datavorous/entrance-exam-dataset huggingface dataset:
A person standing at point P sees angle of elevation of the top of a building, whose base is 50 meters away, to be 60o. Another building whose base is 20 meters away from the base of the first building and is between the observer and first building has height h meters, then the maximum possible height (in meters) of this second building is (Take 3=1.73)

Options:

The correct answer is:
51.9

Solution:
Since initial building can be observed from the point P
âhâ¤303=301.73=51.90 meters

Topic: Mathematics, Heights and Distances, JEE Main
A person standing at the foot of a tower walks a distance of 3 meters from the tower and observes that the angle of elevation of the top of the tower is 30o. He then walks a distance 4 meters perpendicular to the previous direction and observes the angle of elevation to be β. Then, cos⁡2β is equal to

Options:
A 32
B 13
C 23
D 1114

The correct answer is:
1114

Solution:
Let O be the foot of the tower OP
OA=3m,AB=4m
∴OB=5m ∵OA2+AB2=OB2
∴ In ∆AOP,
tan⁡30°=h3 ∴h=3
and in ∆OPB,

tan⁡β=h5 ∴h=5tan⁡β
Thus, h=3=5tan⁡β⇒tan⁡β=35
cos⁡2β=1-tan2⁡β1+tan2⁡β=1-3251+325=1114

Topic: Mathematics, Heights and Distances, JEE Main
A person standing on a spring balance inside a stationary lift measures 60 kg. The weight of that person if the lift descends with uniform downward acceleration of 1.8 m s-2 will be

N. g=10 m s-2

Options:

The correct answer is:
492

Solution:
When lift is at rest

N=mg

⇒60×10=600 N

When lift moves with downward acceleration. In frame of lift pseudo force will be in upward direction.

N'=Mg-a

⇒6010-1.8

N'⇒492 N

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (26 Feb Shift 1)
A person standing on an open ground hears the sound of a jet aeroplane, coming from north at an angle 60o with ground level, but he finds the aeroplane right vertically above his position. If v is the speed of sound, speed of the plane is:

Options:
A 2v3
B v
C v2
D 32v

The correct answer is:
v2

Solution:
dvs=d cos60ova
va=vs2=v2

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2019 (12 Jan Shift 1)
A person standing on the bank of a river observes that the angle of elevation of the top of a tree on the opposite bank of the river is $60^{\circ}$ and when he retires 40 meter away from the tree the angle of elevation becomes $30^{\circ}$. The breadth of the river is

Options:
A
$20 \mathrm{~m}$

B
$30 \mathrm{~m}$

C
$40 \mathrm{~m}$

D
$60 \mathrm{~m}$

The correct answer is:
$20 \mathrm{~m}$

Solution:
$\tan 30^{\circ}=\frac{h}{40+b}$
$\Rightarrow \sqrt{3} h=40+b$
$\tan 60^{\circ}=h / b \Rightarrow h=\sqrt{3} b$
$\Rightarrow b=20 m$

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2004
A person standing on the floor of an elevator drops a coin. The coin reaches the floor in time $t_1$ if the elevator is at rest and in time $t_2$ if the elevator is moving uniformly. The which of the following option is correct?

Options:
A $t_1 < t_2$ or $t_1>t_2$ depending upon whether the lift is going up or down
B $t_1 < t_2$
C $t_1>t_2$
D $t_1=t_2$

The correct answer is:
$t_1=t_2$

Solution:
Let $\mathrm{h}$ be the height through which the coin is dropped. Then, according to the equation of motion, it is given as
$$
\begin{aligned}
\mathrm{h}=\mathrm{ut}+\frac{1}{2} \mathrm{gt}^2 \Rightarrow \mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}} \quad[\because \mathrm{u}=0] \\
\Rightarrow \quad \mathrm{t} \propto \frac{1}{\sqrt{\mathrm{g}}}
\end{aligned}
$$
As the elevator is moving uniformly i.e. its velocity is constant, so the acceleration is zero.
$\therefore$ Relative acceleration of the lift when it is either moving upward or downward is given as, $\mathrm{g}^{\prime}=\mathrm{g} \pm \mathrm{a}=\mathrm{g} \pm 0=\mathrm{g}$
Hence, the time for the coin to reach the floor will remains same i.e. $t_1=t_2$.

Topic: Physics, Motion In One Dimension, NEET, NEET 2019 (Odisha)
A person starts his journey from centre ' $O$ ' of the park and comes back to the same position following path OPQO as shown in the figure. The radius of path taken by the person is $200 \mathrm{~m}$ and he takes $3 \mathrm{~min} 58 \mathrm{sec}$ to complete his journey. The average speed of the person is $\qquad$ $\mathrm{ms}^{-1} \cdot($ take $\pi=3.14)$

Options:
A $2 \mathrm{~m} / \mathrm{s}$
B $3 \mathrm{~m} / \mathrm{s}$
C $4 \mathrm{~m} / \mathrm{s}$
D $5 \mathrm{~m} / \mathrm{s}$

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

Solution:
$\begin{aligned} & V_{a v}=\frac{S_{\text {total }}}{T_{\text {total }}}=\frac{200+200+\frac{2 \pi \times 200}{4}}{238} \\ & =\frac{400+100 \pi}{238}=\frac{714}{238}=3 \mathrm{~m} / \mathrm{s}\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A person suffering from a disease caused by Plasmodium, experiences recurring chill and fever at the time when the

Options:
A sporozoites released from $\mathrm{RBCs}$ are being rapidly killed and broken down inside spleen
B trophozoites reach maximum growth and give out certain toxins
C parasite after its rapid multiplication inside RBCs ruptures them, releasing the stage to enter fresh RBCs
D microgametcytes and megagametocytes are being destroyed by the WBCs

The correct answer is:
parasite after its rapid multiplication inside RBCs ruptures them, releasing the stage to enter fresh RBCs

Solution:
In malaria, chill and fever is due to the release of a toxic substance haemozoin, formed by breakdown of haemoglobin present in RBCs. It will be released after the rupture of RBCs in erythrocytic schizogony.

Topic: Biology, Human Health and Diseases, NEET, NEET 2010 (Mains)
A person suffers punctures in his chest cavity in an accident, without any damage to the lungs its effect could be

Options:
A
reduced breathing rate

B
rapid increase in breathing rate

C
no change in respiration

D
cessation of breathing

The correct answer is:
cessation of breathing

Solution:
The movement of air in and out of the lungs is performed by creating a pressure gradient between the lungs and the surrounding atmosphere. The pressure within the lungs is less than the atmospheric pressure so there is a negative pressure in the lungs with respect to atmospheric pressure. A puncture in the chest affects this pressure gradient maintained by the lungs and thus may cause cessation of breathing.

Topic: Biology, Breathing and Exchange of Gases
A person throws two fair dice. He wins Rs. 15 for throwing a doublet (same numbers on the two dice), wins Rs 12 when the throw results in the sum of 9 , and loses Rs. 6 for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is:

Options:
A 12 loss
B 12 gain
C 2 gain
D 14 loss

The correct answer is:
12 loss

Solution:
win Rs. 15→ number of cases =6
win Rs. 12→ number of cases =4
loss Rs. 6→ number of cases =26
p (expected gain/loss) =15×636+12×436-6×2636=-12

Topic: Mathematics, Probability, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A person travelling on a straight line without changing direction moves with a uniform speed \( v_{1} \) for half distance and next half distance he covers with uniform speed \( v_{2} \). The average speed \( v \) is given by

Options:
A \( v=\frac{2 v_{1} v_{2}}{v_{1}+v_{2}} \)
B \( v=\sqrt{V_{1} V_{2}} \)
C \( \frac{V_{1}+V_{2}}{2} \)
D \( \frac{1}{V}=\frac{1}{V_{1}}+\frac{1}{V_{2}} \)

The correct answer is:
\( v=\frac{2 v_{1} v_{2}}{v_{1}+v_{2}} \)

Solution:
Let the total distance be 2x.

For the first half distance x, let time be t1.

∵ time=distancespeed

∴t1=xv1 ...(i)

For second half distance x, let time be t2.

∵ time=distancespeed

∴t2=xv2 ...(ii)

∴ Average speed for entire journey.

vav=total distancetotal time

⇒vav=2xxv1+xv2

⇒v=2xxv1+v2v1v2 ∵vav=v

⇒v=2v1v2v1+v2.

Note average speed is harmonic mean of given speeds.

Topic: Physics, Motion In One Dimension, JEE Main
A person travels along a straight road for half the distance with velocity $v_1$ and the remaining half distance with velocity ${ }^{v_2}$ The average velocity is given by

Options:
A $v_1 v_2$
B $\frac{v_2^2}{v_1^2}$
C $\frac{v_1+v_2}{2}$
D $\frac{2 v_1 v_2}{v_1+v_2}$

The correct answer is:
$\frac{2 v_1 v_2}{v_1+v_2}$

Solution:
As, the total distance is divided into two equal parts therefore distance averaged speed $=\frac{2 v_1 v_2}{v_1+v_2}$

Topic: Physics, Motion In One Dimension, JEE Main
A person travels x distance with velocity v1 and then x distance with velocity v2 in the same direction. The average velocity of the person is v, then the relation between v, v1 and v2 will be

Options:
A v=v1+v22
B 1v=1v1+1v2
C v=v1+v2
D 2v=1v1+1v2

The correct answer is:
2v=1v1+1v2

Solution:
The time t1 taken by the person to travel distance x with velocity v1 is given by

t1=xv1 ...1

The time t2 taken by the person to travel distance x with velocity v2 is given by

t2=xv2 ...2

The formula to calculate the average speed v of the person is given by

v=2xt1+t2 ...3

Substitute the expressions from equations (1) and (2) into equation (3) to obtain the required relation.

v=2xxv1+xv2⇒xv1+xv2=2xv⇒2v=1v1+1v2

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (10 Apr Shift 2)
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g=9.8 ms-2 :

Options:
A 9.89Ã10-3Â kg
B 12.89Ã10-3 kg
C 2.45Ã10-3Â kg
D 6.45Ã10-3Â kg

The correct answer is:
12.89Ã10-3 kg

Solution:
m=10kg, h=1m, 1000 times
Work done against gravity mgh
=10×1×9.8=98 J
In lifting 1000 times
=1000×98
=9.8Ã104 J
Fat burn with 20% efficiency =3.8Ã107 JÃ0.2
=7.6Ã106 J per kg
m=9.8Ã1047.6Ã106=1.289Ã10-2
=12.89Ã10-3 kg

Topic: Physics, Work Power Energy, JEE Main
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g=9.8 ms-2 :

Options:
A 9.89Ã10-3Â kg
B 12.89Ã10-3 kg
C 2.45Ã10-3Â kg
D 6.45Ã10-3Â kg

The correct answer is:
12.89Ã10-3 kg

Solution:
m=10kg, h=1m, 1000 times
Work done against gravity mgh
=10×1×9.8=98 J
In lifting 1000 times
=1000×98
=9.8Ã104 J
Fat burn with 20% efficiency =3.8Ã107 JÃ0.2
=7.6Ã106 J per kg
m=9.8Ã1047.6Ã106=1.289Ã10-2
=12.89Ã10-3 kg

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2016 (03 Apr)
A person walks along a straight road from his house to a market $2.5 \mathrm{~km}$ away with a speed of $5 \mathrm{~km} / \mathrm{h}$ and instantly turns back and reaches his house with a speed of $7.5 \mathrm{~km} / \mathrm{h}$. The average speed of the person during the time interval 0 to $50 \mathrm{~min}$ is $(\mathrm{in} \mathrm{m} / \mathrm{s})$

Options:
A $4 \frac{2}{3}$
B $\frac{5}{3}$
C $\frac{5}{6}$
D $\frac{1}{3}$

The correct answer is:
$\frac{5}{3}$

Solution:
A according to question, $t_1=\frac{2.5}{5}=\frac{1}{2} \mathrm{~h}=30 \mathrm{~min}$
$$
\text { and } t_2=\frac{2.5}{7.5}=\frac{1}{3} \mathrm{~h}=20 \mathrm{~min}
$$
So, the average speed
$$
\begin{aligned}
& =\frac{\text { Total distance }(\text { in metre })}{\text { Total time }(\text { in second })} \\
& =\frac{(2.5+2.5) \times 1000}{(30+20) \times 60} \\
& =\frac{5 \times 1000}{50 \times 60}=\frac{50}{30}=\frac{5}{3} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, JEE Main
A petroleum fraction having boiling range 70− 200 o  C and containing 6â10 carbon atoms per molecule is called

Options:
A Natural gas
B Gas oil
C Gasoline
D Kerosene

The correct answer is:
Gasoline

Solution:
The bulk of a typical gasoline consists of hydrocarbons with between 4 and 12 carbon atoms per molecule (commonly referred to as C4−C12). It is a mixture of paraffins (alkanes), cycloalkanes (naphthenes), and olefins (alkenes), where the usage of the terms paraffin and olefin is particular to the oil industry. The actual ratio depends on:

The oil refinery that makes the gasoline, as not all refineries have the same set of processing units;

The crude oil feed used by the refinery;

The grade of gasoline, in particular, the octane rating.

Below carbon atoms per molecule are mentioned.

(i) Gasoline C6-C10
(ii) Kerosene C10-C13
(iii) Gas oil C13-C18

Topic: Chemistry, Environmental Chemistry, JEE Main
A. Phenyl methanamine

B. N, N-Dimethylaniline

C. N-Methyl aniline

D. Benzenamine

Choose the correct order of basic nature of the above amines.

Options:
A A>B>C>D
B A>C>B>D
C D>B>C>A
D D>C>B>A

The correct answer is:
A>B>C>D

Solution:
When hydrogen atoms of the amino group of arylamines are replaced by electron donating alkyl groups, the basic character of the resulting amine increases.

B.S. order, Phenyl methanamine > N, N-Dimethylaniline > N-Methyl aniline > Benzenamine.

Topic: Chemistry, Amines, JEE Main, JEE Main 2021 (26 Feb Shift 2)
A photocell is illuminated by a small bright source placed 1 m away. When the same source of light is placed 12 m away, the number of electrons emitted by photocathode would

Options:
A decrease by a factor of 2

B increase by a factor of 2

C decrease by a factor of 4
D increase by a factor of 4

The correct answer is:
increase by a factor of 4

Solution:
I=Pof source4πdistance2=P4πd2
Here, we assume light to spread uniformly in all directions.
The number of photo-electrons emitted from a surface depend on the intensity of light I falling on it. Thus the number of electrons emitted n depends directly on I. P remains constant as the source is the same.

∴ I2I1=n2n1⇒P2P1(d1d2)2=n2n1

∴ n2n1=PP1122=41 (∵ P1 = P2 = P)

Topic: Physics, Wave Optics, NEET
A photocell is illuminated by a small bright source placed 1 m away. When the same source of light is placed 12 m away, the number of electrons emitted by photocathode would

Options:
A decrease by a factor of 2
B increase by a factor of 2
C decrease by a factor of 4
D increase by a factor of 4

The correct answer is:
increase by a factor of 4

Solution:
I=Pof source4πdistance2=P4πd2
Here, we assume light to spread uniformly in all directions.
The number of photo-electrons emitted from a surface depends on the intensity of light I falling on it. Thus the number of electrons emitted n depends directly on I. P remains constant as the source is the same.

∴ I2I1=n2n1⇒P2P1(d1d2)2=n2n1

∴ n2n1=PP1122=41 (∵ P1 =P2 =P)

Topic: Physics, Dual Nature of Matter, JEE Main
A photocell is illuminated by a small bright source placed $1 \mathrm{~m}$ away. When the same source of light is placed $\frac{1}{2} \mathrm{~m}$ away, the number of electrons emitted by photo cathode would

Options:
A
decrease by a factor of 4

B
increase by a factor of 4

C
decrease by a factor of 2

D
increase by a factor of 2

The correct answer is:
increase by a factor of 4

Solution:
$I \propto \frac{1}{r^2}$

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2005
A photoelectric surface is illuminated successively by monochromatic light of wavelengths λ and λ2. If the maximum kinetic energy of the emitted photoelectrons in the second case is 3 times that in the first case, the work function of the surface is :

Options:
A hc2λ
B hcλ
C hc3λ
D 3hcλ

The correct answer is:
hc2λ

Solution:
K.EÎ»=hcÎ»-Ï
K.EÎ»2=hcÎ»2-Ï
K.EÎ»2=3K.EÎ»
â Ï=hc2Î»

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2016 (10 Apr Online)
A photoelectric surface is illuminated successively by monochromatic light of wavelengths Î» and 2Î». If the maximum kinetic energy of the emitted photoelectrons in the first case is 4 times that of the second case, then the work function of the surface is

Options:
A hc3λ
B hc4λ
C hc2λ
D 3hc4λ

The correct answer is:
hc3λ

Solution:
hcλ-ϕ=4hc2λ-ϕ
⇒ 3ϕ=hcλ
ϕ=hc3λ

Topic: Physics, Dual Nature of Matter, JEE Main
A photographic camera with a lens of focal length 5 cm is used for capturing images. The vertical length of the film used is 24 mm in which image of a 1.68 m tall man is to be captured. Find the minimum distance (in m) of the man from the lens such that his complete image can be obtained.

Options:

The correct answer is:
3.55

Solution:
The lens of a camera is convex and the image is inverted

m=hiho=-24 mm1.68 m=-170
m=ff+u=-170

70f=-f-u

-u=71f=355 cm=3.55 m

Topic: Physics, Ray Optics, JEE Main
A photon is emitted in transition from n=4 to n=1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h=4×10-15 eV s)

Options:
A 941 nm
B 974 nm
C 99.3 nm
D 94.1 nm

The correct answer is:
94.1 nm

Solution:
Using the relation between energy of photon and wavelength for the transition in the hydrogen atom, we get

hcλ=13.61n12-1n22

⇒4×10-15×3×108λ=13.61-116⇒λ=12×10-7×1613.6×15=94.1×10-9=94.1 nm

Topic: Physics, Atomic Physics, JEE Main, JEE Main 2023 (24 Jan Shift 2)
A photon of energy 10.2 eV corresponds to light of wavelength Î»0. Due to an electron transition from n=2 to n=1 in a hydrogen atom, the light of wavelength Î» is emitted. If we take into account the recoil of the atom when the photon is emitted, then

Options:
A λ<λ0
B λ>λ0
C λ=λ0
D None of these

The correct answer is:
λ>λ0

Solution:
Without recoil of the atom, the entire energy is carried away by the photon.
hcλ0=10.2 eV
With recoil of the atom, energy is shared between photon and atom. So, the energy of the photon is less, hence its wavelength is more.
hcλ<hcλ0
∴λ0<λ

Topic: Physics, Dual Nature of Matter, JEE Main
A photon of energy $E$ ejects a photoelectron from a metal surface whose work function is $W_0$. If this electron enters into a uniform magnetic field of induction $B$ in a direction perpendicular to the field and describes a circular path of radius $r$, then the radius $r$ is given by, (in the usual notation)

Options:
A $\frac{\sqrt{2 m\left(E-W_0\right)}}{e B}$
B $\sqrt{2 m\left(E-W_0\right) e B}$
C $\frac{\sqrt{2 e\left(E-W_0\right)}}{m B}$
D $\frac{\sqrt{2 m\left(E-W_0\right)}}{e B}$

The correct answer is:
$\sqrt{2 m\left(E-W_0\right) e B}$

Solution:
From Einstein's equation
$\begin{gathered}
E=W_0+\frac{1}{2} m v^2 \\
\sqrt{\frac{2\left(E-W_0\right)}{m}}=v
\end{gathered}$
or A charged particle placed in uniform magnetic field experience a force
$F=\frac{m v^2}{r}$
$\begin{aligned} \text { or } & e v B & =\frac{m v^2}{r} \\ \text { or } & r & =\frac{m v}{e B}\end{aligned}$
$\begin{array}{ll}\text { or } & r=\frac{m \sqrt{\frac{2\left(E-W_0\right)}{m}}}{e B} \\ \Rightarrow & r=\frac{\sqrt{2 m\left(E-W_0\right)}}{e B}\end{array}$

Topic: Physics, Dual Nature of Matter, JEE Main
A photon of wavelength 300 nm interacts with a stationary hydrogen atom in ground state. During the interaction, whole energy the photon is transferred to the electron of the atom. State which possibility is correst. (Consider, Plank constant $=4 \times 10^{-15} eV_{s},$ velocity of light $=3 \times 10^{8} \mathrm{m} / \mathrm{s},$ ionisation energy of hydrogen $=13.6 \mathrm{eV}$ )

Options:
A Electron will be knocked out of the atom
B Electron will go to any excited state of the atom
C Electron will go only to lirst excited state of the atom
D Electron will keep orbiting in the ground state of the atom

The correct answer is:
Electron will keep orbiting in the ground state of the atom

Solution:
The energy of the photon
$$
\begin{aligned}
E &=\frac{h C}{\lambda} \\
&=\frac{4 \times 10^{-15} \mathrm{eVs} \times 3 \times 10^{8} \mathrm{m} / \mathrm{s}}{300 \times 10^{-9} \mathrm{m}} \\
&=\frac{4 \times 10^{2}}{300} \mathrm{eV}=\frac{4}{3} \mathrm{eV}=1.33 \mathrm{eV}
\end{aligned}
$$
The ionisation energy is $13.6 \mathrm{eV}$ which is greater than energy of photon, so atom can not come into excited state and will remain in ground state

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2015
A photon of wavelength $\lambda$ is scattered from an electron, which was at rest. The wavelength shift $\Delta \lambda$ is three times of $\lambda$ and the angle of scattering $\theta$ is $60^{\circ}$. The angle at which the electron recoiled is $\phi$. The value of $\tan \phi$ is : (electron speed is much smaller than the speed of light)

Options:
A
$0.16$

B
$0.22$

C
$0.25$

D
$0.28$

The correct answer is:
$0.22$

Solution:
No Solution Available

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2014 (11 Apr Online)
A physical quantity is measured and its value is found to be $nu$ where $n =$ numerical unit and $u =$ unit. Then which of the following relations is true

Options:
A $n \propto u^2$
B $n \propto u$
C $n \propto \sqrt{u}$
D $n \propto \frac{1}{u}$

The correct answer is:
$n \propto \frac{1}{u}$

Solution:
A physical quantity is measured as a product of numerical value and unit.
\(\mathrm{p}=\mathrm{nu}\)
Where
\(\mathrm{n}=\) numerical value
\(\mathrm{u}=\) unit
from (i) equation, \(p=\) constant
Therefore
\(\mathrm{n} \propto \frac{1}{\mathrm{u}}\)

Topic: Physics, Units and Dimensions, JEE Main
A physical quantity of the dimensions of length that can be formed out of c, G and e24πε0, where [ c is velocity of light, G is universal constant of gravitation, e is charge & ε0 is permittivity of free space]:

Options:
A 1c2Ge24πε012
B c2Ge24ÏÎµ012
C 1c2e2G4ÏÎµ012
D 1cGe24ÏÎµ0

The correct answer is:
1c2Ge24πε012

Solution:
Let e24ÏÎµ0=A=ML3T-2
L=Cx GyAz
[ L ]=LT-1x M-1 L3 T-2y ML3 T-2z
-y+z=0 â y=z ......(i)
x+3y+3z=1 ......(ii)
-x-4z=0 ......(iii)
From (i), (ii) and (iii)
z=y=12, x= -2

Topic: Physics, Units and Dimensions, JEE Main
A physical quantity P is described by the relation P=a12 b2 c3d-4. If the relative errors in the measurement of a, b, c and d respectively, are 2%, 1%, 3% and 5%. Then the relative error in P will be:

Options:
A 12%
B 8%
C 25%
D 32%

The correct answer is:
32%

Solution:
P=a12 b2c3d-4
∆ppmax=12 ∆aa+2∆bb+3∆cc+4∆dd
100×∆PP=12 ∆aa+2∆bb+3∆cc+4∆dd×100
% error in P=12×2+2×1+3 ×3+4 ×5
=32%

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2017 (09 Apr Online)
A physical quantity P is given as P=a2b3cd. The percentage error in the measurement of a, b, c and d are 1%, 2%, 3% and 4% respectively. The percentage error in the measurement of quantity P will be

Options:
A 13%
B 16%
C 12%
D 14%

The correct answer is:
13%

Solution:
The given quantity is P=a2b3cd.

The percentage error for a quantity can be given by the formula ∆xx×100.

Relation of the relative errors will be,

ΔPP=2Δaa+3Δbb+Δcc+12Δdd

Hence, the percentage error is calculated by substituting the values,

ΔPP×100=2×1+3×2+3+12×4=13%.

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2023 (10 Apr Shift 1)
A physical quantity $P$ is related to four observables $a, b, c$ and $d$ as $P=\frac{\sqrt{a b} \cdot d^\alpha}{\sqrt{c}}(\alpha$ is constant) The percentage errors in $a, b, c$ and $d$ are $0.5 \%$ in each. If the percentage error in $P$ is $2 \%$, then $\alpha$ is

Options:
A $\frac{5}{2}$
B $\frac{2}{5}$
C $\frac{3}{4}$
D $\frac{3}{2}$

The correct answer is:
$\frac{5}{2}$

Solution:
As, $P=\frac{a^{1 / 2} \cdot b^{1 / 2} \cdot d^\alpha}{c^{1 / 2}}$
So $\%$ error in $P=\frac{\Delta P}{P} \times 100$
$\begin{aligned}
=\frac{1}{2} & \left(\frac{\Delta a}{a}\right) \times 100+\frac{1}{2}\left(\frac{\Delta b}{b} \times 100\right) \\
& +\alpha\left(\frac{\Delta d}{d} \times 100\right)+\frac{1}{2}\left(\frac{\Delta c}{c} \times 100\right)
\end{aligned}$
$\Rightarrow \%$ error in $P=\frac{1}{2} \times \%$ error in $a+\frac{1}{2} \times \%$ error in
$b+\frac{1}{2} \times \%$ error in $c+\alpha \times \%$ error in $d$
Now given,
\% Error in $P=2 \%$
$\%$ Error in $a=\%$ error in $b$
$=\%$ Error in $c=\%$ error in $d=0.5 \%$
Substituting these values in Eq. (i), we get
$\begin{aligned}
2 & =\frac{1}{2} \times 0.5+\frac{1}{2} \times 0.5+\frac{1}{2} \times 0.5+\alpha \times 0.5 \\
\Rightarrow 2 & =\frac{1}{2}\left(\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\alpha\right) \Rightarrow 4=\frac{3}{2}+\alpha
\end{aligned}$
or $\alpha=4-\frac{3}{2}=\frac{5}{2}$
$\therefore \alpha$ is $5 / 2$.

Topic: Physics, Mathematics in Physics, JEE Main