datavorous/entrance-exam-dataset huggingface dataset:
A mass of $50 \mathrm{~g}$ of water in a closed vessel, with surroundings at a constant temperature takes 2 minutes to cool from $30^{\circ} \mathrm{C}$ to $25^{\circ} \mathrm{C}$. A mass of $100 \mathrm{~g}$ of another liquid in an identical vessel with identical surroundings takes the same time to cool from $30^{\circ} \mathrm{C}$ to $25^{\circ} \mathrm{C}$. The specific heat of the liquid is :
(The water equivalent of the vessel is $30 \mathrm{~g}$.)

Options:
A
$2.0 \mathrm{kcal} / \mathrm{kg}$

B
$7 \mathrm{kcal} / \mathrm{kg}$

C
$3 \mathrm{kcal} / \mathrm{kg}$

D
$0.5 \mathrm{kcal} / \mathrm{kg}$

The correct answer is:
$0.5 \mathrm{kcal} / \mathrm{kg}$

Solution:
As the surrounding is identical, vessel is identical time taken to cool both water and liquid (from $30^{\circ} \mathrm{C}$ to $25^{\circ} \mathrm{C}$ ) is same 2 minutes, therefore
$$
\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\text {water }}=\left(\frac{\mathrm{dQ}}{\mathrm{dt}}\right)_{\text {liquid }}
$$
or, $\frac{\left(\mathrm{m}_{\mathrm{W}} \mathrm{C}_{\mathrm{W}}+\mathrm{W}\right) \Delta \mathrm{T}}{\mathrm{t}}=\frac{\left(\mathrm{m}_{\ell} \mathrm{C}_{\ell}+\mathrm{W}\right) \Delta \mathrm{T}}{\mathrm{t}}$
$(\mathrm{W}=$ water equivalent of the vessel $)$
or, $\mathrm{m}_{\mathrm{w}} \mathrm{C}_{\mathrm{w}}=\mathrm{m}_{\ell} \mathrm{C}_{\ell}$
$\therefore$ Specific heat of liquid, $\mathrm{C}_{\ell}=\frac{\mathrm{m}_{\mathrm{W}} \mathrm{C}_{\mathrm{W}}}{\mathrm{m}_{\ell}}$
$$
=\frac{50 \times 1}{100}=0.5 \mathrm{kcal} / \mathrm{kg}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2013 (25 Apr Online)
A mass of 50 kg is placed at the center of a uniform spherical shell of mass 100 kg and radius 50 m. If the gravitational potential at a point, 25 m from the center is V kg m-1. The value of V is:

Options:
A +2G
B -20G
C -4G
D -60G

The correct answer is:
-4G

Solution:
Use the formula of the gravitational potential in terms of universal gravitational constant, V=-GMR⇒Vp=-G(100)50-G(50)25=-4G.

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (27 Aug Shift 2)
A mass of diatomic gas $(\gamma=1.4)$ at a pressure of $2 \mathrm{~atm}$ is compressed adiabatically so that its temperature rise from $27^{\circ} \mathrm{C}$ to $927^{\circ} \mathrm{C}$. The pressure of the gas is final state is

Options:
A $28 \mathrm{~atm}$
B $68.7 \mathrm{~atm}$
C $256 \mathrm{~atm}$
D $8 \mathrm{~atm}$

The correct answer is:
$256 \mathrm{~atm}$

Solution:
$\begin{aligned}
& T_1=273+27=300 \mathrm{~K} \\
& T_2=273+927=1200 \mathrm{~K}
\end{aligned}$
Gas equation for adiabatic process
$\begin{aligned}
& p V^\gamma=\text { constant } \\
& p\left(\frac{T}{p}\right)^\gamma=\text { constant } \quad(\because p V=R T) \\
& \therefore \quad \frac{p_2}{p_1}=\left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma-1}} \\
& p_2=p_1\left(\frac{T_2}{T_1}\right)^{\frac{\gamma}{\gamma-1}} \\
& p_2=2\left(\frac{1200}{300}\right)^{\frac{1.4}{1.4-1}} \\
& p_2=256 \mathrm{~atm}
\end{aligned}$

Topic: Physics, Thermodynamics, NEET, NEET 2011 (Mains)
A mass of $\mathrm{M} \mathrm{kg}$ is suspended by a weightless string. The horizontal force that is required to displace it until the string makes an angle of $45^{\circ}$ with the initial vertical direction is

Options:
A
$M g(\sqrt{2}-1)$

B
$M g(\sqrt{2}+1)$

C
$M g \sqrt{2}$

D
$\frac{\mathrm{Mg}}{\sqrt{2}}$

The correct answer is:
$M g(\sqrt{2}-1)$

Solution:
$\mathrm{F} \ell \sin 45=M g(\ell-\ell \cos 45)$
$F=M g(\sqrt{2}-1)$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2006
A massage signal of frequency 3 kHz is used to modulate a carrier signal of frequency 1.5 MHz. The bandwidth of the amplitude modulated wave is

Options:
A 6 MHz
B 6 kHz
C 3 MHz
D 3 kHz

The correct answer is:
6 kHz

Solution:
The relation between bandwidth and the signal frequency is given by

Bandwidth=2× signal frequency ...1

Substitute the values of the known parameter into equation (1) to calculate the required bandwidth.

Bandwidth= 2 × 3 kHz= 6 kHz

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (10 Apr Shift 2)
A massive black hole of mass $m$ and radius $R$ is spinning with angular velocity $\omega .$ The power $\mathrm{P}$ radiated by it as gravitational waves is given by $\mathrm{P}=\mathrm{Gc}^{-5} \mathrm{~m}^{\mathrm{x}} \mathrm{R}^{\mathrm{y}} \omega^{\mathrm{z}}$, where $\mathrm{c}$ and $\mathrm{G}$ are speed of light in free space, and the universal gravitational constant, respectively. Then

Options:
A $x=-1, y=2, z=4$
B $x=1, y=1, z=4$
C $x=-1, y=4, z=4$
D $x=2, y=4, z=6$

The correct answer is:
$x=2, y=4, z=6$

Solution:
$\mathrm{P}=\mathrm{ML}^{2} \mathrm{~T}^{-3}, \mathrm{c}=\mathrm{LT}^{-1}, \omega=\mathrm{T}^{-1}, \mathrm{R}=\mathrm{L}, \mathrm{m}=\mathrm{M}$
$\mathrm{G}=\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}$
$\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^{3} \mathrm{~T}^{-2}\right]\left[\mathrm{LT}^{-1}\right]^{-5} \mathrm{M}^{\mathrm{x}} \mathrm{L}^{\mathrm{y}} \mathrm{T}^{-\mathrm{z}}$
solve we get $\mathrm{x}=2, \mathrm{y}=4, \mathrm{z}=6$

Topic: Physics, Units and Dimensions, JEE Main
A massive vertical wall is approaching a man at a speed u. When it is at a distance of 10 m, the man throws a ball with speed 10 m/s at an angle of 37°, which after having a completely elastic collision with the wall, reaches back directly into the hands of the man. The velocity of the wall is

Options:
A 13 3 m/s
B 18 2 m/s
C 26 4 m/s
D 31 5 m/s

The correct answer is:
13 3 m/s

Solution:
As the vertical motion remains unaffected
â T = 2 u sin θ g ⇒ T = 2 × 10 × 3/5 10
T=65s ...(1)
from coefficient of restitutation along the line of collision
e = relative velocity of separation after collision relative velocity of approach before collision

1 = v - u 8 + u
v - 2u = 8 ...(2)
by the time (t1) ball collide with wall, distance travelled by man = 10 - ut1
t 1 = 10 8 + u ...(3)
by the time t 2 ball reach the hand of man t 2 = 10 - u 1 v ...(4)
Since T = t1 + t2 ...(5)
from equation (2), (3), (4), (5)
u = 13 3 m/s

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A massless platform is kept on a light elastic spring, as shown in the figure. When particle of mass 0.1 kg is dropped on the pan from a height of 0.24 m, the particle sticks to the pan, and the spring is compressed by 0.01 m. From what height should the particle be dropped to cause a compression of 0.04 m?

Options:
A 0.96 m
B 2.96 m
C 3.96 m
D 0.48 m

The correct answer is:
3.96 m

Solution:
Let the particle be dropped from a height h and the spring be compressed by y. According to conservation of mechanical energy,
Loss in potential energy of the particle = gain in elastic potential energy of the spring
∴ mg hâ¡ + y = 1 2 k y 2
Now, as the particle and spring remain same, hence
hâ¡ 1 + y 1 hâ¡ 2 + y 2 = y 1 y 2 2
0 · 2 4 + 0 · 0 1 hâ¡ 2 + 0 · 0 4 = 0 · 0 1 0 · 0 4 2
Solving we get, h2 = 3.96 m

Topic: Physics, Work Power Energy, JEE Main
A massless rod BD of length l is suspended by two identical massless strings AB and CD of equal lengths. A block of mass m is suspended from a point P such that BP is equal to x. If the fundamental frequency of the left wire is twice the fundamental frequency of the right wire, then the value of lx is

Options:

The correct answer is:
5

Solution:
f⁡ ∝ v⁡ ∝ T⁡
f⁡ AB = 2 f⁡ CD
∴ T AB = 4 T CD ...(i)
Further Σ τ p = 0
∴ T AB x = T CD ℓ - x
or 4 x = l⁡ - x T AB = 4 T CD
or x = l⁡ / 5

⇒ lx=5

Topic: Physics, Waves and Sound, JEE Main
A massless rod of length l is hung from the ceiling with the help of two identical wires attached at its ends. A block is hung on the case, the frequency of the 1st harmonic of the wire on the left end is equal to the frequency of the 2nd harmonic of the wire on the right. The value of X is

Options:
A l2
B l3
C l4
D l5

The correct answer is:
l5

Solution:
v=λf
T1μT2μ=2lflf
T1T2=4
T1+T2=Mg ; T2=Mg5
ΣtB=0⇒Mg×X=Mg5×l ;
X=l5

Topic: Physics, Waves and Sound, JEE Main
A massless spring k=800 N/m, attached with a mass 500 g is completely immersed in 1 kg of water. The spring is stretched by 2 cm and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely? (Assume that the water container and spring receive negligible heat and specific heat of mass =400 J/kg K, specific heat of water =4184 J/kg K )

Options:
A 10-5 K
B 10-1 K
C 10-3 K
D 10-4 K

The correct answer is:
10-5 K

Solution:
According to the law of conservation of energy,
12kx2=m1s1+m2s2×ΔT
⇒ΔT=12×800×2×10-220.5×400+1×4184=16×10-24384 =3.65×10-5

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2019 (09 Apr Shift 2)
A massless square loop, of wire of resistance 10 Ω, supporting a mass of 1 g, hangs vertically with one of its sides in a uniform magnetic field of 103 G, directed outwards in the shaded region. A dc voltage V is applied to the loop. For what value of V, the magnetic force will exactly balance the weight of the supporting mass of 1 g?

(If sides of the loop = 10 cm, g=10 m s-2)

.

Options:
A 110V
B 100 V
C 1 V
D 10 V

The correct answer is:
10 V

Solution:
For balancing of force,

Magnetic force Fm=weight of supporting mass mg.

∴ILB=mg

∴VRLB=mg ∵From Ohm's law, I=VR

∴V=mgRLB

=1×10-3 kg10 m s-210 Ω0.1 m103×10-4 T=10 V

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2023 (30 Jan Shift 1)
A massless string and a spring connect two blocks A and B to each other. Block B slides over a frictionless inclined plane while block A slides over horizontal surface. Coefficient of friction between block A horizontal surface is μ=0.2.At the instant shown blocks are moving with constant speed. Mass of block A and energy stored in spring are respectively. [ g=10  m/ s 2  ,  m B =2 kg ]

Options:
A 5Â kg,Â 1J
B 10Â kg,Â 0.05Â J
C 5 kg, 0.05 J
D 10Â kg,Â 1J

The correct answer is:
5 kg, 0.05 J

Solution:
2 g sin θ=kx =μ mA g
⇒ m A = 2 g sin θ μ g = 2 sin θ μ = 2 × 1 2 0 . 2 = 5 kg
kx = 2 × 1 0 × 1 2 = 1 0
x = 1 0 1 × 1 0 3 = 1 cm
⇒ Energy = 1 2 × 1 0 3 × 1 0 - 2 2 = 0 . 0 5 Joule

Topic: Physics, Work Power Energy, JEE Main
A material ' $B$ ' has twice the specific resistance of ' $A$ '. A circular wire made of ' $B$ ' has twice the diameter of a wire made of ' $A$ '. Then for the two wires to have the same resistance, the ratio $\ell_A / \ell_B$ of their respective lengths must be

Options:
A
2

B
1

C
$\frac{1}{2}$

D
$\frac{1}{4}$

The correct answer is:
2

Solution:
$\mathrm{R}_1=\frac{\rho_{\mathrm{A}} \ell_{\mathrm{A}}}{\pi \mathrm{R}_{\mathrm{A}}^2} \quad \mathrm{R}_2=\frac{\rho_{\mathrm{B}} \ell_{\mathrm{B}}}{\pi \mathrm{R}_{\mathrm{B}}^2}$
$\frac{\ell_{\mathrm{A}}}{\ell_{\mathrm{B}}}=\frac{\rho_{\mathrm{B}} \mathrm{R}_{\mathrm{A}}^2}{\rho_{\mathrm{A}} \mathrm{R}_{\mathrm{B}}^2}=\frac{2 \rho_{\mathrm{A}} \mathrm{R}_{\mathrm{A}}^2}{\rho_{\mathrm{A}} \cdot 4 \mathrm{R}_{\mathrm{A}}^2} \Rightarrow \frac{\ell_{\mathrm{B}}}{\ell_{\mathrm{A}}}=2$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2006
A material has Poisson's ratio 0.50 . If a uniform rod of it suffers a longitudinal strain of $2 \times 10^{-3}$, then the percentage change in volume is

Options:
A 0.6
B 0.4
C 0.2
D Zero

The correct answer is:
0.4

Solution:
$\frac{d V}{V}=(1+2 \sigma) \frac{d L}{L}$
$\frac{d V}{V}=2 \times 2 \times 10^{-3}=4 \times 10^{-3}\left[\because \sigma=0.5=\frac{1}{2}\right]$
$\therefore$ Percentage change in volume $=4 \times 10^{-1}=0.4 \%$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A material of \( 0.25 \mathrm{~cm}^{2} \) cross-sectional area is placed in a magnetic field of strength (H) \( 1000 \mathrm{~A} \mathrm{~m}^{-1} \). Then the magnetic flux produced is (Susceptibility of material is \( 313 \) ) (Permeability of free space, \( \mu_{0}=4 \pi \times 10^{-7} \mathrm{H} \mathrm{m}^{-1} \) )

Options:
A \( 8.33 \times 10^{-8} \) weber
B \( 1.84 \times 10^{-6} \) weber
C \( 9.87 \times 10^{-6} \) weber
D \( 3.16 \times 10^{-6} \) weber

The correct answer is:
\( 9.87 \times 10^{-6} \) weber

Solution:
B=μH
μr=1+χ
ϕA=μr . μ0H
1+313=314=100π
ϕ=Aμr . μ0H
μ=μr . μ0, π2=10 ≈9.87
=0.25×10-4×314×4π×10-7×1000
=25×10-6×100π×4π×10-4
=25×4×10-6+2+π2×10-4
=100×10×10-8=10-5
≈10×10-6 wb

Topic: Physics, Magnetic Properties of Matter, JEE Main
A matrix whose elements $a_{i j}$ are defined by
$a_{i j}=\frac{1}{3}|i-5 j|, i, j=1,2,3 \text { is }$

Options:
A $\left[\begin{array}{ccc}4 & 3 & \frac{14}{3} \\ 1 & \frac{8}{3} & 13 \\ \frac{2}{3} & \frac{7}{3} & 4\end{array}\right]$
B $\left[\begin{array}{ccc}\frac{4}{3} & 3 & \frac{14}{3} \\ 1 & \frac{8}{3} & \frac{13}{3} \\ \frac{2}{3} & \frac{7}{3} & 4\end{array}\right]$
C $\left[\begin{array}{ccc}\frac{4}{3} & 3 & \frac{10}{3} \\ 1 & \frac{8}{3} & \frac{13}{3} \\ 2 & 7 & 4\end{array}\right]$
D $\left[\begin{array}{lll}4 & 3 & 10 \\ 1 & 8 & 13 \\ 2 & 7 & 4\end{array}\right]$

The correct answer is:
$\left[\begin{array}{ccc}\frac{4}{3} & 3 & \frac{14}{3} \\ 1 & \frac{8}{3} & \frac{13}{3} \\ \frac{2}{3} & \frac{7}{3} & 4\end{array}\right]$

Solution:
The given matrix is defined by
$\mathrm{a}_{\hat{\mathrm{ij}}}=\frac{1}{3}|\mathrm{i}-5 \mathrm{j}|$
So, $\left(\begin{array}{lll}a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}\end{array}\right)=\left(\begin{array}{ccc}\frac{4}{3} & 3 & \frac{14}{3} \\ 1 & \frac{8}{3} & \frac{13}{3} \\ \frac{2}{3} & \frac{7}{3} & 4\end{array}\right)$

Topic: Mathematics, Matrices, JEE Main
A matrix $X$ has $(a+b)$ rows and $(a+2)$ columns; and a matrix $Y$ has $(b+1)$ rows and $(a+3)$ columns. If both $X Y$ and $Y X$ exist, then what are the values of $a, b$ respectively?

Options:
A 3,2
B 2,3
C 2,4
D 4,3

The correct answer is:
2,3

Solution:
The order of a given matrices are
$[X]_{(a+b)\times(a+2)}$ and $[Y]_{(b+1)\times(a+3)}$
As $[X Y]$ and $[Y X]$ exist
$\therefore a+2=b+1$ and $a+3=a+b$
$\Rightarrow a+3=a+b$
$\Rightarrow b=3$
Hence, $a=3+1-2=2$

Topic: Mathematics, Matrices, JEE Main
A mental sphere of radius $r$ and specific heat $S$ is rotated about an axis passing through its centre at a speed of $n$ rotations per second. It is suddenly stopped and $50 \%$ of its energy is used in increasing its temperature. Then, the raise in temperature of the sphere is

Options:
A $\frac{2 \pi^2 n^2 r^2}{5 S}$
B $\frac{1 \pi^2 n^2}{10 r^2 S}$
C $\frac{7}{8} \pi r^2 n^2 S$
D $\frac{5(\pi r n)^2}{14 S}$

The correct answer is:
$\frac{2 \pi^2 n^2 r^2}{5 S}$

Solution:
Rotational KE of sphere $=\frac{1}{2} I \omega^2$
For sphere, $\quad I=\frac{2}{5} m r^2$
$$
\begin{aligned}
& \text { KE } \\
& =\frac{1}{2} \times \frac{2}{5} m r^2(2 \pi n)^2=\frac{1}{5} m r^2 \times 4 \pi^2 n^2 \\
& =\frac{4}{5} m r^2 \pi^2 n^2 \mathrm{~J} \\
& \therefore \text { Heat produced } \Delta Q=\frac{1}{2} \mathrm{KE} \\
& =\frac{1}{2}\left(\frac{4}{5} m r^2 \pi^2 n^2\right) \\
& =\frac{2}{5} m r^2 \pi^2 n^2 \\
& \therefore \quad \Delta Q=m s \Delta t \\
& \therefore \text { In temperature } \\
&
\end{aligned}
$$
$$
\begin{aligned}
\Delta t & =\frac{\Delta Q}{m s}=\frac{2}{5} \frac{m r^2 \pi^2 n^2}{m \cdot s} \\
& =\frac{2}{5} \frac{r^2 \pi^2 n^2}{s}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
A mercury drop of radius 10–3 m is broken into 125 equal size droplets. Surface tension of mercury is 0.45 N m–1 . The gain in surface energy is:

Options:
A 2.26 × 10–5 J
B 28 × 10–5 J
C 17.5 × 10–5 J
D 5 × 10–5 J

The correct answer is:
2.26 × 10–5 J

Solution:
As the volume of the mercury will remain same, therefore

43πR3=12543πr3⇒R=5r

Now, increase in surface energy

∆U=125×S×4πr2-S×4πR2

=0.45×4π×10-3212525-1

=4×0.45×4π×10-6

=2.26×10-5J

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2023 (01 Feb Shift 1)
A merry-go-round is at rest, pivoted on a frictionless axis. A child of mass m runs along the path tangential to the rim with speed v and jumps on to merry-go-round. If R is the radius of the merry-go-round and I is the moment of inertia, then the angular velocity of the merry-go-round and the child is

Options:
A mvR m R 2 +I
B mvR I
C m R 2 +I mvR
D I mvR

The correct answer is:
mvR m R 2 +I

Solution:
By law of conservation of angular momentum
mvR=Isystemω
⇒mvR=I+mR2ω
⇒ω=mvRI+mR2

Topic: Physics, Rotational Motion, JEE Main
A message signal of frequency 5 kHz is used to modulate a carrier signal of frequency 2 MHz. The bandwidth for amplitude modulation is:

Options:
A 5 kHz
B 20 kHz
C 10 kHz
D 2.5 kHz

The correct answer is:
10 kHz

Solution:
We have,

Message frequency fm=5 kHz and carrier wave frequencyfc=2 MHz=2000 KHz

The resultant signal will have bandwidth of frequency given by

fc+fm-fc-fm

=[(2000+5)-(2000-5)] kHz

=10 kHz.

Thus, bandwidth for amplitude modulation is 10 kHz.

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (25 Jan Shift 1)
A message signal of frequency \( 10 \mathrm{kHz} \) and peak voltage \( 10 \mathrm{~V} \) is used to modulate a carrier wave of frequency \( 1 \mathrm{MHz} \) and peak voltage \( 20 \) volts. The bandwidth of signal is

Options:
A \( 10 \mathrm{kHz} \)
B \( 20 \mathrm{kHz} \)
C \( 30 \mathrm{kHz} \)
D \( 40 \mathrm{kHz} \)

The correct answer is:
\( 20 \mathrm{kHz} \)

Solution:
Band width of signal =2fm

Topic: Physics, Communication System, JEE Main
A message signal of frequency 20 kHz and peak voltage of 20 V is used to modulate a carrier wave of frequency 1 MHz and peak voltage of 20 V. The modulation index will be:

Options:

The correct answer is:
1

Solution:
Modulation index

μ=AmAc=2020=1

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (25 Jul Shift 2)
A message signal of frequency 100 MHz and peak voltage 100 V is used execute amplitude modulation on a carrier wave of frequency 300 GHz and peak voltage 400 V . The modulation index and difference between the two side band frequencies are:

Options:
A 4;2×108 Hz
B 0.25;2×108 Hz
C 4;1×108 Hz
D 0.25;1×108 Hz

The correct answer is:
0.25;2×108 Hz

Solution:
Modulating index is given by,
μ=AmAe
=100 V400 V
=0.25
The difference between the maximum frequency of modulated wave and the minimum frequency of modulated wave is given by
Δf=fmax-fmin
=fc+fm-fc-fm
=2fm
=2(100 MHz)
=2×108 Hz

Topic: Physics, Communication System, JEE Main, JEE Main 2019 (10 Apr Shift 1)
A metal A on heating in nitrogen gas gives compound B. B on treatment with H2O gives a colourless gas which when passed through CuSO4 solution gives a dark blue-violet coloured solution. A and B respectively, are:

Options:
A Na and NaNO3
B Na and Mg3N2
C Mg and Mg3N2
D Mg and MgNO32

The correct answer is:
Mg and Mg3N2

Solution:
Magnesium nitride ceramic powder can be used in the preparation of other nitride compound with high hardness, high thermal conductivity, corrosion resistance, wear resistance and high temperature resistance properties; It can be used for preparation of special ceramic material and for special alloy blowing agent

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (08 Jan Shift 2)
A metal ball of mass $0.1 \mathrm{~kg}$ is heated upto $500^{\circ} \mathrm{C}$ and dropped into a vessel of heat capacity $800 \mathrm{JK}^{-1}$ and containing $0.5 \mathrm{~kg}$ water. The initial temperature of water and vessel is $30^{\circ} \mathrm{C}$. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, $4200 \mathrm{Jkg}^{-}$ ${ }^{1} \mathrm{~K}^{-1}$ and $400 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$ ]

Options:
A $15 \%$
B $30 \%$
C $25 \%$
D $20 \%$

The correct answer is:
$20 \%$

Solution:
Assume final temperature $=\mathrm{T}^{\circ} \mathrm{C}$ Heat lass $=$ Heat gain $=\mathrm{ms} \Delta \mathrm{T}$ $\Rightarrow \mathrm{m}_{\mathrm{B}} \mathrm{s}_{\mathrm{B}} \Delta \mathrm{T}_{\mathrm{B}}=\mathrm{m}_{\mathrm{w}} \mathrm{s}_{\mathrm{w}} \Delta \mathrm{T}_{\mathrm{w}}$

$0.1 \times 400 \times(500-\mathrm{T})$

$\quad=0.5 \times 4200 \times(\mathrm{T}-30)+800(\mathrm{~T}-30)$

$\Rightarrow 40(500-\mathrm{T})=(\mathrm{T}-30)(2100+800)$

$\Rightarrow 20000-40 \mathrm{~T}=2900 \mathrm{~T}-30 \times 2900$

$\Rightarrow 20000+30 \times 2900=\mathrm{T}(2940)$

$\mathrm{T}=30.4^{\circ} \mathrm{C}$

$\frac{\Delta \mathrm{T}}{\mathrm{T}} \times 100=\frac{6.4}{30} \times 100=21 \%$

so the closest answer is $20 \%$.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2019 (11 Jan Shift 2)
A metal ball of mass \( 0.1 \mathrm{~kg} \) is heated upto \( 500^{\circ} \mathrm{C} \) and dropped into a vessel of heat capacity \( 800 \mathrm{~J} \mathrm{~K}^{-1} \) and containing \( 0.5 \mathrm{~kg} \) water. The initial temperature of water and vessel is \( 30^{\circ} \mathrm{C} \). What is the approximate percentage increment in the temperature of the water? [Take specific heat capacities of water and metal are, respectively, \( 4200 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} \) and \( 400 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1} \) ]

Options:
A \( 25 \% \)
B \( 15 \% \)
C \( 30 \% \)
D \( 20 \% \)

The correct answer is:
\( 20 \% \)

Solution:
Given:

The mass of the metal ball, mb=0.1kg

The temperature of the ball, Tb=500 °C

Heat capacity of the vessel, C=800 J K-1

Mass of water in vessel, mw=0.5 kg

The initial temperature of water and vessel, Ti=30 °C

We need to find the percentage increment in temperature of the water.

We know from the principle of calorimetry that the total heat absorbed by the vessel and water would be equal to the heat released by the metal ball.

Thus,

Qb=Qv+Qw …1

We know,

Specific heat, c=1mdQdT⇒dQ=mcdT

Let the final temperature of the ball be T.

Then heat lost by the ball is,

Qb=0.1×400500-T …(2)

Heat gained by water,

Qw=0.5×4200T-30 ...3

Heat gained by the vessel is

Qv=C×ΔT

⇒Qv=800×T-30 …4

On substituting the values in the equation 1, we get,

⇒0.1×400500-T=0.5×4200T-30+800T-30

⇒500-T=72.5T-30

⇒T=36.39 °C

Thus, percentage increment in temperature of water,

∆TT×100=36.39 °C-30 °C30×100≈20%

Topic: Physics, Thermal Properties of Matter, JEE Main
A metal ball of mass 2 kg moving with a velocity of 36kmph has a head on collision with a stationary ball of mass 3 kg. After the collision, if both balls move together, the loss in kinetic energy due to collision is

Options:
A 40 J
B 60 J
C 100 J
D 140 J

The correct answer is:
60 J

Solution:
The linear momentum remains conserved before and after the collision, therefore,

⇒2 kg×36×10003600 m s-1+3 kg×0=3 kg+2 kgvv=205=4 m s-1

The loss in kinetic energy is, ∆KE=KEi-KEf=122 kg10 m s-12+0-122 kg+3 kg4 m s-12=100-40=60 kg m2 s-2 =60 J

Topic: Physics, Center of Mass Momentum and Collision, AP EAMCET, AP EAMCET 2021 (19 Aug Shift 1)
A metal ball of surface area 200 cm2, temperature 527 oC is surrounded by a vessel at 27 oC. If the emissivity of metal is 0.4, then the rate of loss of heat is nearly Ï=5.67Ã10-8 Wm-2 K-4

Options:
A 108 W
B 168 W
C 182 W
D 192 W

The correct answer is:
182 W

Solution:
Rate of heat loss per unit time per unit area =eÏAT4-T04
=0.4Ã5.67Ã10-8Ã200Ã10-4Ã8004-3004
=5.67Ã0.4Ã2Ã10-8+2-4Ã108642-92
=5.67Ã0.8Ã10-255Ã73
=18212.04Ã10-2
=182.12 W

Topic: Physics, Thermal Properties of Matter, JEE Main
A metal bar AB can slide on two parallel thick metallic rails separated by a distance l. A resistance R and an inductance L are connected to the rails as shown in the figure. A long straight wire, carrying a constant current I0 is placed in the plane of the rails as shown. The bar AB is held at rest at a distance x0 from the long wire. At t = 0, it is made to slide on the rails away from the wire.

It is observed that at time t=T, the metal bar AB is at a distance of 2x0 from the long wire and the resistance R carries a current i1. The net charge that has flown through resistance R from t=0 to t=T is

Options:
A 1Rμ0l2πln2-Li1
B 1Rμ0I0lπln2-Li1
C 1RÎ¼0I0l2Ïln2-Li1
D 1Rμ0I02πln2-Li1

The correct answer is:
1Rμ0I0l2πln2-Li1

Solution:
Let v be velocity of AB when AB is at a distance x from the wire,
e AB = B . â . v = Î¼ 0 I 0 2 Ï x I . v = d Ï d t
For circuit Loop,
d Ï d t - i R - L d i d t = 0
d Ï d t - d q d t R - L d i d t = 0
â d q = d Ï - L d i R
Î q = Î Ï - L Î I R
= 1 R [ Ï 1 - Ï 2 - L i 1 ]
Îq=1R[Î¼0âI02Ïln2-Li1]

Topic: Physics, Electromagnetic Induction, JEE Main