datavorous/entrance-exam-dataset huggingface dataset:
Consider a cylindrical conductor of length L and area of cross-section A. The specific conductivity varies as σx=σ0Lx where x is the distance along the axis of the cylinder from one of its ends. The resistance of the system along the cylindrical axis is

Options:
A 2L3Aσ0
B 3L2Aσ0
C L3Aσ0
D 4L3Aσ0

The correct answer is:
2L3Aσ0

Solution:
Given, σx=σ01x
∵ Resistance of the system along the cylindrical axis,
R=∫0LρxAdx
= ∫0L1σ0LxAdx ∵ρx=1σx
= ∫0Lxσ0ALdx=1σ0ALx32320L=231σ0ALL32-0
=23⋅1σ0AL×L32, R=23⋅LAσ0

Topic: Physics, Current Electricity, JEE Main
Consider a family of circles passing through two fixed points A3,7 and B6,5. The chord in which the circle x2+y2-4x-6y-3=0 cuts each member of family of circles passes through a fixed point a,b. Then the value of a+3b is

Options:

The correct answer is:
25

Solution:
Family of circles is x-3x-6+y-7y-5+λ
2x+3y-27=0
i.e., x2+y2+x2λ-9+y3λ-12+53-27λ=0
Common chord of family of circles and the given circle is
-5x-6y+56+λ2x+3y-27=0
which represents family of line passing through point of intersection of the lines
−5x−6y+56=0  and 2x+3y−27=0
The point of intersection is 2, 233
a=2
b=233
a+3b=2+23=25

Topic: Mathematics, Circle, JEE Main
Consider a family of circles which are passing through the point $(-1,1)$ and are tangent to $x-$ axis. If $(h, k)$ are the co-ordinates of the centre of the circles, then the set of values of $k$ is given by the interval

Options:
A
$0 < \mathrm{k} < 1 / 2$
B
$k \geq 1 / 2$

C
$-1 / 2 \leq k \leq 1 / 2$

D
$k \leq 1 / 2$

The correct answer is:
$k \geq 1 / 2$

Solution:
Equation of circle $(x-h)^2+(y-k)^2=k^2$
It is passing through $(-1,1)$ then
$\begin{aligned}
& (-1-h)^2+(1-k)^2=k^2 \\
& h^2+2 h-2 k+2=0 \\
& D \geq 0 \\
& 2 k-1 \geq 0 \Rightarrow k \geq 1 / 2.
\end{aligned}$

Topic: Mathematics, Circle, JEE Main, JEE Main 2007
Consider a finite insulated, uncharged conductor placed near a finite positively charged conductor. The uncharged body must have a potential :

Options:
A
less than the charged conductor and more than at infinity.

B
more than the charged conductor and less than at infinity.

C
more than the charged conductor and more than at infinity.

D
less than the charged conductor and less than at infinity.

The correct answer is:
less than the charged conductor and more than at infinity.

Solution:
The potential of uncharged body is less than that of the charged conductor and more than at infinity.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2013 (23 Apr Online)
Consider a force F→=-xi^+yj^ . The work done by this force in moving a particle from point A1,0 to B0,1 along the line segment is : (all quantities are in SI units)

Options:
A 2
B 12
C 1
D 32

The correct answer is:
1

Solution:
It is given that a force F→=-xi^+yj^ acts on a particle. The particle is moved from point A1,0 to B0,1 along the line segment.

Work done by a variable force on the particle,
W=∫F→·dr→=∫F→·dxi^+dyj^
(In two dimension, dr→=dxi^+dyj^
and it is given F→=-xi^+yj^)
W=∫-xi^+yj^·dxi^+dyj^

=∫-xdx+ydy=∫-xdx+∫ydy.
As the particle is displaced from A1,0 to B0,1, x varies from 1 to 0 and y varies from 0 to 1.
So, work will be,

W=-∫10xdx+∫01ydy=-x2210+y2201

=120+12+12-0=1 J.

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2020 (09 Jan Shift 1)
Consider a frame that is made up of two thin massless rods AB and AC as shown in the figure. A vertical force P→ of magnitude 100 N is applied at point A of the frame.

Suppose the force is P→ resolved parallel to the arms AB and AC of the frame. The magnitude of the resolved component along the arm AC is xN. The value of x, to the nearest integer, is ________.

[Given : sin35°=0.573,cos35°=0.819, sin110°=0.939,cos110°=-0.342]

Options:

The correct answer is:
82

Solution:
Component along AC

=100cos35° N

=100×0.819 N

=81.9 N

≈82 N

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Consider a function f:RâR defined by fx=x3+4x+5, then

Options:
A f is one-one but not onto
B f is onto but not one-one
C f is one-one and onto
D f is neither one-one nor onto

The correct answer is:
f is one-one and onto

Solution:
Given, fx=x3+4x+5
Now, f â² x =3 x 2 +4>0,âxâR
â´fx is strictly increasing function
â´ It is one-one
Clearly, fx is a continuous function and also increasing on R
limxâ-âfx=-â and limxââfx=â
â´fx takes every value between -â and â.
Thus, fx is an onto function.

Topic: Mathematics, Functions, JEE Main
Consider a function fx=xx, âxâ1,â. If gx is the inverse function of fx, then the value of g â² 4 is equal to

Options:
A log2⁡e
B 12log2e⁡e
C 14log2e⁡e
D 12loge⁡2e

The correct answer is:
14log2e⁡e

Solution:
fx=xxâlogâ¡fx=xlogâ¡x
Differentiating with respect to x, we get,

f â² x = x x 1+logx
Now, by property g â² f x = 1 f â² x {âµgx is the inverse of fx}
Putting x=2
g'f2=1f'2âg'4=1221+log2=14log2ee

Topic: Mathematics, Differentiation, JEE Main
Consider a function gx=fx-2, âxâR, where fx=1x:xâ¥1ax2+b:x<1. If gx is continuous as well as differentiable for all x, then

Options:
A a=-12,b=32
B a=12,b=32
C a=-12,b=-32
D None of these

The correct answer is:
a=-12,b=32

Solution:
For gx to be continuous and differentiable âxâR, fx must be continuous and differentiable âxâR
Since, f(x) is continuous for all x, therefore, it is continuous at x=1 also.
â´f1=limhâ0+f1-h=1=limhâ0+a1-h2+b
âa+b=1 â¦ 1
Also, fx is differentiable at x=1âf'1â=f'1+
âlimhâ0+f1-h-f1-h=limhâ0+f1+h-f1h
âlimhâ0+a1-h2+b-1-h=limhâ0+11+h-1h
âlimhâ0+a+b-1+h2-2ha-h=limhâ0+1-1-hh1+h
â2a=-1 Using a+b=1
â´a=-12
Hence, a+b=1âb=1-a=32.

Topic: Mathematics, Continuity and Differentiability, JEE Main
Consider a galvanometer shunted with 5 Ω resistance and 2% of current passes through it. What is the resistance of the given galvanometer?

Options:
A 245 Ω
B 344 Ω
C 300 Ω
D 226 Ω

The correct answer is:
245 Ω

Solution:
IG=2100×I (Given)

0.98 I × 5 = 0.02 × RG

RG = 245 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (31 Aug Shift 1)
Consider a horizontal surface moving vertically upward with velocity 2 m s-1. A small ball of mass 2 kg is moving with velocity 2i^â2j^ m s-1 If the coefficient of restitution and coefficient of friction are 12 and 13 respectively, find the horizontal velocity (in m s-1) of the ball after the collision.

Options:

The correct answer is:
0

Solution:
The relative velocity of the particle with respect to the surface, along the normal, is the approach speed of the particle.

vapp=4 m s-1

The separation velocity will be

vsep=evapp=12×4=2 m s-1

∴ v̄m,gr=v̄m,s+v̄s,gr=4 m/s

∫N dt=42+22=12

Now, ∫μN⋅dt=2vxf-2

=-13×12=2vxf-2

vxfinal=0

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n=2 to n= 1 transition has energy 74.8 eV higher than the photon emitted in the n=3 to n=2 transition. Given that the ionization energy of the hydrogen atom is 13.6 eV, what is the value of Z?

Options:

The correct answer is:
3

Solution:
âE2 →1=13.6ÃZ21-14=13.6ÃZ234
âE3 →2=13.6ÃZ214-19=13.6ÃZ2536
âE2 →1=âE3-2+74.8
13.6ÃZ234=13.6ÃZ2536+74.8
13.6ÃZ234-536=74.8
Z2=9
Z=+3

Topic: Physics, Atomic Physics, JEE Main
Consider a hydrogen-like sample of gas whose energy in the nth excited state is given by En=-13.6Z2n2 eV. When the electron makes transitions from an excited state to the ground state, the most energetic photons can have energy Emax=52.224 eV and the least energetic photons can have energy Emin=1.224 eV. If the atomic number of the atom and the state of excitation are Z and n respectively, then what is the value of Z+n?

Options:

The correct answer is:
7

Solution:
52.224=13.6Z2 112-1n2 .....(1)
1.224=13.6z2 1n-12-1n2 .....(2)
On solving Z = 2 n = 5
Hence Z + n = 7

Topic: Physics, Atomic Physics, JEE Main
Consider a hyperbola H : x2-2y2=4. Let the tangent at a point P(4,6) meet the x-axis at Q and latus rectum at Rx1,y1,x1>0. If F is a focus of H which is nearer to the point P, then the area of ΔQFR (in sq. units) is equal to

Options:
A 46
B 6-1
C 76-2
D 46-1

The correct answer is:
76-2

Solution:
x24-y22=1

e=1+b2a2=32

∴ Focus F(ae,0)⇒F(6,0)

equation of tangent at P to the hyperbola is
2x-y6=2

tangent meet x-axis at Q(1,0)

& latus rectum x=6 at R6,266-1

∴ Area of ΔQFR=126-1·266-1

=76-2 sq. units

Topic: Mathematics, Hyperbola, JEE Main
Consider a hyperbola H : x2-2y2=4. Let the tangent at a point P(4,6) meet the x-axis at Q and latus rectum at Rx1,y1,x1>0. If F is a focus of H which is nearer to the point P, then the area of ΔQFR (in sq. units) is equal to

Options:
A 46
B 6-1
C 76-2
D 46-1

The correct answer is:
76-2

Solution:
x24-y22=1

e=1+b2a2=32

∴ Focus F(ae,0)⇒F(6,0)

equation of tangent at P to the hyperbola is
2x-y6=2

tangent meet x-axis at Q(1,0)

& latus rectum x=6 at R6,266-1

∴ Area of ΔQFR=126-1·266-1

=76-2 sq. units

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2021 (18 Mar Shift 2)
Consider a light planet revolving around a massive star in a circular orbit of radius ' $r$ ' with time period ' $T$ '. If the gravitational force of attraction between the planet and the star is proportional to $r^{-\frac{7}{2}}$, then $T^2$ is proportional to

Options:
A $r^{9 / 2}$
B $r^{7 / 2}$
C $r^{5 / 2}$
D $r^{3 / 2}$

The correct answer is:
$r^{9 / 2}$

Solution:
For the planet to orbit around the star, the
centripetal force must be provided by gravitational force. Hence, $\mathrm{F}_{\mathrm{d}}=\mathrm{F}_{\mathrm{a}}$.
$\mathrm{F}_{\mathrm{a}} \propto-\mathrm{r}^{-7 / 2}$
...(Given)
(-ve sign indicates force is towards the centre of orbit)
$$
\begin{array}{ll}
& \text { Hence, } \mathrm{a} \propto-\mathrm{r}^{-7 / 2} \\
\therefore \quad & -\omega^2 \mathrm{r} \propto-\mathrm{r}^{-7 / 2} \\
\therefore \quad & \omega^2 \propto \mathrm{r}^{-9 / 2} \\
\therefore \quad & \frac{4 \pi^2}{\mathrm{~T}^2} \propto \mathrm{r}^{-9 / 2} \\
& \Rightarrow \mathrm{T}^2 \propto \mathrm{r}^{9 / 2}
\end{array}
$$

Topic: Physics, Gravitation, JEE Main
Consider a light ray travelling in air is incident into a medium of refractive index 2n. The incident angle is twice that of refracting angle. Then, the angle of incidence will be

Options:
A sin-1n
B cos-1n2
C sin-12n
D 2cos-1n2

The correct answer is:
2cos-1n2

Solution:
Let angle of incidence be θ, then angle of refraction θ2. From Snell's law, we can write,

1×sinθ=2n×sinθ2, 2sinθ2cosθ2=2nsinθ2

cosθ2=n2 ⇒θ=2cos-1n2

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (27 Jun Shift 1)
Consider a line $\mathrm{L}$ passing through the points $\mathrm{P}(1,2,1)$ and $\mathrm{Q}(2,1,-1)$. If the mirror image of the point $\mathrm{A}(2,2,2)$ in the line $\mathrm{L}$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+6 \gamma$ is equal to _______

Options:

The correct answer is:
6

Solution:
DR's of Line $L \equiv-1: 1: 2$
DR's of $\mathrm{AB} \equiv \alpha-2: \beta-2: \gamma-2$
$\begin{aligned}
& \mathrm{AB} \perp_{\mathrm{ar}} \mathrm{L} \Rightarrow 2-\alpha+\beta-2+2 \gamma-4=0 \\
& 2 \gamma+\beta-\alpha=4
\end{aligned}$
Let $C$ is mid-point of $A B$
$\begin{aligned}
& \mathrm{C}\left(\frac{\alpha+2}{2}, \frac{\beta+2}{2}, \frac{\gamma+2}{2}\right) \\
& \text { DR's of PC }=\frac{\alpha}{2}: \frac{\beta-2}{2}: \frac{\gamma}{2} \\
& \text { line L } \left\lvert\, \mathrm{PC} \Rightarrow \frac{-\alpha}{2}=\frac{\beta-2}{2}=\frac{\gamma}{4}=\mathrm{K}(\text { let) }\right.
\end{aligned}$
$\begin{aligned}
& \alpha=-2 \mathrm{~K} \\
& \beta=2 \mathrm{~K}+2 \\
& \gamma=4 \mathrm{~K}
\end{aligned}$
use in (1) $\Rightarrow K=\frac{1}{6}$
value of $\alpha+\beta+6 \gamma=24 \mathrm{~K}+2=6$

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (04 Apr Shift 2)
Consider a matrix A=0120-30111. If 6A-1=aA2+bA+cI, where a,b,câZ and I is an identity matrix, then a+2b+3c is equal to

Options:
A 10
B -10
C 8
D 0

The correct answer is:
-10

Solution:
A=6≠0 ( A is non sigular)
Given, 6A-1=aA2+bA+cI
⇒6I=aA3+bA2+cA
A2=0120-301110120-30111=2-120901-13
A3=A2A=2-120901-130120-30111=2760-270375
Now, aA3+bA2+cA=2a+2b7a-b+c6a+2b+2c0-27a+9b-3c03a+b+c7a-b+c5a+3b+c
6I=600060006
∵aA3+bA2+cA=6I

⇒a+b=3,
7a-b+c=0,
3a+b+c=0,
-9a+3b-c=2,
5a+3b+c=6
On solving, we get,
a=1,b=2,c=-5
Hence, a+2b+3c=1+4-15=-10

Topic: Mathematics, Matrices, JEE Main
Consider a matrix A=aij3Ã3 where, aij=i+2jâij=even2i-3jâij=odd. If bij is the cofactor of aij in matrix A and Cij=âr=13airbjr, then Cij3Ã3 is

Options:
A 100010001
B -15-746839-3
C 184000184000184
D 231162-152

The correct answer is:
184000184000184

Solution:
A=a11a12a13a21a22a23a31a32a33=-15-746837-3
B=b11b12b13b21b22b23b31b32b33
A=-1-18-56-5-12-24-728-18
=184
Cij=ai1bj1+ai2bj2+ai3bj3
C11=a11b11+a12b12+a13b13=A
C12=a11b21+a12b22+a13b23=0
Similarly,
C13=0,C21=0,C22=A,C23=0,C31=0
C32=0,C33=A
Cij3×3=A000A000A=184000184000184

Topic: Mathematics, Matrices, JEE Main
Consider a mixture of n moles of helium gas and 2n moles of oxygen gas (molecules taken to be rigid) as an ideal gas. Its CPCV value will be:

Options:
A 1913
B 6745
C 4027
D 2315

The correct answer is:
1913

Solution:
γmix =n1cp1+n2cp2n1cv1+n2cv2
=n52R+2n72Rn32R+2n52R
=5+143+10=1913

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Consider a nucleus X3060. It's approximate density is (Take 1 amu=1.6×10-27 kg,Ro=1.2×10-15 m)

Options:
A 1.2×1018 kg m-3
B 8.5×1019 kg m-3
C 3.3×1016 kg m-3
D 2.2×1017 kg m-3

The correct answer is:
2.2×1017 kg m-3

Solution:
Density=massvolume

=A×1.6×10-274π3×R0A133=34π×1.6×10-271.2×10-153=2.2×1017 kg m-3

Topic: Physics, Nuclear Physics, JEE Main
Consider a pair of identical pendulums, which oscillate with equal amplitude independently such that when one pendulum is at its extreme position making an angle of 2o to the right with the vertical, the other pendulum makes an angle of 1o to the left of the vertical. What is the phase difference (in degree) between the pendulums?

Options:

The correct answer is:
120

Solution:
Assuming the two pendulums follow the following functions of their displacement :
θ1 = θ0 sin (ωt + ϕ1) ..(i)
and θ2 = θ0 sin (ωt + ϕ2) ..(ii)
As it is given that amplitude and time period being equal but phases being different.
Now, for first pendulum at any time t
θ1 = θ0 (Right extreme)
From Eq. (i), we get
⇒ θ0 = θ0 sin (ωt + ϕ1)
or 1 = sin (ωt + ϕ1)
or ω t + ϕ 1 = π 2 … iii
Similarly, at the same instant t for pendulum second, we have
θ 2 = - θ 0 2
Note θ0 = 2o is the angular amplitude for the second pendulum the angular displacement is one degree therefore, θ 2 = θ 0 2 and negative sign is taken to show for being left to mean position
From Eq. (ii), then
- θ 0 2 = θ 0 sin ω t + ϕ 2
⇒ sin ω t + ϕ 2 = - 1 2
⇒ ω t + ϕ 2 = - π 6 or 7 π 6
or ω t + ϕ 2 = - π 6 or 7 π 6 … iv
From Eqs. (iv) and (iii), the difference in phases
ω t + ϕ 2 - ω t + ϕ 1 = - 7 π 6 or π 2 = 7 π - 3 π 6 = 4 π 6
or ϕ 2 - ϕ 1 = 4 π 6 = 2 π 3 = 1 2 0 ο

Topic: Physics, Oscillations, JEE Main
Consider a parabola y2=4x with vertex at A and focus at S. PQ is a chord of the parabola which is normal at point P. If the abscissa and the ordinate of the point P are equal, then the square of the length of the diameter of the circumcircle of triangle PSQ is

Options:

The correct answer is:
125

Solution:
t12=2t1 and t2=t1-2t1
∴t1=2 & t2=-3
So, P4,4 and Q9,-6
Slope of PS=43

Slope of QS=-68=-34

⇒PS⊥QS
∴PQ is the diameter of the circle passing through P,S,Q.
Hence, the length of the diameter =25+100=125
∴ square of the length of the diameter=125

Topic: Mathematics, Parabola, JEE Main
Consider a parallel plate capacitor which is maintained at potential $200 \mathrm{~V}$. The separation distance between the plates of capacitor and the area of plates are $1 \mathrm{~mm}$ and $20 \mathrm{~cm}^2$, respectively. Calculate displacement current in $1 \mu \mathrm{s}$.

Options:
A $3.5 \mathrm{~mA}$
B $2.5 \mathrm{~mA}$
C $1.5 \mathrm{~mA}$
D $0.5 \mathrm{~mA}$

The correct answer is:
$3.5 \mathrm{~mA}$

Solution:
Displacement current,
$\begin{aligned} I_d & =\varepsilon_0 \frac{d \phi}{d t}=\varepsilon_0 \frac{\Delta \phi}{\Delta t} \\ & =\varepsilon_0 \frac{\Delta E A}{\Delta t} \quad(\because \phi=E \cdot A)\end{aligned}$
$=\frac{\varepsilon_0 \cdot \Delta V \cdot A}{\Delta t \cdot d} \quad\left(\because E=\frac{V}{d}\right)$
Here, $\Delta V=200 \mathrm{~V}$,
$\begin{aligned} \Delta t & =1 \mu \mathrm{s}=10^{-6} \mathrm{~s}, \\ d & =1 \mathrm{~mm}=1 \times 10^{-3} \mathrm{~m} \\ \text { and } \quad A & =20 \mathrm{~cm}^2=20 \times 10^{-4} \mathrm{~m}^2\end{aligned}$
$\begin{aligned} \therefore \quad I_d & =\frac{8.85 \times 10^{-12} \times 200 \times 20 \times 10^{-4}}{10^{-6} \times 10^{-3}} \\ & =3.5 \times 10^{-3} \mathrm{~A}=3.5 \mathrm{~mA}\end{aligned}$

Topic: Physics, Electromagnetic Waves, TS EAMCET, TS EAMCET 2020 (09 Sep Shift 2)
Consider a parallel plate capacitor with plates in the shape of square and in $X Y$-plane. The gap between the plates is filled with dielectric material. The dielectric constant $k$ varies with $X$-axis as $k(x)=\left[1+\left(\frac{x}{L}\right)^\alpha\right]$, where $\alpha$ is a constant. Let $C_d$ and $C_a$ are capacitance in the presence of dielectric and air, respectively. If the ratio $\frac{C_d}{C_a}=\frac{7}{6}$, then the value of $\alpha$ must be

Options:
A 3
B 5
C 7
D 9

The correct answer is:
5

Solution:
Area of square plate, $A=L^2$
Consider an elemental capacitor at distance $x$ from above of thickness $d x$. Capacitance of elemental capacitor,
$$
d C=\frac{k \varepsilon_0 d A}{d}=\frac{k \varepsilon_0(L \cdot d x)}{d}
$$
All such elemental capacitors will be in parallel and equivalent capacitance will be sum of aII, so
$$
\begin{aligned}
& C_d=\frac{\varepsilon_0 L}{d} \int_0^L 1+\left(\frac{x}{L}\right)^a \cdot d x \\
& C_d=\frac{\varepsilon_0 L}{d}\left[x+\frac{\left(\frac{x}{L}\right)^{1+\alpha}}{(1+\alpha) \cdot \frac{1}{L}}\right]_0^L \\
& C_d=\frac{\varepsilon_0 L}{d}\left[L+\frac{L}{1+\alpha}\right] \Rightarrow C_d=\frac{\varepsilon_0 L^2}{d}\left(\frac{2+\alpha}{1+\alpha}\right)
\end{aligned}
$$
It is given, $\frac{C_d}{C_a}=\frac{7}{6}$
where, $C_a=$ capacitance in absence of dielectric.
$$
\begin{array}{ccc}
\Rightarrow \quad \frac{\frac{\varepsilon_0 L^2}{d}\left(\frac{2+\alpha}{1+\alpha}\right)}{\frac{\varepsilon_0 L^2}{d}}=\frac{7}{6} & \Rightarrow \frac{2+\alpha}{1+\alpha}=\frac{7}{6} \\
\Rightarrow \quad 12+6 \alpha=7+7 \alpha & \Rightarrow \quad \alpha=5
\end{array}
$$

Topic: Physics, Capacitance, JEE Main
Consider a parallelogram constructed on the vectors Aâ=5pâ+2qâ and Bâ=pâ-3qâ. If pâ=2, qâ=5, the angle between pâ and qâ is Ï3 and the length of the smallest diagonal of the parallelogram is k units, then the value of k2 is equal to

Options:

The correct answer is:
109

Solution:
Diagonals of the parallelogram are A→+B→, A→-B→
=6p→-q→,4p→+5q→
A→+B→=36p→2+q→2-12p→⋅q→
=144+25-12×2×5×cosπ3
=109
A→-B→=16p→2+25q→2+40p→⋅q→
=16×4+25×25+40×2×5×cosπ3
=889
So, length of the smaller diagonal is 109 units

Topic: Mathematics, Vector Algebra, JEE Main
Consider a particle moving along the positive direction of X-axis. The velocity of the particle is given by v=αx (α is a positive constant). At time t=0, if the particle is located at x=0, the time dependence of the velocity and the acceleration of the particle are respectively.

Options:
A α22t and α22
B α2t and α2
C α2t and α2
D α24t and α24

The correct answer is:
α22t and α22

Solution:
Given that, v=αx

dxdt=αx ∴v=dxdt

dxx=αdt

∫dxx=α∫dt

⇒ 2x=αt+c

At t=0, x=0

⇒ c=0

∴ 2x=αt

On Squaring,

⇒ 4x=α2t2

⇒ x=α2t24

On differentiating,

dxdt=v=α242t2=α2t2

⇒ a=dvdt=dvdtdxdt=vdvdx

=αxddxαx

=αxαx-1/22=α22

Topic: Physics, Motion In One Dimension, JEE Main
Consider a plane P:2x+y-z=5, a line L:x-32=y+1-3=z-2-1 and a point A3,-4,1. If the line L intersects plane P at B and the xy plane at C, then the area (in sq. units) of âABC is

Options:
A 7
B 8
C 10
D 23

The correct answer is:
10

Solution:
For xy plane, putting z=0, we get,
x-32=y+1-3=2âx=7,y=-7
Coordinates of point C are 7,-7,0
Any general point on the line is 2Î»+3,-3Î»-1,-Î»+2
This must satisfy the given plane
4Î»+6-3Î»-1+Î»-2=5
2Î»+3=5âÎ»=1
So, coordinates of point B are 5,-4,1
ABâ=<2,0,0>ACâ=<4,-3,-1>
Area =12ABâÃACâ=12i^j^k^2004-3-1
=12-2-j^+3k^=j^-3k^=10 sq. units

Topic: Mathematics, Three Dimensional Geometry, JEE Main
Consider a planet in some solar system that has a mass double the mass of earth and density equal to the average density of the earth. If the weight of an object on earth is W, the weight of the same object on that planet will be:

Options:
A 2W
B W
C 213W
D 2 W

The correct answer is:
213W

Solution:
Density is same

M=43πR3ρ, 2m=43πR'3ρ

R'=213R

W=GMmR2

W2=G2MmR'2

W2=213W

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (25 Jul Shift 2)
Consider a point object situated at a distance of $30 \mathrm{~cm}$ from the centre of sphere of radius $2 \mathrm{~cm}$ and refractive
index 1.5 as shown in the figure. If the refractive index of the region surrounding this sphere is 1.4 , then the
position of the image due to refraction by sphere with respect to the centre is

Options:
A $30 \mathrm{~cm}$
B $45 \mathrm{~cm}$
C $\infty$
D $28 \mathrm{~cm}$

The correct answer is:
$30 \mathrm{~cm}$

Solution:
(Note If we takes, $n_2=1.4$ and $n_1=1.5$, then we get option 1 as answer)
For first surface, $n_1=1.5, n_2=1.4$
$$
\begin{array}{rr}
\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} & u=-30 \mathrm{~cm} \\
R=+2 \mathrm{~cm} & \frac{1.4}{v}=\frac{0.1}{2}-\frac{1.5}{30} \\
\frac{1.4}{v}-\frac{1.5}{-30}=\frac{1.5-1.4}{2} & \Rightarrow v=\infty \\
\frac{1.4}{v}=\frac{.3-3}{30} &
\end{array}
$$
For second surface,
$$
\begin{gathered}
\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \quad u=\infty \quad R=-2 \mathrm{~cm} \\
\frac{1.5}{v}-\frac{1.4}{\infty}=\frac{-0.1}{-2} \quad \frac{1.5}{v}=\frac{1}{20} \\
v=20 \times 1.5=30 \mathrm{~cm}
\end{gathered}
$$

Topic: Physics, Ray Optics, JEE Main
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