datavorous/entrance-exam-dataset huggingface dataset:
A transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A the voltage across the secondary coil and the current in the primary coil respectively are:

Options:
A 300Â V,Â 15Â A
B 450 V, 15 A
C 450Â V,Â 13.5Â A
D 600Â V,Â 15Â A

The correct answer is:
450 V, 15 A

Solution:
Current in primary coil =PV=3000200=15A
Pout=Î·% of Pin
V2i2=90100Ã3000
V26=2700
V2=450 volt

Topic: Physics, Alternating Current, NEET, NEET 2014
A transformer having efficiency of $90 \%$ is working on $200 \mathrm{~V}$ and $3 \mathrm{~kW}$ power supply. If the current is the secondary coil is $6 \mathrm{~A}$, the voltage across the secondary coil and the current in the primary coil respectively are

Options:
A 450 V , 12 A
B 600 V , 15 A
C 300 V , 15 V
D 450 V , 15 A

The correct answer is:
450 V , 15 A

Solution:
Primary current, $I_{\mathrm{P}}=\frac{P}{V}=\frac{3000}{200}=15 \mathrm{~A}$
Output power $=90 \%$ of $3000 \mathrm{~W}=2700 \mathrm{~W}$
Therefore, voltage across the secondary coil, $V_{\mathrm{s}}=\frac{2700}{6}=450 \mathrm{~V}$

Topic: Physics, Alternating Current, MHT CET, MHT CET 2022 (11 Aug Shift 1)
A transformer is having 2100 turns in primary and 4200 turns in secondary. An AC source of 120 V, 10 A is connected to its primary. The secondary voltage and current are

Options:
A 240 V, 5 A

B 120 V, 10 A

C 240 V, 10 A

D 120 V, 20 A

The correct answer is:
240 V, 5 A

Solution:
Es=nsnP EP=42002100×120=240 V
is =nsnP iP=21004200×10=5 A

Topic: Physics, Alternating Current, NEET
A transformer of efficiency 90% has turns ratio 10:1. If the voltage across the primary is 220 V and current in the primary is 0.5 A, then the current in secondary is

Options:
A 5.5 A
B 5Â A
C 4Â A
D 4.5 A

The correct answer is:
4.5 A

Solution:
Given, NPNS=101
VP=220 V,âIP=0.5 A
We know that, VPVS=NPNS
Efficiency, Î·=OutputâpowerInputâpower=ISVSIPVP
90100=IS0.5 Ã 1 10
â IS=4.5âA

Topic: Physics, Alternating Current, NEET
A transformer with efficiency $80 \%$ works at $4 \mathrm{~kW}$ and $100 \mathrm{~V}$. If the secondary voltage is $200 \mathrm{~V}$, then the primary and secondary currents are respectively

Options:
A $40 \mathrm{~A}$ and $16 \mathrm{~A}$
B $16 \mathrm{~A}$ and $40 \mathrm{~A}$
C $20 \mathrm{~A}$ and $40 \mathrm{~A}$
D $40 \mathrm{~A}$ and $20 \mathrm{~A}$

The correct answer is:
$40 \mathrm{~A}$ and $16 \mathrm{~A}$

Solution:
$\begin{aligned} & \text { (a) : Here, } \eta=80 \% \\ & P_i=4 \mathrm{~kW}=4000 \mathrm{~W} \\ & V_p=100 \mathrm{~V}, V_s=200 \mathrm{~V} \\ & I_p=\frac{P_i}{V_p}=\frac{4000}{100}=40 \mathrm{~A} \\ & \eta=\frac{V_s I_s}{V_p I_p} ; \frac{80}{100}=\frac{200 I_s}{4000}=\frac{I_s}{20} \\ & \text { or } I_s=\frac{20 \times 80}{100}=16 \mathrm{~A}\end{aligned}$

Topic: Physics, Alternating Current, AIIMS, AIIMS 2017
A transverse harmonic wave on a string is described by $y(x, t)=3.0 \sin \left(36 t+0.018 x+\frac{\pi}{4}\right)$ where $x$ and $y$ in $\mathrm{cm}$ and $t$ is in sec. The positive direction of $x$ is from left to right.

Options:
A
the wave is travelling from right to left

B
the speed of the wave is $20 \mathrm{~m} / \mathrm{s}$

C
frequency of the wave is $5.7 \mathrm{~Hz}$

D
the least distance between two successive crests in the wave is $2.5 \mathrm{~cm}$

The correct answers are:
the wave is travelling from right to left
,
the speed of the wave is $20 \mathrm{~m} / \mathrm{s}$
,
frequency of the wave is $5.7 \mathrm{~Hz}$

Solution:
As given equation
$$
y(x, t)=3.0 \sin \left(36 t+0.018 x+\frac{\pi}{4}\right)
$$
As we know that the standard form of wave propagated from left to right in (+ve) direction.
$$
y=a \sin (\omega t+k x+\phi)
$$
(a) As the given equation involves positive sign with $x$, hence the given wave travelling from right to left. Hence, option (a) is correct.
(b) Compare given equation to standard equation
$$
\begin{aligned}
&\omega=36 \Rightarrow 2 \pi v=36 \\
&v=\text { frequency }=\frac{36}{2 \pi}=\frac{18}{\pi}=5.7 \mathrm{~Hz} \\
&k=0.018 \\
&\Rightarrow \frac{2 \pi}{\lambda}=0.018 \\
&\lambda=\left(\frac{2 \pi}{0.018}\right)
\end{aligned}
$$
We know that,
$$
\begin{aligned}
&v=v \lambda=\frac{18}{\pi} * \frac{2 \pi}{0.018} \\
&=2000 \mathrm{~cm} / \mathrm{sec}=20 \mathrm{~m} / \mathrm{sec}
\end{aligned}
$$
(c) $v$ is the frequency which can be find by compairing given equation to standard equation
So, $v=\frac{36}{2 \pi} \mathrm{Hz}=\frac{18}{\pi}=5.7 \mathrm{~Hz}$
(d) Distance between two successive crests $(\lambda)$
$$
\begin{aligned}
\lambda &=\frac{2 \pi}{0.018} \mathrm{~cm} \\
&=\frac{2000 \pi}{18} \mathrm{~cm}=\frac{20 \pi}{18} \mathrm{~m}=3.48 \mathrm{~m} .
\end{aligned}
$$
Hence, least distance between two successive crests $=\lambda=3.48 \mathrm{~m}$, this is not equal to $2.5 \mathrm{~cm}$.
Hence, option (d) is not verified.

Topic: Physics, Waves and Sound
A transverse section of the trunk of a tree shows concentric rings which are known as growth rings. How are these rings formed? What is the significance of these rings?

Options:

None

Solution:
The concentric growth rings are called annual rings. These rings are formed due to the secondary growth. Secondary growth occurs due to the activity of cambium which is a meristermatic tissue in dicot trees.
The rate of activity of cambium is more in spring so wood formed has larger wider xylem cells, whereas wood formed in autumn has narrower and smaller xylem elements. This results in the formation of two rings called growth rings. By counting these rings, age of the tree can be determined. This branch of science is known as dendrochronology or growth ring analysis.

Topic: Biology, Anatomy of Flowering Plants
A transverse wave given by $\mathrm{y}=2 \sin (0.01 \mathrm{x}+30 \mathrm{t})$ moves on a stretched string from one end to another end in 0.5 second. If ' $x$ ' and ' $y$ ' are in $\mathrm{cm}$ and ' $\mathrm{t}$ ' is in second, then the length of the string is

Options:
A 6m
B 9m
C 12m
D 15m

The correct answer is:
15m

Solution:
Comparing with the standard equation
$$
\mathrm{y}=\mathrm{A} \sin (\mathrm{kx}+\omega \mathrm{t})
$$
we get $\mathrm{k}=0.01 \mathrm{rad} / \mathrm{cm}$ and $\omega=30 \mathrm{rad} / \mathrm{s}$
speed of the wave, $v=\frac{\omega}{\mathrm{k}}=\frac{30}{0.01}=300 \mathrm{~cm} / \mathrm{s}=30 \mathrm{~m} / \mathrm{s}$
$\therefore$ Distance $=\mathrm{vt}=30 \times 0.5=15 \mathrm{~m}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2021 (22 Sep Shift 1)
A transverse wave is represented by the equation $y=2 \sin (30 t-40 x)$ and the measurements of distances are in meters, then the velocity of propagation is

Options:
A $15\ ms{-1}$
B $0.75\ ms{-1}$
C $3.75\ ms{-1}$
D $300\ ms{-1}$

The correct answer is:
$0.75\ ms{-1}$

Solution:
We have $y=2 \sin (30 t-40 x)$
The standard transverse wave is $y=A \sin (\omega t-k x)$
So, $\omega=30$ and $k=40$
On comparing, we get
$v=\frac{\omega}{k}=\frac{30}{40}=0.75 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2015
A transverse wave is represented by the equation
$y=y_0 \sin \frac{2 \pi}{\lambda}(v t-k)$
For what value of $\lambda$ is the particle velocity equal to two times the wave velocity

Options:
A $\lambda=\pi y_0$
B $\lambda=\frac{\pi y_0}{2}$
C $\lambda=\frac{\pi y_0}{3}$
D $\lambda=2 \pi i y_0$

The correct answer is:
$\lambda=\pi y_0$

Solution:
Comparing the given equation of the standard equation of wave
$\begin{aligned} w y_0 & =2 v \\ 2 \pi n y_0 & =2 n \lambda, \\ \lambda & =\pi y_0\end{aligned}$

Topic: Physics, Dual Nature of Matter, JIPMER, JIPMER 2013
A transverse wave propagating along the x-axis is represented by y(x, t)=8.0 sin (0.5 Ïxâ4ÏtâÏ/4) where x is in metres and t is in seconds. The speed of the wave is

Options:
A 8 m s-1
B 4π m s-1
C 0.5Ï m s-1
D π4 m s-1

The correct answer is:
8 m s-1

Solution:
y x t = 8 Â· 0 sin 0 Â· 5 Ï x - 4 Ï t - Ï 4
Compare with a standard wave equation,
y = a sin 2 Ï x Î» - 2 Ï t Tâ¡ + Ï
we get 2 Ï Î» = 0 Â· 5 Ï or, Î» = 2 Ï 0 Â· 5 Ï = 4 m
2 Ï Tâ¡ = 4 Ï or, Tâ¡ = 2 Ï 4 Ï = 1 2 sâ¡
Ï = 1 / T = 2 Hz
Wave velocity, Î½ = Î» Ï = 4 Ã 2 = 8 m s-1

Topic: Physics, Waves and Sound, NEET
A transverse wave propagating along $x$-axis is represented by $y(x, t)=8.0$ $\sin (0.5 \pi x-4 \pi t-\pi / 4)$ where $x$ is in metres and $t$ is in seconds. The speed of the wave is:

Options:
A $8 \mathrm{~m} / \mathrm{s}$
B $4 \pi \mathrm{m} / \mathrm{s}$
C $0.5 \pi \mathrm{m} / \mathrm{s}$
D $\frac{1}{2} \mathrm{~m} / \mathrm{s}$

The correct answer is:
$8 \mathrm{~m} / \mathrm{s}$

Solution:
It is given that:
$$
y\left(x_1 t\right)=8.0 \sin \left(0.5 \pi x-4 \pi t-\frac{\lambda}{9}\right)
$$
Compare this equation with standard equation
$$
y=a \sin \left(\frac{2 \pi x}{\lambda}-\frac{2 \pi t}{\mathrm{~T}}+\phi\right)
$$
we have
$$
\begin{aligned}
\frac{2 \pi}{\lambda} & =0.5 \pi \\
\Rightarrow \quad \lambda & =\frac{2 \pi}{0.5 \pi}=4 \mathrm{~m} \\
\frac{2 \pi}{T} & =4 \pi \\
\Rightarrow \quad T & =\frac{1}{2} \sec \\
U=\frac{1}{T} & =\frac{1}{y_2}=2 \mathrm{H}_2 .
\end{aligned}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2006
A transverse wave propagating on a stretched string of linear density $3 \times 10^{-4} \mathrm{~kg} \mathrm{~m}^{-1}$ is represented by the equation $y=0.2 \sin (15 x+60 t)$
where $x$ is in metres and $t$ is in seconds. The tension in the string (in newton) is

Options:
A 0.24
B 0.48
C 1.20
D 1.80

The correct answer is:
0.48

Solution:
Equation of wave, $y=0.2 \sin (1.5 x+60 t)$ Comparing with standard equation,
$\begin{aligned}
& y=A \sin (k x+\omega t) \\
& k=1.5, \omega=60 \mathrm{rad} / \mathrm{s}
\end{aligned}$
$\therefore$ Velocity of wave, $v=\frac{\omega}{k}=\frac{60}{1.5}=40 \mathrm{~m} / \mathrm{s}$
Velocity of wave in a stretched string,
$v=\sqrt{\frac{T}{m}}$
where $m=$ linear density
$T=\text { tension in the string }$
So,
$\begin{aligned}
T & =v^2 m \\
& =(40)^2 \times 3 \times 10^{-4}=0.48
\end{aligned}$

Topic: Physics, Waves and Sound, AP EAMCET, AP EAMCET 2005
A transverse wave propagating on a stretched string of linear density $3 \times 10^{-4} \mathrm{~kg} \mathrm{~m}^{-1}$ is represented by the equation $y=0.2 \sin (15 x+60 t)$
where $x$ is in metres and $t$ is in seconds. The tension in the string (in newton) is

Options:
A 0.24
B 0.48
C 1.20
D 1.80

The correct answer is:
0.48

Solution:
Equation of wave, $y=0.2 \sin (1.5 x+60 t)$ Comparing with standard equation,
$\begin{aligned}
& y=A \sin (k x+\omega t) \\
& k=1.5, \omega=60 \mathrm{rad} / \mathrm{s}
\end{aligned}$
$\therefore$ Velocity of wave, $v=\frac{\omega}{k}=\frac{60}{1.5}=40 \mathrm{~m} / \mathrm{s}$
Velocity of wave in a stretched string,
$v=\sqrt{\frac{T}{m}}$
where $m=$ linear density
$T=\text { tension in the string }$
So,
$\begin{aligned}
T & =v^2 m \\
& =(40)^2 \times 3 \times 10^{-4}=0.48
\end{aligned}$

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2005
A transverse wave $\mathrm{Y}=2 \sin (0.01 \mathrm{x}+30 \mathrm{t})$ moves on a stretched string from one end to another end in 0.5 second. If $x$ and $y$ are in $\mathrm{cm}$ and $\mathrm{t}$ in second, then the length of the string is

Options:
A $5 \mathrm{~m}$
B $10 \mathrm{~m}$
C $15 \mathrm{~m}$
D $20 \mathrm{~m}$

The correct answer is:
$15 \mathrm{~m}$

Solution:
Given $Y=2 \sin (0.01 \mathrm{x}+30 \mathrm{t})$
Comparing with standard equation,
$$
\begin{aligned}
\mathrm{Y} & =\mathrm{A} \sin (\mathrm{kx}+\omega \mathrm{t}), \\
\therefore \quad \mathrm{k} & =0.01 / \mathrm{cm} \\
\omega & =30 \mathrm{rad} / \mathrm{s}
\end{aligned}
$$
$$
\begin{aligned}
\text { Velocity } \mathrm{v} & =\frac{\omega}{\mathrm{k}}=\frac{30}{0.01}=3000 \mathrm{~cm} / \mathrm{s} \\
\therefore \quad \text { Length } \mathrm{L} & =\mathrm{v} \times \mathrm{t}=30 \times 0.5 \\
& =15 \mathrm{~m}
\end{aligned}
$$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2023 (11 May Shift 1)
A travelling harmonic wave on a string is described by
$$
y(x, t)=7.5 \sin \left(0.0050 x+12 t+\frac{\pi}{4}\right)
$$
(i) What are the displacement and velocity of oscillation of a point at $x=1 \mathrm{~cm}$, and $t=1 \mathrm{~s}$ ? Is this velocity equal to the velocity of wave propagation?
(ii) Locate the points of the string, which have the same transverse displacement and velocity as the $x=1 \mathrm{~cm}$ point at $t=2 \mathrm{~s}, 5 \mathrm{~s}$ and $11 \mathrm{~s}$.

Options:

None

Solution:
Given, $y(x, t)=7.5 \sin (0.0050 x+12 t+\pi / 4)$ $=7.5 \sin \left[0.0050\left\{\frac{12}{0.0050} t+x\right\}+\pi / 4\right] \ldots(1)$
Comparing it with standard equation of travelling wave
$$
y=a \sin \left[\frac{2 \pi}{\lambda}\{v t+x\}+\phi\right]
$$
Comparing equations (1) and (2), we get $a=7.5 \mathrm{~cm}, v=\frac{12}{0.0050} \mathrm{~cm} \mathrm{~s}^{-1}$ and $\frac{2 \pi}{\lambda}=0.0050 \mathrm{~cm}^{-1}$
(i) At $x=1 \mathrm{~cm}$ and $t=1 \mathrm{~s}$, displacement is $y=7.5 \sin (0.0050 \times 1+12 \times 1+\pi / 4)$ $=7.5 \sin 12.79$
$=7.5 \times 0.2222=1.67 \mathrm{~cm}$.
Velocity of oscillation of the particle is
$$
\begin{aligned}
&u=\frac{d y}{d t}=\frac{d}{d t}[7.5 \sin (0.0050 x+12 t+\pi / 4)] \\
&=7.5 \times 12 \cos (0.0050 x+12 t+\pi / 4) \\
&=90 \cos \left(0.0050 x+12 t+\frac{\pi}{4}\right) \\
&\text { At } x=1 \mathrm{~cm} \text { and } t=1 \mathrm{~s}, \\
&u=90 \cos (0.000050 \times 1+12 \times 1+\pi / 4) \\
&=90 \cos 12.79 \\
&=90 \times 0.9751=87.76 \mathrm{~cm} \mathrm{~s}^{-1}
\end{aligned}
$$
Velocity of wave propagation is $v=\frac{12}{0.0050}=2400 \mathrm{~cm} \mathrm{~s}^{-1}$
$\therefore$ Velocity of oscillation of a point is not equal to the velocity of wave propagation.
(ii) As $\frac{2 \pi}{\lambda}-0.0050$
$$
\therefore \lambda=\frac{2 \pi}{0.0050}=1256.64 \mathrm{~cm}
$$
All points located at distance $n \lambda$ (where $n$ is an integer) from the point $x=1 \mathrm{~cm}$ have the same transverse displacement and velocity.

Topic: Physics, Waves and Sound
A triangular plane $\mathrm{ABC}$ with centroid $(1,2,3)$ cuts the coordinate axes at $A, B$, Crespectively. What are the intercepts made by the plane $A B C$ on the axes?

Options:
A $3,6,9$
B $1,2,3$
C $1,4,9$
D $2,4,6$

The correct answer is:
$3,6,9$

Solution:
Centroid $=(1,2,3)$
$(1,2,3)=\left(\frac{x+0+0}{3}, \frac{0+y+0}{3}, \frac{0+0+z}{3}\right)$
$x=3, y=6$ and $z=9$
Intercept made by plane on the axes are 3,6 and 9 . respectively.
$\therefore$ Option (a) is correct.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2015 (Phase 1)
A triangular plane $\mathrm{ABC}$ with centroid $(1,2,3)$ cuts the coordinate axes at $A, B$, Crespectively. What is the equation of plane $A B C$ ?

Options:
A $x+2 y+3 z=1$
B $3 x+2 y+z=3$
C $2 x+3 y+6 z=18$
D $6 x+3 y+2 z=18$

The correct answer is:
$6 x+3 y+2 z=18$

Solution:
The plane passes through the point $\mathrm{A}(3,0,0), \mathrm{B}(0,6,0)$ and $\mathrm{C}(0,0,9)$. So, it should satisfy the equation given in option for all the three points. From option (a) For point $\mathrm{A}(3,0,0)$
$x+2 y+3 z=1$
$\Rightarrow 3+0+0 \neq 1$
option (a) is wrong. From option (b) For point $\mathrm{A}(3,0,0)$
$3 x+2 y+z=3$
$\therefore 3(3)+0+0 \neq 3$
$\therefore \quad$ option (b) is wrong. From option (c) For point $\mathrm{A}(3,0,0)$
$2 x+3 y+6 z=18$
$\therefore \quad 2(3)+0+0 \neq 18$
of option (c) is wrong. From option (d) For point $\mathrm{A}(3,0,0)$ $6 x+3 y+2 z=18$
$\Rightarrow 6(3)+0+0=18$
For point $\mathrm{B}(0,6,0)$
$6 x+3 y+2 z=18$
$\therefore \quad 0+3(6)+0=18$
For point $\mathrm{C}(0,0,9)$
$6 x+3 y+2 z=18$
$0+0+2 \times 9=18$
$\therefore \quad$ Option $(\mathrm{d})$ is correct.

Topic: Mathematics, Three Dimensional Geometry, NDA, NDA 2015 (Phase 1)
(a) Two stable isotopes of lithium ${ }_3^6 \mathrm{Li}$ and ${ }_3^7 \mathrm{Li}$ have respective abundances of $7.5 \%$ and $92.5 \%$. These isotopes have masses $6.01512 \mathrm{u}$ and $7.01600 \mathrm{u}$ respectively. Find the atomic mass of lithium.
(b) Boron has two stable isotopes, ${ }_5^{10} \mathrm{~B}$ and ${ }_5^{11} \mathrm{~B}$. Their respective masses are $10.01294 \mathrm{u}$ and $11.00931 \mathrm{u}$, and the atomic mass of boron is $10.811 \mathrm{u}$. Find the abundances of ${ }_5^{10} \mathrm{~B}$ and ${ }_5^{11} \mathrm{~B}$.

Options:

None

Solution:
(a) The isotope ${ }_3^6 \mathrm{Li}$ has abundance $7.5 \%$ and ${ }_3^7 \mathrm{Li}$ has $92.5 \% . \mathrm{m}\left({ }_3^6 \mathrm{Li}\right)=6.01512 \mathrm{u} \mathrm{m}\left({ }_3^7 \mathrm{Li}\right)=7.01600 \mathrm{u}$ Average mass of Li $=\frac{6.01512 \times 7.5+7.01600 \times 92.5}{7.5+92.5}$ $=\frac{45.1140+648.98}{7.5+92.5}$ $=\frac{694.094}{100}=6.94000 \mathrm{u}$.
(b) $\mathrm{m}\left({ }_5^{10} \mathrm{~B}\right)=10.01294 \mathrm{u} ; \mathrm{m}\left({ }_5^{11} \mathrm{~B}\right)=11.00931 \mathrm{u} ; \mathrm{m}$ (B) $=10.811 \mathrm{u}$.
Let abundance of ${ }_5^{10} \mathrm{~B}$ be $\mathrm{x} \%$ thus abundance of ${ }_5^{11} \mathrm{~B}$ will be $(100-\mathrm{x}) \%$
Average mass of Boron
$$
\begin{aligned}
&=\frac{10.01294 \mathrm{x}+11.00931(100-\mathrm{x})}{100} \\
&=10.811 \mathrm{u}
\end{aligned}
$$
$\Rightarrow 10.01294 \mathrm{x}+1100.931-11.00931 \mathrm{x}=1081.1$
$\Rightarrow \quad 0.996 \mathrm{x}=19.831$
$\Rightarrow \mathrm{x}=19.91 \%=$ abundance of ${ }_5^{10} \mathrm{~B}$
$\quad$ and $(100-\mathrm{x})$
$\quad=(100-19.91)=80.1 \%$ abundance of ${ }_5^{11} \mathrm{~B}$

Topic: Physics, Nuclear Physics
A type of granulocyte (i) and agranulocyte (ii) are phagocytic cells that destroy foreign organisms entering the body. (iii) secrete histamine, serotonin, heparin, etc., and are involved in inflammatory reactions. (iv) resist infections and are also associated with allergic reactions. (v) are responsible for the immune responses of the body.

Options:
A

(i)	(ii)	(iii)	(iv)	(v)

Monocytes	Neutrophils	Basophils	Eosinophils	Lymphocytes

B

(i)	(ii)	(iii)	(iv)	(v)

Neutrophils	Monocytes	Basophils	Eosinophils	Lymphocytes

C

(i)	(ii)	(iii)	(iv)	(v)

Neutrophils	Monocytes	Eosinophils	Basophils	Lymphocytes

D

(i)	(ii)	(iii)	(iv)	(v)

Monocytes	Neutrophils	Eosinophils	Basophils	Lymphocytes

The correct answer is:

(i)	(ii)	(iii)	(iv)	(v)

Neutrophils	Monocytes	Basophils	Eosinophils	Lymphocytes

Solution:
Neutrophils, eosinophils, and basophils are different types of granulocytes, while lymphocytes and monocytes are the agranulocytes. Neutrophils are the most abundant cells (60-65 percent) of the total WBCs and basophils are the least (0.5-1 percent) among them. Neutrophils and monocytes (6-8 percent) are phagocytic cells that destroy foreign organisms entering the body. Basophils secrete histamine, serotonin, heparin, etc., and are involved in inflammatory reactions. Eosinophils (2-3 percent) resist infections and are also associated with allergic reactions. Lymphocytes (20-25 percent) are of two major types â âBâ and âTâ forms. Both B and T lymphocytes are responsible for immune responses of the body.

Topic: Biology, Body Fluids and Circulation, NEET
A typical monocotyledonous root is characterized by ___________.

Options:
A usually more than six xylem bundles.
B large and well developed pith.
C no secondary growth.
D All of the above

The correct answer is:
All of the above

Solution:
Monocotyledonous plants are subgroup of Angiosperms characterized by having one embryonic leaf or one cotyledon in their seed. Anatomically the root system of monocots has certain distinguishing feature even though it is similar to that of dicot plant root system.
The root lacks cambium in it and has no secondary growth. The number of xylem bundles are more than dicot, having higher than six. The pith region of monocots is larger in size and better developed than dicots.

Topic: Biology, Anatomy of Flowering Plants, NEET
A uniform string is vibrating with a fundamental frequency ' $f$ '. The new frequency, if radius \& length both are doubled would be

Options:
A $2 f$
B $3 f$
C $\frac{f}{4}$
D $\frac{f}{3}$

The correct answer is:
$\frac{f}{4}$

Solution:
Fundamental frequency of a vibrating string is expressed as
$$
f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}=\frac{1}{L D} \sqrt{\frac{T}{\pi \rho}}
$$
where,
$$
\begin{aligned}
& D= \text { diameter of string } \\
& \rho= \text { density of the } \\
& \text { material of string }
\end{aligned}
$$
As length $L$ and radius are doubled, the new frequency
$$
f^{\prime}=\frac{1}{(2 L)(2 D)} \sqrt{\frac{T}{\pi \rho}}=\frac{1}{4} f
$$

Topic: Physics, Waves and Sound, AIIMS, AIIMS 2010
A uniform string is vibrating with a fundamental frequency ' $n$ '. If radius and length of string both are doubled keeping tension constant then the new frequency of vibration is

Options:
A $2 \mathrm{n}$
B $3 \mathrm{n}$
C $\frac{n}{4}$
D $\frac{\mathrm{n}}{3}$

The correct answer is:
$\frac{n}{4}$

Solution:
$\begin{aligned} & l_2=2 l_1, \mathrm{R}_2=2 \mathrm{R}_1, \mathrm{~T}_1=\mathrm{T}_2 \\ & \mathrm{n}=\frac{1}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}} \\ & \quad \text { Where, } \mathrm{m}=\text { mass per unit length }=\frac{\left(\mathrm{nR}^2 l\right) \rho}{l} \\ & \therefore \quad \mathrm{m} \propto \mathrm{R}^2 \\ & \therefore \quad \frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{l_1}{l_2} \times \frac{\mathrm{R}_1}{\mathrm{R}_2} \\ & \therefore \quad \frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{l_1}{2 l_1} \times \frac{\mathrm{R}_1}{2 \mathrm{R}_1} \\ & \therefore \quad \mathrm{n}_2=\frac{\mathrm{n}_1}{4}=\frac{\mathrm{n}}{4}\end{aligned}$

Topic: Physics, Waves and Sound, MHT CET, MHT CET 2023 (12 May Shift 1)
A uniform wire of length $\mathrm{L}$, diameter $\mathrm{D}$ and density $\rho$ is stretched under a tension T. The correct relation between its fundamental frequency $f$, the length $L$ and the diameter $D$ is

Options:
A $f \propto \frac{1}{L D}$
B $f \propto \frac{1}{L \sqrt{D}}$
C $\mathrm{f} \propto \frac{1}{\mathrm{D}^{2}}$
D $f \propto \frac{1}{L D^{2}}$

The correct answer is:
$f \propto \frac{1}{L D}$

Solution:
The fundamental frequency is $f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$
$$
f=\frac{1}{2 L} \sqrt{\frac{T}{\rho \pi \frac{D^{2}}{4}}}=\frac{1}{L D} \sqrt{\frac{T}{\pi \rho}}
$$
$\therefore$
$$
f \propto \frac{1}{L D}
$$

Topic: Physics, Waves and Sound, KCET, KCET 2010
A unit vector perpendicular to each of the vectors
$2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ is

Options:
A $\frac{1}{\sqrt{3}} \hat{\mathrm{i}}+\frac{1}{\sqrt{3}} \hat{\mathrm{j}}-\frac{1}{\sqrt{3}} \hat{\mathrm{k}}$
B $\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+\frac{1}{2} \hat{\mathrm{k}}$
C $\frac{1}{\sqrt{3}} \hat{\mathrm{i}}-\frac{1}{\sqrt{3}} \hat{\mathrm{j}}-\frac{1}{\sqrt{3}} \hat{\mathrm{k}}$
D $\frac{1}{\sqrt{3}} \hat{\mathrm{i}}+\frac{1}{\sqrt{3}} \hat{\mathrm{j}}+\frac{1}{\sqrt{3}} \hat{\mathrm{k}}$

The correct answer is:
$\frac{1}{\sqrt{3}} \hat{\mathrm{i}}+\frac{1}{\sqrt{3}} \hat{\mathrm{j}}-\frac{1}{\sqrt{3}} \hat{\mathrm{k}}$

Solution:
$\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & -4 & -1\end{array}\right|$
$=\hat{i}(1+4)-\hat{j}(-2-3)+\hat{k}(-8+3)$
$=5 \hat{i}+5 \hat{j}-5 \hat{k}$
Unit vector $=\frac{5}{5 \sqrt{3}} \hat{i}+\frac{5}{5 \sqrt{3}} \hat{j}-\frac{5}{5 \sqrt{3}} \hat{k}$
$\quad=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}$

Topic: Mathematics, Vector Algebra, NDA, NDA 2018 (Phase 2)