datavorous/entrance-exam-dataset huggingface dataset:
A $500 \Omega$ resistor connected to an external battery is placed inside a thermally insulated cylinder fitted with a frictionless piston. The cylinder contains an ideal gas. A current $i$ of $200 \mathrm{~mA}$ flows through the resistor as shown in the figure. The mass of the piston is $10 \mathrm{~kg}$. Assuming $g=10 \mathrm{~m} / \mathrm{s}^2$, the speed at which the piston will move upward, due to heat dissipated by the resistor, so that the temperature of the gas remains unchanged is

Options:
A $10 \mathrm{~cm} / \mathrm{s}$
B $15 \mathrm{~cm} / \mathrm{s}$
C $20 \mathrm{~cm} / \mathrm{s}$
D $30 \mathrm{~cm} / \mathrm{s}$

The correct answer is:
$20 \mathrm{~cm} / \mathrm{s}$

Solution:
As, temperature of gas remains constant, heat supplied by resistance $=$ piston work
$$
\begin{aligned}
\Rightarrow \quad i^2 R t & =p \times \Delta V \Rightarrow i^2 R t=\frac{F}{A} \times A \times \Delta x \\
\Rightarrow \quad \frac{i^2 R}{F} & =\frac{\Delta x}{t}=\text { velocity of piston } \\
\Rightarrow \quad v_{\text {piston }} & =\frac{\left(200 \times 10^{-3}\right)^2 \times 500}{10 \times 10} \\
& =\frac{4 \times 10^{-2} \times 5 \times 10^2}{10^2} \\
& =20 \times 10^{-2} \mathrm{~ms}^{-1}=20 \mathrm{cms}^{-1}
\end{aligned}
$$

Topic: Physics, Thermodynamics, TS EAMCET, TS EAMCET 2018 (05 May Shift 2)
A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60âgâ mol â1 ) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 gâ cm â3 , molar mass of the substance will be

Options:
A 105.0 g  mol –1
B 210.0 g  mol –1
C 90.0 g  mol –1
D 15.0 g  mol –1

The correct answer is:
210.0 g  mol –1

Solution:
Ïsolution=Ïurea
5.25M=1.560
M=5.25Ã601.5
=210.0 g  mol –1

Topic: Chemistry, Solutions, JEE Main
A $5.25 \%$ solution of a substance is isotonic with a $1.5 \%$ solution of urea $\left(\right.$ molar mass $=60 \mathrm{~g} \mathrm{~mol}^{-1}$ ) in the same solvent. If the densities of both the solutions are assumed to be equal to $1.0 \mathrm{~g} \mathrm{~cm}^{-3}$, molar mass of the substance will be

Options:
A
$90.0 \mathrm{~g} \mathrm{~mol}^{-1}$

B
$115.0 \mathrm{~g} \mathrm{~mol}^{-1}$

C
$105.0 \mathrm{~g} \mathrm{~mol}^{-1}$

D
$210.0 \mathrm{~g} \mathrm{~mol}^{-1}$

The correct answer is:
$210.0 \mathrm{~g} \mathrm{~mol}^{-1}$

Solution:
Solutions with the same osmotic pressure are isotonic
Let the molar mass of the substance be $M$ $\pi_1=\mathrm{C}_1 \mathrm{RT}=\mathrm{C}_2 \mathrm{RT}=\pi_2$
So, $\mathrm{C}_1=\mathrm{C}_2$
As density of the solutions are same
So $\frac{5.25}{M}=\frac{15}{60}$
$M=\frac{5.25 \times 60}{1.5}=210$
Hence (D) is correct

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2007
A 5.25% solution of a substance is isotonic with a 1.5% solution of urea ( molar mass = 60 g mol-1) in the same solvent. If the densities of both the solutions are assumed to be equal to 1.0 g cm-3, molar mass of the substance will be

Options:
A 210 g mol-1
B 90.0 g mol-1
C 115.0 g mol -1
D 105.0 g mol -1

The correct answer is:
210 g mol-1

Solution:
Isotonic solutions have same osmotic pressure.
π 1  =  C 1 RT, π 2  =  C 2 RT
For isotonic solution, Ï 1 = Ï 2
Ï = 5 Â· 2 5 / M 1 0 0 Ã 1 0 RT = 1 Â· 5 / 6 0 1 0 0 Ã 1 0 RT
5 Â· 2 5 M = 1 Â· 5 6 0
M = 6 0 Ã 5 Â· 2 5 1 Â· 5 = 2 1 0 g / mole

Topic: Chemistry, Solutions, JEE Main
A 600 pF capacitor is charged by 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. Electrostatic energy lost in the process is _____ μJ.

Options:

The correct answer is:
6

Solution:
The formula to calculate the initial energy stored in the first capacitor is given by

Ui=12CV2 ...1

The formula to calculate the final energy stored when two capacitors are connected is given by

Uf=12CV22×2=14CV2 ...2

Subtract equation (2) from equation (1) to obtain the energy loss ∆U.

∆U=12CV2-14CV2= CV24 ...3

Substitute the values of the known parameters into equation (3) to calculate the required loss.

∆U=600×10-12×(200)24 J= 6 μJ

Topic: Physics, Capacitance, JEE Main, JEE Main 2023 (08 Apr Shift 2)
A $700 \mathrm{pF}$ capacitor is charged by a $50 \mathrm{~V}$ battery. The electrostatic energy stored by it is

Options:
A $17.0 \times 10^{-8} \mathrm{~J}$
B $13.6 \times 10^{-9} \mathrm{~J}$
C $9.5 \times 10^{-9} \mathrm{~J}$
D $8.7 \times 10^{-7} \mathrm{~J}$

The correct answer is:
$8.7 \times 10^{-7} \mathrm{~J}$

Solution:
$U=\frac{1}{2} C V^2=\frac{1}{2} \times 700 \times 10^{-12}(50)^2=8.7 \times 10^{-7} \mathrm{~J}$

Topic: Physics, Capacitance, JEE Main
A 750 Hz, 20 Vrms source is connected to a resistance of 100 Ω, an inductance of 0.1803 H and a capacitance of 10 μF all in series. The time in which the resistance (heat capacity 2 J/°C ) will get heated by 10°C. (assume no loss of heat to the surroundings) is close to :

Options:
A 418 s
B 245 s
C 365 s
D 348 s

The correct answer is:
348 s

Solution:
The AC circuit is shown below.

Here,R=100, XL=Lω=0.1803×750×2π=850Ω

The capacitive reactance Xc=1Cω=110-5×2π×750=21.23 Ω

Total impedance Z=R2+XL-XC2

=1002+850-21.232=834.77≈835

The heat loss H=irms2Rt=ms∆t

20835×20835×100t=2×10

t=348.61 sec

Topic: Physics, Alternating Current, JEE Main, JEE Main 2020 (03 Sep Shift 1)
A $750 \mathrm{~kg}$ boat is $10 \mathrm{~m}$ long and is floating without motion on still water. A man of mass $80 \mathrm{~kg}$ is at one end and if he runs to another end of the boat and stops, the displacement of boat is

Options:
A 1.8 m in the direction of displacement of man
B 0.96 m in the direction of opposite to the displacement of man
C 0.96 m in the direction of displacement of the man
D 1.8 m in the direction opposite to displacement of man

The correct answer is:
0.96 m in the direction of opposite to the displacement of man

Solution:
Mass of boat, $M=750 \mathrm{~kg}$
Mass of man, $m=80 \mathrm{~kg}$
Length, $L=10 \mathrm{~m}$
Let $C$ be the centre of mass of boat-man system when mass was at point $A$.
When man reaches at point $B$.
Let the boat displaces by $x$ distance in opposite to displacement of man.
Then, new position of centre of mass from the reference line $=x^{\prime} \mathrm{cm}$
By conservation of linear momentum along the boat, distance of centre of mass from reference line remains same.
$$
\begin{aligned}
X_{\mathrm{CM}} & =X^{\prime}{ }_{\mathrm{CM}} \\
\frac{m \times 0+M \frac{L}{2}}{m+M} & =\frac{m(L-x)+M\left(\frac{L}{2}-x\right)}{m+M} \\
\frac{M L}{2} & =m L-m x+\frac{M L}{2}-M x \\
(M+m) x & =m L \\
x & =\frac{m L}{M+m}=\frac{80 \times 10}{750+80}=0.96 \mathrm{~m}
\end{aligned}
$$
Hence, the displacement of boat is $0.96 \mathrm{~m}$ in opposite to the displacement of man.

Topic: Physics, Center of Mass Momentum and Collision, TS EAMCET, TS EAMCET 2021 (06 Aug Shift 1)
A 800 turn coil of the effective area 0.05  m2 is kept perpendicular to the magnetic field 5×10-5  T. When the plane of the coil is rotated by 90o around any of its coplanar axis in 0.1 s, the emf induced in the coil will be:

Options:
A 2 V
B 0.2 V
C 2×10-3 V
D 0.02 V

The correct answer is:
0.02 V

Solution:
By Faraday's law, induced EMF in coil is equal to rate of change of magnetic flux linked with coil so

ε=-∆ϕ∆t=-ϕf-ϕi∆t=-0-NBA∆t
=800×5×10-5×0.050.1
=0.02 V

Topic: Physics, Electromagnetic Induction, JEE Main
A \( 1000 \mathrm{kHz} \) carrier wave is modulated by an audio signal of frequency range \( 600-6000 \mathrm{~Hz} \). What is the bandwidth of the channel?

Options:
A \( 24 \mathrm{kHz} \)
B \( 10 \mathrm{kHz} \)
C \( 48 \mathrm{kHz} \)
D \( 12 \mathrm{kHz} \)

The correct answer is:
\( 12 \mathrm{kHz} \)

Solution:
Maximum frequency in the frequency range of the audio signal, vm=6000 Hz

Bandwidth =2vm (maximum frequency) =2×6×103 Hz=12kHz

Topic: Physics, Communication System, JEE Main
A 2000 kg rocket in free space expels 0.5 kg of gas per second at exhaust velocity 400 m s-1 for 5 s. The increase in the speed of the rocket in this time is

Options:
A 2000 m s-1
B 200 m s-1
C 0.5 m s-1
D zero

The correct answer is:
0.5 m s-1

Solution:
dmdt×u=Mdvdt
0.5 × 400=2000 ×∆v5⇒∆v=0.5 m s-1

Topic: Physics, Laws of Motion, JEE Main
A $3.628 \mathrm{~kg}$ freight car moving along a horizontal rail road spur track at $7.2 \mathrm{~km} /$ hour strikes a bumper whose coil springs experiences a maximum compression of $30 \mathrm{~cm}$ in stopping the car. The elastic potential energy of the springs at the instant when they are compressed $15 \mathrm{~cm}$ is

Options:
A $12.1 \times 10^{4} \mathrm{~J}$
B $121 \times 10^{4} \mathrm{~J}$
C $1.21 \times 10^{4} \mathrm{~J}$
D $1.21 \times 10^{6} \mathrm{~J}$

The correct answer is:
$1.21 \times 10^{4} \mathrm{~J}$

Solution:
\(\begin{aligned} 7.2 \mathrm{~km} / \mathrm{ram} &=\frac{7.2 \times \frac{5}{18} \mathrm{~m} / \mathrm{s}}{7 / 2^{2}} \\ &=\frac{72 \times 5}{10 \times 18}=2 \mathrm{~m} / \mathrm{s} \end{aligned}\)

firt case EC \(\frac{1}{2} k x^{2}=\frac{1}{2} m v^{2}\) \(\Rightarrow \frac{1}{7} k\left(30 \times 10^{-2}\right)^{2}=\frac{1}{2} \times 3.628 \times(2)^{2}\) \(k=\frac{4 \times 3.628}{9 \times 10^{-2}}\)

\(\begin{aligned} \text { EC } & \frac{1}{2} k x^{2}=\frac{1}{2} m v^{2} \\ \Rightarrow & \frac{1}{2} k\left(30 \times 10^{-2}\right)^{2}=\frac{1}{2} \times 3.628 \times(2)^{2} \\ k &=\frac{4 \times 3.628}{9 \times 10^{-2}} \end{aligned}\)

energy potential enrgy when conpressed at \(15 \mathrm{~cm}\)

\(\begin{aligned}\Rightarrow & \frac{1}{2} k\left(15 \times 10^{-2}\right)^{2} \\=& \frac{1}{2} \times \frac{4 \times 3.628}{9 \times 10^{-2}} \times 225 \times 10^{-4} \\=& \frac{1639 \times 10^{-2}}{9} \\=& 182.206 \times 10^{-2} \end{aligned}\)

\(\begin{aligned} &\Rightarrow \frac{1}{2} k\left(15 \times 10^{2}\right. \\ &=\frac{1}{2} \times \frac{\alpha \times 3.628}{9 \times 10^{-2}} \times 225 \times 10^{-4} \\ &=\frac{1639 \times 10^{-2}}{9} \\ &=182.206 \times 10^{-2} \end{aligned}\)

Topic: Physics, Work Power Energy, JEE Main
A 5000 kg rocket is set for vertical firing. The exhaust speed is 800 m s-1. To give an initial upward acceleration of 20 m s-2, the amount of gas ejected per second to supply the needed thrust (in kg s-1) will be (Take g=10 m s-2)

Options:

The correct answer is:
187.5

Solution:
Given : Mass of rocket (m)=5000 Kg
Exhaust speed (v)=800 m s-1
Acceleration of rocket (a)=20 m s-2
Gravitational acceleration (g)=10 m s-2
We know that upward force
F=m(g+a)=5000(10+20)
=5000×30=150000 N
We also know that amount of gas ejected
(dmdt)=Fv=150000800=187.5 kg s-1

Topic: Physics, Laws of Motion, JEE Main
A 5000 kg rocket is set for vertical firing. The exhaust speed is 800 m s-1. To give an initial upward acceleration of 20 m s-2, what is the amount of gas ejected (in kg s-1) to supply the needed thrust? (Take g=10 m s-2)

Options:

The correct answer is:
187.5

Solution:
Given : Mass of rocket (m)=5000 Kg
Exhaust speed (v)=800 m s-1
Acceleration of rocket (a)=20 m s-2
Gravitational acceleration (g)=10 m s-2
We know that upward force
F=m(g+a)=5000(10+20)
=5000×30=150000 N
We also know that amount of gas ejected
(dmdt)=Fv=150000800=187.5 kg s-1

Topic: Physics, Laws of Motion, JEE Main
A $5000 \mathrm{~kg}$ rocket is set for vertical firing. The exhaust speed is $800 \mathrm{~ms}^{-1}$. To give an initial upward acceleration of $20 \mathrm{~ms}^{-2}$, the amount of gas ejected per second to supply the needed thrust will be $\left(g=10 \mathrm{~ms}^{-2}\right)$

Options:
A $127.5 \mathrm{~kg} \mathrm{~s}^{-1}$
B $187.5 \mathrm{~kg} \mathrm{~s}^{-1}$
C $185.5 \mathrm{~kg} \mathrm{~s}^{-1}$
D $137.5 \mathrm{~kg} \mathrm{~s}^{-1}$

The correct answer is:
$187.5 \mathrm{~kg} \mathrm{~s}^{-1}$

Solution:
Given that,
Mass of rocket $\mathrm{m}=5000 \mathrm{Kg}$
Acceleration $\mathrm{a}=20 \mathrm{~m} / \mathrm{s}^2$
Speed $\mathrm{v}=800 \mathrm{~m} / \mathrm{s}$
$\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$
Now, thrust force on the rocket
$\mathrm{F}_{\mathrm{t}}=\mathrm{v}_{\mathrm{r}}\left(\frac{-\mathrm{dm}}{\mathrm{dt}}\right)$
Net force on the rocket
$\begin{aligned}
& F_{\text {net }}=F_t-W \\
& m a=v_r\left(\frac{-d m}{d t}\right)-m g \\
& \left(\frac{-d m}{d t}\right)=\frac{m(g+a)}{v_r}
\end{aligned}$
Rate of gas ejected per second
$\begin{aligned}
& \left(\frac{-\mathrm{dm}}{\mathrm{dt}}\right)=\frac{5000(10+20)}{800} \\
& \frac{-\mathrm{dm}}{\mathrm{dt}}=187.5 \mathrm{~kg} / \mathrm{s}
\end{aligned}$
Hence, the amount of gas ejected per second is $187.5 \mathrm{~kg} / \mathrm{s}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
$\mathbf{a} \times[\mathbf{a} \times(\mathbf{a} \times \mathbf{b})]$ is equal to

Options:
A $(\mathbf{a} \times \mathbf{a}) .(\mathbf{b} \times \mathbf{a})$
B $\mathbf{a} .(\mathbf{b} \times \mathbf{a})-\mathbf{b} .(\mathbf{a} \times \mathbf{b})$
C $[\mathbf{a} .(\mathbf{a} \times \mathbf{b})] \mathbf{a}$
D $(\mathbf{a} \cdot \mathbf{a})(\mathbf{b} \times \mathbf{a})$

The correct answer is:
$(\mathbf{a} \cdot \mathbf{a})(\mathbf{b} \times \mathbf{a})$

Solution:
$\begin{aligned}
& \mathbf{a} \times[\mathbf{a} \times(\mathbf{a} \times \mathbf{b})]=\mathbf{a} \times\{(\mathbf{a} \cdot \mathbf{b}) \mathbf{a}-(\mathbf{a} \cdot \mathbf{a}) \mathbf{b}\} \\
& =(\mathbf{a} \cdot \mathbf{b})(\mathbf{a} \times \mathbf{a})-(\mathbf{a} \cdot \mathbf{a})(\mathbf{a} \times \mathbf{b})=(\mathbf{a} \cdot \mathbf{b}) 0+(\mathbf{a} \cdot \mathbf{a})(\mathbf{b} \times \mathbf{a}) \\
& =(\mathbf{a} \cdot \mathbf{a})(\mathbf{b} \times \mathbf{a})
\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main
A $\triangle A B C$ is right angled at $B$. Then the diameter of the incircle of the triangle is

Options:
A $2(c+a-b)$
B $c+a-2 b$
C $c+a-b$
D None of these

The correct answer is:
None of these

Solution:
$r=\frac{\Delta}{s}-\frac{\left(\frac{1}{2}\right) \mathrm{ac}}{\left(\frac{1}{2}\right)(a+b+c)}$ $=\frac{a c}{(a+b+c)}=\frac{a c(c+a-b)}{(c+a)^2-b^2}=$ $\frac{a c(c+a-b)}{c^2+2 c a+a^2-b^2}$
$=\frac{a c(c+a-b)}{2 c a+b^2-b^2}=\frac{c+a-b}{2} \quad\left(\because a^2+c^2=b^2\right)$

Topic: Mathematics, Properties of Triangles, JEE Main
A air bubble of radius 1 cm in water has an upward acceleration of 9.8 cms–2. The density of water is 1 gm cm–3 and water offers negligible drag force on the bubble. The mass of the bubble is g = 980 cm/s2.

Options:
A 4.51 gm
B 3.15 gm
C 4.15 gm
D 1.52 gm

The correct answer is:
4.15 gm

Solution:
Fb−mg=ma

Fb=mg+a

=43πr3ρg=mg+a

m=43π1g=m980+9.8

m=4.15

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2020 (04 Sep Shift 1)
A alternating current at any instant is given by $i=[6+\sqrt{56} \sin (100 \pi t+\pi / 3)]$ A. The $r m s$ value of the current is ________ A.

Options:

The correct answer is:
8

Solution:
$\begin{aligned} & I_{D N s}=\sqrt{\frac{\int i^2 d t}{\int d t}} \\ & I_{D n s}=\sqrt{(6)^2+\frac{(\sqrt{56})^2}{2}} \\ & =\sqrt{36+28} \\ & =\sqrt{64} \\ & =8 \mathrm{~A}\end{aligned}$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (04 Apr Shift 1)
A. Ammonium salts produce haze in atmosphere.

B. Ozone gets produced when atmospheric oxygen reacts with chlorine radicals.

C. Polychlorinated biphenyls act as cleansing solvents.

D. ‘Blue baby’ syndrome occurs due to the presence of excess of sulphate ions in water.

Choose the correct answer from the options given below:-

Options:
A A, B and C only
B B and C only
C A and D only
D A and C only

The correct answer is:
A and C only

Solution:
A. Ammonium salts produce atmospheric haze (aerosol of fine particles).

B. Ozone is produced by 3 atoms of oxygen. Some of the molecules in the atmosphere break due to sunlight and oxygen atoms are released. The oxygen atoms then react with other oxygen molecules to form ozone. The overall reaction is

O2 →UV O + O

O + O2 → O3

C. Polychlorinated biphenyls have the number of applications including their use as cleansing solvents.

D. 'Blue baby' syndrome occurs due to the presence of excess nitrate ions and not sulphate ions in water.

From the above statement we concluded that Statement A and C are correct.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2023 (25 Jan Shift 2)
A ancient discovery found a sample, where $75 \%$ of the original carbon $\left(\mathrm{C}^{14}\right)$ remains. Then the age of the sample is
$\left(\begin{array}{r}T_{\frac{1}{2}}\left(\mathrm{C}^{14}\right)=5730 \text { years, } \ln 0.5=-0.7 \\ \ln (0.75)=-0.3\end{array}\right)$

Options:
A 2300 years
B 2456 years
C 2546 years
D 3456 years

The correct answer is:
2456 years

Solution:
For a carbon sample $\left(\mathrm{C}^{14}\right)$,
$T_{1 / 2}=5730$ year
Decay constant, $K=\frac{0.693}{T_{1 / 2}}$
$\begin{aligned} & =\frac{0.693}{5730} \\ & =1.209 \times 10^{-4} / \text { year }\end{aligned}$
The rate of counts is proportional to the number of $\mathrm{C}^{14}$ atom in the sample
$N_0=100, N=75$
The age of the sample is given as,
$t=\frac{2.303}{K} \log \frac{N_0}{N}=\frac{2.303}{K} \log \frac{1}{0.75}$
$\begin{aligned} & =\frac{2303}{1.209 \times 10^{-4}} \log \frac{100}{75} \\ & =2456 \times 10^3 \text { years } \\ & =2456 \text { years }\end{aligned}$

Topic: Physics, Nuclear Physics, JEE Main
A ancient discovery found a sample, where $75 \%$ of the original carbon $\left(\mathrm{C}^{14}\right)$ remains. Then the age of the sample is
$\left(\begin{array}{r}T_{\frac{1}{2}}\left(\mathrm{C}^{14}\right)=5730 \text { years, } \ln 0.5=-0.7 \\ \ln (0.75)=-0.3\end{array}\right)$

Options:
A 2300 years
B 2456 years
C 2546 years
D 3456 years

The correct answer is:
2456 years

Solution:
For a carbon sample $\left(\mathrm{C}^{14}\right)$,
$T_{1 / 2}=5730$ year
Decay constant, $K=\frac{0.693}{T_{1 / 2}}$
$\begin{aligned} & =\frac{0.693}{5730} \\ & =1.209 \times 10^{-4} / \text { year }\end{aligned}$
The rate of counts is proportional to the number of $\mathrm{C}^{14}$ atom in the sample
$N_0=100, N=75$
The age of the sample is given as,
$t=\frac{2.303}{K} \log \frac{N_0}{N}=\frac{2.303}{K} \log \frac{1}{0.75}$
$\begin{aligned} & =\frac{2303}{1.209 \times 10^{-4}} \log \frac{100}{75} \\ & =2456 \times 10^3 \text { years } \\ & =2456 \text { years }\end{aligned}$

Topic: Physics, Nuclear Physics, AP EAMCET, AP EAMCET 2022 (05 Jul Shift 1)
$A$ and $B$ are events such that $P(A \cup B)=3 / 4, P(A \cap B)=1 / 4, P(\bar{A})=2 / 3$ then $P(\bar{A} \cap B)$ is

Options:
A
5/12

B
3/8

C
5/8

D
1/4

The correct answer is:
5/12

Solution:
$P(A \cup B)=P(A)+P(B)-P(A \cap B) \Rightarrow \frac{3}{4}=1-P(\bar{A})+P(B)-\frac{1}{4}$
$\Rightarrow 1=1-\frac{2}{3}+P(B) \Rightarrow P(B)=\frac{2}{3} ; \quad$ Now, $P(\bar{A} \cap B)=P(B)-P(A \cap B)=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}$

Topic: Mathematics, Probability, JEE Main, JEE Main 2002
$A$ and $B$ are non-singleton sets and $n(A \times B)=35$. If $B \subset A$, then ${ }^{n(A)} C_{n(B)}$ is equal to

Options:
A 28
B 35
C 42
D 21

The correct answer is:
21

Solution:
Given, $n(A \times B)=35, B \subset A$
$n(A \times B)=35$
$n(A \times B)=7 \times 5$
$n(A \times B)=n(A) \times n(B)$
$\therefore \quad n(A)=7$ and $n(B)=5$
${ }^{n(A)} C_{n(B)}={ }^{7} C_{5}={ }^{7} C_{2}=21$

Topic: Mathematics, Sets and Relations, JEE Main
A and B are subsets of R. Every element x of A is mapped to an element of B by the rule, yx=5xx-3x+3if x≠-1-1if x=-1, then A=

Options:
A R/{-3,+3,-0}
B R/{-3,+3}
C R/{-3,3,0,-1}
D R

The correct answer is:
R/{-3,+3}

Solution:
yx=5xx-3x+3, x≠-1-1, x=-1=fx

fx is a function f: A→B

Here, A is the domain.

For domain of f, x-3≠0, x+3≠0

⇒x≠-3,3

So, A=R/-3,3 or R--3,3

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2021 (05 Aug Shift 1)
\( A \) and \( B \) are symmetric matrices of the same order. If \( X=A B+B A, Y=A B-B A \) and \( (X Y)^{T} \) is equal to

Options:
A \( X Y \)
B \( Y X \)
C \( -Y X \)
D None of these

The correct answer is:
\( -Y X \)

Solution:
Since, A and B are symmetric matrices, hence

AT=A and BT=B

Given that, X=AB+BA and Y=AB-BA

∴ XT=AB+BAT and YT=AB-BAT

We know that P±QT=PT±QT

⇒ XT=ABT+BATand YT=ABT-BAT

Also, we know that PQT=QTPT

⇒ XT=BTAT+ATBT and YT=BTAT-ATBT

⇒ XT=BA+AB and YT=BA-AB

⇒ XT=X and YT=-Y

Now, XYT=YTXT=-YX

Topic: Mathematics, Matrices, JEE Main
$A$ and $B$ are two concentric circular conductors of centre $O$ and carrying currents $i_1$ and $i_2$ as shown in the adjacent figure. If ratio of their radii is $1: 2$ and ratio of the flux densities at $O$ due to $A$ and $B$ is $1: 3$, then the value of $i_1 / i_2$ is

Options:
A $\frac{1}{6}$
B $\frac{1}{4}$
C $\frac{1}{3}$
D $\frac{1}{2}$

The correct answer is:
$\frac{1}{6}$

Solution:
$r_1: r_2=1: 2$ and $B_1: B_2=1: 3$ We know that
$B=\frac{\mu_0}{4 \pi} \cdot \frac{2 \pi n i}{r} \Rightarrow \frac{i_1}{i_2}=\frac{B_1 r_1}{B_2 r_2}=\frac{1 \times 1}{3 \times 2}=\frac{1}{6}$

Topic: Physics, Magnetic Effects of Current, JEE Main
$A$ and $B$ are two events such that $P(A)=0.8, P(B)=0.6$ and $P(A \cap B)=0.5$, then the value of $P(A / B)$ is

Options:
A $\frac{5}{6}$
B $\frac{5}{8}$
C $\frac{9}{10}$
D None of these

The correct answer is:
$\frac{5}{6}$

Solution:
$P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)}=\frac{0.5}{0.6}=\frac{5}{6}$.

Topic: Mathematics, Probability, JEE Main
$A$ and $B$ are two independent events. The probability that both $A$ and $B$ occur is $\frac{1}{6}$ and the probability that neither of them occurs is $\frac{1}{3}$. Then the probability of the two events are respectively

Options:
A $\frac{1}{2}$ and $\frac{1}{3}$
B $\frac{1}{5}$ and $\frac{1}{6}$
C $\frac{1}{2}$ and $\frac{1}{6}$
D $\frac{2}{3}$ and $\frac{1}{4}$

The correct answer is:
$\frac{1}{2}$ and $\frac{1}{3}$

Solution:
$P(A \cap B)=P(A) \cdot P(B)=\frac{1}{6}$
$P(\bar{A} \cap \bar{B})=\frac{1}{3}=1-P(A \cup B)$
$\Rightarrow \frac{1}{3}=1-[P(A)+P(B)]+\frac{1}{6} \Rightarrow P(A)+P(B)=\frac{5}{6}$.
Hence $P(A)$ and $P(B)$ are $\frac{1}{2}$ and $\frac{1}{3}$.

Topic: Mathematics, Probability, JEE Main
$A$ and $B$ are two like parallel forces. A couple of moment $\mathrm{H}$ lies in the plane of $A$ and $B$ and is contained with them. The resultant of $A$ and $B$ after combining is displaced through a distance

Options:
A
$\frac{2 \mathrm{H}}{\mathrm{A}-\mathrm{B}}$

B
$\frac{H}{A+B}$

C
$\frac{H}{2(A+B)}$

D
$\frac{H}{A-B}$

The correct answer is:
$\frac{H}{A+B}$

Solution:
$$
\begin{aligned}
& (A+B)=d=H \\
& d=\left(\begin{array}{c}
H \\
A+B
\end{array}\right) .
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2005
A and B are two matrices such that the order of A is 3×4, if A ′  B and B A ′ are both defined, then

Options:
A order of B ′ is 3×4
B order of B ′ A is 4×4
C order of B ′ A is 3×3
D B ′ A is not defined

The correct answer is:
order of B ′ A is 4×4

Solution:
Let, order of B is m×n
∵ A ′ B is defined so m=3
∴B A ′ is defined so n=4
⇒ order of B is 3×4
∴ order of B ′ A is 4×4

Topic: Mathematics, Matrices, JEE Main
$\mathrm{~A}$ and $\mathrm{B}$ are two metals with threshold frequencies $1.8 \times 10^{14} \mathrm{~Hz}$ and $2.2 \times 10^{4} \mathrm{~Hz}$. Two identical photons of energy $0.825 \mathrm{eV}$ each are incident on them. Then photoelectrons are emitted by $\left(\right.$ Take $\left.\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}-\mathrm{s}\right)$

Options:
A $\mathrm{B}$ alone
B A alone
C Neither A nor B
D Both A and B

The correct answer is:
A alone

Solution:
$\begin{aligned} \text { Threshold energy of A E }_{\mathrm{A}}=\mathrm{hv}_{\mathrm{A}} \\=& 6.6 \times 10^{-34} \times 1.8 \times 10^{14} \\=& 11.88 \times 10^{-20} \mathrm{~J} \\=& \frac{11.88 \times 10^{-20}}{1.6 \times 10^{-19}} \mathrm{eV}=0.74 \mathrm{eV} \end{aligned}$
Similarly, $\mathrm{E}_{\mathrm{B}}=0.91 \mathrm{eV}$
As the incident photons have energy greater than $\mathrm{E}_{\mathrm{A}}$ but less than $\mathrm{E}_{\mathrm{B}}$ So, photoelectrons will be emitted from metal A only.

Topic: Physics, Dual Nature of Matter, JEE Main