datavorous/entrance-exam-dataset huggingface dataset:
Two spherical bodies of mass $M$ and $5 M \&$ radii $R \& 2 R$ respectively are released in free space with initial separation between their centres equal to $12 \mathrm{R}$. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is

Options:
A
$2.5 \mathrm{R}$

B
$4.5 \mathrm{R}$

C
$7.5 \mathrm{R}$

D
$1.5 \mathrm{R}$

The correct answer is:
$7.5 \mathrm{R}$

Solution:
Distance between the surface of the spherical bodies $=12 R-R-2 R=9 R$
Force $\propto$ Mass, $\quad$ Acceleration $\propto$ Mass, $\quad$ Distance $\propto$ Acceleration
$$
\begin{aligned}
& \Rightarrow \frac{\mathrm{a}_1}{\mathrm{a}_2}=\frac{\mathrm{M}}{\mathrm{SM}}=\frac{1}{5} \Rightarrow \frac{\mathrm{S}_1}{\mathrm{~S}_2}=\frac{1}{5} \Rightarrow \mathrm{S}_2=5 \mathrm{~S}_1 \\
& \mathrm{~S}_1+\mathrm{S}_2=9 \Rightarrow 6 \mathrm{~S}_1=9 \Rightarrow \mathrm{S}_1=\frac{9}{6}=1.5,
\end{aligned}
$$
Note: Maximum distance will be travelled by smaller bodies due to the greater acceleration caused by the same gravitational force

Topic: Physics, Gravitation, JEE Main, JEE Main 2003
Two spherical bodies of mass M and 5M and radii R and 2R released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is:

Options:
A 2.5R
B 4.5R
C 7.5R
D 1.5R

The correct answer is:
7.5R

Solution:
Initial position,
For collision, both bodies have to travel distance =12R-3R=9R
Since the bodies move under the mutual forces, centre of mass will remains stationary,
m1x1=m2x2
Mx=5M(9R-x)
x=45R-5x
6x=45R
x=45RR=7.5R

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Two spherical bodies of mass M and 5M and radii R and 2R released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is:

Options:
A 2.5R
B 4.5R
C 7.5R
D 1.5R

The correct answer is:
7.5R

Solution:
Initial position,
For collision, both bodies have to travel distance =12R-3R=9R
Since the bodies move under the mutual forces, centre of mass will remains stationary,
m1x1=m2x2
Mx=5M(9R-x)
x=45R-5x
6x=45R
x=45RR=7.5R

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2015 (Phase 1)
Two spherical bodies of the same mass M, are moving with velocities v1and v2 respectively, along the same line. These bodies collide perfectly inelastically and head-on. The loss in kinetic energy is

Options:
A 12M(v1-v2)
B 12M(v12-v22)
C 14Mv1-v22
D 2M(v12-v22)

The correct answer is:
14Mv1-v22

Solution:
Loss of kinetic energy =12 m1m2m1+m2v1-v22
=12ÃMÃMM+Mv1-v22
=MâM22Mv1-v22
=M4(v1-v2)2

Topic: Physics, Work Power Energy, JEE Main
Two spherical conductor $B$ and $C$ having equal radii and carrying equal charges in them repel each other with a force $F$ when kept apart at some distance. A third spherical conductor having same radius as that of $B$ but uncharged brought in contact with $B$, then brought in contact with $\mathrm{C}$ and finally removed away from both. The new force of repulsion, between $B$ and $C$ is

Options:
A
$\mathrm{F} / 4$

B
$3 F / 4$

C
$\mathrm{F} / 8$

D
$3 \mathrm{~F} / 8$.

The correct answer is:
$3 \mathrm{~F} / 8$.

Solution:
$F^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{(q / 2)(3 q / 4)}{d^2}=\frac{3 F}{8}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2004
Two spherical conductors $A$ and $B$ of radii $1 \mathrm{~mm}$ and $2 \mathrm{~mm}$ are separated by a distance of $5 \mathrm{~cm}$ and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surface of spheres $A$ and $B$ is

Options:
A
$1: 4$

B
$4: 1$

C
$1: 2$

D
$2: 1$

The correct answer is:
$2: 1$

Solution:
$\frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{E}_{\mathrm{B}}}=\frac{\mathrm{r}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{A}}}=\frac{2}{1}$

Topic: Physics, Electrostatics, JEE Main
Two spherical conductors $A$ and $B$ of radii $1 \mathrm{~mm}$ and $2 \mathrm{~mm}$ are separated by a distance of $5 \mathrm{~cm}$ and are uniformly charged. If the spheres are connected by a conducting wire then in equilibrium condition, the ratio of the magnitude of the electric fields at the surface of spheres $A$ and $B$ is

Options:
A
$1: 4$

B
$4: 1$

C
$1: 2$

D
$2: 1$

The correct answer is:
$2: 1$

Solution:
$\frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{E}_{\mathrm{B}}}=\frac{\mathrm{r}_{\mathrm{B}}}{\mathrm{r}_{\mathrm{A}}}=\frac{2}{1}$

Topic: Physics, Electrostatics, JEE Main, JEE Main 2006
Two spherical conductors A and B of radii R and 2R respectively, are separated by a large distance. If some charge is given to both the spheres and later they are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitude of the electric fields at the surface of spheres A and B is

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

The correct answer is:
2 : 1

Solution:
When the spherical conductors are connected by a conducting wire, charge is redistributed and the spheres attain a common potential V.
â´ Intensity EA=14ÏÎµ0QARA2
or EA=1ÃCAV4ÏÎµ0RA2=4ÏÎµ0RAV4ÏÎµ0RA2=VRA
Similarly EB=VRB
â´ EAEB=RBRA=21.

Topic: Physics, Electrostatics, JEE Main
Two spherical conductors of radii 4 m and 5 m are charged to the same potential. If Ï1 and Ï2 be the respective values of the surface density of charge on the two conductors, then the ratio Ï1Ï2 is

Options:
A 2516
B 1625
C 54
D 45

The correct answer is:
54

Solution:
σ1σ2=Q1Q2× R22R12, V=σ R
V=same, σ∝ 1R

σ1σ2=54

Topic: Physics, Electrostatics, JEE Main
Two spherical soap bubbles of radii $a$ and $b$ in vacuum coaleasce under isothermal conditions. The resulting bubbles has a radius given by

Options:
A $\frac{(a+b)}{2}$
B $\frac{a b}{a+b}$
C $\sqrt{a^2+b^2}$
D $a+b$

The correct answer is:
$\sqrt{a^2+b^2}$

Solution:
Since the bubbles coalesce in vacuum and there is no change in temperature, hence its surface energy does not change. This means that the surface area remains unchanged. Hence
$$
4 \pi a^2+4 \pi b^2=4 \pi R^2 \text { or } R=\sqrt{a^2+b^2}
$$

Topic: Physics, Mechanical Properties of Fluids, AIIMS, AIIMS 2010
Two spherical soap bubbles of radii a and b in vacuum coalesce under isothermal conditions. The resulting bubble has a radius given by

Options:
A a+b2
B aba+b
C a2+b2
D a+b

The correct answer is:
a2+b2

Solution:
Since, the bubbles coalesce in vacuum and there is no change in temperature, hence its surface energy does not change. This means that the surface area remains unchanged. Hence,
4 π a2+4 π b2=4π R2 or R=a2+b2

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Two spherical soap bubbles of radii $r_1$ and $r_2$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to:

Options:
A $\frac{r_1+r_2}{2}$
B $\frac{r_1 r_2}{r_1+r_2}$
C $\sqrt{r_1 r_2}$
D $\sqrt{r_1^2+r_2^2}$

The correct answer is:
$\sqrt{r_1^2+r_2^2}$

Solution:
Excess of pressure, inside the first bubble
$p_1=\frac{4 T}{r_1}$
Similarly, $\quad p_2=\frac{4 T}{r_2}$
Let the radius of the large bubble be $R$.
Then, excess of pressure inside the large bubble,
$p=\frac{4 T}{R}$
Under isothermal condition, temperature remains constant.
So, $\quad P V=p_1 V_1+p_2 V_2$
$\frac{4 T}{R}\left(\frac{4}{3} \pi R^3\right)=\frac{4 T}{r_1}\left(\frac{4}{3} \pi r_1^3\right)+\frac{4 T}{r_2}\left(\frac{4}{3} \pi r_2^3\right)$
$R^2=r_1^2+r_2^2$
$\Rightarrow \quad R=\sqrt{r_1^2+r_2^2}$

Topic: Physics, Mechanical Properties of Fluids, TS EAMCET, TS EAMCET 2003
Two spherical soap bubbles of radii $r_1$ and $r_2$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to:

Options:
A $\frac{r_1+r_2}{2}$
B $\frac{r_1 r_2}{r_1+r_2}$
C $\sqrt{r_1 r_2}$
D $\sqrt{r_1^2+r_2^2}$

The correct answer is:
$\sqrt{r_1^2+r_2^2}$

Solution:
Excess of pressure, inside the first bubble
$p_1=\frac{4 T}{r_1}$
Similarly, $\quad p_2=\frac{4 T}{r_2}$
Let the radius of the large bubble be $R$.
Then, excess of pressure inside the large bubble,
$p=\frac{4 T}{R}$
Under isothermal condition, temperature remains constant.
So, $\quad P V=p_1 V_1+p_2 V_2$
$\frac{4 T}{R}\left(\frac{4}{3} \pi R^3\right)=\frac{4 T}{r_1}\left(\frac{4}{3} \pi r_1^3\right)+\frac{4 T}{r_2}\left(\frac{4}{3} \pi r_2^3\right)$
$R^2=r_1^2+r_2^2$
$\Rightarrow \quad R=\sqrt{r_1^2+r_2^2}$

Topic: Physics, Mechanical Properties of Fluids, AP EAMCET, AP EAMCET 2003
Two spherical soap bubbles of radii r1 and r2 in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to:

Options:
A r1r2r1+r2
B r1r2
C r12+r22
D r1+r22

The correct answer is:
r12+r22

Solution:
Number of moles is conserved.

n1+n2=n3

P1 V1+P2 V2=P3 V

4 Sr143πr13+4 Sr243πr23=4 Sr343πr33

r12+r22=r32

r3=r12+r22

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Two spherical soap bubbles of radii r1 and r2 in vacuum combine under isothermal conditions. The resulting bubble has radius equal to

Options:
A r1+r22
B r1 r2r1+r2
C r1r2
D r12+r22

The correct answer is:
r12+r22

Solution:
Excess of pressure, inside the first bubble p1=4Tr1
Similarly, p2=4Tr2
Let the radius of the large bubble be R. then, excess of pressure inside the large bubble p=4TR
Under isothermal condition, temperature remains constant.
So, pV=p1 V1+p2 V2
4TR43πr3=4Tr143πr13+4Tr2 43πr23
R2=r12+r22
⟹ R=r12+r22

Topic: Physics, Mechanical Properties of Fluids, NEET
Two spring of force constant $k_{1}$ and $k_{2}$ are configured as the figure given below
The angular frequency of this configuration is

Options:
A $\frac{k_{1}+k_{2}}{m}$
B $\sqrt{\frac{k_{1}+k_{2}}{m}}$
C $\sqrt{\frac{k_{1}}{k_{2} m}}$
D $\frac{k_{2} m}{k_{1}}$

The correct answer is:
$\sqrt{\frac{k_{1}+k_{2}}{m}}$

Solution:
The equation of motion of this system about mean position is given by
$$
\begin{aligned}
& F &=k_{1} x+k_{2} x & \\
\Rightarrow & m \frac{d^{2} x}{d t^{2}} &=k_{1} x+k_{2} x \quad \Rightarrow \frac{d^{2} x}{d t^{2}}=x\left(\frac{k_{1}+k_{2}}{m}\right) \\
\because & a &=\frac{d^{2} x}{d t^{2}}=\omega^{2} x \\
\therefore & \omega^{2} &=\frac{k_{1}+k_{2}}{m} \text { or } \omega=\sqrt{\frac{k_{1}+k_{2}}{m}}
\end{aligned}
$$

Topic: Physics, Oscillations, JEE Main
Two springs of force constants $300 \mathrm{~N} / \mathrm{m}$ (Spring A) and $400 \mathrm{~N} / \mathrm{m}$ (Spring B) are joined together in series. The combination is compressed by $8.75 \mathrm{~cm}$. The ratio of energy stored in $\mathrm{A}$ and $\mathrm{B}$ is $\frac{E_A}{E_B}$. Then $\frac{E_A}{E_B}$ is equal to:

Options:
A
$\frac{4}{3}$

B
$\frac{16}{9}$

C
$\frac{3}{4}$

D
$\frac{9}{16}$

The correct answer is:
$\frac{4}{3}$

Solution:
Given : $\mathrm{k}_{\mathrm{A}}=300 \mathrm{~N} / \mathrm{m}, \mathrm{k}_{\mathrm{B}}=400 \mathrm{~N} / \mathrm{m}$
Let when the combination of springs is compressed by force F. Spring A is compressed by $x$. Therefore compression in spring $\mathrm{B}$
$$
\begin{aligned}
& x_B=(8.75-x) \mathrm{cm} \\
& \mathrm{F}=300 \times \mathrm{x}=400(8.75-\mathrm{x})
\end{aligned}
$$
Solving we get, $x=5 \mathrm{~cm}$
$$
\mathrm{x}_{\mathrm{B}}=8.75-5=3.75 \mathrm{~cm}
$$
$\frac{\mathrm{E}_{\mathrm{A}}}{\mathrm{E}_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{k}_{\mathrm{A}}\left(\mathrm{x}_{\mathrm{A}}\right)^2}{\frac{1}{2} \mathrm{k}_{\mathrm{B}}\left(\mathrm{x}_{\mathrm{B}}\right)^2}=\frac{300 \times(5)^2}{400 \times(3.75)^2}=\frac{4}{3}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2013 (09 Apr Online)
Two springs, of force constants $k_1$ and $k_2$, are connected to a mass $\mathrm{m}$ as shown. The frequency of oscillation of the mass is $\mathrm{f}$. If both $k_1$ and $k_2$ are made four times their original values, the frequency of oscillation becomes

Options:
A
$\mathrm{f / 2}$

B
$\mathrm{f / 4}$

C
$4 \mathrm{f}$

D
$2 \mathrm{f}$

The correct answer is:
$2 \mathrm{f}$

Solution:
$\begin{aligned} f & =\frac{1}{2 \pi} \sqrt{\frac{k_1+k_2}{m}} \\ f^{\prime} & =\frac{1}{2 \pi} 2 \sqrt{\frac{k_1+k_2}{m}}=2 f\end{aligned}$

Topic: Physics, Oscillations, JEE Main, JEE Main 2007
Two springs of spring constant $k_1$ and $k_2$ are joined in series. The effective spring constant of the combination is given by:

Options:
A $\sqrt{k_1 k_2}$
B $\left(k_1+k_2\right) / 2$
C $k_1+k_2$
D $k_1 k_2 /\left(k_1+k_2\right)$

The correct answer is:
$k_1 k_2 /\left(k_1+k_2\right)$

Solution:
When the spring joined in series the total extension of spring is:
$$
\begin{aligned}
& \Rightarrow y=y_1+y_2=\frac{-F}{k_1}-\frac{F}{k_2} \\
& \Rightarrow \quad y=-\mathrm{F}\left[\frac{1}{k_1}+\frac{1}{k_2}\right] \\
& \therefore \text { Spring constant } k=\frac{k_1 k_2}{k_1+k_2}
\end{aligned}
$$

Topic: Physics, Laws of Motion, JEE Main
Two springs of spring constants ' $\mathrm{K}^{\prime}$ and ' $2 \mathrm{~K}^{\prime}$ are stretched by same force. If ' $\mathrm{w}_{1}{ }^{\circ}$
and ' $\mathrm{w}_{2}$ ' are the energies stored in them respectively then

Options:
A $\mathrm{w}_{1}=2 \mathrm{w}_{2}$
B $\mathrm{w}_{1}=\frac{\mathrm{w}_{2}}{4}$
C $\mathrm{w}_{2}=2 \mathrm{w}_{1}$
D $\mathrm{w}_{1}=\mathrm{w}_{2}$

The correct answer is:
$\mathrm{w}_{1}=2 \mathrm{w}_{2}$

Solution:
Energy stored in spring of
$$
w_{1}=\frac{1}{2} k x^{2} \text { constant } k
$$
Th spring of contant 2k
$$
w_{2}=\frac{1}{2} (2k) x^{2} \text { constant } k
$$
So, $w_{2}=2w_{1}$

Topic: Physics, Work Power Energy, JEE Main
Two square shaped metal plates of side $1 \mathrm{~m}$, kept $0.01 \mathrm{~m}$ apart in air form a parallel plate capacitor. It is connected to a battery of $500 \mathrm{~V}$. The plates of the capacitor are then immersed in an insulting oil by lowering the plates vertically with a speed of $0.001 \mathrm{~ms}^{-1}$. If the dielectric constant of the oil is 11 , then current drawn from the battery during this process is

Options:
A $4.425 \times 10^{-6} \mathrm{~A}$
B $4.425 \times 10^{-5} \mathrm{~A}$
C $4.425 \times 10^{-9} \mathrm{~A}$
D $4.425 \times 10^{-2} \mathrm{~A}$

The correct answer is:
$4.425 \times 10^{-9} \mathrm{~A}$

Solution:
$\begin{aligned} & \text { Capacitance, } C=\frac{(1-\mathrm{x} \cdot 1)}{\mathrm{d}}+\frac{\mathrm{K} \in_0 \mathrm{x}}{\mathrm{d}} \\ & =\frac{\in_0}{\mathrm{~d}}(1-\mathrm{x}+\mathrm{Kx}) \\ & \text { or, } \mathrm{C}=\frac{\in_0}{\mathrm{~d}}[1+(\mathrm{K}-1) \mathrm{x}]\end{aligned}$
$\begin{aligned} \frac{\mathrm{dC}}{\mathrm{dt}} & =\frac{\in_0}{\mathrm{~d}}(\mathrm{~K}-1) \mathrm{V}=\frac{8.85 \times 10^{-12}}{0.01} \times(11-1) \times 0.001 \\ & =8.85 \times 10^{-12} \\ & \because \mathrm{Q}=\mathrm{CV} \text { or, } \mathrm{I}=\mathrm{V}, \frac{\mathrm{dC}}{\mathrm{dt}}=500 \times 8.85 \times 10^{-12} \\ & =4.425 \mathrm{~A}\end{aligned}$

Topic: Physics, Capacitance, JEE Main
Two stars are 10 light years away from the earth. They are seen through a telescope of objective diameter 30 cm. The wavelength of light is 600nm. To see the stars just resolved by the telescope, the minimum distance between them should be 1 light year=9.46×1015m of the order of :

Options:
A 108km
B 1010km
C 1011km
D 106km

The correct answer is:
108km

Solution:
l=Ra1.22×λ
and a = 30 cm
⇒l=9.46×1015×10×1.22×600×10-90.3
=2.30×1011m
⇒ 2.30×108 km

Topic: Physics, Wave Optics, JEE Main, JEE Main 2016 (10 Apr Online)
Two stars each of 1 solar mass $\left(=2 \times 10^{30} \mathrm{~kg}\right)$ are approaching each other for a head on collision. When they are at a distance $10^9 \mathrm{~km}$, their speeds are negligible. What is the speed with which they collide? The radius of each star is $10^4 \mathrm{~km}$. Assume the stars to remain undistorbed until they collide. Use the known value of $\boldsymbol{G}$.

Options:

None

Solution:
Mass of each star $=M=2 \times 10^{30} \mathrm{~kg}$. Radius $=R=10^4$ $\mathrm{km}=10^7 \mathrm{~m}$
Initial distance between 2 stars $=r$ $=10^9 \mathrm{~km}=10^{12} \mathrm{~m}$
Initial P.E. of the system $=\frac{-G M^2}{r}$;
Total K.E. of the system $=\frac{1}{2} M v^2+\frac{1}{2} M v^2=M v^2$ $v=$ speed of collision of each star.
Distance between the centres of the stars when they are about to collide $=r^{\prime}=R+R=2 R$
$\therefore \quad$ Final P.E. of the system $=\frac{-G M^2}{2 R}$
$\because \quad$ Loss in P.E. = Gain in K.E.
$\therefore \quad M V^2=-\frac{G M^2}{r}-\left(-\frac{G M^2}{2 R}\right)$
$$
\begin{aligned}
&2 \times 10^{30} \times V^2=-G M^2\left[\frac{1}{10^{12}}-\frac{1}{2 \times 10^7}\right] \\
&\Rightarrow \quad 2 \times 10^3 \times v^2 \\
&=-6.67 \times 10^{-11} \times\left(2 \times 10^{30}\right)^2\left[\frac{2-10^5}{2 \times 10^{12}}\right] \\
&\quad v^2=\frac{1.334 \times 10^{43}}{2 \times 10^{30}} \\
&\Rightarrow v=2.583 \times 10^6 \mathrm{~m} / \mathrm{s} .
\end{aligned}
$$

Topic: Physics, Gravitation
Two stars of masses 3×1031 kg each, and at distance 2×1011 m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the stars,s rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is ( Take Gravitational constant G=6.67×10-11 N m2 kg-2 )

Options:
A 2.4Ã104 m s-1
B 3.8Ã104 m s-1
C 2.8Ã105 m s-1
D 1.4Ã105 m s-1

The correct answer is:
2.8Ã105 m s-1

Solution:
To escape, the total energy of small particle must be zero.
TE=KE+U
0=12mV2+-GMmd2 Ã2
âV=8GMd
=8Ã6.67Ã10-11Ã3Ã10312Ã1011
=2.8Ã105 m s-1

Topic: Physics, Gravitation, JEE Main, JEE Main 2019 (10 Jan Shift 2)
Two stars of masses 3Ã1031 kg each, and at distance 2Ã1011 m rotate in a plane about their common centre of mass O. A meteorite passes through O moving perpendicular to the stars' rotation plane. In order to escape from the gravitational field of this double star, the minimum speed that meteorite should have at O is ( Take Gravitational constant G=6.67Ã10-11 N m2 kg-2 )

Options:
A 2.4Ã104 m s-1
B 3.8Ã104 m s-1
C 2.8Ã105 m s-1
D 1.4Ã105 m s-1

The correct answer is:
2.8Ã105 m s-1

Solution:
To escape, the total energy of small particle must be zero.
TE=KE+U
0=12mV2+-GMmd2 Ã2
âV=8GMd
=8Ã6.67Ã10-11Ã3Ã10312Ã1011
=2.8Ã105 m s-1

Topic: Physics, Gravitation, JEE Main
Two stars of masses m and 2m at a distance d rotate about their common centre of mass in free space. The period of revolution is

Options:
A 2πd33Gm
B 2π3Gmd3
C 12π3Gmd3
D 12πd33Gm

The correct answer is:
2πd33Gm

Solution:
F=G2mmd2=2mω2d/3

Gmd2=ω2d3

⇒ω2=3Gmd3

⇒ω=3Gmd3

⇒T=2πω=2πd33Gm

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Two statements are given belwo :

Statement I : The melting point of monocarboxylic acid with even number of carbon atoms is higher than that of with odd number of carbon atoms acid immediately below and above it in the series.

Statement II : The solubility of monocarboxylic acids in water decreases with increase in molar mass.

Choose the most appropriate option :

Options:
A Both Statement I and Statement II are correct.
B Both Statement I and Statement II are incorrect.
C Statement I is correct but Statement II is incorrect.
D Statement I is incorrect but Statement II is correct.

The correct answer is:
Statement I is incorrect but Statement II is correct.

Solution:
The melting point of carboxylic acid has no trend as we can observe from the following data of metling point.

Methanoic acid =8 °C, Ethanoic acid =17 °C, Propanoic acid =2 °C, Butanoic acid =-6 °C, Pentonoic acid =-35 °C, Hexonic acid =-20 °C, Heptanoic acid =-8 °C, Octanoic acid =17 °C acid. Water solubility of carboxylic acid decreases as we increase the molar mass & hence hydrophobic part of the chain.

Topic: Chemistry, Carboxylic Acid Derivatives, JEE Main, JEE Main 2022 (24 Jun Shift 1)
Two statements in respect of drug-enzyme interaction are given below

Statement I : Action of an enzyme can be blocked only when an inhibitor blocks the active site of the enzyme.

Statement II : An inhibitor can form a strong covalent bond with the enzyme.

In the light of the above statements. Choose the correct answer from the options given below

Options:
A Both Statement I and Statement II are true
B Both Statement I and Statement II are false
C Statement I is true but Statement II is false
D Statement I is false but Statement II is true

The correct answer is:
Statement I is false but Statement II is true

Solution:
The first function of an enzyme is to hold the substrate for a chemical reaction. Active sites of enzymes hold the substrate molecule in a suitable position, so that it can be attacked by the reagent effectively.

Substrates bind to the active site of the enzyme through a variety of interactions such as ionic bonding, hydrogen bonding, van der Waals interaction or dipole-dipole interaction.

Drugs inhibit activities of enzymes. These can block the binding site of the enzyme and prevent the binding of substrate, or can inhibit the catalytic activity of the enzyme. Such drugs are called enzyme inhibitors

Some drugs do not bind to active sites. These bind to different site of enzyme called allosteric sites.

This binding of inhibitor at allosteric site changes the shape of the active site in such a way that substrate cannot recognise it

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Two stations A and B are 110 km apart on a straight line. One train starts from A at 7 a.m. and travels towards B at 20 kmph. Another train starts from B at 8 a.m. and travels towards A at a speed of 25 kmph. How many hours after the start, the train which starts from station A, will meet the other train which is coming from station B?

Options:

The correct answer is:
3

Solution:
Suppose they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x – 1) hours = 25 (x – 1) km

20x + 25 (x- 1) = 110

45x = 135⇒ x = 3hr

Topic: Physics, Motion In One Dimension, JEE Main
Two steel wires having same length are suspended from a ceiling under the same load. If the ratio of their energy stored per unit volume is 1:4, the ratio of their diameters is:

Options:
A 2:1
B 1:2
C 2:1
D 1:2

The correct answer is:
2:1

Solution:
dudv=12 stress ×stressy
=12F2A2y
dudvα1d4
dudv1dudv2=d24d14=14
d1d2=41/4
d1d2=2:1

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2020 (09 Jan Shift 2)
Two stones are projected from the top of a cliff $\mathrm{h}$ metres high, with the same speed $\mathrm{u}$, so as to hit the ground at the same spot. If one of the stones is projected at an angle $\theta$ to the horizontal then the $\theta$ equals

Options:
A
$\mathrm{u} \sqrt{\frac{2}{\mathrm{gh}}}$

B
$\sqrt{\frac{2 \mathrm{u}}{\mathrm{gh}}}$

C
$2 \mathrm{~g} \sqrt{\frac{\mathrm{u}}{\mathrm{h}}}$

D
$2 \mathrm{~h} \sqrt{\frac{\mathrm{u}}{\mathrm{g}}}$

The correct answer is:
$\mathrm{u} \sqrt{\frac{2}{\mathrm{gh}}}$

Solution:
$\mathrm{R}=\mathrm{u} \sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}=(\mathrm{u} \cos \theta) \times \mathrm{t}$
$t=\frac{1}{\cos \theta} \sqrt{\frac{2 h}{g}}$

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2003
Two stones are thrown up simultaneously from the edge of a cliff 240 m high with an initial speed of 10 m s-1 and 40 m s-1 respectively. Which of the following graph best represents the time variation of the relative position of the second stone with respect to the first? (Assume stones do not rebound after hitting the ground and neglect air resistance, take g=10 ms-2)(the figure are schematic and not drawn to scale)

Options:
A
B
C
D

The correct answer is:

Solution:
S1=10t-12gt2. When S1=-240⇒t=8s. So, at t=8 seconds first stone will reach the ground. S2=40t-12gt2. Till t=8 s, S2-S1=30t. But after 8 s S1 is constant =-240. Relative to stone 1 for (t1>8) s displacements of stone 2 , S2+240. And at t=12 s seconds stone will reach the ground. The corresponding graph of the relative position of second stone with respect to first is,

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2015 (04 Apr)