datavorous/entrance-exam-dataset huggingface dataset:
The velocity of water waves $v$ may depend upon their wavelength $\lambda$, the density of water $\rho$ and the acceleration due to gravity $g$. The method of dimensions gives the relation between these quantities as

Options:
A $v^2 \propto \lambda g^{-1} \rho^{-1}$
B $v^2 \propto g \lambda \rho$
C $v^2 \propto g \lambda$
D $v^2 \propto g^{-1} \lambda^{-3}$

The correct answer is:
$v^2 \propto g \lambda$

Solution:
Let $v^x=k g^y \lambda^z \rho^\delta$. Now by substituting the dimensions of each quantities and equations the powers of $M, L$ and $T$ we get $\delta=0$ and $x=2, y=1, z=1$.

Topic: Physics, Units and Dimensions, JEE Main
The velocity of water waves v may depend upon their wavelength λ , the density of water ρ and the acceleration due to gravity g . The method of dimensions gives the relation between these quantities as

Options:
A v2∝ρg
B v 2 âgÎ»Ï
C v 2 âgÎ»
D v 2 â g â1 Î» â3

The correct answer is:
v 2 ∝gλ

Solution:
Let v x =k g y λ z ρ δ . Now by substituting the dimensions of each quantities and equating the powers of M, L and T we get δ=0 and x=2, y=1, z=1 .

Topic: Physics, Units and Dimensions, JEE Main
The velocity of waves in a string fixed at both ends is $2 \mathrm{~m} / \mathrm{s}$.
The string forms standing waves with nodes $5.0 \mathrm{~cm}$ apart. The frequency of vibration of the string in $\mathrm{Hz}$ is

Options:
A 40
B 30
C 20
D 10

The correct answer is:
20

Solution:
Here the distance between the two nodes is half of the wavelength
\(\frac{\lambda}{2}=5.0 \mathrm{~cm} \Rightarrow \lambda=10 \mathrm{~cm}\)
Hence \(\mathrm{n}=\frac{\mathrm{v}}{\lambda}=\frac{200}{10}=20 \mathrm{~Hz}\)

Topic: Physics, Waves and Sound, JEE Main
The velocity time graphs of a car and a scooter are shown in the figure. (i) The difference between the distance travelled by the car and the scooter in 15 s and (ii) the time at which the car will catch up with the scooter are, respectively.

Options:
A 112.5 m and 22.5 s
B 337.5 m and 25 s
C 225.5 m and 10 s
D 112.5 m and 15 s

The correct answer is:
112.5 m and 22.5 s

Solution:
Distance travelled by car in 15 s =12×45×15×=6752 m. Distance travelled by scooter in 15 s=30 ×15=450. Let car catches scooter in time t
difference between the distance travelled by car and scooter in 15 s in⇒ 450−6752=112.5  m
⇒6752+45t-15=30t
337.5+45t-675=30t ⇒15 t=337.5
⇒t=22.5 s.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2018 (15 Apr)
The velocity $v$ of a particle at time $t$ is given by $v=a t+\frac{b}{t+c}$, where $a, b$ and $c$ are constants. The dimensions of $a, b$ and $c$ are:

Options:
A L, LT and $\mathrm{LT}^2$
B $\mathrm{LT}^{-2}, \mathrm{~L}$ and $\mathrm{T}$
C $\mathrm{L}^2, \mathrm{~T}$ and $\mathrm{LT}^2$
D $\mathrm{LT}^2$, LT and L.

The correct answer is:
$\mathrm{LT}^{-2}, \mathrm{~L}$ and $\mathrm{T}$

Solution:
Using principle of homogeniety.
Dimension of $v=$ Dimension of $a t=$
Dimenstion of $\frac{b}{t+c}$
$\begin{array}{l}
\therefore a t =\left[\mathrm{LT}^{-1}\right] \\
\Rightarrow |a| =\frac{\left[\mathrm{LT}^{-1}\right]}{[\mathrm{T}]}=\left[\mathrm{LT}^{-2}\right] \\
\text {And } \frac{[b]}{[t]} =\left[\mathrm{LT}^{-1}\right] \\
{[b]} =\left[\mathrm{LT}^{-1}\right][\mathrm{T}]=[\mathrm{L}] \\
\text {And } c =\mathrm{T} \\
\Rightarrow c =[\mathrm{T}]
\end{array}$

Topic: Physics, Units and Dimensions, JEE Main
The velocity $(v)$ of a particle (under a force $F$ ) depends on its distance $(x)$ from the origin (with $x>0) v \propto\frac{1}{\sqrt{x}}$. Find how the magnitude
of the force $(F)$ on the particle depends on $x ?$

Options:
A $F \propto \frac{1}{x^{3 / 2}}$
B $F \propto \frac{1}{x}$
C $F \propto \frac{1}{x^{2}}$
D $F \times x$

The correct answer is:
$F \propto \frac{1}{x^{2}}$

Solution:
According to the question.
$$
v \propto \frac{1}{\sqrt{x}}
$$
or $v=\frac{k}{\sqrt{x}}$
$$
\frac{d v}{d t}=\frac{d}{d t} \cdot \frac{K}{\sqrt{x}}=k \cdot \frac{-1}{2} \cdot x^{-\frac{1}{2}-1} \cdot \frac{d x}{d t}
$$
$a=-\frac{k}{2} \cdot x^{-\frac{3}{2}} \cdot \frac{k}{\sqrt{x}}\left[\because \frac{d x}{d t}=v\right.$ and $v=\frac{k}{\sqrt{x}}$ from $\left.\mathrm{Eq}(i)\right]$
$$
=\frac{-k^{2}}{2} \cdot \frac{1}{x^{2}}
$$
$=m \frac{k^{2}}{2} \cdot \frac{1}{x^{2}}$
(magnitue
$F=\frac{k^{2}}{2} \cdot \frac{m}{x^{2}}$
$\therefore$
$F x \frac{1}{x^{2}}$

Topic: Physics, Laws of Motion, JEE Main
The velocity-time graph of a body moving in a straight line is shown in figure.

The ratio of displacement and distance travelled by the body in time 0 to 10 s is

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

The correct answer is:
1:3

Solution:
Area of upper rectangles, which will give us magnitude of the area under the velocity-time graph,

A1=8×2+4×4 =32

Area of lower rectangles, which will give us magnitude of the area under the velocity-time graph,

A2=4×2 +4×2 = 16

Therefore,

Displacement=A1-A2= 32-16 =16 Distance =A1+A2= 32+16=48

Hence, the required ratio would be

=1648=13

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The velocity-time graph of a particle of mass 10 kg is shown in the figure. The net work done on the particle in the first two seconds of the motion is

Options:
A -9300 J
B 12000 J
C -4500 J
D -12000 J

The correct answer is:
-4500 J

Solution:
v=-5t+50
vt=2=40

Work done=∆ K.E. =12402-502×10
=-4500 J

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2016 (10 Apr Online)
The velocity-time graph of a particle of mass 10 kg is shown in the figure. The net work done on the particle in the first two seconds of the motion is

Options:
A -9300 J
B 12000Â J
C -4500 J
D -12000Â J

The correct answer is:
-4500 J

Solution:
v=-5t+50
vat t=2=40
â K.E.=W.D. =12402-502Ã10
=-4500 J

Topic: Physics, Work Power Energy, JEE Main
The velocity-time graphs of a car and a scooter are shown in the figure. (i) the difference between the distance travelled by the car and the scooter in $15 \mathrm{~s}$ and (ii) the time at which the car will catch up with the scooter are, respectively

Options:
A
$337.5 \mathrm{~m}$ and $25 \mathrm{~s}$

B
$225.5 \mathrm{~m}$ and $10 \mathrm{~s}$

C
$112.5 \mathrm{~m}$ and $22.5 \mathrm{~s}$

D
$112.5 \mathrm{~m}$ and $15 \mathrm{~s}$

The correct answer is:
$112.5 \mathrm{~m}$ and $22.5 \mathrm{~s}$

Solution:
Using equation, $a=\frac{v-u}{t}$ and
$$
S=u t+\frac{1}{2} a t^2
$$
Distance travelled by car in $15 \mathrm{sec}$
$=\frac{1}{2} \frac{(45)}{15}(15)^2$
$=\frac{675}{2} \mathrm{~m}$
Distance traveled by scooter in 15 seconds $=30 \times 15=450(\because$ distance $=$ speed $\times$ time $)$ Difference between distance travelled by car and scooter in $15 \mathrm{sec}, 450-337.5=112.5 \mathrm{~m}$ Let car catches scooter in time t;
$\frac{675}{2}+45(t-15)=30 t$
$337.5+45 \mathrm{t}-675=30 \mathrm{t}$
$\Rightarrow \quad 15 \mathrm{t}=337.5$
$\Rightarrow \quad \mathrm{t}=22.5 \mathrm{sec}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
The venturi-meter works on :

Options:
A The principle of perpendicular axes
B Huygen's principle
C Bernoulli's principle
D The principle of parallel axes

The correct answer is:
Bernoulli's principle

Solution:
Venturi Meter works on the principle of Bernoulli's standard which expresses that the distinction in pressure made over the hindrance gives the proportion of the liquid stream rate. The stream rate is relative to the square base of the differential pressure.

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The Vernicr scale of a travelling microscope has 50 divisions which coincides with 49 main scale divisions. If each main scale division is $0.5 \mathrm{~mm}$, then the least count of the microscope is

Options:
A $0.5 \mathrm{~mm}$
B $0.01 \mathrm{~mm}$
C $0.5 \mathrm{~cm}$
D $0.01 \mathrm{~cm}$

The correct answer is:
$0.01 \mathrm{~mm}$

Solution:
Given, reading of 1 main scale division
$$
\begin{aligned}
& =0.5 \mathrm{~mm} \\
& \text { i.e. } \quad 1 \mathrm{MSD}=0.5 \mathrm{~mm} \\
& \text { According to given situation, } \\
& 50 \text { vernier scale division }=49 \text { main scale division } \\
& \Rightarrow 1 \text { vernier scale division (VSD) } \\
& =\frac{49}{50} \text { main scale division (MSD) } \\
& \therefore \text { Least count }=1 \mathrm{MSD}-1 \text { VSD } \\
& =1 \mathrm{MSD}-\frac{49}{50} \mathrm{MSD}=\frac{1}{50} \mathrm{MSD} \\
& =\frac{1}{50} \times 0.5=0.01 \mathrm{~mm} \\
&
\end{aligned}
$$

Topic: Physics, Experimental Physics, KCET, KCET 2022
The Vernier constant of Vernier callipers is 0.1 mm and it has zero error of -0.05 cm. While measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5 . The corrected diameter will be _____×10-2 cm.

Options:

The correct answer is:
180

Solution:
Vernier constant is the least count of the vernier V.C=0.1 mm=0.01 cm

zero error =-0.05 cm

∴correction=0.05 cm

M.S.R.=1.7 cm

V.S.R.=5 V.C.=5×0.01=0.05 cm

The diameter of the sphere will be,

diameter=MSR+VSR+correction

1.7+0.05+0.05=1.8 cm

=180×10-2 cm

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is $0.5 \mathrm{~mm}$, calculate the minimum inaccuracy in the measurement of distance.

Options:

None

Solution:
As we know that inaccuracy will be measured by difference of 1 MSD and 1 VSD
where MSD $=$ main scale division and
VSD $=$ vernier scale division.
As given that $50 \mathrm{VSD}=49 \mathrm{MSD}$
$$
\begin{aligned}
&1 \mathrm{MSD}=\frac{50}{49} \mathrm{VSD} \\
&\begin{aligned}
1 \mathrm{VSD} &=\frac{49}{50} \mathrm{MSD} \\
\text { Inaccuracy } &=1 \mathrm{MSD}-1 \mathrm{VSD} \\
&=1 \mathrm{MSD}-\frac{49}{50} \mathrm{MSD}=\frac{1}{50} \mathrm{MSD}
\end{aligned}
\end{aligned}
$$
Given, Main scales division (1 MSD) $=0.5 \mathrm{~mm}$ Hence, minimum inaccuracy in the measurement of distance
$$
=\frac{1}{50} \times 0.5 \mathrm{~mm}=\frac{1}{100}=0.01 \mathrm{~mm}
$$

Topic: Physics, Units and Dimensions
The vernier scale used for measurement has a positive zero error of 0.2 mm. If while taking a measurement it was noted that 0 on the vernier scale lies between 8.5 cm and 8.6 cm. Vernier coincidence is 6, then the correct value of measurement is cm.

(least count=0.01 cm)

Options:
A 8.36 cm
B 8.54 cm
C 8.58 cm
D 8.56 cm

The correct answer is:
8.54 cm

Solution:
Positive zero error =0.2 mm

Main scale reading =8.5 cm

Vernier scale reading=6×0.01=0.06 cm

Final reading=8.5+0.06-0.02=8.54 cm

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The vertex of common graph of inequalities $2 x+y \geq 2$ and $x-y \leq 3$, is

Options:
A $(0,0)$
B $\left(\frac{5}{3},-\frac{4}{3}\right)$
C $\left(\frac{5}{3}, \frac{4}{3}\right)$
D $\left(-\frac{4}{3}, \frac{5}{3}\right)$

The correct answer is:
$\left(\frac{5}{3},-\frac{4}{3}\right)$

Solution:
So, vertex of common graph is $\left(\frac{5}{3}, - \frac{4}{3}\right)$

Topic: Mathematics, Linear Programming, JEE Main
The vertices A,B and C of âABC are 1,-2, -7,6 and 115,25 respectively. The measure of the interior angle A of the âABC lies in

Options:
A 75o,90o
B 60o,75o
C 45o,60o
D 120o,150o

The correct answer is:
60o,75o

Solution:
We have AB→=-8i^+8j^=8-i^+j^,
AC→=115-1i^+25+2j^
=65i^+125j^=65i^+2j^
∴cos⁡θ=-i^+j^⋅i^+2j^25=-1+210=110
⇒tan⁡θ=3

⇒tan⁡θ∈3,2+3

Hence, θ∈60°,75°

Topic: Mathematics, Vector Algebra, JEE Main
The vertices B and C of a ΔABC lie on the line, x+23=y-10=z4 such that BC=5 units. Then the area (in sq. units) of this triangle, given the point A1, -1, 2, is

Options:
A 6
B 234
C 34
D 517

The correct answer is:
34

Solution:
The points B and C lie on the line x+23=y-10=z4.

Draw perpendicular AD on the line BC.

Clearly area of ΔABC=12⋅AD⋅BC

To find a point on the line, let x+23=y-10=z4=r

⇒x+2=3r, y-1=0, z=4r

⇒x=3r-2, y=1, z=4r

Thus, the point D≡3r-2, 1, 4r

The direction ratios of a line joining two points x1, y1, z1 and x2, y2, z2 are <x2-x1, y2-y1, z2-z1>

Thus, the direction ratios of AD are <3r-2-1, 1+1, 4r-2>=<3r-3, 2, 4r-2>

Since AD is perpendicular to given line, hence the dot product of their direction ratios is zero.

⇒3·3r-3+0·2+4·4r-2=0

⇒9r-9+16r-8=0

⇒25r-17=0

⇒r=1725

⇒D≡3×1725-2, 1, 4×1725=125, 1, 6825

The distance between the points x1, y1, z1 & x2, y2, z2 is x1-x22+y1-y22+z1-z22

⇒AD=125-12+1+12+6825-22

⇒AD=576625+4+324625

⇒AD=13625

⇒AD=2534 units

Hence, the area of ΔABC=12⋅2534⋅5=34 sq units.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2019 (09 Apr Shift 2)
The vertices of a triangle are the points P-26,17, Q30,17 and R10,2. If G and Î be the centroid and incentre of the triangle PQR, then the value of GÎ2 is equal to

Options:
A 2059
B 2053
C 33
D 27

The correct answer is:
2059

Solution:
Coordinates of G are 143,12
PQ=56,QR=25,RP=39
x -coordinate of Ι is -26×25+30×39+10×5656+25+39=9
y -coordinate of Ι is 17×25+17×39+2×5656+25+39=10
⇒GΙ2=143-92+12-102=2059

Topic: Mathematics, Straight Lines, JEE Main
The vertices of the base of an isosceles triangle lie on a parabola $y^{2}=4 x$ and the base is a part of the line $y=2 x-4 .$ If the third vertex of the triangle lies on the $x$-axis, its coordinates are

Options:
A $\left(\frac{5}{2}, 0\right)$
B $\left(\frac{7}{2}, 0\right)$
C $\left(\frac{9}{2}, 0\right)$
D $\left(\frac{11}{2}, 0\right)$

The correct answer is:
$\left(\frac{9}{2}, 0\right)$

Solution:
$$
\begin{array}{l}
(2 x-4)^{2}=4 x \\
(x-2)^{2}=x \\
x^{2}-5 x+4=0 \\
x=1,4
\end{array}
$$
$$
\begin{array}{l}
\mathrm{C}(1,-2) \\
\mathrm{B}(4,4) \quad \because \mathrm{AB}=\mathrm{AC} \\
\sqrt{(\alpha-4)^{2}+16}=\sqrt{(\alpha-1)^{2}+4}
\end{array}
$$
On solving, we get $\alpha=\frac{9}{2}$

Topic: Mathematics, Parabola, JEE Main
The vertices of the triangle ABC are A0,0, B3,0 and C3,4, where A and C are foci of an ellipse and B lies on the ellipse. If the length of the latus rectum of the ellipse is 12p units, then the value of p is ______. If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
7

Solution:
AC=5=2ae
⇒2ae=5…i
Also AB+BC=2a
⇒2a=7 ⇒a=72
∴e=57

⇒1-b2a2=2549⇒b=6
Length of the latus rectum, 2b2a=2×672 units

⇒p=72=3.5

Topic: Mathematics, Ellipse, JEE Main
The vertices of the triangle ABC are A0,0, B3,0 and C3,4, where A and C are foci of an ellipse and B lies on the ellipse. If the length of the latus rectum of the ellipse is 12p units, then the value of p is

Options:

The correct answer is:
3.5

Solution:
AC=5=2ae
⇒2ae=5…i
Also AB+BC=2a
⇒2a=7 ⇒a=72
∴e=57

⇒1-b2a2=2549⇒b=6
Length of the latus rectum, 2b2a=2×672 units

⇒p=72=3.5

Topic: Mathematics, Ellipse, JEE Main
The viability of pollen grains is highly variable. The least pollen viability will be observed in:

Options:
A Soybeans
B Potato
C Peach
D Rice

The correct answer is:
Rice

Solution:
The period for which pollen grains remain viable is highly variable and to some extent depends on the prevailing temperature and humidity. In some cereals such as rice and wheat, pollen grains lose viability within 30 minutes of their release, and in some members of Rosaceae, Leguminoseae and Solanaceae, they maintain viability for months.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET
The viability of seeds is tested by:

Options:
A 2,6 dichlorophenol indophenols
B 2, 3, 5 triphenyl tetrazolium chloride
C DMSO
D Safranine.

The correct answer is:
2, 3, 5 triphenyl tetrazolium chloride

Solution:
Viability of seeds can be known by two methods: (i) ability to germinate,
(ii) testing their ability to respire.
All viable seeds respire. This can be tested by immersing a section of seed containing the embryo in $0.1 \%$ solution of triphenyl tetrazolium chloride. The viable embryo will turn pink due to conversion of colourless triphenyl tetrazolium chloride into insoluble coloured dye called triphenyl formazone due to reduction. Tetrazolium Chloride (TZ) test is also called as quick germination test.
Related Theory
Seed or plant viability is the measure of how many seeds or how much plant material in a lot are alive and could develop into plants that will reproduce under appropriate field conditions. It is very important that seeds or plant material stored in the gene bank are capable of producing plants when sown in the field. They must have high viability at the start of storage and maintain it during storage, as it will stored for a longer period of time.

Topic: Biology, Sexual Reproduction in Flowering Plants, NEET, NEET 2013 (Karnataka)
The vibrating of four air columns are represented in the figure. The ratio of frequencies np:nq:nr:ns is

Options:
A 12:6:3:5
B 1:2:4:3
C 4:2:3:1
D 6:2:3:4

The correct answer is:
1:2:4:3

Solution:
As is clear from figure of question
l=Î»p4, Î»=4l, np=vÎ»p=v4l
l=Î»q2,Î»q=2l, nq=vÎ»q=v2l
l=Î»r, Î»r=l, nr=vÎ»r=vl
l=3Î»s4, Î»s=4l3, ns=vÎ»s=3v4l
â´np:nq:nr:ns=v4l:v2l:vl:3v4l=1:2:4:3

Topic: Physics, Waves and Sound, JEE Main
The viruses which are excellent candidates for species-specific, narrow spectrum, insecticidal applications are

Options:
A Mainly belonging to the genus Nucleopolyhedrovirus
B Not suitable for Integrated Pest Management
C Can cause harm to useful insets as well
D Can cause soil pollution

The correct answer is:
Mainly belonging to the genus Nucleopolyhedrovirus

Solution:
The viruses which are excellent candidates for species-specific, narrow spectrum, insecticidal applications are Baculoviruses. They mainly belong to the genus Nucleopolyhedrovirus. They do not cause any harm to useful insects and are only effective against the target pathogens. They are used in the Integrated Pest Management programs, where useful insects are to be conserved.

Topic: Biology, Microbes in Human Welfare, NEET
The viscosity Î· of a gas depends on the long-range attractive part of the intermolecular force, which varies with molecular separation r according to F = Î¼rân where n is a number and Î¼ is a constant. If Î· is a function of the mass m of the molecules, their mean speed v, and the constant Î¼ then which of following is correct -

Options:
A η∝mn+1vn+3μn-2
B Î·âmn+1n-1 vn+3n-1 Î¼-2n-1
C η∝mnv-nμ-2
D η∝mv μ-n

The correct answer is:
Î·âmn+1n-1 vn+3n-1 Î¼-2n-1

Solution:
Dimension of Î·â¡FrAvâ¡MLT-2[L]L2LT-1
Î·â¡ML-1T-1
Dimensions of Î¼=Frn
Î¼=MLT-2Ln
Î¼=MLn+1T-2
Let âÎ·â depend on mass m mean speed v and constant Î¼ as -
Î·âmavbÎ¼c
ML-1T-1âMaLT-1bMLn+1T-2c
ML-1T-1âMa+cLb+cn+1T-b-2c
Equating dimensions both sides
a+c=1âc=1-a
b+cn+1=-1
âb=-1+cn+1
-b+2c=-1
âb=1-2c
â´1-2c=-1+cn+1
2c-1=1+cn+1
2c-cn+1=2
c2-n-1=2
c1-n=2
=c=-2n-1
a=1-c
a=1+2n-1=n-1+2n-1
a=n+1n-1
b=1-2c
=1+4n-1=n-1+4n-1
b=n+3n-1
â´Î·âmn+1n-1vn+3n-1Î¼-2n-1

Topic: Physics, Units and Dimensions, JEE Main
The vitamin which is essential for blood clotting is

Options:
A vitamin-A
B vitamin-B
C vitamin-C
D vitamin-K

The correct answer is:
vitamin-K

Solution:
Vitamin- K is essential for clotting of blood. It is necessary for the synthesis of various blood clotting factors like prothrombin (II), proconvertin (VII), Stuart power factor (X) times Christmas factor. It is obtained from cabbage, spinach and other green leafy vegetables.Vitamin-A is necessary for synthesis of rhodopsin of rod cells and iodopsin of cone cells of the retina of eye.Vitamin-D promotes absorption of calcium and phosphorus by the intestine. It maintains the normal functioning of parathormone.
Vitamin-B ( $\left.B_1\right)$ serves as a co-enzyme in the conversion of pyruvate into acetyl Co-A and $\alpha$-ketoglutarate.

Topic: Biology, Body Fluids and Circulation, JIPMER, JIPMER 2009
The voltage across the 10 Ω resistor in the given circuit is x volt.

The value of x to the nearest integer is _____.

Options:

The correct answer is:
70

Solution:
Req1=50×2070=1007

Req=1707
v1=1701707×10=70 V

=70.00

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The voltage drop across 15 Ω resistance in the given figure will be _______ V.

Options:

The correct answer is:
6

Solution:
equivalent circuit

i=126=2 A
i1=i2=i/2=1 A
V15Ω=i1×Rab=1×6=6 V

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The voltage gain of an amplifier with $9 \%$ negative feedback is 10 . The voltage gain without feedback will be

Options:
A 100
B 90
C 10
D 1.25

The correct answer is:
100

Solution:
Given $\frac{v_0}{v_i}=10$
and $\left(v_i-0.09 v_0\right) A=v_0$
$$
\Rightarrow \quad A=100
$$

Topic: Chemistry, Solid State, NEET, NEET 2008 (Mains)
The voltage of clouds is 4Ã106 V with respect to ground. In a lightning strike lasting 100 ms , a charge of 4 C is delivered to the ground. The average power of lightning strike is (assume complete discharges )

Options:
A 160 MW
B 80 MW
C 20 MW
D 500 KW

The correct answer is:
80 MW

Solution:
Energy stored between cloud and earth, assuming the capacitor model is
W=qV2
4Ã106Ã42=8Ã106 J
The power of lightning strike is
P=Wt=8Ã1060.1=80Ã106 W
=80 MW

Topic: Physics, Current Electricity, JEE Main
The voltage of domestic ac is 220 volt. What does this represent

Options:
A Mean voltage
B Peak voltage
C Root mean voltage
D Root mean square voltage

The correct answer is:
Root mean square voltage

Solution:
The voltage of domestic ac is $220 \mathrm{~V}$, it represents the root mean square voltage of supply.

Topic: Physics, Alternating Current, JEE Main
The voltage of the cell consisting of Lis and F2g electrodes is 5.92 V at standard condition at 298 K. What is the voltage if the electrolyte consists of 2 M LiF.
(ln2 = 0.693, R = 8.314 J K-1mol-1 and F = 96500 C mol-1 )

Options:
A 5.90 V
B 5.937 V
C 5.88 V
D 4.9 V

The correct answer is:
5.90 V

Solution:
2Lis+F2g→2LiFaq
Ecello=5.92V
Ecell=Ecello-0.05912log[LiF]2
Ecell=5.92-0.0592log22
=5.92-0.059×0.3010=5.90 V

Topic: Chemistry, Electrochemistry, JEE Main
The voltage of the cell consisting of Li s and F2g electrodes is 5.92 V at standard condition at 298 K . What is the voltage if the electrolyte consists of 2 M LiF .

lnâ¡2=0.693, R=8.314JK-1mol-1and F=96500C mol-1

Options:
A 5.90 V
B 5.937 V
C 5.88 V
D 4.9 V
E 4.8 V

The correct answer is:
5.88 V

Solution:
Now, the cell reaction is
Lis+12F2g→Li++F-
We know that,
Ecell=Ecello=RTnFln⁡ProductReactant
=Ecello-2.303RTnFlog⁡Li+F-
=5.92-2.303×8.314×2981×96500log⁡2×2
=5.92-0.0591×2log⁡2
=5.92-0.035=5.887 V

Topic: Chemistry, Electrochemistry, JEE Main
The volume charge density of a sphere of radius 6 m is 2 μC cm-3. The number of lines of force per unit surface area coming out from the surface of the sphere is _____ ×1010 N C-1.

[Given : Permittivity of vacuum ϵ0=8.85×10-12C2 N-1-m-2]

Options:

The correct answer is:
45

Solution:
Given here, ρ=2 μC cm-3 and R=6 m.

Number of lines of force per unit area = electric field at surface=KQR2

E=14πε0ρ43πR3R2

E=ρR3ϵ0=2×63×8.85×10-12=0.45×1012 N C-1

=45×1010 N C-1

Topic: Physics, Electrostatics, JEE Main, JEE Main 2022 (25 Jul Shift 1)
The volume elasticity is possessed by,

Options:
A solids only.
B liquids only.
C gases only.
D all the three states of matter.

The correct answer is:
all the three states of matter.

Solution:
Volume elasticity usually means bulk modulus which means stress required to change fractional change in volume.
Solid, liquid, and gas all substances can attain a definite volume.
Bulk modulus is a quantity possessed by all three states of matter.

Topic: Physics, Mechanical Properties of Solids, JEE Main
The volume in mL of 0.125 M AgNO3 required to quantitatively precipitate chloride ions in 0.3g of CoNH36Cl3 is __________.
CoNH36MCl3=267.46g/mol
AgNOM3=169.87g/mol

Report the nearest integer as the answer.

Options:

The correct answer is:
27

Solution:
ML6Cl30.3g+3AgNO3V ml, 0.125 M→3AgCl
0.3267.46×3=0.125×V×10-3
Or, V=0.3×3×1000267.46×0.125=26.92ml

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The volume (in \( \mathrm{mL} \) ) of \( 0.1 \mathrm{~N} \mathrm{NaOH} \) required to neutralise \( 10 \mathrm{~mL} \) of \( 0.1 \mathrm{~N} \) phosphinic acid is

Options:

The correct answer is:
10

Solution:
Phosphinic acid is hypo phosphorous acid H3PO2.

NaOH+H3PO2 → NaH2PO2+H2O

For neutrization

N1V1acid=N2V2base

0.1×10=0.1×VmLNaOH

VNaOH=10 mL

Topic: Chemistry, Redox Reactions, JEE Main