datavorous/entrance-exam-dataset huggingface dataset:
Two forces are such that the sum of their magnitudes is $18 \mathrm{~N}$ and their resultant is $12 \mathrm{~N}$ which is perpendicular to the smaller force. Then the magnitudes of the forces are

Options:
A
$12 \mathrm{~N}, 6 \mathrm{~N}$

B
$13 \mathrm{~N}, 5 \mathrm{~N}$

C
$10 \mathrm{~N}, 8 \mathrm{~N}$

D
$16 \mathrm{~N}, 2 \mathrm{~N}$

The correct answer is:
$13 \mathrm{~N}, 5 \mathrm{~N}$

Solution:
Use $\tan \alpha=\frac{P \sin \theta}{Q+P \cos \theta} \Rightarrow \tan 90^{\circ}=\frac{P \sin \theta}{Q+P \cos \theta}=\infty$
$\therefore Q+P \cos \theta=0 \Rightarrow P \cos \theta=-Q$
$\mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta} \mathrm{R}=\sqrt{\mathrm{P}^2+\mathrm{Q}^2-2 \mathrm{Q}^2}$ or $\mathrm{R}=\sqrt{\mathrm{P}^2-\mathrm{Q}^2}=12$
$144=(\mathrm{P}+\mathrm{Q})(\mathrm{P}-\mathrm{Q})$ or $\mathrm{P}-\mathrm{Q}=144 / 18=8 \quad \therefore \mathrm{P}=13 \mathrm{~N}$ and $\mathrm{Q}=5 \mathrm{~N}$

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2002
Two forces $\bar{F}_1$ and $\bar{F}_2$ are acting on a body. One force has magnitude thrice that of the other force and the resultant of the two forces is equal to the force of larger magnitude. The angle between $\vec{F}_1$ and $\vec{F}_2$ is $\cos ^{-1}\left(\frac{1}{n}\right)$. The value of $|n|$ is _____.

Options:

The correct answer is:
6

Solution:
$\begin{aligned} & \left|\vec{F}_1\right|=\mathrm{F} \\ & \left|\overrightarrow{\mathrm{F}}_{\mathrm{R}}\right|=\left|\overrightarrow{\mathrm{F}}_2\right|=3 \mathrm{~F} \\ & \mathrm{~F}_{\mathrm{R}}^2=\mathrm{F}_1^2+\mathrm{F}_2^2+2 \mathrm{~F}_1 \mathrm{~F}_2 \cos \theta \\ & 9 \mathrm{~F}^2=\mathrm{F}^2+9 \mathrm{~F}^2+6 \mathrm{~F}^2 \cos \theta \\ & \cos \theta=-\frac{1}{6} \\ & \theta=\cos ^{-1}\left(\frac{1}{-6}\right) \\ & \mathrm{n}=-6 \\ & |\mathrm{n}|=6\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2024 (04 Apr Shift 1)
Two forces, $F_1$ and $F_2$ are acting on a body. One force is double that of the other force and the resultant is equal to the greater force. Then the angle between the two forces is

Options:
A $\cos ^{-1}(1 / 2)$
B $\cos ^{-1}(-1 / 2)$
C $\cos ^{-1}(-1 / 4)$
D $\cos ^{-1}(1 / 4)$

The correct answer is:
$\cos ^{-1}(-1 / 4)$

Solution:
Correct option is 3. $\cos ^{-1}(-1 / 4)$
Given
$F_1=F \quad F_2=2 F$ and
$F_R=2 F$
As we know
$F_R=\sqrt{F_1^2+F_2^2+2 F_1 F_2 \cos A}$
putting the above values, we get
$-F^2=4 F \cos A$
$\cos A=\frac{-1}{4}$
$A=\cos ^{-1}\left(\frac{-1}{4}\right)$

Topic: Physics, Mathematics in Physics, JEE Main
Two forces having magnitude A and A2 are perpendicular to each other. The magnitude of their resultant is:

Options:
A 5A4
B 5A2
C 5A2
D 5A22

The correct answer is:
5A2

Solution:
Using parallelogram law of vector addition, the formula can be written as

Anet=A12+A22+2A1A2cosθ

The data given is

A1→=A, A2→=A2

Since the vectors are perpendicular cos90°=0

⇒Fnet=A2+A24=5A2

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2023 (08 Apr Shift 1)
Two forces P and Q, of magnitude 2F and 3F, respectively, are at an angle θ with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle θ is:

Options:
A 120°
B 60°
C 30°
D 90°

The correct answer is:
120°

Solution:
P→=2F,Q→=3F
R→=P2+Q2+2PQcosθ
=4F2+9F2+12F2cosθ=F13+12cosθ
If Q is doubled
2R→=4F2+36F2+24F2cosθ=F40+24cosθ
413+12cosθ=40+24cosθ
12=-24cosθ
cosθ=-12
θ=120°

Topic: Physics, Mathematics in Physics, JEE Main
Two forces P and Q, of magnitude 2F and 3F, respectively, are at an angle θ with each other. If the force Q is doubled, then their resultant also gets doubled. Then, the angle θ is:

Options:
A 120°
B 60°
C 30°
D 90°

The correct answer is:
120°

Solution:
P→=2F,Q→=3F
R→=P2+Q2+2PQcosθ
=4F2+9F2+12F2cosθ=F13+12cosθ
If Q is doubled
2R→=4F2+36F2+24F2cosθ=F40+24cosθ
413+12cosθ=40+24cosθ
12=-24cosθ
cosθ=-12
θ=120°

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Two forces with equal magnitudes $F$ act on a body and the magnitude of the resultant force is $F / 3$. The angle between the two forces is

Options:
A $\cos ^{-1}\left(-\frac{17}{18}\right)$
B $\cos ^{-1}\left(-\frac{1}{3}\right)$
C $\cos ^{-1}\left(\frac{2}{3}\right)$
D $\cos ^{-1}\left(\frac{8}{9}\right)$

The correct answer is:
$\cos ^{-1}\left(-\frac{17}{18}\right)$

Solution:
$\begin{aligned}
& F_{m t}^2=F_1^2+F_2^2+2 F_1 F_2 \cos \theta \\
& \Rightarrow\left(\frac{F}{3}\right)^2=F^2+F^2+2 F^2 \cos \theta \Rightarrow \cos \theta=\left(-\frac{17}{18}\right)
\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main
Two free parallel wires carrying currents in opposite direction

Options:
A Attract each other
B Repel each other
C Neither attract nor repel
D Get rotated to be perpendicular to each other

The correct answers are:
Repel each other, Get rotated to be perpendicular to each other

Solution:
If two parallel wires carry current in opposite directions, they repel each other whereas if two parallel wires carry current in same direction, they attract each other.

Topic: Physics, Magnetic Effects of Current, JEE Main
Two friends $\mathrm{A}$ and $\mathrm{B}$ are $30 \mathrm{~km}$ apart and they start simultaneously on motorcycles to meet each other. The speed of A is 3 times that of B. The distance between them decreases at the rate of $2 \mathrm{~km}$ per minute. Ten minutes after they start, A's vehicle breaks down and A stops and waits for $\mathrm{B}$ to arrive. After how much time (in minutes) A started riding, does B meet $\mathrm{A}$ ?

Options:
A 15
B 20
C 25
D 30

The correct answer is:
30

Solution:
Speed of $\mathrm{B}=\mathrm{V} \mathrm{km} / \mathrm{hr}$
Speed of $\mathrm{A}=3 \mathrm{~V} \mathrm{~km} / \mathrm{hr}$
Given $4 \mathrm{~V}=2 \times 60 \mathrm{~km} / \mathrm{hr}$
$\Rightarrow \mathrm{V}=30 \mathrm{~km} / \mathrm{hr}$
Distance covered by then after $10 \mathrm{~min}=2 \times 10=20 \mathrm{~km}$
so remaining distance $=(30-20) \mathrm{km}=10 \mathrm{~km}$
Time taken by $\mathrm{B}$ to cover $10 \mathrm{~km}=\frac{10}{30 / 60}=20 \mathrm{~min}$.
Total time $=20+10=30 \mathrm{~min}$

Topic: Mathematics, Basic of Mathematics, KVPY, KVPY 2011 (SA)
Two full turns of the circular scale of a screw gauge cover a distance of $1 \mathrm{~mm}$ on its main scale. The total number of divisions on the circular scale is 50 . Further, it is found that the screw gauge has a zero error of $-0.03 \mathrm{~mm}$ while measuring the diameter of a thin wire, a student notes the main scale reading of $3 \mathrm{~mm}$ and the number of circular scale divisions in line with the main scale as 35 . The diameter of the wire is

Options:
A
$3.32 \mathrm{~mm}$

B
$3.73 \mathrm{~mm}$

C
$3.67 \mathrm{~mm}$

D
$3.38 \mathrm{~mm}$

The correct answer is:
$3.38 \mathrm{~mm}$

Solution:
Diameter $=$ M.S.R. $+$ C.S.R $\times$ L.C. $+$ Z.E. $=3+35 \times(0.5 / 50)+0.03=3.38 \mathrm{~mm}$

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2008
Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of -0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions, in line with the main scale, as 35. The diameter of the wire is

Options:
A 3.38 mm
B 3.32 mm
C 3.73 mm
D 3.67 mm

The correct answer is:
3.38 mm

Solution:
Least count of the screw gauge,
=0.5 mm50=0.01 mm
Main scale reading =3 mm
Vernier scale reading=35
â´ Observed reading =3+0.35=3.35 mm
Zero error=-0.03 mm
â´ Actual diameter of the wire=3.35--0.03=3.38 mm

Topic: Physics, Units and Dimensions, JEE Main
Two functions $f: R \rightarrow R, g: R \rightarrow R$ are defined as follows $f(x)= \begin{cases}0, & x \text { is rational } \\ 1, & x \text { is irrational }\end{cases}$ $g(x)=\left\{\begin{array}{cc}-1, & x \text { is rational } \\ 0, & x \text { is irrational }\end{array}\right.$ Then, $(f \circ g)(\pi)+(g \circ f)(e)$ is equal to

Options:
A $0$
B $-1$
C $2$
D $1$

The correct answer is:
$-1$

Solution:
$\begin{aligned} & (g \circ f)(e)=g(f(e))=g(1)=-1 \\ & (f \circ g)(\pi)=f(g(\pi))=f(-1)=0 \\ & \therefore \quad(f \circ g)(\pi)+(g \circ f)(e)=0-1=-1 \\ & \end{aligned}$

Topic: Mathematics, Functions, JEE Main
Two gaseous equilibria $\mathrm{SO}_{2(g)}+\frac{1}{2} \mathrm{O}_{2(g)} \rightleftharpoons \mathrm{SO}_{3(g)}$ and $2 \mathrm{SO}_{3(g)} \rightleftharpoons \mathrm{SO}_{2(g)}+\mathrm{O}_{2(g)}$ have equilibrium constants $K_1$ and $K_2$ respectively at ${ }^{298} \mathrm{~K}$. Which of the following relationships between ${ }^{K_1}$ and $K_{2 \text { is correct }}$

Options:
A $K_1=K_2$
B $K_2=K_1^2$
C $K_2=\frac{1}{K_1^2}$
D $K_2=\frac{1}{K_1}$

The correct answer is:
$K_2=\frac{1}{K_1^2}$

Solution:
$K_1=\frac{\left[\mathrm{SO}_3\right]}{\left[\mathrm{SO}_2\right]\left[\mathrm{O}_2\right]^{1 / 2}}$ and $\mathrm{K}_2=\frac{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{SO}_3\right]^2} ; K_2=\frac{1}{\mathrm{~K}_1^2}$.

Topic: Chemistry, Chemical Equilibrium, JEE Main
Two gases $A$ and $B$ are contained in two separate, but otherwise identical containers. Gas $A$ consists of monatomic molecules, each with atomic mass of $4 u$ whereas Gas $B$ consists of rigid diatomic molecules, each with atomic mass of $20 u$. If gas $A$ is kept at $27^{\circ} \mathrm{C}$, at what temperature should gas $B$ be kept so that both have the same rms speed?

Options:
A $27^{\circ} \mathrm{C}$
B $54^{\circ} \mathrm{C}$
C $270^{\circ} \mathrm{C}$
D $62^{\circ} \mathrm{C}$

The correct answer is:
$270^{\circ} \mathrm{C}$

Solution:
For monatomic gas $A$,
atomic mass of molecules of gas $A, m_A=4 u$
and temperature of gas, $T_A=27^{\circ} \mathrm{C}$
For diatomic gas $B$,
atomic mass of molecules of gas $B$,
$m_B=2 \times 20 \mathrm{u}=40 \mathrm{u}$
Temperature of gas, $B=\mathrm{T}_B$
Since, $r m s$ speed of the molecules of gas, $A=\mathrm{rms}$ speed of molecules of gas, $B$
$$
\begin{aligned}
\left(v_{\mathrm{rms}}\right)_A & =\left(v_{\mathrm{rms}}\right)_B \\
\Rightarrow \quad \sqrt{\frac{3 R T_A}{m_A}} & =\sqrt{\frac{3 R T_B}{m_B}} \\
\Rightarrow \quad \frac{\mathrm{T}_A}{m_A} & =\frac{\mathrm{T}_B}{m_B} \Rightarrow \frac{27}{4 u}=\frac{\mathrm{T}_B}{40 \mathrm{u}} \\
\mathrm{T}_B & =270^{\circ} \mathrm{C}
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
Two gases $A$ and $B$ having same pressure $P$, volume $V$ and absolute temperature $T$ are mixed. If the mixture has the volume and temperature as $V$ and $T$ respectively, then the pressure of the mixture is

Options:
A $2 P$
B $P$
C $\frac{P}{2}$
D $4 P$

The correct answer is:
$2 P$

Solution:
The total pressure exerted by a mixture of non-reacting gases occupying a vessel is equal to the sum of the individual pressures which each gas exert if it alone occupied the same volume at a given temperature.
For two gases
$P=R_1+P_2=P+P=2 P$

Topic: Physics, Kinetic Theory of Gases, JEE Main
Two gases - argon (atomic radius 0.07nm, atomic weight 40 ) and xenon (atomic radius 0.1nm, atomic weight 140 ) have the same number density and are at the same temperature. The ratio of their respective mean free times is closest to:

Options:
A 3.67
B 1.83
C 2.3
D 4.67

The correct answer is:
1.83

Solution:
Mean free time =12nπd2vrms
tArtXe=dXe2dAr2×m1m2
=0.10.072×40140
=1072×27=1.09
closest value is 1.83

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2020 (09 Jan Shift 2)
Two gases occupy two containers A and B; the gas in A, of volume 0.10 m3, exerts a pressure of 1.40 MPa and that in B, of volume 0.15 m3, exerts a pressure 0.7 MPa. The two containers are joined by a tube of negligible volume and the gases are allowed to intermingle. Then if the temperature remains constant, the final pressure in the containers will be (in MPa)

Options:
A 0.90
B 0.98
C 1.40
D 2.10

The correct answer is:
0.98

Solution:
As the quantity of gas remains constant μ A + μ B = μ
PAVART+PBVBRT=PVA+VBRT
P=PAVA+PBVBVA+VB=1.4×0.1+0.7×0.150.1+0.15
⇒ P=0.98 MPa

Topic: Physics, Thermodynamics, JEE Main
Two gases when mixed give white dense fumes, the gases are

Options:
A $\mathrm{NH}_3$ and $\mathrm{SO}_2$
B $\mathrm{SO}_2$ and steam
C $\mathrm{NH}_3$ and $\mathrm{HCl}$
D $\mathrm{NH}_3$ and $\mathrm{N}_2 \mathrm{O}$

The correct answer is:
$\mathrm{NH}_3$ and $\mathrm{HCl}$

Solution:
Two gases when mixed give white dense fumes, the gases are \(\mathrm{NH}_3\) and \(\mathrm{HCl}\).
It forms \(\mathrm{NH}_4 \mathrm{Cl}\) which gives white dense fumes.
\(\underset{\qquad \qquad \qquad \text{dense white fumes}}{\mathrm{NH}_3+\mathrm{HCl} \rightarrow \mathrm{NH}_4 \mathrm{Cl}}\)

Topic: Chemistry, Practical Chemistry, JEE Main
Two glass bulbs X and Y are connected by a very small tube having a stop-cock. Bulb X has a volume of 100 cm3 and contained the gas; while bulb Y was empty. On opening the stop-cock, the pressure fell down to 60%. The volume of the bulb Y must be

Options:
A 132.32 cm3
B 124 cm3
C 66.66 cm3
D 150 cm3

The correct answer is:
66.66 cm3

Solution:
P1=1 atm
P2=1 atmÃ60100=0.60 atm
V1=100 cm3
V2=?
At constant temperature, P1V1=P2V2
So V2=P1V1P2
-1 atmÃ100 cm30.60 atm
=166.66 cm3
Hence, the volume of bulb Y=(166.66-100) cm3
=66.66 cm3

Topic: Chemistry, States of Matter, JEE Main
Two gold pieces, each of mass 0.035 g, are placed in a box of mass 2.3 g. The total mass of the box with gold pieces is

Options:
A 2.3 g
B 2.4 g
C 2.37 g
D 2.370 g

The correct answer is:
2.4 g

Solution:
(2.3+0.035+0.035) g = 2.37 g
But we have to retain only one decimal place.
So, the total mass is 2.4 g

Topic: Physics, Units and Dimensions, JEE Main
Two guns A and B can fire bullets at speeds 1 km/s and 2 km/s respectively. From a point on a horizontal ground, they are fired in all possible directions. The ratio of maximum areas covered by the bullets fired by the two guns, on the ground is:

Options:
A 1:16
B 1:2
C 1:4
D 1:8

The correct answer is:
1:16

Solution:
A1A2=πRmax12πRmax22=u14u24=116
∵ max. Range=u2g

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2019 (10 Jan Shift 1)
Two half cells have reduction potentials â0.76 V and â 0.13 V respectively. A galvanic cell is made from these two half cells. Which of the following statements is correct?

Options:
A Electrode of half-cell potential â0.76 V acts as cathode
B Electrode of half-cell potential â0.76 V acts as anode
C Electrode of half-cell potential â0.13 V acts as anode
D Electrode of half-cell potential â 0.76 V acts as positive electrode and â0.13 V as negative electrode

The correct answer is:
Electrode of half-cell potential â0.76 V acts as anode

Solution:
According to the electrochemical series, the electrode with high reduction potential will act as a cathode, and the electrode with more negative reduction potential acts as the anode.

Topic: Chemistry, Electrochemistry, JEE Main
Two harmonic travelling waves are described by the equations y1=asinkx-ωt and y2=asin-kx+ωt+ϕ The amplitude of the superposed wave is

Options:
A 2acosϕ2
B 2asinϕ
C 2acosϕ
D 2asinϕ2

The correct answer is:
2asinϕ2

Solution:
y1=asinkx-ωt ...1

y2=asin-kx+ωt+ϕ ...2

On adding 1 and 2,

y1+y2=2asinkx-ωt+-kx+ωt+ϕ2

coskx-ωt+kx-ωt-ϕ2

=2asinϕ2cos2(kx+ωt)-ϕ2

Amplitude, Aϕ is a function of phase difference ϕ between constituent of waves

∴ ϕ=2asinϕ2

Topic: Physics, Waves and Sound, JEE Main
Two hypothetical planets of masses $\mathrm{m}_1$ and $\mathrm{m}_2$ are at rest when they are infinite distance apart. Because of the gravitational force they move towards each other along the line joining their centres. What is their speed when their separation is ' $d$ '?
(Speed of $\mathrm{m}_1$ is $\mathrm{v}_1$ and that of $\mathrm{m}_2$ is $\mathrm{v}_2$ )

Options:
A
$$
\text { } v_1=v_2
$$

B
$$
\begin{aligned}
&\mathrm{v}_1=\mathrm{m}_2 \sqrt{\left.\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right.}\right)} \\
&\mathrm{v}_2=\mathrm{m}_1 \sqrt{\left.\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right.}\right)}
\end{aligned}
$$

C
$$
\begin{aligned}
&\mathrm{v}_1=\mathrm{m}_1 \sqrt{\left.\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right.}\right)} \\
&\mathrm{v}_2=\mathrm{m}_2 \sqrt{\left.\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right.}\right)}
\end{aligned}
$$

D
$$
\begin{aligned}
&\mathrm{v}_1=\mathrm{m}_2 \sqrt{\frac{2 \mathrm{G}}{\mathrm{m}_1}} \\
&\mathrm{v}_2=\mathrm{m}_2 \sqrt{\frac{2 \mathrm{G}}{\mathrm{m}_2}}
\end{aligned}
$$

The correct answer is:
$$
\begin{aligned}
&\mathrm{v}_1=\mathrm{m}_2 \sqrt{\left.\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right.}\right)} \\
&\mathrm{v}_2=\mathrm{m}_1 \sqrt{\left.\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right.}\right)}
\end{aligned}
$$

Solution:
We choose reference point, infinity, where total energy of the system is zero.
So, initial energy of the system $=0$
Final energy
$$
=\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_1^2+\frac{1}{2} \mathrm{~m}_2 \mathrm{v}_2^2-\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}}
$$
From conservation of energy, Initial energy $=$ Final energy
$$
\therefore 0=\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_1{ }^2+\frac{1}{2} \mathrm{~m}_2 \mathrm{v}_2{ }^2-\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}}
$$
or $\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_1^2+\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_2^2=\frac{\mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}} \ldots(1)$
By conservation of linear momentum
$$
\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2=0
$$
or $\frac{\mathrm{v}_1}{\mathrm{v}_2}=-\frac{\mathrm{m}_2}{\mathrm{~m}_1} \Rightarrow \mathrm{v}_2=-\frac{\mathrm{m}_1}{\mathrm{~m}_2} \mathrm{v}_1$
Putting value of $\mathrm{v}_2$ in equation (1), we get
$$
\mathrm{m}_1 \mathrm{v}_1^2+\mathrm{m}_2\left(-\frac{\mathrm{m}_1 \mathrm{v}_1}{\mathrm{~m}_2}\right)^2=\frac{2 \mathrm{Gm}_1 \mathrm{~m}_2}{\mathrm{~d}}
$$
$$
\begin{aligned}
&\frac{m_1 m_2 v_1^2+m_1^2 v_1^2}{m_2}=\frac{2 \mathrm{Gm}_1 m_2}{\mathrm{~d}} \\
&\mathrm{v}_1=\sqrt{\frac{2 \mathrm{Gm}_2^2}{\mathrm{~d}\left(\mathrm{~m}_1+\mathrm{m}_2\right)}}=\mathrm{m}_2 \sqrt{\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right)}} \\
&\text { Similarly } \mathrm{v}_2=-\mathrm{m}_1 \sqrt{\frac{2 \mathrm{G}}{\mathrm{d}\left(\mathrm{m}_1+\mathrm{m}_2\right)}}
\end{aligned}
$$

Topic: Physics, Gravitation, JEE Main, JEE Main 2014 (12 Apr Online)
Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, T1 and T2 . The temperature of the hot reservoir of the first engine is T1 and the temperature of the cold reservoir of the second engine is T2 . T is temperature of the sink of first engine which is also the source for the second engine. How is T related to T1 and T2 , if both the engines perform equal amount of work?

Options:
A T=2T1T2T1+T2
B T=T1+T22
C T=T1T2
D T=0

The correct answer is:
T=T1+T22

Solution:
Q1: Heat input to Ist engine
Q: Heat rejected from Ist engine
Q2: Heat rejected from IInd engine
Work done by Ist engine = work done by IInd engine.
Q-Q2=Q1-Q
2Q=Q1+Q2
2=T1T+T2T
T=T1+T22

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, T1 and T2 . The temperature of the hot reservoir of the first engine is T1 and the temperature of the cold reservoir of the second engine is T2 . T is temperature of the sink of first engine which is also the source for the second engine. How is T related to T1 and T2 , if both the engines perform equal amount of work?

Options:
A T=2T1T2T1+T2
B T=T1+T22
C T=T1T2
D T=0

The correct answer is:
T=T1+T22

Solution:
Q1: Heat input to Ist engine
Q: Heat rejected from Ist engine
Q2: Heat rejected from IInd engine
Work done by Ist engine = work done by IInd engine.
Q-Q2=Q1-Q
2Q=Q1+Q2
2=T1T+T2T
T=T1+T22

Topic: Physics, Thermodynamics, JEE Main
Two ideal monoatomic and diatomic gases are mixed with one another to form an ideal gas mixture. The equation of the adiabatic process of the mixture is PVγ= constant, where γ=117 If n1 and n2 are the number of moles of the monoatomic and diatomic gases in the mixture respectively, find the ratio n1n2

Options:

The correct answer is:
3

Solution:
γmix=n1Cp1+n2Cp2n1Cv1+n2Cv2
117=n152R+n272Rn132R+n252R
n1n2=3

Topic: Physics, Thermodynamics, JEE Main
Two identical antennas mounted on identical towers are separated from each other by a distance of 45 km. What should nearly be the minimum height of receiving antenna to receive the signals in line of sight ? (Assume radius of earth is 6400 km)

Options:
A 19.77 m
B 39.55 m
C 79.1 m
D 158.2 m

The correct answer is:
39.55 m

Solution:
D=22Rh

h=D28R=4528×6400 km≅39.55 m

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Two identical balls A and B, each of mass 0.1 kg, are attached to two identical massless springs. The spring-mass system is constrained to move inside a rigid smooth pipe bent in the form of a circle as shown in the figure. The pipe is fixed in a horizontal plane. The centres of the balls can move in a circle of radius 0.06 m. Each spring has a natural length of 0.06Ï m and spring constant 0.1 N m-1, Initially, both the balls are displaced by an angle Î¸ =Ï6 rad with respect to the diameter PQ of the circle (as shown in the figure) and released from rest.
The frequency of oscillation of balls is

Options:
A 1 2 π Hz
B 1 3 π Hz
C 1 Ï Hz
D 1 4 π Hz

The correct answer is:
1 Ï Hz

Solution:
Given, Mass of each block A and B, m = 0.1 kg
Radius of circle, R=0.06 m
The natural length of spring l 0 = 0.06 Ï = Ï R (Half circle) and spring constant,
k=0.1 N m-1
In the stretched position elongation in each spring
x = R Î¸
Let us draw FBD of A
Spring in lower side is stretched by 2x and on upper side compressed by 2x. Therefore, a force 2kx on each block would be exerted by each spring.
Hence, a restoring force, F = 4kx will act on A in the direction shown in figure.
Restoring torque of this force about origin
Ï = - F. R = - 4 kx R = - 4 kR Î¸ R
or Ï = - 4 kR 2 Â· Î¸ ..... (i)
Since, Ï â - Î¸ , each ball executes angular SHM about origin O.
Equation. (i) can be written as
I Î± = - 4 kR 2 Î¸
or mR 2 Î± = - 4 kR 2 Î¸
or Î± = - 4 k m Î¸
Frequency of oscillation, f = 1 2 Ï acceleration displacement
= 1 2 Ï Î± Î¸
f = 1 2 Ï 4 k m
Substituting the values, we have
f = 1 2 Ï 4 Ã 0.1 0.1 = 1 Ï Hz

Topic: Physics, Oscillations, JEE Main
Two identical balls P and Q are projected with same speeds in a vertical plane from the same point O making projection angles with horizontal 30° and 60°, respectively and they fall directly on the plane AB at points P′ and Q′ respectively. Which of the following statement is true about distances as given in options?

Options:
A APâ²>AQâ²
B AP′<AQ′
C AP′â¤AQ′
D As there are complimentary projection angles AP′=AQ′

The correct answer is:
AP′>AQ′

Solution:
Given,
Projection angle of ball P=30°
Projection angle of ball Q=60°
According to the question,
For complimentary angles ranges, ON is same for P and Q as O and N are on same horizontal plane. From figure AP′>AQ′

Topic: Physics, Motion In Two Dimensions, JEE Main
Two identical bar magnets are fixed with their centres at a distance $d$ apart. A stationary charge $Q$ is placed at $P$ in between the gap of the two magnets at a distance $D$ from the centre $O$ as shown in the figure
The force on the charge $Q$ is

Options:
A zero
B directed along $O P$
C directed along $P O$
D directed perpendicular to the plane of paper

The correct answer is:
zero

Solution:
The force of attraction on $Q$ by the north pole of first magnet is $F$. The force of attraction on the same charge by $n$ the north pole of another magnet must be opposite to that of the first magnet. Thus, they concel the effect of one other and hence force on the charge $Q$ is zero.

Topic: Physics, Magnetic Properties of Matter, JEE Main