datavorous/entrance-exam-dataset huggingface dataset:
A metallic ring of mass 2 kg and radius 1 m with a uniform metallic spoke of same mass 2 kg and length 1 m is rotated about its axis with angular velocity 1 rev sec-1. In a perpendicular uniform magnetic field B of magnitude 10 T as shown in the figure. If the central end of the spoke is connected to the rim of the wheel through a resistor R of magnitude Ï Î© as shown. The resistor does not rotate, it's one end is always at the centre of the ring and another end is always in contact with the ring. A force F as shown is needed to maintain the constant angular velocity of the spoke then, F is equal to (The ring and the spoke has zero resistance)

Options:
A 10 N
B 20 N
C 50 N
D None of these

The correct answer is:
50 N

Solution:
Fb=BIL
Induced current: I=Bωr22R
∴ FB=B Bωr22Rr=B2ωr32R
To maintain constant angular velocity:
F r=FBr2
⇒ F=FB2=B2ωr34R=102×2π×13 4×π=50 N

Topic: Physics, Magnetic Effects of Current, JEE Main
A metallic ring of radius $r$ and cross sectional area $A$ is fitted into a wooden circular disc of radius $R(R>r)$. If the Young's modulus of the material of the ring is $Y$, the force with which the metal ring expands is

Options:
A $\frac{A Y R}{r}$
B $\frac{A Y(R-r)}{r}$
C $\frac{Y(R-r)}{A r}$
D $\frac{Y R}{A R}$

The correct answer is:
$\frac{A Y(R-r)}{r}$

Solution:
$\begin{aligned} & Y=\frac{F}{A} \cdot \frac{L}{l} \\ & \therefore \text { Force } F=\frac{A Y l}{L}=\frac{A Y[2 \pi(R-r)]}{2 \pi r} \\ & \Rightarrow \quad F=\frac{A Y(R-r)}{r}\end{aligned}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A metallic ring (radius R) of negligible resistance has a resistance r connected across its diameter as shown in the figure. It is moving with velocity v0 in a constant magnetic field B0 acting perpendicular to the plane of the paper in the inward direction. The current in the resistance is:

Options:
A 2BRv0r
B BRv0r
C BRv02r
D 0

The correct answer is:
2BRv0r

Solution:
As induced emf is Blv for a moving rod,
â´ the arcs produce of emf of 2B0v0R
Applying KVL in the loop
We get,
2B0v0R=ir
i=2B0v0Rr

Topic: Physics, Electromagnetic Induction, JEE Main
A metallic rod breaks when strain produced is $0.2 \%$. The Young's modulus of the material of the $\operatorname{rod} 7 \times 10^9 \mathrm{~N} / \mathrm{m}^2$. The area of crosssection to support a load of $10^4 \mathrm{~N}$ is

Options:
A $7.1 \times 10^{-6} \mathrm{~m}^2$
B $7.1 \times 10^{-4} \mathrm{~m}^2$
C $7.1 \times 10^{-2} \mathrm{~m}^2$
D $7.1 \times 10^{-8} \mathrm{~m}^2$

The correct answer is:
$7.1 \times 10^{-4} \mathrm{~m}^2$

Solution:
Given, Youngs' modulus, $Y=7 \times 10^9 \mathrm{~N} / \mathrm{m}^2$ Load, $F=10^4 \mathrm{~N}$
We know that.
$$
\begin{aligned}
Y & =\frac{F l}{A \Delta l} \\
\Rightarrow \quad A & =\frac{F l}{Y \Delta l}=\frac{F}{Y} \times \frac{1}{\left(\frac{\Delta l}{l}\right)} \\
& =\frac{10^4}{7 \times 10^9} \times \frac{1}{0.2}=7.1 \times 10^{-4} \mathrm{~m}^2
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A metallic rod of length 1 m, Young's modulus 3Ã1011 N m-2 and density 7.5Ã102 kg m-3 is clamped at its middle. Longitudinal stationary vibrations are produced in the rod with total number of displacement nodes equal to 3. The frequency of vibrations is

Options:
A 30,000 Hz
B 10,000 Hz
C 3000 Hz
D 1500 Hz

The correct answer is:
30,000 Hz

Solution:
l=3Î»2â Î»=2l3
f=Î½Î»=3Î½2l=32lYÏ
= 32Ã13Ã101134Ã103 =324Ã108
= 32Ã2Ã104 Hz
= 30,000 Hz

Topic: Physics, Waves and Sound, JEE Main
A metallic rod of length l is tied to a string of length 2l and made to rotate with angular speed ω on a horizontal table with one end of the string fixed. If there is a vertical magnetic field B in the region, the e.m.f. induced across the ends of the rod is:

Options:
A 4Bωl22
B 5Bωl22
C 2Bωl22
D 3Bωl22

The correct answer is:
5Bωl22

Solution:
e=∫2l3lBVdx
e = ∫ 2 ℓ 3 ℓ B(ωx)dx
e=ωB∫2l3lxdx
e=ωBx222l3l
e=ωB23l2-2l22l3l
e=52ωBl2

Topic: Physics, Electromagnetic Induction, JEE Main
A metallic rod of length l is tied to a string of length 2l and made to rotate with angular speed ω on a horizontal table with one end of the string fixed. If there is a vertical magnetic field B in the region, the e.m.f. induced across the ends of the rod is:

Options:
A 4Bωl22
B 5Bωl22
C 2Bωl22
D 3Bωl22

The correct answer is:
5Bωl22

Solution:
e=∫2l3lBVdx
e = ∫ 2 ℓ 3 ℓ B(ωx)dx
e=ωB∫2l3lxdx
e=ωBx222l3l
e=ωB23l2-2l22l3l
e=52ωBl2

Topic: Physics, Electromagnetic Induction, JEE Main, JEE Main 2013 (07 Apr)
A metallic solid undergoes Frenkel defect. Its original mass, volume and density are \(M_0, V_0\) and \(D_0\) respectively. After Frenkel defect the mass, volume and density are \(M, V\) and \(D\) respectively. The variations of \(M, V\) and \(D\) after Frenkel defect are

Options:
A \(M=M_0, V=V_0, D=D_0\)
B \(M < M_0, V < V_0, D < D_0\)
C \(M>M_0, V=V_0, D>D_0\)
D \(M=M_0, V < V_0, D>D_0\)

The correct answer is:
\(M=M_0, V=V_0, D=D_0\)

Solution:
In Frenkel defect, atoms remain within the solid crystal. Only one vacancy is created. Density remains the same. Thus, \(M=M_0, V=V_0\) and \(D=D_0\). Hence, the correct option is (a).

Topic: Chemistry, Solid State, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 2)
A metallic sphere cools from 50 °C to 40 °C in 300 s. If atmospheric temperature around is 20 °C, then the sphere's temperature after the nest 5 minutes will be close to:

Options:
A 31 °C
B 33 °C
C 28 °C
D 35 °C

The correct answer is:
33 °C

Solution:
Using Newton's Law of cooling

50-405 min=K50+402-20 ......(i)

Next 5 min.

40-θ5=K40+θ2-20 .....(ii)

Dividing (ii)/(i)

40-θ10=40+θ-4050+40-40=θ50

40-θ=θ5

200-5θ=θ

∴ θ=2006=33.3 °C

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2020 (03 Sep Shift 2)
A metallic surface is illuminated with radiation of wavelength λ, the stopping potential is V0. If the same surface is illuminated with radiation of wavelength 2λ, the stopping potential becomes V04. The threshold wavelength for this metallic surface will be

Options:
A 3λ
B 4λ
C 32λ
D λ4

The correct answer is:
3λ

Solution:
Let the threshold frequency be λ0.

By the equation of photoelectric effect, for wavelength λ,

eV0=hc1λ-1λ0 ...(i)

For the wavelength 2λ,

eV04=hc12λ-1λ0 ...(ii)

Dividing (i) by (ii)

4=hc(λ0-λλλ0)hcλ0-2λ2λλ0⇒4=2(λ0-λ)λ0-2λ⇒2λ0=6λ⇒λ0=3λ

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (11 Apr Shift 1)
A metallic V shaped rod OAC is rotated with respect to one of its end in a uniform magnetic field, such that axis of rotation is parallel to the direction of magnetic field. Length of each arm of the rod is L and the angle between the arms is 60° . P is the mid-point of section AC. The magnitude of the magnetic field is B. Then choose the incorrect relation.

Options:
A VA-V0=ωBL22
B VA-VC=ωBL22
C VC-VP=ωBL28
D VA-VP=ωBL28

The correct answer is:
VA-VC=ωBL22

Solution:
OA=OC=L;OP=3L2
VA-V0=ÏBL22 VA-VC=0
VC-V0=ÏBL22 VP-V0=ÏB23L24=3ÏBL28
VC-VP=VC-V0-VP-V0=ÏBL22-3ÏBL28=ÏBL28
VA-VP=VA-V0-VP-V0=ÏBL22-2ÏBL28=ÏBL28

Topic: Physics, Electromagnetic Induction, JEE Main
A metallic wire is folded to form a square loop of side a. It carries a current i and is kept perpendicular to a uniform magnetic field. If the shape of the loop is changed from square to a circle without changing the length of the wire and current, the amount of work done in doing so is

Options:
A iBa2π+2
B iBa2Ï-2
C iBa24/Ï+1
D iBa21-4π

The correct answer is:
iBa21-4π

Solution:
W = Uf â Ui
=-MfB cos 0o--MiB cos 0o
=iBAi-Af
where Ai=a2,Af=ÏR2=4a2Ï
as 2ÏR=4a or R=2aÏ
â´ W=iBa21-4Ï

Topic: Physics, Magnetic Effects of Current, JEE Main
A metallic wire of \( 1 \mathrm{~m} \) length has a mass of \( 10 \times 10^{-3} \mathrm{~kg} \). If a tension of \( 100 \mathrm{~N} \) is applied to a
wire, what is the speed of transverse wave ?

Options:
A \( 100 \mathrm{~ms}^{-1} \)
B \( 10 \mathrm{~ms}^{-1} \)
C \( 200 \mathrm{~ms}^{-1} \)
D \( 0.1 \mathrm{~ms}^{-1} \)

The correct answer is:
\( 100 \mathrm{~ms}^{-1} \)

Solution:
Given, length of wire, \( I=1 \mathrm{~m} ; \) mass of wire, \( m=10 \times 10^{-3} \mathrm{~kg} \) tension, \( T=100 \mathrm{~N} \)
Speed of transverse wave, \( v=\sqrt{\frac{\text { Tension }}{\text { linear density }}} \)
Linear density
\[
\begin{array}{l}
=\frac{\text { mass }}{\text { length }}=\frac{10 \times 10^{-3}}{1}=10 \times 10^{-3} \\
\Rightarrow v=\sqrt{\frac{100}{10 \times 10^{-3}}}=\sqrt{10^{4}}=100 \mathrm{~ms}^{-1}
\end{array}
\]
Therefore, speed of transverse wave \( =100 \mathrm{~ms}^{-1} \)

Topic: Physics, Waves and Sound, JEE Main
A metallic wire with tension $T$ and at temperature $30^{\circ} \mathrm{C}$ vibrates with its fundamental frequency of $1 \mathrm{kHz}$. The same wire with the same tension but at $10^{\circ} \mathrm{C}$ temperature vibrates with a fundamental frequency of $1.001 \mathrm{kHz}$. The coefficient of linear expansion of the wire is

Options:
A $2 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
B $1.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
C $1 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
D $0.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}$

The correct answer is:
$0.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}$

Solution:
Frequency of wire $n=\frac{1}{2 l} \frac{\sqrt{T}}{m}$ $m=$ mass per unit length
In both the condition same wire is kept same tension.
So,
$$
\begin{aligned}
\frac{n_1}{n_2}=\frac{l_2}{l_1} \Rightarrow \frac{1}{1.001} & =\frac{l_2}{l_1} \\
l_2 & =\frac{l_1}{1.001}
\end{aligned}
$$
Fall in temperature $\Delta t=30-10=20^{\circ} \mathrm{C}$
$$
\begin{aligned}
l_2 & =l_1(1-\alpha \Delta t) \\
\frac{l_1}{1.001} & =l_1(1-\alpha \times 20) \\
1-\alpha \times 20 & =\frac{1}{1.001}=\frac{0.001}{1.001} \\
\alpha=\frac{1}{1001 \times 20} & =0.5 \times 10^{-4} /{ }^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
A meter bridge circuit is powered by an ideal battery of emf 5 V and negligible internal resistance. The bridge wire has a resistance per unit length equal to 0.1 Î© cm-1. Unknown resistance X is connected in the left gap and 6 Î© in the right gap. The null point divides the wire in the ratio 2:3. What is the current (in A) drawn from the battery?

Options:

The correct answer is:
1

Solution:
l1l2=23=v1v2
Now v1+v2=v=5 V =IR
For potentiometer wire R1R2=l1l2=23
x6=23⇒xl=4 Ωm
The resistance per unit length of the wire is 0.1Ωcm=1Ωm
∴ x+1=Reff=5
∴ i=VReff=55=1 A

Topic: Physics, Current Electricity, JEE Main
A meter bridge is used to determine the resistance of an unknown wire by measuring the balance point length l. If the wire is replaced by another wire of same material but with double the length and half the thickness, the balancing point is expected to be

Options:
A 18l
B 14l
C 8l
D 16l

The correct answer is:
8l

Solution:
In a meter bridge the ratio of two resistances is
Â Â Â Â Â Â Â RR′=ll′
Where l and l′ are balancing lengths.
Resistance R=ÏlA=ÏlÏr2
In material remains same Ï=Ï′
Given, l′=2l
Â Â Â Â Â Â Â Â Â Â Â Â Â r′=r2
â´Â Â Â R′=Ïl′A′=Ï2lÏr22=8ÏlÏr2
Â Â Â Â Â Â R′=8R
Therefore, the new balancing point is expected to be 8l.

Topic: Physics, Current Electricity, NEET
A metre long narrow bore held horizontally (and closed at one end) contains a $76 \mathrm{~cm}$ long mercury thread, which traps a $15 \mathrm{~cm}$ column of air. What happens if the tube is held vertically with the open end at the bottom ?

Options:

None

Solution:
In horizontal position, the bore will leave an air-column of $9 \mathrm{~cm}$ as shown in figure (a) so as to balance the atmospheric pressure $(=76 \mathrm{~cm}$ of $\mathrm{Hg})$
$$
\therefore P_1=76 \mathrm{~cm} \&
$$
Volume of air column $=15 \mathrm{~cm}^3$
assuming area of cross section of tube $=1 \mathrm{~cm}^3$
Total length of air column in the tube $=24 \mathrm{~cm}$.
In vertical position as shown in figure (b) if $h^{\prime} \mathrm{cm}$ of mercury flows out to balance the atmospheric pressure then new pressure on mercury column $=76-(76-h)=$ $h \mathrm{~cm}$ of $\mathrm{Hg}$
The new volume of air column $=24+h$.
Since the temperature remains constant during the transition from horizontal to vertical position.
$$
\begin{aligned}
&\therefore P_1 V_1=P_2 V_2 \\
&76 \times 15=h \times(24+h) \\
&\text { or } h^2+24 h-1140=0 \\
&\Rightarrow h=23.8 \mathrm{~cm}
\end{aligned}
$$
Hence, $23.8 \mathrm{~cm}$ of mercury flows out from the tube in vertical position.

Topic: Physics, Kinetic Theory of Gases
A metre scale is balanced on a knife edge at its centre. When two coins, each of mass 10 g are put one on the top of the other at the 10.0 cm mark the scale is found to be balanced at 40.0 cm mark. The mass of the metre scale is found to be x×10-2 kg. The value of x is _____ .

Options:

The correct answer is:
6

Solution:
Let the mass of the scale be m.

The centre of mass of the scale will be at 50 cm.

Applying torque balance,

0=mg×50-40-2×10g×40-10⇒m=60 g=6×10-2 kg

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2022 (24 Jun Shift 1)
A metre scale is supported on a wedge at its centre of gravity. A body of weight ' $\mathrm{w}$ ' is suspended from the $20 \mathrm{~cm}$ mark and another weight of 25 gram is suspended from $74 \mathrm{~cm}$ mark balance it and the metre scale remains perfectly horizontal. Neglecting the weight of the metre scale, the weight of the body is

Options:
A 20 gram-wt
B 15 gram-wt
C 33 gram-wt
D 30 gram-wt

The correct answer is:
20 gram-wt

Solution:
Taking moments about the centre of the scale we get
$$
\begin{aligned}
& \mathrm{W} \times 30=25 \times 24 \\
& \therefore \mathrm{W}=\frac{25 \times 24}{30}=20 \mathrm{~g}-\mathrm{wt}
\end{aligned}
$$

Topic: Physics, Rotational Motion, MHT CET, MHT CET 2021 (24 Sep Shift 1)
A metre stick is balanced on a knife-edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass (in gram) of the metre stick?

Options:

The correct answer is:
66

Solution:
Let total mass of the metre stick be M gram
Distance between mid-point E and new centre of gravity (DE),
= 50 - 45 = 5 cm
Distance between 12 cm mark and new centre of gravity (CD),
= 45 - 12 = 33 cm
From principle of moments in equilibrium,
Mg Ã DE = (2 Ã 5g) Ã CD
M Ã 5 = 10 Ã 33
â
or M = 66 g
â´ Mass of the metre stick is 66 g.

Topic: Physics, Rotational Motion, JEE Main
A micelle formed during the cleansing action by soap is

Options:
A a discrete particle of soap
B aggregated particles of soap and dirt
C a discrete particle of dust
D an aggregated particle of dust and water

The correct answer is:
aggregated particles of soap and dirt

Solution:
Soap contains hydrophobic (ie, fat soluble or water repelling) as well as hydrophilic (water soluble) end. Hydrophobic end remains soluble in dust or grease particles while hydrophilic end is attracted towards water. Hence, several hydrophobic pocket surrounds the grease particles. This aggregate of soap and dirt is called micelle. On washing with plenty of water dust is removed along with soap.

Topic: Chemistry, Surface Chemistry, TS EAMCET, TS EAMCET 2010
A micelle formed during the cleansing action by soap is

Options:
A a discrete particle of soap
B aggregated particles of soap and dirt
C a discrete particle of dust
D an aggregated particle of dust and water

The correct answer is:
aggregated particles of soap and dirt

Solution:
Soap contains hydrophobic (ie, fat soluble or water repelling) as well as hydrophilic (water soluble) end. Hydrophobic end remains soluble in dust or grease particles while hydrophilic end is attracted towards water. Hence, several hydrophobic pocket surrounds the grease particles. This aggregate of soap and dirt is called micelle. On washing with plenty of water dust is removed along with soap.

Topic: Chemistry, Surface Chemistry, AP EAMCET, AP EAMCET 2010
A micelle formed during the cleansing action by soap is

Options:
A a discrete particle of soap.
B aggregated particles of soap and dirt.
C a discrete particle of dust.
D an aggregated particle of dust and water.

The correct answer is:
aggregated particles of soap and dirt.

Solution:
The soap contains hydrophobic (i.e. fat soluble or water repelling) as well as hydrophilic (water soluble) end. The hydrophobic end remains soluble in dust or grease particles while the hydrophilic end is attracted towards the water. Hence, several hydrophobic molecules surround the grease particles. This aggregate of soap and dirt is called a micelle. On washing with plenty of water, dust is removed along with soap.

Topic: Chemistry, Surface Chemistry, NEET
A microscope is focused on an object at the bottom of a bucket. If liquid with refractive index 53 is poured inside the bucket, then microscope have to be raised by 30cm to focus the object again. The height of the liquid in the bucket is :

Options:
A 75cm
B 50cm
C 18cm
D 12cm

The correct answer is:
75cm

Solution:
In this case shift is given by, shift=d1-1μ=30 cm

⇒d1-153=30

⇒d=30×52=75 cm

Topic: Physics, Ray Optics, JEE Main, JEE Main 2023 (31 Jan Shift 2)
A microscope is having objective of focal length 1 cm and eyepiece of focal length 6 cm. If tube length is 30 cm and image is formed at the least distance of distinct vision, what is the magnification produced by the microscope. (Take D=25 cm)

Options:
A 6
B 155
C 25
D 125

The correct answer is:
155

Solution:
For a compound microscope,
m=Lf01+Dfe
(Where length of tube L=30 cm, focal length of objective lens f0=1 cm, focal length of eye-piece fe=6 cm, D=25 cm)
âm=3011+256=30Ã6+256
âm=5Ã31
âm=155

Topic: Physics, Ray Optics, JEE Main
A microscope was initially placed in air (refractive index 1). It is then immersed in oil (refractive index 2). For a light whose wavelength in air is $\lambda$, calculate the change of microscope's resolving power due to oil and choose the correct option

Options:
A Resolving power will be 14 in the oil than it was in the air
B Resolving power will be twice in the oil than it was in the air.
C Resolving power will be four times in the oil than it was in the air.
D Resolving power will be 12 in the oil than it was in the air.

The correct answer is:
Resolving power will be twice in the oil than it was in the air.

Solution:
Resolving power of microscope in air is (R.P)air =2μairsinθ1.22λ=2sinθ1.22λ

(R.P)oil =2μoilsinθ1.22λoil =2sinθ21.22λ

R.Poil=R.Pair×2

Thus, resolving power in oil is twice the resolving power in air.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (27 Jul Shift 1)
A microwave of wavelength 2.0 cm falls normally on a slit of width 4.0 cm. The angular spread of the central maxima of the diffraction pattern obtained on a screen 1.5 m away from the slit, will be:

Options:
A 30º
B 15º
C 60º
D 45º

The correct answer is:
60º

Solution:
In diffraction, for first minima a sinθ=λ

⇒sinθ=λa=12

⇒θ=30°

Therefore, angular spread of central maxima will be =2θ=60º.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2024 (01 Feb Shift 2)
A middle aged women visits a doctor complaining of irregular menses, weight loss, and recurrent diarrhea. On testing, a certain hormone comes out to be elevated. On questioning why she is experiencing such symptoms, the doctor informs that the hormone which is elevated regulates the basal metabolic rate and also water and electrolyte imbalance, which in turn affects the menstrual cycle. Which hormone is the doctor talking about?

Options:
A Cortisol
B Aldosterone
C Melatonin
D Thyroid hormones

The correct answer is:
Thyroid hormones

Solution:
Thyroid gland synthesize two hormones: thyroxine (T4) and tri iodothyronine (T3) under the influence of TSH (Thyroid Stimulating hormone). Thyroid hormones regulate Basal metabolic rate. These hormones also support the process of red blood cell formation. Thyroid hormones control the metabolism of carbohydrates, proteins and fats and maintain water and electrolyte balance. Abnormal high levels of hormone leads to a condition called hyperthyroidism which affects the body physiology.

Topic: Biology, Chemical Coordination and Integration, NEET
A milliammeter of range 0-30 mA has internal resistance of 20 Î©. The resistance to be connected in series to convert it into a voltmeter of maximum reading 3 V is

Options:
A 49Â Î©
B 80 Î©
C 40Â Î©
D 30Â Î©

The correct answer is:
80 Î©

Solution:
R=VIg-G=330Ã10-3-20=102-20=80 Î©

Topic: Physics, Current Electricity, JEE Main
A mineral has molecular formula XY2O4 . O2- ions are arranged in ccp and cations of X are present in octahedral voids. Cations of Y are equally distributed among octahedral and tetrahedral voids. The fraction of the octahedral voids occupied will be

Options:
A 16
B 18
C 14
D 12

The correct answer is:
12

Solution:
Number of O2- ions = n
Number of tetrahedral voids = 2n
Number of octahedral voids = n
Number of Y ions =n2 ,
Number of X ions =n4
Occupied octahedral voids =n4+n4=n2
So fraction of occupied octahedral voids =12 .

Topic: Chemistry, Solid State, JEE Main
A mineral is called an ore if

Options:
A Metal present in mineral is precious
B Metal can be extracted from it
C Metal can be extracted profitably from it
D Metal cannot be extracted from it

The correct answer is:
Metal can be extracted profitably from it

Solution:
Ore:
- The entire scientific and technological process used for isolation of the metal from its ores is known as metallurgy.
- Metals that occur naturally in the earth's crust are called minerals.
- Out of many minerals in which a metal may be found, only a few are viable to be used as sources of that metal. Such minerals are known as ores.
- A mineral is called an ore if a metal is profitably extracted from it

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main
A mirror in the first quadrant is in the shape of a hyperbola whose equation is $\mathrm{xy}$ $=1 .$ A light source in the second quadrant emits a beam of light that hits the mirror at the point $(2,1 / 2)$. If the reflected ray is parallel to the $\mathrm{y}$-axis the slope of the incident beam is

Options:
A $\frac{13}{8}$
B $\frac{7}{4}$
C $\frac{15}{8}$
D $2$

The correct answer is:
$\frac{15}{8}$

Solution:
Slope of tangent at $\left(2, \frac{1}{2}\right)$
$$
\begin{array}{l}
\mathrm{m}=-\frac{1}{4} \\
\tan \theta=-\frac{1}{\mathrm{y}} \\
\theta=\phi-90^{\circ} \\
\tan \theta=4
\end{array}
$$
Slope of incident ray $=\mathrm{m}$
$$
\begin{array}{l}
\left|\frac{m-\left(-\frac{1}{4}\right)}{1+m\left(-\frac{1}{4}\right)}\right|=\tan \theta=4 \\
\left|\frac{4 m+1}{4-m}\right|=4 \\
4 m+1=16-4 m \\
m=\frac{15}{8}
\end{array}
$$

Topic: Mathematics, Hyperbola, JEE Main
A miscible mixture of C 6 H 6  +  CHCl 3 can be separated by

Options:
A Sublimation
B Distillation
C Filtration
D Crystallisation

The correct answer is:
Distillation

Solution:
Miscible mixture of C 6 H 6  +  CHCl 3 can be separated by distillation method due to the significant difference in their boiling points. Boiling point of Benzene is 80 .1 o C while that of chloroform is 61 .2 o C .

Topic: Chemistry, Hydrocarbons, JEE Main
A missile is fired for maximum range with a initial velocity of $20 \mathrm{~m} / \mathrm{s}$. If $g=10 \mathrm{~m} / \mathrm{s}^2$, the range of the missile is

Options:
A $50 \mathrm{~m}$
B $60 \mathrm{~m}$
C $20 \mathrm{~m}$
D $40 \mathrm{~m}$

The correct answer is:
$40 \mathrm{~m}$

Solution:
For maximum range of projectile $\theta$ will be $45^{\circ}$ by the law of projectile motion
$$
\therefore \quad R_{\max }=\frac{u^2}{g}
$$
Given, $\quad u=20 \mathrm{~ms}^{-1}$ and $g=10 \mathrm{~ms}^{-2}$
$$
\begin{aligned}
& R_{\max }=\frac{(20)^2}{10}=\frac{400}{10} \\
& R_{\max }=40 \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Motion In Two Dimensions, JEE Main