datavorous/entrance-exam-dataset huggingface dataset:
The resultant of forces $\overrightarrow{\mathrm{P}}$ and $\overrightarrow{\mathrm{Q}}$ is $\overrightarrow{\mathrm{R}}$. If $\overrightarrow{\mathrm{Q}}$ is doubled then $\overrightarrow{\mathrm{R}}$ is doubled. If the direction of $\overrightarrow{\mathrm{Q}}$ is reversed, then $\overrightarrow{\mathrm{R}}$ is again doubled. Then $\mathrm{P}^2: \mathrm{Q}^2: \mathrm{R}^2$ is

Options:
A
$2: 3: 1$

B
$3: 1: 1$

C
$2: 3: 2$

D
$1: 2: 3$

The correct answer is:
$2: 3: 2$

Solution:
$\mathrm{R}^2=\mathrm{P}^2+\mathrm{Q}^2+2 \mathrm{PQ} \cos \theta$
$4 \mathrm{R}^2=\mathrm{P}^2+4 \mathrm{Q}^2+4 \mathrm{PQ} \cos \theta$
$4 \mathrm{R}^2=\mathrm{P}^2+\mathrm{Q}^2-2 \mathrm{PQ} \cos \theta$
$\mathrm{On}(1)+(2), 5 \mathrm{R}^2=2 \mathrm{P}^2+2 \mathrm{Q}^2$
$\mathrm{On}(3) \times 2+(2), 12 \mathrm{R}^2=3 \mathrm{P}^2+6 \mathrm{Q}^2$
$2 \mathrm{P}^2+2 \mathrm{Q}^2-5 \mathrm{R}^2=0$
$3 \mathrm{P}^2+6 \mathrm{Q}^2-12 \mathrm{R}^2=0$
$\frac{\mathrm{P}^2}{-24+30}=\frac{\mathrm{Q}^2}{24-15}=\frac{\mathrm{R}^2}{12-6}$
$\frac{\mathrm{P}^2}{6}=\frac{\mathrm{Q}^2}{9}=\frac{\mathrm{R}^2}{6}$ or $\mathrm{P}^2: \mathrm{Q}^2: \mathrm{R}^2=2: 3: 2$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2003
The resultant of two equal forces inclined at an angle $\alpha$ is half of resultant when they are inclined at the angle $\beta$, then

Options:
A $\cos \alpha=2 \cos \beta$
B $\cos \beta / 2=2 \cos \alpha / 2$
C $\cos \alpha / 2=2 \cos \beta / 2$
D More than one of the above

The correct answer is:
$\cos \beta / 2=2 \cos \alpha / 2$

Solution:
Option $2: \cos \beta / 2=2 \cos \alpha / 2$
Calculation:
According to the questions, resultant of two equal forces inclined at an angle $\alpha$ is half of resultant when they are inclined at the angle $\beta$.
Let $F$ be the force.
When angle between them is $\alpha$.
$\therefore$ Resultant $\mathrm{R}_1=\sqrt{\mathrm{F}^2+\mathrm{F}^2+2 \mathrm{~F}^2 \cos \alpha}$
When angle between them is $\beta$
$\therefore$ Resultant $\mathrm{R}_2=\sqrt{\mathrm{F}^2+\mathrm{F}^2+2 \mathrm{~F}^2 \cos \beta}$
According to the question:
$R_2=2 R_1$
$\Rightarrow R_2^2=4 R_1^2$
$\Rightarrow 2 \mathrm{~F}^2+2 \mathrm{~F}^2 \cos \beta=8 \mathrm{~F}^2+8 \mathrm{~F}^2 \cos \alpha$
$\Rightarrow 2 \mathrm{~F}^2(1+\cos \beta)=4 \times 2 \mathrm{~F}^2(1+\cos \alpha)$
$\Rightarrow(1+\cos \beta)=4(1+\cos \alpha)$
$\Rightarrow 2 \cos ^2(\beta / 2)=4 \times 2 \cos ^2(\alpha / 2)$
$\Rightarrow \cos (\beta / 2)=2 \cos (\alpha / 2)$

Topic: Physics, Mathematics in Physics, JEE Main
The resultant of two forces $\mathrm{P~N}$ and $3 \mathrm{~N}$ is a force of $7 \mathrm{~N}$. If the direction of $3 \mathrm{~N}$ force were reversed, the resultant would be $\sqrt{19} \mathrm{~N}$. The value of $\mathrm{P}$ is

Options:
A
$5 \mathrm{~N}$

B
$6 \mathrm{~N}$

C
$3 \mathrm{~N}$

D
$4 \mathrm{~N}$

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

Solution:
$\begin{aligned}
& 7^2=P^2+3^2+2 \times 3 \times P \cos \theta \quad \dots(1)\\
& (\sqrt{19})^2=P^2+(-3)^2+2 \times(-3) \times P \cos \theta \quad \dots (2)
\end{aligned}$
adding we get
$68=2 P^2+18 \Rightarrow P=5$.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2007
The resultant of two vectors is perpendicular to first vector of magnitude 6 N. If the resultant has magnitude 63 N, then magnitude of second vector is

Options:
A 62 N
B 12 N
C 93 N
D 63 N

The correct answer is:
12 N

Solution:
B Cos Î¸=6
B Sin Î¸=63
B=62+632=12

Topic: Physics, Mathematics in Physics, JEE Main
The resultant $R$ of two forces acting on a particle is at right angles to one of them and its magnitude is one third of the other force. The ratio of larger force to smaller one is

Options:
A
$2: 1$

B
$3: \sqrt{2}$

C
$3: 2$

D
$3: 2 \sqrt{2}$

The correct answer is:
$3: 2 \sqrt{2}$

Solution:
$\mathrm{F}^{\prime}=3 \mathrm{~F} \cos \theta$
$\mathrm{F}=3 \mathrm{~F} \sin \theta$
$\Rightarrow \mathrm{F}^{\prime}=2 \sqrt{2} \mathrm{~F}$
$\mathrm{~F}: \mathrm{F}^{\prime}:: 3: 2 \sqrt{2}$.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2005
The results given in the below table were obtained during kinetic studies of the following reaction: 2A+B→C+D

Experiment	A/molL-1	B/molL-1	Initial rate/molL-1min-1

I	0.1	0.1	6.00×10-3

II	0.1	0.2	2.40×10-2

III	0.2	0.1	1.20×10-2

IV	X	0.2	7.20×10-2

V	0.3	Y	2.88×10-1

X and Y in the given table are respectively :

Options:
A 0.4, 0.4
B 0.4, 0.3
C 0.3, 0.4
D 0.3, 0.3

The correct answer is:
0.3, 0.4

Solution:
Rate =k[A]a[B]b
from Exp 1 & 2 b=2
from Exp 1 & 3 a=1
fromExp 2 & 4⇒3=x0.11 so x=0.3

from Exp1 & 5⇒ 48=31y0.12
42=y0.12 so y=0.4

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2020 (02 Sep Shift 2)
The retardation experienced by a moving motorboat, after its engine is cut off, is given by dvdt=-kv3, where k is a constant. If v0 is the magnitude of the velocity at cut-off, the magnitude of the velocity at time t after the cut-off is

Options:
A v0
B v02
C v0e-kt
D v02v02kt+1

The correct answer is:
v02v02kt+1

Solution:
dvdt=-kv3
â -â«v0vdvkv3=â«0tdt
â1kâ«v0v-v-3dv=t or -1kv-3+1-3+1v0v=t
or 12k1v2-1v02=t or 1v2-1v02=2kt
or 1v2=1v02+2kt or 1v2=1+2v02ktv02
or v=v02v02kt+1

Topic: Physics, Motion In One Dimension, JEE Main
The reverse break down voltage of a Zener diode is 5.6 V in the given circuit.
The current Iz through the Zener is:

Options:
A 10 mA
B 7 mA
C 17 mA
D 15 mA

The correct answer is:
10 mA

Solution:
i1-iz=Vz800=5.6800=7 mA
For loop ABCF,
9-200 i1-Vz=0
i1=0.017 or 17 mA
iz=i1-7 mA=10 mA

Topic: Physics, Semiconductors, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement is incorrect?

Options:
A T3>T1
B T1=T2
C ΔUisothermal>ΔUadiabatic
D Wisothermal>wadiabatic

The correct answer is:
T3>T1

Solution:
In an isothermal process, temperature is constant
T2=T1
Area under P - V graph is more is isothermal
Wisothermal>wadiabatic
Now ΔUisothermal= 0
In adiabatic T1V1γ-1=T3V3γ-1
T3T1=V1V3γ-1
T3<T1 Hence internal energy decreases
ΔUisothermal>ΔUadiabatic

Topic: Chemistry, Thermodynamics (C), JEE Main
The right order of the solubility of sulphates of alkaline earth metals in water is

Options:
A $\mathrm{Be}\gt\mathrm{Ca}\gt\mathrm{Mg}\gt\mathrm{Ba}\gt\mathrm{Sr}$
B $\mathrm{Mg}\gt\mathrm{Be}\gt\mathrm{Ba}\gt\mathrm{Ca}\gt\mathrm{Sr}$
C $\mathrm{Be}\gt\mathrm{Mg}\gt\mathrm{Ca}\gt\mathrm{Sr}\gt\mathrm{Ba}$
D $\mathrm{Mg}\gt\mathrm{Ca}\gt\mathrm{Ba}\gt\mathrm{Be}\gt\mathrm{Sr}$

The correct answer is:
$\mathrm{Be}\gt\mathrm{Mg}\gt\mathrm{Ca}\gt\mathrm{Sr}\gt\mathrm{Ba}$

Solution:
$\mathrm{Be} > \mathrm{Mg} > \mathrm{Ca} > \mathrm{Sr} > \mathrm{Ba}$
On moving down the group lattice energy remains almost constant as the swlphate is so big that small increase in the size of the cations from Be to Ba does not make any difference. However the hydration energy decreases from $B e$ to $B a$. This causes decrease in the solubility of the swphates as the ionic size increases

Topic: Chemistry, s Block Elements, JEE Main
The rms speed of a gas molecule is ' $\mathrm{V}$ ' at pressure ' $\mathrm{P}$ '. If the pressure is increased by two time, then the rms speed of the gas molecule at the same temperature will be

Options:
A $\mathrm{V}$
B $\sqrt{2} \mathrm{~V}$
C $\frac{\mathrm{V}}{3}$
D $\frac{\mathrm{V}}{2}$

The correct answer is:
$\mathrm{V}$

Solution:
Since temperature remains constant, there will be no change in r.m.s velocity.

Topic: Physics, Kinetic Theory of Gases, MHT CET, MHT CET 2021 (24 Sep Shift 1)
The rms speed of \( \mathrm{N}_{2} \) molecules in a gas is \( \mathrm{u} \). If the temperature is doubled and the nitrogen molecules dissociate into nitrogen atoms, the rms speed becomes

Options:
A \( 2 \mathrm{u} \)
B \( 4 \mathrm{u} \)
C \( 14 u \)
D \( \sqrt{2} \mathrm{u} \)

The correct answer is:
\( 2 \mathrm{u} \)

Solution:
u1u2=m2T1m1T2=14×T28×2T=14 =12
∴ u2=2u1
Therefore, If the temperature is doubled and the nitrogen molecules dissociate into nitrogen atoms, the rms speed becomes double.

Topic: Chemistry, States of Matter, JEE Main
The rms speed of N2 molecules in a gas is u. If the temperature is doubled and the nitrogen molecules dissociate into nitrogen atoms, the rms speed becomes

Options:
A 2u
B 4u
C 14u
D 2 u

The correct answer is:
2u

Solution:
u1u2=n1T1n2T2=nÃT2nÃ2T=14 =12
â´ u2=2u1
Therefore, If the temperature is doubled and the nitrogen molecules dissociate into nitrogen atoms, the rms speed becomes double.

Topic: Chemistry, States of Matter, JEE Main
The rms speed of oxygen is $v$ at a particular temperature. If the temperature is doubled and oxygen molecules dissociate into oxygen atoms, the rms speed becomes

Options:
A v
B $\sqrt{2} v$
C $2v$
D $4 v$

The correct answer is:
$2v$

Solution:
As the rms speed is given by
$$
\begin{aligned}
& v_{\text {rms }}=\sqrt{\frac{3 R T}{M}}{V_{\text {ms }}} \propto \sqrt{\frac{T}{M}}
\end{aligned}
$$
When temperature is doubled and molecules dissociates into atoms, then
$$
\begin{aligned}
\frac{v_{\mathrm{rms}_{1}}}{v_{\mathrm{rms}_{2}}} &=\frac{\sqrt{\frac{T}{M}}}{\sqrt{\frac{2 T}{M / 2}}}=\frac{\sqrt{\frac{T}{M}}}{\sqrt{\frac{4 T}{M}}} \\
&=\frac{\sqrt{T}}{\sqrt{4 T}}=\frac{1}{2}
\end{aligned}
$$
If $v_{\mathrm{rms}},$ is $v,$ then $v_{\mathrm{rms}_{2}}$ will be $2 v$.

Topic: Physics, Kinetic Theory of Gases, JEE Main
The R.M.S. speed of oxygen molecules at temperature T (in kelvin) is v m s-1. As the temperature becomes 4T and the oxygen gas dissociates into atomic oxygen, then the speed of atomic oxygen

Options:
A remains the same
B becomes 2v
C becomes 2v
D becomes 22v

The correct answer is:
becomes 22v

Solution:
v∝γ1T
v2∝2×γ2T2=2×γ2×4T ∵T2=4T
v2v1=22γ2γ1
v2v1=25×5×23×7
v2=25021v
v2≅22v

Topic: Physics, Kinetic Theory of Gases, NEET
The R.M.S. speeds of the molecules of Hydrogen, Oxygen, and Carbon dioxide at the same temperature are vH, vO and vC respectively, then:

Options:
A vC>vO>vH
B vH=vO>vC
C vH>vO>vC
D vH=vO=vC

The correct answer is:
vH>vO>vC

Solution:
The formula of root-mean-square speed in terms of absolute temperature and molecular mass of the gas is, vRMS=3RTM⇒vRMS∝1M, here we know that, MH2<MO2<MCO2⇒vH>vO>vC.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The RMS value of AC which when passed through a resistor produces heat, which is twice that produced by a steady current of 1.414 A in the same resistor is

Options:
A 2 A
B 3.46Â A
C 2.818Â A
D 1.732Â A

The correct answer is:
2 A

Solution:
It is given that the amount of heat produced by AC current is twice of the amount of heat produced by DC current.
I=1.414 A
∴ Irms2⋅R=2×I2R
Irms2=2×1.4142=2×2
⇒ Irms=2 A

Topic: Physics, Alternating Current, JEE Main
The rms value of potential difference $V$ shown in the figure is

Options:
A $V_0$
B $\frac{V_0}{\sqrt{2}}$
C $\frac{V_0}{2}$
D $\frac{V_0}{\sqrt{3}}$

The correct answer is:
$\frac{V_0}{\sqrt{2}}$

Solution:
$V_{\mathrm{rms}}=\left[\frac{1}{T} \int_0^{T / 2} V_0^2 d t\right]^{1 / 2}$
$$
\begin{aligned}
& =\left[\frac{V_0^2}{T}[t]_0^{T / 2}\right]^{1 / 2} \\
& =\left[\frac{V_0^2}{T}\left(\frac{T}{2}\right)\right]^{1 / 2} \\
V_{\mathrm{rms}} & =\left[\frac{V_0^2}{2}\right]^{1 / 2} \\
V_{\mathrm{mms}} & =\frac{V_0}{\sqrt{2}}
\end{aligned}
$$

Topic: Physics, Alternating Current, NEET, NEET 2011 (Mains)
The rms value of the electric field of the light coming from the Sun is $720 \mathrm{~N} / \mathrm{C}$. The average total energy density of the electromagnetic wave is

Options:
A
$3.3 \times 10^{-3} \mathrm{~J} / \mathrm{m}^3$

B
$4.58 \times 10^{-6} \mathrm{~J} / \mathrm{m}^3$

C
$6.37 \times 10^{-9} \mathrm{~J} / \mathrm{m}^3$

D
$81.35 \times 10^{-12} \mathrm{~J} / \mathrm{m}^3$

The correct answer is:
$4.58 \times 10^{-6} \mathrm{~J} / \mathrm{m}^3$

Solution:
$U_{a v}=\varepsilon_0 E_{\text {rs }}^2=4.58 \times 10^{-6} \mathrm{~J} / \mathrm{m}^3$
$E_{\text {rms }}=4.58 \times 10^{-6} \mathrm{~J} / \mathrm{m}^3$

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2006
The rms velocity at NTP of the species can be calculated from the expression

Options:
A $\sqrt{\frac{3 P}{d}}$
B $\sqrt{\frac{3 P V}{M}}$
C $\sqrt{\frac{3 R T}{M}}$
D All the above

The correct answer is:
$\sqrt{\frac{3 R T}{M}}$

Solution:
Correct option is C. $\sqrt{\frac{3 R T}{M}}$
The $r m s$ velocity at $N T P$ of the species can be calculated from the expression $\sqrt{\frac{3 R T}{M}}$.
Hence, Option " $\mathrm{C}$ " is the correct answer.

Topic: Chemistry, States of Matter, JEE Main
The r.m.s. voltage of the wave form shown is

Options:
A $10 \mathrm{~V}$
B $7 \mathrm{~V}$
C $6.37 \mathrm{~V}$
D None of these

The correct answer is:
$10 \mathrm{~V}$

Solution:
$V_{\text {ms }}=\sqrt{\frac{1}{T} \int_0^T 10^2 d t}=10 \mathrm{~V}$

Topic: Physics, Alternating Current, JEE Main
The role of depressants in 'Froth Floation method' is to

Options:
A selectively prevent one component of the ore from coming to the froth.
B reduce the consumption of oil for froth formation.
C stabilize the froth.
D enhance non-wettability of the mineral particles.

The correct answer is:
selectively prevent one component of the ore from coming to the froth.

Solution:
In the froth floatation process, the role of the depressants is to separate two sulphide ores by selectively preventing one ore from forming froth.

For example, to separate two sulphide ores, PbS and ZnS,NaCN is utilised. NaCN forms a combination with zinc on the surface of zinc sulphide, Na2ZnCN4, preventing ZnS from foaming selectively.

4NaCN + ZnS⇒Na2ZnCN4+Na2S

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2022 (26 Jun Shift 2)
The root mean square of gas molecules at 25 K and 1.5 bar is 100 ms-1 . If the temperature is raised to 100 K and the pressure to 6.0 bar, the root mean square speed becomes

Options:
A 200 ms-1
B 100Â ms-1
C 400Â ms-1
D 1600Â ms-1

The correct answer is:
200 ms-1

Solution:
u rms u rms ′ = 3RT M 3RT' M = T T'
So 100 m/surmsâ²=25 K100 K=12
or urmsâ²=2Ã100 ms-1
=200 ms-1 .

Topic: Chemistry, States of Matter, JEE Main
The root mean square rms speed of oxygen molecules (O2) at a certain temperature T (on kelvin scale) is V. If the temperature is doubled and oxygen gas dissociates into atomic oxygen, the rms speed becomes

Options:
A two times
B three times
C four times
D unchanged

The correct answer is:
two times

Solution:
V rms = 3RT M
V rms ∝ T M ...(i)
When oxygen gas dissociates into atomic oxygen, its
atomic mass M will become half.
Temperature is doubled.
So, from Eq. (i) Vrms will become two times.
Vrms'Vrms=2TM2TM
= 4 =2
⇒Vrms'=2Vrms

Topic: Physics, Thermodynamics, JEE Main
The root mean square speed of molecules of nitrogen gas at 27°C is approximately: (Given mass of a nitrogen molecule =4.6×10-26 kg and take Boltzmann constant kB=1.4×10-23J K-1 )

Options:
A 27.4 m s-1
B 91 m s-1
C 1260 m s-1
D 523 m s-1

The correct answer is:
523 m s-1

Solution:
The RMS value of a gas is given by

vrms=3RTM=3kBTm

Substituting the given values,

vrms=3×1.4×10-23×3004.6×10-26
≈523 m s-1

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2023 (11 Apr Shift 2)
The root mean square speed of N2 molecules in sample at temperature T is 'x'. If the temperature is doubled, then nitrogen molecules dissociate into atoms, the root mean square speed of nitrogen atoms becomes n times of 'x' find the value of n here?

Options:

The correct answer is:
2

Solution:
vrms=3RTm
x=3RT28 …(i)
vrms=3R×2T14 …(ii)
Dividing equation (i) by equation (ii)
xvrms=14
So, vrms=2x
So, n = 2.

Topic: Chemistry, States of Matter, JEE Main
The root mean square speed of smoke particles of mass 5×10-17 kg in their Brownian motion in air at NTP is approximately.

[Given k=1.38×10-23J K-1]

Options:
A 60 mm s-1
B 12 mm s-1
C 15 mm s-1
D 36 mm s-1

The correct answer is:
15 mm s-1

Solution:
According to Kinetic theory,

Average kinetic energy of a gas molecule =12mvrms2=32kT

Thus, vrms=3kTm

At NTP, T=293 K

Putting the values, we have

=3×1.38×10-23×2935×10-17

≈15 mm s-1

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2022 (29 Jul Shift 2)
The root mean square velocity of molecules of gas is

Options:
A Proportional to square of temperature (T2).
B Inversely proportional to square root of temperature (1T).
C Proportional to square root of temperature T.
D Proportional to temperature (T).

The correct answer is:
Proportional to square root of temperature T.

Solution:
Root mean speed of a gas molecule is stated as vrms=3RTM, where, R is the universal gas constant, T is temperature, and M is the molar mass of the gas.

Thus, vrms∝T.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The root mean square velocity of one mole of a monoatomic having molar mass $\mathrm{M}$ is $\mathrm{U}_{\text {rak. }}$. The relation between the average kinetic energy (E) of the $\mathrm{U}_{\mathrm{mms}}$ is

Options:
A $U_{m s}=\sqrt{\frac{3 E}{2 M}}$
B $U_{\text {ms }}=\sqrt{\frac{2 E}{3 M}}$
C $U_{r m s}=\sqrt{\frac{2 E}{M}}$
D $U_{\mathrm{ms}}=\sqrt{\frac{E}{3 M}}$

The correct answer is:
$U_{r m s}=\sqrt{\frac{2 E}{M}}$

Solution:
$U_{r m s}=\sqrt{\frac{2 E}{M}}$

Topic: Chemistry, States of Matter, JEE Main
The root of the equation

$2(1+i) x^{2}-4(2-i) x-5-3 i=0$ which has

greater modulus is

Options:
A $\frac{3-5 i}{2}$
B $\frac{5-3 \mathrm{i}}{2}$
C $\frac{3-i}{2}$
D none

The correct answer is:
$\frac{3-5 i}{2}$

Solution:
$\begin{aligned} \text { Roots } &=\frac{4(2-i) \pm \sqrt{16(2-i)^{2}+8(1+i)(5+3 i)}}{4(1+i)} \\ &=\frac{4-i}{1+i} \text { or } \frac{-i}{1+i}=\frac{3-5 i}{2} \text { or } \frac{-1-i}{2} \end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
The root-mean-square speed of molecules of a given mass of a gas at 27 °C and 1 atmosphere pressure is 200 m s-1. The root-mean-square speed of molecules of the gas at 127 °C and 2 atmosphere pressure is x3 m s-1. The value of x will be __________.

Options:

The correct answer is:
400

Solution:
vrms=3RTM⇒vrms∝T⇒vrms2vrms1=T2 T1=400300=23

vrms2=23vrms1=23×200⇒vrms2=4003 m s-1.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (24 Feb Shift 2)
The roots of the equation $\left(a^2+b^2\right) t^2-2(a c+b d) t+\left(c^2+d^2\right)=0$ are equal, then

Options:
A $a b=d c$
B $a c=b d$
C $a d+b c=0$
D $\frac{a}{b}=\frac{c}{d}$

The correct answer is:
$\frac{a}{b}=\frac{c}{d}$

Solution:
Accordingly, $\{2(a c+b d)\}^2=4\left(a^2+b^2\right)\left(c^2+d^2\right)$ (as \(D=0\) or \(B^2=4 A C\) for equal roots)
$\Rightarrow 4 a^2 c^2+4 b^2 d^2+8 a b c d=4 a^2 c^2+4 a^2 d^2$ $+4 b^2 c^2+4 b^2 d^2$
$\Rightarrow 4 a^2 d^2+4 b^2 c^2-8 a b c d=0 \Rightarrow 4(a d-b c)^2=0$
$\Rightarrow \quad a d=b c \Rightarrow \frac{a}{b}=\frac{c}{d}$

Topic: Mathematics, Quadratic Equation, JEE Main
The roots of the equation $x^4-4 x^3+6 x^2-4 x+1=0$ are

Options:
A $1,1,1,1$
B $2,2,2,2$
C $3,1,3,1$
D $1,2,1,2$

The correct answer is:
$1,1,1,1$

Solution:
Let $f(x)=x^4-4 x^3+6 x^2-4 x+1=0$
$\Rightarrow(x-1)\left(x^3-3 x^2+3 x-1\right)=0$
$\Rightarrow(x-1)(x-1)^3=0 \Rightarrow x=1,1,1,1$

Topic: Mathematics, Quadratic Equation, JEE Main
The roots of
$\begin{aligned}
(x-a)(x-a-1)+(x-a-1)(x-a-2) & \\
&+(x-a)(x-a-2)=0
\end{aligned}$
$a \in R$ are always

Options:
A equal
B imaginary
C real and distinct
D rational and equal

The correct answer is:
real and distinct

Solution:
$\begin{aligned}(x-a)(x-a-1)+&(x-a-1)(x-a-2) \\ &+(x-a)(x-a-2)=0 \end{aligned}$
Let $x-a=t$, then
$$
\begin{array}{l}
t(t-1)+(t-1)(t-2)+t(t-2)=0 \\
\Rightarrow t^{2}-t+t^{2}-3 t+2+t^{3}-2 t=0 \\
\Rightarrow 3 t^{2}-6 t+2=0 \\
\Rightarrow t=\frac{6 \pm \sqrt{36-24}}{2(3)}=\frac{6 \pm 2 \sqrt{3}}{2(3)} \\
\Rightarrow x-a=\frac{3 \pm \sqrt{3}}{3} \\
\Rightarrow x=a+\frac{3 \pm \sqrt{3}}{3}
\end{array}
$$
Hence, $x$ is real and distinct.

Topic: Mathematics, Quadratic Equation, JEE Main
The roots that originate from the base of the stem are

Options:
A Primary roots
B Prop roots
C Lateral roots
D Fibrous roots

The correct answer is:
Fibrous roots

Solution:
In the majority of dicot plants, the root system is formed by the direct elongation of the radicle into the primary root. Branches of the primary root are called secondary roots. These in turn branch to form tertiary roots. The primary root and its branches constitute the taproot system.

Prop roots: In trees like the banyan and Indian rubber plant, many adventitious roots arise from the horizontal branches of the stem and grow vertically downwards. They become thick pillar-like and provide mechanical support to the huge trees.

Stilt/Fibrous roots: These are small roots growing obliquely from the basal nodes of the main stem. Like prop roots, they also provide mechanical support.

Topic: Biology, Morphology of Flowering Plants, NEET, NEET 2020 (Phase 1)
The rotation of the Earth (of radius $R$ ) about its axis speeds up to a value such that a man at latitude angle $45^{\circ}$ feels weightlessness. The duration of a day in such a case is

Options:
A $\pi \sqrt{\frac{R}{g}}$
B $\pi \sqrt{\frac{2 R}{g}}$
C $\frac{\pi}{2} \sqrt{\frac{R}{g}}$
D $\pi \sqrt{\frac{g}{R}}$

The correct answer is:
$\pi \sqrt{\frac{2 R}{g}}$

Solution:
Given that, angle of latitude, $\lambda=45^{\circ}$
We know that, the expression of gravity at latitude angle is
$$
g_\lambda=g-\omega^2 R \cos ^2 \lambda
$$
The weight of man of mass $m$ at this point
$$
w=m g_\lambda=m\left(g-\omega^2 R \cos ^2 45^{\circ}\right)
$$
According to question, man feels weightlessness
$$
\therefore \quad \begin{aligned}
w & =0 \\
m\left(g-\omega^2 R \cos ^2 45^{\circ}\right) & =0
\end{aligned}
$$
But $m \neq 0$, then $g-\omega^2 R \cos ^2 45^{\circ}=0$
$$
\Rightarrow \quad \omega=\sqrt{\frac{2 g}{R}}
$$
Now, using time period, (duration of day)
$$
=T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{R}{2 g}}=\pi \sqrt{\frac{2 R}{g}}
$$

Topic: Physics, Gravitation, JEE Main
The ruptere and fractionation do not usually occur in water column in vessel/trachieds during the ascent of sap because of

Options:
A lignified thick walls
B cohesion and adhesion
C weak grauitational pull
D transpiration pull

The correct answer is:
cohesion and adhesion

Solution:
The vertical conduction of water from root to arial part of plant is called ascent of sap. The water molecules remain joined to each other due to a force of attraction calledcohesion force.
Attraction between water molecules and the walls of xylem is due to adhesion forces. These forces help to ensure the continuity of water column in xylem.

Topic: Biology, Transport in Plants, JIPMER, JIPMER 2014
The rupture and fractionation do not usually occur in the water column in vessel/tracheids during the ascent of sap because of

Options:
A lignified thick walls
B cohesion and adhesion
C weak gravitational pull
D transpiration pull

The correct answer is:
cohesion and adhesion

Solution:
The vertical conduction of water from root to aerialparts of plant is called ascent of sap. The water molecules remain joined to each other due to a force of attraction called cohesion force. This attraction is due to the presence of hydrogen bonds between them. The magnitude of this force is very high therefore, the continuous water column in the xylem cannot be broken easily due to the force of gravity or other obstructions offered by internal tissues in the upward movement of water. This adhesive property of water, ie, attraction between the water molecules and the walls of xylem ensures the continuity of water column in xylem.
Due to transpiration water evaporates to outer atmosphere through stomata. As a result leaf cells develop low water potential and water from leaf veins moves into leaf cells. The xylem vessel in turn draw water from xylem of main stem. A pressure (pull) is thus, exerted by all the leaves on the stems, which is called transpiration pull. This is strong enough to pull up the column of water to a great height.

Topic: Biology, Anatomy of Flowering Plants, NEET, NEET 2008 (Screening)