datavorous/entrance-exam-dataset huggingface dataset:
A ray of light travelling through glass of refractive index $\sqrt{2}$, is incident on glass - air boundary at an angle of incidence $45^{\circ}$. If refractive index of air is 1 , then the angle of refraction will be $\left[\sin 45^{\circ}=\frac{1}{\sqrt{2}}, \quad \sin 90^{\circ}=1\right]$

Options:
A $30^{\circ}$
B $90^{\circ}$
C $60^{\circ}$
D $45^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
(C)
$\mu=\sqrt{2} \mathrm{i}=45^{\circ}$
$\mu_{i} \sin i=\mu_{r} \sin r$
$\Rightarrow \sqrt{2} \sin 45^{\circ}=\sin \mathrm{r}$
$\therefore \quad \mathrm{r}=90^{\circ}$

Topic: Physics, Ray Optics, MHT CET, MHT CET 2020 (16 Oct Shift 2)
A reaction, $\mathrm{Ni}_{(\mathrm{s})}+\mathrm{Cu}_{(\mathrm{IM})}^{++} \rightarrow \mathrm{Ni}_{(\mathrm{IM} 2}^{++}+\mathrm{Cu}_{(\mathrm{s})}$ occurs in a cell. Calculate $\mathrm{E}_{\mathrm{cell}}^6$ if $\mathrm{E}_{\mathrm{Cu}}^0=0.337 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Ni}}^{\mathrm{o}}=-0.257 \mathrm{~V}$

Options:
A $0.594 \mathrm{~V}$
B $-0.594 \mathrm{~V}$
C $-0.08 \mathrm{~V}$
D $0.08 \mathrm{~V}$

The correct answer is:
$0.594 \mathrm{~V}$

Solution:
The standard cell potential is given by
$\begin{aligned}
\mathrm{E}_{\text {cell }}^o & =\mathrm{E}_{\text {cathode }}^{\circ}-\mathrm{E}_{\text {anode }}^o \\
\mathrm{E}_{\text {cell }}^{\circ} & =\mathrm{E}_{\mathrm{Cu}}^{\circ}-\mathrm{E}_{\mathrm{Ni}}^{\circ} \\
& =(0.337 \mathrm{~V})-(-0.257 \mathrm{~V}) \\
& =0.337 \mathrm{~V}+0.257 \mathrm{~V} \\
& =0.594 \mathrm{~V}
\end{aligned}$

Topic: Chemistry, Electrochemistry, MHT CET, MHT CET 2023 (09 May Shift 2)
A rectangular box with a cover is to have a square base. The volume is to be 10 cubic $\mathrm{cm}$. The surface area of the box in terms of the side $x$ is given by which one of the following functions? $\quad$

Options:
A $f(x)=(40 / x)+2 x^{2}$
B $f(x)=(40 / x)+x^{2}$
C $f(x)=(40 / x)+x$
D $f(x)=(60 / x)+2 x$

The correct answer is:
$f(x)=(40 / x)+2 x^{2}$

Solution:
Let the height of rectangular box be $y \mathrm{~cm}$. $\therefore$ Volume $=x \times x \times y$
$\Rightarrow y=\frac{10}{x^{2}}$
Now, surface area of box $=2\left(x^{2}+x y+y x\right)=2\left(x^{2}+2 x y\right)$
$=2\left(x^{2}+\frac{20}{x}\right) \quad$ From equation (i)
$=2 x^{2}+\frac{40}{x}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 2)
A relation $\mathrm{R}$ is defined on the set $\mathrm{N}$ of natural numbers as $\mathrm{xRy} \Rightarrow \mathrm{x}^{2}-4 \mathrm{xy}+3 \mathrm{y}^{2}=0$, Then which one of the following is
correct?

Options:
A $\mathrm{R}$ is reflexive and symmetric, but not transitive
B $\mathrm{R}$ is reflexive and transitive, but not symmetric
C $\mathrm{R}$ is reflexive, symmetric and transitive
D $\mathrm{R}$ is reflexive, but neither symmetric nor transitive

The correct answer is:
$\mathrm{R}$ is reflexive, but neither symmetric nor transitive

Solution:
Given, $x R Y \Rightarrow x^{2}-4 x y+3 y^{2}=0$
$\Rightarrow x^{2}-x y-3 x y+3 y^{2}=0$
$\Rightarrow x(x-y)-3 y(x-y)=0$
$\Rightarrow(x-y)(x-3 y)=0$
Reflexive property:
$x R x \Rightarrow(x-x)(-3 x)=0$
So, $\mathrm{R}$ is reflexive ...(1)
Symmetric property:
Let us check using an example $(1,2)$ and $(2,1)$ for $(1,2) \Rightarrow(1-2)(1-6)=(-1)(-5)=10$
For $(2,1) \Rightarrow(2-1)(2-3)=(1)(-1)=-1$
So, $\mathrm{R}$ is not symmetric ...(2)
Transitive property :
$\operatorname{For}(9 \mathrm{x}, 3 \mathrm{x}) \Rightarrow(9 \mathrm{x}-3 \mathrm{x})(9 \mathrm{x}-9 \mathrm{x})=0$
for $(3 x, x) \Rightarrow(3 x-3 x)(3 x-9 x)=0$
For $(9 x, x) \Rightarrow(9 x-x)(9 x-3 x)$
So, $(9 x, 3 x) \in R,(3 x, x) \in R$ but $(9 x, x) \notin R$
So, $\mathrm{R}$ is not transitive...(3)
(3) From (1), (2), (3), R is reflexive, but not symmetr ic and transitive.

Topic: Mathematics, Sets and Relations, NDA, NDA 2019 (Phase 1)
A relation $\mathrm{R}$ is defined on the set $\mathrm{Z}$ of integers as follows:
$m R n \Leftrightarrow m+n$ is odd
Which of the following statements is/are true for $\mathrm{R}$ ?
1.R is reflexive
2.R is symmetric
3.$\mathrm{R}$ is transitive Select the correct answer using the code given below:

Options:
A 2 only
B 2 and 3
C 1 and 2
D 1 and 3

The correct answer is:
2 only

Solution:
$\because R$ is a relation defined on the set $Z$ of integers as follows:
$\mathrm{mRn} \Leftrightarrow \mathrm{m}+\mathrm{n}$ is odd
(1) Then, $\mathrm{mRm}=2 \mathrm{~m}$ and $\mathrm{nRn}=2 \mathrm{n}$ are not odd multiples of 2 are not odd. Thus, it is not reflexive.
(2) If $\mathrm{m}$ and $\mathrm{n}$ are numberssuch that $\mathrm{mRn} \Leftrightarrow \mathrm{m}+\mathrm{n}$ is odd. Thus, $\mathrm{n} \mathrm{Rm} \Leftrightarrow \mathrm{n}+\mathrm{m}$ is odd.
This relation is symmetric
(3) $\mathrm{mRn}=\mathrm{m}+\mathrm{n}$, if there is third number $\mathrm{p}$ and $\mathrm{n} \mathrm{Rp}=\mathrm{n}+\mathrm{p}$ is odd. (for $\mathrm{ex}: 2+3=5$ is odd $3+4=7$ is
odd. But, $2+4=6$ is notodd.) Then $\mathrm{mRp}=\mathrm{m}+\mathrm{p}$ may not be odd. So, this relation is not transitive.

Topic: Mathematics, Sets and Relations, NDA, NDA 2006 (Phase 1)
A relation $\mathrm{R}$ is defined over the set of non-negative integers
as $\mathrm{xRy} \Rightarrow \mathrm{x}^{2}+\mathrm{y}^{2}=36$ what is $\mathrm{R}$ ?

Options:
A $\{(0,6)\}$
B $\{(6,0),(\sqrt{11}, 5),(3,3, \sqrt{3})$
C $\{(6,0),(0,6)\}$
D $(\sqrt{11}, 5),(2,4 \sqrt{2}),(5 \sqrt{11)},(4 \sqrt{2,2})$

The correct answer is:
$\{(6,0),(0,6)\}$

Solution:
$\mathrm{R}$ is defined over the set of non negative integers, $x^{2}+y^{2}=36$
$\Rightarrow y=\sqrt{36-x^{2}}=\sqrt{(6-x)(6+x)}, x=0$ or 6
for $x=0, y=6$ and for $x=6, y=0$
$\mathrm{So}, \mathrm{y}$ is 6 or 0
$s o, R=\{(6,0),(0,6)\}$

Topic: Mathematics, Sets and Relations, NDA, NDA 2007 (Phase 1)
A resistance $\mathrm{R}$ is to be measured using a meter bridge, student chooses the standard resistance $S$ to be $100 \Omega$. He findsthenull point at $l_1=2.9 \mathrm{~cm}$. He is told to attempt to improve the accuracy.
Which of the following is a useful way?

Options:
A
He should measure $\mathrm{I}_1$ more accurately

B
He should change 5 to $1000 \Omega$ and repeat the experiment

C
He should change $\mathrm{S}$ to $3 \Omega$ and repeat the experiment

D
He should given up hope of a more accurate measurement with a meter bridge

The correct answer is:
He should change $\mathrm{S}$ to $3 \Omega$ and repeat the experiment

Solution:
Adjusting the blance point near the middle of the bridge, i.e.. when $l_1$ is close to $50 \mathrm{~cm}$. requires a suitable choice of $S, R$ is unknown resistance :
Since, $\quad \frac{\mathrm{R}}{\mathrm{S}}=\frac{\mathrm{R} l_1}{\mathrm{R}\left(100-l_1\right)}$
$\frac{\mathrm{R}}{\mathrm{S}}=\frac{l_1}{100-l_1} \quad$ or $\quad \mathrm{R}=\mathrm{S}\left[\frac{l_1}{100-l_1}\right]$
$\mathrm{R}=\mathrm{S}\left[\frac{2.9}{97.1}\right]$
So, here, $\mathrm{R}: \mathrm{S}=2.9: 97.1$ implies that the $\mathrm{S}$ is nearly 33 times to that of $\mathrm{R}$. In orded to make this ratio $1: 1$ it is necessary to reduce the value of $S$ nearly $\frac{1}{33}$ times i.e., nearly $3 \Omega$,

Topic: Physics, Current Electricity
A rod of length $l$ is held vertically stationary with its lower end located at a point $P$, on the horizontal plane. When the rod is released to topple about $P$, the velocity of the upper end of the rod with which it hits the ground is

Options:
A $\sqrt{\frac{g}{l}}$
B $\sqrt{3 g l}$
C $3 \sqrt{\frac{g}{l}}$
D $\sqrt{\frac{3 g}{l}}$

The correct answer is:
$\sqrt{3 g l}$

Solution:
In this process potential energy of the metre stick will be converted into rotational kinetic energy. PE of metre stick $=\frac{m g l}{2}$
Because its centre of gravity lies at the middle of the rod.
Rotational kinetic energy $E=\frac{1}{2} I \omega^2$
$I=$ moment of inertia of metre stick about point
$A=\frac{m l^2}{3} \text {. }$
By the law of conservation of energy
$\begin{aligned}
m g\left(\frac{l}{2}\right) & =\frac{1}{2} I \omega^2 \\
& =\frac{1}{2} \frac{m l^2}{3}\left(\frac{v_B}{l}\right)^2
\end{aligned}$
By solving, we get $v_B=\sqrt{3 g l}$

Topic: Physics, Rotational Motion, TS EAMCET, TS EAMCET 2009
A rod of length $l$ is held vertically stationary with its lower end located at a point $P$, on the horizontal plane. When the rod is released to topple about $P$, the velocity of the upper end of the rod with which it hits the ground is

Options:
A $\sqrt{\frac{g}{l}}$
B $\sqrt{3 g l}$
C $3 \sqrt{\frac{g}{l}}$
D $\sqrt{\frac{3 g}{l}}$

The correct answer is:
$\sqrt{3 g l}$

Solution:
In this process potential energy of the metre stick will be converted into rotational kinetic energy. PE of metre stick $=\frac{m g l}{2}$
Because its centre of gravity lies at the middle of the rod.
Rotational kinetic energy $E=\frac{1}{2} I \omega^2$
$I=$ moment of inertia of metre stick about point
$A=\frac{m l^2}{3} \text {. }$
By the law of conservation of energy
$\begin{aligned}
m g\left(\frac{l}{2}\right) & =\frac{1}{2} I \omega^2 \\
& =\frac{1}{2} \frac{m l^2}{3}\left(\frac{v_B}{l}\right)^2
\end{aligned}$
By solving, we get $v_B=\sqrt{3 g l}$

Topic: Physics, Rotational Motion, AP EAMCET, AP EAMCET 2009
A salesman has a $70 \%$ chance to sell a product to any customer. The behaviour of successive customers is independent. If two customers $\mathrm{A}$ and $\mathrm{B}$ enter, what is the probability that the salesman will sell the product to customer A or B?

Options:
A $0.98$
B $0.91$
C $0.70$
D $0.49$

The correct answer is:
$0.91$

Solution:
$\quad P(A)=\frac{70}{100}=\frac{7}{10}=P(B)$
$A$ and $B$ are independent.
$\begin{aligned} \Rightarrow P(A \cap B) &=P(A) P(B) \\ \Rightarrow P(A \cup B) &=P(A)+P(B)-P(A \cap B) \\ &=\frac{7}{10}+\frac{7}{10}-\frac{7}{10} \times \frac{7}{10}=0.91 \end{aligned}$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 2)
A secondary amine is

Options:
A a compound with two carbon atoms and an $\mathrm{NH}_2$ group
B a compound with an $\mathrm{NH}_2$ group on the carbon atom in number 2 position
C a compound in which 2 of the hydrogen of $\mathrm{NH}_3$ have been replaced by organic group
D an organic compound with two $\mathrm{NH}_2$ groups

The correct answer is:
a compound in which 2 of the hydrogen of $\mathrm{NH}_3$ have been replaced by organic group

Solution:
A secondary amine is a compound in which 2 of the hydrogen of $\mathrm{NH}_3$ have been replaced by organic group.

Topic: Chemistry, Amines, KCET, KCET 2022
A secondary cell is one

Options:
A can be recharged.
B can be recharged by passing current through it in the same direction.
C can be recharged by passing current through it in the opposite direction.
D can not recharged

The correct answer is:
can be recharged by passing current through it in the opposite direction.

Solution:
In secondary cells electrode reactions can be reversed by external electric energy source, therefore, these cells can
berecharged by passing electric current and used again and again.
The current is passed in the direction opposite to that of the flow of current generated by the cell.

Topic: Chemistry, Electrochemistry, KCET, KCET 2016
A segment of wire vibrates with a fundamental frequency of $450 \mathrm{~Hz}$ under a tension of $9 \mathrm{~kg} \mathrm{wt}$. Then tension at which the fundamental frequency of the same wire becomes $900 \mathrm{~Hz}$ is

Options:
A $36 \mathrm{~kg}-\mathrm{wt}$
B $27 \mathrm{~kg}-\mathrm{wt}$
C $18 \mathrm{~kg}-\mathrm{wt}$
D $72 \mathrm{~kg}-\mathrm{wt}$

The correct answer is:
$36 \mathrm{~kg}-\mathrm{wt}$

Solution:
Fundamental frequency of wire
$\begin{array}{rlrl}f & =\frac{1}{2 \pi} \sqrt{\frac{T}{m}} \\ \text { or } & f & \propto \sqrt{T} \\ \text { or } & \frac{f_2}{f_1} & =\sqrt{\frac{T_2}{T_1}} \\ \text { or } & \frac{900}{450} & =\sqrt{\frac{T_2}{9}} \\ \text { or } & T_2 & =4 \times 9=36 \mathrm{~kg}-\mathrm{wt}\end{array}$

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2007
A segment of wire vibrates with a fundamental frequency of $450 \mathrm{~Hz}$ under a tension of $9 \mathrm{~kg} \mathrm{wt}$. Then tension at which the fundamental frequency of the same wire becomes $900 \mathrm{~Hz}$ is

Options:
A $36 \mathrm{~kg}-\mathrm{wt}$
B $27 \mathrm{~kg}-\mathrm{wt}$
C $18 \mathrm{~kg}-\mathrm{wt}$
D $72 \mathrm{~kg}-\mathrm{wt}$

The correct answer is:
$36 \mathrm{~kg}-\mathrm{wt}$

Solution:
Fundamental frequency of wire
$\begin{array}{rlrl}f & =\frac{1}{2 \pi} \sqrt{\frac{T}{m}} \\ \text { or } & f & \propto \sqrt{T} \\ \text { or } & \frac{f_2}{f_1} & =\sqrt{\frac{T_2}{T_1}} \\ \text { or } & \frac{900}{450} & =\sqrt{\frac{T_2}{9}} \\ \text { or } & T_2 & =4 \times 9=36 \mathrm{~kg}-\mathrm{wt}\end{array}$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2007
A segment of wire vibrates with a fundamental frequency of 450 Hz under a tension of 9 kg-wt. Then tension at which the fundamental frequency of the same wire becomes 900 Hz is

Options:
A 36 kg-wt
B 27 kg-wt
C 18 kg-wt
D 72 kg-wt

The correct answer is:
36 kg-wt

Solution:
Fundamental frequency of wire
f=12ÏTm
Or
fâT
Or
f2f1=T2T1
Or
900450=T29
Or
T2=4Ã9=36 kg-wt

Topic: Physics, Waves and Sound, NEET
A set of annual numerical data, comparable over the year? is given for the last 12 years.
1 . The data is best represented by a broken line graph, each corner (turning point) representing the data of one yean
2 . Such a graph depicts the chronological change and also enables one to make a short-term forecast.
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:
Both 1 and 2

Solution:
The annual numerical data for comparable for last 12 years is represented by broken line graph, where each turning point represent the data of a particular year, while such graph do not depict the chronological change. Option (c) is correct.

Topic: Mathematics, Statistics, NDA, NDA 2015 (Phase 1)
A set of $\mathrm{n}$ values $\mathrm{x}_1, \mathrm{x}_2, \ldots, \mathrm{x}_{\mathrm{n}}$ has standard deviation $\sigma$. What will be the standard deviation of $n$ values $\mathrm{x}_1+\mathrm{k}, \mathrm{x}_2+\mathrm{k}, \ldots, \mathrm{x}_{\mathrm{n}}+\mathrm{k} ?$

Options:

None

Solution:
If each observation is increased by a constant $\mathrm{k}$, then standard deviation is unchanged. So, $\sigma$ will be the standard deviation.

Topic: Mathematics, Statistics
A simple harmonic motion is represented by $\alpha \frac{d^2 x}{d t^2}+\beta x=0$. Its period is

Options:
A $\frac{2 \pi \beta}{\alpha}$
B $2 \pi \sqrt{\frac{\alpha}{\beta}}$
C $2 \pi \sqrt{\frac{\beta}{\alpha}}$
D $\frac{2 \pi \alpha}{\beta}$

The correct answer is:
$2 \pi \sqrt{\frac{\alpha}{\beta}}$

Solution:
The standard equation of SHM is: $\frac{d^2 x}{d t^2}=-\omega^2 x$
On comparing with equation: $\alpha \frac{d^2 x}{d t^2}+\beta x=0$
The angular frequency is given by: $\omega=\frac{\beta}{\alpha}$.
Therefore, the time period of SHM is:
$T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{\alpha}{\beta}}$

Topic: Physics, Oscillations, MHT CET, MHT CET 2022 (10 Aug Shift 1)
A single slit diffraction pattern is formed with white light. For what wavelength of light, the $3^{\text {rd }}$ secondary maximum in diffraction pattern coincides with the $2^{\text {nd }}$ secondary maximum in the pattern of red light of wavelength $6000 Å$ ?

Options:
A $4300 Å$
B $3500 Å$
C $4000 Å$
D $5000 Å$

The correct answer is:
$4300 Å$

Solution:
$\begin{aligned} & \text { Distance of nth maximum } \mathrm{y}_{\mathrm{n}}=\left(\mathrm{n}+\frac{1}{2}\right) \frac{\lambda \mathrm{D}}{\mathrm{a}} \\ & \therefore \mathrm{y}_2=\left(2+\frac{1}{2}\right) \frac{\lambda_1 \mathrm{D}}{\mathrm{a}} \text { and } \mathrm{y}_3=\left(3+\frac{1}{2}\right) \frac{\lambda_2 \mathrm{D}}{\mathrm{a}} \\ & \therefore 2.5 \lambda_1=3.5 \lambda_2 \\ & \therefore 2.5 \times 6000=3.5 \lambda_2 \\ & \therefore \lambda_2=\frac{2.5 \times 6000}{3.5} \simeq 4300 A\end{aligned}$

Topic: Physics, Wave Optics, MHT CET, MHT CET 2021 (22 Sep Shift 1)
A single slit Fraunhofer diffraction pattern is formed with white light. For what wavelength of light the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for red light of wavelength $6500 Å$ ?

Options:
A $4400 Å$
B $4100 Å$
C $4642.8 Å$
D $9100 Å$

The correct answer is:
$4642.8 Å$

Solution:
$x=\frac{(2 n+1) \lambda D}{2 a}$
For red light $x=\frac{(4+1) D}{2 a} \times 6500$
For unknown wavelength of light
$$
x=\frac{(6+1) D}{2 a} \times \lambda
$$
Accordingly,
$$
5 \times 6500=7 \times \lambda=4642.8 Å
$$

Topic: Physics, Wave Optics, KCET, KCET 2007
A single slit Fraunhofer diffraction pattern is formed with white light, for what wavelength of light, the third secondary maximum in the diffraction pattern coincides with the second secondary maximum in the pattern for the red light of wavelength 6500 Aâ?

Options:
A 9100 Aâ
B 4642 A∘
C 4100 A∘
D 4400 A∘

The correct answer is:
4642 A∘

Solution:
∵    x=(2n+1)2aλD
For red light, x=(4+1)2a×6500 A∘×D
For other light, x=(6+1)2aD×λ
∵ x is the same for each
∴5×6500=7λ
λ=57×6500
=4642.8 A∘

Topic: Physics, Wave Optics, NEET
A single slit of width b is illuminated by a coherent monochromatic light of wavelength Î». If the second and fourth minima in the diffraction pattern at a distance 1 cm from the slit are at 3 cm and 6 cm respectively from the central maximum, what is the width of the central maximum? (i.e. distance between first minimum on either side of the central maximum)

Options:
A 4.5 cm
B 1.5 cm
C 6.0 cm
D 3.0 cm

The correct answer is:
3.0 cm

Solution:
For secondary minima dsinâ¡Î¸=nÎ»
sinâ¡Î¸=nÎ»d
For second minima
n=2
sinâ¡Î¸=2Î»d=tanâ¡Î¸1=x1D
For fourth minima n=4
sinâ¡Î¸2=4Î»d=x2D
x2-x1=4Î»d-2Î»d=2Î»d= 6 -3 = 3cm
Width of central maxâ¡=2Î»d=3 cm

Topic: Physics, Wave Optics, NEET
(a) Six lead-acid type of secondary cells each of emf $2.0 \mathrm{~V}$ and internal resistance $0.015 \Omega$ are joined in series to provide a supply to a resistance of $8.5$ $\boldsymbol{\Omega}$. What are the current drawn from the supply and its terminal voltage?
(b) A secondary cell after long use has an emf of 1.9 $V$ and a large internal resistance $380 \Omega$. What maximum current can be drawn from the cell? Could the cell drive the starting motor of a car?

Options:

None

Solution:
(a) Given, number of secondary cells, $\mathrm{N}=6$,
E.m.f. of each cell, $\mathrm{E}=2.0 \mathrm{~V}$
Internal resistance of each cell, $r=0.015 \Omega$,
External resistance, $\mathrm{R}=8.5 \Omega$
By formula, current,
$I=\frac{N E}{R+N r}=\frac{6 \times 2}{8.5+6 \times 0.015}$ $=\frac{12}{8.5+0.09}=1.397 \mathrm{~A}$.
Terminal voltage $\mathrm{V}=\mathrm{IR}=\frac{12 \times 8.5}{8.59}$
$=11.874 \mathrm{~V}$
(b) Given, $\mathrm{E}=1.9 \mathrm{~V}, \mathrm{r}=380 \Omega, \mathrm{I}_{\max }=$ ?
Maximum current can be drawn by short circuit, $\mathrm{I}_{\max }=\frac{\mathrm{E}}{\mathrm{r}}=\frac{1.9}{380}=0.005 \mathrm{~A}$.
Which cannot start a car, because a starter motor requires a large current $(\approx 700 \mathrm{~A}$ ) for few seconds.

Topic: Physics, Current Electricity
A slit of width $a$ is illuminated by red light of wavelength $6500 Å$. If the first diffraction minimum falls at $30^{\circ}$, then the value of $a$ is

Options:
A $6.5 \times 10^{-4} \mathrm{~mm}$
B $1.3$ micron
C $3250 Å$
D $26 \times 10^{-4} \mathrm{~cm}$

The correct answer is:
$1.3$ micron

Solution:
Given, $\lambda=6500 Å=6500 \times 10^{-10} \mathrm{~m}, \theta=30^{\circ}$ According to the diffraction pattern of single slit,
$a \sin \theta=n \lambda$
where, $n=$ order of secondary minima $=1$
$\Rightarrow a \sin 30^{\circ}=1 \times\left(6500 \times 10^{-10}\right)$
$\Rightarrow \quad \frac{a}{2}=6500 \times 10^{-10}$
or
$\begin{aligned}
a &=13000 \times 10^{-10} \\
&=1.3 \times 10^{-6} \mathrm{~m}=1.3 \text { micron }
\end{aligned}$

Topic: Physics, Wave Optics, KCET, KCET 2021
A solution containing $4.5 \mathrm{mM}$ of $\mathrm{MnO}_4^{-}$and $15 \mathrm{mM}$ of $\mathrm{Mn}^{2+}$ shows $\mathrm{pH}$ of 2 . The potential of half-cell reaction is ...... .
(Given, $\log 15=1.76$ and $\log 45=1.653$ and standard potential of $\mathrm{MnO}_4^{-} \longrightarrow \mathrm{Mn}^{2+}$ ) is $1.51 \mathrm{~V}$ )

Options:
A $1.51 \mathrm{~V}$
B $1.31 \mathrm{~V}$
C $1.71 \mathrm{~V}$
D $1.04 \mathrm{~V}$

The correct answer is:
$1.31 \mathrm{~V}$

Solution:
$$
\begin{aligned}
& \text { (b) } \mathrm{MnO}_4^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_2 \mathrm{O} \\
& E=E_{\mathrm{Mn}^{\circ}}{ }^{7+} \mathrm{Mn}^{2+} /+\frac{0.059}{5} \log \frac{\left[\mathrm{MnO}_4^{-}\right]\left[\mathrm{H}^{+}\right]^8}{\left[\mathrm{Mn}^{2+}\right]} \\
& E=1.51+\frac{0.059}{5} \log \frac{\left[\frac{4.5}{1000}\right]\left[10^{-2}\right]^8}{\left[\frac{15}{1000}\right]} \\
& \Rightarrow \quad E=1.31 \mathrm{~V}
\end{aligned}
$$
Hence, the correct option is (2).

Topic: Chemistry, Electrochemistry, AP EAMCET, AP EAMCET 2020 (22 Sep Shift 1)
A sonometer wire 100 cm long has a fundamental frequency of 300 Hz . The velocity of propagation of transverse waves on the wire is

Options:
A 330 m s-1
B 660 m s-1
C 990 m s-1
D 550 m s-1

The correct answer is:
660 m s-1

Solution:
Frequency
n=vλ=v2l
∴ v=n2l=330×2=660 m s-1

Topic: Physics, Waves and Sound, NEET
A sonometer wire ' $A$ ' of diameter ' $\mathrm{d}$ ' under tension ' $T$ ' having density ' $\rho_1$ ' vibrates with fundamental frequency ' $n$ '. If we use another wire 'B' which vibrates with same frequency under tension ' $2 \mathrm{~T}$ ' and diameter ' $2 \mathrm{D}$ ' then density ' $\rho_2$ ' of wire ' $B$ ' will be

Options:
A $\rho_2=2 \rho_1$
B $\rho_2=\rho_1$
C $\rho_2=\frac{\rho_1}{2}$
D $\rho_2=\frac{\rho_1}{4}$

The correct answer is:
$\rho_2=\frac{\rho_1}{2}$

Solution:
The formula for frequency of a sonometer is $\mathrm{f}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\pi \rho \mathrm{D}^2}}$
Here $l$ is length, $\mathrm{T}$ is tension, $\mathrm{D}$ is diameter and $\rho$ is density.
The frequency of both the wires is same.
The frequency of the wire $\mathrm{A}$ is $\mathrm{f}_{\mathrm{A}}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\pi \rho_1 \mathrm{D}^2}}$
The frequency of the wire $\mathrm{B}$ is $f_{\mathrm{B}}=\frac{1}{2 l} \sqrt{\frac{2 \mathrm{~T}}{\pi \rho_2(2 \mathrm{D})^2}}$
Equating both the frequencies
$\begin{aligned}
& \frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\pi \rho \mathrm{D}^2}}=\frac{1}{2 l} \sqrt{\frac{2 \mathrm{~T}}{\pi \rho_2(2 \mathrm{D})^2}} \\
& \sqrt{\frac{1}{\rho_1}}=\sqrt{\frac{1}{2 \rho_2}} \\
& \frac{1}{\rho_1}=\frac{1}{2 \rho_2} \\
& \therefore \quad \rho_2=\frac{\rho_1}{2}
\end{aligned}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2023 (11 May Shift 2)
A sonometer wire has a length of $114 \mathrm{~cm}$, between two fixed ends. Where should two bridges be placed so as to divide the wire into three segments (in $\mathrm{cm}$ ) whose fundamental frequencies are in the ratio $1: 3: 4$ ?

Options:
A $l_1, l_2, l_3=18,24,72$
B $l_1, l_2, l_3=24,18,72$
C $l_1, l_2, l_3=72,18,24$
D $l_1, l_2, l_3=72,24,18$

The correct answer is:
$l_1, l_2, l_3=72,24,18$

Solution:
If length of the wire between the two bridges is $l$, the frequency of vibration
$n=\frac{1}{2 l} \sqrt{\frac{T}{m}}$
$\therefore \quad n \propto \frac{1}{l}$
Therefore, $n_1: n_2: n_3=\frac{1}{l_1}: \frac{1}{l_2}: \frac{1}{l_3}$
$n_1: n_2: n_3=\frac{1}{\frac{114 \times 1}{8}}: \frac{1}{\frac{114 \times 3}{8}}: \frac{1}{\frac{114 \times 4}{8}}$
or $\quad n_1: n_2: n_3=8: \frac{8}{3}: 2$
or $\quad n_1: n_2: n_3=72: 24: 18$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2010