datavorous/entrance-exam-dataset huggingface dataset:
The densities of a liquid at $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$ are respectively 1.0127 and 1 . A specific gravity bottle is filled with $300 \mathrm{~g}$ of the liquid at $0^{\circ} \mathrm{C}$ upto the brim and it is heated to $100^{\circ} \mathrm{C}$. Then, the mass of the liquid expelled in grams is: (Coefficient of linear expansion of glass $\left.=9 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right)$

Options:
A $\frac{3}{10.1}$
B $\frac{3}{1.01}$
C $\frac{3.81}{1.0127}$
D $\frac{3.81}{0.0127}$

The correct answer is:
$\frac{3}{1.01}$

Solution:
Density at $0^{\circ} \mathrm{C}, \rho_0=1.0127$
Density at $100^{\circ} \mathrm{C}, \rho_{100}=1$
Coefficient of real expansion of liquid,
$\gamma_{\text {real }}=\frac{\rho_0-\rho_{100}}{\rho_{100} \times \Delta t}$
$=\frac{1.0127-1}{1 \times 100}=0.0127 \times 10^{-2}$
$=1.27 \times 10^{-4}$
$\gamma_{\text {real }}=\gamma_{\text {app }}+\gamma_g$
$\gamma_g=$ coefficient of volume expansion of glass $=3 \alpha$
$\therefore \quad 1.27 \times 10^{-4}=\gamma_{\mathrm{app}}+3 \alpha$
$1.27 \times 10^{-4}=\gamma_{\text {app }}+3 \times 9 \times 10^{-6}$
$\gamma_{\mathrm{app}}=1.27 \times 10^{-4}-27 \times 10^{-6}$
$=1.27 \times 10^{-4}-0.27 \times 10^{-4}$
$=1 \times 10^{-4}$
$\therefore \quad \gamma_{\text {app }}=\frac{\text { mass expelled }}{\text { remaining mass } \times \Delta t}$
$1 \times 10^{-4}=\frac{m_1-m_2}{m_2 \times 100}$
$\left(\frac{m_1}{m_2}-1\right)=1 \times 10^{-4} \times 100=10^{-2}$
$\frac{m_1}{m_2}=1+10^{-2}=1.01$
$m_2=\frac{m_1}{1.01}=\frac{300}{1.01}$
Mass expelled $=m_1-m_2$
$=300-\frac{300}{1.01}$
$=\frac{3}{1.01}$

Topic: Physics, Thermal Properties of Matter, JEE Main
The densities of a liquid at $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$ are respectively 1.0127 and 1 . A specific gravity bottle is filled with $300 \mathrm{~g}$ of the liquid at $0^{\circ} \mathrm{C}$ upto the brim and it is heated to $100^{\circ} \mathrm{C}$. Then, the mass of the liquid expelled in grams is: (Coefficient of linear expansion of glass $\left.=9 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right)$

Options:
A $\frac{3}{10.1}$
B $\frac{3}{1.01}$
C $\frac{3.81}{1.0127}$
D $\frac{3.81}{0.0127}$

The correct answer is:
$\frac{3}{1.01}$

Solution:
Density at $0^{\circ} \mathrm{C}, \rho_0=1.0127$
Density at $100^{\circ} \mathrm{C}, \rho_{100}=1$
Coefficient of real expansion of liquid,
$\gamma_{\text {real }}=\frac{\rho_0-\rho_{100}}{\rho_{100} \times \Delta t}$
$=\frac{1.0127-1}{1 \times 100}=0.0127 \times 10^{-2}$
$=1.27 \times 10^{-4}$
$\gamma_{\text {real }}=\gamma_{\text {app }}+\gamma_g$
$\gamma_g=$ coefficient of volume expansion of glass $=3 \alpha$
$\therefore \quad 1.27 \times 10^{-4}=\gamma_{\mathrm{app}}+3 \alpha$
$1.27 \times 10^{-4}=\gamma_{\text {app }}+3 \times 9 \times 10^{-6}$
$\gamma_{\mathrm{app}}=1.27 \times 10^{-4}-27 \times 10^{-6}$
$=1.27 \times 10^{-4}-0.27 \times 10^{-4}$
$=1 \times 10^{-4}$
$\therefore \quad \gamma_{\text {app }}=\frac{\text { mass expelled }}{\text { remaining mass } \times \Delta t}$
$1 \times 10^{-4}=\frac{m_1-m_2}{m_2 \times 100}$
$\left(\frac{m_1}{m_2}-1\right)=1 \times 10^{-4} \times 100=10^{-2}$
$\frac{m_1}{m_2}=1+10^{-2}=1.01$
$m_2=\frac{m_1}{1.01}=\frac{300}{1.01}$
Mass expelled $=m_1-m_2$
$=300-\frac{300}{1.01}$
$=\frac{3}{1.01}$

Topic: Physics, Thermal Properties of Matter, TS EAMCET, TS EAMCET 2003
The densities of hydrogen and oxygen are 0.09 and $1.44 \mathrm{~g} \mathrm{~L}^{-1}$. If the rate of diffusion of hydrogen is 1 then that of oxygen in the same units will be

Options:
A 4
B $1 / 4$
C 16
D $1 / 16$

The correct answer is:
$1 / 4$

Solution:
$r_O=r_H \sqrt{\frac{d_H}{d_O}}=1 \sqrt{\frac{0.09}{1.44}}=\sqrt{\frac{1}{16}}=\frac{1}{4}$

Topic: Chemistry, States of Matter, JEE Main
The density and breaking stress of a wire are $6 \times 10^4 \mathrm{~kg} / \mathrm{m}^3$ and $1.2 \times 10^8 \mathrm{~N} / \mathrm{m}^2$ respectively. The wire is suspended from a rigid support on a planet where acceleration due to gravity is $\frac{1}{3}^{\text {rd }}$ of the value on the surface of earth. The maximum length of the wire with breaking is _____ m (take, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ ).

Options:

The correct answer is:
600

Solution:
$\begin{aligned} & \mathrm{T}=\mathrm{mg} \\ & \sigma=\frac{\mathrm{T}}{\mathrm{A}}=\frac{\mathrm{mg}}{\mathrm{A}} \\ & \frac{(\sigma \mathrm{A} \ell) \mathrm{g}}{\mathrm{A}} \\ & \Rightarrow \ell=\frac{\sigma}{\rho \mathrm{g}}=\frac{1.2 \times 10^8 \times 3}{6 \times 10^4 \times 10}=600\end{aligned}$

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2024 (05 Apr Shift 1)
The density (in $\mathrm{g} \mathrm{~mL}^{-1}$ ) of a $3.60 \mathrm{~M}$ sulphuric acid solution that is $29 \% ~\mathrm{H}_2 \mathrm{SO}_4$ (Molar mass $=98 \mathrm{~g}$ $\mathrm{mol}^{-1}$ ) by mass will be

Options:
A
$1.64$

B
$1.88$

C
$1.22$

D
$1.45$

The correct answer is:
$1.22$

Solution:
Let the density of solution be ' $\mathrm{d}$ '
Molarity of solution given $=3.6$
i.e. $1$ litre of solution contains $3.6$ moles of $\mathrm{H}_2 \mathrm{SO}_4$
or $1$ litre of solution contains $3.6 \times 98 \mathrm{~gms}$ of $\mathrm{H}_2 \mathrm{SO}_4$
Since, the solution is $29 \%$ by mass.
$100 \mathrm{~gm}$ solution contains $29 \mathrm{~gm} \mathrm{~H}_2 \mathrm{SO}_4$
$\frac{100}{\mathrm{~d}} \mathrm{ml}$ solution contains $29 \mathrm{~gm}$ of $\mathrm{H}_2 \mathrm{SO}_4$
$1000 \mathrm{~ml}$ solution contains $3.6 \times 98 \mathrm{~gm}$ of $\mathrm{H}_2 \mathrm{SO}_4$ $\therefore 3.6 \times 98=\frac{29 \times \mathrm{d}}{100} \times 1000$
$d=1.22$
Hence, (C) is correct.

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2007
The density of $1 \mathrm{M}$ solution of a compound ' $X$ ' is $1.25 \mathrm{~g} \mathrm{~mL}^{-1}$. The correct option for the molality of solution is (Molar mass of compound $X=85 \mathrm{~g}$ )

Options:
A $1.165 \mathrm{~m}$
B $0.858 \mathrm{~m}$
C $0.705 \mathrm{~m}$
D $1.208 \mathrm{~m}$

The correct answer is:
$0.858 \mathrm{~m}$

Solution:
Relation between Molality ( $m$ ), Molarity (M) and density (d) is
$m=\frac{1000 M}{1000 \times d-M \times M_{\text {solute }}}$
Msolute denotes molar mass of solute.
$m=\frac{1000 \times 1}{(1000 \times 1.25)-(1 \times 85)}=\frac{1000}{1250-85}=\frac{1000}{1165}$
$\mathrm{m}=0.858 \mathrm{~m}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
The density of 3M solution of NaCl is 1.0 g mL-1. Molality of the solution is ______×10-2 m (Nearest integer).

Given: Molar mass of Na and Cl is 23 and 35.5 g mol-1 respectively.

Options:

The correct answer is:
364

Solution:
Formula unit mass of sodium chloride = 23 + 35.5 gmol-1 = 58.5 gmol-1
∴m(molality)=1000×M1000×d-M×M.W of solute ...(M = Molarity of solution)

=1000×31000×1-(3×58.5)=3.64

=364×10-2

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2023 (01 Feb Shift 1)
The density of $3 \mathrm{M}$ solution of sodium chloride is $1.252 \mathrm{~g} \mathrm{~mL}^{-1}$. The molality of the solution will be :
$\left(\right.$ molar mass, $\left.\mathrm{NaCl}=585 \mathrm{~g} \mathrm{~mol}^{-1}\right)$

Options:
A
$260 \mathrm{~m}$

B
$2.18 \mathrm{~m}$

C
$2.79 \mathrm{~m}$

D
$3.00 \mathrm{~m}$

The correct answer is:
$2.79 \mathrm{~m}$

Solution:
The relation between molarity (M) and molality $(m)$ is
$d=\mathrm{M}\left(\frac{1}{m}+\frac{\mathrm{M}_2}{1000}\right), \mathrm{M}_2=$ Mol. mass of solute
On putting value
$$
1.252=3\left(\frac{1}{m}+\frac{58.5}{1000}\right)
$$
on solving $m=2.79$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2013 (22 Apr Online)
The density of a cube is measured by measuring its mass and length of its sides. If the maximum errors in the measurement of its mass and length are 4% and 3%, respectively, then, the maximum error in density is

Options:
A 1%
B 7%
C 5%
D 13%

The correct answer is:
13%

Solution:
ρ = MassL3
% error in ρ = % error in mass + 3[% error in length]
= 4% + 3[3%] = 13%

Topic: Physics, Units and Dimensions, JEE Main
The density of a liquid contained in a tank of area 0.5 m2 is 1200 kg m-3. The container is closed by a massless, frictionless and movable piston and there is a hole of area 1 cm2 near the bottom of the container. A load of 30 kg is placed on the piston and the height of liquid level is 75 cm above the bottom. What is the initial flow velocity (in m s-1) of the liquid? [Take, g=10 m s-2]

Options:

The correct answer is:
4

Solution:
Let the velocity of efflux at point 2 is v.
Given, a=1 cm2, A=0.5 m2=0.5Ã104 cm2
Applying continuity equation,
AV=av
V=aAv
Since, aâªA,Vâªv, V2 is negligible.
Pressure at (1),
p1=p0+mgA
=p0+30Ã100.5
=p0+600
Pressure at (2) ,
p2=p0
where, p0 is atmospheric pressure.
Applying Bernoulli's equation between (1) and (2),
p1+Ïgh+12ÏV2=Ï2+12Ïv2
V2 is negligible, then
p0+600+1200Ã10Ã0.75
=p0+12Ã1200v2
600v2=9600
â v2=16
â v=4 m s-1

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The density of a material, in the shape of a cube, is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are 1.5% and 1%, respectively, the maximum error in determining the density is:

Options:
A 6 %
B 2.5 %
C 3.5 %
D 4.5 %

The correct answer is:
4.5 %

Solution:
ρ=ML3⇒Δρρ=ΔMM+3ΔLL⇒Δρρ×100=1.5+3×1
% relative error,=4.5 %

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2018 (08 Apr)
The density of a material in the shape of a cube is determined by measuring three sides of the cube and its mass. If the relative errors in measuring the mass and length are respectively $1.5 \%$ and $1 \%$, the maximum error in determining the density is

Options:
A $3.5 \%$
B $4.5 \%$
C $6 \%$
D $2.5 \%$

The correct answer is:
$4.5 \%$

Solution:
Density $\rho=\mathrm{m} / \mathrm{V}=\mathrm{m} / \mathrm{l}^3$
The maximum percentage error in density $(\Delta \rho / \rho) x$ $100=(\Delta \mathrm{m} / \mathrm{m}) \times 100 \times 3(\Delta \mathrm{I} / \mathrm{l}) \times 100$
$\Rightarrow(\Delta \rho / \rho) \times 100=1.5+3 \times 1=4.5 \%$
Answer: (2) $4.5 \%$

Topic: Physics, Mathematics in Physics, JEE Main
The density of a rod gradually decreases from one end to the other. It is pivoted at an end, so that it can move about a vertical axis through the pivot. A horizontal force \( F \) is applied on the free end in a direction perpendicular to the rod. The quantities that do not depend on which end of the rod is pivoted are

Options:
A angular acceleration.
B angular velocity when the rod completes one rotation.
C angular momentum when the rod completes one rotation.
D torque of the applied force.

The correct answer is:
torque of the applied force.

Solution:
Density of the rod is increasing from one end to other, lets say it increases from end A to end B.

Therefore, the rod element near end B is heavier than end A.

So, the moment of inertia of rod about the axis passing through the end A is greater than about the axis passing through the end B,

i.e., I A>IB

Torque about an end, τ=r⊥F

where, r⊥=l.

So, torque about both the ends is same.

Angular acceleration, α=τI

∴αAαB=IBIA<1

All quantities are independent of time,

∴ α is constant.

ωAωB=αAαB=IBIA<1

Angular momentum, L=Iω

∴LALB=IAIBωAωB=IAIB

So, only torque is same about both ends.

Topic: Physics, Rotational Motion, JEE Main
The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is

Options:
A 0.9%
B 2.4%
C 3.1%
D 4.2%

The correct answer is:
3.1%

Solution:
Least count =0.550=0.01 mm
Diameter of ball D=2.5 mm+200.01
D=2.7 mm
ρ=Mvol=M43πD23
∆ρρmax=∆MM+3∆DD; ∆ρρmax=2%+3 0.012.7×100%
∆ρρ=3.1%

Topic: Physics, Units and Dimensions, NEET
The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is $0.5 \mathrm{~mm}$ and there are 50 divisions on the circular scale. The reading on the main scale is $2.5 \mathrm{~mm}$ and that on the circular scale is $20$ divisions. If the measured mass of the ball has a relative error of $2 \%$, the relative percentage error in the density is

Options:
A $0.9 \%$
B $2.4 \%$
C $3.1 \%$
D $4.2 \%$

The correct answer is:
$3.1 \%$

Solution:
Least count $=($ pitch $/$ number of division on circular scale $)=(0.5 / 50)=0.01 \mathrm{~mm}$
Diameter of solid ball $=2.5+(0.5 / 50) \times 20=2.7 \mathrm{~mm}$
As density $\rho=$ mass $/$ volume $=M /\left[(4 \pi / 3)(D / 2)^3\right]$
The relative percentage error in density is
$(\Delta \rho / \rho) \times 100=\{(\Delta M / M)+(3 \Delta D / D)\} \times 100$
Percentage of error $=\{(\Delta M / M)+(3 \Delta D / D)\} \times 100 \times 100$
$=(2+3 \times(0.01 / 2.7) \times 100)=3.1 \%$
Answer:(3) $3.1 \%$

Topic: Physics, Mathematics in Physics, JEE Main
The density of a solid metal sphere is diameter. The maximum error in the density of the sphere is x100%. If the relative errors in measuring the mass and the diameter are 6.0% and 1.5% respectively, the value of x is -

Options:

The correct answer is:
1050

Solution:
ρ=m43πd23

ρ=6mπd3

taking log

ℓnρ=ℓnm-3ℓnd

Maximum error by differentiation

dρρ=dmm+3ddd

dpρ=6+3×1.5

=10.5%=1050100%

Ans. x = 1050

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2020 (06 Sep Shift 1)
The density of a solution prepared by dissolving $120 \mathrm{~g}$ of urea (mol. Mass $=60 \mathrm{u}$ ) in $1000 \mathrm{~g}$ of water is $1.15 \mathrm{~g} / \mathrm{mL}$. The molarity of this solution is :

Options:
A
$0.50 \mathrm{~M}$

B
$1.78 \mathrm{~M}$

C
$1.02 \mathrm{~M}$

D
$2.05 \mathrm{~M}$

The correct answer is:
$2.05 \mathrm{~M}$

Solution:
Total weight of solution $=1000+120=1120 \mathrm{~g}$
$\text { Molarity }=\frac{120}{60} \times \frac{1000}{1120 / 1.15}=2.05 \mathrm{~M}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2012 (Offline)
The density of a sphere is measured by measuring the mass and diameter. If it is known that the maximum percentage errors in the measurement of mass and diameter are 2% and 3% respectively then the maximum percentage error in the measurement of density is

Options:

The correct answer is:
11

Solution:
The density can be computed by

ρ=mv=mπd36

or ρ ∝ md3

∴∆ρρ×100%=∆mm×100% + 3∆dd×100%

or ∆ρρ×100=2+3×3=11

Hence, the percentage error in measurement of density is 11%

Topic: Physics, Units and Dimensions, JEE Main
The density of a substance at 0 â is 10 g/cc and at 100 â, its density is 9.7 g/cc. The coefficient of linear expansion of the substance is

Options:
A 10-4 â-1
B 10-2 ℃-1
C 10-3 ℃-1
D 102 ℃-1

The correct answer is:
10-4 â-1

Solution:
As Ï=Ï0(1-Î³âT)
â´ 9.7=10(1-Î³Ã100)
9.710 =1-Î³Ã100
Î³Ã100 =1-9.710=0.310=3Ã10-2
Î³ =3Ã10-4 â´ Î±=13 Î³=10-4 â-1

Topic: Physics, Thermal Properties of Matter, JEE Main
The density of alkali metals is in the order

Options:
A K<Cs<Na<Rb
B Na<Rb<K<Cs
C Na<K<Cs<Rb
D K<Na<Rb<Cs

The correct answer is:
K<Na<Rb<Cs

Solution:
As we down in a group then its size increases and also volume of the atom increases as the number of electrons and protons increases. But the increase in mass is greater than that of increase in volume so the density which is defined as mass per unit volume will increase generally. Density of alkali metals increases down the group except potassium. Therefore, correct order of density of given alkali metals is K<Na<Rb<Cs

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (12 Apr Shift 1)
The density of copper is 8.94 g mL-1 . Find the charge needed to plate an area of 10Ã10 cm2 to a thickness of 10-2 cm using a CuSO4 solution as electrolyte (atomic weight of Cu = 63.6 g/mol).

Options:
A 2.7×104C
B 8.8×104C
C 18.3×104C
D 1.7×104C

The correct answer is:
2.7×104C

Solution:
W=MnFIt
or ρ×V=MnFQ.
Q=ρ×area×thicknessM×nF
=8.94×10×10×10-2×2×9650063.6
=2.7×104C .

Topic: Chemistry, Electrochemistry, JEE Main
The density of electron-hole pair in a pure germanium is 3Ã1016 m-3 at room temperature. On doping with aluminium, the hole density increases to 4.5Ã1022 m-3. Now, the electron density in m-3 in doped germanium will be

Options:
A 1×1010
B 2×1010
C 0.5×1010
D 4×1010

The correct answer is:
2×1010

Solution:
ni=3Ã1016 m-3
nh=4.5Ã1022 m-3
ne=?
nenh=ni2
â ne=ni2nh=3Ã101624.5Ã1022=9Ã10324.5Ã1022=2Ã1010

Topic: Physics, Semiconductors, JEE Main
The density of KBr is 2.75 g cm-3 , length of the unit cell is 654 pm. K = 39, Br = 80, then what can be true about the predicted nature of the solid?
(Given: N A =6.023Ã 10 23 )

Options:
A Solid has face centred cubic system with co-ordination number = 6
B Solid has simple cubic system with co-ordination number = 8
C Solid has face centred cubic system with co-ordination number =12
D None of the above

The correct answer is:
Solid has face centred cubic system with co-ordination number = 6

Solution:
Given that âÏ=2.75âgâc m â3
we need to find type of unit cell

â´âââÏ= ZÃM N A a 3 âµâââââZ= ÏÃ N A Ã a 3 M âââââââââââ=â 2.75Ã6.023Ã 10 23 Ã ( 654Ã 10 â10 ) 3 119
â´âââZâ4ââSoâfccâunitâcell
In FCC 1 atm surrounded by 6 atoms so coordination number is 6.

Topic: Chemistry, Solid State, JEE Main
The density of material in CGS system of units is $4 \mathrm{~g} / \mathrm{cm}^3$. In a system of units in which unit of lengths is $10 \mathrm{~cm}$ and unit of mass is $100 \mathrm{~g}$, the value of density of material will be

Options:
A 0.4
B 40
C 400
D 0.04

The correct answer is:
40

Solution:
Mass,
$$
\begin{aligned}
& M=d V \\
& d=\frac{M}{L^3} \\
& d=4 \mathrm{~g} / \mathrm{cm}^3
\end{aligned}
$$
If unit of mass is $100 \mathrm{~g}$ and unit of distance is $10 \mathrm{~cm}$.
So, density
$$
\begin{aligned}
y & =\frac{4\left(\frac{100 \mathrm{~g}}{100}\right)}{\left(\frac{10}{10} \mathrm{~cm}\right)^3} \\
& =\frac{\left(\frac{4}{100}\right)}{\left(\frac{1}{10}\right)^3} \cdot \frac{(100 \mathrm{~g})}{(10 \mathrm{~cm})^3} \\
& =40 \mathrm{~g} \mathrm{~cm}^{-3}
\end{aligned}
$$
Aliter
$$
\begin{aligned}
u_2 & =u_1\left(\frac{M_1}{M_2}\right)\left(\frac{L_1}{L_2}\right)^{-3} \\
& =4\left(\frac{1}{100}\right)\left(\frac{1}{10}\right)^{-3} \\
& =40
\end{aligned}
$$

Topic: Physics, Units and Dimensions, NEET, NEET 2011 (Mains)
The density of NaOH solution is 1.2 g cm-3. The molality of this solution is m (Round off to the Nearest Integer):
[Use : Atomic masses : Na:23.0u O:16.0u
H:1.0u
Density of H2O:1.0 g cm-3

Options:

The correct answer is:
5

Solution:
Consider 1ℓ solution
mass of solution =(1.2×1000)g
=1200 gm
Neglecting volume of NaOH Mass of water =1000 gm
⇒ Mass of NaOH=(1200-1000)gm
=200 gm
⇒ Moles of NaOH=200 g40 g/mol=5 mol
⇒ molality =5 mol1 kg=5 m

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2021 (27 Jul Shift 1)
The density of O2 is 16 at NTP. At what temperature(in oC) its density will be 14? Consider that the pressure remains constant.

Options:

The correct answer is:
39

Solution:
As d=PMRT
d1T1=d2T2
When p remains constant
d1=16;d2=14;T1=273 K, T2=?
d1T1=d2T2
16Ã273=14ÃT2
T2=312 K
T2=312-273=39â

Topic: Chemistry, States of Matter, JEE Main
The density of the material of a cube can be estimated by measuring its mass and the length of one of its sides. If the maximum error in the measurement of mass and length are $0.3 \%$ and $0.2 \%$ respectively, the maximum error in the estimation of the density of the cube is approximately.

Options:
A $1.1 \%$
B $0.5 \%$
C $0.9 \%$
D $0.7 \%$

The correct answer is:
$0.9 \%$

Solution:
Given,
Maximum error in the measurement of mass $=0.3 \%$
Maximum error in the measurement of length $=0.2 \%$
We know that,
Error in density is given as,
$\text {Density, } \rho=\frac{\text { mass }(m)}{\text { volume }(V)}=\frac{m}{L^{3}}$
where, $L=$ side of cube
Error in density is given as,
$\left(\frac{\Delta \rho}{\rho}\right)=\frac{\Delta m}{m}+\frac{3 \Delta L}{L}$
or $\left(\frac{\Delta \rho}{\rho}\right) \times 100=\left(\frac{\Delta m}{m}+\frac{3 \Delta L}{L}\right) \times 100$
Substituting the given values, we get
$\left(\frac{\Delta \rho}{\rho}\right)_{\max }=(0.3 \%+3(0.2) \%=0.3 \%+0.6 \%$
$\therefore$ Maximum percentage error measurement of density $\left(\frac{\Delta \rho}{\rho}\right)_{\max }=0.9 \%$

Topic: Physics, Mathematics in Physics, JEE Main
The density of water at $20^{\circ} \mathrm{C}$ in $998 \mathrm{~kg} / \mathrm{m}^3$ and at a $40^{\circ} \mathrm{C} 992 \mathrm{~kg} / \mathrm{m}^3$. The cocfficient of volume expansion of water is:

Options:
A $3 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
B $2 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
C $6 \times 10^{-4} /{ }^{\circ} \mathrm{C}$
D $10^{-3} \rho^{\circ} \mathrm{C}$

The correct answer is:
$3 \times 10^{-4} /{ }^{\circ} \mathrm{C}$

Solution:
According to the question
$$
\begin{aligned}
& T_1=20^{\circ} \mathrm{C}, T_2=40^{\circ} \mathrm{C} \\
& \rho_r=998 \mathrm{~kg} / \mathrm{m}^3 \rho_{40}=992 \mathrm{~kg} / \mathrm{m}^3 \\
& \text { So, } \rho_{\mathrm{T}_2}=\frac{\rho_{\mathrm{T}_1}}{(1+\mathrm{v} \Delta T)}=\frac{\rho_{\mathrm{T}_1}}{1+\mathrm{v}\left(T_2-T_1\right)}
\end{aligned}
$$
Putting the values we get
$$
\begin{aligned}
& 992=\frac{998}{1+v(40-20)} \\
& 992=\frac{998}{1+20 \mathrm{v}} \\
& \Rightarrow 992(1+20 \mathrm{v})=998 \\
& \Rightarrow \quad 1+20 \mathrm{v}=\frac{998}{992} \\
& \Rightarrow \quad 20 \mathrm{v}=\frac{6}{992} \\
& \Rightarrow \mathrm{v}=\frac{6}{992} \times \frac{1}{20}=3 \times 10^{-4} /{ }^{\circ}
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main
The denticity of an organic ligand, biuret is :

Options:
A 2
B 3
C 4
D 6

The correct answer is:
2

Solution:
Structure of biuret is

In biuret, donar atoms are 2 of the three nitrogen atoms. So it is a bidentate ligand.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The denticity of the ligand present in the Fehling's reagent is _____ .

Options:

The correct answer is:
4

Solution:
Denticity is defined as the number of electron pairs donated by a ligand to a central atom. Denticity of the ligand present in fehling's solution(Copper tartarate complex) is 2 based on the given diagram. But the maximum denticity is 4.

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2023 (29 Jan Shift 2)
The dependence of acceleration due to gravity $g$ on the distance $r$ from the centre of the earth, assumed to be a sphere of radius $R$ of uniform density is as shown in figures below
(1)
(2)
(3)
(4)
The correct figure is

Options:
A (4)
B (1)
C (2)
D (3)

The correct answer is:
(4)

Solution:
$\because g \propto r$
and $g \propto \frac{1}{\mathbf{r}_2}$
(if $r < R$ )
(if $r>R$ )
So, the correct figure will be

Topic: Physics, Gravitation, NEET, NEET 2010 (Mains)
The dependency of speed of water surface waves (capillary waves) on the density of water (Ï) their wavelength (Î») and surface tension (Î³) is -

Options:
A γaρ
B γρλ
C γρλ13
D γρλ

The correct answer is:
γρλ

Solution:
Velocity of surface waves depends on density of water, its own wavelength of surface tension.
vâÏaÎ»bÎ³c
LT-1âML-3aLbMLT-2Lc
LT-1âMa+cL-3a+bT-2c
a+c=0 ..........(i)
-3a+b=1 .............(ii)
-2c=-1 .............(iii)
Solving
a=-12 , b=-12 , c=12 ,
vâÏ-12Î»-12Î³12
vâÎ³ÏÎ»

Topic: Physics, Units and Dimensions, JEE Main
The depletion layer in the $P-N$ junction region is caused by

Options:
A Drift of holes
B Diffusion of charge carriers
C Migration of impurity ions
D Drift of electrons

The correct answer is:
Diffusion of charge carriers

Solution:
Due to the large concentration of electrons in $N$-side and holes in $P$-side, they diffuses from their own side to other side. Hence depletion region produces.

Topic: Physics, Semiconductors, JEE Main