datavorous/entrance-exam-dataset huggingface dataset:
In a meter bridge, as shown in the figure, it is given that resistance $Y=12.5 \Omega$ and that the balance is obtained at a distance $39.5 \mathrm{~cm}$ from end $A$ (by jockey J). After interchanging the resistances $X$ and $Y$, a new balance point is found at a distance $l_2$ from end $A$. What are the values of $X$ and $l_2$ ?

Options:
A
$19.15 \Omega$ and $39.5 \mathrm{~cm}$

B
$8.16 \Omega$ and $60.5 \mathrm{~cm}$

C
$19.15 \Omega$ and $60.5 \mathrm{~cm}$

D
$8.16 \Omega$ and $39.5 \mathrm{~cm}$

The correct answer is:
$8.16 \Omega$ and $60.5 \mathrm{~cm}$

Solution:
For a balanced meter bridge,
$$
\begin{aligned}
&\frac{\mathrm{X}}{39.5}=\frac{\mathrm{Y}}{(100-39.5)} \\
&\Rightarrow \mathrm{Y}=39.5=\mathrm{X} \times(100-39.5) \\
&or \ldots \mathrm{X}=\frac{12.5 \times 39.5}{60.5}=8.16 \Omega
\end{aligned}
$$
When $\mathrm{X}$ and $\mathrm{Y}$ are interchanged $l_1$ and (100 $-l_1$ ) will also interchange so, $l_2=60.5 \mathrm{~cm}$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
In a meter bridge experiment null point is obtained at $l \mathrm{~cm}$ from the left end. If the meter bridge wire is replaced by a wire of same material but twice the area of across-section, then the null point is obtained at a distance

Options:
A $2 l \mathrm{~cm}$ from left end.
B $l \mathrm{~cm}$ from left end.
C $l / 2 \mathrm{~cm}$ from left end.
D $l / 4 \mathrm{~cm}$ from left end.

The correct answer is:
$l \mathrm{~cm}$ from left end.

Solution:
At null condition, $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\mathrm{X}}{100-\mathrm{X}}$
This condition is independent of the crosssectional area of the wire used.
$\therefore \quad$ The null point will remain the same. (i.e., $l \mathrm{~cm}$ from left end)

Topic: Physics, Experimental Physics, MHT CET, MHT CET 2023 (09 May Shift 1)
In a meter Bridge experiment resistance x is connected in the right gap. When R1 and R2 are connected in the left gap separately the balance points are 40 cm and 50 cm respectively from the left end. Now if both R1 & R2 are connected in series in the left gap with x in the right gap then what will be the new balance point from the left end (in cm)?

Options:

The correct answer is:
62.5

Solution:
R1x= 4060 = 23....1, R2x= 5050 = 1 ..... 2
R1+R2x= l100–l .... (3) Add eq (1) & (2)
R1+R2x=23 + 1 = 53 ... (4) Placing eq (4) in (3)
5/3 =l100–l⇒ 500–5l = 3l
⇒ 8l=500∴l= 62.5 cm

Topic: Physics, Experimental Physics, JEE Main
In a meter bridge experiment, resistances are connected as shown in figure. The balancing length l1= 55 cm . Now, an unknown resistance x is connected in series with P and the new balancing length is found to be 75 cm . What is the value of The value of 11x (in ohm)?(Given, P=4 Î© )

Options:

The correct answer is:
64

Solution:
Given in the meter bridge,
In I st case, PQ=l1l2
⇒4Q=55100-55
⇒Q=4×4555 ...(i)
In 2 nd case, P+xQ=7525
⇒4+x3=Q ...(ii)
From Equations (i) and (ii), we get
4+x 3 = 36 11
4+x= 108 11
x= 108−44 11
=6411

⇒11x=64

Topic: Physics, Experimental Physics, JEE Main
In a meter bridge experiment resistances are connected as shown in the figure. Initially resistance P=4 Ω and the neutral point N is at 60 cm from A . Now an unknown resistance R is connected in series to P and the new position of the neutral point is at 80 cm from A . The value of unknown resistance R is -

Options:
A 335 Î©
B 6Â Î©
C 203 Î©
D 7 Î©

The correct answer is:
203 Î©

Solution:
Given initially a resistance P = 4 Ω is connected in the left gap and let Q is connected in the right gap of meter bridge and the null point is obtained at 60 cm,

From the balanced condition of the Wheatstone bridge

PQ=RS ...1

Let ρ is the resistance per unit length of meter bridge wire than

For wire AN, P=ρl=40 ρ ∵ l = 40 cm

And for NB wire

Q = ρ100-lQ= ρ100-60=40 ρ

On substituting the values from equation 1

460=Q40

⇒Q=166=83 Ω

Now an unknown resistance R is connected in series with the 4 Ω resistance and the null point is obtained at l=80 cm

Now, P = ρ l =80 l

and Q = ρ100-lQ= ρ100-80= 20 ρ

Then from equation 1

4+R80=Q20⇒ 4+R80=8320 ∵ Q=83⇒4+R80=860 ⇒ R = 203 Ω

Topic: Physics, Current Electricity, JEE Main, JEE Main 2017 (09 Apr Online)
In a meter bridge, if resistance \( R_{1} \) in the resistance box is \( 300 \Omega \), then the balanced length is found to be \( 75.0 \mathrm{~cm} \) from the end \( A \). The diameter of the unknown wire is \( 1 \mathrm{~mm} \) and the length of the unknown wire is \( 31.4 \mathrm{~cm} \). Find the specific resistance of the unknown wire.

Options:
A \( 250 \times 10^{-6} \Omega \mathrm{m} \)
B \( 350 \times 10^{-6} \Omega \mathrm{m} \)
C \( 150 \times 10^{-6} \Omega \mathrm{m} \)
D \( 450 \times 10^{-6} \Omega \mathrm{m} \)

The correct answer is:
\( 250 \times 10^{-6} \Omega \mathrm{m} \)

Solution:
Given

R1=300 Ω, balanced length, l=75 cm, the length and diameter of the unknown wire, x=31.4 cm and d=2r=10-1 cm.

Recall the principle of Wheatstone bridge in the arrangement of meter bridge,

RS=l100-l,

where S is the unknown resistance of the unknown wire,

⇒S=100-llR=100-7575300=100 Ω

Now, remember the relation between resistance and specific resistance ρ,

⇒S=ρxA=ρx(πd2/4)

∴ ρ=22710-3210040.314

=2.5×10-4 Ω m=250×10-6 Ω m

Topic: Physics, Current Electricity, JEE Main
In a metre bridge experiment null point is obtained at $20 \mathrm{~cm}$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X < Y$, then where will be the new position of the null point from the same end, if one decides to balance a resistance of $4 X$ against Y?

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

B
$80 \mathrm{~cm}$

C
$40 \mathrm{~cm}$

D
$70 \mathrm{~cm}$

The correct answer is:
$50 \mathrm{~cm}$

Solution:
We have from meter bridge experiment, $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{\ell_1}{\ell_2}$, where $\ell_2=\left(100-\ell_1\right) \mathrm{cm}$ In the first case, $X / Y=20 / 80$ In the second case $\frac{4 X}{Y}=\frac{\ell}{100-\ell}$ $\ell=50 \mathrm{~cm}$.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2004
In a metre bridge experiment null point is obtained at $40 \mathrm{~cm}$ from one end of the wire when resistance $\mathrm{X}$ is balanced against another resistance $\mathrm{Y}$. If $\mathrm{X} < \mathrm{Y}$, then the new position of the null point from the same end, if one decides to balance a resistance of $3 \mathrm{X}$ against $\mathrm{Y}$, will be close to :

Options:
A
$80 \mathrm{~cm}$

B
$75 \mathrm{~cm}$

C
$67 \mathrm{~cm}$

D
$50 \mathrm{~cm}$

The correct answer is:
$67 \mathrm{~cm}$

Solution:
From question, $\frac{x}{y}=\frac{40}{100-40}=\frac{2}{3}$
$$
\Rightarrow \mathrm{x}=\frac{2}{3} \mathrm{y}
$$
Again, $\frac{3 x}{y}=\frac{Z}{100-Z}$
or $\frac{3 \times \frac{2 y}{3}}{y}=\frac{Z}{100-Z}$
Solving we get $\mathrm{Z}=67 \mathrm{~cm}$
Therefore new position of null point
$$
\cong 67 \mathrm{~cm}
$$

Topic: Physics, Current Electricity, JEE Main, JEE Main 2013 (09 Apr Online)
In a metre-bridge when a resistance in the left gap is 2 Ω and unknown resistance in the right gap, the balance length is found to be 40 cm. On shunting the unknown resistance with 2 Ω, the balance length changes by:

Options:
A 22.5 cm
B 20 cm
C 62.5 cm
D 65 cm

The correct answer is:
22.5 cm

Solution:
In first circuit:

For balanced condition, we get 240=X60⇒X=3Ω

In second case:

Net resistance of the right side due to shunting becomes, X'=2×32+3=1.2 Ω

Again for balanced condition,

2ℓ=1.2100-ℓ

⇒200-2ℓ=1.2ℓ

⇒ℓ=2003.2=62.5 cm

Clearly, balance length changes by 22.5 cm.

Topic: Physics, Current Electricity, JEE Main, JEE Main 2024 (01 Feb Shift 2)
In a moderately asymmetrical distribution the mode and mean are 7 and 4 respectively. The median is

Options:
A $4$
B $5$
C $6$
D $7$

The correct answer is:
$5$

Solution:
For a moderately Skewed distribution, Mode $=3$ median -2 mean $\Rightarrow 7=3$ median $-2 \times 4$
$\Rightarrow 15=3$ median $\therefore$ Median $=5$.

Topic: Mathematics, Statistics, JEE Main
In a molecule of pyrophosphoric acid, the number of P-O H, P=0 and P-O-P bonds/moiety(ies) respectively are :

Options:
A 2, 4 and 1
B 3, 3 and 3
C 4, 2 and 0
D 4, 2 and 1

The correct answer is:
4, 2 and 1

Solution:
No. of P=0 bond =2

P-OH bond =4

P-O-P bond =1.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main, JEE Main 2020 (03 Sep Shift 1)
In a monoclinic unit cell, the relation of sides and angles are respectively:

Options:
A
$\mathrm{a}=\mathrm{b} \neq \mathrm{c}$ and $\alpha=\beta=\gamma=90^{\circ}$

B
a $\neq$ b $\neq$ c and $\alpha=\beta=\gamma=90^{\circ}$

C
$\mathrm{a} \neq \mathrm{b} \neq \mathrm{c}$ and $\beta=\gamma=90^{\circ} \neq \alpha$

D
$\mathrm{a} \neq \mathrm{b} \neq \mathrm{c}$ and $\alpha \neq \beta \neq \gamma \neq 90^{\circ}$

The correct answer is:
$\mathrm{a} \neq \mathrm{b} \neq \mathrm{c}$ and $\beta=\gamma=90^{\circ} \neq \alpha$

Solution:
No Solution Available

Topic: Chemistry, Solid State, JEE Main, JEE Main 2014 (12 Apr Online)
In a motor, the electrical power input is 500 W and the mechanical power output is 0.54 horse power. The heat developed in the motor in 1 hour is (assuming that all the electric energy which is not converted to mechanical energy is converted to heat) is

Options:
A 4.18×104 cal
B 3.6×105 cal
C 8.6×104 cal
D 1.28×105 cal

The correct answer is:
8.6×104 cal

Solution:
Efficiency = 0.54 × 7 4 6 5 0 0 = 0.80 or 80%

(0.5 kW = 500 W and 0.54 hp = 0.54×746 W)

∴ 80% of the electrical energy is converted to mechanical energy and the rest 20% is converted to heat energy.
∴ 2 0 1 0 0 × 5 0 0 = 1 0 0 W of power is converted to heat
∴ Heat produced in 1h (or 3600 s)
= 100×3600 = 36×104 J
= 3 6 × 1 0 4 4.18 cal = 8.6 × 1 0 4 cal

Topic: Physics, Thermal Properties of Matter, JEE Main
In a multi – electron atom, which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic acid and electric fields?
(a) $n=1, l=0, m=0$
(b) $n=2, l=0, m=0$
(c) $n=2, l=1, m=1$
(d) $n=3, l=2, m=1$
(e) $n=3, l=2, m=0$

Options:
A
(a) and (b)

B
(b) and (c)

C
(c) and (d)

D
(d) and (e)

The correct answer is:
(d) and (e)

Solution:
n = same

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2005
In a nerve, if sodium pump is blocked, which of the following is most likely to happen?

Options:
A Na+ Outside the cell will increase
B K+ inside the nerve will increase
C Na+ Inside the nerve will increase
D Na+ and K+ will increase outside the cell

The correct answer is:
Na+ Inside the nerve will increase

Solution:
Sodium potassium pump pushes 3 Na+ out of the nerve for 2 K+ that it pushes inside the nerve. This helps in maintaining the resting membrane potential. If it is blocked Na+ won't be able to leave the nerve and will get accumulated within the nerve and K+ will not be able to enter the nerve and will accumulate outside the nerve.

Topic: Biology, Neural Control and Coordination, NEET
In a new system of units, the unit of force is 100 N, unit of length is 10 m and unit of time is 100 s. Unit of mass in this system can be expressed as

Options:
A 10 5 kg
B 10 6 kg
C 10 2 kg
D 10 3 kg

The correct answer is:
10 5 kg

Solution:
Here, force (F) = [MLTâ2] = 100 N ...(i)
Length (L) = [L] = 10 m ...(ii)
Time (t) = [T] = 100 s ...(iii)
Substituting values of L and T from Equations. (ii) and (iii) in Equation. (i), we get
M Ã 10 Ã (100)â2 = 100
or Mâ¡ Ã 1 0 1 0 0 Ã 1 0 0 = 1 0 0
or M = 100 Ã 1000 kg
M= 10 5 kg

Topic: Physics, Units and Dimensions, JEE Main
In a normal pregnant woman, the amount of total gonadotropin activity was assessed. The result expected was

Options:
A High level of circulating FSH and LH in the uterus to stimulate implantation of the embryo
B High level of circulating HCG to stimulate endometrial thickening
C High levels of FSH and LH in uterus to stimulate endometrial thickening
D High level of circulating HCG to stimulate oestrogen and progesterone synthesis

The correct answer is:
High level of circulating HCG to stimulate oestrogen and progesterone synthesis

Solution:
In a pregnant female, high levels of HCG will maintain corpus luteum and stimulate it to secrete oestrogen and progesterone.

Topic: Biology, Human Reproduction, NEET, NEET 2012 (Screening)
In a nuclear fission process, a high mass nuclide (A≈236) with binding energy 7.6 MeV/Nucleon dissociated into two middle mass nuclides (A≈118), having binding energy of 8.6 MeV/Nucleon. The energy released in the process would be _______ MeV.

Options:

The correct answer is:
236

Solution:
The formula to calculate the energy released in the given process can be written as

Q=BEProduct -BEReactant ...1

From equation (1), it follows that

Q=2(118)(8.6)-236(7.6) MeV=236 MeV

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2024 (27 Jan Shift 1)
In a nuclear fission reaction of an isotope of mass M, three similar daughter nuclei of same mass are formed. The speed of a daughter nuclei in terms of mass defect ΔM will be :

Options:
A 2cΔMM
B ΔMc23
C c2ΔMM
D c3ΔMM

The correct answer is:
c2ΔMM

Solution:
Let the reaction be, (P)→(X)+(Y)+(Z)

Let mass of each daughter nuclei be m, then mass defect can be written as ∆M=M-3m.

Now the energy released will be in the form of kinetic energy of the daughter nuclei. Therefore,

ΔMc2=12mV2+12mV2+12mV2⇒V=2ΔMc23m=2ΔMc2M-∆M≈c2ΔMM

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2024 (30 Jan Shift 2)
In a nuclear fission, when a mass of 0.5 g of uranium is consumed, the energy obtained in kWh is nÃ107. Find the value of n.

Options:

The correct answer is:
1.25

Solution:
E=Δmc2
=0.5×10-3×3×1082
=4.5×1013
E=4.5×10133.6×106
=1.25×107 kWh

Topic: Physics, Nuclear Physics, JEE Main
In a nuclear reactor, the main purpose of the moderator is to

Options:
A Initiate the fission process by giving away its neutron
B Slow down the fast neutrons
C Cool down the excess of heat generated in the reactor

D Absorb excess of neutrons and control the reaction rate

The correct answer is:
Slow down the fast neutrons

Solution:
The moderator helps slow down the neutrons produced by fission to sustain the chain reaction. Control rods can then be inserted into the reactor core to reduce the reaction rate or withdrawn to increase it.

Topic: Physics, Semiconductors, JEE Main
In a nuclear reactor, the main purpose of the moderator is to

Options:
A Initiate the fission process by giving away its neutron
B Slow down the fast neutrons
C Cool down the excess of heat generated in the reactor

D Absorb excess of neutrons and control the reaction rate

The correct answer is:
Slow down the fast neutrons

Solution:
The moderator helps slow down the neutrons produced by fission to sustain the chain reaction. Control rods can then be inserted into the reactor core to reduce the reaction rate or withdrawn to increase it.

Topic: Physics, Semiconductors, TS EAMCET, TS EAMCET 2021 (04 Aug Shift 1)
In a nucleophilic substitution reaction : R - Br + Cl - ⟶ DMF R - Cl + Br - , which one of the following undergoes complete inversion of configuration ?

Options:
A C6H5CHC6H5Br
B C6H5CCH3C6H5Br
C C6H5CH2Br
D C6H5CH2CH2Br

The correct answer is:
C6H5CH2CH2Br

Solution:
In SN2 (nucleophilic bimolecular substitution) inversion takes place
SN2∝1stearic hindrance
Rate of R–X is 1° > 2° > 3°

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2014 (09 Apr Online)
In a one dimensional collision between two identical particles A and B, B is stationary and A has momentum p before impact. During impact B gives an impulse J to A. Then the coefficient of restitution between the two is

Options:
A 2JP-1
B 2JP+1
C JP+1
D JP-1

The correct answer is:
2JP-1

Solution:
Let p1 and p2 be the momenta of A and B after collision.
Then applying impulse = change in linear momentum for the two particles
For B :J=p1 â¦(i)
For A :J=p-p2
Or p2=p-J â¦(ii)
Coefficient of restitution e=P1-P2P
=p1-p+Jp
=J-p+Jp=2Jp-1

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
In a $p-n$ junction:

Options:
A high potential is at $n$-side and low potential at $p$-side
B high potential is at $p$-side and low potential at $n$-side
C $p$ and both are at same potential
D undetermined

The correct answer is:
high potential is at $n$-side and low potential at $p$-side

Solution:
In a p-n junction high potential is at $n$-side and law potential at $p$-side.

Topic: Physics, Semiconductors, JEE Main
In a p-n junction, an electric field of 5×105 V m exists in the depletion region. The minimum kinetic energy of a conduction electron, in order to diffuse from n-side to the p-side, is found to be 3.2×10-20 J. The width of the depletion region is,

Options:
A 2×10-4 cm
B 8×10-5 cm
C 5×10-6 cm
D 4×10-5 cm

The correct answer is:
4×10-5 cm

Solution:
The kinetic energy in terms of electric potential, K.E.=eV, it means the value of electric potential, V=K.E.e=3.2×10-201.6×10-19, now the relation between electric field strength and electric potential in terms of the width of the depletion region,
d=VE=3.2×10-201.6×10-195×105=4×10-5 cm.

Topic: Physics, Semiconductors, JEE Main
In a \(p\) - \(n-p\) transistor working as a common base amplifier, when the current gain is 0.96 and emitter current is \(7.2 \mathrm{~mA}\), the base current is

Options:
A \(0.4 \mathrm{~mA}\)
B \(0.2 \mathrm{~mA}\)
C \(0.29 \mathrm{~mA}\)
D \(0.35 \mathrm{~mA}\)

The correct answer is:
\(0.29 \mathrm{~mA}\)

Solution:
In common base amplifier for \(p-n-p\) transistor, current gain, \(\alpha=0.96\)
Emitter current, \(I_E=7.2 \mathrm{~mA}\)
We know that,
\(\begin{aligned}
& \alpha & =\frac{I_C}{I_E} \Rightarrow 0.96=\frac{I_C}{7.2} \\
\Rightarrow \quad & I_C & =0.96 \times 7.2=6.912 \mathrm{~mA}
\end{aligned}\)
\(\therefore\) Base current,
\(\begin{aligned}
I_B & =I_E-I_C=7.2-6.912 \\
& =0.288 \mathrm{~mA} \simeq 0.29 \mathrm{~mA}
\end{aligned}\)

Topic: Physics, Semiconductors, JEE Main
In a parallel plate capacitor with air between the plates, the distance ' $\mathrm{d}$ ' between the plates is changed and the space is filled with dielectric constant 8 . The capacity of the capacitor is increased 16 times, the distance between the plates is

Options:
A $2 \mathrm{~d}$
B $4 \mathrm{~d}$
C $\frac{\mathrm{d}}{2}$
D $\frac{\mathrm{d}}{4}$

The correct answer is:
$\frac{\mathrm{d}}{2}$

Solution:
$\mathrm{C}_1=\frac{\mathrm{A} \varepsilon_0}{\mathrm{~d}}$ ...(i)
$\mathrm{C}_2=\frac{8 \mathrm{~A} \varepsilon_0}{\mathrm{~d}^{\prime}} \Rightarrow 16 \mathrm{C}_1=\frac{8 \mathrm{~A} \varepsilon_0}{\mathrm{~d}^{\prime}}$
...(iii)
where $\mathrm{d}^{\prime}$ the unknown separation.
Dividing (i) by (ii)
$\frac{C_1}{16 C_1}=\frac{A \varepsilon_0}{d} \times \frac{d^{\prime}}{A \varepsilon_0}$
$\frac{8 \mathrm{~d}}{16}=\mathrm{d}^{\prime}$
$\therefore \quad \mathrm{d}^{\prime}=\frac{\mathrm{d}}{2}$

Topic: Physics, Capacitance, JEE Main
In a parallelogram ABCD, AB→=a, AD→=b & AC→=c. DB→·AB→ has the value:

Options:
A 12a2+b2+c2
B 14a2+b2-c2
C 13b2+c2-a2
D 12a2-b2+c2

The correct answer is:
12a2-b2+c2

Solution:
AB→=a
AD→=b
AC→=b→+a→=c
Squaring both side b→2+a→2-2b→·a→=c2
b2+a2-c2=2b→·a→…1
DB→.AB→=a→-b→·a→
=a→2-a→·b→
=12(a2-b2+c2)

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2015 (11 Apr Online)
In a particular system, the unit of length, mass and time are chosen to be 10 cm, 10 g and 0.1 s respectively. The unit of force in this system will be equivalent to -

Options:
A 110 N
B 1 N
C 10 N
D 100 N

The correct answer is:
110 N

Solution:
Dimensionally
F=MLT-2
In C.G.S. system
1 dyne=1g 1 cm 1s-2
In the new system
1x=10 g10 cm0.1 s-2
1 dyne1 x=1 g10 gÃ1 cm10 cm10 s1 s-2
1 dyne=110,000Ã1x
104 dyne=1x
10x=105 dyne=1 N
x=110 N

Topic: Physics, Units and Dimensions, JEE Main
In a period, atom with smallest radius generally is:

Options:
A Chalcogen
B Halogen
C Aerogen
D Pnicogen

The correct answer is:
Halogen

Solution:
In a period on moving left to right Zeffective increases, so atomic size decreases.
For inert gases, Van der waal radius is considered, so it is bigger in size as compared to halogens.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
In a permanent magnet at room temperature

Options:
A domains are partially aligned.
B magnetic moment of each molecule is zero.
C domains are all perfectly aligned.
D the individual molecules have non-zero magnetic moment which are all perfectly aligned.

The correct answer is:
domains are partially aligned.

Solution:
(A)
Even in the case of a permanent magnet all the domains are not perfectly aligned due to thermal agitations

Topic: Physics, Magnetic Effects of Current, KCET, KCET 2019
In a permanent magnet at room temperature

Options:
A magnetic moment of each molecule is zero
B the individual molecules have non-zero magnetic moment which are all perfectly aligned
C domains are partially aligned
D domains are all perfectly aligned

The correct answer is:
domains are all perfectly aligned

Solution:
At room temperature, all domains of permanent magnet are perfectly aligned like ferromagnetic material.

Topic: Physics, Magnetic Properties of Matter, KCET, KCET 2020
In a permanent magnet at room temperature.

Options:
A
magnetic moment of each molecule is zero

B
the individual molecules have non-zero magnetic moment which are all perfectly aligned

C
domains are partially aligned

D
domains are all perfectly aligned

The correct answer is:
domains are all perfectly aligned

Solution:
We know that a permanent magnet is a substance which at room temperature retain ferromagnetic property for a long period of time. The individual atoms in a ferromagnetic material possess a dipole moment as in a paramagnetic material. However, they interact with one another in such a way that they spontaneously align themselves in a common direction over a macroscopic volume i.e., domain.
Hence, in a permanent magnet at room temperature, domains are all perfectly aligned.

Topic: Physics, Magnetic Properties of Matter
In a photo electric experiment, if both the intensity and frequency of the incident light are doubled, then the saturation photoelectric current

Options:
A is halved
B is doubled
C becomes four times
D remains constant

The correct answer is:
is doubled

Solution:
In photoelectric effect, saturation photocurrent is directly proportional to the intensity of incident radiation but independent of frequency of incident radiation. Hence, saturation current in photoelectric effect becomes double when both intensity and frequency of the incident light are doubled.

Topic: Physics, Dual Nature of Matter, KCET, KCET 2022
In a photodiode, the conductivity increases when the material is exposed to light. It is found that conductivity changes only if the wavelength is less than \( 600 \mathrm{~nm} \). What is the bandgap?

Options:
A \( 2.06 \mathrm{eV} \)
B \( 1.2 \mathrm{eV} \)
C \( 3.4 \mathrm{eV} \)
D \( 4.2 \mathrm{eV} \)

The correct answer is:
\( 2.06 \mathrm{eV} \)

Solution:
Given,
The maximum wavelength of light, which increases the conductivity, λ=600 nm
Hence the energy of this photon must be equal to band gap ( Eg).
∴ Eg=hcλ=6.6×10-34×3×108600×10-9 J
=6.6×3×10-17600×1.6×10-19 eV=2.06 eV

Topic: Physics, Semiconductors, JEE Main
In a photoelectric effect experiment, stopping potential changes by 30 V, if we change the frequency of the radiation. The magnitude of change in the frequency is (h=6Ã10â34J s)

Options:
A 4× 10 −15 s -1
B 8Ã1015 s-1
C 10 16 s -1
D 18× 10 15 s -1

The correct answer is:
8Ã1015 s-1

Solution:
eV 0 = h Ï - h Ï 0
V 0 = h e Ï - h e Ï 0
Slope m = h e = Î V 0 Î Ï

Î Ï = e h Ã Î V Ï
= 1.6 Ã 1 0 - 1 9 6 Ã 1 0 - 3 4 Ã 3 0 = 8 Ã 1 0 1 5 s - 1

Topic: Physics, Dual Nature of Matter, JEE Main
In a photoelectric effect experiment, stopping potential changes by 30 Volt if we change frequency of the radiation. The magnitude of change in the frequency is (h=6Ã10â34J s)

Options:
A 4× 10 −15 s -1
B 8Ã1015 s-1
C 10 16 s -1
D 18× 10 15 s -1

The correct answer is:
8Ã1015 s-1

Solution:
eV 0 = h Ï - h Ï 0
V 0 = h e Ï - h e Ï 0
Slope m = h e = Î V 0 Î Ï

Î Ï = e h Ã Î V Ï
= 1.6 Ã 1 0 - 1 9 6 Ã 1 0 - 3 4 Ã 3 0 = 8 Ã 1 0 1 5 s - 1

Topic: Physics, Dual Nature of Matter, JEE Main