datavorous/entrance-exam-dataset huggingface dataset:
A focus of an ellipse having eccentricity 12 is at 0, 0 and a directrix is the line x=4. Then the equation of one such ellipse is

Options:
A 9x264+3y216=1
B (2x+1)232+y216=1
C (3x+4)264+y232=1
D (3x+4)2+12y2=64

The correct answer is:
(3x+4)2+12y2=64

Solution:
A focus of an ellipse is at origin.

Directrix is x=4 and eccentricity is e=12

Major axis is along x-axis ,

⇒ae-ae=4

⇒a12-a2=4

⇒a=83

By the help of options we get the idea that this case is of a>b

Then we can say that e=1-b2a2

⇒14=1-9b264⇒b2=163

By this time we can say that either the correct option is 1 or 4

The standard form of equation of an ellipse is given by x-h2a2+y-k2b2=1

where centre is h,k

The foci are h-a2-b2,k & h+a2-b2,k

By this we can say that h+a2+b2=0

Now putting all values, we get

h=-649-163h=-43

And k=0

So by this we can say that the required equation of ellipse is 3x+42+12y2=64

Topic: Mathematics, Ellipse, TS EAMCET, TS EAMCET 2021 (04 Aug Shift 1)
A force $\overrightarrow{\mathrm{F}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}$ is applied at the point $(1,-1,2)$
What is the moment of the force about the point $(2,-1,3)$ ?

Options:
A $\hat{\mathrm{i}}+4 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
B $2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
C $2 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
D $2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-\hat{\mathrm{k}}$

The correct answer is:
$2 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$

Solution:
Let point $P$ is $(1,-1,2)$
and point $Q$ is $(2,-1,3)$ $\Rightarrow$ Position vector of $P$ w.r.t. $Q$ is
$\vec{r}=(1-2) \hat{i}+(-1+1) \hat{j}+(2-3) \hat{k}$
$\Rightarrow \vec{r}=-\hat{i}+0 \hat{j}-\hat{k}$ and $\vec{F}=3 \hat{i}+2 \hat{j}-4 \hat{k}$
$\Rightarrow$ Moment $=\vec{r} \times \vec{F}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -1 & 0 & -1 \\ 3 & 2 & -4\end{array}\right|$
$=\hat{i}(0+2)-\hat{j}(4+3)+\hat{k}(-2+0)=2 \hat{i}-7 \hat{j}-2 \hat{k}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2016 (Phase 2)
A force $\vec{F}=3 \hat{i}+4 \hat{j}-3 \hat{k}$ is applied at the point $\mathrm{P}$, whose
position vector is $\vec{r}=\widehat{2} i-2 \hat{j}-3 \hat{k}$. What is the magnitude of the moment of the force about the origin?

Options:
A 23 units
B 19 units
C 18 units
D 21 units

The correct answer is:
23 units

Solution:
Moment of force, $\mathrm{m}=\overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$
$\mathrm{m}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & -2 & -3 \\ 3 & 4 & -3\end{array}\right|$
$=\hat{\mathrm{i}}(6+12)-\hat{\mathrm{j}}(-6+9)+\hat{\mathrm{k}}(8+6)$
$=18 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+14 \hat{\mathrm{k}}$
$=\sqrt{(18)^{2}+(-3)^{2}+(14)^{2}}$
$=\sqrt{529}=23$ units.

Topic: Mathematics, Vector Algebra, NDA, NDA 2015 (Phase 1)
A force $\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$ acts on a particle to displace it from
the point $\mathrm{A}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})$ to the point $\mathrm{B}(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+5 \hat{\mathrm{k}})$. The
work done by the force will be

Options:
A 5 units
B 7 units
C 9 units
D 10 units

The correct answer is:
9 units

Solution:
$\overrightarrow{\mathrm{F}}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
$\overrightarrow{\mathrm{AB}}=(3-1) \hat{\mathrm{i}}+(-1-2) \hat{\mathrm{j}}+(5-(-3)) \hat{\mathrm{k}}$
$=2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}$
Work done $=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{AB}}=(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+8 \hat{\mathrm{k}})$
$=(1 \times 2)+(3 \times(-3))+(2 \times 8)$
$=2-9+16=9$ units.

Topic: Mathematics, Vector Algebra, NDA, NDA 2017 (Phase 2)
A force $m \hat{i}-3 \hat{j}+\hat{k}$ acts on a point and so the point moves
from $(20,3 \mathrm{~m}, 0)$ to $(0,0,7)$. If the work done by the force is $-48$ unit, what is the value of $\mathrm{m}$ ?

Options:
A 5
B 3
C 2
D 1

The correct answer is:
5

Solution:
Force, $\overrightarrow{\mathrm{F}}=\mathrm{m} \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}$
Due to this force, point moves from $\mathrm{A}(20,3 \mathrm{~m}, 0)$ to $\mathrm{B}(0,0,$,7). So, the displacement vector $\overrightarrow{\mathrm{AB}}$ is given by $\overrightarrow{\mathrm{AB}}=-20 \hat{\mathrm{i}}-3 \mathrm{~m} \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$
Work done $=\overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{AB}}$
$=(\mathrm{m} \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot(-20 \hat{\mathrm{i}}-3 \mathrm{~m} \hat{\mathrm{j}}+7 \hat{\mathrm{k}})$
$=(-20 \mathrm{~m}+9 \mathrm{~m}+7)$ unit
But work done $=-48$ unit, as given $\Rightarrow-11 \mathrm{~m}+7=-48$
$\Rightarrow-11 \mathrm{~m}=-55$
$\Rightarrow \mathrm{m}=5$

Topic: Mathematics, Vector Algebra, NDA, NDA 2006 (Phase 2)
A fowl consumes grains. What will we call the rate at which new organic matter is produced by this fowl?

Options:
A Primary Productivity Rate
B Gross Primary Productivity
C Net Primary Productivity
D Secondary productivity

The correct answer is:
Secondary productivity

Solution:
The amount of biomass or organic matter produced by plants by photosynthesis per unit area and over a period of time is called primary productivity. It is divided into gross and net primary productivity. The rate of production of organic matter by plants through photosynthesis is called gross primary productivity and the net primary productivity is the availability of biomass for consumption by heterotrophs. The rate at which the new organic matter is produced by consumers is called secondary productivity. Here fowl or chicken is a consumer.

Topic: Biology, Ecosystem, NEET
A Fraunhofer diffraction pattern due to a narrow slit is obtained on a screen placed at a distance \(D\) from the slit whose slit width is \(a\). The distance of first secondary maximum from the central maximum is

Options:
A \(\frac{3 D \lambda}{a}\)
B \(\frac{3 D \lambda}{2 a}\)
C \(\frac{2 D \lambda}{3 a}\)
D \(\frac{2 D \lambda}{a}\)

The correct answer is:
\(\frac{3 D \lambda}{2 a}\)

Solution:
In Fraunhofer diffraction pattern, the direction of secondary maximum is given as
\(\begin{aligned}
\theta & =(2 n+1) \frac{\lambda}{2 a}=(2 \times 1+1) \frac{\lambda}{2 a} \\
\Rightarrow \theta & =\frac{3 \lambda}{2 a}
\end{aligned}\)
\(\therefore\) Distance of first secondary maximum from the central maximum is given by
\(x=\theta D=\frac{3 \lambda}{2 a} \cdot D=\frac{3 D \lambda}{2 a}\)

Topic: Physics, Wave Optics, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 2)
A function $\mathrm{f}: \mathrm{A} \rightarrow \mathrm{R}$ is defined by the equation $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}-4 \mathrm{x}+5$
where $\mathrm{A}=(1,4)$. What is the range of the function?

Options:
A $(2,5)$
B $(1,5)$
C $[1,5)$
D $[1,5]$

The correct answer is:
$[1,5)$

Solution:
$\begin{aligned} f(x) &=x^{2}-4 x+5=(x-2)^{2}+1 \\ f(1) &=2, f(4)=5 \end{aligned}$
$\therefore$ Range $f(x)=[1,5)$

Topic: Mathematics, Functions, NDA, NDA 2018 (Phase 2)
A function $\mathrm{f}$ defined by $\mathrm{f}(\mathrm{x})=\ln \left(\sqrt{x^{2}+1}-x\right)$ is

Options:
A an even function
B an odd function
C Both even and odd function
D Neither even nor odd function

The correct answer is:
an odd function

Solution:
$\begin{aligned} & f(x)=\ln \left(\sqrt{x^{2}+1}-x\right) \\ & f(-x)=\ln \left(\sqrt{(-x)^{2}+1}-(-x)\right)=\ln \left(\sqrt{x^{2}+1}+x\right) \\ &=\ln \left(\frac{\left(\sqrt{x^{2}+1}+x\right)\left(\sqrt{x^{2}+1}-x\right)}{\sqrt{x^{2}+1}-x}\right) \\ &=\ln \left(\frac{x^{2}+1-x^{2}}{\sqrt{x^{2}+1}-x}\right)=\ln \left(\frac{1}{\sqrt{x^{2}+1}-x}\right) \\ &=-\ln \left(\sqrt{x^{2}+1}-x\right)=-f(x) \end{aligned}$
So, $\mathrm{f}(\mathrm{x})$ is odd function.

Topic: Mathematics, Functions, NDA, NDA 2019 (Phase 1)
A function $\mathrm{f}$ is defined as follows
$\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{p}} \cos \left(\frac{1}{\mathrm{x}}\right), \mathrm{x} \neq 0$
$f(0)=0$
What conditions should be imposed on p so that $\mathrm{f}$ may be continuous at $x=0$ ?

Options:
A $\mathrm{p}=0$
B $\mathrm{p}>0$
C $\mathrm{p} < 0$
D No value of p

The correct answer is:
$\mathrm{p}>0$

Solution:
Given function is defined as:
$f(x)=\left\{\begin{array}{ll}x^{p} \cos \left(\frac{1}{x}\right) & x \neq 0 \\ 0, & x=0\end{array}\right.$
For continuity:
LHS $: \lim _{x \rightarrow 0} f(x)=$ RHS $\lim _{x \rightarrow 0} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x^{p} \cos \left(\frac{1}{x}\right)=0$
$\Rightarrow \lim _{x \rightarrow 0} x^{p} \cos \left(\frac{1}{x}\right)=0$
$\cos \left(\frac{1}{\mathrm{x}}\right)$ is always a finite quantity if $\mathrm{x} \rightarrow 0$
$\Rightarrow \mathrm{x}^{\mathrm{p}}=0$
which is possible only if $\mathrm{p}>0$.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2007 (Phase 2)
A function $f$ is defined by $f(x)=x+\frac{1}{x}$. Consider the
following.
(1) $\quad(f(x))^{2}=f\left(x^{2}\right)+2$
(2) $(f(x))^{3}=f\left(x^{3}\right)+3 f(x)$
Which of the above 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:
$f\left(x^{2}\right)+2=x^{2}+\frac{1}{x^{2}}+2$
$=\left(x+\frac{1}{x}\right)^{2}=\{f(x)\}^{2}$
and $f\left(x^{3}\right)+3 f(x)$
$=x^{3}+\frac{1}{x^{3}}+3\left(x+\frac{1}{x}\right)=\left(x+\frac{1}{x}\right)^{3}=\{f(x)\}^{3}$
Thus, both 1 and 2 are correct.

Topic: Mathematics, Functions, NDA, NDA 2008 (Phase 2)
A function $f$ is such that $f^{\prime}(x)=6-4 \sin 2 x$ and $f(0)=3$. What is $f(x)$ equal to?

Options:
A $6 x+2 \cos 2 x$
B $6 x-2 \cos 2 x$
C $6 x-2 \cos 2 x+1$
D $6 x+2 \cos 2 x+1$

The correct answer is:
$6 x+2 \cos 2 x+1$

Solution:
Given,
$f^{\prime}(x)=6-4 \sin 2 x$ and $f(0)=3$
Consider $f^{\prime}(x)=6-4 \sin 2 x$
Integrate both sides w.r.t $x$
$\int f^{\prime}(x) d x=\int(6-4 \sin 2 x) d x$
$f(x)=6 x-\frac{4(-\cos 2 x)}{2}+c$
Where ' $c$ ' is constant of integration $\Rightarrow f(x)=6 x+2 \cos 2 x+c$
By using $f(0)=3$, we have $3=f(0)=6.0+2 \cos 0+c$
$\Rightarrow 3=2+c \Rightarrow c=1$
Hence, $f(x)=6 x+2 \cos 2 x+1$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 2)
A function $f: R \rightarrow R$ is defined as $f(x)=x^{2}$ for $x \geq 0$ and $f(x)=-x$ for $x < 0 .$
Consider the following statements in respect of the above function:
1- The function is continuous at $x=0$.
2- The function is differentiable at $x=0$.
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:
1 only

Solution:
$\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}, \mathrm{f}(\mathrm{x})=\left\{\begin{array}{ll}\mathrm{x}^{2} & , \quad \mathrm{x} \geq 0 \\ -\mathrm{x} & , \quad \mathrm{x} < 0\end{array}\right.$
For continuity at $\mathrm{x}=0$ $\mathrm{f}(0-0)=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(0-\mathrm{h})$
$=\lim _{\mathrm{h} \rightarrow 0}[(0-\mathrm{h})]=\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h}=0$
$\mathrm{f}(0+0)=\lim _{\mathrm{h} \rightarrow 0} \mathrm{f}(0+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}(0+\mathrm{h})^{2}=0$ and $\mathrm{f}(0)=0$
$\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$
For differentiability at $\mathrm{x}=0$ $\lim _{h \rightarrow 0} \frac{-(-h)-0}{-h}=\lim _{h \rightarrow 0}=\frac{h}{-h}=-1$
and $\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \rightarrow 0} h=0$
$\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}-0$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2013 (Phase 2)
A function is defined as follows:
$f(\mathrm{x})=\left\{\begin{array}{cl}-\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}}} & , \mathrm{x} \neq 0 \\ 0, & \mathrm{x}=0\end{array}\right.$
Which one of the following is correct in respect of the above function?

Options:
A $f(\mathrm{x})$ is continuous at $\mathrm{x}=0$ but not differentiable at $\mathrm{x}=0$
B $f(\mathrm{x})$ is continuous as well as differentiable at $\mathrm{x}=0$
C $f(\mathrm{x})$ is discontinuous at $\mathrm{x}=0$
D None of the above

The correct answer is:
$f(\mathrm{x})$ is discontinuous at $\mathrm{x}=0$

Solution:
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}\frac{-\mathrm{x}}{\sqrt{\mathrm{x}^{2}}}, \mathrm{x} \neq 0 \\ 0, \mathrm{x}=0\end{array}\right.$
$=\left\{\begin{array}{l}\frac{-\mathrm{x}}{|\mathrm{x}|}, \mathrm{x} \neq 0 \\ 0, \mathrm{x}=0\end{array}\right.$
$\mathrm{f}(0+\mathrm{h})=\frac{-\mathrm{h}}{\mathrm{h}}=-1 ; \mathrm{f}(0-\mathrm{h})=\frac{\mathrm{h}}{\mathrm{h}}=1$
So, $\mathrm{f}(\mathrm{x})$ is discontinuous at $\mathrm{x}=0$.

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2017 (Phase 2)
A function is defined in $(0, \infty)$ by $\quad[2017-I I]$ $f(x)=\left\{\begin{array}{ccc}1-x^{2} & \text { for } & 0 < x \leq 1 \\ \operatorname{In} & x & \text { for } 1 < x \leq 2 \\ \text { In } 2-1+0.5 x & \text { for } & 2 < x < \infty\end{array}\right.$
Which one of the following is correct in respect of the derivative of the function, i.e., $f^{\prime}(\mathrm{x}) ?$

Options:
A $f^{\prime}(x)=2 x$ for $0 < x \leq 1$
B $f^{\prime}(x)=-2 x$ for $0 < x \leq 1$
C $f^{\prime}(x)=-2 x$ for $0 < x < 1$
D $f^{\prime}(x)=0$ for $0 < x < \infty$

The correct answer is:
$f^{\prime}(x)=-2 x$ for $0 < x < 1$

Solution:
$\mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}1-\mathrm{x}^{2} \text { for } 0 < \mathrm{x} \leq 1 \\ \ln \mathrm{x} \text { for } 1 < \mathrm{x} \leq 2 \\ \ell \mathrm{n} 2-1+0.5 \mathrm{x} \text { for } 2 < \mathrm{x} < \infty\end{array}\right.$
$\mathrm{f}^{\prime}(\mathrm{x})=-2 \mathrm{x}$ for $0 < \mathrm{x} < 1$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2017 (Phase 2)
A fungus with hyphae containing nuclei from different genomes, the nuclei do not fuse and are without sex organ but produces sexual spore belongs to

Options:
A Phycomycetes
B Zygomycetes
C Deuteromycetes
D Basidiomycetes

The correct answer is:
Basidiomycetes

Solution:
The intermediate phase between two stages (plasmogamy and karyogamy) is known as dikaryophase.
It gives rise to secondary mycelium. It has multiple septate hyphae. The septa have spores and hook shaped outgrowths. They are known as clamp connections. Septa mycelia multiply by formation of many spores including uredospores and chlamydospores. In basidia, karyogamy and meiosis occur which leads to the production of basidiospores.

Topic: Biology, Biological Classification, NEET
A galvanic cell is set up from electrodes $A$ and $B$ Electrode $A: \mathrm{Cr}_2 \mathrm{O}_7^{2-} / \mathrm{Cr}^{3+}, E_{\text {red }}^{\circ}=+1.33 \mathrm{~V}$ Electrode $B: \mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}, \quad E_{\text {red }}^{\circ}=0.77 \mathrm{~V}$ Which of the following statements is false ?

Options:
A Standard e.m.f of the cell is $0.56 \mathrm{~V}$
B Current will flow from electrode $A$ to $B$ in the external circuit
C $A$ will act as cathode and have positive polarity
D None of these

The correct answer is:
None of these

Solution:
No Solution Available

Topic: Chemistry, Electrochemistry, VITEEE, VITEEE 2021
A gas deviates from ideal behaviour at a high pressure because its molecules

Options:
A attract one another
B show the Tyndall effect
C have kinetic energy
D are bound by covalent bonds

The correct answer is:
attract one another

Solution:
At high pressure, the volume is decreased appreciably, so the attractive forces become large and the molecules are crowded together. Thus, pressure correction is necessary and the gas deviates more from ideal behaviour.

Topic: Chemistry, States of Matter, KCET, KCET 2008
A gas deviates from ideal behaviour at a high pressure because its molecules

Options:
A attract one another
B show the Tyndall effect
C have kinetic energy
D are bound by covalent bonds

The correct answer is:
attract one another

Solution:
A gas deviates from ideal behaviour at high pressure because its molecules attract one another as their intermolecular distance decreases.

Topic: Chemistry, States of Matter, COMEDK, COMEDK 2017
A genetically engineered micro-organism used successfully in bioremediation of oil spills is a species of:

Options:
A Trichoderma
B Xanthomonas
C Bacillus
D Pseudomonas.

The correct answer is:
Pseudomonas.

Solution:
A genetically engineered microorganism called Pseudomonas putida is used in thebioremediation of oil spills. It is also known as super bug and was discovered by Ananda Mohan Chakrabarty. Cyclosporin is obtained from Trichoderma polysporum which is used as immunosuppressive agent. Xanthomonas is a disease causing gram negative bacteria.
Related Theory
Bioremediation is a biotechnical process which involves the removal of complex material or contamination by degrading environmental pollutants using living microorganisms. The ability of micro-organisms could be enhanced by providing optimum growth conditions to micro-organism or by employing the genetically modified microorganisms (GMMs). This technology has been implemented to eradicate environmentally hazardous chemicals and detoxify them into nontoxic forms.
Caution
Bioaugmentation involves the injection of a small amount of oil-degrading microbes into an affected area. Biostimulation is the addition of nutrients to stimulate the growth of innate oil-degrading microbes to increase the rate of remediation.

Topic: Biology, Biological Classification, NEET, NEET 2007
A given quantity of metal is to be cast into a half cylinder (i,e, with a rectangular base and semicircular ends). If the total surface area is to be minimum, then the ratio of the height of the half cylinder to the diameter of the semicircular ends is

Options:
A $\pi:(\pi+2)$
B $(\pi+2): \pi$
C $1: 1$
D None of the above

The correct answer is:
$\pi:(\pi+2)$

Solution:
Volume of half cylinder with height ${ }^{\prime} 1^{\prime}$ and radius ' $\mathrm{r}$ '
$=\frac{1}{2} \pi r^{2} l$
Surface area of half cylinder.
$S=l \times 2 r+2\left(\frac{1}{2} \pi r^{2}\right)+2\left(\frac{1}{2} \pi r l\right)$
$S=2 r l+\pi r^{2}+\pi r l$
$S=\frac{4 V}{\pi r}+\frac{\pi \cdot 2 V}{\pi r}+\pi r^{2} \quad($ from $(1))$
$\frac{d s}{d r} \Rightarrow \frac{-4 V}{\pi r^{2}}-\frac{2 V}{r^{2}}+2 \pi r=0$
$\Rightarrow 2 \pi r=\frac{4 V}{\pi r^{2}}+\frac{2 V}{r^{2}}$
$\Rightarrow 2 \pi r=\frac{4 . \pi r^{2} l}{2 \pi r^{2}}+\frac{2}{r^{2}} \cdot \frac{1}{2} \pi r^{2} l$
$\Rightarrow \frac{l}{2 r}=\frac{\pi}{\pi+2}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2019 (Phase 1)
A glass tube is open at both the ends. A tuning fork of frequency $f$ resonates with the air column inside the tube. Now the tube is placed vertically inside water so that half the length of the tube is filled with water. Now the air column inside the tube is in unison with another fork of frequency $f^{\prime}$. Then

Options:
A $f^{\prime}=f$
B $f^{\prime}=4 f$
C $f^{\prime}=2 f$
D $f^{\prime}=\frac{f}{2}$

The correct answer is:
$f^{\prime}=f$

Solution:
For fundamental frequency of vibration of air column
$$
l=\frac{\lambda}{2} \Rightarrow \lambda=2 l
$$
Frequency, $f=\frac{v}{\lambda}$
$\Rightarrow \quad f=\frac{v}{2 l}$
When tube is placed inside water with half length immersed in water then
Frequency, $f^{\prime}=\frac{v}{\lambda^{\prime}}=\frac{v}{2 l}$
From Eqs. (i) and (ii), we get
$$
f^{\prime}=f
$$

Topic: Physics, Waves and Sound, KCET, KCET 2007
A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics $(c, e$, mass of electron, mass of proton) and the gravitational constant $\mathrm{G}$, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe ( $\sim 15$ /billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of ). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?

Options:

None

Solution:
Using basic constants such as speed of light (c), charge on electron $(e)$, mass of electron $\left(\mathrm{m}_e\right)$, mass of proton $\left(\mathrm{m}_v\right)$ and gravitational constant $(\mathrm{G})$, we can construct the quantity,
$$
\tau=\left(\frac{e^2}{4 \pi \varepsilon_0}\right)^2 \times \frac{1}{m_p m_e^2 c^3 \mathrm{G}}
$$
$$
\begin{aligned}
\text { Now }\left[\frac{e^2}{4 \pi \varepsilon_0}\right] &=\left[\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2} r^2\right]=\left[\mathrm{Fr}^2\right] \\
&=\left[\mathrm{MLT}^{-2} \cdot \mathrm{L}^2\right]=\left[\mathrm{ML}^3 \mathrm{~T}^{-2}\right] \\
\therefore[\tau] &=\frac{\left[\mathrm{ML}^3 \mathrm{~T}^{-2}\right]^2}{[\mathrm{M}][\mathrm{M}]^2\left[\mathrm{LT}^{-1}\right]^3\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]}=[\mathrm{T}]
\end{aligned}
$$
Clearly, the quantity $\tau$ has the dimensions of time.
Put $\mathrm{G}=6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}$,
$$
\begin{aligned}
&c=3 \times 10^8 \mathrm{~m} / \mathrm{s} \\
&e=1.6 \times 10^{-19} \mathrm{C}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg} \\
&\mathrm{~m}=1.67 \times 10^{-27} \mathrm{~kg} \\
&\text { and } \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{Nm}^2 \mathrm{C}^2 \\
&\therefore \tau=\frac{\left[9 \times 10^9 \times\left(1.6 \times 10^{-19}\right)^2\right]^2}{1.67 \times 10^{-27} \times\left(9.1 \times 10^{-31}\right)^2 \times\left(3 \times 10^8\right)^3 \times 6.67 \times 10^{-11}}
\end{aligned}
$$
$$
\begin{aligned}
&=2.13 \times 10^{16} \mathrm{~s} \\
&=\frac{2.13 \times 10^{16}}{3.156 \times 10^7} \text { years }=0.667 \times 10^9 \text { years. } \\
&=0.667 \text { billion years. }
\end{aligned}
$$
This time is slightly less than the age of the universe ( $\approx 15$ billion years). It implies that the values of the basic constants of physics should change with time because the age of the universe increases with time.

Topic: Physics, Units and Dimensions
A group consists of 5 men and 5 women. If the number of different five-person committees containing $\mathrm{k}$ men and $(5-\mathrm{k})$ women is 100, what is the value of $\mathrm{k}$ ?

Options:
A 2 only
B 3 only
C 2 or 3
D 4

The correct answer is:
2 or 3

Solution:
$\mathrm{K}$ men selected out of 5 and $5-\mathrm{k}$ women out of 5 . These are ${ }^{5} \mathrm{C}_{\mathrm{k}}$ and ${ }^{5} \mathrm{C}_{5-\mathrm{k}}$
According to problem:
${ }^{5} \mathrm{C}_{\mathrm{k}} \times{ }^{5} \mathrm{C}_{5-\mathrm{k}}=100$
$\Rightarrow \frac{5 !}{\mathrm{k} !(5-\mathrm{k}) !} \times \frac{5 !}{(5-\mathrm{k}) ! 5 !}=100$
$\Rightarrow\left(\frac{5}{k !(5-k) !}\right)^{2}=100$
$\Rightarrow \frac{5 !}{\mathrm{k} !(5-\mathrm{k}) !}=10$
This is true for $\mathrm{k}=2$ or 3 .

Topic: Mathematics, Permutation Combination, NDA, NDA 2008 (Phase 1)
A hollow pipe of length $0.8 \mathrm{~m}$ is closed at one end. At its open end, a $0.8 \mathrm{~m}$ long uniform string is vibrating in its second harmonic and it resonates with the fundamental frequency of pipe. If the tension in the sting is $50 \mathrm{~N}$ and speed of sound in air is $320 \mathrm{~m} / \mathrm{s}$, the mass of the strings is

Options:
A $10 \mathrm{~g}$
B $20 \mathrm{~g}$
C $5 \mathrm{~g}$
D $40 \mathrm{~g}$

The correct answer is:
$10 \mathrm{~g}$

Solution:
For a closed organ pipe, the fundamental frequency is given $f=\frac{v}{4 L}$, where $v=320 \mathrm{~m} / \mathrm{s}$ is the velocity of sound in the medium of organ pipe and $L=0.8 \mathrm{~m}$ being the length of pipe.
Now, we are given second harmonic frequency of wire is equal and in resonance with fundamental frequency of pipe, thus;
$f=\frac{v}{4 L}=\frac{1}{l} \sqrt{\frac{T}{\mu}}$
Putting, $T=50 \mathrm{~N}$ and $l=0.8 \mathrm{~m}$
$\begin{aligned} & \frac{320}{4(0.8)}=\frac{1}{0.5} \sqrt{\frac{50}{\mu}} \\ & \Rightarrow \sqrt{\frac{\mu}{50}}=\frac{1}{50}\end{aligned}$
The length of string $=0.5 \mathrm{~m}$
Thus, mass of string $=0.02 \times 0.5=0.01 \mathrm{~kg}=10 \mathrm{~g}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2022 (10 Aug Shift 2)