datavorous/entrance-exam-dataset huggingface dataset:
A steam volatile organic compound which is immiscible with water has a boiling point of $250^{\circ} \mathrm{C}$. During steam distillation, a mixture of this organic compound and water will boil :

Options:
A above $100^{\circ} \mathrm{C}$ but below $250^{\circ} \mathrm{C}$
B above $250^{\circ} \mathrm{C}$
C at $250^{\circ} \mathrm{C}$
D close to but below $100^{\circ} \mathrm{C}$

The correct answer is:
close to but below $100^{\circ} \mathrm{C}$

Solution:
If one of the substances in the mixture is water and the other, a water insoluble substance, then the mixture will boil close to but below, $373 \mathrm{~K}\left(100^{\circ} \mathrm{C}\right)$.

Topic: Chemistry, General Organic Chemistry, NEET, NEET 2024 (Re-NEET)
A steel ball mass 0.1 kg falls freely from a height of 10 m and bounces to a height of 5.4 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is (specific heat of steel = 460 K kg-1 Â°C-1, g=10 m s-2)

Options:
A 0.01 °C

B 0.1 Â°C
C 1 Â°C
D 1.1 Â°C

The correct answer is:
0.1 Â°C

Solution:
Loss in energy = mg(h - h')
= 0.1×10×(10 - 5.4)
= 4.6 J
Now,. 4.6 J = ms Δ θ
= 0.1 × 4 6 0 × Δ θ
∴ Δ θ = 0.1 ∘ C

Topic: Physics, Thermal Properties of Matter, NEET
A steel rod of length 2l, cross sectional area A and mass M is set rotating in a horizontal plane about an axis passing through the centre. If Y is the Young’s modulus for steel, find the extension in the length of the rod.

Options:
A $\delta =\frac{2}{3YA}\mu {{\omega }^{2}}{{l}^{2}}$
B $2\delta =\frac{2}{3YA}\mu {{\omega }^{2}}{{l}^{2}}$
C $2\delta =\frac{2}{5YA}\mu {{\omega }^{2}}{{l}^{2}}$
D $\delta =\frac{2}{5YA}\mu {{\omega }^{2}}{{l}^{2}}$

The correct answer is:
$2\delta =\frac{2}{3YA}\mu {{\omega }^{2}}{{l}^{2}}$

Solution:
Correct Option is : (B)
$2\delta =\frac{2}{3YA}\mu {{\omega }^{2}}{{l}^{2}}$

Topic: Chemistry, Mechanical Properties of Solids, NEET
A steel rod (Y = 2.0 × 1011Nm–2; and α = 10–50C–1) of length 1 m and area of cross-section 1 cm2is heated from 0°C to 200°C, without being allowed to extend or bend. What is the tension produced in the rod?

Options:
A 2 × 104N
B 3 × 104N
C 4 × 104N
D 5 × 104N

The correct answer is:
4 × 104N

Solution:
Correct Option is : (C)
4 × 104N

Topic: Chemistry, Mechanical Properties of Solids, NEET
A steel scale is to be prepared such that the millimeter intervals are to be accurate within 6Ã10-5 mm. The maximum temperature variation from the temperature of calibration during the reading of the millimeter marks is (linear coefficient of expansion for steel is given as Î±=12Ã10-6 Â°C-1)

Options:
A 4.0°C
B 4.5°C
C 5.0Â°C
D 5.5°C

The correct answer is:
5.0°C

Solution:
We know that thermal extension is given by,

∆l =l0α∆T
6×10-5=1×12×10-6×Δ⊤

=6×10-512×10-6

ΔT=5°C

Topic: Physics, Thermal Properties of Matter, NEET
A steel wire can withstand a load up to 2940 N. A load of 150 kg is suspended from a rigid support. The maximum angle through which the wire can be displaced from the mean position, so that the wire does not break when the load passes through the position of equilibrium, is

Options:
A 30°
B 60°
C 80°
D 85°

The correct answer is:
60°

Solution:
While crossing the equilibrium position
T=mg+mv2r=mg+ml2gl1-cos⁡θ
=mg+2mg-2mgcos⁡θ=μγ3-2cos⁡θ
∴3-2cos⁡θ=29401509.8=2
⇒2cos⁡θ=1⇒cos⁡θ=12⇒θ=60°

Topic: Physics, Laws of Motion, NEET
A steel wire of length l has a magnetic moment M. It is bent in L - shape (Figure). The new magnetic moment is

Options:
A M
B M2
C M2
D 2M

The correct answer is:
M2

Solution:
Magnetic moment, M=ml
Ml=m, where m is the pole strength. Therefore distance between poles
= l22+l22=l2
So, M′=ml2=M2

Topic: Physics, Magnetic Properties of Matter, NEET
A steel wire of length l has a magnetic moment M. It is then bent into a semicircular arc. The new magnetic moment is

Options:
A Mπ
B 2Mπ
C 3Mπ
D 4Mπ

The correct answer is:
2Mπ

Solution:
Let pole strength =m
So, M=ml
When wire is in form of arc, then the distance between poles = 2l π
So, M ′ = m2l π = 2M π

Topic: Physics, Magnetic Properties of Matter, NEET
A steel wire, of uniform area 2 m2, is heated up to 50Â°C and is stretched by tying its ends rigidly. The change in tension, when the temperature falls from 50 Â°C to 30 Â°C is (Take Y=2Ã1011 N m-2, Î±=1.1 Ã10-5 Â°C-1)

Options:
A 1.5 Ã1010 N
B 5 N
C 88 N
D 2.5Ã1010 N

The correct answer is:
88 N

Solution:
âL= Î±LâT=FALY
â´ F= Î±âTAY
=1.1Ã10-5Ã50-30Ã2Ã10-6Ã2Ã1011

=88 N

Topic: Physics, Thermal Properties of Matter, NEET
A steel wire, of uniform area 2 m2, is heated up to 50Â°C and is stretched by tying its ends rigidly. The change in tension, when the temperature falls from 50Â°C to 30Â°C is (Take Y=2Ã1011N m-2, Î±=1.1 Ã10-5Â°C-1)

Options:
A 1.5 Ã1010 N
B 5 N
C 88 N
D 2.5Ã1010 N

The correct answer is:
88 N

Solution:
âL= Î±LâT=FALY
â´ F= Î±âTAY
=1.1Ã10-5Ã50-30Ã2Ã10-6Ã2Ã1011

=88 N

Topic: Physics, Thermal Properties of Matter, NEET
A stone falls freely under gravity. It covers distances $h_1, h_2$ and $h_3$ in the first 5 seconds, the next 5 seconds and the next 5 seconds respectively. The relation between $h_1, h_2$ and $h_3$ is

Options:
A $h_1=2 h_2=3 h_3$
B $h_1=\frac{h_2}{3}=\frac{h_3}{5}$
C $h_2=3 h_1$ and $h_3=3 h_2$
D $h_1=h_2=h_3$

The correct answer is:
$h_1=\frac{h_2}{3}=\frac{h_3}{5}$

Solution:
The relation between $h_1, h_2$ and $h_3$ is
$$
\begin{aligned}
h_1 & =\frac{h}{3}=\frac{h_3}{5} \\
\Rightarrow h_1: h_2: h_3 & =1: 3: 5
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, NEET, NEET 2013 (All India)
A stone hanging from a massless string of length 15 m is projected horizontally with speed 147 m s-1 . Then the speed of the particle, at the point where tension in string equals the weight of particle, is

Options:
A 10 m s-1
B 7 m s-1
C 12 m s-1
D none of these

The correct answer is:
7 m s-1

Solution:
T-mgcos⁡θ=mv2l
and v2-u2=-2gl1-cos⁡θ
also T=mg

Topic: Physics, Laws of Motion, NEET
A stone is accelerated upwards by a cord whose breaking strength is three times the weight of the stone. The maximum acceleration with which the stone can be moved up without breaking the cord is

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

The correct answer is:
2g

Solution:
3mg=mg+ma
or 2mg=ma or a=2g

Topic: Physics, Laws of Motion, NEET
A stone is dropped from a height of 45 m on a horizontal level ground. There is horizontal wind blowing due to which horizontal acceleration of the stone becomes 10â m/s 2. (Take g = 10â m/s 2 ). The time taken t by stone to reach the ground and the net horizontal displacement x of the stone from the time it is dropped and till it reaches the ground are respectively

Options:
A t = 3 s, x = 45 m
B t = 4 s, x = 54 m
C t = 3 s, x = 54 m
D t = 4 s, x = 45 m

The correct answer is:
t = 3 s, x = 45 m

Solution:
Taking motion in vertical direction,
u=0, g=10 m/s2, h=45 m
h=ut+ 1 2 g t 2
⇒ h=0+12 gt2
⇒t=2hg=2×4510
⇒ t = 3 sec.
(b) Taking motion in horizontal direction,
u=0, a=10 m/s2, t=3 sec
x=ut+12at2
⇒x=0+121032⇒x=45 m

Topic: Physics, Motion In Two Dimensions, NEET
A stone is projected vertically upward to reach maximum height h. The ratio of its kinetic energy to its potential energy at a height 4 5 h,â¡ will be

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

The correct answer is:
1:4

Solution:
At a height 4 5 h,⁡
Potential energy = mg × 4 h⁡ 5 = 4 5 mgh
Total energy = mgh
∴ K E at that height = mgh - 4 5 mgh⁡ = 1 5 mgh⁡
∴ At a height 4 5 h⁡ ,⁡ the ratio of
KE PE = 1 5 mgh 4 5 mgh = 1 4

Topic: Physics, Work Power Energy, NEET
A stone is released from an aeroplane which is rising with upward acceleration of 5 m s-2. Here g=10 m s-2. Two seconds after the release, the separation between stone and aeroplane will be

Options:
A 10 m
B 20Â m
C 30 m
D 25Â m

The correct answer is:
30 m

Solution:
∴ Acceleration of aeroplane with respect to the earth,
aA→E=5 m s-2
Acceleration of stone with respect to the earth,
aS→E =-10 m s-2
or aE→S =-aS→E =10 m s-2
Thus, the acceleration of aeroplane with respect to the stone,
aA→S =aA→E+aE→S=5+10=15 m s-2
∴ The relative separation between the stone and aeroplane in 2 s,
s=12aA→St2=12×15×22=30 m

Topic: Physics, Motion In One Dimension, NEET
A stone is thrown vertically upward. When the particle is at a height half of its maximum height, its speed is 10 m s-1, then the maximum height attained by the particle is
[g=10 m s-2]

Options:
A 8 m

B 10 m

C 15 m
D 20 m

The correct answer is:
10 m

Solution:
Let h be the maximum height to be attained by stone
The velocity with which particle is thrown
u=2gh
v2= u2–2gH
When H=h2, u=2gh, v=10 m s-1
100=2gh-2gh2⇒h=10 m

Topic: Physics, Motion In One Dimension, NEET
A stone is tied to a string of length ' $l$ ' and is whirled in a vertical circle with the other end of the string as the centre. At a certain instant of time, the stone is at its lowest position and has a speed ' $u$ '. The magnitude of change in velocity as it reaches a position where the string is horizontal ( $g$ being acceleration due to gravity)is:

Options:
A $\sqrt{2\left(u^2-g l\right)}$
B $\sqrt{u^2-g l}$
C $u-\sqrt{u^2-2 g l}$
D $\sqrt{2 g l}$

The correct answer is:
$\sqrt{2\left(u^2-g l\right)}$

Solution:
According to the question
Total energy at $\mathrm{A}=$ Total energy at $\mathrm{B}$

$\begin{array}{l}
\Rightarrow \frac{1}{2} m u^2 =\frac{1}{2} m v^2+m g l \\
\Rightarrow v^2 =\sqrt{u^2-2 g l}
\end{array}$
$\therefore$ Change in magnitude of velocity
$=\sqrt{u^2+v^2}=\sqrt{2\left(u^2-g l\right)}$

Topic: Physics, Work Power Energy, NEET, NEET 2004
A stone is tied to one end of a string and is rotated in a horizontal circle with a uniform angular velocity. Let T be the tension in the string. If the length of the string is halved and the angular velocity of the stone is doubled, the tension in the string will be

Options:
A 2T
B 4T
C T
D 8T

The correct answer is:
2T

Solution:
Tension in the string
Initial T 1 = m Ï 2 r
Finally T 2 = m r 2 Â· 2 Ï 2
= 2 mr Ï 2
T 2 T 1 = 2

Topic: Physics, Laws of Motion, NEET
A stone of density 2000 kg m-3 completely immersed in a lake is allowed to sink from rest. If the effect of friction is neglected, then after 4 seconds, the stone will reach a depth of

Options:
A 78.4 m
B 39.2Â m
C 19.6Â m
D 9.8Â m

The correct answer is:
39.2 m

Solution:
Given, density of stone Ï=2000 kg m-3
Density of water Ï=1000 kg m-3
According to question,
gâ²=g1-ÏÏ
gâ²=g1-10002000
gâ²=g2
We know that,
S=ut+12gâ²t2
S=12gâ²t2
S=12Ãg2Ã42
S=12Ã9.82Ã42
S=9.8Ã164
S=39.2 m

Topic: Physics, Motion In One Dimension, NEET
A stone of mass 0.3 kg attached to a 1.5 m long string is whirled around in a horizontal circle at a speed of 6 m/s. The tension in the string is

Options:
A 10 N

B 20 N

C 7.2 N

D None of these

The correct answer is:
7.2 N

Solution:
The force of tension in the string provides required centripetal force to keep the stone in circular motion.
â´T=mÏ2R=0.3Ã621.5=7.2 N

Topic: Physics, Laws of Motion, NEET
A stone of mass 1 kg is tied at one end of a string of length 1 m. It is whirled in a vertical circle at a constant speed of 4 m s-1. The tension in the string is 6 N when the stone is at (g=10 m s-2)

Options:
A Top of the circle
B Bottom of the circle
C Half way down
D None of these

The correct answer is:
Top of the circle

Solution:
When the string makes an angle θ with the vertical, then
T-mgcosθ=mv2r
Substituting the values, we obtain
6 - 1 1 0 cos θ = 4 2 . 1 1 = 16
cos θ = - 1
θ = 1 8 0 o

Topic: Physics, Laws of Motion, NEET
A stone of mass 1 kg is tied to a string 4 m long and is rotated at a constant speed of 40 m s-1 in a vertical circle. The ratio of the tension at the top and the bottom is

Options:
A 11 : 12

B 39 : 41

C 41 : 39

D 12 : 11

The correct answer is:
39 : 41

Solution:
Ttop=mv2r-mg ...(i)
Tbottom=mv2r+mg ...(ii)
TtopTbottom=v2r-gv2r+g
=400-10400+10=390410=3941

Topic: Physics, Laws of Motion, NEET
A stone of mass 1 kg tied to the end of a string of length 1 m, is whirled in a horizontal circle with a uniform angular velocity 2 rad s-1. The tension of the string is (in newton)

Options:
A 2
B 13
C 4
D 14

The correct answer is:
4

Solution:
Tension in string will provide required centripetal force.
The tension of the string,
T=mrÏ2
=1Ã1Ã(2)2=4 N

Topic: Physics, Laws of Motion, NEET
A stone of mass 2 kg is projected upward with the kinetic energy of 98 J. The height at which the kinetic energy of the stone becomes half of its original value is given by

Options:
A 5 m

B 2.5 m

C 1.5 m
D 0.5 m

The correct answer is:
2.5 m

Solution:
At the time of projection kinetic energy of the stone,
K⁡ = 1 2 mu 2
where m is the mass of the stone and u is the velocity of the projection
or u⁡ 2 = 2 K⁡ m⁡ = 2 × 9 8 2 = 9 8
Using, v2 = u2 – 2gh
∴ h⁡ = u⁡ 2 2 g ∵ v = 0 … i
h⁡ = 9 8 2 × 9 · 8 = 5 m
Also, K⁡ = 1 2 m⁡ 2 gh Using i
K⁡ ′ = 1 2 m⁡ v ′ 2 = 1 2 m⁡ × 2 gh ′
∴ K⁡ ′ K⁡ = h⁡ ′ h⁡
According to the problem
K⁡ ′ = K⁡ 2
K⁡ 2 K⁡ = h⁡ ′ h⁡
h⁡ ′ = h⁡ 2 = 5 2 m = 2 · 5 m

Topic: Physics, Work Power Energy, NEET
A stone of mass 0.3 kg attached to a 1.5 m long string is whirled around in a horizontal circle on a frictionless table at a speed of 6 m s-1. The tension in the string is

Options:
A 10 N
B 20 N
C 7.2 N
D none of these

The correct answer is:
7.2 N

Solution:
The force of tension in the string provides the required centripetal force to keep the stone in circular motion.
â´ T = mv 2 R = 0 . 3 Ã 6 2 1 . 5 = 7 . 2 N

Topic: Physics, Laws of Motion, NEET
A stone of mass 500 g is dropped from the top of a tower of 100 m height and simultaneously other stone of mass 1 kg is thrown horizontally with a speed of 10 m/s from same point. The height of the centre of mass of the above two stone system after 3 s is g=10 m/s2

Options:
A 45Â m
B 35Â m
C 55 m
D none of these

The correct answer is:
55 m

Solution:
If height of stones is h1 and h2 after t=3 s then,
h1=100-12gt2=55 m
h2=100-12gt2=55 m
hCM=m1h1+m2h2m1+m2=m(55+55)2m

hCM=55 m

Topic: Physics, Center of Mass Momentum and Collision, NEET
A stone of mass 500 g is dropped from the top of a tower of 100 m height. Simultaneously another stone of mass 1 kg is thrown horizontally with a speed of 10 m s-1 from same point. The height of the centre of mass of the above two stone system after 3 sec is g=10 m s-2

Options:
A 45 m
B 35 m
C 55 m
D none of these

The correct answer is:
55 m

Solution:
If height of stones is h1 and h2 after t=3 sec then,
h1=100-12gt2=55 m
h2=100-12gt2=55 m
∴ height of centre of mass of the two stone system,
hCM=m1h1+m2h2m1+m2=55 m

Topic: Physics, Center of Mass Momentum and Collision, NEET
A stone of mass m is tied to an elastic string of negligible mass and spring constant k. The unstretched length of the string is L and has negligible mass. The other end of the string is fixed to a nail at a point P. Initially the stone is at the same level as the point P. The stone is dropped vertically from point P. Find the distance y from the top when the mass comes to rest for an instant, for the first time

Options:
A $y=\frac{\left( kL-mg \right)\pm \sqrt{mgkL+{{m}^{2}}{{g}^{2}}}}{k}$
B $y=\frac{\left( kL+mg \right)\pm \sqrt{mgkL+{{m}^{2}}{{g}^{2}}}}{k}$
C $y=\frac{\left( kL+mg \right)\pm \sqrt{2mgkL+{{m}^{2}}{{g}^{2}}}}{k}$
D $y=\frac{\left( kL-mg \right)\pm \sqrt{2mgkL+{{m}^{2}}{{g}^{2}}}}{k}$

The correct answer is:
$y=\frac{\left( kL+mg \right)\pm \sqrt{2mgkL+{{m}^{2}}{{g}^{2}}}}{k}$

Solution:
Correct Option is : (C)
$y=\frac{\left( kL+mg \right)\pm \sqrt{2mgkL+{{m}^{2}}{{g}^{2}}}}{k}$

Topic: Chemistry, Mechanical Properties of Solids, NEET
A stone of mass m is tied to an elastic string of negligible mass and spring constant k. The unstretched length of the string is L and has negligible mass. The other end of the string is fixed to a nail at a point P. Initially the stone is at the same level as the point P. The stone is dropped vertically from point P. What shall be the nature of the motion after the stone has reached its lowest point?

Options:
A ${{y}_{o}}=L+\frac{2m}{k}g$
B ${{y}_{o}}=L+\frac{m}{k}g$
C ${{y}_{o}}=L-\frac{2m}{k}g$
D ${{y}_{o}}=L-\frac{m}{k}g$

The correct answer is:
${{y}_{o}}=L+\frac{m}{k}g$

Solution:
Correct Option is : (B)
${{y}_{o}}=L+\frac{m}{k}g$

Topic: Chemistry, Mechanical Properties of Solids, NEET
A stone of mass m is tied to an elastic string of negligible mass and spring constant k. The unstretched length of the string is L and has negligible mass. The other end of the string is fixed to a nail at a point P. Initially the stone is at the same level as the point P. The stone is dropped vertically from point P. What is the maximum velocity attained by the stone in this drop?

Options:
A v=(2gL+2mg2/k)1/2
B v=(gL+2mg2/k)1/2
C v=(gL+mg2/k)1/2
D v=(2gL+mg2/k)1/2

The correct answer is:
v=(2gL+mg2/k)1/2

Solution:
Correct Option is : (D)
v=(2gL+mg2/k)1/2

Topic: Chemistry, Mechanical Properties of Solids, NEET
A stone tied to the end of a string of 1 $\mathrm{m}$ long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of acceleration of the stone?

Options:
A $\pi^2 \mathrm{~ms}^{-2}$ and direction along the radius towards the centre
B $\pi^2 \mathrm{~ms}^{-2}$ and direction along the radius away from the centre
C $\pi^2 \mathrm{~ms}^{-2}$ and direction along the tangent to the circle
D $\pi^2 / 4 \mathrm{~ms}^{-2}$ and direction along the radius towards the centre

The correct answer is:
$\pi^2 \mathrm{~ms}^{-2}$ and direction along the radius towards the centre

Solution:
\(a=\frac{v^2}{r}=\omega^2 r=4 \pi^2 n^2 r=4 \pi^2\left(\frac{22}{44}\right)^2 \times 1=\pi^2 \mathrm{~m} / \mathrm{s}^2\)
and its direction is always along the radius and towards the centre.

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2005
A stop codon will always have a

Options:
A uracil nucleotide at is 5' end
B adenine nucleotide at is 5' end
C uracil nucleotide at is 3' end
D adenine nucleotide at is 3' end

The correct answer is:
uracil nucleotide at is 5' end

Solution:
The codons or triplets of genetic code that signal the termination of protein chain during translation process are called stop codons. UAA, UAG and UGA are the stop codons and it always has uracil nucleotide at is 5' end.

Topic: Biology, Molecular Basis of Inheritance, NEET
A straight conductor carrying current $i$ splits into two parts as shown in the figure. The radius of the circular loop is $R$. The total magnetic field at the centre $P$ at the loop is

Options:
A Zero
B $3 \mu_0 i / 32 R$, outward
C $3 \mu_0 i / 32 R$, inward
D $\frac{\mu_0 i}{2 R}$, inward

The correct answer is:
Zero

Solution:
The magnetic field at the centre of an arc subtended at an angle $\theta$ is given by
$\mathrm{B}=\frac{\mu_0 \mathrm{i}}{2 \mathrm{R}} \times \frac{\theta}{2 \pi}$

Then, the magnetic field due to larger arc $A B$ is
$\mathrm{B}_1=\frac{\mu_0 \mathrm{i}_1}{2 \mathrm{R}} \times \frac{270}{2 \pi}$
which acts in inward direction according to right hand thumb rule.
And magnetic field due to smaller arc $A B$ is
$\mathrm{B}_2=\frac{\mu_0 \mathrm{i}_2}{2 \mathrm{R}} \times \frac{90}{2 \pi}$
which acts in outward direction.
The resultant magnetic field
$\begin{aligned}
& \mathrm{B}_{\mathrm{R}}=\mathrm{B}_1+\mathrm{B}_2 \\
& =-\frac{\mu_0 \mathrm{i}_1 \times 270}{4 \pi \mathrm{R}}+\frac{\mu_0 \mathrm{i}_2 \times 90}{4 \pi \mathrm{R}} \text { [From Eq. (i) and (ii)] }
\end{aligned}$
which acts in inward direction as $\mathrm{B}_1>\mathrm{B}_2$.
Two arcs can also be seen as the two resistances in parallel combination.
So, the potential across them will be same i.e.
$\begin{aligned}
\mathrm{V}_1 & =\mathrm{V}_2 \\
\mathrm{i}_1 \mathrm{R}_1 & =\mathrm{i}_2 \mathrm{R}_2
\end{aligned}$
where, $\mathrm{R}_1$ and $\mathrm{R}_2=$ Resistance of respective segments
The wire is uniform so
$\begin{aligned}
& \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{L}_1}{\mathrm{~L}_2}=\frac{\mathrm{R} \times 270}{\mathrm{R} \times 90} [\because \text { length of arc }=\text { radius } \times \\
& \text { angle] }
\end{aligned}$
From Eq. (iv), we get
$\Rightarrow \frac{\mathrm{i}_1}{\mathrm{i}_2}=\frac{\mathrm{R}_2}{\mathrm{R}_1}=\frac{90}{270}=\frac{1}{3}$
or $3 i_1=i_2$
From Eq. (iii) and (v), we get
$\begin{aligned}
\mathrm{B}_{\mathrm{R}} & =\frac{\mu_0}{4 \pi \mathrm{R}}\left(-270 \mathrm{i}_1+90 \mathrm{i}_2\right) \\
& =\frac{\mu_0}{4 \pi R}\left[-270 \mathrm{i}_1+90\left(3 \mathrm{i}_1\right)\right] \\
& =\frac{\mu_0}{4 \pi \mathrm{R}}\left(-270 \mathrm{i}_1+270 \mathrm{i}_1\right)=0
\end{aligned}$

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2019 (Odisha)
A straight conductor of length 0.4 m is moving with a speed of 7 ms-1 perpendicular to the magnetic field of intensity of 0.9 Wb m-2. The induced emf across the conductor will be

Options:
A 7.25 V
B 5.52 V
C 1.25 V
D 2.52 V

The correct answer is:
2.52 V

Solution:
If a rod of length l is moved with velocity vâ at an angle Î¸ to the length of the rod in a filed Bâ which is perpendicular to the plane of motion, the flux linked with the area generated by the motion of rod in time t.
Ï=Blv sinÎ¸t
So, e=dÏdt=Bv l sinÎ¸
This will be maximum when
sinÎ¸=max=1
i.e., the rod is moving perpendicular to its length and then
emax=Bvl
â´e=0.9Ã7Ã0.4=2.52 V

Topic: Physics, Electromagnetic Induction, NEET
A straight fertilizer is the one, which is

Options:
A Absorbed by roots directly
B Absorbed by the plants from aerial spray
C Having only one primary nutrient
D Not easily leached

The correct answer is:
Having only one primary nutrient

Solution:
Fertilizers are classified in several ways. They are classified according to whether they provide a single nutrient (e.g., K, P, or N), in which case they are classified as "straight fertilizers." "Multinutrient fertilizers" (or "complex fertilizers") provide two or more nutrients, for example N and P. A fertilizer, which contains only one nutrient is known as straight fertilizer or simple fertilizer.

Topic: Biology, Microbes in Human Welfare, NEET
A straight wire carrying current I is made into a circular loop. If M is the magnetic moment associated with the loop, then the length of the wire is

Options:
A L=M4πI
B L=M2πL
C L=4πMI
D L=4πM⋅I

The correct answer is:
L=4πMI

Solution:
Let the length of the wire =L,
then L=2Ïr
Where, r= radius of the loop.
â´r=L2Ï
â´Â Area of the loop, A=Ïr2
=Ï⋅L24Ï2=L24Ï
Thus, magnetic moment of the loop,
M=IA=I⋅L24Ï
â´Â Â L=4ÏMI

Topic: Physics, Magnetic Effects of Current, NEET
A straight wire of diameter 0.5 mm carrying a current of 1 A is replaced by another wire of 1 mm diameter carrying the same current. The strength of the magnetic field far away is

Options:
A twice the earlier value
B same as the earlier value
C one - half of the earlier value
D one - quarter of the earlier value

The correct answer is:
same as the earlier value

Solution:
The current is same in both the wire, hence magnetic field induced will be same.

Topic: Physics, Magnetic Effects of Current, NEET
A straight wire of length 0.5 m and carrying a current of 1.2 A is placed in a uniform magnetic field of induction 2 T. The magnetic field is perpendicular to the length of the wire. The force on the wire is

Options:
A 2.4 N
B 1.2 N
C 3.0 N
D 2.0 N

The correct answer is:
1.2 N

Solution:
F=Bil=2 ×1.2 ×0.5=1.2 N

Topic: Physics, Magnetic Effects of Current, NEET
A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field B. the magnitude of B (in tesla) is

Options:
A 0.65
B 0.55
C 0.75
D 0.45

The correct answer is:
0.65

Solution:
Given, the mass of the wire =200 g=0.2 kg
Length of the wire 1.5 m
Current i=2 A
Magnetic field B=?
The force acting on the current-carrying wire in a uniform magnetic field
F=Bilsin⁡θ
F=Bil ∵ θ=90o
Weight of the wire w=mg=0.2×9.8 N
In the position of the suspension
Bil=mg
B=mgil=0.2×9.82×1.5=0.65 T

Topic: Physics, Magnetic Effects of Current, NEET