datavorous/entrance-exam-dataset huggingface dataset:
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, TS EAMCET, TS EAMCET 2010
A source emits electromagnetic waves of wavelength 3 m . One beam reaches the observer directly and other after reflection from a water surface, travelling 1.5 m extra distance and with intensity reduced to $1 / 4$ as compared to intensity due to the direct beam alone. The resultant intensity will be

Options:
A (1/4) fold
B (3/4) fold
C (5/4) fold
D (9/4) fold

The correct answer is:
(9/4) fold

Solution:
We know that a phase change of $\pi$ occurs when the reflection takes place at the boundary of denser medium. This is equivalent to a path difference of $\lambda / 2$.
$\therefore$ Total phase difference $=\pi-\pi=0$
Thus, the two waves superimpose in phase.
$\begin{aligned}
& \text { Resultant amplitude }=\sqrt{I}+\sqrt{\left(\frac{I}{4}\right)}=\frac{3}{2} \sqrt{I} \\
& \text { Resultant intensity }=\left(\frac{3}{2} \sqrt{I}\right)^2=\frac{9}{4} I=\frac{9}{4} \text { fold. }
\end{aligned}$

Topic: Physics, Wave Optics, JIPMER, JIPMER 2017
A spacecraft at $\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ is subjected to a force $\lambda \hat{\mathrm{k}}$ by firing a rocket. The spacecraft is subjected to a moment of magnitude

Options:
A $\lambda$
B $\sqrt{3} \lambda$
C $\sqrt{5} \lambda$
D None of these

The correct answer is:
$\sqrt{5} \lambda$

Solution:
$\begin{aligned} & \vec{\tau}=\vec{r} \times \overrightarrow{\mathrm{F}}=(\hat{i}+2 \hat{j}+3 \hat{k}) \times \lambda \hat{k} \\ &=2 \lambda \hat{i}-\lambda \hat{j} \\ \Rightarrow &|\vec{\tau}|=\sqrt{5} \lambda \end{aligned}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2018 (Phase 2)
A special dice with numbers $1,-1,2,-2,0$ and 3 is thrown thrice. What is the probability that the sum of the numbers occurring on the upper face is zero?

Options:
A $1 / 72$
B $1 / 8$
C $7 / 72$
D $25 / 216$

The correct answer is:
$25 / 216$

Solution:
Total no. of elementary events $=6^{3}$. Favourable no. of elementary events
$=$ coefficient of $x^{0}$ in $\left[x+x^{-1}+x^{0}+x^{-2}+x^{2}+x^{3}\right]^{3}$
$=$ coeff. of $x^{0}$ in $\left[\frac{1+x+x^{2}+x^{3}+x^{4}+x^{5}}{x^{2}}\right]^{3}$
= coeff. of $x^{6}$ in $\left[1+x+x^{2}+x^{3}+x^{4}+x^{5}\right]^{3}$
$=$ coeff. of $x^{6}$ in $\left[\frac{1-x^{6}}{1-x}\right]^{3}$
$=$ coeff. of $x^{6}$ in $\left[1-x^{6}\right]^{3}[1-x]$
$=$ coeff. of $x^{6}$ in $\left[1-{ }^{3} C_{1} x^{6}+\ldots .\right][1-x]^{-3}$
$=$ coeff. of $x^{6}$ in $(1-x)^{-3} \cdot{ }^{3} C_{1}$ coeff. of $x^{0}$ in $(1-x)^{-3}$
$={ }^{6+3-1} C_{3-1}-{ }^{3} C_{1}$
$={ }^{8} C_{2}-{ }^{3} C_{1}=\frac{8 !}{6 ! 2 !}-\frac{3 !}{2 !}$
$=\frac{8 \times 7}{2}-3=25$
Required probability $=\frac{25}{216}$

Topic: Mathematics, Probability, NDA, NDA 2016 (Phase 2)
A sphere of constant radius $\mathrm{r}$ through the origin intersects the coordinate axes in $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. What is the locus of the centroid of the triangle ABC?

Options:
A $x^{2}+y^{2}+z^{2}=r^{2}$
B $x^{2}+y^{2}+z^{2}=4 r^{2}$
C $9\left(x^{2}+y^{2}+z^{2}\right)=4 r^{2}$
D $3\left(x^{2}+y^{2}+z^{2}\right)=2 r^{2}$

The correct answer is:
$9\left(x^{2}+y^{2}+z^{2}\right)=4 r^{2}$

Solution:
Let the sphere passing through points $\mathrm{A}(\mathrm{a}, 0,0), \mathrm{B}(0,$, $\mathrm{b}, 0), \mathrm{C}(0,0, \mathrm{c})$
Equation of sphere is $\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{z}^{2}-\mathrm{ax}-\mathrm{by}-\mathrm{cz}=0$
radius, $\mathrm{r}=\frac{1}{2} \sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}$
$\Rightarrow a^{2}+b^{2}+c^{2}=4 r^{2}$
(Squaring on both sides)
Let $(\alpha, \beta, \gamma)$ be centroid of sphere. $(\alpha, \beta, \gamma)=\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}\right)$
$=\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$
$\alpha^{2}+\beta^{2}+\gamma^{2}=\frac{\mathrm{a}^{2}}{9}+\frac{\mathrm{b}^{2}}{9}+\frac{\mathrm{c}^{2}}{9}$
$=\frac{a^{2}+b^{2}+c^{2}}{9}$
$($ from $(1))$
$\left.z^{2}\right)=4 r^{2}$

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2018 (Phase 1)
A spherical balloon of radius $\mathrm{r}$ subtends an angle $\alpha$ at the eye of an observer, while the angle of elevation of its centre is $\beta$. What is the height of the centre of the balloon (neglecting the height of the observer)?

Options:
A $\frac{r \sin \beta}{\sin \left(\frac{\alpha}{2}\right)}$
B $\frac{r \sin \beta}{\sin \left(\frac{\alpha}{4}\right)}$
C $\frac{\mathrm{r} \sin \left(\frac{\beta}{2}\right)}{\sin \alpha}$
D $\frac{r \sin \alpha}{\sin \left(\frac{\beta}{2}\right)}$

The correct answer is:
$\frac{r \sin \beta}{\sin \left(\frac{\alpha}{2}\right)}$

Solution:
Let 'A' be the position of eye. Let ' $\mathrm{O}$ ' be the centre of spherical balloon. Let ${ }^{\circ} \mathrm{h}$ ' be the height of centre of balloon.
From figure, in $\Delta \mathrm{OAD}, \sin \frac{\alpha}{2}=\frac{\mathrm{OD}}{\mathrm{OA}}=\frac{\mathrm{r}}{\mathrm{OA}}$
$\Rightarrow \mathrm{OA}=\frac{\mathrm{r}}{\sin \frac{\alpha}{2}}$
In $\Delta \mathrm{OAB}, \sin \beta=\frac{\mathrm{OB}}{\mathrm{OA}}=\frac{\mathrm{h}}{\mathrm{OA}}$
$\Rightarrow \mathrm{h}=\mathrm{OA} \cdot \sin \beta$
$=\frac{\mathrm{r} \cdot \sin \beta}{\sin \frac{\alpha}{2}} \quad(\operatorname{from}(1))$

Topic: Mathematics, Heights and Distances, NDA, NDA 2018 (Phase 1)
A spoon to be electroplated with gold should be

Options:
A cathode
B anode
C electrolyte
D none of these.

The correct answer is:
cathode

Solution:
Standard electrode potential for Gold is
$$
\mathrm{Au}^{3+}+3 e^{-} \rightleftharpoons \mathrm{Au}
$$
is 1.50 volt. therefore, gold get easily reduced. So to electroplate gold on the spoon, spoon should be made cathode as reduction takes place on cathode.

Topic: Chemistry, Electrochemistry, AIIMS, AIIMS 2001
A square is drawn by joining mid points of the sides of a square. Another square is drawn inside the second square in the same way and the process is continued in definitely. If the side of the first square is $16 \mathrm{~cm}$, then what is the sum of the areas of all the squares?

Options:
A $256 \mathrm{sq} \mathrm{cm}$
B $512 \mathrm{sq} \mathrm{cm}$
C $1024 \mathrm{sq} \mathrm{cm}$
D $512 / 3 \mathrm{sq} \mathrm{cm}$

The correct answer is:
$512 \mathrm{sq} \mathrm{cm}$

Solution:
Let ABCD, EFGH and IJKL be squares. Let side of square $\mathrm{ABCD}=16$ Now, Area of $\mathrm{ABCD}=(16)^{2}$
Area of EFGH $=\frac{(16)^{2}}{2}$,
Area of IJKL $=\frac{(16)^{2}}{4}$ So on.
Required sum,
$=16^{2}+\frac{1}{2}(16)^{2}+\frac{1}{4}(16)^{2}+\ldots . \infty$
$=(16)^{2}\left\{1+\frac{1}{2}+\frac{1}{4}+\ldots \infty\right\}$
Now, $1+\frac{1}{2}+\frac{1}{4}+\ldots \ldots \infty$ is a G.P.
$\therefore \quad$ Sum $=\frac{a}{1-r}$ where $a=1$ and $r=\frac{1}{2}$
$\therefore \quad 1+\frac{1}{2}+\frac{1}{4}+\ldots \ldots+\infty=\frac{1}{1-\frac{1}{2}}=2$
$=256 \times 2=512$ sq. $\mathrm{cm}$

Topic: Mathematics, Straight Lines, NDA, NDA 2010 (Phase 2)
A square matrix $\left[\mathrm{a}_{\mathrm{ij}}\right]$ such that $\mathrm{a}_{\mathrm{ij}}=0$ for $\mathrm{i} \neq \mathrm{j}$ and $\mathrm{a}_{\mathrm{ij}}=\mathrm{k}$ where
$\mathrm{k}$ is a constant for $1=\mathrm{j}$ is called:

Options:
A diagonal matrix, but not scalar matrix
B scalar matrix
C unit matrix
D None of the above

The correct answer is:
scalar matrix

Solution:
Scalar Matrix. We know that, $\mathrm{A}=\left[\mathrm{a}_{\mathrm{ij}}\right]_{\mathrm{nxn}}$ is called a scalar matrix ifa $_{\mathrm{i}}$ $=0$ for $i \neq j$ and $a_{i j}=k$ for $i=j$ [where $k$ is constant $]$

Topic: Mathematics, Matrices, NDA, NDA 2012 (Phase 2)
A square matrix A is called orthogonal if

Options:
A $\mathrm{A}=\mathrm{A}^{2}$
B $\mathrm{A}^{\prime}=\mathrm{A}^{-1}$
C $\mathrm{A}=\mathrm{A}^{-1}$
D $\mathrm{A}=\mathrm{A}^{\prime}$
Where $A^{\prime}$ is the transpose of $A$.

The correct answer is:
$\mathrm{A}^{\prime}=\mathrm{A}^{-1}$

Solution:
If $\mathrm{A}^{-1}=\mathrm{A}^{\mathrm{T}}$, then $\mathrm{A}$ is orthogonal matrix.

Topic: Mathematics, Matrices, NDA, NDA 2018 (Phase 2)
A standard hydrogen electrode has a zero potential because

Options:
A hydrogen can be most easily oxidised
B hydrogen has only one electron
C the electrode potential assumed to be zero
D hydrogen is the lightest element

The correct answer is:
the electrode potential assumed to be zero

Solution:
According to convention, the standard hydrogen electrode is assigned a zero potential at all temperatures.

Topic: Chemistry, Electrochemistry, NEET
A standing wave in a pipe with a length $L=1.2 \mathrm{~m}$ is described by
$y(x, t)=y_{0} \sin [(2 \pi / L) x] \sin [(2 \pi / \mathrm{L}) x+\pi / 4]$
Based on above information, which one of the following statements is incorrect. (Speed of sound in air is $300 \mathrm{~m} \mathrm{~s}^{-1}$ )-

Options:
A The pipe is closed at hoth ends
B The wavelength of the wave could be $1.2 \mathrm{~m}$
C There could be a node at $\mathrm{x}=0$ and antinode at $\mathrm{x}=\mathrm{L} / 2$
D The frequency of the fundamental mode of vibrations is $137.5 \mathrm{~Hz}$

The correct answer is:
The frequency of the fundamental mode of vibrations is $137.5 \mathrm{~Hz}$

Solution:
$\begin{array}{l}
\frac{2 \pi}{\lambda} \mathrm{x}=\frac{2 \pi}{\mathrm{L}} \mathrm{x} \\
\therefore \lambda=\mathrm{L}=1.2 \mathrm{~m} \\
\text { at } \mathrm{x}=0, \mathrm{x}=\mathrm{L}, \mathrm{y}=0 \\
\mathrm{v}=\frac{\mathrm{v}}{\lambda}=\frac{300}{1.2}=250 \mathrm{~Hz}
\end{array}$

Topic: Physics, Waves and Sound, KVPY, KVPY 2012 (SB/SX)
A stationary wave is represented by $y=10 \sin \left(\frac{\pi x}{4}\right) \cos (20 \pi t)$ where $x$ and $y$ are in $\mathrm{cm}$ and $t$ in second. The distance between two consecutive nodes is

Options:
A $1 \mathrm{~cm}$
B $8 \mathrm{~cm}$
C $4 \mathrm{~cm}$
D $2 \mathrm{~cm}$

The correct answer is:
$4 \mathrm{~cm}$

Solution:
Given, $y=10 \sin \left(\frac{\pi x}{4}\right) \cos (\pi t)$
Compare with standard equation formed after superposing
$\begin{aligned} & y=A \sin (k x+\omega t) \text { and } y=A \sin (k x-\omega t) \\ & y=2 A \sin (k x) \cos (\omega t)\end{aligned}$
Therefore, $k=\frac{\pi}{4}$
We know, the propagation constant is:
$k=\frac{2 \pi}{\lambda}=\frac{\pi}{4}$
$\Rightarrow \lambda=8 \mathrm{~cm}$
The distance between two consecutive nodes is given by half of wavelength:
$\frac{\lambda}{2}=4 \mathrm{~cm}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2022 (10 Aug Shift 2)
A steel rod $100 \mathrm{~cm}$ long is clamped at its middle. The fundamental frequency of longitudinal vibrations of the rod are given to be $2.53 \mathrm{kHz}$. What is the speed of sound in steel?

Options:

None

Solution:
When the rod is clamped at the middle, a node is forrmed at the middle and 2 antinodes are forrmed at the two ends.
Here, $L=100 \mathrm{~cm}=1 \mathrm{~m}, v=2.53 \mathrm{kHz}=2.53 \times 10^3 \mathrm{~Hz}$
$$
\begin{aligned}
&\therefore \quad L=\frac{\lambda}{4}+\frac{\lambda}{4}=\frac{\lambda}{2} \Rightarrow \lambda=2 L=2 \mathrm{~m} \\
&\mathrm{v}=\mathrm{v} \lambda=2.53 \times 10^3 \times 2=5.06 \times 10^3 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Waves and Sound
A steel rod of length $2 l$, cross-sectional area $A$ and mass $M$ is set rotating in a horizontal plane about an axis passing through the centre. If $Y$ is the Young's modulus for steel, find the extension in the length of the rod. (assume the rod is uniform)

Options:

None

Solution:
Let an element of width $d r$ at $r$ as shown in the figure.
Let $T(r)$ and $T(r+d r)$ are the tensions at $r$ and $r+d r$ respectively.
Net centrifugal force due to rotation on the element = $\omega^2 r d m$
(where $\omega$ is angular velocity of the rod)
$=\omega^2 r \mu d r \quad(\because \mu=$ mass/length $)$
Centrifugal force on element $(d r)$ due to tension difference
$$
\begin{aligned}
=& T(r+d r)-T(r)=-d T \\
\Rightarrow \quad & T(r+d r)-T(r)=\mu \omega^2 r d r \\
& \quad-d T=\mu \omega^2 r d r
\end{aligned}
$$
Integrating both sides,
$$
\begin{aligned}
&\therefore \quad-\int_{T=0}^T d T=\int_{r=r}^{r=l} \mu \omega^2 r d r=\mu \omega^2\left[\frac{r^2}{2}\right]_r^l \\
&{[\because T=0 \text { at } r=l]}
\end{aligned}
$$
Tension in rod at distance $r$ from centre so, limits will varies $r$ to $l$.
$$
\begin{aligned}
&T(r)=\frac{\mu \omega^2}{2}\left(l^2-r^2\right) \\
&\Rightarrow T(r)=\frac{-\mu \omega^2}{2}\left(l^2-r^2\right)
\end{aligned}
$$
Let the increase in length of the element $d r$ at distance $r$ from centre is $\delta r$ then
$$
\begin{aligned}
&\text { So, Young's modulus } Y=\frac{\text { Stress }}{\text { Strain }}=\frac{T(r) / A}{\delta r} \\
&\therefore \quad \frac{\delta r}{d r}=\frac{T(r)}{A}=\frac{-\mu \omega^2}{2 Y A}\left(l^2-r^2\right) \\
&\therefore \quad \delta r=\frac{-1}{Y A} \frac{\mu \omega^2}{2}\left(l^2-r^2\right) d r \\
&\therefore \delta r=\frac{1}{Y A} \frac{\mu \omega^2}{2} \int_0^l\left(l^2-r^2\right) d r \\
&\int_0^\delta \delta r=\left(\frac{1}{Y A}\right) \frac{\mu \omega^2}{2}\left[l^3-\frac{l^3}{3}\right]=\frac{1}{3 Y A} \mu \omega^2 l^3
\end{aligned}
$$
(for rod one from centre)
$\therefore$ Total extension in rod both sides
$$
=2 \delta=\frac{2}{3 Y A} \mu \omega^2 l^3
$$

Topic: Physics, Mechanical Properties of Solids
A steel wire of length $1 \mathrm{~m}$, mass $0.1 \mathrm{~kg}$ and uniform area of cross section $10^{-6} \mathrm{~m}^2$ is rigidly fixed at both the ends without any tension. Its temperature is lowered by $20^{\circ} \mathrm{C}$ and transverse waves are set up by plucking the wire at the middle. The frequency of the fundamental mode is
$$
\left(\mathrm{Y}=200 \mathrm{GPa}, \alpha=1.21 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)
$$

Options:
A $21 \mathrm{~Hz}$
B $42 \mathrm{~Hz}$
C $11 \mathrm{~Hz}$
D $22 \mathrm{~Hz}$

The correct answer is:
$11 \mathrm{~Hz}$

Solution:
No solution. Refer to answer key.

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2018 (24 Apr Shift 2)
A steel wire of length $1.25 \mathrm{~m}$ is stretched between two rigid supports. The tension in the wire produces an elastic strain of $0.14 \%$. The fundamental frequency of the wire is (Density and Young's modulus of steel are $7.7 \times 10^3 \mathrm{kgm}^{-3}$ and $2.2 \times 10^{11} \mathrm{Nm}^{-2}$ respectively)

Options:
A $20 \mathrm{~Hz}$
B $40 \mathrm{~Hz}$
C $80 \mathrm{~Hz}$
D $160 \mathrm{~Hz}$

The correct answer is:
$80 \mathrm{~Hz}$

Solution:
Given, length of wire $=1.25 \mathrm{~m}$
strain produced $=0.14 \%$
density of wire, $d=7.7 \times 10^3 \mathrm{kgm}^{-3}$
young's modulus, $Y=2.2 \times 10^{11} \mathrm{Nm}^{-2}$
Fundamental frequency of vibration over a string is given by
$$
f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}
$$
Where, $T=$ Tension in string,
$L=$ Length of string and
$\mu=$ mass per unit length of string.
So, we can express frequency as
$$
f=\frac{1}{2 L} \sqrt{\frac{T / A}{M / l A}}
$$
From young's modulus of elasticity we have,
$$
\begin{aligned}
& & Y=\frac{T / A}{\Delta l / l} \\
\Rightarrow & \frac{T}{A} & =2.2 \times 10^{11} \times \frac{0.14}{100} \\
\Rightarrow & \frac{T}{A} & =0.308 \times 10^9
\end{aligned}
$$
Substituting for $L, \frac{T}{A}$ and $d$ in eq. (i) we get, Fundamental frequency,
$$
\begin{aligned}
f & =\frac{1}{2 \times 1.25} \times \sqrt{\frac{0.308 \times 10^9}{7.7 \times 10^3}} \\
& =\frac{1}{25} \times \sqrt{0.04 \times 10^6} \\
& =\frac{1}{25} \times \sqrt{4 \times 10^4}=\frac{2}{25} \times 10^2=80 \mathrm{~Hz}
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, AP EAMCET, AP EAMCET 2022 (07 Jul Shift 2)
A step down transformer has \( 50 \) turns on secondary and \( 1000 \) turns on primary winding. If a
transformer is connected to \( 220 \mathrm{~V} 1 \mathrm{AA} \).C. source, what is output current of the transformer?

Options:
A \( \frac{1}{20} A \)
B \( 20 \mathrm{~A} \)
C \( 100 \mathrm{~A} \)
D \( 2 \mathrm{~A} \)

The correct answer is:
\( 20 \mathrm{~A} \)

Solution:
Given, number of turns on secondary, $N_{s}=50 ;$ number of turns on primary, $N_{P}=1000 ;$ source voltage $=220 \mathrm{~V} ;$
current $=1 \mathrm{~A}$.
We know $\frac{N_{s}}{N_{p}}=\frac{V_{s}}{V_{p}}=\frac{I_{p}}{I_{s}}$
Therefore $\frac{N_{s}}{N_{p}}=\frac{I_{p}}{I_{s}} \Rightarrow I_{s}=\frac{N_{p}}{N_{s}} I_{p}$
Substituting the values, we get
$I_{s}=\frac{1000}{50} \times 1=20 \mathrm{~A}$
Thus, output current of transformer $=20 \mathrm{~A}$

Topic: Physics, Mathematics in Physics, KCET, KCET 2014
A step-down transformer has $N_{\mathrm{S}}$ to $N_{\mathrm{p}}$ ratio of $20: 1$. If $8 \mathrm{~V}$ are developed across $0.4 \Omega$ secondary, the primary current will be

Options:
A $1 \mathrm{~A}$
B $0.5 \mathrm{~A}$
C $4 \mathrm{~A}$
D $2 \mathrm{~A}$

The correct answer is:
$1 \mathrm{~A}$

Solution:
For a step-down transformer:
$\frac{N_{\mathrm{P}}}{N_{\mathrm{S}}}=\frac{20}{1}=\frac{V_p}{V_{\mathrm{s}}}=\frac{I_s}{I_p}$
$\begin{aligned} & \therefore I_s=\frac{8 \mathrm{~V}}{0.4 \mathrm{~A}}=20 \mathrm{~A} \\ & \Rightarrow I_p=I_s \times \frac{1}{20}=1 \mathrm{~A}\end{aligned}$

Topic: Physics, Alternating Current, MHT CET, MHT CET 2022 (06 Aug Shift 2)
A step-down transformer has turns ratio 20: 1 . If $8 \mathrm{~V}$ is applied across 0.4 ohm secondary coil, then the primary current will

Options:
A 2 A
B 1 A
C 0.5 A
D 4 A

The correct answer is:
1 A

Solution:
$\begin{aligned} & \mathrm{V}_{\mathrm{s}}=8 \mathrm{~V}, \mathrm{R}_{\mathrm{s}}=0.4 \Omega \\ & \mathrm{I}_{\mathrm{s}}=\frac{\mathrm{V}_{\mathrm{s}}}{\mathrm{R}_{\mathrm{s}}}=\frac{8}{0.4}=20 \mathrm{~A} \\ & \frac{\mathrm{I}_{\mathrm{p}}}{\mathrm{I}_{\mathrm{s}}}=\frac{\mathrm{N}_{\mathrm{s}}}{\mathrm{N}_{\mathrm{p}}}=\frac{1}{20} \\ & \therefore \mathrm{I}_{\mathrm{p}}=\frac{\mathrm{I}_{\mathrm{s}}}{20}=\frac{20}{20}=1 \mathrm{~A}\end{aligned}$

Topic: Physics, Alternating Current, MHT CET, MHT CET 2021 (22 Sep Shift 2)
A step-down transformer is used on a 1000 V line to deliver 20 A at 120 V at the secondary coil. If the efficiency of the transformer is 80%, the current drawn from the line is

Options:
A 3 A

B 30 A
C 0.3 A
D 2.4 A

The correct answer is:
3 A

Solution:
80100=120Ã201000ÃIp
Ip=120Ã201000Ã0.8=3 A

Topic: Physics, Alternating Current, NEET
A step-down transformer is used on a 1000 V line to deliver 20 A at 120 V at the secondary coil. If the efficiency of the transformer is 80%, the current drawn from the line is

Options:
A 3 A

B 30 A
C 0.3 A
D 2.4 A

The correct answer is:
3 A

Solution:
η%=80100=VsIsVpIp=120×201000×Ip
Ip=120×201000×0.8=3 A

Topic: Physics, Alternating Current, NEET
A step-up transformer has 300 turns of primary winding and 450 turns of secondary winding. A primary is connected to 150 volt and the current flowing through it is $9 \mathrm{~A}$. The current and voltage in the secondary are

Options:
A $13.5 \mathrm{~A}, 225 \mathrm{~V}$
B $13.5 \mathrm{~A}, \quad 100 \mathrm{~V}$
C $4.5 \mathrm{~A}, \quad 100 \mathrm{~V}$
D $6.0 \mathrm{~A}, \quad 225 \mathrm{~V}$

The correct answer is:
$6.0 \mathrm{~A}, \quad 225 \mathrm{~V}$

Solution:
$6.0 A, 225 V$
Given that, number of turns in primary winding, $N_{p}=300$
Number of turns in secondary winding, $N_{S}=450$
Primary voltage, $\mathrm{V}_{\mathrm{p}}=150 \mathrm{~V}$
Primary current, $I_{p}=9 \mathrm{~A}$
For step-up transformer,
$\begin{array}{l}
\frac{V_{S}}{V_{P}}=\frac{N_{S}}{N_{P}} \\
\frac{V_{S}}{150}=\frac{450}{300} \\
\Rightarrow V_{S}=\frac{450}{300} \times 150 \\
\Rightarrow V_{S}=225 \mathrm{~V}
\end{array}$
$\begin{array}{l}
\text { Again, } \mathrm{V}_{\mathrm{p}} \mathrm{I}_{\mathrm{p}}=\mathrm{V}_{\mathrm{s}} \mathrm{I}_{\mathrm{S}} \\
150 \times 9=225 \times \mathrm{I}_{\mathrm{S}} \\
\mathrm{I}_{\mathrm{s}}=\frac{1350}{225}=6.0 \mathrm{~A}
\end{array}$

Topic: Physics, Alternating Current, MHT CET, MHT CET 2020 (16 Oct Shift 2)
A step-up transformer is used on 120 V line to provide a potential difference of 2400 V. If the number of turns in the primary is 75, then the number of turns in the secondary shall be

Options:
A 25
B 150
C 1500
D 500

The correct answer is:
1500

Solution:
NsNp=VsVp

Ns75=2400120
Ns=1500

Topic: Physics, Alternating Current, NEET
A step-up transformer operates on $220 \mathrm{~V}$ and supplies current of $2 \mathrm{~A}$. The ratio of primary and secondary windings is $1: 20$. The current in the primary is

Options:
A $5 A$
B $2 A$
C $40 A$
D $20 A$

The correct answer is:
$40 A$

Solution:
$\begin{aligned} & \frac{N_{\mathrm{s}}}{N_p}=20=\frac{I_p}{I_s} \\ & \therefore I_p=20 I_S \\ & =20 \times 2=40 \mathrm{~A}\end{aligned}$

Topic: Physics, Alternating Current, MHT CET, MHT CET 2021 (23 Sep Shift 1)
A step-up transformer operates on a \( 230 \mathrm{~V} \) line and a load current of \( 2 \mathrm{~A} \). The ratio of primary
and secondary windings is \( 1: 25 \). Then the current in the primary is

Options:
A \( 25 \) A
B \( 50 \mathrm{~A} \)
C \( 15 \mathrm{~A} \)
D \( 12.5 \mathrm{~A} \)

The correct answer is:
\( 50 \mathrm{~A} \)

Solution:
Transformer operates on \( 230 \mathrm{~V} \) line; load current \( =2 \mathrm{~A} \)
Ratio of primary and secondary windings, \( \frac{N_{P}}{N_{S}}=\frac{1}{25} \)
Since, \( \frac{N_{s}}{N_{p}}=\frac{I_{p}}{I_{s}} \)
\( \Rightarrow I_{p}=\frac{N_{s}}{N_{p}} \times I_{s} \) \( \Rightarrow I_{p}=\frac{25}{1} \times 2=50 \)
Therefore, current in primary \( =50 \mathrm{~A} \)

Topic: Physics, Alternating Current, KCET, KCET 2018
A step-up transformer operates on a $230 \mathrm{~V}$ line and supplies a load of 2 ampere. The ratio of the primary and secondary windings is $1: 25$. The current in the primary is

Options:
A $15 \mathrm{~A}$
B $50 \mathrm{~A}$
C $25 \mathrm{~A}$
D $12.5 \mathrm{~A}$

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

Solution:
No Solution Available

Topic: Physics, Alternating Current, VITEEE, VITEEE 2021
A stone thrown vertically upward satisfies the equation $\mathrm{s}=64 \mathrm{t}-16 \mathrm{t}^{2}$, where s is in meter and $\mathrm{t}$ is in second. What is the time required toreach the maximumheight?

Options:
A $1 \mathrm{~s}$
B $2 \mathrm{~s}$
C $3 \mathrm{~s}$
D $4 \mathrm{~s}$

The correct answer is:
$2 \mathrm{~s}$

Solution:
Given equation is $s=64 t-16 t^{2}$ $\therefore$ On differentiating w.r.t. $t$, we get $\frac{d s}{d t}=64-32 t$
Put $\frac{d s}{d t}=0$ for maximumheight $\Rightarrow 64-32 t=0 \quad \Rightarrow t=2$
Now, $\frac{d^{2} s}{d t^{2}}=-32$
At $t=2, \frac{d^{2} s}{d t^{2}}=-32$
Since, $\left(\frac{d^{2} s}{d t^{2}}\right)_{t=2} < 0$
- Required time $=2$ second

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 1)
A straight line cuts off an intercept of 2 units on the positive direction of $\mathrm{x}$ -axis and passes through the point $(-3,5)$. What is the foot of the perpendicular drawn from the point $(3,3)$ on this line?

Options:
A $(1,3)$
B $(2,0)$
C $(0,2)$
D $(1,1)$

The correct answer is:
$(1,1)$

Solution:
The given line passes through $(-3,5)$ and $(2,0)$. Its equation is
$y-y_{1}=\left(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)\left(x-x_{1}\right)$
$\Rightarrow(y-5)=\left(\frac{0-5}{2+3}\right)(x+3)$
$\Rightarrow y=-x+2$ ...(1)
Slope $=m=-1$
and slope of perpendicular line $=-\frac{1}{m}=1$
Equation of this line passing through $(3,3)$ is:
$(y-3)=1(x-3)$
$\Rightarrow \quad y=x .$
From eq. (1) we get, $x=-x+2$
$\Rightarrow x=1$ and $y=1 .$

Topic: Mathematics, Straight Lines, NDA, NDA 2016 (Phase 2)