datavorous/entrance-exam-dataset huggingface dataset:
Two cylindrical conductors with equal cross-sections and different resistivities Ï1 and Ï2 are put end to end. Find the charge at the boundary of the conductors if a current I flows from conductor 1 to conductor 2.â

Options:
A ε0ρ2-ρ1I
B ε02ρ2+ρ1I
C 2ε0ρ1-ρ2I
D ε0ρ1+ρ2I

The correct answer is:
ε0ρ2-ρ1I

Solution:
According to ohm's law,Eâ=ÏJâ
The problem is solved by the following steps:
Step I: Determine the electric field in both conductors.
The current density in the first conductor is
J1=IÏR2(Along the length of the conductor)
Similarly, J2=IÏR2 (along the length of the conductor)
The electric field inside the first conductor is
E2=Ï2IÏR2 (along the length of the conductor)
Step II: Consider small cylindrical region near the boundary and apply Gauss's law:
We consider a small cylindrical region at the boundary
The area of the cylindrical region is
S = ÏR2â
The net flux through this region is
Ï = Ï 1 + Ï 2 = E â 1 . S â + E â 2 S â
= E1S cos 180Â° + E2S cos0Â°= - E1S + E2S = (E2 - E1) S
According to Gauss's Law,
Ï = q Îµ 0
or E 2 - E 1 = q S Îµ 0 = Ï Îµ 0
â´ The surface charge density at the boundary is
Ï = (E2 - E1) Îµ0
Putting the value of E1 and E2, we get
â´ Ï = (Ï2j - Ï1j) or Ï = (Ï2 - Ï1)j
â´ Ï = Îµ 0 Ï 2 - Ï 1 I Ï R 2
or ÏÏR2= Îµ0(Ï2 - Ï1)I
or q = Îµ0(Ï2 - Ï1)I
Charge at the boundary q = Îµ0(Ï2 - Ï1)I

Topic: Physics, Current Electricity, JEE Main
Two cylindrical rods of same cross-section area and same length are connected in series to an ideal cell as shown. The resistivity of the left rod is Ï and that of the right rod is 2 Ï. Then, the variation of potential at any point P at a distance x from the left end of the combined system is most appropriately shown in which of the plots?

Options:
A
B
C
D

The correct answer is:

Solution:
From relation, E = Ï J, the magnitude of the electric field is greater in the right rod as compared to the left rod. Therefore the magnitude of the potential gradient in the right rod is greater. (Remember that potential is continuous).
Therefore the variation is shown by the following figure.

Topic: Physics, Current Electricity, JEE Main
Two cylindrical vessels of equal cross-sectional area 16 cm2 contain water upto heights 100 cm and 150 cm respectively. The vessels are interconnected so that the water levels in them become equal. The work done by the force of gravity during the process, is [Take density of water =103 kg m-3 and g=10 ms-2]

Options:
A 0.25 J
B 1 J
C 8 J
D 12 J

The correct answer is:
1 J

Solution:
Initially, the centre of mass of the right column will be at height of 75 cm and the centre of mass of the left column will be at 50 cm. After opening of the valve both columns will have height of 125 cm and their centre of mass will be at height 1252 cm. Therefore,

W=-∆U=-m1+m2gh1+h24-m1gh12-m2gh22

Now, m1=ρAh1 and m2=ρAh2

Putting ρ=1000 kg m-3, g=10 m s-2, h1=100 cm, h2=150 cm and A=16×10-4 m2, we get,

W=1 J

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Two cylindrical wires A and B have the same resistance. The ratio of their specific resistances and diameters are 1:2 each, then what is the ratio of the length of B to the length of A?

Options:

The correct answer is:
2

Solution:
Resistance of a wire R = Ï l Ï r 2 = Ï l Ã 4 Ï D 2
âµ R A = R B
â´ 4 Ï A L A Ï D A 2 = 4 Ï B L B Ï D B 2
or L B L A = Ï A Ï B D B D A 2
= Ï A 2 Ï A 2 D A D A 2 = 4 2 = 2 1

Topic: Physics, Current Electricity, JEE Main
Two data sets each of size 10 has the variance as 4 and k and the corresponding means as 2 and 4 respectively. If the variance of the combined data set is 5â5, then the value of k is equal to

Options:
A 5
B 6
C 4
D 3

The correct answer is:
5

Solution:
Combined mean, x-=2×10+4×1020=3
d1=x1--x-=-1
d2=x2--x-=1
Combined variance =n1σ12+n2σ2+n1d12+n2d22n1+n2
⇒112=10×4+10×k+10×1+10×120
⇒k=5

Topic: Mathematics, Statistics, JEE Main
Two deuterons undergo nuclear fusion to form a Helium nucleus. The energy released in this process is (given binding energy per nucleon for deuteron=1.1 MeV and for helium=7.0 MeV)

Options:
A 23.6 MeV
B 30.2 MeV
C 25.8 MeV
D 32.4 MeV

The correct answer is:
23.6 MeV

Solution:
Q=∑B.E. of products- ∑B.E. of reactants
The equation for two deuterons combining to form Helium nucleus is given as H2+1H2→2H4
Energy released=Q=4B.E.2H4 -4B.E.1H2
=4×7−4×1.1

=28−4.4

The energy released in this process is =23.6 MeV

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2017 (08 Apr Online)
Two dice are thrown at a time, the probability that the absolute value of the difference of number on dice is 3, is

Options:
A 34
B 56
C 27
D 16

The correct answer is:
16

Solution:
Total case =6×6=36
Favourable case =1,4,2,53,6,4,15,26,3=6 cases
Hence, the required probability =16

Topic: Mathematics, Probability, JEE Main
Two dice are thrown independently. Let A be the event that the number appeared on the 1st die is less than the number appeared on the 2nd die, B be the event that the number appeared on the 1st die is even and that on the second die is odd, and C be the event that the number appeared on the 1st die is odd and that on the 2nd is even. Then

Options:
A The number of favourable cases of the event (A∪B)∩C is 6
B A and B are mutually exclusive
C The number of favourable cases of the events A, B and C are 15,6 and 6 respectively
D B and C are independent

The correct answer is:
The number of favourable cases of the event (A∪B)∩C is 6

Solution:
Given, A: No. on 1st die < No. on 2nd die

A=(1,2),(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)

∴ n(A)=15

B: No. on 1st die =even & No. of 2nd die = odd

B=(2,1),(2,3),(2,5),(4,1),(4,3),(4,5), (6,1),(6,3),(6,5)

∴ n(B)=9

C: No. on 1st die =odd & No. on 2nd die =even

C=(1,2),(1,4),(1,6),(1,2),(1,4),(1,6),(5,2),(5,4),(5,6)

∴ n(C)=9

Now,
n(A∩B)=3, n(A∩C)=6, n(B∩C)=0

n(A∩B∩C)=0

Since (4,5)∈A and (4,5)∈B

∴ A and B are not exclusive events

Now,

n((A∪B)∩C)=n(A∩C)+n(B∩C)-n(A∩B∩C)

=6

Since, P(B)=936, P(C)=936, P(B∩C)=0

⇒P(B).P(C)≠P(B∩C)

∴B and C are not independent

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (01 Feb Shift 2)
Two dice are thrown simultaneously. What is the probability of obtaining a multiple of 2 on one of them and a multiple of 3 on the other

Options:
A $\frac{5}{36}$
B $\frac{11}{36}$
C $\frac{1}{6}$
D $\frac{1}{3}$

The correct answer is:
$\frac{11}{36}$

Solution:
Fayourable cases for one are three i.e. 2,4 and 6 and for other are two i.e. 3,6 .
Hence required probability
$=\left[\left(\frac{3 \times 2}{36}\right) 2-\frac{1}{36}\right]=\frac{11}{36}$
\{As same way happen when dice changes numbers among themselves\}

Topic: Mathematics, Probability, JEE Main
Two dice are thrown. The probability that the sum of numbers appearing is more than 10 , is

Options:
A $\frac{1}{18}$
B $\frac{1}{12}$
C $\frac{1}{6}$
D None of these

The correct answer is:
$\frac{1}{12}$

Solution:
Required probability
$=\frac{2+1}{36}=\frac{1}{12}$.

Topic: Mathematics, Probability, JEE Main
Two dice are thrown. What is the probability that the sumof the faces equals or exceeds $10 ?$

Options:
A $1 / 12$
B $1 / 4$
C $1 / 3$
D $1 / 6$

The correct answer is:
$1 / 6$

Solution:
Let $\mathrm{E}$ be the sum of the faces equals or exceeds
Then, $\mathrm{E}=\{(5,5),(4,6),(6,4),(5,6),(6,5),(6,6)\}$
$\therefore \quad n(E)=6$
Hence, $P(E)=\frac{n(E)}{n(S)}=\frac{6}{36}=\frac{1}{6}$

Topic: Mathematics, Probability, JEE Main
Two dices are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is:

Options:
A 49
B 1736
C 512
D 12

The correct answer is:
1736

Solution:
nS=36

The possible ordered pairs are 1,1, 1,2, 1,3, 1,5, 1,7,2,1, 2,2, 2,3, 2,5, 3,1, 3,2, 3,3, 3,5, 5,15,2,5,3,7,1

nE=5+4+4+3+1=17

So, PE=1736

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (17 Mar Shift 1)
Two different adiabatic paths for the same gas intersect two isothermal curves as shown in P-V diagram. The relation between the ratio $\frac{V_a}{V_d}$ and the ratio $\frac{V_b}{V_c}$ is:

Options:
A $\frac{V_a}{V_d} \neq \frac{V_b}{V_c}$
B $\frac{V_a}{V_d}=\frac{V_b}{V_c}$
C $\frac{V_a}{V_d}=\left(\frac{V_b}{V_c}\right)^{-1}$
D $\frac{V_a}{V_d}=\left(\frac{V_b}{V_c}\right)^2$

The correct answer is:
$\frac{V_a}{V_d}=\frac{V_b}{V_c}$

Solution:
For adiabatic process
$\mathrm{TV}^{\gamma-1}=$ constant
$\begin{aligned} & \mathrm{T}_{\mathrm{a}} \cdot \mathrm{V}_{\mathrm{a}}^{\gamma-1}=\mathrm{T}_{\mathrm{d}} \cdot \mathrm{V}_{\mathrm{d}}^{\gamma-1} \\ & \left(\frac{\mathrm{V}_{\mathrm{a}}}{\mathrm{V}_{\mathrm{d}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{d}}}{\mathrm{T}_{\mathrm{a}}} \\ & \mathrm{T}_{\mathrm{b}} \cdot \mathrm{V}_{\mathrm{b}}^{\gamma-1}=\mathrm{T}_{\mathrm{c}} \cdot \mathrm{V}_{\mathrm{c}}^{\gamma-1} \\ & \left(\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{c}}}{\mathrm{T}_{\mathrm{b}}} \\ & \frac{\mathrm{V}_{\mathrm{a}}}{\mathrm{V}_{\mathrm{d}}}=\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}} \quad\left(\begin{array}{c}\because \mathrm{T}_{\mathrm{d}}=\mathrm{T}_{\mathrm{c}} \\ \mathrm{T}_{\mathrm{a}}=\mathrm{T}_{\mathrm{b}}\end{array}\right)\end{aligned}$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Two different coils have self inductance $8 \mathrm{mH}$ and $2 \mathrm{mH}$. The current in both coils are increased at same constant rate. The ratio of the induced emf's in the coil is

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

The correct answer is:
$4: 1$

Solution:
The induced emf, $e=\frac{L d i}{d t}$
Here, $L_{1}=8 \mathrm{mH}=8 \times 10^{-3} \mathrm{H}$
and $L_{2}=2 \mathrm{mH}=2 \times 10^{-3} \mathrm{H}$
So, ratio of induced emfs, $\frac{e_{1}}{e_{2}}=\frac{L_{1} \frac{d i}{d t}}{L_{2} \frac{d i}{d t}}=\frac{L_{1}}{L_{2}}$
(since, current is increased at same rate)
$\Rightarrow$$\frac{\mathrm{q}}{e_{2}}=\frac{8 \times 10^{-3} \mathrm{H}}{2 \times 10^{-3} \mathrm{H}}=\frac{4}{1}$

Topic: Physics, Electromagnetic Induction, JEE Main
Two different families A and B are blessed with equal number of children. There are 3 tickets to be distributed amongst the children of these families so that no child gets more than one ticket. If the probability that all the tickets go to the children of the family B is 112, then the number of children in each family is

Options:
A 6
B 5
C 3
D 4

The correct answer is:
5

Solution:
Let n number of children be there in each family.

112= nC3·3! 2nC3·3!

⇒ nC3 2nC3=112

⇒nn-1n-22n2n-12n-2=112

⇒n-22n-1=13

⇒3n-6=2n-1

⇒n=5

Topic: Mathematics, Probability, JEE Main, JEE Main 2018 (16 Apr Online)
Two different metal bodies A and B of equal mass are heated at a uniform rate under similar conditions. The variation of temperature of the bodies is graphically represented as shown in the figure. The ratio of specific heat capacities is:

Options:
A 83
B 38
C 34
D 43

The correct answer is:
38

Solution:
ΔQΔtA=ΔQΔtB

mSAΔTΔtA=mSBΔTΔtB

SASB=ΔTΔtAΔTΔtB=90/6120/3=1540=38

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2021 (25 Jul Shift 1)
Two different radioactive elements with half-lives ' $\mathrm{T}_{1}$ ' and ' $\mathrm{T}_{2}$ ' have undecayed
atoms ' $\mathrm{N}_{1}{ }^{\prime}$ and ${ }^{\prime} \mathrm{N}_{2}{ }^{\prime}$ respectively, present at a given instant. The ratio of their
activities at this instant is

Options:
A $\frac{\mathrm{T}_{1} \mathrm{~T}_{2}}{\mathrm{~N}_{1} \mathrm{~N}_{2}}$
B $\frac{\mathrm{N}_{1} \mathrm{~N}_{2}}{\mathrm{~T}_{1} \mathrm{~T}_{2}}$
C $\frac{\mathrm{N}_{1} \mathrm{~T}_{1}}{\mathrm{~N}_{2} \mathrm{~T}_{2}}$
D $\frac{\mathrm{N}_{1} \mathrm{~T}_{2}}{\mathrm{~N}_{2} \mathrm{~T}_{1}}$

The correct answer is:
$\frac{\mathrm{N}_{1} \mathrm{~T}_{2}}{\mathrm{~N}_{2} \mathrm{~T}_{1}}$

Solution:
Activity, $\mathrm{R}\left(=\frac{\mathrm{dN}}{\mathrm{dt}}\right)=\lambda \mathrm{N}$
Decay constant $\lambda=\frac{\log _{e} 2}{\mathrm{~T}}$
$\therefore$ Activity $R=\frac{\left(\log _{\mathrm{e}} 2\right) \mathrm{N}}{\mathrm{T}}$
$\therefore \mathrm{R}_{1}=\frac{\left(\log _{e} 2\right) \mathrm{N}_{1}}{\mathrm{~T}_{1}}, \mathrm{R}_{2}=\frac{\left(\log _{e} 2\right) \mathrm{N}_{2}}{\mathrm{~T}_{2}}$
For two elements $\frac{R_{1}}{R_{2}}=\frac{N_{1}}{T_{1}} \times \frac{T_{2}}{N_{2}}=\left(\frac{N_{1}}{N_{2}}\right)=\left(\frac{T_{2}}{T_{1}}\right)$

Topic: Physics, Nuclear Physics, JEE Main
Two different wires having lengths L1 and L2 and respective temperature coefficient of linear expansion α1 and α2, are joined end-to-end. Then the effective temperature coefficient of linear expansion is :

Options:
A α1L1+α2L2L1+L2
B 2α1α2
C α1+α22
D 4α1α2α1+α2L2L1L2+L12

The correct answer is:
α1L1+α2L2L1+L2

Solution:
The wire are joined end-to-end, so they are in series combination, then total length of wire is given by
L=L1+L2 ...(1)

When the temperature is increase by ∆θ, then its final length is given by using concept of linear expansion,
L'=L(1+α∆θ) ...(2)

Using equation (1) and (2) for the series combination of wires, final length is given by

L'=L1'+L2' ...(3)

⇒L1+L2(1+αeff∆θ)=L1(1+α1∆θ)+L2(1+α2∆θ)

⇒αeff=L1α1+L2α2L1+L2

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2020 (05 Sep Shift 2)
Two diodes are connected in the following fashion. Provision is made to connect either $+5 \mathrm{~V}$ or ground $(0 \mathrm{~V})$ to the points $A$ to $B$. The output $Q$ will act as

Options:
A OR gate
B AND gate
C XOR gate
D NAND gate

The correct answer is:
AND gate

Solution:
From circuit diagram, when both or either terminal of $A$ and $B$ are at low $(0 \mathrm{~V})$ then, $5 \mathrm{~V}$ of supply from input appears across resistor $R$ in opposite sense of voltage $V_2$ which eliminated the voltage across $R$. Therefore, the final output across $R$ is zero (low).
When both the terminal of $A$ and $B$ at $5 \mathrm{~V}$ (high), then both diodes become in reverse bias, hence no voltage appears across due to input voltage, therefore the voltage across $R$ appears as $5 \mathrm{~V}$ high due to $V_2$.
This situation is obtained only is AND gate, hence output $Q$ will be act as AND gate.

Topic: Physics, Semiconductors, JEE Main
Two discs are rotating about their axes, normal to the plane of the discs and passing through the centre of the discs. Disc $D$ has $2 \mathrm{~kg}$ mass and $0.2 \mathrm{~m}$ radius and initial angular velocity of $50 \mathrm{rad} / \mathrm{s}$. Disc $D_2$ has $4 \mathrm{~kg}$ mass, $0.1 \mathrm{~m}$ radius and initial angular velocity of $200 \mathrm{rad} / \mathrm{s}$. The two discs are brought in contact face to face with their axes of rotation coincident. The final angular velocity (in rad/s) of the system is

Options:
A $50 \mathrm{rad} / \mathrm{s}$
B $20 \mathrm{rad} / \mathrm{s}$
C $100 \mathrm{rad} / \mathrm{s}$
D $200 \mathrm{rad} / \mathrm{s}$

The correct answer is:
$100 \mathrm{rad} / \mathrm{s}$

Solution:
MOI of disc D, about an axis passing through its centre and normal to its plane.
$\mathrm{I}_1=\frac{M R^2}{2}=\frac{2 \times(0.2)^2}{2}=0.04 \mathrm{~kg}-\mathrm{m}^2$
Initial angular velocity of disc $\mathrm{D}_1$, $\omega_1=50 \mathrm{rad} / \mathrm{s}$
MOI of disc $\mathrm{D}_2$ about an axis passing through its centre and normal to the plane.
$\begin{aligned}& 1_2=\frac{M R^2}{2}=\frac{4 \times(0.1)^2}{2} \\& =0.02 \mathrm{~kg}-\mathrm{m}^2
\end{aligned}$
Initial angular velocity of disc $\mathrm{D}_2$, $\omega_2=200 \mathrm{rad} / \mathrm{s}$
Total angular momentum of the two discs, initially is
$L_i=I_1 \omega_1+I_2 \omega_2$
When both discs are brought in contact, axes of rotation coincide.
Consider both discs as a system. Hence, final angular momentum of the system is,
$\mathrm{L}_t=\left(\mathrm{K}_1+\mathrm{I}_2\right) \omega$
Here, $\omega$ is the final angular speed of the system.
According to the law of conservation of angular momentum, we get,
$\begin{aligned}& L_1=L_{+} \\& 1_1 \omega_1+l_2 \omega_2=\left(1_1+1_2\right) \omega \\& \rightarrow \omega=\frac{(0.04) \times(50)+(0.02) \times(200)}{(0.04+0.02)} \\& \rightarrow \omega=100 \mathrm{rad} / \mathrm{s}
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
Two discs of moment of inertia I1=4 kg m2 and I2=2 kg m2 about their central axes & normal to their planes, rotating with angular speeds 10 rad s-1 & 4 rad s-1 respectively are brought into contact face to face with their axe of rotation coincident. The loss in kinetic energy of the system in the process is _________J.

Options:

The correct answer is:
24

Solution:
Applying conservation of angular momentum, we get

I1ω1+I2ω2=I1+I2ω0⇒4×10+2×4=4+2ω0⇒ω0=8 rad s-1

Now, initial rotational kinetic energy is E1=12I1ω12+12I2ω22=216 J

and final rotational kinetic energy is

E2=12I1+I2ω02=192 J

Required, energy loss ΔE=E1-E2=24 J

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2024 (30 Jan Shift 2)
Two discs of moments of inertia $I_1$ and $I_2$ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed $\omega_1$ and $\omega_2$ are brought into contact face to face with their axes of rotation coincident.
(a) Does the law of conservation of angular momentum apply to the situation? Why?
(b) Find the angular speed of the two discs system.
(c) Calculate the loss in kinetic energy of the system in the process.
(d) Account for this loss.

Options:

None

Solution:
According to question draw the diagram shown below :
(a) The law of conservation of angular momentum can be applied (in this situation). Because, there is no net external torque on the system of the two discs.
External forces, gravitation and normal reaction, act through the axis of rotation. These will not produce any effect hence, net torque is zero.
(b) By law of conservation of angular momentum
$$
\begin{aligned}
&L_f=L_i \\
&I \omega=I_1 \omega_1+I_2 \omega_2 \\
&\omega=\frac{I_1 \omega_1+I_2 \omega_2}{I}=\frac{I_1 \omega_1+I_2 \omega_2}{I_1+I_2}\left(\because I=I_1+I_2\right)
\end{aligned}
$$
(c) Final kinetic energy = Rotational kinetic energy+ Translational Kinetic energy
$$
(K . E .)_f=(K . E .)_R+(K . E .)_T
$$
As there is no translational energy. $\therefore(K . E .)_T=0$
$$
\begin{aligned}
&(K \cdot E \cdot)_f=(K \cdot E \cdot)_R=\frac{1}{2} I \omega^2 \\
=& \frac{1}{2}\left(I_1+I_2\right) \frac{\left(I_1 \omega_1+I_2 \omega_2\right)^2}{\left(I_1+I_2\right)^2} \\
=& \frac{1}{2} \frac{\left(I_1 \omega_1+I_2 \omega_2\right)^2}{\left(I_1+I_2\right)} \\
K_i=&\left.K \cdot E_{\cdot 1 R}+K \cdot E_{\cdot 2 R}+K \cdot E_{\cdot 1 T}+K \cdot E_{\cdot 2 T}\right)
\end{aligned}
$$
As no translational motion in the discs so $K . E_{1 T}$ and $K . E_{2 T}$ are zero, so
$$
\begin{aligned}
&K_i=K . E_{1 R}+K . E_{2 R} \\
&\text { So, } K_i=\frac{1}{2}\left(I_1 \omega_1^2+I_2 \omega_2^2\right) \\
&\Delta K=K_f-K_i=-\frac{I_1 I_2}{2\left(I_1+I_2\right)}\left(\omega_1-\omega_2\right)^2 < 0
\end{aligned}
$$
(d) Negative sign shows that $K_f < K_i$ as the energy is lost. Hence, there is loss in $K E$ of the system. The loss in kinetic energy is mainly due to the work against the friction between moving surface of discs.

Topic: Physics, System of Particles and Rotational Motion
Two discs of same mass and different radii are made of different materials such that their thicknesses are 1 cm and 0.5 cm respectively. The densities of materials are in the ratio 3:5. The moment of inertia of these discs respectively about their diameters will be in the ratio of x6. The value of x is ______.

Options:

The correct answer is:
5

Solution:
Mass of the disc in terms of density can be written as, m=ρπR2t.

⇒R2=mρπt

Now moment of inertia of the disc about the diameter is, mR24.

Therefore,

I=mR24=m24ρπt⇒I∝1ρt

Required ratio, I1I2=ρ2t2ρ1t1=53×0.51=56

Hence, x=5.

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Two discs of same moment of inertia ( I ) are rotating in same sense about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities ω1 and ω2. They are brought into contact face to face coinciding the axis of rotation. The expression for loss in energy during this process is

Options:
A 12Iω1+ω22
B 14Iω1-ω22
C Iω1-ω22
D 18 ω1-ω22

The correct answer is:
14Iω1-ω22

Solution:
The initial angular momentum of two discs is I1ω1+I2ω2=Iω1+ω2.

The final angular momentum of system is I1+I2ω=2Iω

From the conservation of angular momentum, ω=ω1+ω22.

Initial kinetic energy, KEi=12Iω12+ω22

and final kinetic energy, KEf=122Iω1+ω224

Loss in kinetic energy is ΔKE=12I1I2I1+I2 ω1-ω22
=12I22I ω1-ω22
=14Iω1-ω22

Topic: Physics, Rotational Motion, JEE Main
Two distinct numbers are chosen from 1, 3, 5, 7........151, 153, 155 and multiplied. The probability that the product is a multiple of 5 is

Options:
A 10203003
B 11123003
C 10113003
D 11223003

The correct answer is:
11123003

Solution:
Given numbers 1, 3, 5........, 153, 155
Tn=a+n-1d⇒155=1+n-12⇒n=78
Total number of ways =78C2=3003
The numbers which are multiple of 5 are 5, 15, 25, ...., 155
155=5+n-110⇒n=16
The number of ways in which product is a multiple of 5= (Both two numbers from 5,15,25,......,155 ) or (one number from 5,15,25,......,145 and one number from the remaining numbers).
=16C2+16C1×78-16C1
=816×152+16×62
=120+992=1112

Topic: Mathematics, Probability, JEE Main
Two distinct polynomial $\mathrm{f}(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ are defined as follows:
$$
f(x)=x^{2}+a x+2 ; \quad g(x)=x^{2}+2 x+a
$$
If the equation $f(x)=0$ and $g(x)=0$ have a common root then the sum of the roots of the equation $f(x)+g(x)=0$ is

Options:
A $-\frac{1}{2}$
B 0
C $\frac{1}{2}$
D 1

The correct answer is:
$\frac{1}{2}$

Solution:
$$
\begin{array}{l}
f(x)=x^{2}+a x+2 \\
g(x)=x^{2}+2 x+a
\end{array}
$$
Here a common Roat then
$$
\left|\begin{array}{ll}
1 & a \\
1 & 2
\end{array}\right| \quad\left|\begin{array}{ll}
a & 2 \\
2 & a
\end{array}\right|=\left|\begin{array}{cc}
2 & 1 \\
a & 1
\end{array}\right|^{2}
$$
$$
\begin{array}{l}
=\quad a=2,-3 \\
f(x)+g(x)=2 x^{2}+(a+2) x+a+2
\end{array}
$$
Sum of roots $=\frac{-(\mathrm{a}+2)}{2}$ if $\mathrm{a}=-3$ then sum $=\frac{1}{2}$

Topic: Mathematics, Quadratic Equation, JEE Main
Two drops of equal radius coalesce to form a bigger drop. What is ratio of surface energy of bigger drop to smaller one?

Options:
A $2^{1 / 2}: 1$
B $1: 1$
C $2^{2 / 3}: 1$
D None of these

The correct answer is:
$2^{2 / 3}: 1$

Solution:
Volume remains constant after coalescing. Thus,
$$
\frac{4}{3} \pi R^{3}=2 \times \frac{4}{3} \pi r^{3}
$$
where $R$ is radius of bigger drop and $r$ is radius of each smaller drop.
$\therefore$
$$
R=2^{1 / 3} r
$$
Now, surface energy per unit surface area is the surface tension. So,
Surface energy, $W=T \Delta A$
Or
$$
W=4 \pi R^{2} T
$$
Therefore, surface energy of bigger drop
$$
\begin{aligned}
W_{1} &=4 \pi\left(2^{1 / 3} r\right)^{2} T \\
&=\left(2^{2 / 3}\right) 4 \pi r^{2} T
\end{aligned}
$$
Surface energy of smaller drop
$$
W_{2}=4 \pi r^{2} T
$$
Hence, required ratio
$$
\frac{W_{1}}{W_{2}}=2^{2 / 3}: 1
$$

Topic: Physics, Thermodynamics, JEE Main
Two drops of equal radius coalesce to form a bigger drop. What is ratio of surface energy of bigger drop to smaller one?

Options:
A $2^{1 / 2}: 1$
B $1: 1$
C $2^{2 / 3}: 1$
D None of these

The correct answer is:
$2^{2 / 3}: 1$

Solution:
Volume remains constant after coalescing. Thus,
$$
\frac{4}{3} \pi R^{3}=2 \times \frac{4}{3} \pi r^{3}
$$
where $R$ is radius of bigger drop and $r$ is radius of each smaller drop.
$\therefore$
$$
R=2^{1 / 3} r
$$
Now, surface energy per unit surface area is the surface tension. So,
Surface energy, $W=T \Delta A$
Or
$$
W=4 \pi R^{2} T
$$
Therefore, surface energy of bigger drop
$$
\begin{aligned}
W_{1} &=4 \pi\left(2^{1 / 3} r\right)^{2} T \\
&=\left(2^{2 / 3}\right) 4 \pi r^{2} T
\end{aligned}
$$
Surface energy of smaller drop
$$
W_{2}=4 \pi r^{2} T
$$
Hence, required ratio
$$
\frac{W_{1}}{W_{2}}=2^{2 / 3}: 1
$$

Topic: Physics, Thermodynamics, MHT CET, MHT CET 2007