruits produced will be determined by the genotype of

Options:
A Stock
B Scion
C Both stock and scion
D Neither stock nor scion.

The correct answer is:
Scion

Solution:
The main rooted part of the plant is called stock and the bud of stem grafted is called scion. Flowers and thus fruits are produced on scion thus the genotype of fruit is determined by scion.

Topic: Biology, Reproduction in Organisms, NEET
A scooter accelerates from rest for time t1 at constant rate a1 and then retards at constant rate a2 for time t2 and comes to rest. The correct value of t1t2 will be :

Options:
A

a2a1

B a1a2
C a1+a2a1
D a1+a2a2

The correct answer is:

a2a1

Solution:
Draw vt curve

tanθ1=a1=vmaxt1
& tanθ2=a2=vmaxt2
÷ above

t1t2=a2a1

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (26 Feb Shift 2)
A scooter going due east at $10 \mathrm{~m} / \mathrm{s}$ turns right through an angle of $90^{\circ}$. If the speed of the scooter remains unchanged in taking turn, the change is the velocity of the scooter is

Options:
A $20.0 \mathrm{~m} / \mathrm{s}$ south eastern direction.
B Zero
C $10.0 \mathrm{~m} / \mathrm{s}$ in southern direction.
D $14.14 \mathrm{~m} / \mathrm{s}$ in south-west direction.

The correct answer is:
$14.14 \mathrm{~m} / \mathrm{s}$ in south-west direction.

Solution:
If the magnitude of vector remains same, only direction change by $\theta$ then
$\vec{\Delta v}=\vec{v_2}-\vec{v_1}, \vec{\Delta v}=\vec{v_2}+\left(-\vec{v_1}\right)$
Magnitude of change in vector $|\overrightarrow{\Delta v}|=2 v \sin \left(\frac{\theta}{2}\right)$
$|\overrightarrow{\Delta v}|=2 \times 10 \times \sin \left(\frac{90^{\circ}}{2}\right)=10 \sqrt{2}=14.14 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Mathematics in Physics, JEE Main
A screen is at a distance D=80 cm from a diaphragm having two narrow slits S1 and S2 which are d=2 mm apart. The slit S1 is covered by a transparent sheet of thickness t1=2.5 Î¼m and the slit S2 by another sheet of thickness t2=1.25 Î¼m âas shown in the figure. Both sheets are made of the same material having a refractive index Î¼m=1.40. Water is filled in space between the diaphragm and the screen. A monochromatic light beam of wavelength Î»=5000 Ã is incident normally on the diaphragm. Assuming the intensity of the beam to be uniform and slits of equal width, calculate the ratio of intensity at C to the maximum intensity of interference pattern obtained on the screen, where C is the foot of the perpendicular bisector of S1S2. [Refractive index of water, Î¼w=43]

Options:
A 3 4
B 12
C 13
D 35

The correct answer is:
3 4

Solution:
Optical path = (Refractive Index) Ã (Geometrical Path Length)
Path Difference at point C on the screen
Î x = Î¼ m t 1 + Î¼ w S 1 C - t 1 - Î¼ m t 2 + Î¼ w S 2 C - t 2
Î x = Î¼ m t 1 - t 2 - Î¼ w t 1 + Î¼ w t 2 ( âµ s 1 c= s 2 c)
Î x = Î¼ m t 1 - t 2 - Î¼ w t 1 - t 2
Î x = Î¼ m - Î¼ w t 1 - t 2
Î x = 1.4 - 4 3 2.5 Ã 10 - 6 - 1.25 Ã 10 - 6
Î x = 0.2 3 Ã 1.25 Ã 10 - 6
Î x = 2.5 3 Ã 1 0 - 7 m
Î x = 2500 3 Ã= 5000 6 Ã
Î x = Î» 6 ,
Phase difference Ï = Ï 3 = 60 Î¿
at any point on the screen I net = I 1 + I 2 +2 I 1 I 2 âcosÏ
Resultant intensity I C = I + I + 2 I Ã I cos 6 0
I C = 3I
& I max =I+I+2I when Ï= 0 o ââcosÏ=+1
I max = 4I
I C I max = 3 4

Topic: Physics, Wave Optics, JEE Main
A screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading : $0 \mathrm{~mm}$
Circular scale reading : 52 divisions
Given that $1 \mathrm{~mm}$ on main scale corresponds to 100 divisions of the circular scale.
The diameter of wire from the above date is :

Options:
A
$0.052 \mathrm{~cm}$

B
$0.026 \mathrm{~cm}$

C
$0.005 \mathrm{~cm}$

D
$0.52 \mathrm{~cm}$

The correct answer is:
$0.052 \mathrm{~cm}$

Solution:
Diameter of wire $=\frac{1}{100} \times 52=0.52 \mathrm{~mm}=0.052 \mathrm{~cm}$

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2011
A screw gauge gives the following reading when used to measure the diameter of a wire.
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that $1 \mathrm{~mm}$ on main scale corresponds to 100 divisions of the circular scale.
The diameter of wire from the above data is :

Options:
A $0.026 \mathrm{~cm}$
B $0.005 \mathrm{~cm}$
C $0.52 \mathrm{~cm}$
D $0.052 \mathrm{~cm}$

The correct answer is:
$0.052 \mathrm{~cm}$

Solution:
Here, the pitch of the screw gauge, $p=1 \mathrm{~mm}$
Number of circular divisions, $\mathrm{n}=100$
Thus, least count $\mathrm{LC}=\mathrm{p} / \mathrm{n}=1 / 100=0.01 \mathrm{~mm}=0.001 \mathrm{~cm}$
Diameter of the wire $=M S R+(C S R \times L C)=0+$ $(52 \times 0.001 \mathrm{~cm})=0.052 \mathrm{~cm}$
Answer: (4) $0.052 \mathrm{~cm}$

Topic: Physics, Mathematics in Physics, JEE Main
A screw gauge gives the following reading when used to measure the diameter of a wire.

Main scale reading : 0 mm

Circular scale reading: 52 divisions

Given that 1 mm on the main scale corresponds to

100 divisions of the circular scale.

The diameter of the wire from the above data is

Options:
A 0.052 cm
B 0.026 cm
C 0.005 cm
D 0.52 cm

The correct answer is:
0.052 cm

Solution:
Diameter of wire,
d=MSR+CSR ÃLC
=0+52Ã1100
=0.52 mm=0.052 cm.

Topic: Physics, Units and Dimensions, NEET
A screw gauge gives the following readings when used to measure the diameter of a wire
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is:

Options:
A 0.52 cm
B 0.026 cm
C 0.26 cm
D 0.052 cm

The correct answer is:
0.052 cm

Solution:
L.C.=1100mm

=0.01 mm

Now,

Diameter of wire =52×0.01

=0.52 mm

=0.052 cm

Topic: Physics, Experimental Physics, NEET, NEET 2021
A screw gauge with a pitch of 1 mm has 100 divisions on its circular scale. When it is used to measure the diameter of a thin wire, the main scale reading is 3 mm and the circular scale reading is 45. If D is the diameter of the wire in mm, then find the value of D.

Options:

The correct answer is:
3.45

Solution:
Least count =PitchNo. of division=1mm100=0.01 mm
Diameter of wire = main scale + circular scale reading
=3 mm+L.CÃNo. of division
=3 mm+0.01Ã45
D=3.45 mm

Topic: Physics, Units and Dimensions, JEE Main
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?

Options:
A 0.70 mm
B 0.50 mm
C 0.75 mm
D 0.80 mm

The correct answer is:
0.80 mm

Solution:
LC=0.550=0.01 mm
Zero error =0.50-0.45=-0.05
Thickness =0.5+25×0.01+0.05
=0.50+0.25+0.05
=0.80 mm

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2016 (03 Apr)
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of aluminium. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line?

Options:
A 0.70 mm
B 0.50 mm
C 0.75 mm
D 0.80 mm

The correct answer is:
0.80 mm

Solution:
LC=0.550=0.01 mm
Zero error =0.50-0.45=-0.05
Thickness =0.5+25Ã0.01+0.05
=0.50+0.25+0.05
=0.80 mm

Topic: Physics, Units and Dimensions, JEE Main
A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of aluminum. Before starting the measurement, it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness (in mm) of the sheet if the main scale reading is 0.5 mm and the 30th division coincides with the main scale line?

Options:

The correct answer is:
0.85

Solution:
Least count, LC=0.550=0.01 mm

Zero error, ZE=50-45×0.01=-0.05 mm

Main Scale Reading MSR=0.5 mm
Thickness, t=0.5 mm +30×0.01 mm - -0.05 mm

t=0.85 mm

Topic: Physics, Experimental Physics, JEE Main
A scrubber in the exhaust of a chemical industrial plant removes:

Options:
A Gases like sulphur dioxide
B Particulate matter of the size 5 micrometer or above
C Gases like ozone and methane
D Particulate matter of the size 2.5 micrometer or less

The correct answer is:
Gases like sulphur dioxide

Solution:
A scrubber can remove gases like sulphur dioxide. In a scrubber, the exhaust is passed through a spray of water or lime. Sulphur dioxide is a major pollutant of chemical industrial plants and the dry scrubbing method is mainly used to remove this pollution.

Topic: Biology, Environmental Issues, NEET, NEET 2014