datavorous/entrance-exam-dataset huggingface dataset:
One end of a metal rod of length $1.0 \mathrm{~m}$ and area of cross section $100 \mathrm{~cm}^2$ is maintained at $100^{\circ} \mathrm{C}$. If the other end of the rod is maintained at $0^{\circ} \mathrm{C}$, the quantity of heat transmitted through the rod per minute is (Coefficient of thermal conductivity of material of rod $=100 \mathrm{~W} / \mathrm{m}-\mathrm{K}$ )

Options:
A $3 \times 10^3 J$
B $6 \times 10^3 \mathrm{~J}$
C $9 \times 10^3 J$
D $12 \times 10^3 \mathrm{~J}$

The correct answer is:
$6 \times 10^3 \mathrm{~J}$

Solution:
$\begin{gathered}\frac{Q}{t}=\frac{K A\left(\theta_1-\theta_2\right)}{l}=\frac{100 \times 100 \times 10^{-4}(100-0)}{1} \\ \Rightarrow \frac{Q}{t}=100 \mathrm{Joule} / \mathrm{sec}=6 \times 10^3 \mathrm{Joule} / \mathrm{min}\end{gathered}$

Topic: Physics, Thermal Properties of Matter, JEE Main
One end of a metal wire is fixed to a ceiling and a load of 2 kg hangs from the other end. A similar wire is attached to the bottom of the load and another load of 1 kg hangs from this lower wire. Then the ratio of longitudinal strain of upper wire to that of the lower wire will be

[Area of cross section of wire =0.005 cm2, Y=2×1011 N m-2 and g=10 m s-2]

Options:

The correct answer is:
3

Solution:
As we know, change in length is given by ΔL=FLAY

Therefore, strain required ΔLL=FAY

Now the ratio will be, ΔL1 L1ΔL2 L2=F1 F2=T1 T2=3010=3

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2024 (01 Feb Shift 2)
One end of a rod of length $L=1 \mathrm{~m}$ is fixed to a point on the circumference of a wheel of radius $\mathrm{R}=1 / \sqrt{3} \mathrm{~m}$. The other end is sliding freely along a straight channel passing through the center $\mathrm{O}$ of the wheel as shown in the figure below. The wheel is rotating with a constant angular velocity $\omega$ about $\mathrm{O}$.

Options:
A $\frac{2 \omega}{3}$
B $\frac{\omega}{3}$
C $\frac{2 \omega}{\sqrt{3}}$
D $\frac{\omega}{\sqrt{3}}$

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

Solution:
$\tan 60^{\circ}=\frac{\mathrm{L}}{\mathrm{R}}=\frac{1}{1 / \sqrt{3}}=\sqrt{3}=\tan \theta$
$\therefore \quad \mathrm{x}=\sqrt{1+\frac{1}{3}}=\frac{2}{\sqrt{3}}$
$\cos \theta=\frac{\mathrm{R}^{2}+\mathrm{x}^{2}-\mathrm{L}^{2}}{2 \mathrm{Rx}}$
$\Rightarrow \mathrm{R}^{2}+\mathrm{x}^{2}-\mathrm{L}^{2}=2 \mathrm{Rx} \cos \theta$
$\begin{aligned} \Rightarrow 2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{R}\left[\mathrm{x}(-\sin \theta)+\cos \theta \frac{\mathrm{dx}}{\mathrm{dt}}\right] \\ & \frac{\mathrm{dx}}{\mathrm{dt}}[\mathrm{x}-\mathrm{R} \cos \theta]=-\mathrm{Rx} \sin \theta \frac{\mathrm{d} \theta}{\mathrm{dt}} \\ \therefore \quad-&{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v} \& \frac{\mathrm{d} \theta}{\mathrm{dt}}=\omega\end{aligned}$
$\Rightarrow \mathrm{v}=\frac{\mathrm{R} \mathrm{x} \sin \theta \omega}{\mathrm{x}-\mathrm{R} \cos \theta}$
$\begin{aligned} &=\frac{\frac{1}{\sqrt{3}} \frac{2}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} \cdot \omega}{\frac{2}{\sqrt{3}}-\frac{1}{\sqrt{3}} \cdot \frac{1}{2}}=\frac{\frac{\omega}{\sqrt{3}}}{\frac{1}{\sqrt{3}} \frac{3}{2}}=\frac{2 \omega}{3} \\ \mathrm{v}=& \frac{2 \omega}{3} \end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
One end of a spring of force constant K is fixed to a vertical wall and other to a body of mass m resting on smooth horizontal surface. There is another wall at a distance of 3x0 from the body. If all the collisions are elastic and spring is compressed by 2x0 and released, then the time period of oscillation is

Options:
A 5π3mK
B 5π4mK
C 3π2mK
D 4π3mK

The correct answer is:
5π3mK

Solution:
T=2πmk
t=T2+2×T6=5T6

Topic: Physics, Oscillations, JEE Main
One end of a thermally insulated rod is kept at a temperature $T_1$ and the other at $T_2$. The rod is composed of two sections of lengths $\ell_1$ and $\ell_2$ and thermal conductivities $k_1$ and $k_2$ respectively. The temperature at the interface of the two sections is

Options:
A
$\left(\mathrm{k}_2 \ell_2 \mathrm{T}_1+\mathrm{k}_1 \ell_1 \mathrm{T}_2\right) /\left(\mathrm{k}_1 \ell_1+\mathrm{k}_2 \ell_2\right)$

B
$\left(\mathrm{k}_2 \ell_1 \mathrm{T}_1+\mathrm{k}_1 \ell_1 \mathrm{T}_2\right) /\left(\mathrm{k}_2 \ell_1+\mathrm{k}_1 \ell_2\right)$

C
$\left(\mathrm{k}_1 \ell_2 \mathrm{T}_1+\mathrm{k}_2 \ell_1 \mathrm{T}_2\right) /\left(\mathrm{k}_1 \ell_2+\mathrm{k}_2 \ell_1\right)$

D
$\left(\mathrm{k}_1 \ell_1 \mathrm{T}_1+\mathrm{k}_2 \ell_2 \mathrm{T}_2\right) /\left(\mathrm{k}_1 \ell_1+\mathrm{k}_2 \ell_2\right)$

The correct answer is:
$\left(\mathrm{k}_1 \ell_2 \mathrm{T}_1+\mathrm{k}_2 \ell_1 \mathrm{T}_2\right) /\left(\mathrm{k}_1 \ell_2+\mathrm{k}_2 \ell_1\right)$

Solution:
$\frac{\left(\mathrm{T}_1-\mathrm{T}\right) \mathrm{k}_1}{\ell_1}=\frac{\left(\mathrm{T}-\mathrm{T}_2\right) \mathrm{k}_2}{\ell_2}$
$\mathrm{~T}=\frac{\mathrm{T}_1 \mathrm{k}_1 \ell_2+\mathrm{T}_2 \mathrm{k}_2 \ell_1}{\mathrm{k}_1 \ell_2+\mathrm{k}_2 \ell_1}$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2007
One end of a thermally insulated rod is kept at a temperature T 1 and the other at T 2 . The rod is composed of two sections of lengths l 1 and l 2 and thermal conductivities K 1 and K 2 respectively. The temperature at the interface of the two sections is

Options:
A ( K 1 l 1 T 1 + K 2 l 2 T 2 ) ( K 1 l 1 + K 2 l 2 )
B ( K 2 l 2 T 1 + K 1 l 1 T 2 ) ( K 1 l 1 + K 2 l 2 )
C ( K 2 l 1 T 1 + K 1 l 2 T 2 ) ( K 2 l 1 + K 1 l 2 )
D ( K 1 l 2 T 1 + K 2 l 1 T 2 ) ( K 1 l 2 + K 2 l 1 )

The correct answer is:
( K 1 l 2 T 1 + K 2 l 1 T 2 ) ( K 1 l 2 + K 2 l 1 )

Solution:
k1Al1T1-T=k2Al2T-T2
k1l2+k2+l1T=k1l2T1+k2l1T2
Or T=k1l2T1+k2kl2T2 k1l2+k2l!

Topic: Physics, Thermal Properties of Matter, JEE Main
One end of a uniform metal rod of length $100 \mathrm{~cm}$ is placed in ice and the other end is placed in boiling water. A point of the rod which is at a distance of $60 \mathrm{~cm}$ from the ice end is maintained at a constant temperature of $325^{\circ} \mathrm{C}$. If $2 \mathrm{~g}$ of water is converted into steam per second, the mass of ice melted per second in steady state is
(Latent heat of steam $=6.75$ times latent heat of fusion of ice)

Options:
A $13 \mathrm{~g}$
B $4 \mathrm{~g}$
C $6.75 \mathrm{~g}$
D $8 \mathrm{~g}$

The correct answer is:
$13 \mathrm{~g}$

Solution:
No solution. Refer to answer key.

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2017 (25 Apr Shift 2)
One end of a uniform wire of length $L$ and of weight $W$ is attached rigidly to a point in the roof and a weight ${W}^1$ is suspended from its lower end. If $S$ is the area of cross-section of the wire, the stress in the wire at a height $3 L / 4$ from its lower end is

Options:
A $\frac{W_1}{s}$
B $\frac{W_1+(W / 4)}{S}$
C $\frac{W_1+(3 W / 4)}{S}$
D $\frac{W_1+W}{S}$

The correct answer is:
$\frac{W_1+(3 W / 4)}{S}$

Solution:
Total force at height $3 L / 4$ from its lower end $=$ Weight suspended + Weight of $3 / 4$ of the chain
$=W_1+(3 W / 4)$
Hence stress$=\frac{W_1+(3 W / 4)}{S}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
One end of a U-tube containing mercury is connected to a suction pump and the other end to atmosphere. $A$ small pressure difference is maintained between the
two columns. Show that, when the suction pump is removed, the column of mereury in the U-tube executes simple harmonic motion.

Options:

None

Solution:
The suction pump creates the pressure difference, thus mercury rises in one limb of the U-tube. When it is removed, a net force acts on the liquid column due to the difference in levels of mercury in the two limbs and hence the liquid column executes S.H.M. which can be expressed as:
Consider the mercury contained in a vertical U-tube upto the level $\rho$ and $Q$ in its two limbs.
Let $\rho=$ density of the mercury.
$L=$ Total length of the mercury column in both the limbs.
$A=$ internal cross-sectional area of U-tube.
$m=$ mass of mercury in U-tube = LAP.
Assume, the mercury be depressed in left limb to $P^{\prime}$ by a small distance $y$, then it rises by the same amount in the right limb to position $Q^{\prime}$,
$\therefore \quad$ Difference in levels in the two limbs
If $W=$ weight of liquid contained in the column of length $2 y$.
Then $W=m g=A \times 2 y \times \rho \times \mathrm{g}$
This weight produces the restoring force $(F)$ which tends to bring back the mercury to its equilibrium position.
If $a=$ acceleration produced in the liquid column, Then
$$
a=\frac{F}{m}=-\frac{(2 A \rho g) y}{L A \rho}=g=\frac{-2 g}{2 h} y(\because L=2 h)
$$
Where $h=$ height of mercury in each limb. Now from eqn. (i), it is clear that $a \alpha y$ and -ve sign shows that it acts opposite to $y$, so the motion of mercury in $\mathrm{U}-$ tube is simple harmonic in nature having time period (T) given by
$$
T=2 \pi \sqrt{\frac{y}{a}}=2 \pi \sqrt{\frac{2 h \rho}{2 \rho g}}=2 \pi \sqrt{\frac{h \rho}{\rho g}}
$$
$$
T=2 \pi \sqrt{\frac{h}{g}}
$$

Topic: Physics, Oscillations
One fourth length of a uniform rod is placed on a rough horizontal surface and rod starts rotating about the edge. At an angle θ between the rod and the horizontal, rod begins to slide on the edge. If coefficient of friction between rod and surface is μ , and x tan θ = 4 μ . Find x. (Take g = 10 m/s2)

Options:
A 13
B 11
C 9
D 7

The correct answer is:
13

Solution:
One fourth length
mg â 4 sin θ = 1 2 I ω 2
α = mg â 4 cos θ I

μ N - mg sin θ = m ω 2 â 4
mg cos θ - N = m â 4 α
Solving , 1 3 tan θ = 4 μ

Topic: Physics, Rotational Motion, JEE Main
One $\mathrm{gm}$ mol of a diatomic gas $(\gamma=1.4)$ is compressed adiabatically so that its temperature rises from $27^{\circ} \mathrm{C}$ to $127^{\circ} \mathrm{C}$. The work done will be

Options:
A 2077.5 joules
B 207.5 joules
C $207.5 \mathrm{ergs}$
D None of the above

The correct answer is:
2077.5 joules

Solution:
$\begin{gathered}W=\frac{R}{\gamma-1}\left(T_1-T_2\right) \\ =\frac{8.31 \times\{(273+27)-(273+127)\}}{1.4-1}=-2077.5 \text { joules }\end{gathered}$

Topic: Physics, Thermodynamics, JEE Main
One hundred identical coins each with probability \( p \) of showing up heads are tossed once. If \( 0 < p < 1 \) and the probability
of heads showing on \( 50 \) coins is equal to that of heads showing on \( 51 \) coins, then the value of \( p \) is

Options:
A \( \frac{1}{2} \)
B \( \frac{49}{101} \)
C \( \frac{50}{101} \)
D \( \frac{51}{101} \)

The correct answer is:
\( \frac{51}{101} \)

Solution:
We have 100C50p501-p50= 100C51p511-p49
or 1-pp=100!51! ⋅49!×50! ⋅50!100!=5051 or 51-51p=50p
⇒ p=51101

Topic: Mathematics, Probability, JEE Main
One important component of cytoskeleton is

Options:
A flagellin
B chitin
C cartilage
D microtubules

The correct answer is:
microtubules

Solution:
Eukaryotic cells are composed of a network of fibrous proteins that are called cytoskeleton and it is present in the cytoplasm. The cytoskeletal filaments are microfilaments, microtubules, and intermediate filaments and they help in maintaining the shape of the cell. It is a dynamic structure because microtubules and microfilaments can depolymerize and repolymerize easily.

Topic: Biology, Cell - The Unit of Life, NEET
One kg of a diatomic gas is at a pressure of $8 \times 10^4 \mathrm{~N} / \mathrm{m}^2$. The density of the gas is $4 \mathrm{~kg} / \mathrm{m}^{-3}$. What is the energy of the gas due to its thermal motion?

Options:
A
$3 \times 10^4 \mathrm{~J}$

B
$5 \times 10^4 \mathrm{~J}$

C
$6 \times 10^4 \mathrm{~J}$

D
$7 \times 10^4 \mathrm{~J}$

The correct answer is:
$5 \times 10^4 \mathrm{~J}$

Solution:
Thermal energy corresponds to internal energy
Mass $=1 \mathrm{~kg}$
density $=8 \mathrm{~kg} / \mathrm{m}^3$
$\Rightarrow$ Volume $=\frac{\text { mass }}{\text { density }}=\frac{1}{8} \mathrm{~m}^3$
Pressure $=8 \times 10^4 \mathrm{~N} / \mathrm{m}^2$
$\therefore$ Internal Energy $=\frac{5}{2} \mathrm{P} \times \mathrm{V}=5 \times 10^4 \mathrm{~J}$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2009
One $\mathrm{Kg}$ of copper is drawn into a wire of $1 \mathrm{~mm}$ diameter and a wire of $2 \mathrm{~mm}$ diameter. The resistance of the two wires will be in the ratio

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

The correct answer is:
$16: 1$

Solution:
Mass $=\left(\pi \mathrm{r}_1^2 \ell_1\right) \sigma$ (Ist wire)
$\begin{aligned} & \text { Mass }=\left(\pi \mathrm{r}_1^2 \ell_2\right) \sigma(2 \text { nd wire) } \\ & \left(\pi \mathrm{r}_1^2 \ell_1\right) \sigma=\left(\pi \mathrm{r}_2^2 \ell_2\right) \sigma\end{aligned}$
$\begin{aligned} & \left(\pi \mathrm{r}_1^2 \ell_1\right) \sigma=\left(\pi \mathrm{r}_2^2 \ell_2\right) \sigma \\ & \frac{\ell_1}{\ell_2}=\left(\frac{r_2}{r_1}\right)^2\end{aligned}$
$\frac{R_1}{R_2}=\frac{\rho \frac{\ell_1}{A_1}}{\rho \frac{\ell_2}{A_2}}=\frac{\ell_1}{\ell_2} \times \frac{A_2}{A_1}=\frac{\ell_1}{\ell_2} \times\left(\frac{r_2}{r_1}\right)^2$
$\begin{aligned} & =\left(\frac{r_2}{r_1}\right)^4 \\ & \Rightarrow 16: 1\end{aligned}$

Topic: Physics, Current Electricity, JEE Main
One kilogram of ice at $0^{\circ} \mathrm{C}$ is mixed with one kilogram of water at $80^{\circ} \mathrm{C}$. The final temperature of the mixture is
(Take : specific heat of water $=4200 \mathrm{~J}^{-1} \mathrm{~K}^{-1}$, latent heat of ice $=336 \mathrm{~kJ}^{-1} \mathrm{~kg}^{-1}$ )

Options:
A $40^{\circ} \mathrm{C}$
B $60^{\circ} \mathrm{C}$
C $0^{\circ} \mathrm{C}$
D $50^{\circ} \mathrm{C}$

The correct answer is:
$0^{\circ} \mathrm{C}$

Solution:
$\begin{aligned} & \theta_{\text {mix }}=\frac{m_w \theta_w-\frac{m_i L_i}{c_w}}{m_i+m_w} \\ & \Rightarrow m_i=m_w \Rightarrow \theta_{\text {mix }}=\frac{\theta_w-\frac{L_i}{c_w}}{2}=\frac{80-\frac{336}{4.2}}{2}=0^{\circ} \mathrm{C}\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JEE Main
One litre of 1âM solution of an acid HA K a â=â 10 â4 at 25 â o C has pHâ=â2. It is diluted by water so the new pH becomes double. The solution was diluted to yÃ10z ml . The value of y+z2 is:

Options:

The correct answer is:
5.5

Solution:
pH=2âH+1=10-2=c1Î±1=1ÃÎ±1
Î±1=10-2
pH=4âH+2=10-4=1Ã1000VÃÎ±2
ka=10-4=Î±22C21-Î±2=Î±2ÃÎ±2ÃC21-Î±2
10-4=Î±2Ã10-41-Î±2âÎ±2=0.5
H+2=1Ã1000VÃÎ±2=10-4
1Ã1000VÃ0.5=10-4
V=5Ã106 ml
â x+y2=5+62=112=5.5

Topic: Chemistry, Ionic Equilibrium, JEE Main
One litre of $0.15 \mathrm{M} \mathrm{Na}_2 \mathrm{SO}_3$ aqueous solution is mixed with $500 \mathrm{~mL}$ of $0.2 \mathrm{M} \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ aqueous solution in acid medium. What is the number of moles of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ remaining in the solution after the reaction?

Options:
A 0.1
B 0.0125
C 0.025
D 0.05

The correct answer is:
0.05

Solution:
When $0.15 \mathrm{M} \mathrm{Na}_2 \mathrm{CO}_3$ (1 L) mixed with $0.2 \mathrm{M}$ $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ (5L) than total molarity $(M)$.
$\begin{aligned}
M & =\frac{M_1 V_1+M_2 V_2}{V_3} \\
M & =\frac{0.15 \times 1+0.2 \times 0.5}{1.5} \\
M & =0.15 \mathrm{~mol} / \mathrm{L}
\end{aligned}$
Remaining mole of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7=0.2-0.15$
$=0.05 \mathrm{~mol} / \mathrm{L}$

Topic: Chemistry, Redox Reactions, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 1)
One litre of oxygen at a pressure of $1 \mathrm{~atm}$ and two litres of nitrogen at a pressure of $0.5 \mathrm{~atm}$, are introduced into a vessel of volume $1 \mathrm{~L}$. If there is no change in temperature, the final pressure of the mixture of gas (in atm) is

Options:
A $1.5$
B $2.5$
C $2$
D $4$

The correct answer is:
$2$

Solution:
Ideal gas equation is given by
$$
p V=n R T
$$
For oxygen, $p=1 \mathrm{~atm}, V=1 \mathrm{~L}, n=n_{\mathrm{O}}$
Therefore Eq. (i) becomes
$$
\begin{aligned}
& \therefore & 1 \times 1 & =n_{\mathrm{O}_2} R T \\
& \Rightarrow & n_{\mathrm{O}_2} & =\frac{1}{R T}
\end{aligned}
$$
For nitrogen $p=0.5 \mathrm{~atm}, V=2 \mathrm{~L}, n=n_{\mathrm{N}}$
$$
\begin{array}{rlrl}
& \therefore & 0.5 \times 2 & =n_{\mathrm{N}_2} R T \\
\Rightarrow & n_{\mathrm{N}_2} & =\frac{1}{R T}
\end{array}
$$
For mixture of gas
$$
\begin{array}{rlrl}
& & p_{\text {mix }} V_{\text {mix }} & =n_{\text {mix }} R T \\
\text { Here, } & n_{\text {mix }} & =n_{\mathrm{O}}+n_{\mathrm{N}_2} \\
& \therefore \quad & \frac{p_{\text {mix }} V_{\text {mix }}}{R T} & =\frac{1}{R T}+\frac{1}{R T} \\
& \Rightarrow & p_{\text {mix }} V_{\text {mix }} & =2
\end{array}
$$
Here,

Topic: Physics, Thermodynamics, JEE Main
One main scale division of a vernier caliper is equal to $m$ units. If $\mathrm{n}^{\text {th }}$ division of main scale coincides with $(n+1)^{\text {th }}$ division of vernier scale, the least count of the vernier caliper is :

Options:
A $\frac{n}{(n+1)}$
B $\frac{1}{(n+1)}$
C $\frac{m}{(n+1)}$
D $\frac{m}{n(n+1)}$

The correct answer is:
$\frac{m}{(n+1)}$

Solution:
$\begin{aligned} & \mathrm{n} M S D=(\mathrm{n}+1) \mathrm{VSD} \\ & \Rightarrow 1 \mathrm{VSD}=\frac{\mathrm{n}}{\mathrm{n}+1} \mathrm{MSD} \\ & \mathrm{L} \cdot \mathrm{C}=1 \mathrm{MSD}-1 \mathrm{VSD} \\ & \mathrm{L} \cdot \mathrm{C}=\mathrm{m}-\mathrm{m}\left(\frac{\mathrm{n}}{\mathrm{n}+1}\right) \\ & \mathrm{L} \cdot \mathrm{C}=\mathrm{m}\left(\frac{\mathrm{n}+1-\mathrm{n}}{\mathrm{n}+1}\right) \\ & \Rightarrow \mathrm{L} \cdot \mathrm{C}=\left(\frac{\mathrm{m}}{\mathrm{n}+1}\right)\end{aligned}$

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2024 (09 Apr Shift 1)
One main scale division of a vernier callipers is a cm and nth division of the vernier scale coincide with n-1th division of the main scale. The least count of the callipers in mm is :

Options:
A 10nan-1
B 10an-1
C n-110n
D 10an

The correct answer is:
10an

Solution:
n-1a=na'

a'=n-1an

∴ L.C.=1MSD-1VSD

=a-a' cm

=a-n-1an

=na-na+an=an cm

=10an mm

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2021 (16 Mar Shift 1)
One mL of an organic compound was dissolved in ethanol and a very small drop of dilute alkali was added to it and them a drop of phenolphthalein was added. The red colour of the indicator was seen. The mixture was then heated. The colour disappeared in a few minutes. The organic compound is most likely to be

Options:
A an aldehyde
B a ketone
C a carboxylic acid
D an ester

The correct answer is:
an ester

Solution:
Phenolphthalein does not give any colour with a carboxylic acid containing a drop of dilute alkali.
The red colour of the indicator indicates that alkali remains un-reacted before heating the compound.
After heating, disappearance of red colour indicates the consumption of NaOH. This because the ester hydrolyses in presence of base to form alcohol and sodium salt of carboxylic acid.
RCOOR'+NaOH→ΔRCOONa+R'OH

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
One mol of an ideal gas was taken from A→B as shown in given figure. Magnitude of work involved in process is R=253Jmol K :

Options:
A 5 kJ
B 7.5 kJ
C 2.5 kJ
D None of these

The correct answer is:
2.5 kJ

Solution:
w=-nRΔT
=-1×253×300
=-2500 J

Topic: Chemistry, Thermodynamics (C), JEE Main
One mole crystal of a metal halide of the type MX with molecular weight 119 g having face centered cubic structure with unit cell length 6.58 Å was recrystallized. The density of the recrystallized crystal was found to be 2.44gcm-3. The type of defect introduced during the recrystalization may be

Options:
A additional M+ and X- ions at interstitial sites
B Schottky defect
C F-centre
D Frenkel defect

The correct answer is:
Schottky defect

Solution:
Calculated density d=ZÃmoler massNaÃvolume=2.78 gcm-3
Is greater then actual density 2.44 gcm-3 therefore density decreases and defect is Schottky defect

Topic: Chemistry, Solid State, JEE Main
One mole of 4-nitrocatechol (4-nitro-1,2-dihydroxybenzene) on treatment with an excess of NaH followed by one mole of methyl iodide and H 2 O respectively gives

Options:
A 4-nitro-1, 2-diamethoxybenzene
B 4-nitro-5methy-1, 2-dimethoxybenzene
C 2-methoxy-5 nitrophenol
D 2-methoxy-4nitrophenol

The correct answer is:
2-methoxy-4nitrophenol

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
One mole of a certain ideal gas obtains an amount of heat Q = 1.60 kJ when its temperature is increased by ÎT = 72 K, keeping its pressure constant. The value of CPCV for the gas is

Options:
A 1.60
B 1.40
C 1.50
D 1.30

The correct answer is:
1.60

Solution:
By the first law of thermodynamics
Δ Q = Δ U + Δ W
In an isobaric process
Δ Q = C p Δ T
and Δ U = C v Δ T (always)
∴ C p Δ T = C v Δ T + Δ W
or Δ W = C p - C v Δ T
or Δ W = R Δ T âµ C p - C v = R
∴ Δ W = 8.3 × 7 2 = 597.6 J
Δ Q = C p Δ T
∴ 1.6 × 1 0 0 0 = C p × 7 2
⇒ C p = 1.6 × 1 0 0 0 7 2 = 22.2 J mol - 1 K - 1
Δ U = Δ Q = Δ W = 1.6 × 1 0 0 0 - 597.6 J = 1002.4 J
But Δ U = C v Δ T
∴ C v = 1002.4 7 2 = 13.9 J mol - 1 K - 1
∴ γ = C p C v = 22.2 13.9 = 1.60

Topic: Physics, Thermodynamics, JEE Main
One mole of a diatomic gas does a work $\frac{\mathrm{Q}}{3}$, when the amount of heat supplied is
'Q'. In this process, the molar heat capacity of the gas is

Options:
A $\frac{15 \mathrm{R}}{4}$
B $\frac{9 \mathrm{R}}{4}$
C $\frac{7 \mathrm{R}}{4}$
D $\frac{3 \mathrm{R}}{4}$

The correct answer is:
$\frac{15 \mathrm{R}}{4}$

Solution:
The amount of heat required to increase the internal energy is
$\left(\mathrm{Q}-\frac{\mathrm{Q}}{3}\right)=\frac{2}{3} \mathrm{Q}$
For a diatomic gas, the amount of heat required to increase the internal energy is $C_{v}=\frac{5}{2} R$
$\begin{array}{l}
\therefore \frac{2}{3} Q=\frac{5}{2} R \\
\therefore Q=\frac{15}{4} R
\end{array}$

Topic: Physics, Kinetic Theory of Gases, JEE Main
One mole of a diatomic gas undergoes a process P=P0[1+(V/V0)3], where P0 and V0 are constants. The translational kinetic energy of the gas when V=V0 is given by

Options:
A 5P0V04
B 3P0V02
C 35P0V02
D 5P0V02

The correct answer is:
3P0V02

Solution:
P = P 0 1 + V / V 0 3 = P 0 2
T = P 0 V 0 2 R
Therefore, translational kinetic energy is equal to
3 2 RT = 3 R 2 P 0 V 0 2 R = 3 P 0 V 0 4

Topic: Physics, Thermodynamics, JEE Main
One mole of $\mathrm{A}(\mathrm{g})$ is heated to $\mathrm{T}(\mathrm{K})$ till the following equilibrium is obtained
$\mathrm{A}(\mathrm{g}) \stackrel{\mathrm{T}(\mathrm{K})}{\rightleftharpoons} \mathrm{B}(\mathrm{g})$
The equilibrium constant of this reaction is $10^{-1}$. After reaching the equilibrium, 0.5 moles of $\mathrm{A}(\mathrm{g})$ is added and heated. The equilibrium is again established. The value of $\frac{[\mathrm{A}]}{[\mathrm{B}]}$ is

Options:
A $10^{-1}$
B 10
C $10^{-2}$
D 100

The correct answer is:
10

Solution:
$\mathrm{K}_{\mathrm{C}}=\frac{[\mathrm{B}]}{[\mathrm{A}]}=10^{-1}$ or $\frac{1}{10}$
Since $\mathrm{K}_{\mathrm{C}}$ remains constant at a given temperature, the value of $\frac{[\mathrm{B}]}{[\mathrm{A}]}$ will still be the same.
$\Rightarrow \frac{[\mathrm{B}]}{[\mathrm{A}]}=\frac{1}{10} \Rightarrow \frac{[\mathrm{A}]}{[\mathrm{B}]}=10$

Topic: Chemistry, Chemical Equilibrium, AP EAMCET, AP EAMCET 2023 (17 May Shift 2)
One mole of a monatomic gas at pressure 2âatm,â279âK taken to final pressure 4âatm by a reversible path described by P/V= constant. Calculate the magnitude of Î E w for the process.

Options:

The correct answer is:
3

Solution:
dE = dq + dw (F L T)
P 1 = K â² V 1 and P 2 = K â² V 2 Also V 1 = P 1 K â² , V 2 = P 2 K â²
T 2 = T 1 P 2 V 2 P 1 V 1 = T 1 P 2 P 2 / K â² P 1 P 1 / K â² = T 1 P 2 2 P 1 2
T 2 = T 1 4 2 2 2 = 4 T 1
Î E = C v Î T = C v 4 T 1 - T 1 = 3 C v T 1
= 3 3 2 R T 1 = 9 R T 1 2
w = - â« v 1 v 2 pdv = - â« v 1 v 2 K â² v dv = - K â² v 2 2 v 1 v 2
K â² v 2 = K â² v v = P v = RT
w =- RT 2 T 1 T 2 = - R 2 T 2 - T 1 = - R 2 3 T 1
Î E w = 9 R T 1 / 2 - 3 R T 1 / 2 = - 3

Topic: Chemistry, Thermodynamics (C), JEE Main