datavorous/entrance-exam-dataset huggingface dataset:
Three vectors $\mathbf{A}=a \mathbf{i}+\mathbf{j}+\mathbf{k} ; \mathbf{B}=\mathbf{i}+b \mathbf{j}+\mathbf{k}$
and $\mathbf{C}=\mathbf{i}+\mathbf{j}+c \mathbf{k}$ are mutually perpendicular (i, $\mathbf{j}$ and $\mathbf{k}$ are unit vectors along $X, Y$ and $Z$ axes respectively). The respective values of $a$, $b$ and $c$ are

Options:
A 0,0,0
B $-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}$
C 1,-1,1
D $\frac{1}{2}, \frac{1}{2}, \frac{1}{2}$

The correct answer is:
$-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}$

Solution:
Three vectors are mutually perpendicular, so the scalar or dot product of two vectors will zero. So,
$\mathbf{A} \cdot \mathbf{B}=0, \mathbf{B} \mathbf{C}=0,$
So, $\mathbf{A} \cdot \mathbf{B}=(a \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+b \hat{\mathbf{j}}+\hat{\mathbf{k}})=0$
$\Rightarrow a+b+1=0$
$\mathbf{B} \cdot \mathbf{C}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \cdot \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}=$
$\Rightarrow 1+b+c=0$
$\mathbf{A} \cdot \mathbf{C}=(a \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}+c \hat{\mathbf{k}})=0$
$\mathbf{A} \cdot \mathbf{C}=0$
$\Rightarrow \quad a+1+c=0$
(iii Addition of Eqs. (i), (ii) and (iii),
$\begin{array}{l}
2(a+b+c)+3=0 \\
a+b+c=-\frac{3}{2}
\end{array}$
$\Rightarrow \quad-1+c=-\frac{3}{2}$
$\begin{aligned}
c &=-\frac{3}{2}+1 \\
&=-\frac{1}{2}=0.5 \\
a &=-\frac{1}{2} \\
b &=-\frac{1}{2} \\
c &=-\frac{1}{2}
\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main
Three vectors $\overrightarrow{\mathrm{OP}}, \overrightarrow{\mathrm{OQ}}$ and $\overrightarrow{\mathrm{OR}}$ each of magnitude $\mathrm{A}$ are acting as shown in figure. The resultant of the three vectors is $\mathrm{A} \sqrt{x}$. The value of $x$ is ________ .

Options:

The correct answer is:
3

Solution:
$\begin{aligned} & \vec{R}=\left(A+\frac{A}{\sqrt{2}}\right) \hat{i}+\left(A-\frac{A}{\sqrt{2}}\right) \hat{j} \\ & |\vec{R}|=\sqrt{\left(A+\frac{A}{\sqrt{2}}\right)^2+\left(A-\frac{A}{\sqrt{2}}\right)^2}=\sqrt{3} A\end{aligned}$

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2024 (08 Apr Shift 1)
Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. First and third plates are maintained at absolute temperatures $2 T$ and $3 T$ respectively.
Temperature of the middle plate in steady state is

Options:
A $\left(\frac{65}{2}\right)^{\frac{1}{4}} \mathrm{~T}$
B $\left(\frac{97}{4}\right)^{\frac{1}{4}} \mathrm{~T}$
C $\left(\frac{97}{2}\right)^{\frac{1}{4}} \mathrm{~T}$
D $(97)^{\frac{1}{4}} \mathrm{~T}$

The correct answer is:
$\left(\frac{97}{2}\right)^{\frac{1}{4}} \mathrm{~T}$

Solution:
Let assume temperature of middle plate is $T_0$.
As temperatures are constant,
$\therefore$ Heat gained by middle surface $=$ Heat lost by middle surface.
So, $\sigma A\left[(3 T)^4-T_0^4\right]=\sigma A\left[T_0^4-(2 T)^4\right]$
$$
\Rightarrow \quad T_0^4=\frac{97}{2} T^4 \quad \Rightarrow T_0=\left(\frac{97}{2}\right)^{1 / 4} T
$$

Topic: Physics, Thermal Properties of Matter, AP EAMCET, AP EAMCET 2018 (22 Apr Shift 1)
Three very large plates of the same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high thermal conductivity. The first and third plates are maintained at temperatures 2T and 3T respectively. The temperature of the middle plate under steady state condition is

Options:
A 6521/4T
B 9741/4T
C 9721/4T
D 971/4T

The correct answer is:
9721/4T

Solution:
σ A 2 T 4 + σ A 3 T 4 = σ 2 A T ′ 4
1 6 T 4 + 8 1 T 4 = 2 T ′ 4
9 7 T 4 = 2 T ′ 4
T ′ 4 = 9 7 2 T 4
∴ T ′ = ( 9 7 2 ) 1 / 4 T

Topic: Physics, Thermal Properties of Matter, JEE Main
Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and third contains uranium hexafluoride (polyatomic). Arrange these on the basis of their root mean square speed vrms and choose the correct answer from the options given below:

Options:
A vrms(mono)>vrms(dia)>vrms(poly)
B Vrms(mono)=Vrms(dia)=Vrms(poly)
C vrms(mono)<vrms(dia)<vrms(poly)
D vrms(dia)<vrms(poly)<vrms(mono)

The correct answer is:
vrms(mono)>vrms(dia)>vrms(poly)

Solution:
The RMS speed is given vrms =3RTM.

The temperature and pressure is same for all the gases. So the speed is inversely proportional to the square root of the molar mass of the gases.

vrms(mono)=3RT4×10-3vrms(dia)=3RT71×10-3vrms(poly)=3RT352×10-3

vrms ∝1M

Hence,

vrms(poly)<vrms(dia)<vrms(mono).

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2023 (11 Apr Shift 1)
Three voltmeters, all having different internal resistances are joined as shown in figure. When some potential difference is applied across A and B, their readings are V1,V2 and V3. Choose the correct option.

Options:
A V1=V2
B V1≠V3-V2
C V1+V2>V3
D V1+V2=V3

The correct answer is:
V1+V2=V3

Solution:
Let's consider the following diagram:

According to the Kirchhoff's voltage law, the algebraic sum of the voltages in a closed loop is zero.

From the above figure, it can be written that

V1+V2-V3=0⇒V1+V2=V3

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Three waves of amplitudes 7 mm, 3 mm and 11 mm with a successive phase difference of Ï2 are super -imposed. The amplitude of the resulting wave will be

Options:
A 5 mm
B 12 mm
C 8 mm
D 6 mm

The correct answer is:
5 mm

Solution:
Let us say that the first wave has a zero phase, then the second one is ahead of the first by a phase of Ï2 and so the third one is ahead of the first one by a phase of Ï. So the first and the third are opposite to each other. Hence,

Ar=(11â7)2+(3)2

Ar=5 mm

Topic: Physics, Waves and Sound, JEE Main
Three weights $W, 2 W$ and $3 W$ are connected to identical springs suspended from a rigid horizontal rod. The assembly of the rod and the weights fall freely. The positions of the weights from the rod are such that

Options:
A $3 W$ will be farthest
B $W$ will be farthest
C All will be at the same distance
D $2 W$ will be farthest

The correct answer is:
All will be at the same distance

Solution:
For $W, 2 W, 3 W$ apparent weight will be zero because the system is falling freely. So the distances of the weight from the rod will be same.

Topic: Physics, Laws of Motion, JEE Main
Thron of Bougainvillea and tendril of cucurbita are example of

Options:
A Retrogressive evolution
B Analogous organs
C Homologous organs
D Vestigial organs

The correct answer is:
Homologous organs

Solution:
No Solution Available

Topic: Biology, Evolution, NEET, NEET 2008 (Mains)
Through a coil of resistance R charge is flowing in such a manner that in every time interval t0 the amount of charge flow is reduced by a factor of 12. If a total of q charge flows through the coil in very long time, then the total amount of heat generated in the coil will be

Options:
A 14t0q2Rloge2
B 3Â Â 2q2Rloge2t0
C q2Rloge22 t0
D 3q2Rloge24t0

The correct answer is:
q2Rloge22 t0

Solution:
I(t)=I0e-λt, λ=loge2T12=loge2T0
q=∫dq=∫0∞I(t)⋅dt=∫0∞I0⋅e-λtdt=I0(-λ)⋅e-λt0∞
⇒ q=I0λ ⇒I0=qλ
H=∫dH=∫I2Rdt=R∫0∞I02e-2λtdt
H=RI0212λ=R2λ×q2λ2=q2Rλ2=q2R2×loge2t0
⇒ H=q2Rloge22t0

Topic: Physics, Current Electricity, JEE Main
Through the point (4, 5), a straight line is drawn making positive intercepts on the coordinate axes. The area of the triangle thus
formed is least, when the ratio of the intercepts on the X and Y axes is

Options:
A 1 : 1
B 3 : 4
C 4 : 5
D 2 : 3

The correct answer is:
4 : 5

Solution:
Since, intercept form of line is $\frac{x}{a}+\frac{y}{b}=1$
where X-intercept = a, Y-intercept = b
Hence, equation of line AB
$$
\frac{x}{m}+\frac{y}{n}=1
$$
$\because$ Line (i) passes through $(4,5)$.
$$
\begin{aligned}
\therefore \quad \frac{4}{m} & +\frac{5}{n}=1 \\
\frac{4}{m} & =1-\frac{5}{n} \\
m & =\frac{4 n}{n-5}
\end{aligned}
$$
$\begin{gathered}\because \text { Area of } \triangle O A B=\frac{1}{2} \times O A \times O B \\ A=\frac{1}{2} m n=\frac{1}{2}\left(\frac{4 n}{n-5}\right) n \\ A=\frac{2 n^2}{n-5} \\ \therefore \frac{d A}{d n}=2\left\{\frac{(n-5) \cdot 2 n-n^2(1)}{(n-5)^2}\right\} \\ \frac{d A}{d n}=2\left\{\frac{2 n^2-10 n-n^2}{(n-5)^2}\right\} \\ =\frac{2 n^2-20 n}{(n-5)^2}\end{gathered}$
$\because$ Area is least for some value of $n$.
$$
\begin{array}{ll}
\therefore \quad & \frac{d A}{d n}=0 \\
& 2 n^2-20 n=0 \\
& 2 n(n-10)=0 \\
\Rightarrow \quad \quad \quad n=0,10 \\
\text { At } & n=10, m=\frac{4 n}{n-5}=\frac{4 \times 10}{10-5}=\frac{40}{5} \\
\quad & m=8 \\
\therefore m & n=8: 10=4: 5
\end{array}
$$

Topic: Mathematics, Application of Derivatives, JEE Main
Through the vertex $\mathrm{O}$ of a parabola $\mathrm{y}^{2}=4 \mathrm{x}$, chords OP and $O Q$ are drawn at right angles to one another. The locus of the middle point of $\mathrm{PQ}$ is

Options:
A $y^{2}=2 x+8$
B $\quad y^{2}=x+8$
C $y^{2}=2 x-8$
D $\quad y^{2}=x-8$

The correct answer is:
$y^{2}=2 x-8$

Solution:
Given parabola is $\mathrm{y}^{2}=4 ...(1)

\mathrm{x}$ Let $\mathrm{P} \equiv\left(\mathrm{t}_{1}^{2}, 2 \mathrm{t}_{1}\right)$ and $\mathrm{Q} \equiv\left(\mathrm{t}_{2}^{2}, 2 \mathrm{t}_{2}\right)$

Slope of $\mathrm{OP}=\frac{2 \mathrm{t}_{1}}{\mathrm{t}_{1}^{2}}=\frac{2}{\mathrm{t}_{1}}$ and slope of $\mathrm{OQ}=\frac{2}{\mathrm{t}_{2}}$

Since $\mathrm{OP} \perp \mathrm{OQ}, \therefore \frac{4}{\mathrm{t}_{1} \mathrm{t}_{2}}=-1$ or $\mathrm{t}_{1} \mathrm{t}_{2}=-4 \ldots(2)$

Let $\mathrm{R}(\mathrm{h}, \mathrm{k})$ be the middle point of $\mathrm{PQ},$ then $\mathrm{h}=\frac{\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}}{2} \quad \ldots(3) \quad$ and $\quad \mathrm{k}=\mathrm{t}_{1}+\mathrm{t}_{2} \ldots(4)$

From $(4), \mathrm{k}^{2}=\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+2 \mathrm{t}_{1} \mathrm{t}_{2}=2 \mathrm{~h}-8 \quad[\mathrm{From}$

(2) and (3)] Hence locus of $\mathrm{R}(\mathrm{h}, \mathrm{k})$ is $\mathrm{y}^{2}=2 \mathrm{x}-8$

Topic: Mathematics, Parabola, JEE Main
Through their effect on plant growth regulators, what do the temperature and light control in the plants?

Options:
A Apical dominance
B Flowering
C Closure of stomata
D Fruit elongation

The correct answer is:
Flowering

Solution:
There are three factors which govern the flowering in angiosperms. These factors are completion of juvenile phase and attainment of mature or adult phase, suitable light and proper temperature.

Topic: Biology, Plant - Growth and Development, NEET, NEET 2012 (Mains)
Throwing a biased die, a person will get 5 Rupees if he throws the number 5 and will get 8 Rupees for any other number, then the expected income (in Rupees) per throw is (it is given that the number 5 will appear 5 times as frequently as any other number)

Options:
A 6.5
B 7
C 5
D 5.5

The correct answer is:
6.5

Solution:
Let the probability of getting any number other than 5 is P, then the probability of getting 5 is 5P.
âµP+P+P+P+P+5P=1âP=110
Expected income per throw =510Ã5+110Ã85
=6510=6.5

Topic: Mathematics, Probability, JEE Main
$\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$ is a $3 d^{2}$ system, the value of crystal field stabilisation energy decreases by

Options:
A $4 \times 10^{-1} \triangle_{o}$
B $4.0 \Delta_{o}$
C $8.0 \Delta_{o}$
D $8 \times 10^{-1} \Delta_{o}$

The correct answer is:
$8 \times 10^{-1} \Delta_{o}$

Solution:
For a $3 d^{2}$ system, electrons will go into more stable $t_{2 g}$ orbitals and each electron going into a $t_{2 g}$ $t_{2 g}$ orbital is stabilised by $0.4 \Delta_{o}$ i.e., 2 electrons will be stabilised by $0.8 \Delta_{o}$ or $8 \times 10^{-1} \Delta_{o}$.

Topic: Chemistry, Coordination Compounds, JEE Main
TiH2O63+ absorbs light of wavelength 498 nm during a d-d transition. The octahedral splitting energy for the above complex is _______ ×10-19 J. (Round off to the Nearest Integer). h=6.626×10-34 Js; c=3×108 ms-1.

Options:

The correct answer is:
4

Solution:
λabsorbed=498 nm (given)

The octahedral splitting energy

Δ0 or E=hcλ

=0.0399×10-17 J

=3.99×10-19 J

=4.00×10-19 J (round off)

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2021 (16 Mar Shift 2)
Tibu and Babu are playing a game in which both the players roll a pair of dice. Tibu wins if the sum of the numbers appearing on the dice is a prime number while Babu wins if the product of the numbers appearing on the dice is prime. The chance that no one wins is (both players are allowed to win simultaneously)

Options:
A 1936
B 12
C 1736
D 49

The correct answer is:
1736

Solution:
Psum is prime=P2+P3+P5+P7+P11

=1+2+4+6+236=1536=PA (say)
Pproduct is prime=P12,21,13,31,15,51=636=PB (say)
PA∩B=236
Required probability =1-PA∪B=1-1536+636-236=1736

Topic: Mathematics, Probability, JEE Main
Tiger is not a resident in which one of the following National Park ?

Options:
A Jim Corbett
B Ranthambhor
C Sunderbans
D Gir

The correct answer is:
Gir

Solution:
No Solution Available

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2009 (Mains)
Time period of a simple pendulum in a stationary lift is T. If the lift accelerates with g6 vertically upwards then the time period will be
(Where g= acceleration due to gravity)

Options:
A 65T
B 56T
C 67T
D 76T

The correct answer is:
67T

Solution:
The time period of a simple pendulum is given by T=2πlg

In an upward accelerating frame the effective geff=g+a

⇒T2=2πlg+a=2πlg+g6

TT2=g+g6g=76

⇒T2=67T

Topic: Physics, Oscillations, JEE Main, JEE Main 2022 (26 Jun Shift 1)
Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration g2, the time period of pendulum will be :

Options:
A 3T
B T3
C 32T
D 23T

The correct answer is:
23T

Solution:
When lift is stationary

T=2πLg

When lift is moving upwards

⇒ Pseudo force acts downwards

⇒geff=g+g2=3g2

⇒ New time period

T'=2πLgeff=2π2L3g

T'=23T

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (16 Mar Shift 1)
Time period of a simple pendulum is T. The time taken to complete 58 oscillations starting from mean position is α12T. The value of α is ______.

Options:

The correct answer is:
7

Solution:
58th of oscillation =12+18th of oscillation

π+θ=ωt

π+π6=2πTt

7π6=2πTt

t=7T12

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (26 Feb Shift 2)
Time period of a spring-mass system is T. If this spring is cut into two parts whose lengths are in the ratio 1:3 and the same mass is attached to the longer part, the new time period will be

Options:
A 32 T
B T3
C 3T2
D 3T

The correct answer is:
3T2

Solution:
T = 2 π m k
When spring is cut in the ratio 1 : 3, the force constant of the longer part is 4 k 3 .
∴ T'=2π3m4k=32T

Topic: Physics, Oscillations, JEE Main
Time period of simple pendulum on earth's surface is ' $\mathrm{T}$ '. Its time period becomes ' $\mathrm{xT}$ ' when taken to a height $\mathrm{R}$ (equal to earth's radius) above the earth's surface. Then the value of ' $x$ ' will be

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

The correct answer is:
2

Solution:
$$
\mathrm{T}=2 \pi \sqrt{\frac{l}{\mathrm{~g}}}
$$
At a height 'h' from earth's surface,
$$
\begin{array}{ll}
& \mathrm{xT}=2 \pi \sqrt{\frac{l}{g_{\mathrm{h}}}} \\
\therefore \quad & \mathrm{x}=\sqrt{\frac{\mathrm{g}}{\mathrm{g}_{\mathrm{h}}}} \quad \ldots . .(\mathrm{i}) \\
& \text { Now, } \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2} \\
\therefore \quad & \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{GM}}{4 \mathrm{R}^2} \\
\therefore \quad & \mathrm{g}_{\mathrm{h}}=\frac{\mathrm{g}}{4} \\
\therefore \quad & \text { From equations (i) and (ii), } \\
& \mathrm{x}=\sqrt{\frac{\mathrm{g}}{\mathrm{g} / 4}}=\sqrt{4}=2
\end{array}
$$
$\therefore \quad$ From equations (i) and (ii),
$$
x=\sqrt{\frac{g}{g / 4}}=\sqrt{4}=2
$$

Topic: Physics, Oscillations, JEE Main
Time periods of oscillation of the same simple pendulum measured using four different measuring clocks were recorded as $4.62 \mathrm{~s}, 4.632 \mathrm{~s}, 4.6 \mathrm{~s}$ and $4.64 \mathrm{~s}$. The arithmetic mean of these readings in correct significant figure is :

Options:
A $5 \mathrm{~s}$
B $4.623 \mathrm{~s}$
C $4.6 \mathrm{~s}$
D $4.62 \mathrm{~s}$

The correct answer is:
$4.6 \mathrm{~s}$

Solution:
Sum of number by considering significant digi
$\text {sum }=4.6+4.6+4.6+4.6=18.4$
Arithmetic Mean $=\frac{\text { sum }}{4}=\frac{18.4}{4}=4.6$

Topic: Physics, Mathematics in Physics, JEE Main, JEE Main 2024 (05 Apr Shift 1)
Time required for 100 per cent completion of a zero order reaction is

Options:
A $\frac{2 k}{a}$
B $\frac{\mathrm{a}}{2 \mathrm{k}}$
C $\frac{\mathrm{a}}{\mathrm{k}}$
D ak

The correct answer is:
$\frac{\mathrm{a}}{\mathrm{k}}$

Solution:
For zero order reactions,
$$
[\mathrm{A}]=-\mathrm{kt}+[\mathrm{A}]_{0}
$$
where, $[\mathrm{A}]_{0}=$ initial concentration $=\mathrm{a}$
$[\mathrm{A}]=$ remaining concentration $=\mathrm{a}-\mathrm{a}=0$
On putting values of $[\mathrm{A}]_{0}$ and $[\mathrm{A}]$, we get
$$
\mathrm{t}=\frac{\mathrm{a}}{\mathrm{k}}
$$

Topic: Chemistry, Chemical Kinetics, KCET, KCET 2010
Time required for \(99.9 \%\) completion of a first order reaction is _______ times the time required for completion of \(90 \%\) reaction.(nearest integer)

Options:

The correct answer is:
3

Solution:
$\begin{aligned} & \mathrm{K}=\frac{1}{\mathrm{t}_{99.9 \%}} \ln \left(\frac{100}{0.1}\right)=\frac{1}{\mathrm{t}_{90 \%}} \ln \left(\frac{100}{10}\right) \\ & \mathrm{t}_{99.9 \%}=\mathrm{t}_{90 \%} \frac{\ell \mathrm{n}\left(10^3\right)}{\ell \mathrm{n} 10} \\ & \mathrm{t}_{99.9 \%}=\mathrm{t}_{90 \%} \times 3\end{aligned}$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Time T, velocity C and angular momentum h are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be:

Options:
A M=T-1C-2h
B [M]=[T−1C2h]
C M=T-1C-2h-1
D [M]=[TC−2h]

The correct answer is:
M=T-1C-2h

Solution:
M∝TxCyhz
M1L0T0=T1xL1T-1yM1L2T-1z
[M1 L0T0]=[Mz Ly+2zTx-y-z]]
⟹z=1
y+2z=0 , x-y-z=0
solving
y=-2
and x=-1
M⇒ T-1 C-2 h1

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2017 (08 Apr Online)
Time T, velocity C and angular momentum h are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be:

Options:
A M=T-1C-2h
B [M]=[T−1C2h]
C M=T-1C-2h-1
D [M]=[TC−2h]

The correct answer is:
M=T-1C-2h

Solution:
M∝Tx Cy hz
M1L0T0=T1xL1 T-1y M1 L2 T-1z
[M1 L0T0]=[Mz Ly+2zTx-y-z]
⟹z=1
y+2z=0 , x-y-z=0
solving
y=-2
and x=-1
M ⇒ T-1 C-2 h1

Topic: Physics, Units and Dimensions, JEE Main
Time taken by a $836 \mathrm{~W}$ heater to heat one litre of water from $10^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ is

Options:
A
$50 \mathrm{~s}$

B
$100 \mathrm{~s}$

C
$150 \mathrm{~s}$

D
$200 \mathrm{~s}$

The correct answer is:
$150 \mathrm{~s}$

Solution:
Let $\mathrm{t}$ be the time taken, then
$$
\begin{aligned}
& \frac{836 \times t}{4.2}=1000 \times 1 \times(40-10)[\text { using } Q=m s t] \\
& \Rightarrow t=150 \text { sec. }
\end{aligned}
$$

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2004
Time taken by an object falling from rest to cover the height of ${ }^{h_1}$ and ${ }^{h_2}$ is respectively ${ }^{t_1}$ and ${ }^{t_2}$ then the ratio of ${ }^{t_1}$ to ${ }^{t_2}$ is

Options:
A $h_1: h_2$
B $\sqrt{h_1}: \sqrt{h_2}$
C $h_1: 2 h_2$
D $2 h_1: h_2$

The correct answer is:
$2 h_1: h_2$

Solution:
$t=\sqrt{\frac{2 h}{g}} \Rightarrow \frac{t_1}{t_2}=\sqrt{\frac{h_1}{h_2}}$

Topic: Physics, Motion In One Dimension, JEE Main
Time taken by the projectile to reach from A to B is t. Then the distance AB is equal to

Options:
A ut3
B 3ut2
C 3ut
D 2ut

The correct answer is:
ut3

Solution:
Horizontal component of velocity
u H = u cos 60 o = u 2
â´ AC = u H t = u t 2
and AB = AC sec 30Â°
= ut 2 2 3 = ut 3

Topic: Physics, Motion In Two Dimensions, JEE Main
Time taken for an electron to complete one revolution in the Bohr orbit of hydrogen atom is

Options:
A $\frac{4 \pi^2 m r^2}{n h}$
B $\frac{n h}{4 \pi^2 m r}$
C $\frac{n h}{4 \pi^2 m r^2}$
D $\frac{h}{2 \pi m r}$

The correct answer is:
$\frac{4 \pi^2 m r^2}{n h}$

Solution:
It will take $\frac{4 \pi^2 m r^2}{n h}$

Topic: Chemistry, Structure of Atom, JEE Main
Tincture of iodine is a solution of _______

Options:
A 2 to 3% percent solution of iodine in water
B 3 to 6% percent solution of iodine in alcohol and water
C 2 to 3% percent solution of iodine in potassium iodide
D 2 to 3% percent solution of iodine in alcohol and water

The correct answer is:
2 to 3% percent solution of iodine in alcohol and water

Solution:
The correct answer is the $\mathbf{2}$ to $3 \%$ percent solution of iodine in alcohol and water
Tincture of iodine is an Antiseptic Solution.
It is prepared by dissolving lodine in a mixture of Ethanol and Water. Therefore option 4 is correct.
A tincture of lodine is used to disinfect the region of any cut or wound in the body.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Tischenko reaction is a modification of:

Options:
A
Aldol condensation

B
Claisen condensation

C
Cannizzaro reaction

D
Pinacol-pinacolon reaction

The correct answer is:
Cannizzaro reaction

Solution:
Tishchenko reaction is a modification of Canizzaro's reaction. This reaction involves disproportionation of an aldehyde lacking a hydrogen atom in the alpha position in the presence of an alkoxide. The reaction product is an ester. Catalysts are aluminium alkoxide or sodium alkoxide.
In Cannizzaro's reaction the base is sodium hydroxide and the oxidation product is a carboxylic acid and the reduction product is an alcohol.

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2014 (11 Apr Online)
Tissue that is mainly composed of fat-storing cells

Options:
A Dense regular
B Adipose tissue
C Areolar
D Dense irregular

The correct answer is:
Adipose tissue

Solution:
Adipose tissue is modified areolar connective tissue specialized for fat storage. A few largeand spherical cells occurred in areolar tissue singly or clusters around small blood vesselsand called as adipocytes. Each cell contains a large globule of fat surrounded by a thinperipheral layer of cytoplasm having a nucleus.

Topic: Biology, Structural Organisation in Animals, NEET
To a 10 ml, 1M aqueous solution of Br2 , excess of NaOH is added so that all Br2 is disproportionated to Br- and BrO3- . The resulting solution is made free of Br- by extraction and excess of OH- is neutralised by acidifying the solution. The resulting solution is sufficient to react with 2g of impure CaC2O4 (M = 128 g/mole) sample. Which of the following is the correct % purity of oxalate sample

Options:
A 84.3%
B 32.5%
C 60%
D 64%

The correct answer is:
64%

Solution:
6H++BrO3-+3C2O42-âBr-+6CO2+3H2O
10/3 m mole BrO3- reacts with 10 m. mole C2O42-
Mass of CaC 2 O 4 â=â10âÃâ 10 â3 âÃâ128â=â1.28âg
% purity =  Weight of substance Weight of compound  × 100
%âpurityâ= 1.28 2 âÃâ100=64%

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main
To a 25 mL H2O2 solution, excess of acidified solution of potassium iodide was added. The iodine liberated required 20 mL of 0.3 N sodium thiosulphate solution. Calculate the volume strength of H2O2 solution and report your answer by multiplying it with 1000.

Options:

The correct answer is:
1334

Solution:
Meq of H2O2 = Meq of I2 = Meq of Na2S2O3.
If N is normality of H2O2, then

N x 25 = 0.3 x 20, N H 2 O 2 = 0 . 2 4 N

M H 2 O 2 = 0.12 M
N = 0.24
Volume strength = 0.12 Ã 11.2
=1.334 Vol
Final answer =1.334×1000=1334

Topic: Chemistry, Redox Reactions, JEE Main
To a stationary man, rain appears to be falling at his back at an angle 30o with the vertical. As he starts moving forward with a speed of 0.5 m s-1, he finds that rain falls vertically. The speed of rain with respect to stationary man is

Options:
A 0.5 m s-1
B 1 m s-1
C 32 m s-1
D 13 m s-1

The correct answer is:
1 m s-1

Solution:
For stationary person
For running person
Speed of rain with respect to a stationary man =ν
v→=v sin30o i^−v cos30o j^
The velocity of rain with respect to man,
v→RM=v→R−v→M
=(v sin30o i^−v cos30o j^)−(0.5 i^)
Since V→RM is vertically downward, its x-component will become zero.
∴ v sin30o−0.5=0
⇒v=0.5sin30o=1 m s-1

Topic: Physics, Motion In Two Dimensions, JEE Main