datavorous/entrance-exam-dataset huggingface dataset:
A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance 1Ω (Fig. 3.16). Determine the equivalent resistance of the network

Options:
A ⅔ Ω
B ⅙ Ω
C ⅚ Ω
D ⅓ Ω

The correct answer is:
⅚ Ω

Solution:
Correct Option is : (C)
⅚ Ω

Topic: Chemistry, Current Electricity, NEET
A battery of emf 12 V and internal resistance 2 Î© is connected in series with a tangent galvanometer of resistance 4 Î©. The deflection is 60o when the plane of the coil is along the magnetic meridian. To get a deflection of 30o, the resistance to be connected in series with the tangent galvanometer is

Options:
A 12 Ω
B 20 Ω
C 10 Ω
D 5 Ω

The correct answer is:
12 Ω

Solution:
I=ktanâ¡Î¸,k being reduction factor and Iâ1R
âÂ Â R+rR+r+R′=tanâ¡Î¸′tanâ¡Î¸
âÂ Â Â 4+26+R′=tanâ¡30otanâ¡60o
âÂ Â R′=12Â Î©

Topic: Physics, Experimental Physics, NEET
A battery of internal resistance 4 Î© is connected to the network of resistances, as shown in the figure. In order that the maximum power can be delivered to the network, the value of R in Î© should be

Options:
A 49
B 2
C 83
D 18

The correct answer is:
2

Solution:
The given figure is a balanced form of Wheatstone bridge therefore, 6R resistance can be removed.
â´ 1RP=1R1+1R2=13R+16R=12R
â RP=2R.
The power delivered to the network is maximum when internal resistance = External resistance.
4=2R
â R=2 Î©

Topic: Physics, Current Electricity, NEET
A battery of internal resistance 4 Î© is connected to the network of resistances as shown in the figure. To deliver maximum power to the network, the magnitude of resistance R (in Î©) should be x 2 1. Find x.

Options:
A 16
B 17
C 19
D 23

The correct answer is:
19

Solution:
The battery has an internal resistance of 4 Ω.
From maximum power transfer theorem, we know that for maximum power, the equivalent resistance of the network should be equal to the internal resistance of the battery.
Req=4 Ω
⇒7R12R19R=84R19=4 Ω
⇒ R=1921 Ω

Topic: Physics, Current Electricity, NEET
$A B C$ is an equilateral triangle with $O$ as its centre. $\mathrm{F}_1, \mathrm{~F}_2$ and $\mathrm{F}_3$ represent three forces acting along the sides $A B, B C$ and $A C$ respectively. If the total torque about $O$ is zero then the magnitude of $\mathrm{F}_3$ is

Options:
A $F_1+F_2$
B $F_1-F_2$
C $\frac{F_1+F_2}{2}$
D $2\left(F_1+F_2\right)$

The correct answer is:
$F_1+F_2$

Solution:
If we take clockwise torque
$\begin{aligned}
\tau & =\tau_{F_1}+\tau_{F_2}+\tau_{F_3} \\
0 & =F_1 r+F_2 r+F_3 r \\
F_3 & =F_1+F_2
\end{aligned}$

Topic: Physics, Rotational Motion, NEET, NEET 2012 (Screening)
A bead of mass m can slide without friction on a fixed circular horizontal ring of radius 3R having a centre at the point C. The bead is attached to one of the ends of spring of spring constant k. Natural length of spring is R and the other end of the spring is fixed at point O as shown in the figure. If the bead is released from position A, then the kinetic energy of the bead when it reaches point B is

Options:
A 252kR2
B 92kR2
C 8kR2
D 12kR2

The correct answer is:
8kR2

Solution:
Given,
Mass of bead =m
The radius of circular horizontal ring =3R
The natural length of spring =R
According to the conservation of energy,
KEi+PEi=KEf+PEf
⇒ 0+12kOA-R2=KEf+12kOB-R2
⇒ 0+12k5R-R2=KEf+12kR-R2
⇒ KEf=8kR2

Topic: Physics, Work Power Energy, NEET
A bead of mass m can slide without friction on a fixed circular horizontal ring of radius 3R having a centre at the point C. The bead is attached to one of the ends of spring with spring constant k. The natural length of spring is R and the other end of the spring is fixed at point O as shown in the figure. If the bead is released from position A, then the kinetic energy of the bead when it reaches point B is

Options:
A 252kR2
B 92kR2
C 8kR2
D 12kR2

The correct answer is:
8kR2

Solution:
Given,
Mass of bead =m
The radius of circular horizontal ring =3R
The natural length of spring =R
According to the conservation of energy,
KEi+PEi=KEf+PEf
⇒ 0+12kOA-R2=KEf+12kOB-R2
⇒ 0+12k5R-R2=KEf+12kR-R2
⇒ KEf=8kR2

Topic: Physics, Work Power Energy, NEET
A beam of cathode rays is subjected to crossed Electric ( $E$ ) and Magnetic fields $(B)$. The fields are adjusted such that the beam is not deflected. The specific charge of the cathode rays is given by
(where $V$ is the potential difference between cathod and anode)

Options:
A $\frac{\mathrm{B}^2}{2 \mathrm{VE}^2}$
B $\frac{2 \mathrm{VB}^2}{\mathrm{E}^2}$
C $\frac{2 \mathrm{VE}^2}{\mathrm{~B}^2}$
D $\frac{\mathrm{E}^2}{2 \mathrm{VB}^2}$

The correct answer is:
$\frac{\mathrm{E}^2}{2 \mathrm{VB}^2}$

Solution:
As the electron beam is not deflected, then or
$$
\begin{aligned}
\mathrm{F}_{\mathrm{m}} & =\mathrm{F}_{\mathrm{e}} \\
\mathrm{Bev} & =\mathrm{Ee} \\
\mathrm{v} & =\frac{\mathrm{E}}{\mathrm{B}}
\end{aligned}
$$
As the electron moves from cathode to anode, its potential energy at the cathode appears as its kinetic energy at the anode. If $V$ is the potential difference between the anode and cathode, then potential energy of the electron at cathode $=\mathrm{eV}$. Also, kinetic energy of the electron at anode $=\frac{1}{2} \mathrm{mv}^2$. According to law of conservation of energy
$$
\begin{aligned}
\frac{1}{2} \mathrm{mv}^2 & =\mathrm{eV} \\
\mathrm{v} & =\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}
\end{aligned}
$$
From Eqs. (i) and (ii), we have
$$
\begin{aligned}
\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}} & =\frac{E}{\mathrm{~B}} \\
\frac{\mathrm{e}}{\mathrm{m}} & =\frac{\mathrm{E}^2}{2 \mathrm{VB}^2}
\end{aligned}
$$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2010 (Screening)
A beam of light (Î»=600 nm) from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is

Options:
A 1.2 cm
B 1.2 mm
C 2.4 cm
D 2.4 mm

The correct answer is:
2.4 mm

Solution:
Distance between 1st order dark fringes = width of principle maximum
X=2λDd
=2×600×10-9×210-3
=2400×10-6
=2.4×10-3 m
=2.4mm

Topic: Physics, Wave Optics, NEET
A beam of light composed of red and green ray is incident obliquely at a point on the face of rectangular glass slab. When coming out on the opposite parallel face, the red and green ray emerge from:

Options:
A two points preparing in two different non parallel directions
B two points propagating in two different parallel directions
C one point propagating in two different directions
D one point propagating in the same directions

The correct answer is:
two points propagating in two different parallel directions

Solution:
Red and green rays emerge from two points propagating in two different parallel direction.

Topic: Physics, Ray Optics, NEET, NEET 2004
A beam of light consisting of red, green and blue colours is incident on a right-angled prism. The refractive index of the material of the prism for the above red, green and blue wavelengths are 1.39, 1.44 and 1.47 respectively.
The prism will:

Options:
A Separate all the three colours from one another
B Not separate the three colours at all
C Separate the red colour part from the green and blue colours
D Separate the blue colour part from the red and green colours

The correct answer is:
Separate the red colour part from the green and blue colours

Solution:
For Total internal reflection, \(i > i_{\mathrm{c}}\)
so, \(\sin i > \sin i_c\)
\(\sin 45^{\circ} > \frac{1}{\mu}\)
⇒μ=2
⇒μ=1.414
Since μ of green and violet is greater than 1.414, so they will undergo total internal reflection. But red colour will be refracted. So the prism will separate red colour from the green and blue colours.

Topic: Physics, Ray Optics, NEET, NEET 2015 (Phase 2)
A beam of light consisting of red, green and blue colours is incident on an isosceles right angled prism as shown in the figure. The refractive indices of the material of the prism for red, green and blue colours are 1.39, 1.43 and 1.47 respectively. The prism will

Options:
A separate red colour from green and blue colours.
B separate blue colour from red and green colours
C separate green colour from, red and blue colours.
D separate all the three colours from one another

The correct answer is:
separate red colour from green and blue colours.

Solution:
Refractive index and critical angle are related as,
μ=1sin⁡c
If the critical angle is 45°,
μ=1sin⁡45°=2
μ=1.414
Therefore green and blue undergo total internal reflection while red gets transmitted through the prism.

Topic: Physics, Ray Optics, NEET
A beam of light from a source L is incident normally on a plane mirror fixed at a certain distance x from the source. The beam is reflected back as a spot on a scale placed just above the source L. When the mirror is rotated through a small angle θ, the spot of the light is found to move through a distance y on the scale. The angle θ is given by

Options:
A y2x
B yx
C x2y
D xy

The correct answer is:
y2x

Solution:
When mirror is rotated by θ angle reflected ray will be rotated by 2θ .
for small θ, yx=2θ
θ=y2x

Topic: Physics, Ray Optics, NEET, NEET 2017
A beam of light is travelling from region II to region III (see the figure). The refractive index in the region I, II and III are n0,n02, and n02 respectively. The angle of incidence Î¸ for which the beam just misses entering region III is

Options:
A 30°
B 45Â°
C 60Â°
D sin-1â¡2

The correct answer is:
30°

Solution:
The critical angle for region II and III

sinâ¡C=μIIIμII=n02n02=12
ie. ∠ C=45°
The ray, if incident at 45° at the interface of II and III it will be totally internally reflected.
Now, from Snell's lay in region I and II.
n0sinâ¡θ=n02sinâ¡C
Orsinâ¡θ=12×sinâ¡45°
Orsinâ¡θ=12×12
Orsinâ¡θ=12
Or θ=30°

Topic: Physics, Ray Optics, NEET
A beam of light of λ=600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between first dark fringes on either side of the central bright fringe is:

Options:
A 1.2Â cm
B 1.2Â mm
C 2.4Â cm
D 2.4 mm

The correct answer is:
2.4 mm

Solution:
Angular width of 1st maxima
2Î¸=2Î»a
Linear width of 1st maxima =D2Î¸=2Î»Da=2Ã600Ã10-921Ã10-3=2.4 mm

Topic: Physics, Wave Optics, NEET, NEET 2014
A beam of light of a wavelength 600 nm from a distant source falls on a single slit of 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is

Options:
A 1.2 cm

B 1.2 mm

C 2.4 cm
D 2.4 mm

The correct answer is:
2.4 mm

Solution:
2 D Î» a = 2 Ã 6 0 0 Ã 1 0 - 9 Ã 2 1 0 - 3
= 2.4 Ã 1 0 - 3 m = 2.4 mm

Topic: Physics, Wave Optics, NEET
A beam of light of intensity I is normally incident on a plate of area A. If plate absorbs fraction η of incident light and reflects rest of it back, then the force exerted by the incident light on the plate is given by [c=speed of light]

Options:
A IAc1-η
B IAcη+1
C IAc2η-1
D IAc2-Î·

The correct answer is:
IAc2-η

Solution:
Force exerted by the fraction of ,ight that is absorbed is given as, F1=ηIAc.

Force exerted by the fraction of light that is totally reflected is given as, F2=1-η×2IAc.

Hence, total force will be given as, F=F1+F2=2-ηIAc.

Topic: Physics, Dual Nature of Matter, NEET
A beam of light of wavelength 600 nm from a distant source falls on a single slit 1mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of central bright fringe is

Options:
A 1.2Â cm
B 1.2Â mm
C 2.4Â cm
D 2.4 mm

The correct answer is:
2.4 mm

Solution:
The distance between the first dark fringes on either side of central bright fringe is
â2y=2DÎ»a
â2y=2Ã2Ã6Ã10-71Ã10-3
â2y=24Ã10-4 m
â2y=2.4 mm

Topic: Physics, Wave Optics, NEET
A beam of light of wavelength 600 nm from a distant source falls on a single slit 1mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is

Options:
A 1.2Â cm
B 1.2Â mm
C 2.4Â cm
D 2.4 mm

The correct answer is:
2.4 mm

Solution:
The distance between the first dark fringes is equal to width of central bright
2Dλa=2×600×10-9×210-3=2.4×10-3 m=2.4 mm

Topic: Physics, Wave Optics, NEET
A beam of metal supported at the two ends is loaded at the centre. The depression at the centre is proportional to

Options:
A Y2
B Y
C 1Y
D 1Y2

The correct answer is:
1Y

Solution:
Î´ = Depression in the beam
Y = Young's Modulus
W = Load Applied
L = Length of the beam
b = width of the beam
d = thickness of the beam
Î´=WL34Ybd3
â´ Î´â1Y

Topic: Physics, Mechanical Properties of Solids, NEET
A beam of natural light falls on a system of 5 nicol prisms, whose transmission axis are turned, each through an angle Î± = 30o, with respect to the previous prism. The fraction of the intensity that passed through the system is

Options:
A 0.24
B 0.16
C 0.12
D 0.32

The correct answer is:
0.16

Solution:
I=I02Ãcos2304=I02Ã344
= I02Ã81256=I0Ã81512â0.16

Topic: Physics, Wave Optics, NEET
A beam of α- particle is incident on a gold foil. Corresponding to the incident beams A,B and C , the emergent beams A′,B′,and C′ . The transmission and deflection of α- particles through the foil take place such that

Options:
A the number of Î±- particle in A′ is maximum and in B′ minimum
B the number of Î±- particles in A′ is minimum and in C′ maximum
C the number of Î±- particles in A′ , B′ and in C′ is the same.
D the number of Î±- particles in B′ is minimum and in C′ maximum

The correct answer is:
the number of Î±- particle in A′ is maximum and in B′ minimum

Solution:
From figure, we can see that B is reflected totally, so transmission of B is minimum.
But A is passing through foil with minimum deviation.
Therefore, transmission of A is maximum.

Topic: Physics, Atomic Physics, NEET
A beam of unpolarized light of intensity $I_0$ is passed through a polaroid $A$, then through another polaroid $B$, oriented at $60^{\circ}$ and finally through another polaroid $C$, oriented at $45^{\circ}$ relative to $B$ as shown. The intensity of emergent light is:

Options:
A $\frac{I_0}{16}$
B $\frac{I_0}{4}$
C $\frac{I_0}{2}$
D $\frac{I_0}{32}$

The correct answer is:
$\frac{I_0}{16}$

Solution:
$I_1=\frac{I_0}{2}$
$I_B=\left(I_1\right) \cos ^2 60^{\circ}$
$=\frac{I_0}{2}\left(\frac{1}{2}\right)^2=\frac{I_0}{8}$
$I_C=I_B \cos ^2 45^{\circ}$
$=\frac{I_0}{8} \times\left(\frac{1}{\sqrt{2}}\right)^2=\frac{I_0}{16}$

Topic: Physics, Wave Optics, NEET, NEET 2024 (Re-NEET)
A Beckmann thermometer is used to measure

Options:
A High temperature
B Low temperature
C Normal temperature
D All temperature

The correct answer is:
Low temperature

Solution:
A Beckmann thermometer is a device used to measure small differences of temperature, but not absolute temperature values.
It's length is usually 40 – 50 cm. The temperature scale typically covers about 5 °C and it is divided into hundredths of a degree. With a magnifier it is possible to estimate temperature changes to 0.001 °C.

Topic: Chemistry, Thermodynamics (C), NEET
A biconvex lens has a radius of curvature of magnitude $20 \mathrm{~cm}$. Which one of the following options describe best the image formed of an object of height $2 \mathrm{~cm}$ placed $30 \mathrm{~cm}$ from the lens?

Options:
A Virtual, upright, height $=0.5 \mathrm{~cm}$
B Real, inverted, height $=4 \mathrm{~cm}$
C Real, inverted, height $=1 \mathrm{~cm}$
D Virtual, upright, height $=1 \mathrm{~cm}$

The correct answer is:
Real, inverted, height $=4 \mathrm{~cm}$

Solution:
In general we have assumed $\mu=1.5$
So, $f=20 \mathrm{~cm}$
$$
\begin{aligned}
\frac{1}{f} & =\frac{1}{y}+\frac{1}{u} \\
\frac{1}{20} & =\frac{1}{y}+\frac{1}{30} \\
\frac{1}{y} & =\frac{1}{20}-\frac{1}{30}=\frac{10}{600} \\
y & =60 \mathrm{~cm} \\
\frac{h_i}{h_0} & =2 \\
h_i & =2 \times \mid h_o \\
h_i & =4 \mathrm{~cm}
\end{aligned}
$$
Here, Image is real, inverted, magnified field and height of image is $4 \mathrm{~cm}$.

Topic: Physics, Ray Optics, NEET, NEET 2011 (Screening)
A biconvex lens has radii of curvature 20 cm each. If the refractive index of the material of the lens is 1.5, the power of the lens is:

Options:
A +20 D
B +5 D
C infinity
D +2 D

The correct answer is:
+5 D

Solution:
Power of the lens is,

P=1f=μ-11R1-1R2⇒P=1.5-110.2-1-0.2=0.5×10⇒P=+5 D

Topic: Physics, Ray Optics, NEET, NEET 2022 (Phase 1)
A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is

Options:
A Virtual and at a distance of 16 cm from the mirror
B Real and at a distance of 16 cm from the mirror

C Virtual and at a distance of 20 cm from the mirror
D None of the above

The correct answer is:
Real and at a distance of 16 cm from the mirror

Solution:
Object is placed at distance 2f from the lens. So first image I will be formed at distance 2f on other side. This image I1 will behave like a virtual object for mirror. The second image I2 will be formed at distance 20 cm in front of the mirror, or at distance 10 cm to the left hand side of the lens.
Now applying lens formula
1v-1u=1f
â´1v-1+10=1+15
or v=16 cm
Therefore, the final image is at distance 16 cm from the mirror. But, this image will be real.
This is because ray of light is travelling from right to left.

Topic: Physics, Ray Optics, NEET
A biconvex lens of focal length 40 cm is placed in front of an object at a distance of 20 cm. Now a slab of refractive index 43 is placed somewhere in between the lens and the object. The shift in the image formed after the introduction of slab equals (thickness of the slab is 2 mm)

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

The correct answer is:
2 mm

Solution:
Before the slab is introduced,
f=40 cm, u=-10 cm
1v-1u=1f
⇒1v+120=140
⇒v=-40 cm
Shift in image of object due to slab =t1-1μS
=21-34
=12 mm=0.5 mm=du
Also, dvdu=v2u2
dv=v2u2du
40202du=4×0.5 mm
=2 mm

Topic: Physics, Ray Optics, NEET
A biconvex lens of focal length f forms a circular image of sun of radius r in focal plane. Then?

Options:
A πr2 ∝ f
B Ïr2 âf2
C If lower half part is covered by black sheet, then area of the image is equal to Ïr22
D If f is doubled, intensity will increase

The correct answer is:
πr2 ∝f2

Solution:
r = radius of sun's image
tanâ¡Î¸=rf=const
â´ Ïr2 âf2
Intensity only depends on the number of rays forming the image. Also only two rays are sufficient to form an image.

Topic: Physics, Ray Optics, NEET
A biconvex lens of focal length f forms a circular image of the radius r of the sun in the focal plane. Then which option is correct?

Options:
A Ïr2âf
B ÏrÂ²âf2
C If lower half part is covered by black sheet, then area of the image is equal to Ï r 2 2
D If f is doubled, intensity will increase

The correct answer is:
πr²∝f2

Solution:
r=ftanθ
Or r∝f
∴ πr2∝f2
small angled parallel rays meet at the focal plane.

Topic: Physics, Ray Optics, NEET
A big circular coil of 1000 turns and average radius 10 m is rotating about its horizontal diameter at 2 rad s-1. If the vertical component of earth's magnetic field at that place is 2×10-5 T and electrical resistance of the coil is 12.56 Ω, then the maximum induced current in the coil will be:

Options:
A 1.5 A
B 1 A
C 2 A
D 0.25 A

The correct answer is:
1 A

Solution:
The magnetic flux is given by,

ϕ=BAcosωt. Therefore, induced EMF,

ε=-dϕdt=-BAωsinωt

For maximum current sinωt=1

⇒imax=BAωR=BNπr2ωR=2×10-5×1000×3.14×100×212.56=1 A

Topic: Physics, Electromagnetic Induction, NEET, NEET 2022 (Phase 1)
A big explosion on the moon cannot be heard on the earth because

Options:
A the explosion produces high frequency sound waves which are inaudible.
B sound waves require a material medium for propagation.
C sound waves are absorbed in the atmosphere of moon.
D sound waves are absorbed in the earth's atmosphere.

The correct answer is:
sound waves require a material medium for propagation.

Solution:
As the sound waves are mechanical waves they requires medium for propagation.

Topic: Physics, Waves and Sound, NEET
A big particle of mass 3+m kg blasts into 3 pieces, such that a particle of mass 1 kg moves along x -axis, with velocity 2 m s-1 and a particle of mass 2 kg moves with velocity 1 m s-1 perpendicular to direction of 1 kg particle. If the third particle moves with velocity 2 m s-1 , then m is

Options:
A 2 kg
B 1Â kg
C 22Â kg
D none of these

The correct answer is:
2 kg

Solution:
conservation of linear momentum
0=1Ã2i^+2Ã1j^+mvâ
âmvâ=-2i^-2j^
âmv=22
â m2=22m=2 kg

Topic: Physics, Rotational Motion, NEET
A big water drop is formed by the combination of n small water drops of equal radii. The ratio of the surface energy of n drops to the surface energy of the big drop is

Options:
A n2Â :1
B nÂ :1
C nÂ :1
D n3 :1

The correct answer is:
n3 :1

Solution:
Volume of big drop = n (Volume of small drop)
43ÏR3=n.43Ïr3
R3=nr3
R=n13.r
E2=surface energy of n drops=nÃ4Ïr2ÃT
E1=surface of energy of big drop=4ÏR2T
∴ E 2 E 1 = n r 2 R 2 = n r 2 ( n 1 3 .r ) 2 = n r 2 n 2 3 . r 2 ∴ E 2 E 1 = n 1 3 = n 3 :1

Topic: Physics, Mechanical Properties of Fluids, NEET
A billiard ball moving at a speed of 6.6 m s-1 strikes an identical stationary ball elastically. After the collision, one ball is found to be moving at a speed of 3.3 m s-1 in a direction making an angle of 60Â° with the original line of motion. The velocity of the other ball is

Options:
A 4.4 m s-1
B 6.6 m s-1
C 3.3 m s-1
D 5.7 m s-1

The correct answer is:
5.7 m s-1

Solution:
Applying conservation of momentum
m6.6i^+0=m3.3cosâ¡60oi^+3.3sinâ¡60oj^+mvâ
vâ=4.95i^-2.85j^
v=5.7 ms-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
A bimetallic is made of two strips A and B having coefficients of linear expansion Î±A and Î±B. If Î±A< Î±B, then on heating, the strip will

Options:
A Bend with A on outer side
B Bend with B on outer side
C Not bend at all
D None of the above

The correct answer is:
Bend with B on outer side

Solution:
AsαB> αA, therefore, strip B will appear on outer side.

Topic: Physics, Thermal Properties of Matter, NEET
A bimetallic strip consists of metals X and Y . It is mounted rigidly at the base, as shown in the figure. The metal X has a higher coefficient of expansion as compared to that for metal Y . When the bimetallic strip is placed in a cold bath

Options:
A it will bend towards the right
B it will bend towards the left
C it will not bend but shrink
D it will neither bend nor shrink

The correct answer is:
it will bend towards the left

Solution:
As Î±x>Î±y, on cooling X contracts more than Y . So, the strip bends towards X , i.e., towards left.

Topic: Physics, Waves and Sound, NEET
A binary solid (A+B-) has zinc blende structure with B- ions constituting the lattice and A+ ions occupying 25% tetrahderal holes. The formula of solid is

Options:
A AB
B A2B
C AB2
D AB4

The correct answer is:
AB2

Solution:
Suppose number of B- ions =N
So number of tetrahedral holes =2N
As 25% tetrahedral holes are occupied by A+ ions.
So number of A+ ions =2N4=N2
So ratio of A+:B-=N2:N
1:2
So the formula is AB2

Topic: Chemistry, Solid State, NEET
A binary solution contains x1 and x2 mole fraction of two components having vapour pressure p1o and p2o in their pure states. The total vapour pressure above the solution is

Options:
A (p1o-p2o)x1+p2o
B (p2o-p1o)x1+p2o
C (p1o-p2o)x1+p1o
D (p2o-p1o)x1+p1o

The correct answer is:
(p1o-p2o)x1+p2o

Solution:
According to Raoult's law, we have
p1=p1ox1 and p2=p2ox2
So p= p 1 + p 2 = p 1 o x 1 + p 2 o 1â x 1
∵ [x1+x2=1]
p=(p1o-p2o)x1+p2o

Topic: Chemistry, Solutions, NEET
A biocatalyst

Options:
A decreases the activation energy and decreases the energy yield
B increases the activation energy and increases the energy yield
C decreases the activation energy and has no effect on the energy yield
D increases the energy yield and decreases the activation energy

The correct answer is:
decreases the activation energy and has no effect on the energy yield

Solution:
Enzyme is a catalyst that reduces the activation energy for a reaction and helps in increasing the concentration of product without any change in energy yield.

Topic: Chemistry, Biomolecules, NEET