datavorous/entrance-exam-dataset huggingface dataset:
A concave lens of glass of refractive index 1.5, has both surfaces of the same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a

Options:
A convergent lens of focal length 3.5R
B convergent lens of focal length 3.0R
C divergent lens of focal length 3.5R
D divergent lens of focal length 3.0R

The correct answer is:
convergent lens of focal length 3.5R

Solution:
Focal length of equiconcave lens
f=-R2μ-1
Where μ=μlensμmedium
Therefore focal length in the medium
fmed=-R21.51.75-1
fmed=+R20.251.75
fmed=+3.5R
Positive sign implies convergent sign.

Topic: Physics, Ray Optics, NEET
A concave mirror for face viewing has a focal length of 0.4 m. The distance at which you hold the mirror from your face in order to see your image upright with a magnification of 5 is

Options:
A 1.60Â m
B 0.16Â m
C 0.32 m
D 0.24Â m

The correct answer is:
0.32 m

Solution:
m=5=-vu ⇒v=-5u
f=-0.4 m

From mirror formula,
1v+1u=1f
1-5u+1u=1-0.4
-1+55u=1-0.4
u=45-0.4=-0.32 m
So, distance of object from mirror=0.32 m

Topic: Physics, Ray Optics, NEET
A concave mirror of focal length $f_1$ is placed at a distance of $d$ from a convex lens of focal length $f_2$. A beam of light coming from infinity and falling on this convex lens concave mirror combination returns to infinity. The distance $d$ must be equal

Options:
A $f_1+f_2$
B $-f_1+f_2$
C $2 f_1+f_2$
D $-2 f_1+f_2$

The correct answer is:
$2 f_1+f_2$

Solution:
$d=2 f_1+f_2$

Topic: Physics, Ray Optics, NEET, NEET 2012 (Screening)
A concave mirror of focal length f produces an image P time the size of the object. If the image is real, then the distance of the object from the mirror is

Options:
A P-1f
B P+1f
C P-1Pf
D P+1Pf

The correct answer is:
P+1Pf

Solution:
Given, Magnification, M=IO=-vu=-P
Or v=uP.
Using mirror's formula, 1v+1u=1f
Here, the image of the real object is also real, therefore, v.u.f, all are on the negative side.
Hence, 1-uP+1-u=1-f
â 1u1P+1=1f
â u=P+1Pf

Topic: Physics, Ray Optics, NEET
A condenser of capacitance 10 Î¼F has been charged to 100 V. It is now connected to another uncharged condenser in parallel. The common potential becomes 40 V. The capacitance of another condenser is

Options:
A 15 Î¼F
B 5 Î¼F
C 10 Î¼F
D 16 Î¼F

The correct answer is:
15 Î¼F

Solution:
15 Î¼F
C1=10 Î¼F C2=?
V1=100 V V2=0
Vcommon=40 V
Vcommon=C1V1+C2V2C1+C2
40=10100+010 +C2
C2=15 Î¼F

Topic: Physics, Electrostatics, NEET
A condenser of capacitance C is fully charged by a 200 V supply. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat 250 J kg-1 K-1 and of mass 100 g. If the temperature of the block rise by 0.4 K, then the value of C is

Options:
A 300 μF
B 200 μF
C 400 μF
D 500 μF

The correct answer is:
500 μF

Solution:
Given that
V = 200 V
m = 100g = 0.1 kg
s=250 J kg-1K-1
∆θ=0.4 K
C =?
The energy stored in the capacitor is given by
U=12CV2
=12C×2002
=2C×104 J
The energy is used to heat up the block. Let ∆θ be the rise in temperature, then heat energy is given by
O=ms∆θ=0.1×250×0.4
= 10 J
Now,
2C×104=10
⇒ C=102×104=5×10-4
∴ C=500 μF

Topic: Physics, Current Electricity, NEET
A condition where an individual heterozygous for two pairs of linked genes (AaBb) possesses the two dominant genes on one homologous chromosome pair and two recessive on the other it is said to be

Options:
A Cis-arrangement
B Trans-arrangement
C Partly cis partly trans
D More than one option is correct

The correct answer is:
Cis-arrangement

Solution:
Genetic linkage is the tendency of genes that are located proximal to each other on a chromosome to be inherited together during meiosis. Genes whose loci are nearer to each other are less likely to be separated onto different chromatids during chromosomal crossover, and are therefore said to be genetically linked. Cis arrangement means one chromosome has both dominant alleles and its homologous has both recessive alleles. Hence a condition where an individual heterozygous for two pairs of linked genes (AaBb) possesses the two dominant genes on one homologous chromosome pair and two recessive on the other it is said to be Cis arrangement.

Topic: Biology, Principles of Inheritance and Variation, NEET
A conducting circular loop is placed in a uniform magnetic field $0.04 \mathrm{~T}$ with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at $2 \mathrm{mms}^{-1}$. The induced emf in the loop when the radius is $2 \mathrm{~cm}$ is

Options:
A $3.2 \pi \mu \mathrm{V}$
B $4.8 \pi \mu \mathrm{V}$
C $0.8 \pi \mu \mathrm{V}$
D $1.6 \pi \mu \mathrm{V}$

The correct answer is:
$3.2 \pi \mu \mathrm{V}$

Solution:
Key Idea According to Faraday's second law of electromagnetic induction the induced emf is given by rate of change of magnetic flux linked with the circuit.
Here, $B=0.04 \mathrm{~T}$ and $\frac{-\mathrm{dr}}{\mathrm{dt}}=2 \mathrm{mms}^{-1}$
Induced emf, $\mathrm{e}=\frac{-\mathrm{d} \phi}{\mathrm{dt}}=\frac{-\mathrm{BdA}}{\mathrm{dt}}=-\mathrm{B} \frac{\mathrm{d}\left(\pi \mathrm{r}^2\right)}{\mathrm{dt}}$
$$
=-\mathrm{B} \pi 2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}
$$
Now, if $\mathrm{r}=2 \mathrm{~cm}$
$$
\begin{aligned}
\mathrm{e} & =-0.04 \times \pi \times 2 \times 2 \times 10^{-2} \times 2 \times 10^{-3} \\
& =3.2 \pi \mu \mathrm{V}
\end{aligned}
$$

Topic: Physics, Electromagnetic Induction, NEET, NEET 2009 (Screening)
A conducting circular loop is placed in a uniform magnetic field, $B=0.025 \mathrm{~T}$ with its plane perpendicular to the loop. The radius of the loop is made to shrink at a constant rate of $1 \mathrm{mms}^{-1}$. The induced emf when the radius is $2 \mathrm{~cm}$, is

Options:
A $2 \pi \mu \mathrm{V}$
B $\pi \mu \mathrm{V}$
C $\frac{\pi}{2} \mu \mathrm{V}$
D $2 \mu \mathrm{V}$

The correct answer is:
$\pi \mu \mathrm{V}$

Solution:
Given: The magnetic field in the region is, \(\mathrm{B}=0.025 \mathrm{~T}\).
Radius of the loop, \(\mathrm{r}=2 \mathrm{~cm}=2 \times 10^{-2}\)
Constant rate at which radius of the loop shrimks, \(\frac{\mathrm{dr}}{\mathrm{dt}}=1 \times 10^{-3} \mathrm{~ms}^{-1}\)
To find: The induced emf in the loop when radius is 2 cm.
The magnetic flux linked with the loop is given by:
\(\begin{aligned}
& \phi=\mathrm{BA} \cos \theta \\
& =\mathrm{B}\left(\pi \mathrm{r}^2\right) \cos \mathrm{O}^{\circ} \\
& \Rightarrow \mathrm{B} \pi \mathrm{r}^2
\end{aligned}\)
The magnitude of induced emf is given as:
\(\begin{aligned}
& |\epsilon|=\frac{\mathrm{d} \phi}{\mathrm{dt}} \\
& =\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{B} \pi \mathrm{r}^2\right) \\
& =\mathrm{B} \pi 2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}
\end{aligned}\)
Substitute the values.
\(\begin{aligned}
& |\epsilon|=0.025 \times \pi \times 2 \times 2 \times 10^{-2} \times 1 \times 10^{-3} \\
& =\pi \times 10^{-6} \mathrm{~V} \\
& \Rightarrow \pi \mu \mathrm{V}
\end{aligned}\)

Topic: Physics, Electromagnetic Induction, NEET, NEET 2010 (Screening)
A conducting circular loop of radius r carries a constant current i. It is placed in a uniform magnetic field B0 which is perpendicular to the plane of the loop. The magnetic force acting on the loop is

Options:
A irB0
B 2πirB0
C zero
D πirB0

The correct answer is:
zero

Solution:
The magnetic field is perpendicular to the plane of the paper. Let us consider two diametrically opposite elements. By Flemingâs Left hand rule on element AB the direction of force will be Leftwards and the magnitude will be
dF=Idl Bsin90o=IdlB
On element CD, the direction of force will be towards right on the plane of the paper and the magnitude will be dF=IdlB.

Topic: Physics, Magnetic Effects of Current, NEET
A conducting loop (as shown) has total resistance R. A uniform magnetic field B=γt is applied perpendicular to plane of the loop where γ is a constant and t is time. The induced current flowing through loop is

Options:
A b2+a2γtR
B b2-a2γR
C b2-a2γtR
D b2+a2γR

The correct answer is:
b2-a2γR

Solution:
Given,
Uniform magnetic field B=Î³t
Total flux, B1â. A1â+B2â.A2â
=Bb2cosâ¡0o+Ba2cosâ¡180o
=Bb2-Ba2
Ï=B(b2-a2)
Ï=Î³t(b2-a2)
We know that,
Induced current (i) =eR=dÏdtR
From Equation (i),
i=ddt[Î³tb2-a2]R
i=b2-a2Î³R

Topic: Physics, Electromagnetic Induction, NEET
A conducting loop carrying a current I is placed in a uniform magnetic field pointing into the plane of the paper as shown. The loop will have a tendency to

Options:
A Contract
B Expand
C Move towards +ve x-axis
D Move towards -ve x-axis

The correct answer is:
Expand

Solution:
Net force on a current carrying loop in uniform magnetic field is zero. Hence the loop can't translate. So, options (c) and (d) are wrong. From Fleming's left hand rule we can see that if magnetic field is perpendicular to paper inwards and current in the loop is clockwise (as shown) the magnetic force Fm on each element of the loop is radially outwards, or the loops will have a tendency to expand.

Topic: Physics, Magnetic Effects of Current, NEET
A conducting loop carrying a current I is placed in a uniform magnetic field pointing into the plane of the paper as shown. Then, the loop will have a tendency to

Options:
A Expand
B Move towards +ve x-axis
C Contract
D Move towards - ve x-axis

The correct answer is:
Expand

Solution:
The net force on a current-carrying loop in a uniform magnetic field is zero. Hence, the loop cannot translate.
From Fleming's left-hand rule, we can see that if the magnetic field is perpendicular to paper inwards and current in the loop is clockwise the magnetic force Fm on each element of the loop is radially outwards or the loop will have the tendency of expanding.

Topic: Physics, Magnetic Effects of Current, NEET
A conducting rod AB moves parallel to x-axis in a uniform magnetic field, pointing in the positive z-direction. The end A of the rod gets

Options:
A positively charged
B negatively charged
C neutral
D first positively charged and then negatively charged

The correct answer is:
positively charged

Solution:
When the conducting rod AB moves parallel to x-axis in a uniform magnetic field pointing in the positive z direction, then according to Fleming's left hand rule, the electrons will experience a force towards B. Hence end A will be positive.

Topic: Physics, Electromagnetic Induction, NEET
A conducting rod AB of length l=1 m is moving at a velocity Î½=4m s-1 making an angle of 30Â° with its length. A uniform magnetic field B=2 T exists in a direction perpendicular to the plane of motion, then potential difference is

Options:
A VA-VB=8 V
B VA-VB=4 V
C VB-VA=8 V
D VB-VA=4 V

The correct answer is:
VA-VB=4 V

Solution:
e=BÎ½sinâ¡30Â°l=4 V
Using Fleming's right hand rule, induced current is from B to A
â VA-VB=4 V

Topic: Physics, Electromagnetic Induction, NEET
A conducting rod AB of length l=1 m is moving at a velocity v=4 m s-1 making an angle of 30Â° with its length. A uniform magnetic field B=2 T exists in a direction perpendicular to the plane of motion, then

Options:
A vA-vB=8 v
B vA-vB=4 v
C vB-vA=8 v
D vB-vA=4 v

The correct answer is:
vA-vB=4 v

Solution:
e=Bv sin30ol=4 volt
Fleming right hand Iind B to A
⇒VA-VB=4 volt

Topic: Physics, Electromagnetic Induction, NEET
A conducting rod of length l is moving in a transverse magnetic field of strength B with velocity v. The resistance of the rod is R. The current in the rod is

Options:
A BlvR
B Blv
C Zero
D B2v2l2R

The correct answer is:
Zero

Solution:
Since the rod is moving in transverse magnetic field, so it will cut no flux passing through the field and hence no induced emf is produced. So, no current will flow through the rod.

Topic: Physics, Electromagnetic Induction, NEET
A conducting rod PQ of length L=1.0 m is moving with a uniform speed v=2.0 m s-1 in a uniform magnetic field = 4.0 T directed into the paper. A capacitor of capacity C=10 Î¼F is connected as shown in the figure. Then,

Options:
A qA=-80 μC and qB=+80 μC
B qA=+80 Î¼C and qB=-80 Î¼C
C qA=0=qB
D Charge stored in the capacitor increases expotentially with time

The correct answer is:
qA=+80 Î¼C and qB=-80 Î¼C

Solution:
Motional emf across PQ
V=Blv=41 2= 8 V
This is the potential to which the capacitor is charged.
As q=CV
∴ q=10×10-68=10-5C=80 μC
As the magnetic force on an electron in the conducting rod PQ is towards Q, therefore, A is positively charged and B is negatively charged
ie, qA=+80 μC and qB=-80 μC

Topic: Physics, Electromagnetic Induction, NEET
A conducting sphere of radius R is given a charge Q. The electric potential and the electric field at the center of the sphere respectively are:

Options:
A Zero and Q4ÏÎµ0 R2
B Q4ÏÎµ0 R and Zero
C Q4ÏÎµ0 R and Q4ÏÎµ0 R2
D Both are zero

The correct answer is:
Q4ÏÎµ0 R and Zero

Solution:
This is a learning based question where we have to choose the option which is correct for the electric potential and electric field at the centre of the sphere.

For a conducting sphere
Electric field at centre =0
Potential at centre =kQR=Q4πε0 R

So Option (2) will be correct.

Topic: Physics, Electrostatics, NEET, NEET 2014
A conducting square frame of side a and a long straight wire carrying current I are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity V. The e.m.f induced in the frame (when the centre of the frame is at a distance x from the wire) will be proportional to :

Options:
A 1x2
B 12x-a2
C 12x+a2
D 12x-a2x+a

The correct answer is:
12x-a2x+a

Solution:
The potential difference across AB is
VA-VB=B1 a.V.
⇒ μi2π x-a2 aV
The potential difference across CD is
VC-VD=B2 a.V
B2=μ0i2πx+a2
VC-VD= μ0i2πx+a2 aV
Net Potential difference =μi aV2 π 1x - a2- 1x + a2
( V A − V B )−( V C − V D )= μia 2π ( 2a x 2 − a 2 4 ) ∝ 14x2-a2 ∝ 12x+a(2x-a)

Topic: Physics, Electromagnetic Induction, NEET, NEET 2015 (Phase 1)
A conducting wheel in which there are four rods of length 25 cm as shown in the figure is rotating with constant angular velocity 20 rad/s in a uniform magnetic field 8 T. The induced potential difference between its centre and rim will be

Options:
A 4 V

B 5 V

C â6 V
D ​5 V

The correct answer is:
​5 V

Solution:
V (induced) = 1 2 B ω ℓ 2
= 12  ×  8  × 20  × (25100)2 =  5 V

Topic: Physics, Electromagnetic Induction, NEET
A conducting wire frame is placed in a magnetic field, which is directed into the paper. The magnetic field is increasing at a constant rate. The directions of induced currents in wires AB and CD are

Options:
A A to B and C to D
B B to A and C to D
C A to B and D to C
D B to A and D to C

The correct answer is:
B to A and D to C

Solution:
In the given question the magnetic field directed into the paper is increasing at a constant rate.
â´ Induced current should produce a magnetic field directed out of the paper. Thus current in both loops must be anticlockwise considering individually.
Induced emf on right side of loop will be more as area of its loop on right side is more.
âµe=-dÏdt=-AdBdt
eâA
Hence the net current induced in the complete loop will be along DCBAD .

Topic: Physics, Electromagnetic Induction, NEET
A conduction rod of 1 m length and 1 kg mass is suspended by two vertical wires through its ends. An external magnetic field of 2 T is applied normal to the rod. Now the current to be passed through the rod so as to make the tension in the wires zero is Take g=10 m s-2

Options:
A 15 A
B 5 A
C 1.5 A
D 2.5 A

The correct answer is:
5 A

Solution:
Magnetic force on the rod = BIl
Weight of the rod = mg
For no tension in the wire, BIl=mg
or I=mgBl=1×102×1=5 A

Topic: Physics, Magnetic Effects of Current, NEET
A conductor is carrying a current i. The magnetic field intensity at the point O which is the common centre for three arcs is

Options:
A 5Âµ0iÎ¸24ÏR
B µ0iθ24πR
C 11µ0iθ24πR
D zero

The correct answer is:
5µ0iθ24πR

Solution:
Find B due to each arc at O and add them using vector addition.
Since magnetic field at the centre of an arc is equal to B=Âµ0I4ÏrÎ¸
Hence, net B=Âµ0i4Ï 1R-12R+13RÎ¸=5Âµ0iÎ¸24ÏR

Topic: Physics, Magnetic Effects of Current, NEET
A conductor of length L is placed along the x-axis, with one of its ends at x=0 and the other at x=L. If the rate of flow of heat energy through the conductor is constant and its thermal resistance per unit length is also constant, then which of the following graphs is/are correct?

Options:
A
B
C
D

The correct answer is:

Solution:
We know that heat is flowing dQdt=-KAT-T0x
T=-dQdtxKA+T0
Therefore the graph of the negative slope but positive intercept

Topic: Physics, Thermal Properties of Matter, NEET
A conductor of resistance 3 Î© is stretched uniformly till its length is doubled. The wire is now bent in the form of an equilateral triangle. The effective resistance between the ends of any side of the triangle in ohms is

Options:
A 92
B 83
C 2
D 1

The correct answer is:
83

Solution:
R=3 Ω
When stretched R′=4×3=12 Ω
Therefore each side of the triangle has resistance 4 Ω
RAB=8×48+4=3212=83 Ω

Topic: Physics, Current Electricity, NEET
A conductor with a cavity is charged positively and its surface charge density is σ. If E and V represent the electric field and potential, then inside the cavity

Options:
A E=0 and V=0
B E=0 and σ= constant
C V=0 and σ= constant
D E=0 and V= constant

The correct answer is:
E=0 and V= constant

Solution:
Surface charge density =σ
Inside the surface, the electric field is zero.
⇒ Einside=0
∵ Vinside=Vsurface
The potential inside the surface of charged sphere is constant
∴ E=0, V= constant.

Topic: Physics, Electrostatics, NEET
A conductor with rectangular cross-section has dimensions a×2a×4a as shown in the figure. Resistance across AB is x, across CD is y and across EF is z. Then -
.

Options:
A x=y=z
B x>y>z
C y>z>x
D x>z>y

The correct answer is:
x>z>y

Solution:
We know that R=ÏlA
x=Ï4aaÃ2a=2Ïa
y=Ïa2aÃ4a=Ï8a
z=Ï2aaÃ4a=Ï2a
â´ x>z>y

Topic: Physics, Current Electricity, NEET
A constant pressure air thermometer gave a reading of 47.5 units of volume when immersed in ice cold water and 67 units in a boiling liquid. The boiling point of the liquid will be

Options:
A 135 °C
B 125 °C
C 112 °C
D 100 °C

The correct answer is:
112 °C

Solution:
Initial volume, V1=47.5 units
Temperature of ice cold water, T1=0 °C=273 K
Final volume, V2=67 units

Applying Charles's law, we have V1T1=V2T2 (where temperature T2 is the boiling point)
⇒T2=V2V1×T1=67×27347.5=385 K=112 °C

Topic: Physics, Thermal Properties of Matter, NEET
A constant torque of $100 \mathrm{~N} \mathrm{~m}$ turns a wheel of moment of inertia $300 \mathrm{~kg} \mathrm{~m}^2$ about an axis passing through its centre. Starting from rest, its angular velocity after $3 \mathrm{~s}$ is

Options:
A $10 \mathrm{rad} / \mathrm{s}$
B $15 \mathrm{rad} / \mathrm{s}$
C $1 \mathrm{rad} / \mathrm{s}$
D $5 \mathrm{rad} / \mathrm{s}$

The correct answer is:
$1 \mathrm{rad} / \mathrm{s}$

Solution:
About fixed axis $\rightarrow$ Torque $=I \alpha$
$\begin{aligned} & \alpha=\frac{100}{300} \\ & \alpha=\frac{1}{3} \mathrm{rad} / \mathrm{s}^2\end{aligned}$
$\begin{aligned} & \Rightarrow \omega_i=0 \\ & \omega_f=\omega_i+\alpha t \\ & \text { at } t=3 \mathrm{~s} \\ & \omega_f=\frac{1}{3} \times 3=1 \mathrm{rad} / \mathrm{s}\end{aligned}$

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2023 (Manipur)
A container of volume $200 \mathrm{~cm}^3$ contains 0.2 mole of hydrogen gas and 0.3 mole of argon gas. The pressure of the system at temperature $200 \mathrm{~K}\left(R=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$ will be

Options:
A $4.15 \times 10^5 \mathrm{~Pa}$
B $4.15 \times 10^6 \mathrm{~Pa}$
C $6.15 \times 10^5 \mathrm{~Pa}$
D $6.15 \times 10^4 \mathrm{~Pa}$

The correct answer is:
$4.15 \times 10^6 \mathrm{~Pa}$

Solution:
We know that
$\begin{aligned} & P V=n R T \\ & P_1=\frac{n_1}{V} R T\end{aligned}$
$\begin{gathered}P_2=\frac{n_2}{V} R T \\ P_{\text {net }}=P_1+P_2\end{gathered}$
$\begin{aligned} & =\left(n_1+n_2\right) \frac{R T}{V} \\ & =(0.2+0.3) \times \frac{8.3 \times 200}{200 \times 10^{-6}}\end{aligned}$
$=8.3 \times 0.5 \times 10^6=4.15 \times 10^6 \mathrm{~Pa}$

Topic: Physics, Thermodynamics, NEET, NEET 2023 (Manipur)
A converging lens has a focal length of 0.12 m. To get an image of unit magnification the object should be placed at what distance from the converging lens?

Options:
A 0.24 m
B 0.12Â m
C 0.06Â m
D 0.4Â m

The correct answer is:
0.24 m

Solution:
Given: f=+0.12 m (converging lens),
To get a unit magnification from a converging lens, i.e., convex lens, the object must be placed at 2f.
u=2Ã0.12 m=0.24 m

Topic: Physics, Ray Optics, NEET
A converging lens having magnitude of focal length as f1 is kept coaxially in contact with a diverging lens having magnitude of focal length as f2. The focal length of the combination would be:

Options:
A f1f2f1-f2
B

f 1 + f 2 f 1 â f 2

C

f 1 − f 2 f 1   f 2

D

f 1   f 2 f 1 + f 2

The correct answer is:
f1f2f1-f2

Solution:
We have to consider sign convention as only magnitude of focal length is given.

If f 1 is focal length of convex (converging) lens and f 2 is focal length of concave (diverging) lens. Then their equivalent focal length F would be

1F=1f1+1-f2=f2âf1f1f2 [âµ1F=1f1+1f2]â´ F=f1f2f2âf1

Topic: Physics, Ray Optics, NEET
A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept along the same axis with a distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then the distance d (in cm) will be

Options:
A 20
B 15
C 30
D 50

The correct answer is:
15

Solution:
1F=1f1+1f2-df1f2
â´1â=120+1-5-d20-5
â0=120-15+d100
â´d100=15-120=20-5100
â´ d=15 cm

Topic: Physics, Ray Optics, NEET
A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept along the same axis with a distance d between them. If a parallel beam of light falling on A leaves B as a parallel beam, then the distance d in cm will be :

Options:
A 50
B 15
C 25
D 30

The correct answer is:
15

Solution:
1F=1f1+1f2-df1f2
∴1∞=120+1-5-d20-5
⇒0=120-15+d100
∴d100=15-120=20-5100
∴ d=15 cm

From figure it is clear that separation between lenses, d=20−5=15 cm

Topic: Physics, Ray Optics, NEET, NEET 2021
A convex lens and a concave lens, each having same focal length of $25 \mathrm{~cm}$, are put in contact to form a combination of lenses. The power in diopters of the combination is:

Options:
A zero
B 25
C 50
D infinite

The correct answer is:
zero

Solution:
Focal length of lens
$$
f_1=25 \mathrm{~cm}
$$
focal length of concave lens
$$
f_2=25 \mathrm{~cm}
$$
Power of combination in dioptres
$$
\begin{aligned}
P=P_1+P_2 & =\frac{100}{f_1}+\frac{100}{f_2} \\
& =\frac{100}{25}+\frac{100}{-25}=0 .
\end{aligned}
$$

Topic: Physics, Ray Optics, NEET, NEET 2006
A convex lens is in contact with concave lens. The magnitude of the ratio of their focal length is 2/3. Their equivalent focal length is 30 cm. What are their individual focal lengths?

Options:
A -75, 50
B -10, 15
C 75, 50
D -15, 10

The correct answer is:
-15, 10

Solution:
Let focal length of convex lens is +f then focal length of concave lens would be -32f.
From the given condition,
130=1f-23f=13f
∴ f=10 cm
Therefore, focal length of convex lens = + 10 cm and that of concave lens = -15 cm.

Topic: Physics, Ray Optics, NEET
A convex lens is used to obtain a magnified image of an object on a screen. The object is at a distance 10 m from the lens. If the magnification is 19. the focal length of the lens is

Options:
A 9.5 cm
B 0.95 cm
C 9.5 m
D 0.95 m

The correct answer is:
9.5 m

Solution:
Here u=â10 m, m=19

â´m=âÏuâ19=âÏâ10âV=190 m

Using the lens formula, 1 f = 1 v â 1 u

1 f = 1 190 â 1 â10 â 1 f = 1 190 + 1 10 â 1 f = 1+19 190 âf= 190 20 =9.5 m

Hence, the focal length of lens = 9.5 m

Topic: Physics, Ray Optics, NEET
A convex mirror forms an image one-fourth the size of the object. If the object is at a distance of 0.5 m from the mirror, the focal length of the mirror is

Options:
A 0.17 m
B -1.5 m

C 0.4 m
D -0.4 m

The correct answer is:
0.17 m

Solution:
ff-u=14=ff-(-0.5)
Or 4f=f+0.5 or 3f=0.5
Or f=0.53 m=0.17 m

Topic: Physics, Ray Optics, NEET