datavorous/entrance-exam-dataset huggingface dataset:
The temperature of a wire is doubled. The Young's modulus of elasticity

Options:
A
will also double

B
will become four times

C
will remain same

D
will decrease

The correct answer is:
will decrease

Solution:
As we know that, length of a wire when the temperature increased
$$
L_t=L_0(1+\alpha \Delta T)
$$
where $\Delta T$ is change in the temperature.
$L_0$ is original length,
$\alpha$ is coefficient of linear expansion and
$L_t$ is length at temperature $T$.
Now, $\Delta L=L_t-L_0=L_0 \alpha \Delta T$
Now, Young's modulus $(Y)=\frac{\text { Stress }}{\text { Strain }}$ $=\frac{F L_0}{A \times \Delta L}=\frac{F L_0}{A L_0 \alpha \Delta T} \propto \frac{1}{\Delta T}$
As, $Y \propto \frac{1}{\Delta T}$
So, if temperature increass $\Delta T$ increases, hence Young's modulus of elasticity $(Y)$ decreases.

Topic: Physics, Mechanical Properties of Solids
The temperature of a wire of length 1 metre and area of cross-section $1 \mathrm{~cm}^2$ is increased from $0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$. If the rod is not allowed to increase in length, the force required will be $\left(\alpha=10^{-5} /{ }^{\circ} \mathrm{C}\right.$ and $\left.Y=10^{11} \mathrm{~N} / \mathrm{m}^2\right)$

Options:
A $10^3 \mathrm{~N}$
B $10^4 \mathrm{~N}$
C $10^5 \mathrm{~N}$
D $10^9 \mathrm{~N}$

The correct answer is:
$10^4 \mathrm{~N}$

Solution:
$F=$ force developed $=Y A \propto(\Delta \theta)$
$=10^{11} \times 10^{-4} \times 10^{-5} \times 100=10^4 \mathrm{~N}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
The temperature of an ideal gas in three dimensions is 300 K. The corresponding de-Broglie wavelength of the electron approximately at 300 K is:

me=mass of electron =9×10-31 kg, h=Planck constant =6.6×10-34 J s, kB=Boltzmann constant=1.38×10-23 J K-1

Options:
A 2.26 nm
B 3.25 nm
C 8.46 nm
D 6.26 nm

The correct answer is:
6.26 nm

Solution:
k=1.38×10-23 J K-1

Mass of electron =9×10-31 kg

⇒λD=h2m×32×kT=h3mkT

⇒λD=6.6×10-343×9×10-31×1.38×10-23×300=6.6×10-34105.72×10-27=62 Aο

Thus x=62 Aο

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2021 (01 Sep Shift 2)
The temperature of an ideal gas is increased from $27^{\circ} \mathrm{C}$ to $127^{\circ} \mathrm{C}$, then percentage increase in $\mathrm{v}_{\mathrm{rms}}$

is

Options:
A $37 \%$
B $11 \%$
C $33 \%$
D $15.5 \%$

The correct answer is:
$15.5 \%$

Solution:
We know, $\mathrm{V}_{\mathrm{mms}}=\sqrt{\frac{3 \mathrm{RT}}{\mathrm{M}}}$

$\Rightarrow \%$ increase in

$$

\begin{array}{l}

\mathrm{V}_{\mathrm{rms}}=\frac{\sqrt{\frac{3 \mathrm{R} \mathrm{T}_{2}}{\mathrm{M}}}-\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}}}}{\sqrt{\frac{3 \mathrm{RT}_{1}}{\mathrm{M}}}} \times 100 \\

=\frac{\sqrt{\mathrm{T}_{2}}-\sqrt{\mathrm{T}_{1}}}{\sqrt{\mathrm{T}_{1}}} \times 100 \\

=\frac{\sqrt{400}-\sqrt{300}}{\sqrt{300}} \times 100 \\

=\frac{20-17.32}{17.32} \times 100=15.5 \%

\end{array}

$$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The temperature of an ideal gas is increased from $27^{\circ} \mathrm{C}$ to $927^{\circ} \mathrm{C}$. The root mean square speed of its molecules becomes

Options:
A Twice
B Half
C Four times
D One-fourth

The correct answer is:
Twice

Solution:
$v_{r m s} \propto \sqrt{T}, \frac{v_2}{v_1}=\sqrt{\frac{T_2}{T_1}} \Rightarrow v_2=\sqrt{\frac{(273+927)}{(273+2)}} \Rightarrow v_2=2 v_1$

Topic: Physics, Kinetic Theory of Gases, JEE Main
The temperature of an ideal gas is increased from 100 K to 400 K . If the rms speed of the gas molecule is V at 100 K , then at 400 K it becomes

Options:
A 2V
B 4V
C 0.5V
D V

The correct answer is:
2V

Solution:
âµ rms speed of molecule VrmsâT
for T=100 K
Vrms1â100
For T=400 K
Vrms2â400
therefore, Vrms1Vrms2=100400
or VVrms2=12
or Vrms2=2V

Topic: Physics, Kinetic Theory of Gases, JEE Main
The temperature of an ideal gas is increased from 200 K to 800 K. If r.m.s. speed of gas at 200 K is v0. Then, r.m.s. speed of the gas at 800 K will be:

Options:
A v04
B v0
C 4v0
D 2v0

The correct answer is:
2v0

Solution:
At 200 K, the rms speed is given by

v0=3R×200M ...(1)

At 800 K, the rms speed is given by

v'=3R×800M ...2

Divide equation (2) by equation (1) and solve to calculate the required speed.

v'v0=3R×800M3R×200M=800200= 2⇒v'=2v0

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2023 (06 Apr Shift 2)
The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1×105 Pa. If ni and nf are the number of molecules in the room before and after heating, then nf-ni will be:

Options:
A -2.5×1025
B -1.61Ã1023
C 1.38Ã1023
D 2.5Ã1025

The correct answer is:
-2.5×1025

Solution:
Ti=273+17=290K
Tf=273+27=300K
From ideal gas equation, PV=nRT ⇒n=PVRT

nf-ni=PVR 1Tf-1Ti= -2.5×1025

Topic: Physics, Kinetic Theory of Gases, JEE Main
The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1×105 Pa. If ni and nf are the number of molecules in the room before and after heating, then nf-ni will be:

Options:
A -2.5×1025
B -1.61Ã1023
C 1.38Ã1023
D 2.5Ã1025

The correct answer is:
-2.5×1025

Solution:
Ti=273+17=290K
Tf=273+27=300K
From ideal gas equation, PV=nRT ⇒n=PVRT

nf-ni=PVR 1Tf-1Ti= -2.5×1025

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2017 (02 Apr)
The temperature of equal masses of three different liquids $A, B$ and $C$ are $12^{\circ} \mathrm{C}, 19^{\circ} \mathrm{C}$ and $28^{\circ} \mathrm{C}$ respectively. The temperature when $A$ and $B$ are mixed is $16^{\circ} \mathrm{C}$ and when $B$ and $C$ are mixed is $23^{\circ} \mathrm{C}$. The temperature when $A$ and $C$ are mixed is

Options:
A $18.2^{\circ} \mathrm{C}$
B $22^{\circ} \mathrm{C}$
C $20.2^{\circ} \mathrm{C}$
D $25.2^{\circ} \mathrm{C}$

The correct answer is:
$20.2^{\circ} \mathrm{C}$

Solution:
Heat gain = heat lost
$C_A(16-12)=C_B(19-16) \Rightarrow \frac{C_A}{C_B}=\frac{3}{4}$
$\Rightarrow \frac{C_A}{C_C}=\frac{15}{16}$ ...(i)
If $\theta$ is the temperature when $A$ and $C$ are mixed then,
$C_A(\theta-12)=C_C(28-\theta) \Rightarrow \frac{C_A}{C_C}=\frac{28-\theta}{\theta-12}$ ...(ii)
On solving equation (i) and (ii) $\theta=20.2^{\circ} \mathrm{C}$.

Topic: Physics, Thermal Properties of Matter, JEE Main
The temperature of equal masses of three different liquids A, B and C are 12℃, 19℃ and 28℃ respectively. The temperature when A and B are mixed is 16℃ and when B and C are mixed is 23℃. The temperature when A and C are mixed is

Options:
A 18.2 â
B 22 â
C 20.2 â
D 24.2 â

The correct answer is:
20.2 â

Solution:
1st case A and B mixed
ms A t - t A = ms B t - t B
sA(16-12)=sB(19-16)
4sA=3sB
2nd case B and C mixed
msBt-tB=msC(tC-t)
sB23-19=sC(28-23)
4sB=5sC
3sB=154sC
∴ 4sA=3sB=154sC
⇒ 16sA=12sB=15sC=k
sA:sB:sC=116:112:115
sA=k16 , sC=k15
3rd Case When A and C are mixed
msAt-tA=msC(tC-t)
k16t-12=k15(28-t)
15t-180=448-16t
31t=628
⇒ t=20.2 ℃

Topic: Physics, Thermal Properties of Matter, JEE Main
The temperature of equal masses of three different liquids A, B and C are 12 oC, 19 oC and 28 oC respectively. The temperature when A and B are mixed is 16 oC and when B and C are mixed is 23 oC. The temperature when A and C are mixed is

Options:
A 18.2 oC
B 22 oC
C 20.2 oC
D 25.2 oC

The correct answer is:
20.2 oC

Solution:
Heat gain = heat lost
C A ( 16â12 )= C B ( 19â16 )â C A C B = 3 4 and C B ( 23â19 )= C C ( 28â23 )â C B C C = 5 4
â C A C C = 15 16 â¦(i)
If Î¸ is the temperature when A and C are mixed then,
C A ( Î¸â12 )= C C ( 28âÎ¸ )
â C A C C = 28âÎ¸ Î¸â12 â¦ (ii)
On solving equations (i) and (ii) Î¸= 20.2 o C

Topic: Physics, Thermodynamics, JEE Main
The temperature of equal masses of three different liquids x, y and z are 10°C, 20°C and 30°C respectively. The temperature of mixture when x is mixed with y is 16°C and that when y is mixed with z is 26°C. The temperature of mixture when x and z are mixed will be :

Options:
A 25.62°C
B 20.28°C
C 28.32°C
D 23.84°C

The correct answer is:
23.84°C

Solution:
When A&B are mixed

m1 S1(16-10)=m2 S2(20-16)

6 m1 S1=4 m2 S2 .......(1)

When B&C are mixed

m2 S2(26-20)=m3 S3(30-26)

6 m2 S2=4 m3 S3 ...........(2)

From equation (1) and (2)

9 m1 S1=4 m3 S3

When A&C are mixed

m1 S1( T-10)=m3 S3(30-T)

T=30m3S3+10m1S1m1S1+m3S3

T=103m3S3+m1S1m3S3+m1S1

T=103m3 S3+49 m3 S3m3 S3+49 m3 S3

T=10×3113

=23.84°C

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The temperature of n moles of an ideal gas is increased from T to 4T through a process for which pressure P=a T â1 , where a is a constant. The work done by the gas is

Options:
A nRT
B 4nRT
C 2nRT
D 6nRT

The correct answer is:
6nRT

Solution:
According to ideal gas equation
PV=nRT
or Vâ¡ = nRTâ¡ Pâ¡
âµ Pâ¡ = a Tâ¡ Given â¦ i
âµ Vâ¡ = n Râ¡ Tâ¡ 2 a
â dV = 2nRT a dT
Work done by the gas, dW = PdV
or W = â« Tâ¡ 4 Tâ¡ a Tâ¡ 2 n Râ¡ Tâ¡ a dTâ¡ â¦ Using (i) and (ii)
= 2 nâ¡ Râ¡ Tâ¡ Tâ¡ 4 Tâ¡ = 6 n RTâ¡

Topic: Physics, Thermodynamics, JEE Main
The temperature of two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity $K$ and $2 K$ and thickness $x$ and $4 x$, respectively are $T_2$ and $T_1\left(T_2>T_1\right)$. The rate of heat transfer through the slab, in a steady state is $\left(\frac{A\left(T_2-T_1\right) K}{x}\right) f$, with $f$ equal to

Options:
A
1

B
$1 / 2$

C
$2 / 3$

D
$1 / 3$

The correct answer is:
$1 / 3$

Solution:
$\Delta \mathrm{q}=\frac{\mathrm{kA}}{\mathrm{x}}\left[\mathrm{T}_2-\frac{2 \mathrm{~T}_2-\mathrm{T}_1}{3}\right]$
$=\frac{\mathrm{kA}}{3 \mathrm{x}}\left[\mathrm{T}_2-\mathrm{T}_1\right]$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2004
The temperature on Celsius scale is $25^{\circ} \mathrm{C}$. What is the corresponding temperature on the Fahrenheit scale

Options:
A $40^{\circ} \mathrm{F}$
B $77^{\circ} \mathrm{F}$
C $50^{\circ} \mathrm{F}$
D $45^{\circ} \mathrm{F}$

The correct answer is:
$77^{\circ} \mathrm{F}$

Solution:
$77^{\circ} \mathrm{F}$
$F=(C \times 1.8)+32$

Topic: Physics, Thermal Properties of Matter, JEE Main
The temperature Tt of a body at time t=0 is 160° F and it decreases continuously as per the differential equation dTdt=−KT−80, where K is positive constant. If T15=120° F, then T45 is equal to

Options:
A 85° F
B 95° F
C 90° F
D 80° F

The correct answer is:
90° F

Solution:
Given: dTdt=−KT−80

⇒dTT−80=−Kdt

⇒∫160TdTT−80=∫0t−Kdt

⇒logT−80160T=−Kt

⇒logT−80−log80=−Kt

⇒logT−8080=−Kt

⇒T=80+80e−Kt

Now, using the value T15=120° we get,

⇒120=80+80e−K·15

⇒4080=e−15k

⇒e−15k=12

∴T45=80+80e−45k

⇒T45=80+80e−15k3

⇒T45=80+80×18

⇒T45=90° F

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (31 Jan Shift 2)
The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

Options:
A
$1 / 2$

B
$1 / 4$

C
$1 / 3$

D
$2 / 3$

The correct answer is:
$1 / 3$

Solution:
$\eta=\frac{\Delta \mathrm{W}}{\mathrm{Q}_{\mathrm{BC}}}=\frac{\frac{\mathrm{S}_0 \mathrm{~T}_0}{2}}{\frac{3 \mathrm{~S}_0 \mathrm{~T}_0}{2}}=1 / 3$

Topic: Physics, Thermodynamics, JEE Main
The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

Options:
A
$1 / 2$

B
$1 / 4$

C
$1 / 3$

D
$2 / 3$

The correct answer is:
$1 / 3$

Solution:
$\eta=\frac{\Delta \mathrm{W}}{\mathrm{Q}_{\mathrm{BC}}}=\frac{\frac{\mathrm{S}_0 \mathrm{~T}_0}{2}}{\frac{3 \mathrm{~S}_0 \mathrm{~T}_0}{2}}=1 / 3$

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2005
The temporary hardness of a water sample is due to compound X. Boiling this sample converts X to compound Y. X and Y, respectively, are:

Options:
A MgHCO32 and MgCO3
B CaHCO32 and CaOH2
C MgHCO32 and MgOH2
D CaHCO32 and CaO

The correct answer is:
MgHCO32 and MgOH2

Solution:
Both MgHCO32 and CaHCO32 are responsible for temporary hardness.
MgHCO32X→Boiling2MgCO3↓+CO2↑+H2O
CO3-2+H2O→HCO3-+OH-
Mg+2+2OH-→MgOH2(Y)

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The temporary hardness of water is due to:

Options:
A Na2SO4
B NaCl
C CaCl2
D CaHCO32

The correct answer is:
CaHCO32

Solution:
Cl- ions are the cause of permanent hardness & HCO3- ions are the cause of temporary Hardness.

Topic: Chemistry, Hydrogen, JEE Main, JEE Main 2019 (09 Jan Shift 2)
The tendency of population to remain in genetic equilibrium may be disturbed by

Options:
A random mating
B lack of migration
C lack of mutations
D lack of random mating

The correct answer is:
lack of random mating

Solution:
According to Hardy-Weirberg principle, allele frequencies in a population are are stable and is constant from generation to generation allele frequencies in a population will remain constant over generations only if the following condition are met
(i) There is no mutation no gene flow and all mating is random.
(ii) All genotypes reproduce equally well (i.e., no natural selection). But, all there conditions rarely met in nature.

Topic: Biology, Evolution, NEET, NEET 2013 (All India)
The tension in the spring is

Options:
A Zero
B $2.5 \mathrm{~N}$
C $5 \mathrm{~N}$
D $10 \mathrm{~N}$

The correct answer is:
$5 \mathrm{~N}$

Solution:
$5 \mathrm{~N}$ force will not produce any tension in spring without support of other $5 \mathrm{~N}$ force. So here the tension in the spring will be $5 \mathrm{~N}$ only.

Topic: Physics, Laws of Motion, JEE Main
The tenth elements in the periodic table resembles with the

Options:
A First period
B Second period
C Fourth group
D Ninth group

The correct answer is:
Second period

Solution:
Each period consists of a series of elements whose atom have the same principal quantum no. of the outer most shell i.e. In second period, this shell has four orbitals (one $2 \mathrm{~s}$ and three ) which can have eight electrons, hence second period contains 8 elements from atomic no. 3 to 10 .

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
The term anomers of glucose refers to

Options:
A
isomers of glucose that differ in configurations at carbons one and four $(C-1$ and $C-4)$

B
a mixture of $(D)$-glucose and $(\mathrm{L})$-glucose

C
enantiomers of glucose

D
isomers of glucose that differ in configuration at carbon one $(\mathrm{C}-1)$

The correct answer is:
isomers of glucose that differ in configuration at carbon one $(\mathrm{C}-1)$

Solution:
No Solution Available

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2006
The term independent of $x$ in $\left[\sqrt{\left(\frac{x}{3}\right)}+\sqrt{\left(\frac{3}{2 x^{2}}\right)}\right]^{10}$ is

Options:
A $\frac{5}{4}$
B $\frac{7}{4}$
C $\frac{9}{4}$
D None of these

The correct answer is:
None of these

Solution:
Given, expansion is $\left[\sqrt{\left(\frac{x}{3}\right)}+\sqrt{\left(\frac{3}{2 x^{2}}\right)}\right]^{10}$
Let $t_{r+1}$ be general term, then
$$
\begin{aligned}
t_{r+1} &={ }^{10} C_{r}\left(\sqrt{\frac{x}{3}}\right)^{10-r}\left(\sqrt{\frac{3}{2 x^{2}}}\right)^{r} \\
&={ }^{10} C_{r} \frac{x \frac{10-r}{2}}{\frac{10-r}{2}} \cdot \frac{3^{i / 2}}{2^{r / 2} \cdot x^{r}}={ }^{10} C_{r} x-\frac{r}{2}-r \cdot \frac{3^{r-5}}{2^{r / 2}}
\end{aligned}
$$
For the term independent of $x$.
Puts $5-\frac{r}{2}-r=0$
$$
\Rightarrow \quad 5-\frac{3}{2} r=0 \Rightarrow \frac{3}{2} r=5 \Rightarrow r=\frac{10}{3}
$$
Fractional value of $r$ is not possible. So, no term is independent of $x$.

Topic: Mathematics, Binomial Theorem, JEE Main
The term independent of x in the binomial expansion of 1-1x+3x5 2x2-1x8 is

Options:
A - 496
B -400
C 496
D 400

The correct answer is:
400

Solution:
The term independent of x will come from
1-1x+3x52x2-1x8
General term of 2x2-1x8 is C 8 r ( − 1 x ) r 2x28-r= C 8 r 2 8−r (−1) r x 16−3r
From expansion
12x2-1x8-1x2x2-1x8+3x52x2-1x8
12x2-1x8→ no constant term
-1x2x2-1x8→ for constant term
r=5 ⇒ − C 8 3 2 3 (−1) 5 =+448
3x52x2-1x8→ for constant term
r=7⇒ C 38 7 2 1 (−1) 7 =48
Hence, term independent from x=448-48=400

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2015 (11 Apr Online)
The term independent of x in the expansion of 160-x881.2x2-3x26 is equal to

Options:
A -72
B 36
C -108
D -36

The correct answer is:
-36

Solution:
To find the coefficient of term independent of x in given expression, first let's find constant term and coefficient of x-8 in the expansion of 2x2-3x26
Tr+1= 6Cr 2x26-r× -3x2r
Tr+1= 6Cr 26-r× -3r × x12-4r
For constant term: r=3 , hence constant term = 6C3 23×-33×x0
For coefficient of x-8 : r=5 , hence coefficient of x-8= 6C5×21-35×x-8
Given expression can be written as
160-x881 6C3×23×-33×x0+…+ 6C5×21×-35×x-8+…
Hence, coefficient of term independent of x
=160× 6C3×23-33-181× 6C5×21×-35
= -72+36= -36

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (12 Apr Shift 2)
The term independent of x in the expansion of (1-x)2(x+1x)10 is

Options:
A 10C5
B 10C6
C 11C5
D 17C7

The correct answer is:
11C5

Solution:
We have, (1-x)2(x+1x)10
= 1â2x+ x 2 x 10 + C 10 1 x 9 Â· 1 x + C 10 2 x 8 Ã 1 x 2 +...+ 1 x 10
= 1â2x+ x 2 x 10 + C 10 1 x 8 + C 10 2 x 6 +... + 10 C 9 Ã 1 x 8 + 1 x 10
So, the term independent of x is
C 10 5 + â2 Ã0+ C 10 6 = C 10 5 + C 10 6
=11C5

Topic: Mathematics, Binomial Theorem, JEE Main
The term independent of \( x \) in the expansion of \( \left(\frac{x+1}{x^{2 / 3}-x^{1 / 3}+1}-\frac{x-1}{x-x^{1 / 2}}\right)^{10} \), where \( x \neq 0,1 \) is equal to

Options:

The correct answer is:
210

Solution:
x+1x2/3-x1/3+1-x-1x-x1/210

=x1/3+1-x1/2+1x1/210

=x1/3-1x1/210

Now the r+1th Term

Tr+1=Cr10x1/310-r·-1x1/2r

For independent term

10-r3-r2=0⇒r=4

⇒T5=C410=210

Topic: Mathematics, Binomial Theorem, JEE Main
The term independent of x in the expansion of the expression 1+x+2x332x2-13x9 is equal to

Options:
A 1363
B 1945
C 1754
D 2336

The correct answer is:
1754

Solution:
Coefficient of x0 in 1+x+2x332x2-13x9= Coefficient of x0 in 32x2-13x9+ Coefficient of x-1 in 32x2-13x9+2× coefficient of x-3 in 32x2-13x9
For 32x2-13x9→Tr+1=Cr932x29-r-13xr=Cr9329-r-13rx18-3r
18-3r=0⇒r=6,18-3r=-1⇒3r=19
18-3r=-3⇒3r=21⇒r=7
Required coefficient =C96323-136+C97322-137×2
=322-136C6932+C79·2·-13=1754

Topic: Mathematics, Binomial Theorem, JEE Main
The term independent of x in the expansion of x+1x2/3-x1/3+1-x-1x-x1/210 is

Options:
A 210
B 310
C 4
D 120

The correct answer is:
210

Solution:
We know, x2/3-x1/3+1x1/3+1=x+1
⇒x+1x2/3-x1/3+1=x1/3+1
Also, x-1x-x=x-1x+1xx-1=x+1x=1+x-1/2
Hence, the given binomial becomes x1/3-x-1/210.
General term =10Cr-1rx10-r3x-r2
Hence, for term independent of x.
We get 10-r3=r2
20-2r=3r
r=4
∴ Term Independent of x=10C4=210

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2013 (07 Apr)
The term independent of x in the expansion of x+1x2/3-x1/3+1-x-1x-x1/210, where x≠0,1 is equal to

Options:

The correct answer is:
210

Solution:
x+1x2/3-x1/3+1-x-1x-x1/210

x1/3+1x2/3-x1/3+1x2/3-x1/3+1-x1/2+1x1/2-1x1/2x1/2-110

=x1/3+1-x1/2+1x1/210

=x1/3-1x1/210

Now the r+1th Term

Tr+1=Cr10x1/310-r·-1x1/2r

Tr+1=Cr10x1/310-r·x-r/2-1r

For independent term

10-r3-r2=0⇒r=4

⇒T5=C410-14

=10×9×8×74×3×2×1=210

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (25 Jul Shift 1)
The term independent of x in the expansion of x+1x2/3-x1/3+1-x-1x-x1/210, x≠1, is equal to ___.

Options:

The correct answer is:
210

Solution:
Given x+1x2/3-x1/3+1-x-1x-x1/210

⇒x1/3+1-x+1x10

⇒x1/3-x-1/210

General term is Tr+1=10Crx1/310-r-x-1/2r

∴ 10-r3-r2=0⇒20-2r-3r=0

⇒r=4

Then, T5=C410=10×9×8×74×3×2×1=210

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2021 (18 Mar Shift 2)
The term independent of x in the expansion of x-1x4x+1x3 is

Options:
A -3
B 0
C 1
D 3

The correct answer is:
0

Solution:
x-1xx-1x3x+1x3
=x-1xx2-1x23
=x-1xx6-3x2+31x2-1x6
There is no term containing 1x and x in the second bracket.
So, the expansion will not contain any term independent of x.

Topic: Mathematics, Binomial Theorem, JEE Main
The term independent of $x(x>0, x \neq 1)$ in the expansion of $\left[\frac{(x+1)}{\left(x^{2 / 3}-x^{1 / 3}+1\right)}-\frac{(x-1)}{(x-\sqrt{x})}\right]^{10}$

Options:
A $105$
B $210$
C $315$
D $420$

The correct answer is:
$210$

Solution:
$\begin{aligned} & {\left[\frac{(x+1)}{\left(x^{2 / 3}-x^{1 / 3}+1\right)}-\frac{(x-1)}{(x-\sqrt{x})}\right]^{10}} \\ & \quad=\left[\frac{\left(x^{1 / 3}\right)^3+1^3}{\left(x^{2 / 3}-x^{1 / 3}+1\right)}-\frac{\left\{(\sqrt{x})^2-1\right\}}{\sqrt{x}(\sqrt{x}-1)}\right]^{10} \\ & \quad=\left[\frac{\left(x^{1 / 3}+1\right)\left(x^{2 / 3}-x^{1 / 3}+1\right)}{\left(x^{2 / 3}-x^{1 / 3}+1\right)}-\frac{\left\{(\sqrt{x})^2-1\right\}}{\sqrt{x}(\sqrt{x}-1)}\right]^{10} \\ & \quad=\left[\left(x^{1 / 3}+1\right)-\frac{(\sqrt{x}+1)}{\sqrt{x}}\right]^{10}=\left(x^{1 / 3}-x^{-1 / 2}\right)^{10}\end{aligned}$
The general term is
$$
\begin{aligned}
T_{r+1} & { }^{10} C_r\left(x^{1 / 3}\right)^{10-r}\left(-x^{-1 / 2}\right)^r \\
& ={ }^{10} C_r(-1)^r \cdot x^{\frac{10-r}{3}-\frac{r}{2}}
\end{aligned}
$$
For the term independent of $x$, put
$$
\begin{array}{cc}
& \frac{10-r}{3}-\frac{r}{2}=0 \\
\Rightarrow & 20-2 r-3 r=0 \\
\Rightarrow & 20=5 r \Rightarrow r=4 \\
\therefore & T_5={ }^{10} C_4=\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}=210
\end{array}
$$

Topic: Mathematics, Binomial Theorem, JEE Main
The term 'polyadelphous' is related to ------------

Options:
A gynoecium
B androecium
C corolla
D calyx

The correct answer is:
androecium

Solution:
The stamens in a flower may either remain free (polyandrous) or may be united in varying degrees. The stamens may be united into one bunch or one bundle (monoadelphous) as in China rose, or two bundles (diadelphous) as in pea, or into more than two bundles (polyadelphous) as in Citrus.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2016 (Phase 2)
The terminal velocity of a copper ball of radius $5 \mathrm{~mm}$ falling through a tank of oil at room temperature is $10 \mathrm{~cm} \mathrm{~s}^{-1}$. If the viscosity of oil at room temperature is $0.9 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-1}$, the viscous drag force is:

Options:
A $8.48 \times 10^{-3} \mathrm{~N}$
B $8.48 \times 10^{-5} \mathrm{~N}$
C $4.23 \times 10^{-3} \mathrm{~N}$
D $4.23 \times 10^{-6} \mathrm{~N}$

The correct answer is:
$8.48 \times 10^{-3} \mathrm{~N}$

Solution:
By Stobal's law, $\mathrm{F}=6 \pi \eta r v$
(where, the symbols have their usual meanings)
$\begin{aligned}
\mathrm{F} & =6 \times 3.14 \times 0.9 \times 5 \times 10^{-3} \times 10 \times 10^{-2} \\
\mathrm{~F} & =84.78 \times 10^{-4} \\
& =8.478 \times 10^{-3} \mathrm{~N}=8.48 \times 10^{-3} \mathrm{~N}
\end{aligned}$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The terminal velocity of a small sphere of radius $a$ in a viscous liquid is proportional to

Options:
A
$a^2$

B
$a^3$

C
$a$

D
$a^{-1}$

The correct answer is:
$a^2$

Solution:
Terminal velocity in a viscous medium is given by:
$$
\begin{aligned}
& V_T=\frac{2 a^2(\rho-\sigma) g}{9 \eta} \\
& \therefore V_T \propto a^2
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2012 (26 May Online)
The terminal velocity vt of the spherical rain drop depends on the radius r of the spherical rain drop as

Options:
A r
B r2
C 1r
D 1r2

The correct answer is:
r2

Solution:
The terminal velocity of a sphere of radius r and density ρ, immersed in a liquid of density σ and viscosity η is given by vT=29ρ-σr2gη

Hence, for spherical raindrop, vT∝r2.

Topic: Physics, Mechanical Properties of Fluids, JEE Main, JEE Main 2022 (25 Jun Shift 1)