datavorous/entrance-exam-dataset huggingface dataset:
The vapour pressure of water at T (K) is 20 mm Hg. The following solutions are prepared at T (K)
I. 6 g of urea (molecular weight = 60) is dissolved in 178.2 g of water.
II. 0.01 mol of glucose is dissolved in 179.82 g of water.
III. 5.3 g of Na2CO3 (molecular weight = 106) is dissolved in 179.1 g of water.
Identify the correct order in which the vapour pressure of solutions increases

Options:
A III < I < II
B II < III < I
C I < II < III
D I < III < II

The correct answer is:
III < I < II

Solution:
I. XB=nBnA+nB
=660178.218+660=0.01
ΔPP0=XB=0.01
II. XB=nBnA+nB
=0.01179.8218+0.01=0.001
ΔPP0=0.001
III. XB=nBnA+nB
=5.3106179.118+5.3106=0.005
ΔPP0=iXB
=3×0.005=0.015
So vapour pressure of solution will increase in the following sequence.
III < I < II

Topic: Chemistry, Solutions, JEE Main
The vapour pressure vs. temperature curve for a solution solvent system is shown below.

The boiling point of the solvent is _____°C.

Options:

The correct answer is:
82

Solution:
We know that the temperature at which vapour pressure of a liquid becomes 1 atm. is called the boiling point of the liquid.

From the figure, we can clearly see that Graph I is for pure solvent whereas graph II is for solution. Hence the boiling point of solvent and solution are 82°C and 83°C respectively.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2023 (08 Apr Shift 1)
The vapour pressures of A and B at 25°C are 90 mmHg and 15 mm Hg respectively. If A and B are mixed such that the mole fraction of A in the mixture is 0.6, then the mole fraction of B in the vapour phase is x×10-1. The value of x is (Nearest integer)

Options:

The correct answer is:
1

Solution:
Given PA°=90 mmHg, at 25°C

PB=∘15 mmHg

and XA=0.6XB=0.4PT=XAPAo+XBPBo

=(0.6×90)+(0.4×15)

=54+6=60 mm

Now mol fraction of B in the vapour phase

i.e. YB=PBPT=XBPBo60=0.1=1×10-1

therefore: x=1

Topic: Chemistry, Solutions, JEE Main, JEE Main 2021 (20 Jul Shift 2)
The vapour pressures of pure liquids A and Bare 400 and 600 mm Hg respectively at 298 K. On mixing the two liquids, the sum of their volumes is equal to the volume of the final mixture. The mole fraction of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fractions of components A and B in the vapour phase, respectively are

Options:
A 500 mm Hg, 0.5, 0.5
B 450 mm Hg, 0.4, 0.6
C 450 mm Hg, 0.5, 0.5
D 500 mm Hg, 0.4, 0.6

The correct answer is:
500 mm Hg, 0.4, 0.6

Solution:
Raoult’s law for ideal solution,

PTotal=PA+ PB

PTotal=xA×PA∘+ xB×PB∘

PTotal=12×400+12×600=500 mm of Hg

Dalton’s law,

yA×PTotal=PA

yA=xA×PA0PTotal=12×400500=0.4

yB=1-0.4=0.6

Topic: Chemistry, Solutions, JEE Main, JEE Main 2019 (08 Apr Shift 1)
The vapour pressures of pure liquids A and B are 400 and 600 mmHg respectively at 298 K. On mixing the two liquids, the sum of their initial volumes is equal to the volume of the final mixture. The mole fractional of liquid B is 0.5 in the mixture. The vapour pressure of the final solution, the mole fraction of components A and B in vapour phase, respectively are

Options:
A 500 mmHg, 0.5, 0.5
B 450 mmHg, 0.4, 0.6
C 450 mmHg, 0.5, 0.5
D 500 mmHg, 0.4, 0.6

The correct answer is:
500 mmHg, 0.4, 0.6

Solution:
Let XA and XB be the mole fraction of liquid A and B in the mixture.
Given, PAÂ°=400 mm Hg, PBÂ°=600 mm Hg
P Total = X A . P A Â° + X B . P B Â°
=0.5Ã400+0.5Ã600
=500 mm Hg
Now, mole fraction of A in vapour,
YA=PAPtotal=0.5Ã400500=0.4 âµ[PA=XAPAÂ°]
and mole fraction of B in vapour, YB=1-0.4=0.6

Topic: Chemistry, Solutions, JEE Main
The vapour pressures of two volatile liquids A and B at 25 °C are 50 Torr and 100 Torr, respectively. If the liquid mixture contains 0.3 mole fraction of A, then the mole fraction of liquid B in the vapour phase is x17. The value of x is

Options:

The correct answer is:
14

Solution:
Since,

yA=p°A×xAp°A×xA+p°B×xB

and xA=0.3 and xB=0.7

yA=50×0.350×0.3+100×0.7

yA=1585.

YB1-yB=PB0PA0XB1-XB

⇒yB1-yB=100500.70.3=143

⇒yB=1417

Topic: Chemistry, Solutions, JEE Main, JEE Main 2022 (28 Jun Shift 1)
The variance and mean of 15 observations are respectively 6 and 10 . If each observation is increase by 8 then the new variance and new mean of resulting observations are respectively

Options:
A 14,10
B 14,18
C 6,18
D 6,10

The correct answer is:
6,18

Solution:
The new mean will be increased by 8
i.e., new mean becomes $10+8=18$
But, new variance will remain unchanged
Hence, new variance $=6$

Topic: Mathematics, Statistics, JEE Main
The variance and mean of 15 observations are respectively 6 and 10 . If each observation is increase by 8 then the new variance and new mean of resulting observations are respectively

Options:
A 14,10
B 14,18
C 6,18
D 6,10

The correct answer is:
6,18

Solution:
The new mean will be increased by 8
i.e., new mean becomes $10+8=18$
But, new variance will remain unchanged
Hence, new variance $=6$

Topic: Mathematics, Statistics, MHT CET, MHT CET 2022 (06 Aug Shift 2)
The variance of 20 observations is 5 . If each one of the observations is multiplied by 2 , then the variance of the resulting observations is

Options:
A 40
B 80
C 20
D 10

The correct answer is:
20

Solution:
Variance $=5$ and $n=20$
$\because \quad$ Variance $=5=\frac{1}{n} \Sigma(x i-\bar{x})^2$
$$
\Rightarrow \quad \Sigma(x i-\bar{x})^2=100...(1)
$$
Let after multiplication, the new observations be $y_1, y_2 \ldots y_{20}$ Where $y_i=2(x i)$ for some $i$...(2)
New variance $=\frac{1}{2} \Sigma\left(y_i-y\right)^2$ where $\bar{y}=\frac{1}{n} \Sigma y_i$
$$
\Rightarrow \bar{y}=\frac{1}{20} \Sigma 2 x i=2\left(\frac{1}{20} \Sigma x i\right)=2 \bar{x}...(3)
$$
$\because \quad \Sigma(x i-\bar{x})^2=100$
\{from (1)\}
$\Rightarrow \quad \Sigma\left(\frac{1}{2} y_i-\frac{1}{2} \bar{y}\right)^2=100$
$\Rightarrow\left(\frac{1}{2}\right)^2 \Sigma\left(y_i-y\right)^2=100 \Rightarrow \Sigma\left(y_i-y\right)^2=400$
$\therefore$ New variance $=\frac{1}{n} \Sigma(y i-\bar{y})^2=\frac{1}{20} \times 400=20$

Topic: Mathematics, Statistics, JEE Main
The variance of first 10 multiples of 3 is

Options:
A $74.15$
B $73.15$
C $74.25$
D $70.15$

The correct answer is:
$74.25$

Solution:
Variance
$\begin{aligned} & =\frac{\sum x_i^2}{n}-(\bar{x})^2=\frac{3^2+6^2+9^2+\ldots . .+30^2}{10}-\left(\frac{3+6+9+\ldots \ldots+30}{10}\right)^2 \\ & =\frac{3^2\left(1^2+2^2+3^2 \ldots .0^2\right)}{10}-3^2\left(\frac{1+2+3+\ldots . .+10}{10}\right)^2 \\ & =\frac{9 \times 10 \times 11 \times 21}{6 \times 10}-9 \times\left(\frac{10 \times 11}{2 \times 10}\right)^2=\frac{33 \times 21}{2}-\frac{9 \times 121}{4} \\ & =\frac{99}{4}(14-11) \\ & =\frac{99 \times 3}{4}=24.75 \times 3=74.25\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
The variance of first 20 natural numbers is

Options:
A $133 / 4$
B $279 / 12$
C 133/2
D 399/4

The correct answer is:
$133 / 4$

Solution:
Variance of $n$ natural numbers $=\frac{n^{2}-1}{12}=\frac{(20)^{2}-1}{12} \quad$ $[\because n=20]$,
$=\frac{400-1}{12}=\frac{399}{12}=\frac{133}{4}$

Topic: Mathematics, Statistics, JEE Main
The variance of the data $2,4,6,8,10$ is

Options:
A $6$
B $7$
C $8$
D None of these

The correct answer is:
$8$

Solution:
Here, $\bar{x}=\frac{2+4+6+8+10}{5}=6$
Hence, variance $=\frac{1}{n} \Sigma\left(x_i-\bar{x}\right)^2$
$=\frac{1}{5}\left\{(2-6)^2+(4-6)^2+(6-6)^2+(8-6)^2+(10-6)^2\right\}$
$=\frac{1}{5}\left\{(16+4+0+4+16\}=\frac{1}{5}\{40\}=8\right.$.

Topic: Mathematics, Statistics, JEE Main
The variance of the first 20 positive integral multiples of 4 is equal to

Options:
A 532
B 133
C 266
D 600

The correct answer is:
532

Solution:
The numbers are 4,8,12........80
Mean, (xÂ¯)=41+2+3+......2020=42
Î£xi2=4212+22+......202
=16Ã20Ã21Ã416=45920
So, the variance is =Î£xi220-(x-)2=2296-1764=532

Topic: Mathematics, Statistics, JEE Main
The variance of the first 50 even natural numbers is :

Options:
A 4 3 7
B 4 3 7 4
C 8 3 3 4
D 8 3 3

The correct answer is:
8 3 3

Solution:
Variance of a set of n numbers x1, x2 ...xi, ...xn is given
σ2=1nΣi= 1nxi-x¯2, where x¯ is the average of the numbers.
We want the variance of 2, 4, 6, 8 ..... 48, 50, 52, ..... 98, 100
Here x¯=51.
xi-x¯ will be symmetric about 51,
∴σ2=25012+32+52+... 492
Now, 12 + 32 + 52 + ... + 492 = 12 + 22 + 32 + ... + 492 + 502 - 22 + 42 + 62 + ... + 482 + 502
=50511016-412+22+
=50511016-42526516
= 42925 - 22100
= 20825
∴σ2=250×20825=833

Topic: Mathematics, Statistics, JEE Main, JEE Main 2014 (06 Apr)
The variance of the first $n$ natural numbers is

Options:
A $\frac{n^2-1}{12}$
B $\frac{n^2-1}{6}$
C $\frac{n^2+1}{6}$
D $\frac{n^2+1}{12}$

The correct answer is:
$\frac{n^2-1}{12}$

Solution:
Variance $=(\text { S.D. })^2=\frac{1}{n} \Sigma x^2-\left(\frac{\Sigma x}{n}\right)^2,\left(\because \bar{x}=\frac{\Sigma x}{n}\right)$
$=\frac{n(n+1)(2 n+1)}{6 n}-\left(\frac{n(n+1)}{2 n}\right)^2=\frac{n^2-1}{12}$.

Topic: Mathematics, Statistics, JEE Main
The variance of the following data is
$$
\begin{array}{cccccccc}
x_i & 6 & 10 & 14 & 18 & 24 & 28 & 30 \\
f_i & 2 & 4 & 7 & 12 & 8 & 4 & 3
\end{array}
$$

Options:
A 33.4
B 34.3
C 43.4
D 44.3

The correct answer is:
43.4

Solution:
$\begin{aligned} & \text { Now, variance }=\frac{\sum f_i d_i^2}{N}-\left(\frac{\sum f_i d_i}{N}\right)^2 \\ & =\frac{1}{40} \times 1776-1=\frac{1736}{40}=43.4\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
The variance of the following frequency distribution is

Options:
A $\frac{295}{16}$
B $\frac{304}{16}$
C $\frac{37}{4}$
D $\frac{97}{4}$

The correct answer is:
$\frac{295}{16}$

Solution:
Given data,
Here, $\quad A=10, h=4$
$$
\begin{aligned}
& \sigma^2=h^2\left(\frac{\Sigma f_i d_i^2}{\Sigma f_i}-\left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)^2\right)=16\left(\frac{19}{16}-\frac{9}{256}\right) \\
& \sigma^2=16\left(\frac{304-9}{256}\right)=\frac{295}{16}
\end{aligned}
$$

Topic: Mathematics, Statistics, JEE Main
The variance of the numbers 2,3,11 and x is 494, then the values of x are

Options:
A 6 or143
B 6 or 145
C 6 or 163
D 4 or 135

The correct answer is:
6 or143

Solution:
From the given data, we make the following table

x	x2

2	4

3	9

11	121

x	x2

âx=16+x	âx2=134+x2

But we know that, variance =âx2n-âxn2
â134+x24-16+x42=494 (given)
â134+x24-256+x2+32x16=494
â3x2-32x+28016=494
â3x2-32x+280=196â3x2-32x+84=0
âx-63x-14=0âx=6,x=143
Therefore, the values of x are 6 or 143

Topic: Mathematics, Statistics, JEE Main
The variation of acceleration due to gravity g with distance d from the centre of the earth is best represented by (R= Earth's radius):

Options:
A
B
C
D

The correct answer is:

Solution:
We know that acceleration due to gravity is given by,
g= GMrR3,0≤r<RGMr2, r≥R
∴ The desired graph is,

Topic: Physics, Gravitation, JEE Main, JEE Main 2017 (02 Apr)
The variation of applied potential and current flowing through a given wire is shown in figure. The length of wire is 31.4 cm. The diameter of wire is measured as 2.4 cm. The resistivity of the given wire is measured as x×10-3 Ω cm. The value of x is _____ .

[Take π=3.14]

Options:

The correct answer is:
144

Solution:
Since, slope of V−I graph represents the resistance, so R=tan45°=1 Ω

Now, resistance of wire is given by R=ρLA

So, resistivity of the wire is ρ=ARL=πr2RL=π1.22×10-431.4×10-2=1.44×10-3 Ω m=144×10-3Ω cm

Topic: Physics, Current Electricity, JEE Main, JEE Main 2022 (29 Jun Shift 1)
The variation of decay rate with number of active nuclei is correctly shown in graph

Options:
A D
B C
C A
D B

The correct answer is:
A

Solution:
By law of radioactive decay
$$
-\frac{\mathrm{dN}}{\mathrm{dt}} \propto \lambda \mathrm{N}
$$
$\mathrm{N}$ is the number of radioactive nuclei in the sample
So, $\frac{\mathrm{dN}}{\mathrm{dt}}$ can be negative
The variation of decay rate with number of active nuclei is shown by the graph

Topic: Physics, Nuclear Physics, JEE Main
The variation of equilibrium constant with temperature is given below :

TemperatureEquilibriumConstantT1=25°CK1=10T2=100°CK2=100

The values of ΔH°, ΔG° at T1 and ΔG° at T2 (in kJ mol-1 ) respectively, are close to [use R=8.314JK-1mol-1]

Options:
A 28.4,-7.14 and -5.71
B 0.64,-7.14 and -5.71
C 28.4,-5.71 and -14.29
D 0.64,-5.71 and -14.29

The correct answer is:
28.4,-7.14 and -5.71

Solution:
T1=323K T2=373K
k1=10 k2=100
logk2k1=ΔH2.303R1T1-1T2

log10010=ΔH2.303×8.3141298-1373

log10=ΔH2.303×8.31475298×373

ΔH=2.303×8.314×298×37375=28.4KJ

At T1=25°C=298K,K1=10

ΔG=-2.303RT1logk 1

=-2.303×8.314×298×log(10)

=-2.303×8.314×298×1

=-5.7KJ

At T2=100°C=373K K2=100

ΔG=-2.303RT2logK2

=-2.303×8.314×373×log(10)2

=-2.303×2×8.314×373×1

=-14283.7J

=-14.29KJ

Topic: Chemistry, Thermodynamics (C), JEE Main
The variation of equilibrium constant with temperature is given below :

TemperatureEquilibriumConstantT1=25°CK1=10T2=100°CK2=100

The values of ΔH°, ΔG° at T1 and ΔG° at T2 (in kJ mol-1 ) respectively, are close to [use R=8.314JK-1mol-1]

Options:
A 28.4,-7.14 and -5.71
B 0.64,-7.14 and -5.71
C 28.4,-5.71 and -14.29
D 0.64,-5.71 and -14.29

The correct answer is:
28.4,-7.14 and -5.71

Solution:
T1=323K T2=373K
k1=10 k2=100
logk2k1=ΔH2.303R1T1-1T2

log10010=ΔH2.303×8.3141298-1373

log10=ΔH2.303×8.31475298×373

ΔH=2.303×8.314×298×37375=28.4KJ

At T1=25°C=298K,K1=10

ΔG=-2.303RT1logk 1

=-2.303×8.314×298×log(10)

=-2.303×8.314×298×1

=-5.7KJ

At T2=100°C=373K K2=100

ΔG=-2.303RT2logK2

=-2.303×8.314×373×log(10)2

=-2.303×2×8.314×373×1

=-14283.7J

=-14.29KJ

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2020 (06 Sep Shift 1)
The variation of induced \( \operatorname{emf}(\varepsilon) \) with time ( \( t \) ) in a coil if a short bar magnet is moved along its axis with a constant velocity is best represented as

Options:
A
B
C
D

The correct answer is:

Solution:
consider the coil to have long length and short bar magnet can influence only for short distaance.

so now as magnet approaching coil flux gradually increases and attains a peak vaalue and starts decreasing. so direction of emf induced will be in in opposite opposing change.it increases gradually attains a maximum value and again decreases in magnitude.

same happens to emf when magnet is moving out but now direction of emf will be in opposite.

Topic: Physics, Electromagnetic Induction, JEE Main
The variation of intensity of magnetisation (I) and the applied magnetic field
intensity (H) for three magnetic materials ' $\mathrm{X}^{\prime}$, ${ }^{\prime} \mathrm{Y}^{\prime}$ and ' $\mathrm{Z}$ ' are shown in the graph as
OX, OY and OZ respectivley. The materials ' $\mathrm{X}^{\prime},{ }^{\prime} \mathrm{Y}^{\prime}$ and ' $\mathrm{Z}$ ' respectively are

Options:
A paramagnetic, diamagnetic, ferromagnetic
B diamagnetic, paramagnetic, ferromagnetic
C ferromagnetic, diamagnetic, paramagnetic
D diamagnetic, ferromagnetic, paramagnetic

The correct answer is:
diamagnetic, ferromagnetic, paramagnetic

Solution:
\(\mathrm{X}\) - diamagnetic, \(\mathrm{Z}\) - paramagnetic, \(\mathrm{Y}\) - ferromagnetic
As manetization is the property of material and,
\(\mathrm{I}_{\text {diamagnetic }}<\mathrm{I}_{\text {paramagnetic }}<\mathrm{I}_{\text {ferromagnetic }}\)

Topic: Physics, Magnetic Properties of Matter, JEE Main
The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x) starting from mean position to extreme position (A) is given by

Options:
A
B
C
D

The correct answer is:

Solution:
The formula to calculate the kinetic energy of a particle executing SHM is given by

K=12mω2A2-x2 ...1

Substitute 0 for x into equation (1) to obtain the kinetic energy at the mean position.

K=12mω2A2-02= 12mω2A2

Substitute Afor x into equation (1) to obtain the kinetic energy at the end points.

K=12mω2A2-A2= 0

Hence, the variation can be shown as follows-

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (11 Apr Shift 1)
The variation of molar conductively with concentration of an electrolyte (X) in aqueous solution is shown in the given figure.

The electrolyte X is :

Options:
A HCl
B NaCl
C KNO3
D CH3COOH

The correct answer is:
CH3COOH

Solution:
Graph represent variation of λMC with respect to C for weak electrolyte.

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The variation of potential energy of a harmonic oscillator is as shown in the figure. Then, find the spring constant.

Options:
A \(1 \times 10^2 \mathrm{Nm}^{-1}\)
B \(150 \mathrm{Nm}^{-1}\)
C \(0.667 \times 10^2 \mathrm{Nm}^{-1}\)
D \(3 \times 10^2 \mathrm{Nm}^{-1}\)

The correct answer is:
\(150 \mathrm{Nm}^{-1}\)

Solution:
According to figure, when \(y=20 \mathrm{~mm}\)
\(=2 \times 10^{-2} \mathrm{~m}\)
then \(U_{\max }=0.04 \mathrm{~J}=4 \times 10^{-2} \mathrm{~J}\)
When \(y=0\)
then \(U_{\min }=0.01 \mathrm{~J}=1 \times 10^{-2} \mathrm{~J}\)
\(\therefore\) The change in potential energy,
\(\begin{array}{ll}
& U_{\max }-U_{\min }=\frac{1}{2} K y^2 \\
\Rightarrow & 4 \times 10^{-2}-1 \times 10^{-2}=\frac{1}{2} K \times\left(2 \times 10^{-2}\right)^2 \\
\Rightarrow & 3 \times 10^{-2}=K \times 2 \times 10^{-4} \\
\Rightarrow & \quad K=\frac{3 \times 10^{-2}}{2 \times 10^{-4}}=1.5 \times 10^2=150 \mathrm{Nm}^{-1}
\end{array}\)

Topic: Physics, Oscillations, JEE Main
The variation of solubility of four different gases (G1, G2, etc.) in a given solvent with pressure at a constant temperature is shown in the plot.
The gas with the highest value of Henry's law constant is

Options:
A G4
B G2
C G3
D $\mathrm{G} 1$

The correct answer is:
$\mathrm{G} 1$

Solution:
$\because \mathrm{P}=\mathrm{K}_{\mathrm{H}} \cdot \mathrm{X}_{(\mathrm{g})}$
$\mathrm{X}_{(g)}=\mathrm{P} \times \frac{1}{\mathrm{~K}_{\mathrm{H}}}$
slope $=\frac{1}{\mathrm{~K}_{\mathrm{H}}}$

Topic: Chemistry, Solutions, JEE Main
The variation of stopping potential V0 as a function of the frequency (ν) of the incident light for a metal is shown in figure. The work function of the surface is

Options:
A 2.98 eV
B 2.07 eV
C 1.36 eV
D 18.6 eV

The correct answer is:
2.07 eV

Solution:
The relation between the stopping potential and the frequency of the incident light is given by

eV0=hν-ϕ ...1

When V0 = 0,

ϕ=hν ...2

Substitute the value of the frequency corresponding to zero stopping potential into equation (2) to calculate the required work function of the material.

ϕ=6.626×10-34 J·s×5×1014 Hz1.6×10-19 J/eV=6.626×516≈2.07eV

Topic: Physics, Dual Nature of Matter, JEE Main, JEE Main 2023 (10 Apr Shift 2)
The variation of susceptibility $(\chi)$ with absolute temperature $(T)$ for a paramagnetic material is represented as:

Options:
A

B

C

D

The correct answer is:

Solution:
Curie law $\chi_m \propto \frac{1}{T}$ (Paramagnetic)
Susceptibility $\chi$ varies inversely with absolute temperature.

Topic: Physics, Magnetic Properties of Matter, JEE Main
The variation of the rate of an enzyme catalyzed reaction with substrate concentration is correctly represented by graph

(a)

(b)

(c)

(d)

Options:
A b
B c
C d
D a

The correct answer is:
c

Solution:
The substrate concentration influences enzyme activity. As the substrate concentration increases the rate of reaction also increases. This is because the more substrate molecules will interact with enzyme molecules, the more products will be formed.

However, after a certain concentration, further increase in substrate concentration will have no effect on the rate of reaction, since the substrate concentration will no longer be the limiting factor. At this stage, enzyme molecules become saturated and work at their maximum possible rate.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The variation of velocity of a particle with time moving along a straight line is illustrated in the following figure. The distance travelled by the particle in four seconds is

Options:
A 60 m
B 55 m
C 25 m
D 30 m

The correct answer is:
55 m

Solution:
Distance $=$ Area under $v-$ t graph $=A_1+A_2+A_3+A_4$
$\begin{aligned} & =\frac{1}{2} \times 1 \times 20+(20 \times 1)+\frac{1}{2}(20+10) \times 1+(10 \times 1) \\ & =10+20+15+10=55 \mathrm{~m}\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main
The vector $\vec{a}$ and $\vec{b}$ are not perpendicular and $\vec{c}$ and $\vec{d}$ are two vectors satisfying: $\vec{b} \times \vec{c}=\vec{b} \times \vec{d}$ and $\vec{a} \cdot \vec{d}=0$. Then the vector $\vec{d}$ is equal to

Options:
A
$\vec{c}+\left(\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right) \vec{b}$

B
$\vec{b}+\left(\frac{\vec{b} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right) \vec{c}$

C
$\vec{c}-\left(\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right) \vec{b}$

D
$\vec{b}-\left(\frac{\vec{b} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right) \vec{c}$

The correct answer is:
$\vec{c}-\left(\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right) \vec{b}$

Solution:
$\overline{\mathrm{b}} \times \overline{\mathrm{c}}=\overline{\mathrm{b}} \times \overline{\mathrm{d}}$
$\Rightarrow \overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{d}})$
$\Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}}=(\overline{\mathrm{a}} \cdot \overline{\mathrm{d}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{d}}$
$\Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}}=-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{d}}$
$\therefore \overline{\mathrm{d}}=\overline{\mathrm{c}}-\left(\frac{\overline{\mathrm{a}} \cdot \overline{\bar{c}}}{\overline{\mathrm{a}} \overline{\mathrm{b}}}\right) \overline{\mathrm{b}}$

Topic: Mathematics, Vector Algebra, JEE Main
The vector $\vec{a}$ and $\vec{b}$ are not perpendicular and $\vec{c}$ and $\vec{d}$ are two vectors satisfying: $\vec{b} \times \vec{c}=\vec{b} \times \vec{d}$ and $\vec{a} \cdot \vec{d}=0$. Then the vector $\vec{d}$ is equal to

Options:
A
$\vec{c}+\left(\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right) \vec{b}$

B
$\vec{b}+\left(\frac{\vec{b} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right) \vec{c}$

C
$\vec{c}-\left(\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right) \vec{b}$

D
$\vec{b}-\left(\frac{\vec{b} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right) \vec{c}$

The correct answer is:
$\vec{c}-\left(\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}\right) \vec{b}$

Solution:
$\overline{\mathrm{b}} \times \overline{\mathrm{c}}=\overline{\mathrm{b}} \times \overline{\mathrm{d}}$
$\Rightarrow \overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{c}})=\overline{\mathrm{a}} \times(\overline{\mathrm{b}} \times \overline{\mathrm{d}})$
$\Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}}=(\overline{\mathrm{a}} \cdot \overline{\mathrm{d}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{d}}$
$\Rightarrow(\overline{\mathrm{a}} \cdot \overline{\mathrm{c}}) \overline{\mathrm{b}}-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{c}}=-(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}) \overline{\mathrm{d}}$
$\therefore \overline{\mathrm{d}}=\overline{\mathrm{c}}-\left(\frac{\overline{\mathrm{a}} \cdot \overline{\bar{c}}}{\overline{\mathrm{a}} \overline{\mathrm{b}}}\right) \overline{\mathrm{b}}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2011