datavorous/entrance-exam-dataset huggingface dataset:
Three particles are projected in the air with the minimum possible speeds, such that the first goes from A to B, the second goes from B to C and the third goes from C to A. Points A and C are at the same horizontal level. The two inclines make the same angle α with the horizontal, as shown. The relation among the projection speeds of the three particles is

Options:
A u3=u1+u2
B u32=2u1u2
C 1u3=1u1+1u2
D u32=u12+u22

The correct answer is:
u32=2u1u2

Solution:
For a given speed,

the maximum range of a projectile on a horizontal level is

Rmax=u2g

the maximum range up and down the incline are

Rupmax=u2g1+sinαRdownmax=u2g1-sinα

by substituting the values of the velocities given in the problem, we get

R=u22g1+sin⁡α⇒u22=Rg1+sin⁡α …(1)
2Rcos⁡α= u32g⇒u32 =2Rgcos⁡α …(2)
Now, u1u2=Rg1-sin2⁡α=Rg cos α=u322
∴u32 =2u1u2

Topic: Physics, Motion In Two Dimensions, JEE Main
Three particles, each having a charge of $10 \mu \mathrm{C}$ are placed at the corners of an equilateral triangle of side 10 cm. The electrostatic potential energy of the system is $\left(\right.$ Given $\left.\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{~N}-\mathrm{m}^{2} \mathrm{C}^{2}\right)$

Options:
A Zero
B $\infty$
C $27 \mathrm{~J}$
D $100 \mathrm{~J}$

The correct answer is:
$27 \mathrm{~J}$

Solution:
For a pair of charges
$$
\begin{array}{c}
U=\frac{1}{4 \pi \varepsilon_{0}} \frac{q_{1} q_{2}}{r} \\
U_{\text {system }}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100}\right. \\
+\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100} \\
\left.+\frac{10 \times 10^{-6} \times 10 \times 10^{-6}}{10 / 100}\right] \\
=3 \times 9 \times 10^{9} \times \frac{100 \times 10^{-12} \times 100}{10} \\
=27 \mathrm{~J}
\end{array}
$$

Topic: Physics, Electrostatics, JEE Main
Three particles each of mass m=1 kg, are located at the vertices of an equilateral triangle of side a=4 m. The speed at which they must move to revolve in a circular orbit, under the influence of mutual gravitational force, circumscribing and preserving the equilateral triangle is v=nG, then the value of n is _______. If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
1

Solution:
The net gravitational force on any particle due to the other two is
Fg=2Gm2a2cos30Â°=3Gm2a2
Now, r=a3

For circular motion,

Fg=3Gm2a2=mv2r

Solving it we obtain,
v=0.50G

Topic: Physics, Gravitation, JEE Main
Three particles each of mass m=1 kg, are located at the vertices of an equilateral triangle of side a=4 m. The speed at which they must move to revolve in a circular orbit, under the influence of mutual gravitational force, circumscribing and preserving the equilateral triangle is v=nG, then the value of n is

Options:

The correct answer is:
0.50

Solution:
The net gravitational force on any particle due to the other two is
Fg=2Gm2a2cos30Â°=3Gm2a2
Now, r=a3

For circular motion,

Fg=3Gm2a2=mv2r

Solving it we obtain,
v=0.50G

Topic: Physics, Gravitation, JEE Main
Three particles, each of mass $m$ gram, are situated at the vertices of an equilateral triangle $A B C$ of side $f$. $\mathrm{cm}$ (as shown in the figure). The moment of inertia of the system about a line $A X$ perpendicular to $A B$ and in the plane of $A B C$, in gram-cm2 units will be

Options:
A $\frac{3}{2} m \ell^2$
B $\frac{3}{4} m \ell^2$
C $2 m \ell^2$
D $\frac{5}{4} m \ell^2$

The correct answer is:
$\frac{5}{4} m \ell^2$

Solution:
Perpendicular distance of mass at $A$ from $A X=r_A=0$
Perpendicular distance of mass at $B$ from $A X=r_B=1$
Perpendicular distance of mass at $C$ from $A X=r_C=I \sin 30^{\circ}$
The moment of inertia of the system. $I=I_A+I_B+I_C$
$\begin{aligned}
& \mathrm{I}=\mathrm{m}_A r_A^2+\mathrm{m}_B r_B^2+\mathrm{m}_C r_C^2 \\
& \mathrm{I}=\mathrm{m}_A(0)^2+m(I)^2+m\left(I \sin 30^{\circ}\right)^2 \\
& \mathrm{I}=m \mathrm{~m}^2+\left.m\right|^2 \times\left(\frac{1}{4}\right)=\left(\frac{5}{4}\right) m l^2
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
Three particles, each of mass $m$ gram, are situated at the vertices of an equilateral triangle $\mathrm{ABC}$ side $l \mathrm{~cm}$ (as shown in the figure). The moment of inertia of the system about a line $\mathrm{AX}$ perpendicular to $A B$ and in the plane of $A B C$, in gram $\mathrm{cm}^2$ units will be:

Options:
A $\frac{3}{4} m l^2$
B $2 \mathrm{ml}$
C $\frac{5}{4} m l^2$
D $\frac{3}{2} m l^2$

The correct answer is:
$\frac{5}{4} m l^2$

Solution:
According to the question
$$
\begin{aligned}
I_{A X} & =m l^2+m\left(\frac{l}{2}\right)^2 \\
& =m l^2+\frac{5}{4} \\
& =\frac{5}{4} m l^2
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main
Three particles, each of mass ' $m$ ' grams situated at the vertices of an equilateral $\triangle A B C$ of side ' $a$ ' $\mathrm{cm}$ (as shown in the figure). The moment of inertia of the system about a line $\mathrm{AX}$ perpendicular to $\mathrm{AB}$ and in the plane of $\mathrm{ABC}$ in $\mathrm{g}-\mathrm{cm}^2$ units will be

Options:
A $2 \mathrm{ma}^2$
B $\frac{3}{2} \mathrm{ma}^2$
C $\frac{3}{4} \mathrm{ma}^2$
D $\frac{5}{4} \mathrm{ma}^2$

The correct answer is:
$\frac{5}{4} \mathrm{ma}^2$

Solution:
$\mathrm{I}=\mathrm{~ma}^2+\mathrm{m}\left(\mathrm{a} \cos 60^{\circ}\right)^2=m \mathrm{a}^2+\frac{\mathrm{ma}^2}{4}=\frac{5 \mathrm{ma}^2}{4}$

Topic: Physics, Rotational Motion, JEE Main
Three particles, each of mass M, situated at the vertices of an equilateral triangle of side length L. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original separation L. The initial speed that should be given to each particle is

Options:
A 2GML
B GM2L
C GML
D 3GML

The correct answer is:
GML

Solution:
As the particles move under gravitational influence, the gravitational force will provide the centripetal force required.

Gravitational force between two bodies of masses m1 and m2 separated by distance d is given by, Gm1m2d2, where G is the universal gravitational constant.

Net force on the particle at A due to particles at B and C is,

F →net=F →AC+F→AB

F→net=GM2L2AC^+GM2L2AB^.

Here, AB^ and AC^ are unit vectors along AB and AC.

On simplifying vector additiontaking F=GM2L2, we get,

Fnet=2Fcos30°=2GM2L232=3GM2L2

From geometry, radius of circle, R=L3.

Since the force Fnet provides the centripetal force required, so,

⇒Mv2R=3GM2L2

⇒ v2=3GMRL2=3GML3L2=GML

⇒ v=GML.

Topic: Physics, Gravitation, JEE Main
Three particles, each of mass M, situated at the vertices of an equilateral triangle of side length L. The only forces acting on the particles are their mutual gravitational forces. It is desired that each particle moves in a circle while maintaining the original separation L. The initial speed that should be given to each particle is

Options:
A 2GML
B GM2L
C GML
D 3GML

The correct answer is:
GML

Solution:
As the particles move under gravitational influence, the gravitational force will provide the centripetal force required.

Gravitational force between two bodies of masses m1 and m2 separated by distance d is given by, Gm1m2d2, where G is the universal gravitational constant.

Net force on the particle at A due to particles at B and C is,

F →net=F →AC+F→AB

F→net=GM2L2AC^+GM2L2AB^.

Here, AB^ and AC^ are unit vectors along AB and AC.

On simplifying vector additiontaking F=GM2L2, we get,

Fnet=2Fcos30°=2GM2L232=3GM2L2

From geometry, radius of circle, R=L3.

Since the force Fnet provides the centripetal force required, so,

⇒Mv2R=3GM2L2

⇒ v2=3GMRL2=3GML3L2=GML

⇒ v=GML.

Topic: Physics, Gravitation, TS EAMCET, TS EAMCET 2022 (18 Jul Shift 1)
Three particles having charges in the ratio of \( 2: 3: 5 \) produce the same point on the photographic film in Thomson experiment. Their masses are in the ratio of

Options:
A \( 2: 3: 5 \)
B \( 5: 3: 2 \)
C \( 15: 10: 6 \)
D \( 3: 5: 2 \)

The correct answer is:
\( 2: 3: 5 \)

Solution:
Particles having the same charge to mass ratio are deflected to the same extent in the Thomson experiment. Hence, their masses also have the same ratio as the charges have, so they will meet at the same point on the photographic plate.

Topic: Physics, Dual Nature of Matter, JEE Main
Three particles of masses 50 g, 100 g and 150 g are placed at the vertices of an equilateral triangle of side 1 m (as shown in the figure). The x,ây coordinates of the centre of mass will be:

Options:
A 712 m, 34 m
B 712 m, 38 m
C 34 m, 512 m
D 38 m, 712 m

The correct answer is:
712 m, 34 m

Solution:
Coordinates of A :0, 0
B:12 m, 32 m
C :(1 m, 0)
Coordinates of center of mass
xcm,ycm=50×0+150×12+100×150+150+100 m, 50×0+150×32+100×050+150+100 m
=712 m, 34 m

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2019 (12 Apr Shift 2)
Three particles of masses 50 g, 100 g and 150 g are placed at the vertices of an equilateral triangle of side 1 m (as shown in the figure). The x,ây coordinates of the centre of mass will be:

Options:
A 712 m, 34 m
B 712 m, 38 m
C 34 m, 512 m
D 38 m, 712 m

The correct answer is:
712 m, 34 m

Solution:
Coordinates of A :0, 0
B:12 m, 32 m
C :(1 m, 0)
Coordinates of center of mass
xcm,ycm=50×0+150×12+100×150+150+100 m, 50×0+150×32+100×050+150+100 m
=712 m, 34 m

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Three particles P, Q and R are moving along the vectors A→=i^+j^, B→=j^+k^ and C→=-i^+j^, respectively. They strike on a point and start to move in different directions. Now particle P is moving normal to the plane which contains vector A→ and B→. Similarly particle Q is moving normal to the plane which contains vector A→ and C→. The angle between the direction of motion of P and Q is cos-11x. Then the value of x is .

Options:

The correct answer is:
3

Solution:
Direction of P, v^1=±A→×B→|A→×B→|=±i^-j^+k^3

Direction of Q, v^2=±A→×C→|A→×C→|=±2k^2=±k^

Angle between v^1 and v^2,

v^1·v^2v^1v^2=±13(1)(1)=±13

⇒x=3

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Three particles P, Q and R are placed on a straight line as shown in the figure. The masses of P, Q and R are 3 m , 3 m and m respectively. The gravitational force on a fourth particle S of mass m is equal to

Options:
A 3 Gm 2 2 d 2 in ST direction only
B 3 Gm 2 2 d 2 in SQ direction and 3 Gm 2 2 d 2 in SU direction
C 3 Gm 2 2 d 2 in SQ direction only
D 3 Gm 2 2 d 2 in SQ direction and 3 Gm 2 2 d 2 in ST direction

The correct answer is:
3 Gm 2 2 d 2 in SQ direction only

Solution:
Along the direction parallel to TU
Net force = G 3 mm 1 2 d 2 cos 30 â - Gm 2 4 d 2 cos 60 â
= Gm 2 8 d 2 - Gm 2 8 d 2 = 0
Along the direction perpendicular to TU
Net force = G 3 m 2 1 2 d 2 cos 60 â + G 3 m 2 3 d 2 + Gm 2 4 d 2 cos 30 â
= 3 Gm 2 2 4 d 2 + 3 Gm 2 3 d 2 + 3 Gm 2 8 d 2
= 3 Gm 2 d 2 1 + 8 + 3 2 4 = 3 Gm 2 2 d 2 along SQ

Topic: Physics, Gravitation, JEE Main
Three perfect gases at absolute temperatures $T_1, T_2$ and $T_3$ are mixed. The masses of molecules are $m_1$, $m_2$ and $m_3$ and the number of molecules are $n_1, n_2$ and $n_3$ respectively. Assuming no loss of energy, the final temperature of the mixture is :

Options:
A
$\frac{n_1 T_1+n_2 T_2+n_3 T_3}{n_1+n_2+n_3}$

B
$\frac{n_1 T_1+n_2 T_2^2+n_3 T_3^2}{n_1 T_1+n_2 T_2+n_3 T_3}$

C
$\frac{n_1^2 T_1^2+n_2^2 T_2^2+n_3^2 T_3^2}{n_1 T_1+n_2 T_2+n_3 T_3}$

D
$\frac{\left(T_1+T_2+T_3\right)}{3}$

The correct answer is:
$\frac{n_1 T_1+n_2 T_2+n_3 T_3}{n_1+n_2+n_3}$

Solution:
Data $\Rightarrow n, k, t_1+n_2 k T_2+n_3 k T_3=\left(n_1+n_2+n_3\right) k T$
$\therefore T=\frac{n_1 T_1+n_2 T_2+n_3 T_3}{n_1+n_2+n_3}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2011
Three persons P, Q and R independently try to hit a target. If the probabilities of their hitting the target are 34,12 and 58 respectively, then the probability that the target is hit by P or Q but not by R is:

Options:
A 3964
B 2164
C 964
D 1564

The correct answer is:
2164

Solution:
Let probability of hitting target by P is E1 ,Q is E2 ,and by R is E3

From the question Required probability is P(E1)P(E2)P(E3¯)+P(E1¯)P(E2)P(E3¯)+P(E1)P(E2¯)P(E3¯)=34 12 38+14 1238+341238
=12+964
=2164

Topic: Mathematics, Probability, JEE Main, JEE Main 2017 (08 Apr Online)
Three point charges $1 \mathrm{C},-2 \mathrm{C}$ and $-2 \mathrm{C}$ are placed at the vertices of an equilateral triangle of side 1 metre. The work done by an external force to increase the separation of the charges to 2 metre in joule is ( $\varepsilon_0=$ permittivity of air)

Options:
A $\frac{1}{4 \pi \varepsilon_0}$
B $\frac{1}{8 \pi \varepsilon_0}$
C $\frac{1}{16 \pi \varepsilon_0}$
D $zero$

The correct answer is:
$zero$

Solution:
Work done by the external force must be zero.

Topic: Physics, Electrostatics, JEE Main
Three point charges of magnitude 5 μC, 0.16 μC and 0.3 μC are located at the vertices A,B,C of a right angled triangle whose sides are AB=3 cm, BC=32 cm and CA=3 cm and point A is the right angle corner. Charge at point A experiences _____ N of electrostatic force due to the other two charges.

Options:

The correct answer is:
17

Solution:
Using Coulomb's law the electrostatic force on A due to C

FAC=KqAqCrAC2=k×5×0.3×10-129×10-4=9×109×5×0.3×10-129×10-4

=1.5×10=15 N

Similarly, the force on A due to B

FAB=KqAqBrAB2=9×109×5×0.16×10-129×10-4=8 N

We can see that the forces are perpendicular to each other. Therefore, the net force experienced by charge at A is,

F=FAC2+FAB2=152+82=289=17 N

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Three point charges q,-2q and 2q are placed on x axis at a distance x=0,x=34R and x=R respectively from origin as shown. If q=2×10-6 C and R=2 cm, the magnitude of net force experienced by the charge -2q is _____ N.

Options:

The correct answer is:
5440

Solution:
The net force on the given charge will be the resultant of the forces due to the other two charges.

Hence, the net force Fn can be calculated as follows:

Fn= 14πε04q2R42-14πε02q23R42=9×109×16q2R24-29=9×109×16×2×10-6 C20.02 m2×349=5440 N

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Three point charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown in the figure. If the net electrostatic energy of the configuration is zero, find the value of Qq is [Take 2=1.4] If your answer is K, mark absolute value of 10K as answer after rounding off to nearest integer.

Options:

The correct answer is:
6

Solution:
Net electrostatic energy of the configuration will be
U = K [ q · q a + Q · q 2 a + Q · q a ] Here, K = 1 4 π ε 0
Putting U = 0 we get, Qq=-22+2×2-22-2=4-222
=-2-2=-0.6

Topic: Physics, Electrostatics, JEE Main
Three point charges Q, +q and +q are placed at the vertices of a right-angled isosceles triangle as shown in the figure. If the net electrostatic energy of the configuration is zero, find the value of Qq is [Take 2=1.4]

Options:

The correct answer is:
-0.6

Solution:
Net electrostatic energy of the configuration will be
U = K [ q · q a + Q · q 2 a + Q · q a ] Here, K = 1 4 π ε 0
Putting U = 0 we get, Qq=-22+2×2-22-2=4-222
=-2-2=-0.6

Topic: Physics, Electrostatics, JEE Main
Three point masses 1.0 kg, 1.5 kg and 2.5 kg are placed at the vertices of a right-angle triangle of sides 4.0 cm, 3.0 cm and 5.0 cm, as shown in the figure. The centre of mass of the system is

Options:
A 0.6 cm right and 2.0 cm above 1 kg mass
B 2.0 cm right and 0.9 cm above 1 kg mass
C 0.9 cm right and 2.0 cm above 1 kg mass
D 1.5 cm right and 1.2 cm above1 kg mass

The correct answer is:
0.9 cm right and 2.0 cm above 1 kg mass

Solution:
Let us assume that the origin of the coordinate system is at the location of 1kg mass, then

xcm=1×0+1.5×3+2.5×05=0.9 cm
ycm=1×0+1.5×0+2.5×45=2 cm

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Three point particles of masses 1.0 kg, 1.5 kg and 2.5 kg are placed at three corners of a right angle triangle of sides 4.0 cm, 3.0 cm and 5.0 cm as shown in the figure. The centre of mass of the system is at a point:

Options:
A 0.6 cm right and 2.0 cm above 1 kg mass.
B 1.5 cm right and 1.2 cm above 1 kg mass.
C 2.0 cm right and 0.9 cm above 1 kg mass.
D 0.9 cm right and 2.0 cm above 1 kg mass.

The correct answer is:
0.9 cm right and 2.0 cm above 1 kg mass.

Solution:
Take 1 kg mass at origin

Xcm=1×0+1.5×3+2.5×05=0.9 cm
Ycm=1×0+1.5×0+2.5×45=2 cm

Topic: Physics, Center of Mass Momentum and Collision, JEE Main, JEE Main 2020 (07 Jan Shift 1)
Three point sized bodies each of mass \( M \) are fixed at three corners of light triangular frame of side length \( L \). About an axis perpendicular to the plane of frame and passing through centre of frame the moment of inertia of three bodies is \( \mathrm{ML}^{2} \). In above problem about an axis passing through any side of frame the moment of inertia of three bodies is

Options:
A \( \mathrm{ML}^{2} \)
B \( \frac{3 \mathrm{ML}^{2}}{2} \)
C \( \frac{3 \mathrm{ML}^{2}}{4} \)
D \( \frac{2 \mathrm{ML}^{2}}{3} \)

The correct answer is:
\( \frac{3 \mathrm{ML}^{2}}{4} \)

Solution:
Moment of inertia of a body with discrete particles is given by

I=∑MR2

Here, M is the mass of each particle and R is the distance of the axis from the particle.

Given, the axis is passing through a side of the triangular frame.

So, the moment of inertia due to the two particles present on the axis and at the corners of the frame will be zero due to the zero distance.

Therefore, moment of inertia will be only due to one particle which is not on the axis.

I=MR2=ML2-L22

⇒ I=3ML24

Topic: Physics, Rotational Motion, JEE Main
Three positive acute angles α, β and γ satisfy the relation tan⁡β2=13cot⁡α2 and cot⁡γ2=123tan⁡α2+cot⁡α2. Then, the value of α+β+γ is equal to

Options:
A π
B 2π
C π2
D 3π2

The correct answer is:
π

Solution:
We have, tanâ¡Î±2tanâ¡Î²2=13
â1-tanâ¡Î±2tanâ¡Î²2=23
Now, tanâ¡Î±+Î²2=tanâ¡Î±2+tanâ¡Î²21-tanâ¡Î±2tanâ¡Î²2=tanâ¡Î±2+tanâ¡Î²223
âtanâ¡Î±+Î²2=32tanâ¡Î±2+tanâ¡Î²2
=32tanâ¡Î±2+123tanâ¡Î²2=32tanâ¡Î±2+12cotâ¡Î±2
=cotâ¡Î³2
âtanâ¡Î±+Î²2=cotâ¡Î³2=1tanâ¡Î³2
â1-tanâ¡Î±+Î²2âtanâ¡Î³2=0
âÎ±+Î²2+Î³2=Ï2âÎ±+Î²+Î³=Ï

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
Three positive charges of equal value $q$ are placed at vertices of an equilateral triangle. The resulting lines of force should be sketched as in

Options:
A

B

C

D

The correct answer is:

Solution:
Electric lines of force due to a positive charge is spherically symmetric.
All the charges are positive and equal in magnitude. So repulsion takes place. Due to which no lines of force are present inside the equilateral triangle and the resulting lines of force obtained as shown:

Topic: Physics, Electrostatics, JEE Main, JEE Main 2012 (26 May Online)
Three positive numbers form an increasing G.P. If the middle term in this G.P. is doubled, the new numbers are in A.P. Then the common ratio of the G.P. is :

Options:
A 2 - 3
B 2 + 3
C 2 + 3
D 3 + 2

The correct answer is:
2 + 3

Solution:
Let the three numbers of the GP be a, ar and ar2 where r is the common ratio.
According to the given condition, a, 2ar and ar2 are in AP.
∴ 2ar=a+ar22
∴ a+ar2=4ar
∴ r2+1=4r
∴ r2-4r+1=0
∴ r=4±16-42
= 4 ± 2 3 2 = 2 ± 3
As it is an 'increasing' GP, r > 1.
∴ r=2+3.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2014 (06 Apr)
Three prisms 1,2 and 3 have the prism angle $A=60^{\circ}$, but their refractive indices are respectively 1.4, 1.5 and 1.6. If $\delta_1, \delta_2, \delta_3$ be their respective angles of deviation then

Options:
A $\delta_3\gt\delta_2\gt\delta_1$
B $\delta_1\gt\delta_2\gt\delta_3$
C $\delta_1=\delta_2=\delta_3$
D $\delta_2\gt\delta_1\gt\delta_3$

The correct answer is:
$\delta_3\gt\delta_2\gt\delta_1$

Solution:
$\delta=(\mu-1) \mathrm{A}($ for small $\mathrm{A})$
So, $\delta_1=\left(\mu_1-1\right) \mathrm{A}$
$=(1.4-1) \mathrm{A}$
$=0.4 \mathrm{~A}$
$\delta_2=\left(\mu_2-1\right) \mathrm{A}$
$=(1.5-1) \mathrm{A}$
$=0.5 \mathrm{~A}$
$\delta_3=\left(\mu_3-1\right) \mathrm{A}$
$=(1.6-1) \mathrm{A}$
$=0.6 \mathrm{~A}$
Clearly, $\delta_3\gt\delta_2\gt\delta_1$

Topic: Physics, Ray Optics, JEE Main
Three rays of light, namely red R, green (G) and blue (B) are incident on the face PQ of a right angled prism PQR as shown in figure.

The refractive indices of the material of the prism for red, green and blue wavelength are 1.27,1.42 and 1.49 respectively. The colour of the ray(s) emerging out of the face PR is :

Options:
A green
B red
C blue and green
D blue

The correct answer is:
red

Solution:
Assuming that the right angled prism is an isoceles prism, so the other angles will be 45° each.

⇒ Each incident ray will make an angle of 45° with the normal at face PR.

⇒ The wavelength corresponding to which the incidence angle is less than the critical angle, will pass through PR.

⇒θC= critical angle

⇒θC=sin-11μ

⇒ If θC≥45°

the light ray will pass

⇒θCRed=sin-111.27=51.94°

Red will pass.

⇒θCGreen=sin-111.42=44.76°

Green will not pass

⇒θCBlue=sin-111.49=42.15°

Blue will not pass

⇒ So only red will pass through PR.

Topic: Physics, Ray Optics, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Three reactions involving $\mathrm{H}_2 \mathrm{PO}_4^{-}$are given below :
(i) $\mathrm{H}_3 \mathrm{PO}_4+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{H}_2 \mathrm{PO}_4^{-}$
(ii) $\mathrm{H}_2 \mathrm{PO}_4^{-}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{HPO}_4^{2-}+\mathrm{H}_3 \mathrm{O}^{+}$
(iii) $\mathrm{H}_2 \mathrm{PO}_4^{-}+\mathrm{OH}^{-} \rightarrow \mathrm{H}_3 \mathrm{PO}_4+\mathrm{O}^{2-}$
In which of the above does $\mathrm{H}_2 \mathrm{PO}_4^{-}$act as an acid ?

Options:
A
(ii) only

B
(i) and (ii)

C
(iii) only

D
(i) only

The correct answer is:
(ii) only

Solution:
(i) $\mathrm{H}_3 \mathrm{PO}_4+\mathrm{H}_2 \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\underset{\text { acriugate base }}{\mathrm{H}_2 \mathrm{PO}_4^{-}}$
(ii) $\mathrm{H}_2 \mathrm{PO}_4^{-}+\mathrm{H}_2 \mathrm{O} \rightarrow \underset{\text { aconiugate base }}{\mathrm{HPO}_4^{-2}}+\mathrm{H}_3 \mathrm{O}^{+}$
(iii) $\mathrm{H}_2 \mathrm{PO}_4^{-}+\underset{\text { acid }}{\mathrm{OH}^{-}} \rightarrow \underset{\text { conjugate acid }}{\mathrm{H}_3 \mathrm{PO}_4}+\mathrm{O}^{-2}$
Only in reaction (ii) $\mathrm{H}_2 \mathrm{PO}_4^{-}$acids as 'acid'.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
Three reactions involving $\mathrm{H}_2 \mathrm{PO}_4^{-}$are given below :
(i) $\mathrm{H}_3 \mathrm{PO}_4+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{H}_2 \mathrm{PO}_4^{-}$
(ii) $\mathrm{H}_2 \mathrm{PO}_4^{-}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{HPO}_4^{2-}+\mathrm{H}_3 \mathrm{O}^{+}$
(iii) $\mathrm{H}_2 \mathrm{PO}_4^{-}+\mathrm{OH}^{-} \rightarrow \mathrm{H}_3 \mathrm{PO}_4+\mathrm{O}^{2-}$
In which of the above does $\mathrm{H}_2 \mathrm{PO}_4^{-}$act as an acid ?

Options:
A
(ii) only

B
(i) and (ii)

C
(iii) only

D
(i) only

The correct answer is:
(ii) only

Solution:
(i) $\mathrm{H}_3 \mathrm{PO}_4+\mathrm{H}_2 \rightarrow \mathrm{H}_3 \mathrm{O}^{+}+\underset{\text { acriugate base }}{\mathrm{H}_2 \mathrm{PO}_4^{-}}$
(ii) $\mathrm{H}_2 \mathrm{PO}_4^{-}+\mathrm{H}_2 \mathrm{O} \rightarrow \underset{\text { aconiugate base }}{\mathrm{HPO}_4^{-2}}+\mathrm{H}_3 \mathrm{O}^{+}$
(iii) $\mathrm{H}_2 \mathrm{PO}_4^{-}+\underset{\text { acid }}{\mathrm{OH}^{-}} \rightarrow \underset{\text { conjugate acid }}{\mathrm{H}_3 \mathrm{PO}_4}+\mathrm{O}^{-2}$
Only in reaction (ii) $\mathrm{H}_2 \mathrm{PO}_4^{-}$acids as 'acid'.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2010
Three resistances of one ohm each are connected in parallel. Such connection is again connected with $2 / 3 \Omega$ resistor in series. The resultant resistance will be

Options:
A $\frac{5}{3} \Omega$
B $\frac{3}{2} \Omega$
C $1 \Omega$
D $\frac{2}{3} \Omega$

The correct answer is:
$1 \Omega$

Solution:
Resistance of 1 ohm group $=\frac{R}{n}=\frac{1}{3} \Omega$
This is in series with $\frac{2}{3} \Omega$ resistor.
$\therefore$ Total resistance $=\frac{2}{3}+\frac{1}{3}=\frac{3}{3} \Omega=1 \Omega$

Topic: Physics, Current Electricity, JEE Main