datavorous/entrance-exam-dataset huggingface dataset:
All of the following statements apply to proteins except

Options:
A
Proteins generally have no definite melting point

B
Proteins contain the grouping - $\mathrm{CONH}-$

C
Proteins have high molecular weight

D
Proteins can only contain the elements $\mathrm{C}$, H, O and N.

The correct answer is:
Proteins can only contain the elements $\mathrm{C}$, H, O and N.

Solution:
Statement (d) is not correct. Some proteins also contain $\mathrm{S}$, with $\mathrm{C}, \mathrm{H}, \mathrm{O}$ and $\mathrm{N}$.

Topic: Chemistry, Biomolecules, JEE Main, JEE Main 2012 (07 May Online)
All of the following statements are correct for blue green algae, except

Options:
A they have chlorophyll - a, which is also found in higher plants.
B they may be filamentous, colonial, or unicellular.
C they have heterocysts which are clear cells without chlorophyll.
D they are connecting link between plants and animals.

The correct answer is:
they are connecting link between plants and animals.

Solution:
Cyanobacteria have autotrophic nutrition and can perform photosynthesis with the help of chlorophyll-a which is similar in green plants. They may be unicellular, colonial, or filamentous forms. Nitrogen fixation in cyanobacteria mainly occurs in specialized cells called heterocysts. Heterocysts are large-sized, pale-colored, non-motile, mucilage free thick-walled cells that are impermeable to oxygen. Euglena is the connecting link between plants and animals.

Topic: Biology, Biological Classification, NEET
All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such numbers in which the odd digits occupy even places is

Options:
A 175
B 162
C 180
D 160

The correct answer is:
180

Solution:
Odd digits: 1, 1, 3
Even digits: 2, 2, 2, 2, 4, 4
The number of ways of placing odd digits at even position = C34×3!2!=4×3=12
The number of ways of placing even digits =6!4! 2!=15
∴ Total number of ways =12×15=180

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2019 (08 Apr Shift 1)
All possible numbers are formed using the digits 1, 1, 2, 2, 2, 2, 3, 4, 4 taken all at a time. The number of such numbers in which the odd digits occupy even places is

Options:
A 175
B 162
C 180
D 160

The correct answer is:
180

Solution:
Odd digits: 1, 1, 3
Even digits: 2, 2, 2, 2, 4, 4
The number of ways of placing odd digits at even position = C34×3!2!=4×3=12
The number of ways of placing even digits =6!4! 2!=15
∴ Total number of ways =12×15=180

Topic: Mathematics, Permutation Combination, JEE Main
All possible values of θ∈0,2π for which sin2θ+tan2θ>0 lie in :

Options:
A 0,π2∪π,3π2
B 0,π2∪π2,3π4∪π,7π6
C 0,π4∪π2,3π4∪π,5π4∪3π2,7π4
D 0,π4∪π2,3π4∪3π2,11π6

The correct answer is:
0,π4∪π2,3π4∪π,5π4∪3π2,7π4

Solution:
sin2θ+tan2θ>0

⇒sin2θ+sin2θcos2θ>0

⇒sin2θcos2θ+1cos2θ>0⇒tan2θ2cos2θ>0

Note: cos2θ≠0

⇒1-2sin2θ≠0⇒sinθ≠±12

Now, tan2θ1+cos2θ>0

⇒tan2θ>0 (as cos2θ+1>0)

⇒2θ∈0,π2∪π,3π2∪2π,5π2∪3π,7π2

⇒θ∈0,π4∪π2,3π4∪π,5π4∪3π2,7π4

As sinθ≠±12; which has been already considered

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2021 (25 Feb Shift 1)
All possible values of scalar \( \mathrm{k} \) so that the matrix \( \mathrm{A}^{-1} \) - \( \mathrm{kI} \) is singular where \( \mathrm{A}=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 1 & 0 & 0\end{array}\right] \)

Options:
A \( \frac{-1}{2}, 1 \)
B \( -1, \frac{1}{2} \)
C \( \frac{1}{2}, \frac{-1}{2} \)
D \( -1,1 \)

The correct answer is:
\( -1, \frac{1}{2} \)

Solution:
|A-1 - kI| = 0
|A|A-1 - kI| = 0 (|A| ≠ 0)
|I - kA| = 0
I k - A = 0 ⇒ A - 1 k · I = 0
⇒ |A - λI| = 0 where λ = 1 k
= 1 - λ 0 2 0 2 - λ 1 1 0 - λ = 0
=(1 - λ) (-λ) (2 - λ) + 2(0 - (2 - λ)) = 0
= -λ3 + 3λ2 - 2λ - 4 + 2λ = 0
= λ3 - 3λ2 + 4 = 0 ⇒ λ = 2, 2,-1 λ k = - 1 , 1 2

Topic: Mathematics, Matrices, JEE Main
All sulphates of alkaline earth metals are not soluble in water. Which is the right order for their solubility

Options:
A Mg>Ca>Ba>Be>Sr
B Be>Mg>Ca>Sr>Ba
C Be>Ca>Mg>Ba>Sr
D Mg>Be>Ba>Ca>Sr

The correct answer is:
Be>Mg>Ca>Sr>Ba

Solution:
Be > Mg > Ca > Sr > Ba
On moving down the group lattice energy remains almost constant as the sulphate is so big that small increase in the size of the cations from Be to Ba does not make any difference. However the hydration energy decreases from Be+2 to Ba+2 . This causes decrease in the solubility of the sulphates as the ionic size increases.

Topic: Chemistry, s Block Elements, JEE Main
All surfaces shown in figure are assumed to be frictionless and the pulleys and the string are light. The acceleration of the block of mass 2 kg is:

Options:
A g
B g3
C g2
D g4

The correct answer is:
g3

Solution:
For 4 kg block:

40-2T=4a⇒T=20-2a

For 2 kg block:

T-10=4a⇒20-2a-10=4a⇒10=6a⇒a=106=g6 m s-2

Acceleration of the block of mass 2 kg, 2a=g3 m s-2.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2024 (30 Jan Shift 1)
All surfaces shown in the figure are smooth. The system is released with the spring unstretched. In equilibrium, compression in the spring will be -

Options:
A mg2k
B 2mgk
C M+mg2k
D mgk

The correct answer is:
mgk

Solution:
Let N be the normal reaction between m and M ,
Equilibrium of M
N sin⁡45∘=kx ... (i)
Equilibrium of m in vertical direction gives
N cos⁡ 45∘=mg ... (ii)
From Eqs. (i) and (ii), we get
x=mgk

Topic: Physics, Laws of Motion, JEE Main
All the arrangements, with or without meaning, of the word FARMER are written excluding any word that has two R appearing together. The arrangements are listed serially in the alphabetic order as in the English dictionary. Then the serial number of the word FARMER in this list is _____ .

Options:

The correct answer is:
77

Solution:
Number of such words without both R together

= total possible words - total possible words with both R together

Number of such words start with A=5!2!-4!=36

Number of such words start with E=5!2!-4!=36

Number of such words start with FAE=3!2!-2=1

Number of such words start with FAM=3!2!-2=1

Number of such words start with FARE=2!=2

Number of such words start with FARMER=1

Total =77

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2021 (01 Sep Shift 2)
All the bulbs in the given figure are identical. Which bulb lights up most brightly?
(Et

Options:
A \( 1 \)
B \( 2 \)
C \( 3 \)
D \( 4 \)

The correct answer is:
\( 1 \)

Solution:
The bulb which draws the maximum power will light up most brightly.

Since, all the bulbs have the same resistance, the bulb having maximum current will light up the brightest.

Maximum current flows through bulb (1) because other bulbs are in parallel connection. Hence, the current will distribute in them according to Kirchoff's first law.

Therefore, bulb 1 will light up most brightly.

Topic: Physics, Current Electricity, JEE Main
All the components of the nodal tissue are autoexcitable. Why does the SA node act as the normal pacemaker?

Options:
A SA node has the lowest rate of depolarisation
B SA node is the only component to generate the threshold potential
C Only SA node can convey the action potential to the other components
D SA node has the highest rate of depolarisation

The correct answer is:
SA node has the highest rate of depolarisation

Solution:
The nodal musculature has the ability to generate action potentials without any external stimuli, i.e. it is autoexcitable. However, the number of action potentials that could be generated in a minute vary at different parts of the nodal system. The SAN (Sino-Atrial Node) can generate the maximum number of action potentials, i.e $70-75 \mathrm{~min}$, i.e. the highest rate of depolarisation and is responsible for initiating and maintaining the rhythmic contractile activity of the heart. Therefore, it is called pacemaker.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2019 (Odisha)
All the edges of a block with parallel faces are unequal. Its largest edge is twice its shortest edge. What is the ratio of the maximum to the minimum resistance between parallel faces?

Options:

The correct answer is:
4

Solution:
Let the length of various edges in increase order be l,x,2l respectively
Rmax=Ï2lxl=2Ïx;Rmin=Ïl2lx=Ï2x
â´ RmaxRmin=4

Topic: Physics, Current Electricity, JEE Main
All the energy released from the reaction XâY,ÎrGo=-193 kJ mol-1 is used for oxidizing M+ as M+âM3++2e-,Eo=-0.25V.
Under standard conditions, the number of moles of M+ oxidized when one mole of X is converted to Y is [F = 96500 C mol-1 ]

Options:

The correct answer is:
4

Solution:
ΔGo=-nFEo
-n=193×10396500×0.25=8
So No. of moles of M+ oxidized by 1 moles of X=82=4

Topic: Chemistry, Electrochemistry, JEE Main
All the events of the menstrual cycle stop and there is no menstruation during pregnancy is due to

Options:
A Negative feedback of FSH & LH
B Negative feedback of Oxytocin
C Negative feedback of Prolactin
D Negative feedback of Progesterone

The correct answer is:
Negative feedback of Progesterone

Solution:
After fertilization the corpus luteum and later on placenta also secretes large amount of progesterone which is essential to maintain endometrium for implantation and other pregnancy conditions.

Topic: Biology, Human Reproduction, NEET
All the following are correct statement about a small pond ecosystem, except:

Options:
A This is fairly a self-sustainable unit.
B All the four basic components of an ecosystem are well exhibited.
C The abiotic component is the water with all the dissolved organic and inorganic substances and the rich soil deposit at the bottom of the pond.
D This ecosystem performs all the functions of any ecosystem except the unidirectional flow of energy.

The correct answer is:
This ecosystem performs all the functions of any ecosystem except the unidirectional flow of energy.

Solution:
An ecosystem can be of several types like terrestrial or aquatic, larger or smaller etc. In small pond also an ecosystem can be maintained. Like other ecosystems, this small pond ecosystem is also self-sustainable unit, it carries all four basic components of an ecosystem are well exhibited, It also performs all the functions of any ecosystem and even unidirectional flow of energy.

Topic: Biology, Ecosystem, NEET
All the following species are strong oxidizing agents. Their strength as oxidizing agents in acidic solution is such that

Options:
A S2O82->Cr2O72->MnO4-
B MnO4->Cr2O72->S2O82-
C S2O82->MnO4->Cr2O72-
D MnO4->S2O82->Cr2O72-

The correct answer is:
S2O82->MnO4->Cr2O72-

Solution:
The standard reduction potentials of the above species in acidic medium are as given below
S2O82-(aq)+2e-→2SO42-(aq);E°=+2.01V
MnO4-(aq)+8H++5e-→Mn2+(aq)+4H2O;E°=+1.77V
Cr2O72-(aq)+14H++6e-→2Cr3++7H2O;E°=+1.33V
The more the standard reduction potential, the more the tendency of the species to undergo reduction and more strongly it (species) would work as oxidizing agent.
As a result, their strength as oxidizing agent in acidic solution decreases in the following order
S2O82->MnO4->Cr2O72-

Topic: Chemistry, Redox Reactions, JEE Main
All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it up.

Options:
A
B
C
D

The correct answer is:

Solution:
The graph 3 initial slope is zero which is not possible since the initial velocity is non-zero in all other three graphs.

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2018 (08 Apr)
All the letters of the word' ABRACADABRA' are arranged in different possible ways. Then the number of such arrangements in which the vowels are together is

Options:
A 1200
B 1240
C 1220
D 1260

The correct answer is:
1260

Solution:
The word ABRACADABRA has
A: 5 times, B: 2 times, R: 2 times, C, D: 1 time each When all vowels are together, we have to arrange 7 elements. $\therefore$ No. of arrangements $\frac{7 !}{2 ! 2 !}=1260$

Topic: Mathematics, Permutation Combination, JEE Main
All the letters of the word GTWENTY are written in all possible ways with or without meaning and these words are written as in a dictionary. The serial number of the word GTWENTY IS

Options:

The correct answer is:
553

Solution:
We have to find the rank of word GTWENTY.

Words starting from E=6!2!=7202=360

Words starting with GE=5!2=60

Words starting with GN=5!2=60

Words starting with GTE=4!=24

Words starting with GTN=4!=24

Words starting with GTT=4!=24

Then the next word will be GTWENTY and its rank will be 360+60+60+24+24+24+1=553

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2024 (29 Jan Shift 1)
All the letters of the word PUBLIC are written in all possible orders and these words are written as in a dictionary with serial numbers. Then the serial number of the word PUBLIC is

Options:
A 576
B 578
C 580
D 582

The correct answer is:
582

Solution:
The given word is PUBLIC

Arranging the letters alphabetically, we get

BCILPU

When the word starts with any of the letters B/C/I/L, the number of possibilities=5!×4=480

Now when the word starts with PB, then the number of possibilities=4!=24

Now when the word starts with PC, then the number of possibilities =4!=24

Now when the word starts with PI, then the number of possibilities =4!=24

Now when the word starts with PL, then the number of possibilities =4!=24

Now when the word starts with PUBC, then the number of possibilities=2!=2

Now when the word starts with PUBI, then the number of possibilities =2!=2

Now when the word starts with PUBLC, then the number of possibilities=1

Now when the word starts with PUBLIC, then the number of possibilities =1

Rank =480+24×4+2×2+1×2=582

Hence, rank of the work PUBLIC is 582.

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (06 Apr Shift 2)
All the pairs (x, y), that satisfy the inequality 2sin2x-2sinx+5·14sin2y≤1 also satisfy the equation:

Options:
A 2sinx=siny
B sinx=2siny
C sinx=siny
D 2sinx=3siny

The correct answer is:
sinx=siny

Solution:
Given inequality is
2sin2⁡x-2sin⁡x+5 ≤4sin2⁡y
⇒sin2⁡x-2sin⁡x+1+4≤2sin2⁡y
⇒sin⁡x-12+4≤2sin2⁡y
∵2sin2⁡y∈0, 2 & sin⁡x-12+4∈2, 22
∵ Only equality holds
⇒sin⁡x-12+4=2sin2⁡y=2
⇒sin⁡x-12+4=4sin4⁡y=4
⇒sin2⁡y=1 ⇒ sin⁡y=1 & sin⁡x-12=0 ⇒sin⁡x=1
∴sin⁡y=sin⁡x

Topic: Mathematics, Trigonometric Equations, JEE Main, JEE Main 2019 (10 Apr Shift 1)
All the points in the set S=α+iα-i, α∈R, i=-1 lie on a

Options:
A straight line whose slope is -1
B circle whose radius is 2
C circle whose radius is 1
D straight line whose slope is 1

The correct answer is:
circle whose radius is 1

Solution:
Given z=α+iα-i , since α∈R

⇒z=α+iα-i=α+iα-i

⇒z=α2+12α2+-12

⇒z=1

Hence, locus of z is a circle with radius 1.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2019 (09 Apr Shift 1)
All the s–block elements of the periodic table are placed in the groups …

Options:
A $I A$ and $II A$
B $IIIA$ and $IVA$
C $B$ sub groups
D $\mathrm{VA}$ to $\mathrm{VII} A$

The correct answer is:
$I A$ and $II A$

Solution:
$I A$ and $IIA$

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main
All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 10 to each of the students. Which of the following statistical measures will not change even after the grace marks were given ?

Options:
A mode
B variance
C mean
D median

The correct answer is:
variance

Solution:
As we know that with the change in origin, the variance does not change.

Therefore, variance does not change even after giving grace marks of 10 to each of the students.

Whereas, mean, mode and median changes upon changing the origin.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2013 (07 Apr)
All the students of a class performed poorly in Mathematics. The teacher decided to give grace marks of 12 to the entire class. Which of the following statistical measures will not change even after the grace marks were given?

Options:
A Median
B Mode
C Variance
D Mean

The correct answer is:
Variance

Solution:
Variance is not changed by the change of origin.
Ï=âx-x-2n.
Therefore, y=x+12ây-+x-+12
Ï1=ây+12-y--122n=ây-y-2n=Ï.

Topic: Mathematics, Statistics, JEE Main
All the students of a class performed poorly in physics. The teacher decided to give grace marks of 15 to the entire class. Which of the following statistical measures will not change even after the grace marks were given?

Options:
A median
B mode
C variance
D mean

The correct answer is:
variance

Solution:
Variance is not changed by the change of origin.
Ï=âx-x-2n. Therefore, y=x+15ây-=x-+15
Ï1=ây+15-y--152n=ây-y-2n=Ï.

Topic: Mathematics, Statistics, JEE Main
All the values of $m$ for which both roots of the equations $x^2-2 m x+m^2-1=0$ are greater than $-2$ but less than 4 , lie in the interval

Options:
A
−2 < m < 0
B
m > 3

C
−1 < m < 3
D
1 < m < 4

The correct answer is:
−1 < m < 3

Solution:
Equation $x^2-2 m x+m^2-1=0$
$$
\begin{aligned}
& (x-m)^2-1=0 \\
& (x-m+1)(x-m-1)=0 \\
& x=m-1, m+1 \\
& -2 < m-1 \text { and } m+1 < 4
\end{aligned}
$$
$m>-1$ and $m < 3$
$-1 < m < 3$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2006
All the words that can be formed using alphabets A, $H, L, U$ and $R$ are written as in a dictionary (no alphabet is repeated). Rank of the word RAHUL is

Options:
A 71
B 72
C 73
D 74

The correct answer is:
74

Solution:
No. of words starting with $\mathrm{A}$ are $4 !=24$

No. of words starting with $\mathrm{H}$ are $4 !=24$

No. of words starting with $L$ are $4 !=24$

These account for 72 words Next word is RAHLU and the $74^{\text {th }}$ word RAHUL

Topic: Mathematics, Permutation Combination, JEE Main
All these pairs of plants and their structures are correctly matched, except

Options:
A Rhizophora - Pneumaotphores
B Ficus benghalensis - Stilt roots
C Bryophyllum - Leaf buds
D Pumpkin -Tendrils

The correct answer is:
Ficus benghalensis - Stilt roots

Solution:
Modification of roots and leaf takes place in order to perform various functions. In marshy areas, roots of plants such as Rhizophora grow vertically upward in order to get oxygen for respiration. In plants like maize and sugarcane, they develop roots from the basal nodes of the main stem that help in supporting the plant. These are known as Stilt roots. In the Banyan (Ficus benghalensis) tree, roots which give support to growing branches are called prop roots

Topic: Biology, Morphology of Flowering Plants, NEET
All those elements belong to f - block whose atomic numbers are

Options:
A 58 to 71
B 90 to 103
C Both (a) and (b)
D None

The correct answer is:
Both (a) and (b)

Solution:
58 to 71 and 90 to 103 (Lanthanides \& actinides).

Topic: Chemistry, d and f Block Elements, JEE Main
All trigonometric functions have inverse over their respective domains.

Options:
A
True

B
False

The correct answer is:
False

Solution:
No Solution Available

Topic: Mathematics, Inverse Trigonometric Functions
All vertebrates are chordates but all chordates are not vertebrates, why?

Options:
A Notochord is replaced by vertebral column in adult of some chordates.
B Ventral hollow nerve cord remains throughout life in some chordates.
C All chordates possess vertebral column.
D All chordates possess notochord throughout their life.

The correct answer is:
Notochord is replaced by vertebral column in adult of some chordates.

Solution:
Chordates are characterized by the presence of rod-like supporting notochord extending throughout the length of the body. Chordates are divided into three sub-phylum, i.e. Urochordates, Cephalocordates and Vertebrates.

In the vertebrates, the notochord in adults is replaced by a cartilaginous or bony vertebral column. However, in Urochordates, the notochord is found only in the tail region, while in cephalochordates the notochord extends above the head in adults.

Thus, all vertebrates are chordates but all chordates are not vertebrates.

Topic: Biology, Animal Kingdom, NEET, NEET 2020 (Phase 2)
All words, with or without meaning, are made using all the letters of the word MONDAY. These words are written as in a dictionary with serial numbers. The serial number of the word MONDAY is

Options:
A 327
B 328
C 324
D 326

The correct answer is:
327

Solution:
The given word is MONDAY

Arranging the letters alphabetically, we get

ADMNOY

When the word starts with any of the letters A/D, the number of possibilities=5!×2=240

Now when the word starts with MA, then the number of possibilities=4!=24

Now when the word starts with MD, then the number of possibilities =4!=24

Now when the word starts with MN, then the number of possibilities =4!=24

Now when the word starts with MOA, then the number of possibilities=3!=6

Now when the word starts with MOD, then the number of possibilities =3!=6

Now when the word starts with MONA, then the number of possibilities =2!=2

Now when the word starts with MONDAY, then the number of possibilities=1

Rank =240+24×3+6×2+2+1=327

Hence, rank of the word MONDAY is 327.

SHORTCUT METHOD

Another method is a shortcut, first number the alphabet and then check how many number are there on right hand side less that number and then start number in factorial in decreasing order and then multiply and add them and in last add one to find rank,

354216MONDAY2321005!4!3!2!1!0!

∴ Rank =(2×5!+3×4!+2×3!+1×2!)

=240+72+12+2+1=327

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2023 (13 Apr Shift 2)
Allyl phenyl ether can be prepared by heating :

Options:
A CH2 = CH - Br + C6H5 - CH2 - ONa
B C6H5 - CH = CH - Br + CH3 - ONa
C CH2 =â CH - CH2 - Br + C6H5ONa
D C6H5Br + CH2 = CH - CH2 - ONa

The correct answer is:
CH2 =â CH - CH2 - Br + C6H5ONa

Solution:
It is a Williamson's ether synthesis reaction. It is a SN2 reaction. Aryl halides will not undergo nucleophilic substitution reactions readily. Hence, allyl-phenyl ether can be prepared by reaction of allyl halide and phenoxide ion as follows

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main, JEE Main 2014 (09 Apr Online)
Along a streamline,

Options:
A
the velocity of a fluid particle remains constant

B
the velocity of all fluid particles crossing a given position is constant

C
the velocity of all fluid particles at a given instant is constant

D
the speed of a fluid particle remains constant

The correct answer is:
the velocity of all fluid particles crossing a given position is constant

Solution:
In streamline flow, the speed of liquid of each particle at a point in a particular cross-section is constant, between two cross-section of a tube of flow because $A V=$ constant (law of continuity).

Topic: Physics, Mechanical Properties of Fluids
Alpha particles are____

Options:
A twice the mass of beta particles
B negatively charged
C just like helium nuclei
D composite particles consisting of one protons and one neutrons

The correct answer is:
just like helium nuclei

Solution:
The correct answer is just like helium nuclei.
Alpha particles consist of two protons and two neutrons bound together into a particle identical to a helium nucleus.
An alpha particle is like a Helium nucleus that has no electrons and has a valency $2+$.
Alpha particles are emitted by radioactive nuclei such as uranium or radium, in a process known as alpha decay.

Topic: Physics, Atomic Physics, JEE Main
Alternating current can not be measured by dc ammeter because

Options:
A ac cannot pass through dc ammeter
B Average value of complete cycle is zero
C ac is virtual
D ac changes its direction

The correct answer is:
Average value of complete cycle is zero

Solution:
Alternating current cannot be measured by D.C ammeter because A.C. current have alternating signs and average value of current for complete cycle is zero and practical frequencies are like \(50-60 \mathrm{~Hz}\) that means it changes signs more than 100 times a second which either is not visible on D.C. ammeters.

Topic: Physics, Alternating Current, JEE Main
Alternating current can not be measured by D.C. ammeter because

Options:
A
A.C. cannot pass through D.C.

B
A.C. changes direction

C
average value of current for complete cycle is zero

D
D.C. ammeter will get damaged.

The correct answer is:
average value of current for complete cycle is zero

Solution:
No Solution Available

Topic: Physics, Alternating Current, JEE Main, JEE Main 2004