datavorous/entrance-exam-dataset huggingface dataset:
A double convex lens has power P and same radii of curvature r of both the surfaces. The radius of curvature of a surface of a plano-convex lens made of the same material with power 1.5 P is :

Options:
A 2R
B R2
C 3R2
D R3

The correct answer is:
R3

Solution:
1P=(n-1)1R-1-R

1P=(n-1)2R .........1

11.5P=(n-1)1R-1∞ .......ii

23P=n-1R

From i and ii

32=R'2R

⇒ R'=R3

Topic: Physics, Ray Optics, JEE Main, JEE Main 2020 (06 Sep Shift 2)
A double convex thin lens made of the refractive index 1.6 has radii of curvature 15 cm each. What is the focal length (in cm) of this lens when immersed in a fluid of refractive index 1.63?

Options:

The correct answer is:
-407.5

Solution:
Focal length of lens in given by
1fw= Î¼g-11R1-1R2
=1.61.63-1115+115
=-0.03Ã21.63Ã15=.061.63Ã15
so, fw= -8152
=-407.5 cm

Topic: Physics, Ray Optics, JEE Main
A double ordinate $\mathrm{PQ}$ of the hyperbola $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ is such that $\Delta \mathrm{OPQ}$ is equilateral, $\mathrm{O}$ being the centre of the hyperbola. Then the eccentricity e satisfies the relation

Options:
A $1 < \mathrm{e} < \frac{2}{\sqrt{3}}$
B $\mathrm{e}=\frac{2}{\sqrt{3}}$
C $\mathrm{e}=\frac{\sqrt{3}}{2}$
D $\mathrm{e}>\frac{2}{\sqrt{3}}$

The correct answer is:
$\mathrm{e}>\frac{2}{\sqrt{3}}$

Solution:
Hint:
$P \equiv(a \sec \theta, b \tan \theta) \angle \mathrm{POM}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{\mathrm{b}}{\mathrm{a}} \sin \theta$
$\Rightarrow \sin \theta=\frac{a}{b \sqrt{3}}$
$\Rightarrow \frac{b^{2}}{a^{2}}>\frac{1}{3} \quad \Rightarrow 1+\frac{b^{2}}{a^{2}}>\frac{4}{3} \quad \Rightarrow e>\frac{2}{\sqrt{3}}$

Topic: Mathematics, Hyperbola, JEE Main
A double slit experiment is performed with light of wavelength $500 \mathrm{~nm}$. A thin film of thickness $2 \mu \mathrm{m}$ and refractive index 1.5 is introduced in the path of the upper beam. The location of the central maximum will

Options:
A remain unshifted
B shift downward by nearly two fringes
C shift upward by nearly two fringes
D shift downward by 10 fringes.

The correct answer is:
shift upward by nearly two fringes

Solution:
Wavelength $\lambda=500 \mathrm{~nm}=500 \times 10^{-9} \mathrm{~m}$
Thickness of the film, $t=2 \mu \mathrm{m}=2 \times 10^{-6} \mathrm{~m}$ Refractive index $\mu=1.5$
When there is no thin film placed in the path of any one of the two beams, the path difference between
them is given by $d y / D$ (considering the two beams meetings at a point $P)$. In case we put a thin film in the path of one of the beams, then the optical path
of that beam gets longer.
Now for the central maximum, path difference in absence of the film is $\Delta x=0$. But when we put the film, the path difference becomes
$$
\begin{aligned}
\Delta x & =\mu t-t=(\mu-1) t \\
& =(1.5-1) \times 2 \times 10^{-6}=10^{-6} \mathrm{~m}=1 \mathrm{~mm} .
\end{aligned}
$$
Now, $\Delta x=\frac{d y}{D} \Rightarrow y=\frac{D}{d} \Delta x=\frac{D}{d} \times 1 \mu \mathrm{m}$
And the fringe width is given by
$$
\begin{aligned}
& W=\frac{D \lambda}{d}=\frac{D}{d} \times 500 \times 10^{-9} \mathrm{~m}=\frac{D}{d} \times 0.5 \times 1 \mu \mathrm{m} \\
\therefore \quad & y=\frac{D}{d} \times 1 \mu \mathrm{m}=2 \times \frac{D}{d} \times \frac{1}{2} \times 1 \mu \mathrm{m}=2 W .
\end{aligned}
$$
As the film is placed in the path of the upper beam, the central maximum will shift upward by nearly two fringes.

Topic: Physics, Wave Optics, AIIMS, AIIMS 2003
A doubly ionised $\mathrm{Li}$ atom is excited from its ground $\operatorname{state}(n=1)$ to $n=3$ state. The wavelengths of the spectral lines are given by $\lambda_{32}, \lambda_{31}$ and $\lambda_{21}$. The ratio $\lambda_{32} / \lambda_{31}$ and $\lambda_{21} / \lambda_{31}$ are, respectively

Options:
A
$8.1,0.67$

B
$8.1,1.2$

C
$6.4,1.2$

D
$6.4,0.67$

The correct answer is:
$6.4,1.2$

Solution:
$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$ where $R=$ Rydberg
$$
\begin{aligned}
& \text { constant } \\
& \frac{1}{\lambda_{32}}=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36} \\
& \Rightarrow \lambda_{32}=\frac{36}{5}
\end{aligned}
$$
Similarly solving for $\lambda_{31}$ and $\lambda_{21}$
$$
\begin{aligned}
& \lambda_{31}=\frac{9}{8} \text { and } \lambda_{21}=\frac{4}{3} \\
& \therefore \quad \frac{\lambda_{32}}{\lambda_{31}}=6.4 \text { and } \frac{\lambda_{21}}{\lambda_{31}} \simeq 1.2
\end{aligned}
$$

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2012 (12 May Online)
A driver applies the brakes on seeing the red traffic single $400 \mathrm{~m}$ ahead. At the time of applying the brakes, the vehicle was moving with $15 \mathrm{~m} / \mathrm{s}$ and retarding at $0.3 \mathrm{~m} / \mathrm{s}^2$. The distance covered by the vehicle from the traffic light 1 minute after the application of brakes is

Options:
A 25 m
B 360 m
C 40 m
D 375 m

The correct answer is:
40 m

Solution:
$$
\begin{aligned}
& \mathrm{u}=15 \mathrm{~m} / \mathrm{s}, \mathrm{a}=-0.3 \mathrm{~m} / \mathrm{s}^2, \mathrm{t}=60 \mathrm{~s} \\
& \mathrm{~s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 \\
& =15 \times 60+\frac{1}{2} \times(-0.3) \times(60)^2 \\
& =900-540=360 \mathrm{~m}
\end{aligned}
$$
Distance from traffic light $=400-360=40 \mathrm{~m}$

Topic: Physics, Motion In One Dimension, JEE Main
A driver in a car, approaching a vertical wall notices that the frequency of his car horn has changed from 440 Hz to 480 Hz, when it gets reflected from the wall. If the speed of sound in air is 345 m s-1, then the speed of the car is:

Options:
A 54 km/hr
B 36 km/hr
C 18 km/hr
D 24 km/hr

The correct answer is:
54 km/hr

Solution:
Given,

Frequency of sound after reflection from the wall is

f''=480 Hz

Frequency of sound emitted by the horn of car is

f=440 Hz

Speed of sound is v=345 m s-1

The diver in the car (as a source) approaching the vertical wall which is the observer in this case, The frequency of sound received by the stationary wall is given by

fr'=vv-vcf ...(1)

As the frequency f' is received by wall will act as a source now, and driver in the car is an observer, the frequency of sound received by driver is ,

fr''=v+vcvf' ...(2)

Equating equation (1) and (2) we get,

fr''=v+vcv-vcf ...(3)

Substitute the given value in the equation (3), we get
⇒480=345+vc345-vc440

⇒1211=345+vc345-vc

Velocity of car is given by

⇒vc=54 km hr-1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2020 (05 Sep Shift 2)
A drop of each of the following, is placed separately on four slides. Which of them will not coagulate?

Options:
A blood serum
B sample from the thoracic duct of lymphatic system
C whole blood from pulmonary vein
D blood plasma.

The correct answer is:
blood serum

Solution:
Serum is the component of blood which does not play a role in clotting. Serum includes all proteins not used in blood clotting; all electrolytes, antibodies, antigens, hormones; and any exogenous substances (e.g., drugs or microorganisms).
Blood serum $=$ Blood $-($ cell + clotting factors)
Caution
Serum is the liquid that remains after the blood has clotted. Plasma is the liquid that remains when clotting is prevented with the addition of an anticoagulant.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2007
A drop of liquid of density ρ is floating half immersed in a liquid of density σ and surface tension 7.5×10-4 N cm-1. The radius of drop in cm will be : (Take : g=10 ms-2)

Options:
A 152ρ-σ
B 15ρ-σ
C 32ρ-σ
D 3202ρ-σ

The correct answer is:
152ρ-σ

Solution:
Balancing the forces on drop

Buoyant force + surface tension =mg

σV2g+2πRT=ρVg

2πRT=2ρ-σ2·43πR3 g; As V=43πR3

⇒R3=3T2ρ-σg⇒R=3×7.5×10-2 N-m-12ρ-σ×10

R=3202ρ-σ m=152ρ-σ cm

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2022 (25 Jul Shift 2)
A drop of liquid of density ' $\rho$ ' is floating half immersed in a liquid of density ' $d$ '. If ' $T$ ' is the surface tension, then the diameter of the drop of the liquids is

Options:
A $\sqrt{\frac{6 T}{g(2 \rho-d)}}$
B $\sqrt{\frac{T}{g(2 \rho-d)}}$
C $\sqrt{\frac{2 T}{g(2 \rho-d)}}$
D $\sqrt{\frac{12 T}{g(2 \rho-d)}}$

The correct answer is:
$\sqrt{\frac{12 T}{g(2 \rho-d)}}$

Solution:
The drop is in equilibrium under the action of the following forces:
Weight of the liquid, $\mathrm{W}=\mathrm{Mg}=\frac{4}{3} \pi \mathrm{r}^3 \rho \mathrm{g}$ (downwards)
Upthrust $=$ weight of the liquid displaced
$$
\therefore \mathrm{F}_{\mathrm{t}}=\frac{2}{3} \pi \mathrm{r}^3 \mathrm{dg}
$$
(upwards)
Force due to surface tension, $\mathrm{F}=2 \pi \mathrm{rT}$ (upwards)
$$
\begin{aligned}
& \therefore \mathrm{Mg}=\mathrm{F}+\mathrm{F}_{\mathrm{t}} \\
& \therefore \mathrm{F}=\mathrm{Mg}-\mathrm{F}_{\mathrm{t}} \\
& \therefore 2 \pi \mathrm{r} T=\frac{4}{3} \pi \mathrm{r}^3 \rho g-\frac{2}{3} \pi \mathrm{r}^3 \mathrm{dg} \\
& \therefore \mathrm{T}=\frac{2}{3} \mathrm{r}^2 \rho g-\frac{1}{3} \mathrm{r}^2 \mathrm{dg} \\
& =\mathrm{r}^2 g\left(\frac{2}{3} \rho-\frac{1}{3} d\right) \\
& =\mathrm{r}^2 g\left(\frac{2 \rho-d}{3}\right) \\
& \therefore \mathrm{r}=\sqrt{\frac{3 \mathrm{~T}}{\mathrm{~g}(2 \rho-\mathrm{d})}} \\
& \therefore \mathrm{D}=\sqrt{\frac{12 \mathrm{~T}}{\mathrm{~g}(2 \rho-d)}}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A drop of water of mass m, surface tension T, is trapped between two parallel glass plates, separated by distance d. The force needed to separate the plates, to pull them apart, perpendicular to the plane of plates, is

Options:
A 2Tmπd2ρ
B 2Tmd2ρ
C Tm2d2ρ
D Tm2πd2ρ

The correct answer is:
2Tmd2ρ

Solution:
Pressure inside the liquid film is smaller than outside by
ΔP=Tr=Tcosθd2=2Tcosθd,
F=ΔP×A=A×2Tcosθd
m=A×d×ρ⇒ A=md×ρ
⇒ F=2Tcosθd×mdρ=2Tmcosθd2ρ
⇒ F=2Tmd2⋅ρ

Topic: Physics, Mechanical Properties of Fluids, JEE Main
A drum of radius $R$ and mass $M$ rolls down without slipping along an inclined plane of angle $\theta$. The frictional force:

Options:
A dissipates energy as heat
B decreases the rotational motion'
C decreases the rotational and translational motion
D converts translational energy to rotational energy

The correct answer is:
converts translational energy to rotational energy

Solution:
Required frictional force converts translational energy into rotational energy.

Topic: Physics, Rotational Motion, JEE Main
A drum of radius R and mass M rolls down without slipping along an inclined plane of angle Î¸. The frictional force

Options:
A Converts translational energy to rotational energy
B Dissipates energy as heat
C Decreases the rotational motion
D Decreases the rotational and translational motion

The correct answer is:
Converts translational energy to rotational energy

Solution:
When a body rolls down without slipping along an inclined plane of inclination θ, it rotates about a horizontal axis through its centre of mass and also its centre of mass moves. Therefore, rolling motion may be regarded as a rotational motion about an axis through its centre of mass plus a translational motion of the centre of mass. As it rolls down, it suffers loss in gravitational potential energy provided translational energy due to frictional force is converted into rotational energy.

Topic: Physics, Rotational Motion, JEE Main
A dynamometer D, which is a device used to measure force, is attached to two blocks of masses 6 kg and 4 kg. Forces of 20 N and 10 N are applied on the blocks as shown in the figure. The reading of the dynamometer is

Options:
A 10 N
B 20 N
C 6 N
D 14 N

The correct answer is:
14 N

Solution:
Make a free body diagram then write force equation for the blocks involving spring force.
20-T=6a
T-10=4a
â´ a=1 and T=14 N

Topic: Physics, Laws of Motion, JEE Main
A electric dipole is placed at an angle of $30^{\circ}$ to a non-uniform electric field. The dipole will experience

Options:
A
a torque only

B
a translational force only in the direction of the field

C
a translational force only in a direction normal to the direction of the field

D
a torque as well as a translational force

The correct answer is:
a torque as well as a translational force

Solution:
A torque as well as a translational force

Topic: Physics, Electrostatics, JEE Main, JEE Main 2006
(a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of $500 \mathrm{~V}$ with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its $\mathrm{e} / \mathrm{m}$ is given to be $1.76 \times 10^{11} \mathrm{C} \mathrm{kg}^{-1}$.
(b) Use the same formula you employ in (a) to obtain electron speed for collector potential of $10 \mathrm{MV}$. Do you see what is wrong? In what way is the formula to be modified?

Options:

None

Solution:
Given: $\mathrm{V}=500 \mathrm{~V}, \frac{\mathrm{e}}{\mathrm{m}}=1.76 \times 10^{11} \mathrm{C} / \mathrm{kg}$.
(a) The accelerated $\mathrm{e}^{-}$has k.E given by, $\frac{1}{2} \mathrm{mv}^2=\mathrm{eV}$
$$
\Rightarrow \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}=\sqrt{2 \times 1.76 \times 10^{11} \times 500}
$$
$=1.327 \times 10^7 \mathrm{~m} / \mathrm{s}$
(b) $\mathrm{V}=10 \mathrm{MV}=10 \times 10^6 \mathrm{~V}$.
$$
\begin{aligned}
&\Rightarrow \text { total energy is } \mathrm{E}=\frac{\mathrm{m}_0 \mathrm{c}^2}{\sqrt{1-\mathrm{v}^2 / \mathrm{c}^2}} \\
&\Rightarrow \sqrt{1-\frac{\mathrm{v}^2}{\mathrm{c}^2}}=\frac{\mathrm{m}_0 \mathrm{c}^2}{\mathrm{E}}=\frac{0.51 \mathrm{MeV}}{10 \mathrm{MeV}}=0.051
\end{aligned}
$$
(Calculate $\mathrm{m}_0 \mathrm{c}^2$ from $\mathrm{m}_0=9.1 \times 10^{-31} \mathrm{~kg}$,
$$
\begin{aligned}
&\left.\mathrm{c}=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right) \\
&\Rightarrow 1-\frac{\mathrm{v}^2}{\mathrm{c}^2}=(0.051)^2=0.0026 \\
&\Rightarrow \frac{\mathrm{v}^2}{\mathrm{c}^2}=1-0.0026 . \Rightarrow \mathrm{v}^2=0.9974 \mathrm{c}^2 \\
&\Rightarrow \mathrm{v}=0.999 \mathrm{c}(\text { never can } \mathrm{v}>>\mathrm{c})
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter
A factory is operating in two shifts, day and night, with 70 and 30 workers, respectively.If per day mean wage of the day shift workers is, ₹54 and per day mean wage of all the workers is ₹60, then per day mean wage of the night shift workers (in ₹) is :

Options:
A 75
B 74
C 69
D 66

The correct answer is:
74

Solution:
n1 x¯1 +n2 x¯2n1+n2 = x¯
70⋅54+30 x¯270+30=60
⇒3780+30 x¯2 =6000
∴ x¯2= 6000-378030
= 222030
=74

Topic: Mathematics, Statistics, JEE Main, JEE Main 2015 (10 Apr Online)
A fair coin is tossed a fixed number of times. If the probability of getting 7 heads is equal to probability of getting 9 heads, then the probability of getting 2 heads is

Options:
A 15213
B 15214
C 15212
D 1528

The correct answer is:
15213

Solution:
Let the coin be tossed n-times

PH=PT=12

P(7 heads)=C7n12n-7127=C7n2n

P(9 heads)=C9n12n-9129=C9n2n

P(7 heads)=P (7heads)

C7n=C9n⇒n=16

P(2 heads)=C2161214122=15×8216

P(2 heads)=15213

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (26 Feb Shift 1)
A fair coin is tossed n- times such that the probability of getting at least one head is at least 0.9. Then the minimum value of n is _______.

Options:

The correct answer is:
4

Solution:
When a coin is tossed, then probability of head, PHead=PTail=12

Therefore, 1-P (All tail)≥0.9

⇒1-12n≥0.9

⇒12n≤110

⇒nmin=4.

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (25 Jul Shift 2)
A fair coin is tossed once. If it shows head, then 2 fair dice are thrown simultaneously otherwise 3 fair dice are thrown simultaneously. The probability that all the dice show different numbers is k, then 180k is equal to

Options:

The correct answer is:
125

Solution:
Hâa fair coin shows head
Tâa fair coin shows tail
Aâ2 fair dice show unequal numbers
Bâ3 fair dice show unequal numbers
Probability if all dice shows unequal number
=PH.PAH+PTPBT
=12Ã6Ã56Ã6+12Ã6Ã5Ã46Ã6Ã6
=512+58=15+1036=2536

Topic: Mathematics, Probability, JEE Main
A fair coin is tossed repeatedly until two consecutive heads are obtained. If the probability that 2 consecutive heads occur on fourth and fifth toss is p, then 30p is equal to

Options:

The correct answer is:
320

Solution:
Fourth and fifth toss turns up head hence tail must occur on third toss.

So, HT,TH or TT occurs on the first and second toss.
Hence, the required probability =12×12×12×12×12⏟HTTHH+12×12×12×12×12⏟THTHH+12×12×12×12×12⏟TTTHH
=332

Topic: Mathematics, Probability, JEE Main
A fair die is thrown n number of times. If the probability of always getting a number greater than the previous number is 554, then the value of n is equal to n≤6

Options:

The correct answer is:
3

Solution:
Total cases =6n
Favourable cases =6Cn
Required probability =6Cn6n=554
by observation n=3 satisfies

Topic: Mathematics, Probability, JEE Main
A fair die is thrown until 2 appears. Then the probability, that 2 appears in even number of throws, is

Options:
A 56
B 16
C 511
D 611

The correct answer is:
511

Solution:
Let the probability of 2 appearing in a throw be given by P2 and the probability of 2 not appearing be given by P2'.

⇒P2=16, P2'=56

Now, we want 2 to appear in 2nd throw or 4th throw or 6th throw ... and so on

So, required probability is given by,

PE=56×16+563×16+565×16+...

⇒PE=16×561-2536

⇒PE=16×561136

⇒PE=56×3611×16

⇒PE=511

Topic: Mathematics, Probability, JEE Main, JEE Main 2024 (29 Jan Shift 1)
A fair die is tossed until six is obtained on it. Let X be the number of required tosses, then the conditional probability P(X⩾5∣X>2) is :

Options:
A 2536
B 56
C 1136
D 125216

The correct answer is:
2536

Solution:
Px≥5x>2=P(x≥5∩x>2)P(x>2)

=56416+56516+.......∞562·16+56316+……….∞

=564·161-56562161-56=562=2536

Topic: Mathematics, Probability, JEE Main, JEE Main 2021 (26 Aug Shift 2)
A fair n n>1 faces die is rolled repeatedly until a number less than n appears. If the mean of the number of tosses required is n9, then n is equal to

Options:

The correct answer is:
10

Solution:
Given,

A fair n n>1 faces die is rolled repeatedly until a number less than n appears,

xi 1 2 3 4 ...pi n-1n 1n·n-1n 1n2·n-1n 1n3·n-1n ...

Now we know that, mean is given by,

Mean =∑i=1∞pixi=1·n-1n+2n·n-1n+3n2 n-1n+...

⇒n9=1-1nS ........1

Where, S=1+2n+3n2+4n3+... ......2

1nS=1n+2n2+3n3+... ......3

Now subtracting equation 2-3 we get,

1-1nS=1+1n+1n2+1n3+...

⇒1-1nS=11-1n

Now putting the value of S in equation 1 we get,

⇒n9=1-1n×11-1n2=nn-1

⇒n=10

Topic: Mathematics, Probability, JEE Main, JEE Main 2023 (12 Apr Shift 1)
A fall in glomerular filtration rate (GFR) activates

Options:
A juxta glomerular cells to release renin
B adrenal cortex to release aldosterone
C adrenal medulla to release adrenaline
D posterior pituitary to release vasopressin

The correct answer is:
juxta glomerular cells to release renin

Solution:
Juxtaglomerular apparatus consists of macula densa and juxtaglomerular cells. The apparatus becomes active when there is decrease in renal blood pressure or decrease in glomerular filtration rate. There is increased absorption of $\mathrm{NaCl}$ in ascending limb or loop of Henle. The decreased $\mathrm{NaCl}$ content of filtrate activates cells of macula densa to increase secretion of renin from juxta glomerular cells. Renin convert angiotensinogen to angiotensin peptides.

Topic: Biology, Excretory Products and their Elimination, NEET, NEET 2012 (Mains)
A fan is making 600 revolutions per minute. If after some time it makes 1200 revolutions per minute, then increase in its angular velocity

Options:
A $10 \pi \mathrm{rad} / \mathrm{sec}$
B $20 \pi \mathrm{rad} / \mathrm{sec}$
C $40 \pi \mathrm{rad} / \mathrm{sec}$
D $60 \pi \mathrm{rad} / \mathrm{sec}$

The correct answer is:
$20 \pi \mathrm{rad} / \mathrm{sec}$

Solution:
$20 \pi \mathrm{rad} / \mathrm{sec}$
$\omega=600 \times \frac{2 \pi}{60}=20 \pi \mathrm{rad} / \mathrm{s}$
$\omega$ is doubled if it takes $1200 \mathrm{rpm}$
$\omega_0=\frac{1200 \times 2 \pi}{60}=40 \pi \mathrm{rad} / \mathrm{s}$
$\therefore \Delta \omega=\omega_0-\omega=20 \pi \mathrm{rad} / \mathrm{s}$

Topic: Physics, Laws of Motion, JEE Main
A farmer F1 has a land in the shape of a triangle with vertices at P(0, 0),Q(1, 1) and R(2, 0). From this land, a neighbouring farmer F2 takes away the region which lies between the side PQ and a curve of the form y=xn;(n>1). If the area of the region taken away by the farmer F2 is exactly 30% of the area of ∆PQR, then the value of n is

Options:

The correct answer is:
4

Solution:
Area =â«01x-xndx=310
x22-xn+1n+101=310
12-1n+1=310 â´ n+1=5
â n=4

Topic: Mathematics, Area Under Curves, JEE Main
A farmer was facing the problem of low yield from his farm. He was advised to keep a beehive in the vicinity. Why? How would the beehive help in enhancing yield?

Options:

None

Solution:
Apiculture or bee keeping is an important enterprise of agriculture dealilng with the maintenance of honeybee hives for the commercial production of honey and wax. Bee keeping is a cottage industry that has low investment, is less problematic and highly profitable.

Topic: Biology, Strategies for Enhancement in Food Production
A faulty thermometer reads 5°C in melting ice and 95°C in steam. The correct temperature on absolute scale will be ______ K when the faulty thermometer reads 41°C.

Options:

The correct answer is:
313

Solution:
As per working of thermometers, X-LFPUFP-LFP=constant.

Therefore, we can write

41°-5°95°-5°=C-0°100°-0°

⇒C=3690×100=40°C=40+273 K=313 K

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2023 (30 Jan Shift 2)
A female is having excessive weight gain in spite of having adequate balanced diet. She also complains of fatigue and tiredness. Her menstrual cycle is also irregular since past few months. Which of the following conditions is she suffering from?

Options:
A Hyperthyroidism
B Hypothyroidism
C Tetany
D Diabetes mellitus

The correct answer is:
Hypothyroidism

Solution:
Thyroid hormones T3 and T4 are responsible for maintaining basal metabolic rate. Hypothyroidism leads to decreased BMR which causes excessive weight gain, fatigue, tiredness, palpitations and also irregular menstrual cycle. Tetany is caused due to deficiency of parathormone, and diabetes mellitus is caused due to deficiency of insulin

Topic: Biology, Chemical Coordination and Integration, NEET
A fern differs from a moss in having

Options:
A swimming archegonia
B swimming antherozoids
C independent gametophytes
D independent sporophytes

The correct answer is:
independent sporophytes

Solution:
A fern (pteridophyte) differs from a moss (bryophyte) in having independent gametophyte, while in moss the sporophyte is simpler than the gametophyte and remains attached to the parent gametophyte throughout its life. This sporophyte is dependent upon gametophyte partially or wholly for its nutrition.

Topic: Biology, Plant Kingdom, JIPMER, JIPMER 2012
A ferromagnetic material is heated above its curie temperature. Which one of the following is the correct statement?

Options:
A Ferromagnetic domains are perfectly arranged
B Ferromagnetic domains become random
C Ferromagnetic domains are not influenced
D Ferromagnetic material changes into diamagnetic material

The correct answer is:
Ferromagnetic domains become random

Solution:
Beyond curie temperature, ferromagnetic material turns into paramagnetic material, as if ferromagnetic domains become random.

Topic: Physics, Magnetic Properties of Matter, JEE Main
A ferromagnetic material is placed in an external magnetic field. The size of magnetic domains

Options:
A decrease only
B increases only
C some increase while other decrease
D remains unchanged

The correct answer is:
some increase while other decrease

Solution:
Some domains, which are in the direction of applied field, increase in size, while others decrease

Topic: Physics, Magnetic Properties of Matter, NEET
A few spherical equipotential surfaces are shown in the figure. The electric field at any point, at a distance x from the centre, is

Options:
A 6x2 perpendicular to the plane of paper
B 600x2 perpendicular to the plane of paper
C 6x2 radially
D 600x2 radially

The correct answer is:
6x2 radially

Solution:
V=q4Ïâ0x â q=4ÏÎµ0Vx=60Ã 0.1Ã4ÏÎµ0=24ÏÎµ0
â´ E=q4Ïâ0x2=24Ïâ04Ïâ0x2=6x2
Direction of electric field is radially outward.

Topic: Physics, Electrostatics, JEE Main