datavorous/entrance-exam-dataset huggingface dataset:
A bullet of mass $4 \mathrm{~g}$ is fired horizontally with a speed of $300 \mathrm{~m} / \mathrm{s}$ into $0.8 \mathrm{~kg}$ block of wood at rest on a table. If the coefficient of friction between the block and the table is $0.3$, how far will the block slide approximately?

Options:
A
$0.19 \mathrm{~m}$

B
$0.379 \mathrm{~m}$

C
$0.569 \mathrm{~m}$

D
$0.758 \mathrm{~m}$

The correct answer is:
$0.379 \mathrm{~m}$

Solution:
Given, $\mathrm{m}_1=4 \mathrm{~g}, \mathrm{u}_1=300 \mathrm{~m} / \mathrm{s}$ $\mathrm{m}_2=0.8 \mathrm{~kg}=800 \mathrm{~g}, \mathrm{u}_2=0 \mathrm{~m} / \mathrm{s}$ From law of conservation of momentum, $\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2$ Let the velocity of combined system $=\mathrm{v} \mathrm{m} / \mathrm{s}$ then, $4 \times 300+800 \times 0=(800+4) \times v$ $\mathrm{v}=\frac{1200}{804}=1.49 \mathrm{~m} / \mathrm{s}$
Now, $\mu=0.3$ (given)
$$
\begin{aligned}
&\mathrm{a}=\mu \mathrm{g} \\
&\mathrm{a}=0.3 \times 10\left(\text { take } \mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\right) \\
&=3 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
$$
then, from $\mathrm{v}^2=\mathrm{u}^2+2 \mathrm{as}$
$$
(1.49)^2=0+2 \times 3 \times s
$$
$\mathrm{s}=\frac{\left(1.49^2\right)}{6} ; \mathrm{s}=\frac{2.22}{6}=0.379 \mathrm{~m}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2014 (12 Apr Online)
A bullet of mass 5 gram, travelling with a speed of 210 m s-1 strikes a fixed wooden target. One half of its kinetic energy is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is 0.030 gram °C-1 1 calorie=4.2×107 ergs close to :

Options:
A 87.5°C
B 83.3°C
C 119.2°C
D 38.4°C

The correct answer is:
87.5°C

Solution:
As per given condition Half of kinetic energy is transferred into heat energy and raises the temperature of bullet

12×12mv2=msΔT

Δt=V24s

=210×2104×4.2×0.3×1000=87.5°C

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2020 (05 Sep Shift 1)
A bullet of mass 10 g and speed 500 m s-1 is fired into a door and gets embedded exactly at the centre of the door. The door is 1.0 m wide and weighs 12 kg. It is hinged at one end and rotates about a vertical axis practically without friction. If Ï is the angular speed (in rad s-1) of the door just after the bullet embeds into it, then find the value of 10Ï. Ignore the mass of the bullet as compared to the door.

Options:

The correct answer is:
6.25

Solution:
Given, mass of bullet (m) = 10 g = 0.01 kg
Speed of bullet (v) = 500 m s-1
Width of the door (l) = 1.0 m
Mass of the door (M) = 12 kg
As bullet gets embedded exactly at the centre of the door, therefore its distance from the hinged end of the door,
â (r) = lâ¡ 2 = 1 2 m
Angular momentum transferred by the bullet to the door,
(L) = mv Ã r
= 0.01 Ã 500 Ã 1 2
= 2.5 J-s
Moment of inertia of the door about the vertical axis at one of its end,
(I) = M lâ¡ 2 3 = 12 Ã (1) 2 3 = 4 kg-m 2
But angular momentum, (L) = IÏâ
â´ 2.5 = 4 Ã Ï
or Ï = 2.5 4
Ï = 0.625 rad s-1

Topic: Physics, Rotational Motion, JEE Main
A bullet of mass $10 \mathrm{~g}$ and speed $500 \mathrm{~m} / \mathrm{s}$ is fired into a door and gets embedded exactly at the centre of the door. The door is $1.0 \mathrm{~m}$ wide and weighs $12 \mathrm{~kg}$. It is hinged at one end and rotates about a vertical axis practically without friction. The angular speed of the door just after the bullet embeds into it will be:

Options:
A $6.25 \mathrm{rad} / \mathrm{sec}$
B $0.625 \mathrm{rad} / \mathrm{sec}$
C $3.35 \mathrm{rad} / \mathrm{sec}$
D $0.335 \mathrm{rad} / \mathrm{sec}$

The correct answer is:
$0.625 \mathrm{rad} / \mathrm{sec}$

Solution:
Angular momentum imparted by bullet on the door \(=\mathrm{mvr}\)
\(=\left(10 \times 10^{-3}\right) \times 500 \times 0.5 \mathrm{kgm}^2 / \mathrm{s}\)
Moment of inertia of the door, \(\mathrm{I}=\mathrm{ML}^2 / 3=\frac{1}{3} \times 12 \times 1^2=4 \mathrm{kgm}^2\) Angular momentum of the system after the bullet gets embedded \(\approx \mathrm{I} \omega\)
From conservation of angular momentum about the rotation axis,
\(\begin{aligned}
\mathrm{mvr} & =\mathrm{I} \omega \\
\Rightarrow \omega & =0.625 \mathrm{rad} / \mathrm{s}
\end{aligned}\)

Topic: Physics, Rotational Motion, JEE Main
A bullet of mass $10 \mathrm{~g}$ and speed $500 \mathrm{~m} / \mathrm{s}$ is fired into a door and gets embedded exactly at the centre of the door. The door is $1.0 \mathrm{~m}$ wide and weighs $12 \mathrm{~kg}$. It is hinged at one end and rotates about a vertical axis practically without friction. The angular speed of the door just after the bullet embeds into it will be :

Options:
A
$6.25 \mathrm{rad} / \mathrm{sec}$

B
$0.625 \mathrm{rad} / \mathrm{sec}$

C
$3.35 \mathrm{rad} / \mathrm{sec}$

D
$0.335 \mathrm{rad} / \mathrm{sec}$

The correct answer is:
$0.625 \mathrm{rad} / \mathrm{sec}$

Solution:
No solution is given.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2013 (09 Apr Online)
A bullet of mass 20 g has an initial speed of 1 m s-1, just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistance of 2.5Ã10-2 N, the speed of the bullet after emerging from the other side of the wall is close to:

Options:
A 0.7 m s-1
B 0.3Â mÂ s-1
C 0.1Â mÂ s-1
D 0.4Â mÂ s-1

The correct answer is:
0.7 m s-1

Solution:
a=-2.5×10-220×10-3=-54 m s-2
From v2-u2=2as
v2=u2+2as
v=12-2×54×20×10-2
=0.5
=0.7 m s-1

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2019 (10 Apr Shift 2)
A bullet of mass $20 \mathrm{~g}$ moving with $500 \mathrm{~ms}^{-1}$ is pierced $1 \mathrm{~cm}$ into a wooden block, then the retarding force experienced by the bullet is

Options:
A $125 \times 10^3 \mathrm{~N}$
B $750 \times 10^3 \mathrm{~N}$
C $500 \times 10^3 \mathrm{~N}$
D $250 \times 10^3 \mathrm{~N}$

The correct answer is:
$250 \times 10^3 \mathrm{~N}$

Solution:
Mass, $\mathrm{m}=20 \mathrm{~g}=20 \times 10^{-3} \mathrm{~kg}$
velocity, $\mathrm{u}=500 \mathrm{~m} / \mathrm{s}$
distance, $\mathrm{d}=1 \mathrm{~cm}=0.0 / \mathrm{m}$
final velocity, $v=0$
Using Equation of Motion, $v^2-u^2=2 a d$
$\begin{aligned} & -(500)^2=2 \times \mathrm{a} \times 0.01 \\ & \mathrm{a}=-\frac{500 \times 50000}{2 \times 0.01}=-1.25 \times 10^7 \mathrm{~m} / \mathrm{s}^2\end{aligned}$
using newton's second law,
$\begin{aligned} & \mathrm{F}=\mathrm{ma} \\ & \mathrm{F}=20 \times 10^{-3} \times 1.25 \times 10^7 \\ & \mathrm{~F}=250 \times 10^3 \mathrm{~N}\end{aligned}$

Topic: Physics, Laws of Motion, JEE Main
A bullet of mass $4.2 \times 10^{-2} \mathrm{kg},$ moving at a speed of $300 \mathrm{ms}^{-1}$, gets stuck into a block with a mass 9 times that of the bullet. If the block is free to move without any kind of friction, the heat generated in the process will be

Options:
A $45 \mathrm{cal}$
B $405 \mathrm{cal}$
C $450 \mathrm{cal}$
D $1701 \mathrm{cal}$

The correct answer is:
$405 \mathrm{cal}$

Solution:
Let mass of bullet = $m$
Mass of block $=\mathrm{M}$
Velocity of bullet $=v=300 \mathrm{m} / \mathrm{s}$
Velocity of combined system $M+m=V$
Here, fhom momentum conservation $\frac{M+m}{m} V=v$
$v=\frac{300 \times 42 \times 10^{-4}}{4.2 \times 10^{-4}+9\left(4.2 \times 10^{-4}\right)}$
$30 \mathrm{ms}$
Now, heat produced = Loss in kinetic energy of bullet
$$
\begin{array}{l}
=\frac{1}{2} m^{2}-\frac{1}{2}(M+m) V^{2} \\
=\frac{1}{2} \times 4.2 \times 10^{-4}(300)^{2}-\frac{1}{2}\left(4.2 \times 10^{-2}\right. \\
+9 \times 4.2 \times 1)(30)^{2} \\
=63 \times 270 \\
=1701 \mathrm{J} \\
=\frac{1701}{4.2} \mathrm{Cal}
\end{array}
$$
$=405 \mathrm{Cal}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A bullet of mass $50 \mathrm{~g}$ is fired with a speed $100 \mathrm{~m} / \mathrm{s}$ on a plywood and emerges with $40 \mathrm{~m} / \mathrm{s}$. The percentage loss of kinetic energy is :

Options:
A $84 \%$
B $16 \%$
C $32 \%$
D $44 \%$

The correct answer is:
$84 \%$

Solution:
$\begin{aligned} \mathrm{K}_{\mathrm{i}} & =\frac{1}{2} \mathrm{~m}(100)^2 \\ \mathrm{~K}_{\mathrm{f}} & =\frac{1}{2} \mathrm{~m}(40)^2 \\ \% \text { loss } & =\frac{\left|\mathrm{K}_{\mathrm{f}}-\mathrm{K}_{\mathrm{i}}\right|}{\mathrm{K}_{\mathrm{i}}} \times 100 \\ & =\frac{\left|\frac{1}{2} \mathrm{~m}(40)^2-\frac{1}{2} \mathrm{~m}(100)^2\right|}{\frac{1}{2} \mathrm{~m}(100)^2} \times 100 \\ & =\frac{|1600-100 \times 100|}{100}=84 \%\end{aligned}$

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2024 (06 Apr Shift 1)
A bullet of mass 200 g having initial kinetic energy 90 J is shot inside a long swimming pool as shown in the figure. If it's kinetic energy reduces to 40 J within 1 s, the minimum length of the pool, the bullet has to travel so that it completely comes to rest is

Options:
A 45m
B 90m
C 125m
D 25m

The correct answer is:
45m

Solution:
Let us assume that the motion of the bullet is of constant retardation inside the water.

Initial kinetic energy 12mv02=90 J

⇒v0=30 m s-1

Let the retardation is a, then after 1 s

12m30-a×12=40 J

⇒30-a=20

⇒a=10 m s-2

Stopping distance for the constant retardation is given by,

l=u22a=30220=45 m.

Therefore, the minimum length of the pool required is 45 m.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (28 Jul Shift 2)
A bullet of mass $a$ and velocity $b$ is fired into a large block of mass $c$. The final velocity of the system is

Options:
A $\frac{c}{a+b} \cdot b$
B $\frac{a}{a+c} \cdot b$
C $\frac{a+b}{c} \cdot a$
D $\frac{a+c}{a} \cdot b$

The correct answer is:
$\frac{a}{a+c} \cdot b$

Solution:
Initially bullet moves with velocity b and after collision bullet get embedded in block and both move together with common velocity. By the conservation of momentum
$\Rightarrow a \times b+0=(a+c) V \Rightarrow V=\frac{a b}{a+c}$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A bullet of mass $\mathrm{m}$ moving with velocity 'v' is fired into a wooden block of mass 'M'. If the bullet remains embedded in the block, the final velocity of the system is

Options:
A $\frac{V}{m(M+m)}$
B $\frac{m+M}{m}$
C $\frac{\mathrm{M}+\mathrm{m}}{\mathrm{mV}}$
D $\frac{\mathrm{mV}}{\mathrm{m}+\mathrm{M}}$

The correct answer is:
$\frac{\mathrm{mV}}{\mathrm{m}+\mathrm{M}}$

Solution:
Since there is no extra force other than the action and reaction force so the linear momentum should be conserved.
Suppose the system moves with velocity V then momentum before collision is mv and that after collision will be $(M+$ $\mathrm{m}) \mathrm{V}$
Equating both we get $m v=(M+m) V$ or $V=\frac{m}{m+M}^{v}$

Topic: Physics, Center of Mass Momentum and Collision, MHT CET, MHT CET 2020 (16 Oct Shift 1)
A bullet of mass $m$ moving with velocity $v$ strikes a block of mass $M$ at rest and gets embedded into it. The kinetic energy of the composite block will be

Options:
A $\frac{1}{2} m v^2 \times \frac{m}{(m+M)}$
B $\frac{1}{2} m v^2 \times \frac{M}{(m+M)}$
C $\frac{1}{2} m v^2 \times \frac{(M+m)}{M}$
D $\frac{1}{2} M v^2 \times \frac{m}{(m+M)}$

The correct answer is:
$\frac{1}{2} m v^2 \times \frac{m}{(m+M)}$

Solution:
By conservation of momentum, $\quad m v+M \times 0=(m+M) v$ Velocity of composite block $V=\left(\frac{m}{m+M}\right) v$ K.E. of composite block $=\frac{1}{2}(M+m) V^2$ $=\frac{1}{2}(M+m)\left(\frac{m}{M+m}\right)^2 v^2=\frac{1}{2} m v^2\left(\frac{m}{m+M}\right)$

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A bullet of mass m strikes an obstruction and deviates off at 60Â° to this original direction. If its speed also changed from u to v, find the magnitude of the impulse acting on the bullet.

Options:
A mu2-uv+v2
B mu2+uv-v2
C mu4-uv+v4
D mu4+uv-v4

The correct answer is:
mu2-uv+v2

Solution:
impulse = Δ P x i ^ + Δ P y j ^
Δ P x = mv cos 60 o - mu ⇒ Δ P X = ( mv 2 - u )
Δ P y = mv sin 60 o - 0 ⇒ Δ P y = mv 3 2
Δ P → = P x 2 + P y 2 ⇒ Δ P → = m ( v 2 - u ) 2 + ( 3 v 2 ) 2
Δ P → = m v 2 4 + u 2 - uv + 3 v 2 4
impulse = m v 2 + u 2 - uv

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A bus is moving with a speed of $10 \mathrm{~ms}^{-1}$ on a straight road. A scooterist wishes to overtake the bus in $100 \mathrm{~s}$. If the bus is at a distance of $1 \mathrm{~km}$ from the scooterist, with what speed should the scooterist chase the bus ?

Options:
A $10 \mathrm{~ms}^{-1}$
B $20 \mathrm{~ms}^{-1}$
C $40 \mathrm{~ms}^{-1}$
D $25 \mathrm{~ms}^{-1}$

The correct answer is:
$20 \mathrm{~ms}^{-1}$

Solution:
$$
\begin{aligned}
\mathrm{S}_{\mathrm{r}} & =\mathrm{v}_{\mathrm{r}} \mathrm{t} \\
1000 & =(\mathrm{v}-10) \times 100 \\
\mathrm{v} & =20 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, JEE Main
A bus is moving with a speed of $10 \mathrm{~ms}^{-1}$ on a straight road. A scooterist wishes to overtake the bus in $100 \mathrm{~s}$. If the bus is at a distance of $1 \mathrm{~km}$ from the scooterist, with what speed should the scooterist chase the bus ?

Options:
A $10 \mathrm{~ms}^{-1}$
B $20 \mathrm{~ms}^{-1}$
C $40 \mathrm{~ms}^{-1}$
D $25 \mathrm{~ms}^{-1}$

The correct answer is:
$20 \mathrm{~ms}^{-1}$

Solution:
$$
\begin{aligned}
\mathrm{S}_{\mathrm{r}} & =\mathrm{v}_{\mathrm{r}} \mathrm{t} \\
1000 & =(\mathrm{v}-10) \times 100 \\
\mathrm{v} & =20 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, NEET, NEET 2009 (Mains)
A bus is moving with a velocity of 5 m s-1 towards a huge wall. The driver sounds a horn of frequency 165 Hz. If the speed of sound in air is 335 m s-1, the number of beats heard per second by a passenger inside the bus will be

Options:
A 3
B 4
C 5
D 6

The correct answer is:
5

Solution:
Apparent frequency
v′=vv-vLv-vs
=165355+5355-5
∵vL=-5 m s-1, vs=5 m s-1
=165×340330=170 Hz
Therefore, the number of beats heard
= 170 - 165 = 5

Topic: Physics, Waves and Sound, JEE Main
A bus moving along a straight highway with speed of $72 \mathrm{~km} / \mathrm{h}$ is brought to halt within $4 s$ after applying the brakes. The distance travelled by the bus during this time (Assume the retardation is uniform) is _____m.

Options:

The correct answer is:
40

Solution:
$\begin{aligned} & \text { Initial velocity }=u=72 \mathrm{~km} / \mathrm{h}=20 \mathrm{~m} / \mathrm{s} \\ & \mathrm{v}=\mathrm{u}+\text { at } \\ & \Rightarrow 0=20+\mathrm{a} \times 4 \\ & \mathrm{a}=-5 \mathrm{~m} / \mathrm{s}^2 \\ & \mathrm{v}^2-\mathrm{u}^2=2 \text { as } \\ & \Rightarrow 0^2-20^2=2(-5) . \mathrm{s} \\ & \mathrm{s}=40 \mathrm{~m}\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (04 Apr Shift 2)
A bus starts from rest and moves with constant acceleration \( 8 \mathrm{~m} \mathrm{~s}^{-2} \). At the same time, a car travelling with constant velocity \( 16 \mathrm{~m} \mathrm{~s}^{-1} \) overtakes and passes the bus. At what distance, bus overtakes the car?

Options:
A \( 24 \mathrm{~m} \)
B \( 32 \mathrm{~m} \)
C \( 60 \mathrm{~m} \)
D \( 64 \mathrm{~m} \)

The correct answer is:
\( 64 \mathrm{~m} \)

Solution:
Let the bus overtakes the car after time t and at
distance d.

Bus, S=ut+12at2

d=12×8×t2 …i

Car: S=vt

d= 16t …ii

by i and ii

t=sec, d=64 m.

Topic: Physics, Motion In Two Dimensions, JEE Main
A butterfly is flying with a velocity 42 m s-1 in north-east direction. Wind is slowly blowing at 1 m s-1 from north to south. The resultant displacement of the butterfly in 3 seconds is:

Options:
A 3 m
B 20 m
C 122 m
D 15 m

The correct answer is:
15 m

Solution:
V→BW=42cos45i^+42sin45j^
=4i^+4j^
V→w=-j^
V→B=V→BW+V→w=4i^+3j^
S→B=V→B×t=(4i^+3j^)×3=12i^+9j^
S→B=(12)2+(9)2=15 m

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2021 (20 Jul Shift 1)
AC4H8Cl2→373 KHydrolysisBC4H8O

B reacts with Hydroxyl amine but does not give Tollen's test. Identify A and B.

Options:
A 2,2-Dichlorobutane and Butanal
B 2,2-Dichlorobutane and Butan-2-one
C 1,1-Dichlorobutane and 2-Butanone
D 1,1-Dichlorobutane and Butanal

The correct answer is:
2,2-Dichlorobutane and Butan-2-one

Solution:
Aldehydes and ketones react with hydroxyl amine to form oximes but ketones do not react with Tollens reagent.

Compound B'' does not gives Tollen's test due to presence of kenotic group but react with hydroxyl amine

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2021 (26 Feb Shift 1)
A CE amplifier, has a power gain of 40 dB . The input resistance and output load resistance are 500 Ω and 2 kΩ respectively. Find the value of common-emitter current gain.
[Take gain in dB=10logPoutPin]

Options:

The correct answer is:
50

Solution:
Given that,
Power gain (AP)=40 dB
â´10logP0Pin=40
â´logP0Pm=4
â´P0Pin=104
Also, AP=Î²2Ã Resistance gain
=Î²2ÃRoutRin
=Î²2Ã2000500
=Î²2Ã4
â´Î²2=1044
â´Î²=1002=50

Topic: Physics, Semiconductors, JEE Main
Aâcold KMnO4 Mesobutane-2,3-diol. A is

Options:
A Cis 2-butane
B Trans 2-butene
C 1-butene
D Iso-butene

The correct answer is:
Cis 2-butane

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
AC voltage V(t)=20 sinωt of frequency 50 Hz is applied to a parallel plate capacitor. The separation between the plates is 2 mm and the area is 1 m2. The amplitude of the oscillating displacement current for the applied AC voltage is Take ε0=8.85×10-12 F m-1

Options:
A 21.14 μA
B 83.37 μA
C 27.79 μA
D 55.58 μA

The correct answer is:
27.79 μA

Solution:
From the given information,

C=∈0 A d=ϵ0×12×10-3 F
∴ XC=1ωC=2×10-32×50π×∈0=2×10-325×4π∈0 Ω
∴ XC=2×10-325×9×109=1825×106 Ω
∴ i0=V0XC=20×2518×10-6 A=27.47 μA.

The value of amplitude of displacement current will be same as value of amplitude of conventional current.
Hence option 3.

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (20 Jul Shift 1)
A cable of resistance $10 \Omega$ carries electric power from a generator producing $250 \mathrm{~kW}$ at $10000 \mathrm{volt}$. The current in the cable is

Options:
A $25 \mathrm{~A}$
B $250 \mathrm{~A}$
C $100 \mathrm{~A}$
D $1000 \mathrm{~A}$

The correct answer is:
$25 \mathrm{~A}$

Solution:
Since $P=V I \Rightarrow I=\frac{P}{V}=\frac{250000}{10000}=25 \mathrm{~A}$

Topic: Physics, Current Electricity, JEE Main
A calorimeter contains 0.2 kg of water at 30 Â°C, in which 0.1 kg of water at 60 Â°C is added. The mixture is well stirred and the resulting temperature is found to be 35 Â°C. The thermal capacity of the calorimeter is

Options:
A 6300 J K-1
B 1260 J K-1
C 4200 J K-1
D 3200 J K-1

The correct answer is:
1260 J K-1

Solution:
Let X be the thermal capacity of the calorimeter and specific heat of water = 4200 J kg-1 K-1.
Heat lost by 0.1 kg of water = Heat gained by water in calorimeter + Heat gained by calorimeter
⇒ 0.1 × 4200 × (60 - 35)
= 0.2 × 4200 × (35 - 30) + X(35 - 30)
10500 = 4200 + 5X ⇒ X = 1260 J K-1

Topic: Physics, Thermal Properties of Matter, JEE Main
A calorimeter of negligible heat capacity contains 100 g water at 40 Â°C. The water cools to 35 Â°C in 5 min. If water is now replaced by a liquid of the same volume as that of water at the same initial temperature it cools to 35 Â°C in 2 min. Given specific heats of water and liquid are 4200 J kg-1 Â°C-1 and 2100 J kg-1 Â°C-1 respectively. Find the density of liquid give the answer in kg m-3. [Assume Newton law of cooling is applicable]

Options:

The correct answer is:
8

Solution:
-dTdt=K100ÃSwT-T0
â« 40 35 âdT Tâ T 0 = â« 0 5 K 100Ã S w dt
â« 40 35 âdT Tâ T 0 = â« 0 2 dt 100Ã Ï l s l
5K100Sw=2K100ÃÏlsl
ρ l = 4 5 g/ cm 3 = 4× 10 −3 kg 5× 10 −6 m 3 = 4 5 × 10 3
= 40 5 × 10 2 =800 kg/ m 3

Topic: Physics, Thermal Properties of Matter, JEE Main
A camphor sample melts at $176^{\circ} \mathrm{C} . K_f$ for camphor is $40 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. A solution of $0.02 \mathrm{~g}$ of a hydrocarbon in $0.8 \mathrm{~g}$ of camphor melts at $156.77^{\circ} \mathrm{C}$. The hydrocarbon is made up of $92.3 \%$ of carbon. What is the molecular formula of the hydrocarbon?

Options:
A $\mathrm{C}_6 \mathrm{H}_6$
B $\mathrm{C}_{12} \mathrm{H}_{12}$
C $\mathrm{C}_4 \mathrm{H}_4$
D $\mathrm{C}_8 \mathrm{H}_8$

The correct answer is:
$\mathrm{C}_4 \mathrm{H}_4$

Solution:
Melting point of camphor $=176^{\circ} \mathrm{C}$
$\left(K_f\right)$ for camphor $=40 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Mass of hydrocarbon $\left(w_B\right)=0.02 \mathrm{~g}$
Mass of camphor $\left(w_A\right)=0.8 \mathrm{~g}$
Temperature of camphor (at which it melts)
$=156.77^{\circ} \mathrm{C}$
Thus, $\Delta T_f$ for camphor $=176-156.77$
$=19.23^{\circ} \mathrm{C}=19.23 \mathrm{~K}$
$\therefore$ Depression in freezing point,
$\begin{aligned} \Delta T_f & =\frac{K_f \times w_B}{M_B} \times \frac{1000}{W_A} \\ 19.23 & =\frac{40 \times 0.02 \times 1000}{M_B \times 0.8}\end{aligned}$
Molar mass of solute $\left(M_B\right)$
$=\frac{40 \times 0.02 \times 1000}{19.23 \times 0.8}=52$
$\therefore$ Empirical formula $=\mathrm{CH}$
Empirical formula weight $=12+1=13$
Multiple $(n)=\frac{\text { Molecular weight }}{\text { Empirical formula weight }}=\frac{52}{13}=4$
Thus, molecular formula $=(\mathrm{CH})_4=\mathrm{C}_4 \mathrm{H}_4$

Topic: Chemistry, Solutions, JEE Main
A candidate is required to answer 6 out of 12 questions which are divided into two parts A and $B$, each containing 6 questions and he/she is not permitted to attempt more than 4 questions from any part. In how many different ways can he/she make up hisher choice of 6 questions?

Options:
A 850
B 800
C 750
D 700

The correct answer is:
850

Solution:
$$
\begin{array}{|c|c|}
\hline A & B \\
\hline 4 & 2 \\
\hline 3 & 3 \\
\hline 2 & 4 \\
\hline
\end{array}
$$
$\therefore$ Total number of ways $={ }^{6} C_{4} \times{ }^{6} C_{2}+{ }^{6} C_{3} \times{ }^{6} C_{3}+{ }^{6} C_{2} \times{ }^{6} C_{4}$
$=\frac{6 \times 5}{2 \times 1} \times \frac{6 \times 5}{2 \times 1}+\frac{6 \times 5 \times 4}{3 \times 2 \times 1} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1}+\frac{6 \times 5}{2 \times 1} \times \frac{6 \times 5}{2 \times 1}$
$=15 \times 15+20 \times 20+15 \times 15$
$=225+400+225$
$=850$

Topic: Mathematics, Permutation Combination, JEE Main
A candle flame 0.5 cm high is kept between a wall and a concave mirror of focal length 1.5 m such that its image 1.5 cm high is formed on the wall. Find the distance of candle from the wall and also the distance between the wall and the mirror.

Options:
A 4 m, 6 m

B 6 m, 7 m

C 6 m, 8 m

D 9 m, 7 m

The correct answer is:
4 m, 6 m

Solution:
Image of the flame formed by the concave mirror on the wall will be real and inverted, hence magnification
mâ¡ = Iâ¡ Oâ¡ = - 1 . 5 0 Â· 5 = - 3
mâ¡ = - vâ¡ uâ¡ = - 3
â´ v = 3u
Using 1 vâ¡ + 1 uâ¡ = 1 fâ¡
1 3 uâ¡ + 1 uâ¡ = - 1 1 . 5 fâ¡ = - 1 . 5 m
4 3 uâ¡ = - 1 1 . 5
â´ uâ¡ = - 4 Ã 1 . 5 3 = - 2 m
Thus the flame is at distance 2m from the mirror and v = 3u = â6 m
Hence the distance between wall and mirror is 6 m. It implies that the distance between the flame and the wall will be 4m.

Topic: Physics, Ray Optics, JEE Main
A candle of diameter $d$ is floating on a liquid in a cylindrical container of diameter $D(D>>d)$ as shown in figure. If it is burning at the rate of $2 \mathrm{~cm} /$ hour then the top of the candle will

Options:
A remain at the same height
B fall at the rate of $1 \mathrm{~cm} /$ hour
C fall at the rate of $2 \mathrm{~cm} /$ hour
D go up at the rate of $1 \mathrm{~cm} /$ hour.

The correct answer is:
fall at the rate of $1 \mathrm{~cm} /$ hour

Solution:
Initial weight of the candle = weight of liquid displaced
$$
\rho_C V_C g=\rho_L \text { (volume displaced) } g
$$
$$
\Rightarrow \rho_C \pi\left(\frac{d}{2}\right)^2 2 L g=\rho_L \pi \frac{d^2}{2} L g \Rightarrow \frac{\rho_C}{\rho_L}=\frac{1}{2}
$$
When $2 \mathrm{~cm}$ has been burnt, total length $=2 L-2$ But $\rho_C(2 L-2)=\rho_L(L-x)$
$$
\begin{aligned}
& \rho_C 2(L-1)=2 \rho_C(L-x) \\
\therefore \quad & x=1 \mathrm{~cm} .
\end{aligned}
$$
[Using eqn. (i)]
Outside also it has decreased $1 \mathrm{~cm}$ as the total decrease is $2 \mathrm{~cm}$. The level of the candle comes down at half the rate of burning.

Topic: Physics, Mechanical Properties of Fluids, AIIMS, AIIMS 2005