datavorous/entrance-exam-dataset huggingface dataset:
In a potentiometer experiment, a cell in the secondary circuit gives null point at 2.5 m when the length of the potentiometer wire is 10 m. If the length of the potentiometer wire is increased by 1 m without changing the cell in the primary circuit, then the position of the null point will be

Options:
A 3.5 m
B 3 m
C 2.75 m
D 2 m

The correct answer is:
2.75 m

Solution:
The resistance of a potentiometer wire is given as,

R=ÏÃlA

R'=2.5ÏAÃ10

Potential V'=IÃR',

=Iâ¡2.5ÏAâ¡Ã10

Now, again the length of potentiometer wire is increased by 1 m, then the resistance of null position wire.

R''=ÏÃl11ÃA

V"=IR"

and, V'=V"

IÃ2.5ÏAÃ10=IÃÏÃl11ÃA

2.5Ã1110=l=2.75 m

Topic: Physics, Current Electricity, NEET
In a potentiometer experiment a cell of e.m.f 1.5 V connected in the secondary circuit gives a balancing length of 165 cm of the wire. If a resistance of 5 Î© is connected in parallel to the cell, the balancing length of the wire is 150 cm . The internal resistance of the cell is

Options:
A 5 Ω
B 1.5 Ω
C 1 Ω
D 0.5 Ω

The correct answer is:
0.5 Ω

Solution:
In a potentiometer experiment internal resistance is given by,
r=l1l2–1R
r=165150–1×5
r=0.5 Ω

Topic: Physics, Current Electricity, NEET
In a potentiometer experiment, when a cell in the secondary circuit is shunted by a resistance ' $R$ ', balancing length is ' $L_1$ '. On doubling the shunt resistance, balancing length increases to ' $L_2$ '. Internal resistance of the cell is

Options:
A $\frac{2 \mathrm{R}\left(\mathrm{L}_2-\mathrm{L}_1\right)}{\mathrm{L}_1-2 \mathrm{~L}_2}$
B $\frac{2 \mathrm{R}\left(\mathrm{L}_2-\mathrm{L}_1\right)}{\left(2 \mathrm{~L}_1-\mathrm{L}_2\right)}$
C $\frac{\mathrm{R}\left(\mathrm{L}_2-\mathrm{L}_1\right)}{\mathrm{L}_1-2 \mathrm{~L}_2}$
D $\frac{\mathrm{R}\left(\mathrm{L}_2-\mathrm{L}_1\right)}{\left(2 \mathrm{~L}_1-\mathrm{L}_2\right)}$

The correct answer is:
$\frac{2 \mathrm{R}\left(\mathrm{L}_2-\mathrm{L}_1\right)}{\left(2 \mathrm{~L}_1-\mathrm{L}_2\right)}$

Solution:
No solution. Refer to answer key.

Topic: Physics, Current Electricity, AP EAMCET, AP EAMCET 2017 (24 Apr Shift 1)
In a random arrangement of the letters of the word 'UNIVERSITY', what is the probability that two I's do not come together?

Options:
A $4 / 5$
B $1 / 5$
C $1 / 10$
D $9 / 10$

The correct answer is:
$4 / 5$

Solution:
Total number of words formed byletters of UNIVERSITY
$=\frac{10 !}{2 !} \quad(\because \mathrm{I}$ is repeated $)$
Taking two $\mathrm{I}_{\mathrm{s}}$ together, number of ways to arrange letters of UNIVERSITY $=9 !$ $\therefore$ Probability of two $\mathrm{I}_{\mathrm{s}}$ coming together
$=\frac{9 !}{\frac{10 !}{2 !}}=\frac{9 ! \times 2 !}{10 !}=\frac{2}{10}=\frac{1}{5}$
$\therefore$ Probability of two $\mathrm{I}_{\mathrm{s}}$ not coming together
$=1-\frac{1}{5}=\frac{4}{5}$

Topic: Mathematics, Probability, NDA, NDA 2011 (Phase 2)
In a relayrace, there are six teams $A, B, C, D, E$ and $F$. What is the probability that $A, B, C$ finish first, second, third respectively?

Options:
A $\frac{1}{2}$
B $\frac{1}{12}$
C $\frac{1}{60}$
D $\frac{1}{120}$

The correct answer is:
$\frac{1}{120}$

Solution:
Probability of $\mathrm{A}=\frac{1}{6}$ Probability of $\mathrm{B}=\frac{1}{5}$ Probability of $\mathrm{C}=\frac{1}{4}$ Hence, required probability $=\frac{1}{6} \times \frac{1}{5} \times \frac{1}{4}=\frac{1}{120}$

Topic: Mathematics, Probability, NDA, NDA 2013 (Phase 2)
In a right-angled triangle $\mathrm{ABC}$, if the hypotenuse $\mathrm{AB}=\mathrm{p}$, then what is $\overrightarrow{\mathrm{AB}}$. $\overrightarrow{\mathrm{AC}}+\overrightarrow{\mathrm{BC}} . \overrightarrow{\mathrm{BA}}+\overrightarrow{\mathrm{CA}} \cdot \overrightarrow{\mathrm{CB}}$ equal to?

Options:
A $\mathrm{p}$
B $\mathrm{p}^{2}$
C $2 \mathrm{p}^{2}$
D $\frac{\mathrm{p}^{2}}{2}$

The correct answer is:
$\mathrm{p}^{2}$

Solution:
$\begin{aligned} \overrightarrow{A B} \cdot \overrightarrow{A C}+\overrightarrow{B C} \cdot \overrightarrow{B A}+\overrightarrow{C A} \cdot \overrightarrow{C B} \\=&(A B \cdot A C \cdot \cos \theta)+(B C \cdot B A \cdot \cos (90-\theta)) \\ & \quad+(C A \cdot C B \cdot \cos 90) \\=&(p \cdot x \cdot \cos \theta)+(y \cdot p \cdot \sin \theta)+0 \\=& p[x \cos \theta+y \sin \theta] \end{aligned}$
By projection formula :
$p=x \cos \theta+y \cos (90-\theta)$
$=x \cos \theta+y \sin \theta$
$\therefore \quad p[x \cos \theta+y \sin \theta]=p \times p=p^{2} .$

Topic: Mathematics, Vector Algebra, NDA, NDA 2016 (Phase 2)
In a $S_{\mathrm{N}} 2$ substitution reaction of the type $\mathrm{R}-\mathrm{Br}+\mathrm{Cl}^{-} \stackrel{\mathrm{DMF}}{\longrightarrow} \mathrm{R}-\mathrm{Cl}+\mathrm{Br}^{-}$, which one of the following has the highest relative rate?

Options:
A $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$
B

C

D $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}$

The correct answer is:
$\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}$

Solution:
Key Idea: The relative reactivity of alkyl halides towards $\mathrm{S}_{\mathrm{N}} 2$ reactions is as follows:
Primary $>$ Secondary $>$ Tertiary
However, if the primary alkyl halide or the nucleophile/base is sterically hindered the nucleophile will have difficulty to getting the back side of the $\alpha$-carbon as a result of this, the elimination product will be predominant. Here $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br}$ is the least hindered, hence it has the highest relative rate towards $\mathrm{S}_{\mathrm{N}} 2$ reaction.

Topic: Chemistry, Haloalkanes and Haloarenes, NEET, NEET 2008 (Screening)
In a school, all the students play at least one of three indoor games - chess, carrom and table tennis, 60 play chess, 50 play table tennis, 48 play carrom, 12 play chess and carrom, 15 play carrom and table tennis, 20 play table tennis and chess.
What can be the minimum number of students in the school?

Options:
A 123
B 111
C 95
D 63

The correct answer is:
111

Solution:
Number of students who play chess, $\mathrm{n}(\mathrm{A})=60$ Number of students who play tennis, $\mathrm{n}(\mathrm{B})=50$ Number of students who play carrom, $\mathrm{n}(\mathrm{C})=48$ Given, $n(A \cap B)=20$
$n(B \cap C)=15$
$n(A \cap C)=12$
$\mathrm{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})+\mathrm{n}(\mathrm{C})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})-\mathrm{n}(\mathrm{B} \cap$
C) $-n(A \cap C)+n(A \cap B \cap C)$
$=60+50+48-20-15-12+n(A \cap B \cap C)$
$=111+\mathrm{n}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})$
So, minimum number of students $=111$

Topic: Mathematics, Sets and Relations, NDA, NDA 2019 (Phase 1)
In a school, all the students play at least one of three indoor games - chess, carrom and table tennis, 60 play chess, 50 play table tennis, 48 play carrom, 12 play chess and carrom, 15 play carrom and table tennis, 20 play table tennis and chess.
What can be the maximum number of students in the school?

Options:
A 111
B 123
C 125
D 135

The correct answer is:
123

Solution:
$\mathrm{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=111+\mathrm{n}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})$
Maximum number of students $=111+12=123$

Topic: Mathematics, Sets and Relations, NDA, NDA 2019 (Phase 1)
In a series of 3 one-day cricket matches between teams A and $\mathrm{B}$ of a college, the probability of team A winning or drawing are $1 / 3$ and $1 / 6$ respectively. If a win, loss or draw gives 2,0 and 1 point respectively, then what is the probability that team A will score 5 points in the series?

Options:
A $\frac{17}{18}$
B $\frac{11}{12}$
C $\frac{1}{12}$
D $\frac{1}{18}$

The correct answer is:
$\frac{1}{18}$

Solution:
$\begin{array}{l}\text { Req. Prob. }=\mathrm{P}(5 \text { points })=\mathrm{P}(\text { two wins and one draw) } \\ = & \mathrm{P}(\mathrm{WWD})+\mathrm{P}(\mathrm{WDW})+\mathrm{P}(\mathrm{DWW}) \\ = & \frac{1}{3} \times \frac{1}{3} \times \frac{1}{6}+\frac{1}{3} \times \frac{1}{6} \times \frac{1}{3}+\frac{1}{6} \times \frac{1}{3} \times \frac{1}{3} \\ = & \frac{1}{18}\end{array}$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 2)
In a single-slit diffraction pattern, the position of first secondary maximum is at 30∘, then what will be the angular position of second minima?

Options:
A sin-1 2/3
B sin-1⁡(1)
C sin-1⁡1/2
D None

The correct answer is:
sin-1 2/3

Solution:
For first secondary maxima,
sinθ1=1+12λa=3λ2a⇒sin30o=3λ2a

⇒λa=13
For second minima,
sin⁡θ=2λa=23 or θ=sin-1 2/3

Topic: Physics, Wave Optics, NEET
In a standard ECG, which one of the following alphabets is the correct representation of the respective activity of the human heart?

Options:
A R-repolarisation of ventricles
B S-start of systole
C T-end of diastole
D P-depolarisation of the atria

The correct answer is:
P-depolarisation of the atria

Solution:
In ECG, P-wave indicates impulse of contraction generated by SA-node and it spreads in atria causing atrial depolarisation.
T-wave is the most constant and conspicuous wave having the tallest amplitude. It is the first positive deflection during ventricular depolarisation.
S-wave is the next downward deflection constant but often inconspicuous.
T-wave is repolarisation wave of ventricular relaxation.

Topic: Biology, Body Fluids and Circulation, NEET, NEET 2009 (Screening)
In a standard ECG which one of the following alphabets is the correct representation of the respective activity of the human heart?

Options:
A S - start of systole
B T - end of diastole
C P - depolarisation of the atria
D R - repolarisation of ventricles

The correct answer is:
P - depolarisation of the atria

Solution:
A normal electrogram (ECG) is composed of a P wave, a QRS wave (complex) and a $\mathrm{T}$ wave. The P-wave represents the electrical excitation or depolarisation of the atria. The QRS complex represents the depolarisation of the ventricles which initiates the ventricular contraction. The contraction starts shortly after Q and marks the beginning of the systole. The T-wave represents the repolarisation of ventricles. The end of the T-wave marks the end of systole.

Topic: Biology, Body Fluids and Circulation, AIIMS, AIIMS 2014
In a standard ECG, which one of the following is the correct representation of the respective activity of the human heart?

Options:
A S - start of systole
B T - end of diastole
C P - depolarisation of the atria
D R - repolarisation of ventricles

The correct answer is:
P - depolarisation of the atria

Solution:
In a standard ECG, the P-wave is a small upward wave that indicates the depolarization of the atria. This is caused by the activation of the SA node.

Topic: Biology, Body Fluids and Circulation, NEET
In a state with a population of $75 \times 10^{\circ}, 45 \%$ of them know Hindi, $22 \%$ know English, $18 \%$ know Sanskrit, $12 \%$ know Hindi and English, $8 \%$ know English and Sanskrit, $10 \%$ know Hindi and Sanskrit and $5 \%$ knowall the three languages.
What is the number of people who know only one language

Options:
A $3 \times 10^{6}$
B $4 \times 10^{6}$
C $3 \times 10^{7}$
D $4 \times 10^{7}$

The correct answer is:
$4 \times 10^{6}$

Solution:
Number of people who know only one language $=a+b+c=21 \times 10^{6}+5.25 \times 10^{6}+3.75 \times 10^{6}$
$=30 \times 10^{6}=3 \times 10^{7} .$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 1)
In a state with a population of $75 \times 10^{\circ}, 45 \%$ of them know Hindi, $22 \%$ know English, $18 \%$ know Sanskrit, $12 \%$ know Hindi and English, $8 \%$ know English and Sanskrit, $10 \%$ know Hindi and Sanskrit and $5 \%$ knowall the three languages.
What is the number of people who know only two $\begin{array}{ll}\text { languages ? } \end{array}$

Options:
A $11.25 \times 10^{3}$
B $11.25 \times 10^{6}$
C $12 \times 10^{5}$
D $12.5 \times 10^{5}$

The correct answer is:
$12 \times 10^{5}$

Solution:
Number of people who know only two language $=\mathrm{d}+\mathrm{e}+\mathrm{f}=3.75 \times 10^{6}+2.25 \times 10^{6}+5.25 \times 10^{6}$
$=11.25 \times 10^{6}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 1)
In a state with a population of $75 \times 10^{\circ}, 45 \%$ of them know Hindi, $22 \%$ know English, $18 \%$ know Sanskrit, $12 \%$ know Hindi and English, $8 \%$ know English and Sanskrit, $10 \%$ know Hindi and Sanskrit and $5 \%$ knowall the three languages.
What is the number of people who know Hindi only?

Options:
A $21 \times 10^{6}$
B $25 \times 10^{6}$
C $28 \times 10^{6}$
D $3 \times 10$

The correct answer is:
$3 \times 10$

Solution:
Number of people who knowonly Hindi=a $=21 \times 10^{6}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 1)
In a state with a population of $75 \times 10^{\circ}, 45 \%$ of them know Hindi, $22 \%$ know English, $18 \%$ know Sanskrit, $12 \%$ know Hindi and English, $8 \%$ know English and Sanskrit, $10 \%$ know Hindi and Sanskrit and $5 \%$ knowall the three languages.
What is the number of people who know English only?

Options:
A $5 \times 10^{6}$
B $45 \times 10^{5}$
C $4 \times 10^{6}$
D None of the above

The correct answer is:
$4 \times 10^{6}$

Solution:
Number of people who know only English $=5.25 \times 10^{6}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 1)
In a state with a population of $75 \times 10^{\circ}, 45 \%$ of them know Hindi, $22 \%$ know English, $18 \%$ know Sanskrit, $12 \%$ know Hindi and English, $8 \%$ know English and Sanskrit, $10 \%$ know Hindi and Sanskrit and $5 \%$ knowall the three languages.
What is the number of people who know Sanskrit only?

Options:
A $5 \times 10^{6}$
B $45 \times 10^{5}$
C $4 \times 10^{6}$
D None of the above

The correct answer is:
None of the above

Solution:
Number of people who knowonly Sanskrit $\mathrm{c}=3.75 \times 10^{6}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2013 (Phase 1)
In a step up transformer, which one of the following statements is correct?

Options:
A Number of turns in the secondary coil is less than in primary coil
B Voltage in secondary coil is less than in primary coil
C Current in the primary coil is more than current in the secondary coil
D Current in the primary coil is equal to current in the secondary coil

The correct answer is:
Current in the primary coil is more than current in the secondary coil

Solution:
In a step-up transformer, the current in the secondary is less than the current in the primary.

Topic: Physics, Alternating Current, MHT CET, MHT CET 2021 (21 Sep Shift 2)
In a step-down transformer, the number of turns in

Options:
A primary are less
B primary are more
C primary and secondary are equal
D secondary are infinite

The correct answer is:
primary are more

Solution:
In a step-down transformer, the number of turns of the primary coil are more than the number of turns of the secondary.

Topic: Physics, Alternating Current, MHT CET, MHT CET 2012
In a step-up transformer, the ratio of turns of primary to secondary is 1:10 and the primary voltage is 230 V. If the load current is 2 A, then the current in the primary is

Options:
A 20 A

B 10 A

C 2 A

D 1 A

The correct answer is:
20 A

Solution:
NpNs=110
Vp= 230 V, Is = 2 A,
IpIs=NsNp
Ip2=101
⇒ Ip=20 A​

Topic: Physics, Alternating Current, NEET
In a step-up transformer, the voltage in the primary is 220 V and the current is 5 A. The secondary voltage is found to be 22000 V. The current in the secondary coil (neglect losses) is

Options:
A 5 A
B 50 A
C 500 A
D 0.05 A

The correct answer is:
0.05 A

Solution:
VsVp=IpIs
⇒ Is=Ip×VpVs
=5×22022000=0.05 A

Topic: Physics, Alternating Current, NEET
In a study of two groups, the following results wer obtained $b$ \begin{array}{lcc}
\hline & Group & Group \\
\hline & \mathbf{A} & \mathbf{B} \\
Sample Size & 20 & 25 \\
Sample mean & 22 & 23 \\
Sample standard deviation & 10 & 12 \\
\hline
\end{array} Which of the following statements is correct?

Options:
A Group A is less variable than Group B because Group A's standard deviation is smaller.
B Group A is less variable than Group B because Group A's sample size is smaller.
C Group A is less variable than Group B because Group is A's sample mean is smaller.
D Group A is less variable than group B because Group A's coefficient of variation is smaller.

The correct answer is:
Group A is less variable than group B because Group A's coefficient of variation is smaller.

Solution:
For Group A:
Coefficient of variation
$C V_{A}=\frac{\text { S.D. }}{\text { Mean }}=\frac{10}{22}=0.4545$
For Group B:
$C V_{B}=\frac{12}{23}=0.522$
$\Rightarrow$ Group $\mathrm{A}$ is less variable.

Topic: Mathematics, Statistics, NDA, NDA 2016 (Phase 2)
In a study on the relationship between investment $(X)$ and profit $(Y)$, the following two regression equations were beained hased on the data on $X$ and $Y$ $3 X+Y-12=0$
$X+2 Y-14=0$
What is the mean $\bar{X}$ ?

Options:
A 6
B 5
C 4
D 2

The correct answer is:
2

Solution:
Given lines of regression are $3 X+Y-12=0$ and $X+2 Y-14=0$
Since, lines of regression passes through $(\bar{X}, \bar{Y})$.
therefore $(\bar{X}, \bar{Y})$ satisfies the given equations.
$\therefore 3 \bar{X}+\bar{Y}-12=0$
and $\bar{X}+2 \bar{Y}-14=0$
Multiply equation (ii) by 3 and subtract from (i), we get
$(3 \bar{X}+\bar{Y}-12)-(3 \bar{X}+6 \bar{Y}-42)=0$
$\Rightarrow 5 \bar{Y}+30=0 \Rightarrow \bar{Y}=6$
Thus, $\bar{X}=14-2 \bar{Y}=14-12=2$
Hence, mean, $\mathrm{X}=2$

Topic: Mathematics, Statistics, NDA, NDA 2009 (Phase 2)
In a survey of 25 students, it was found that 15 had taker Mathematics, 12 had taken Physics and 11 had taken Chemistry, 5 had taken Mathematics and Chemistry, 9 had taken Mathematics and Physics, 4 had taken Physics and Chemistry and 3 had taken all the three subjects.
onsider the following statements
1.The number of students who had taken only one
subject is equal to the number of students who had taken only two subjects
2.The number of students who had taken at least two subjects is four times the number of students who had taken all the three subjects
Which of the above statements is/are correct?

Options:
A 1 only
B 2 only
C Both 1 and 2
D Neither 1 nor 2

The correct answer is:
2 only

Solution:
Statement 1:
Students, who had taken only one subject $=2+5+4=11$
Students, who had taken only two subjects $=6+2+1=9$
$1 \neq 9$
State me nt $2:$
Students who had taken atleast two subject $=1+2+6+3=12$
Students who had taken all three subjects
$4 \times 3=12$

Topic: Mathematics, Sets and Relations, NDA, NDA 2014 (Phase 1)
In a survey of 25 students, it was found that 15 had taker Mathematics, 12 had taken Physics and 11 had taken Chemistry, 5 had taken Mathematics and Chemistry, 9 had taken Mathematics and Physics, 4 had taken Physics and Chemistry and 3 had taken all the three subjects.
The number of students who had taken only two subjects
IS

Options:
A 7
B 8
C 9
D 10

The correct answer is:
9

Solution:
Only two subjects $=6+2+1=9$ 99

Topic: Mathematics, Sets and Relations, NDA, NDA 2014 (Phase 1)
In a survey of 25 students, it was found that 15 had taker Mathematics, 12 had taken Physics and 11 had taken Chemistry, 5 had taken Mathematics and Chemistry, 9 had taken Mathematics and Physics, 4 had taken Physics and Chemistry and 3 had taken all the three subjects.
The number of students who had taken only physics is

Options:
A 2
B 3
C 5
D 6

The correct answer is:
2

Solution:
Only Physics $=12-(1+3+6)=2$ 198

Topic: Mathematics, Sets and Relations, NDA, NDA 2014 (Phase 1)
In a system of units if force $(F)$, acceleration $(A)$, and time ( $T$ ) are taken as fundamental units, then the dimensional formula of energy is :

Options:
A $F A^2 T$
B $F A T^2$
C $F^2 A T$
D FAT

The correct answer is:
$F A T^2$

Solution:
Let energy $E=F^x A^y T^3$
Writing the dimensions on both sides
$\begin{aligned} & {\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]=\left[\mathrm{MLT}^{-2}\right]^x\left[\mathrm{LT}^{-2}\right]^y[\mathrm{~T}]^2} \\ & {\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]=\left[\mathrm{M}^x \mathrm{~L}^{x+y} \mathrm{~T}^{-2 x-2 y+z}\right]}\end{aligned}$
On comparing both sides
$\begin{aligned} & x=1 \\ & x+y=2 \\ & \Rightarrow \quad y=2-x=2-1=1 \\ & \text { and } \quad-2 x-2 y+z=-2 \\ & \Rightarrow-2 \times 1-2 \times 1+z=-2 \\ & \Rightarrow \quad z=-2+2+2 \\ & =2 \\ & \end{aligned}$
$\therefore$ Dimensional formula of energy
$=F A T^2$

Topic: Physics, Units and Dimensions, JIPMER, JIPMER 2006
In a test in Mathematics, $20 \%$ of the students obtained "first class". If the data are represented by a Pie-Chart, what is the central angle corresponding to "first class"

Options:
A $20^{\circ}$
B $72^{\circ}$
C 36
D 144

The correct answer is:
36

Solution:
Central angle $=20 \%$ of $360^{\circ}$
$=\frac{20 \times 60}{100}$
$=72^{\circ}$

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 1)
In a thermochemical equation, what does the standard state refer to?

Options:
A The state of matter when a substance is in its liquid phase.
B The state of matter when a substance is in its gaseous phase.
C The pure form of a substance at 0°C.
D The most stable state of a substance at a specified temperature and pressure.

The correct answer is:
The most stable state of a substance at a specified temperature and pressure.

Solution:
The standard state in a thermochemical equation refers to the most stable state of a substance under specific conditions of temperature and pressure.

Topic: Chemistry, Thermodynamics (C), NEET
In a titration experiment, 10 mL of an FeCl2 solution consumed 25 mL of a standard K2Cr2O7 solution to reach the equivalent point. The standard K2Cr2O7 solution is prepared by dissolving 1.225 g of K2Cr2O7 in 250 mL water. The concentration of the FeCl2 solution is closest to

[Given : molecular weight of K2Cr2O7=294 g mol-1]

Options:
A 0.25 N
B 0.50 N
C 0.10 N
D 0.04 N

The correct answer is:
0.25 N

Solution:
In titration at equivalence point, number of equivalents of both reactants must be same.

∴ N1V1=N2V2

N (normality) of FeCl2=N×V of K2Cr2O7V of FeCl2

N (normality) of K2Cr2O7= Mass ×1000 Equivalent mass × Volume of solution prepared

Equivalent mass = Molar mass n factor

=2946

=49 g/equi.

(K2Cr2O7 as oxidising agent is reduced to 2Cr3+).

Oxidation state of 2 Cr atoms change from +6 to +3.

Total change is3×2=6, (∴n-factor =6)

∴ N (of K2Cr2O7) =1225×100049×250=0.1N

∴ N (of FeCl2) =0.1×2510

=0.25 N

Topic: Chemistry, Redox Reactions, KVPY, KVPY 2019 (SB/SX)
In a transformer number of turns in primary circuit is 500 and in secondary circuit number of turns is 10 . The load resistance is $10 \Omega$ and voltage of secondary coil is $50 \mathrm{~V}$. Then find the current in primary circuit.

Options:
A $0.2 \mathrm{~A}$
B $0.3 \mathrm{~A}$
C $0.4 \mathrm{~A}$
D $0.1 \mathrm{~A}$

The correct answer is:
$0.1 \mathrm{~A}$

Solution:
Given, $N_1=500, N_2=10, R_L=10 \Omega$
$V_2=50 \mathrm{~V}$
Current in the secondary circuit is
$\begin{aligned} & i_2=\frac{V_2}{R_L}=\frac{50}{10}=5 \mathrm{~A} \\ & \frac{i_1}{i_2}=\frac{N_2}{N_1} \Rightarrow \frac{i_1}{5}=\frac{10}{500} \Rightarrow i_1=0.1 \mathrm{~A}\end{aligned}$

Topic: Physics, Alternating Current, AIIMS, AIIMS 2019 (25 May)
In a transformer, the number of turns in the primary coil is 140 and that in the secondary coil is 280. If the current in the primary coil is 4 A, then that in the secondary coil is

Options:
A 4 A
B 2 A
C 6 A
D 10 A

The correct answer is:
2 A

Solution:
I 2 N 2 = I 1 N 1
∴ I 2 = I 1 N 1 N 2 = 4 × 140 280 = 2 A.

Topic: Physics, Alternating Current, NEET
In a triangle $A B C, a=(1+\sqrt{3}) \mathrm{cm}, b=2 \mathrm{~cm}$ and angle $C=60^{\circ} .$ Then the other two angles are

Options:
A $45^{\circ}$ and $75^{\circ}$
B $30^{\circ}$ and $90^{\circ}$
C $105^{\circ}$ and $15^{\circ}$
D $100^{\circ}$ and $20^{\circ}$

The correct answer is:
$45^{\circ}$ and $75^{\circ}$

Solution:
Now as a $>$ b $\angle \mathrm{A}>\angle \mathrm{B}$
Now from Sine Rule, $\frac{\sin A}{a}=\frac{\sin B}{b}$
$\frac{\sin A}{1+\sqrt{3}}=\frac{\sin B}{2}$
From option
(a) $\frac{\sin 75}{1+\sqrt{3}}=\frac{\sin 45}{2}$
$\frac{\sqrt{6}+\sqrt{2}}{4(1+\sqrt{3})}=\frac{1}{2 \sqrt{2}}$
$2 \sqrt{12}+4=4+4 \sqrt{3}$
$4+4 \sqrt{3}=4+4 \sqrt{3}$
$\therefore \quad$ Option (a) is correct.

Topic: Mathematics, Properties of Triangles, NDA, NDA 2015 (Phase 1)