datavorous/entrance-exam-dataset huggingface dataset:
Consider a quadratic equation $a x^2+b x+c=0$, where $2 a+3 b+6 c=0$ and let $g(x)=a \frac{x^3}{3}+b \frac{x^2}{2}+c x$.
Statement 1: The quadratic equation has at least one root in the interval $(0,1)$.
Statement 2: The Rolle's theorem is applicable to function $g(x)$ on the interval $[0,1]$.

Options:
A
Statement 1 is false, Statement 2 is true.

B
Statement 1 is true, Statement 2 is false.

C
Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.

D
Statement 1 is true, Statement 2 is true, , Statement 2 is a correct explanation for Statement 1.

The correct answer is:
Statement 1 is true, Statement 2 is true, , Statement 2 is a correct explanation for Statement 1.

Solution:
Let $g(x)=\frac{a x^3}{3}+b \cdot \frac{x^2}{2}+c x$
$$
\mathrm{g}^{\prime}(x)=a x^2+b x+c
$$
Given: $a x^2+b x+c=0$ and $2 a+3 b+6 c=0$
Statement-2:
(i) $g(0)=0$ and $g(1)$
$$
\begin{aligned}
& =\frac{a}{3}+\frac{b}{2}+c=\frac{2 a+3 b+6 c}{6} \\
& =\frac{0}{6}=0 \\
& \Rightarrow g(0)=g(1)
\end{aligned}
$$
(ii) $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$
$\therefore$ By Rolle's theorem $\exists k \in(0,1)$ such that $g^{\prime}(k)=0$
This holds the statement 2. Also, from statement-2, we can say $a x^2+b x+c=0$ has at least one root in $(0,1)$.
Thus statement-1 and 2 both are true and statement-2 is a correct explanation for statement-1.

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2012 (19 May Online)
Consider a radioactive nuclide which follow decay rate given by $A(t)=A_0 2^{\left(t / t_0\right\}}$, where $A(t)$ is the fraction of radioactive material remaining after time $t$ from the initial $A_0$ at zero time. Let $A_1$ be the fraction of original activity which remains after 120 hours. Likewise $A_2$ is the fraction of original activity remaining after 200 hours. If $\frac{A_2}{A_2}=1.6$, then the half-life $\left(t_0\right)$ will be

Options:
A 10 hours
B 20 hours
C 40 hours
D 60 hours

The correct answer is:
20 hours

Solution:
Given, decay rate, $A=A_0 2^{-\left(\frac{t}{t_0}\right)}$
Let $\frac{t}{t_0}=n$
where, $t_0=$ half life.
As,
$\begin{aligned}
& \frac{A_1}{A_2}=16 \quad \Rightarrow \frac{A_0 \cdot 2^{-n_1}}{A_0 \cdot 2^{-n_2}}=16 \\
& \Rightarrow \quad 2^{i_2-n_1}=16=2^4 \quad \Rightarrow \quad n_2-n_1=4
\end{aligned}$
$\Rightarrow \quad \frac{t_2}{t_0}-\frac{t_1}{t_0}=4$
As, $t_1=120 \mathrm{~h}$ and $t_2=200 \mathrm{~h}$,
$\Rightarrow \quad \frac{200}{t_0}-\frac{120}{t_0}=4 \quad \Rightarrow \quad \frac{80}{t_0}=4$
$\Rightarrow \quad t_0=\frac{80}{4}=20 \mathrm{~h}$

Topic: Physics, Nuclear Physics, JEE Main
Consider a radioactive nuclide which follow decay rate given by $A(t)=A_0 2^{\left(t / t_0\right\}}$, where $A(t)$ is the fraction of radioactive material remaining after time $t$ from the initial $A_0$ at zero time. Let $A_1$ be the fraction of original activity which remains after 120 hours. Likewise $A_2$ is the fraction of original activity remaining after 200 hours. If $\frac{A_2}{A_2}=1.6$, then the half-life $\left(t_0\right)$ will be

Options:
A 10 hours
B 20 hours
C 40 hours
D 60 hours

The correct answer is:
20 hours

Solution:
Given, decay rate, $A=A_0 2^{-\left(\frac{t}{t_0}\right)}$
Let $\frac{t}{t_0}=n$
where, $t_0=$ half life.
As,
$\begin{aligned}
& \frac{A_1}{A_2}=16 \quad \Rightarrow \frac{A_0 \cdot 2^{-n_1}}{A_0 \cdot 2^{-n_2}}=16 \\
& \Rightarrow \quad 2^{i_2-n_1}=16=2^4 \quad \Rightarrow \quad n_2-n_1=4
\end{aligned}$
$\Rightarrow \quad \frac{t_2}{t_0}-\frac{t_1}{t_0}=4$
As, $t_1=120 \mathrm{~h}$ and $t_2=200 \mathrm{~h}$,
$\Rightarrow \quad \frac{200}{t_0}-\frac{120}{t_0}=4 \quad \Rightarrow \quad \frac{80}{t_0}=4$
$\Rightarrow \quad t_0=\frac{80}{4}=20 \mathrm{~h}$

Topic: Physics, Nuclear Physics, TS EAMCET, TS EAMCET 2018 (07 May Shift 1)
Consider a reaction $a G+b H \rightarrow$ products. When concentration of both the reactants $G$ and $H$ is doubled, the rate increases by eight times. However, when concentration of $G$ is doubled keeping the concentration of $H$ fixed, the rate is doubled. The overall order of the reaction is

Options:
A
0

B
1

C
2

D
3

The correct answer is:
3

Solution:
$a G+b H \longrightarrow$ Product
Suppose order of reaction $=n$
When concentration of both $G$ and $H$ is double than rate increase by eight times.
$$
\begin{aligned}
\text { rate } & =k(\text { reactants })^n \\
(2)^8 & =k(2)^n \\
(2)^3 & =k(2)^n \\
n & =3
\end{aligned}
$$
when concentration of $G$ is doubled than concentration $H$ is fixed, the rate is double.
than $\quad$ rate $\propto[G]^1[H]^2$

Topic: Chemistry, Chemical Kinetics, JEE Main
Consider a rectangle whose length is increasing at the uniform rate of $2 \mathrm{~m} / \mathrm{sec}$, breadth is decreasing at the uniform rate of $3 \mathrm{~m} / \mathrm{sec}$ and the area is decreasing at the uniform rate of $5 \mathrm{~m}^2 / \mathrm{sec}$. If after some time the breadth of the rectangle is $2 \mathrm{~m}$ then the length of the rectangle is

Options:
A
$2 \mathrm{~m}$

B
$4 \mathrm{~m}$

C
$1 \mathrm{~m}$

D
$3 \mathrm{~m}$

The correct answer is:
$3 \mathrm{~m}$

Solution:
Let $A$ be the area, $b$ be the breadth and $\ell$ be the length of the rectangle.
Given: $\frac{d A}{d t}=-5, \frac{d \ell}{d t}=2, \frac{d b}{d t}=-3$
We know, $A=\ell \times b$
$$
\begin{aligned}
& \Rightarrow \frac{d A}{d t}=\ell \cdot \frac{d b}{d t}+b \cdot \frac{d \ell}{d t}=-3 \ell+2 b \\
& \Rightarrow-5=-3 \ell+2 b .
\end{aligned}
$$
When $b=2$, we have
$$
-5=-3 \ell+4 \Rightarrow \ell=\frac{9}{3}=3 m
$$

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2012 (12 May Online)
Consider a region of space in which constant, uniform, and mutually perpendicular fields Eâand Bâ exists. Now a particle of charge q enters this region with a velocity vâ which is perpendicular to both the fields. If the velocity of the charge remains unchanged, then

Options:
A v→=B→×E→E2
B vâ=EâÃBâB2
C v→=B→×E→B2
D v→=E→×B→E2

The correct answer is:
v→=E→×B→B2

Solution:
when E→ and B→ are perpendicular and velocity has no change then qE=qvB i.e., v=EB.
The two forces oppose each other so, v is along v→×B→ i.e., v→=E→× B→B2
As E→ and B→ are perpendicular to each other
E→×B→B2=EBsin⁡90oB2=EB

Topic: Physics, Magnetic Effects of Current, NEET
Consider a region of space in which constant, uniform, and mutually perpendicular fields Eâand Bâ exists. Now a particle of charge q enters this region with a velocity vâ which is perpendicular to both the fields. If the velocity of the charge remains unchanged, then

Options:
A v→=B→×E→E2
B v→=E→×B→B2
C v→=B→×E→B2
D v→=E→×B→E2

The correct answer is:
v→=E→×B→B2

Solution:
when Eâ and Bâ are perpendicular and velocity has no change then qE=qvB i.e., v=EB. The two forces oppose each other so, v is along vâÃBâ i.e.,vâ=EâÃ BâB2
As Eâ and Bâ are perpendicular to each other
EâÃBâB2=EBsinâ¡90oB2=EB

Topic: Physics, Magnetic Effects of Current, JEE Main
Consider a region R=x, y∈R2:x2≤y≤2x. If a line y=α divides the area of region R into two equal parts, then which of the following is true ?

Options:
A α3-6α2+16=0
B 3α2-8α3/2+8=0
C 3α2-8α+8=0
D α3-6α3/2-16=0

The correct answer is:
3α2-8α3/2+8=0

Solution:
∫04y-y2dy=2∫0αy-y2dy

2y3/23-y2204=22y3/23-y240α

⇒163-4=223α3/2-α24

⇒ 43=223α3/2-α24

⇒ 23=8α3/2-3α212

⇒ 8α3/2-3α2=8

⇒3α2-8α3/2+8=0

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2020 (02 Sep Shift 2)
Consider a regular 10-gon with its vertices on the unit circle. With one vertex fixed, draw straight lines to the other 9 vertices. Call them L1, L2,…,L9 and denote their lengths by l1,l2,…,l9 respectively. Then the product l1,l2,…,l9 is

Options:
A 10
B 103
C 503
D 20

The correct answer is:
10

Solution:
Let α=ei2π10=eiπ5

Now, z10-1=(z-1)(z-α)…z-α9….(1)
so ,ℓ1ℓ2…ℓ9=|1-α|1-α2…1-α9
=(1-α)1-α2…1-α9
=limz→1z10-1z-1=10

Topic: Mathematics, Complex Number, JEE Main
Consider a relation R defined as aRb if 2+ab>0 where a,b are real numbers. Then, the relation R is

Options:
A reflexive and symmetric
B symmetric and transitive
C transitive and reflexive
D None of these

The correct answer is:
reflexive and symmetric

Solution:
Relation R is reflexive, since 2+a×a>0, ∀ real number a.

It is symmetric, since 2+ab>0⇒2+ba>0.
However, R is not transitive.

Consider three real numbers 2,-16 and -2 . We have
2+2×-16=53>0
and 2+-16-2=73>0
So, 2R-16 and -16R-2
But, 2R-2 since 2+2-2=-2≯0

Topic: Mathematics, Sets and Relations, JEE Main
Consider a reversible engine of efficiency $\frac{1}{6}$. When the temperature of the sink is reduced by $62^{\circ} \mathrm{C}$, its efficiency gets doubled. The temperature of the source and sink respectively are

Options:
A $372 \mathrm{~K}$ and $310 \mathrm{~K}$
B $273 \mathrm{~K}$ and $300 \mathrm{~K}$
C $99^{\circ} \mathrm{C}$ and $10^{\circ} \mathrm{C}$
D $200^{\circ} \mathrm{C}$ and $37^{\circ} \mathrm{C}$

The correct answer is:
$372 \mathrm{~K}$ and $310 \mathrm{~K}$

Solution:
Given
$$
\eta_1=\frac{1}{6}
$$
According to the question,
$$
\begin{gathered}
\eta_1=\frac{T_1-T_2}{T_1} \\
\frac{T_1-T_2}{T_1}=\frac{1}{6} \\
\eta_2=\frac{T_1-\left(T_2-62\right)}{T_1} \Rightarrow 2 \times \eta_1=\frac{T_1-T_2+62}{T_1} \\
2 \times \frac{1}{6}=\frac{T_1-T_2+62}{T_1} \Rightarrow \frac{1}{3}=\frac{T_1-T_2+62}{T_1}
\end{gathered}
$$
(ii)
From Eq. (i), we get
$$
T_1-T_2=\frac{T_1}{6}
$$
Substituting this value in Eq. (ii), we get
$$
\begin{gathered}
\frac{1}{3}=\frac{\frac{T_1}{6}+62}{T_1} \Rightarrow \frac{1}{3}=\frac{T_1+372}{6 T_1} \\
6 T_1=3 T_1+1116 \Rightarrow 3 T_1=1116 \Rightarrow T_1=372 \mathrm{~K}
\end{gathered}
$$
From Eq. (i), we get
$$
\begin{aligned}
\frac{372-T_2}{372} & =\frac{1}{6} \\
6\left(372-T_2\right) & =372
\end{aligned}
$$
$$
\begin{aligned}
2232-6 T_2 & =372 \\
6 T_2 & =2232-372 \\
6 T_2 & =1860 \\
T_2 & =\frac{1860}{6} \\
T_2 & =310 \mathrm{~K}
\end{aligned}
$$

Topic: Physics, Thermodynamics, JEE Main
Consider a rigid square $\mathrm{ABCD}$ as in the figure with $\mathrm{A}$ and $\mathrm{B}$ on the $\mathrm{x}$ and y axis respectively.
When A and B slide along their respective axes, the locus of $\mathrm{C}$ forms a part of

Options:
A a circle
B a parabola
C a hyperbola
D an ellipse which is not a circle

The correct answer is:
an ellipse which is not a circle

Solution:
$\mathrm{C}(\mathrm{h}, \mathrm{k})$
$\mathrm{h}=\mathrm{a} \sin \theta \Rightarrow \sin \theta=\frac{\mathrm{h}}{\mathrm{a}}......(1)$
$\mathrm{k}=\mathrm{a} \sin \theta+\mathrm{a} \cos \theta$
$\mathrm{k}=\mathrm{h}+\mathrm{a} \cos \theta$
$\frac{\mathrm{k}-\mathrm{h}}{\mathrm{a}}=\cos \theta ......(2)$
$\begin{array}{l}
\sin ^{2} \theta+\cos ^{2} \theta=1 \\
\frac{\mathrm{h}^{2}}{\mathrm{a}^{2}}+\frac{(\mathrm{k}-\mathrm{h})^{2}}{\mathrm{a}^{2}}=1 \\
\mathrm{~h}^{2}+(\mathrm{k}-\mathrm{h})^{2}=\mathrm{a}^{2}
\end{array}$
locus $x^{2}+(y-x)^{2}=a^{2}$
solving we get $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{x}^{2}-2 \mathrm{xy}=\mathrm{a}^{2}$
$2 \mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{xy}=\mathrm{a}^{2}$
check $\mathrm{h}^{2}-\mathrm{ab} < 0$
ellipse

Topic: Mathematics, Ellipse, JEE Main
Consider a rubber ball freely falling from a height $h=4.9 \mathrm{~m}$ onto a horizontal elastic plate. Assume that the duration of collision is negligible and the collision with the plate is totally elastic. Then the velocity as a function of time the height as function of time will be

Options:
A

B

C

D

The correct answer is:

Solution:
$\mathrm{h}=\frac{1}{2} \mathrm{gt}^2, \quad \mathrm{v}=-\mathrm{gt}$ and after the collision, $\mathrm{v}=\mathrm{gt}$.
(parabolic) (straight line)
Collision is perfectly elastic then ball reaches to same height again and again with same velocity.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2009
Consider a sample of oxygen behaving like an ideal gas. At 300 K, the ratio of root-mean-square (RMS) velocity to the average velocity of the gas molecule would be :

(Molecular weight of oxygen is 32 g mol-1; R=8.3 J K-1 mol-1)

Options:
A 33
B 83
C 3π8
D 8π3

The correct answer is:
3π8

Solution:
vrms=3RTM

vavg=8πRTM

vrmsvavg=3π8

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Consider a series of measurements of the length of a box in an experiment. The readings are $2.4 \mathrm{~m}, 2.5 \mathrm{~m}, 2.6 \mathrm{~m}, 2.8 \mathrm{~m}$, $3.0 \mathrm{~m}$. What would be the relative error?

Options:
A 0.110
B 0.089
C 0.079
D 0.072

The correct answer is:
0.072

Solution:
Given that, number of observations, $n=5$
$\begin{aligned}
& a_1=2.4 \mathrm{~m}, a_2=2.5 \mathrm{~m}, a_3=2.6 \mathrm{~m}, a_4=2.8 \mathrm{~m}, \\
& a_5=3.0 \mathrm{~m}
\end{aligned}$
Mean value of observations,
$\begin{aligned}
\bar{a} & =\frac{a_1+a_2+a_3+a_4+a_5}{5} \\
\bar{a} & =\frac{2.4+2.5+2.6+2.8+3.0}{5} \\
& =\frac{13.3}{5}=2.66
\end{aligned}$
Absolute errors in individual observed value
$\begin{aligned}
& \left|\Delta a_1\right|=\left|a_1-\bar{a}\right|=|2.4-2.66|=|-0.26|=0.26 \\
& \left|\Delta a_2\right|=\left|a_2-\bar{a}\right|=|2.5-2.66|=|-0.16|=0.16 \\
& \left|\Delta a_3\right|=\left|a_3-\bar{a}\right|=|2.6-2.66|=|-0.06|=0.06 \\
& \left|\Delta a_4\right|=\left|a_4-\bar{a}\right|=|2.8-2.66|=0.14 \\
& \left|\Delta a_5\right|=\left|a_5-\bar{a}\right|=|3.0-2.66|=0.34
\end{aligned}$
Mean absolute error,
$\begin{aligned}
\Delta \bar{a} & =\frac{\left|\Delta a_1\right|+\left|\Delta a_2\right|+\left|\Delta a_3\right|+\left|\Delta a_4\right|+\left|\Delta a_5\right|}{n} \\
\Delta \bar{a} & =\frac{0.26+0.16+0.06+0.14+0.34}{5}=\frac{0.96}{5} \\
\Delta \bar{a} & =0.192
\end{aligned}$
$\begin{aligned}
\text {Relative error } & =\frac{\text { Mean absolute error }}{\text { Mean value }} \\
& =\frac{\Delta \bar{a}}{\bar{a}}=\frac{0.192}{2.66}=0.072
\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main
Consider a short bar magnet of magnetic length 10 cm. Its geometric length is

Options:
A 12 cm
B 5 cm
C 3 cm
D 4 cm

The correct answer is:
12 cm

Solution:
Geometric length of a magnet is 65 times its magnetic length.
∴ Geometric length = 65×10=12 cm

Topic: Physics, Magnetic Effects of Current, JEE Main
Consider a situation in which a ring, a solid cylinder and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and having identical diameter. The correct statement for this situation is

Options:
A The sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.
B The ring has the greatest and the cylinder has the least velocity of the centre of mass at the bottom of the inclined plane.
C All of them will have same velocity.
D The cylinder has the greatest and the sphere has the least velocity of the centre of mass at the bottom of the inclined plane.

The correct answer is:
The sphere has the greatest and the ring has the least velocity of the centre of mass at the bottom of the inclined plane.

Solution:
a=gsinθ1+ImR2
Iring >Isolidcylinder >Isolidsphere
⇒aring <asolidcylinder <asolidsphere
⇒vring <vsolidcylinder <vsolidsphere

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Consider a situation in which reverse biased current of a particular P-N junction increases when it is exposed to a light of wavelength ≤621 nm. During this process, enhancement in carrier concentration takes place due to generation of hole-electron pairs. The value of band gap is nearly.

Options:
A 2 eV
B 4 eV
C 1 eV
D 0.5 eV

The correct answer is:
2 eV

Solution:
Band gap =hcλ0,
where λ0- threshold wavelength

Band gap =1242 ev-nm621 nm=2 eV

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (22 Jul Shift 1)
Consider a skew-symmetric matrix A= a b-b c such that a, b and c are selected from the set S=0, 1, 2, 3,........12. If A is divisible by 3, then the number of such possible matrices is

Options:
A 4
B 5
C 6
D 12

The correct answer is:
5

Solution:
âµ A is skew symmetric ⇒a=c=0
A=b2
âµA is divisible by 3
⇒b can be 0,3,6,9,12 (five possibilities)

Topic: Mathematics, Matrices, JEE Main
Consider a solenoid carrying current supplied by a DC source with a constant emf containing iron core inside it When the core is pulled out of the solenoid, the change in current will

Options:
A remain same
B decrease
C increase
D modulate

The correct answer is:
remain same

Solution:
When the core is pulled out the solenoid, the change in current will remain same.

Topic: Physics, Electromagnetic Induction, TS EAMCET, TS EAMCET 2017
Consider a solid cube made up of insulating material having a uniform volume charge density. Assuming the electrostatic potential to be zero at infinity, the ratio of the potential at a corner of the cube to that at the centre will be

Options:
A 1:1
B 1:2
C 1:4
D 1:8

The correct answer is:
1:2

Solution:
Ï-Charge density of the cube
Vlcorner- Potential at the corner of a cube of side l
Vlcentre- Potential at the centre of a cube of side l
Vl/2centre- Potential at the center of a cube of side l2
Vl/2corner- Potential at the corner of a cube of side l2
By dimensional analysis
VlcornerâQl=Ïl2
Vlcorner=4 Vl/2corner
But by superposition Vlcentre=8Vl/2corner
Because of the centre of the larger cube lies at a corner of the eight smaller cubes of which it is made
Therefore, VlcornerVlcentre=4Vl/2corner8Vl/2corner=12

Topic: Physics, Electrostatics, JEE Main
Consider a solid sphere of radius R and mass density ρr=ρ01-r2R2,0<r≤R. The minimum density of a liquid in which it will float is:

Options:
A ρ03
B ρ05
C 2ρ05
D 2ρ03

The correct answer is:
2ρ05

Solution:
The density of any point inside the sphere is

ρ=ρ01-r2R2   0<r≤R

weight = buoyant force
mg = B

mg=ρLVg
∫ρ4πr2drg=ρL43πR3g
∫0Rρ01-r2R24πr2dr=ρL43πR3
∫0Rρ04πr2-r4R2dr=ρ04πr33-r55R20R=ρL43πR3
25ρ0=ρL

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Consider a sphere of radius R which carries a uniform charge density ρ . If a sphere of radius R2 is carved out of it, as shown, the ratio EA→EB→ of magnitude of electric field EA→ and EB→ , respectively, at points A and B due to the remaining portion is:

Options:
A 2134
B 1834
C 1754
D 1854

The correct answer is:
1834

Solution:
E=ρr3ε0
EA=-ρR23ε0
EA=ρR6ε0
Electric filed at point B=EB=E1A+E2A
E1A= Electric Filed Due to solid sphere of radius R at point B=ρR3ε0
E2A= Electric Filed Due to solid sphere of radius R2 (which having charge density –ρ ) at point B
=-KQ'×49R2=-ρR54ε0
EB=E1A+E2A=ρR3ε0-ρR54ε0=17ρR54ε0
EAEB=917

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (09 Jan Shift 1)
Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume u=UV∝T4 and pressure p=13UV . If the shell now undergoes an adiabatic expansion the relation between T and R is:

Options:
A T∝1R3
B Tâe-R
C Tâe-3R
D Tâ1R

The correct answer is:
T∝1R

Solution:
âµ in an adiabatic process.
Q=0
So by first law of thermodynamics
Q=dU+W
â0=dU+W
âW=-dU
â´W=PdV
âPdV=-dU ....(i)
Given that UVâT4âU=kVT4
âdU=kdVT4=KT4dV+4T3V dT
Also, P=13UV=13kVT4V=KT43
Putting these values in equation (i)
âKT43dV=-kT4dV+4T3V dT
âTdV3=-TdV-4VdT
â4T3dV=-4VdT
â13dVV=-dTT
â13lnâ¡V=-lnâ¡T âlnâ¡V=lnâ¡T-3
âVT3=constant
43ÏR3 T3=constant
RT=constant
âTâ1R

Topic: Physics, Thermodynamics, JEE Main
Consider a spherical shell of radius R at temperature T. The black body radiation inside it can be considered as an ideal gas of photons with internal energy per unit volume u=UV∝T4 and pressure p=13UV . If the shell now undergoes an adiabatic expansion the relation between T and R is:

Options:
A T∝1R3
B Tâe-R
C Tâe-3R
D Tâ1R

The correct answer is:
T∝1R

Solution:
âµ in an adiabatic process.
Q=0
So by first law of thermodynamics
Q=dU+W
â0=dU+W
âW=-dU
â´W=PdV
âPdV=-dU ....(i)
Given that UVâT4âU=kVT4
âdU=kdVT4=KT4dV+4T3V dT
Also, P=13UV=13kVT4V=KT43
Putting these values in equation (i)
âKT43dV=-kT4dV+4T3V dT
âTdV3=-TdV-4VdT
â4T3dV=-4VdT
â13dVV=-dTT
â13lnâ¡V=-lnâ¡T âlnâ¡V=lnâ¡T-3
âVT3=constant
43ÏR3 T3=constant
RT=constant
âTâ1R

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2015 (04 Apr)
Consider a spongy block of mass $m$ floating on a flowing river. The maximum mass of the block is related to the speed of the river flow $v$, acceleration due to gravity $g$ and the density of the block $\rho$ such that $m_{\max }=k v^x g^y \rho^z(k$ is constant $)$. The values of $x, y$ and $z$ should then respectively be
(Mass of the spongy block is assumed to vary due to absorption of water by it)

Options:
A $6,3,2$
B $6,-3,1$
C $3,6,1$
D $6,1,3$

The correct answer is:
$6,-3,1$

Solution:
Since, the maximum mass of the block
floating on river depends, speed of flow of the river $=v$, acceleration due to gravity $=g$ and density of the block $=\rho$,
$m_{\max }=k v^x g^y \rho^z$
Compairing the dimensions of $\mathrm{M}, \mathrm{L}$ and $\mathrm{T}$ on both
sides, we get
$z=1$
$\begin{aligned} x+y-3 z & =0 \\ -x-2 y & =0 \\ x+y-3 \times 1 & =0[\text { [From Eq. (i) and (ii)] } \\ x+y & =3\end{aligned}$
From Eqs. (iii) and (iv), we get
$-y=3 \Rightarrow y=-3$
From Eq. (iv), we have
$\begin{aligned} & & x-3 & =3 \\ \Rightarrow & & x & =6\end{aligned}$
Hence, the value of $x, y$ and $z$ will be $(6,-3,1)$.

Topic: Physics, Units and Dimensions, JEE Main
Consider a square matrix A of order 2 which has its elements as 0,1,2,4. If the absolute value of A is least then, the absolute value of adjadjA is equal to

Options:
A 0
B 2
C 1
D 4

The correct answer is:
2

Solution:
For absolute least value of A
A=0214 or 2401
⇒A=±2
adjadjA=A(2-1)2=A=±2
So, absolute value is equal to 2

Topic: Mathematics, Determinants, JEE Main
Consider a system of blocks $X, A$ and $B$ as shown in the figure. The blocks $A$ and $B$ have equal mass and are connected by a massless string through a massless pulley. The coefficient of friction between block $A$ and $X$ or $B$ and $X$ is 0.5 . If block $X$ moves on the horizontal frictionless surface what should be Its minimum acceleration such that blocks $A$ and $B$ remain stationary. $(g=$ acceleration due to gravity.)

Options:
A $\frac{g}{3}$
B $3 g$
C $\frac{g}{4}$
D $\frac{3 g}{4}$

The correct answer is:
$\frac{g}{3}$

Solution:
Given, Mass of block $A$ and $B, m_A=m_B=m$. Coefficient of friction between each block and surface $\mu=0.5$.
Block $\not X$ moves with acceleration $a$ such that block $A$ and $B$ remains stationary. Block $B$ is stationary only when, Tension force on block $B$ is equal to friction force on $B$.
i.e.,
$$
\begin{aligned}
& T=\mu R \\
& T=\mu m a \quad \text {... (i) }[\because R=m a]
\end{aligned}
$$
(i) $[\because R=m a]$
Block $A$ is stationary only when friction force on block $A$ is equal to the sum of tension force on block $A$ and applied force on block $A$.
$$
\begin{aligned}
\mu R & =T+m a \\
\mu m g & =\mu m a+m a
\end{aligned}
$$
[From Eq. (i),]
$$
\begin{aligned}
\mu g & =\mu a+a \\
\mu g & =a(\mu+1) \\
a=\frac{\mu g}{\mu+1} & =\frac{0.5 g}{0.5+1}=\frac{0.5 g}{1.5}=\frac{g}{3} \\
\therefore \quad & =\frac{g}{3}
\end{aligned}
$$
Thus, $g / 3$ should be its minimum acceleration such that blocks $A$ and $B$ remains stationary.

Topic: Physics, Laws of Motion, JEE Main