datavorous/entrance-exam-dataset huggingface dataset:
A transformer is used to light a $100 \mathrm{~W}$ and $110 \mathrm{~V}$ lamp from a $220 \mathrm{~V}$ mains. If the main current is $0.5 \mathrm{amp}$, the efficiency of the transformer is approximately.

Options:
A $50 \%$
B $90 \%$
C $10 \%$
D $30 \%$

The correct answer is:
$90 \%$

Solution:
The efficiency of transformer
$\begin{aligned}
\eta & =\frac{V_s I_s}{V_p I_p} \times 100 \\
& =\frac{100}{220 \times 0.5} \times 100=90 \%
\end{aligned}$

Topic: Physics, Alternating Current, NEET, NEET 2007
A transformer of frequency 60 Hz and 120 V input has 8:1 turn ratio. The frequency of the output is

Options:
A 40 Hz
B 480 Hz
C 2 Hz
D 60 Hz

The correct answer is:
60 Hz

Solution:
In a transformer, frequency remains same for input & output.

Topic: Physics, Alternating Current, NEET
A transformer operating at primary voltage 8kV and secondary voltage 160 V serves a load of 80 kW. Assuming the transformer to be ideal with purely resistive load and working on unity power factor, the loads in the primary and secondary circuit would be

Options:
A 800 Ω and 1.06 Ω
B 10 Ω and 500 Ω
C 800 Ω and 0.32 Ω
D 1.06 Ω and 500 Ω

The correct answer is:
800 Ω and 0.32 Ω

Solution:
Since the transformer is assumed to be purely resistive, the power of the transformer can be calculated with P=V2R. Therefore,

For the primary coil,

P=8×1032RP=80×103 W

⇒RP=800 Ω

For the secondary coil,

P=1602RS=80×103 W

⇒RS=0.32Ω

Topic: Physics, Alternating Current, JEE Main, JEE Main 2022 (28 Jul Shift 2)
A transformer with efficiency 80% works at 4 kW and 100 V. If the secondary voltage is 200 V, then the primary and secondary currents are respectively

Options:
A 40 A and 16 A
B 16 A and 40 A
C 20 A and 90 A
D 40 A and 20 A

The correct answer is:
40 A and 16 A

Solution:
Here, η=80%
Pi=4 kW=4000 W
Vp=100 V, Vs=200 V
Ip=PiVp=4000100=40 A
η=VsIsVpIp;80100=200Is4000=Is20
or Is=20×80100=16 A

Topic: Physics, Alternating Current, JEE Main
A transgeneic food crop which may help in solving the problem of night blindness in developing countries is

Options:
A Golden rice
B Flavr Savr tomatoes
C Starlink maize
D Bt Soybean

The correct answer is:
Golden rice

Solution:
Golden rice is transgenic rice in which the genes for synthesis of enzymes involved in $\beta-$ carotene formation are transferred. So golden rice is rich in vitamin A, preventing night blindness.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2008 (Mains)
A transgenic food crop, which may help in solving the problem of nightblindness in developing countries is

Options:
A Flavr savr tomatoes
B Starlink maize
C Bt soyabean
D Golden rice

The correct answer is:
Golden rice

Solution:
Vitamin-A deficiency causes nightblindness among children. Vitamin-A deficiency often occurs, where rice is the staple food since, rice grain does not contain $\beta$-carotene. Three transgenes providing phytoene synthase, phytoene desaturase, $\beta$ carotene desaturase and lycopene cyclase activities were transferred into rice by Agrobacterium mediated transformation. The resulting transgenic rice popularly called Golden rice contains good quantities of $\beta$-carotene.
Flavr savr is genetically modified tomato, which remains fresh and retains their flavour much longer than normal tomato due to blocking of synthesis of fruit softening enzyme polygalacturonase.

Topic: Biology, Biotechnology and Its Applications, JIPMER, JIPMER 2012
A transgenic food crop, which may help in solving the problem of nightblindness in developing countries is

Options:
A Ftavr savr tomatoes
B Starlink maize
C Bt soybean
D Golden rice

The correct answer is:
Golden rice

Solution:
Vitamin-A deficiency causes nighthlindness among children. Vitamin-A deficiency often occurs where rice is the staple food since, rice grain does not contain $\beta$ carotene. Three transgenes providing phytoene synthase, phytoene desaturase, zeta carotene desaturase and lycopene cyclase activities were transferred into rice by Agrobacterium mediated transformation. The resulting transgenic rice popularly called Golden rice contains good quantities of $\beta$-carotene.
Flav sav is genetically modified tomato, which remains fresh and retains their flavour much longer than normal tomato due to blocking of synthesis of fruit softening enzyme polygalacturonase.

Topic: Biology, Biotechnology and Its Applications, NEET, NEET 2008 (Screening)
A transistor, connected in common emitter configuration, has input resistance R in = 2 k Î© and load resistance of 5 k Î© . If Î² = 6 0 and an input signal 12ââmV is applied, find the power gain.

Options:
A 9000
B 4000
C 6000
D 8000

The correct answer is:
9000

Solution:
Power gain =Î²2RLRBE
= 6 0 2 5 2
= 1 8 0 0 0 2 = 9 0 0 0

Topic: Physics, Semiconductors, JEE Main
A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V and the voltage drop across a resistor of 1000 Ω in the collector circuit is 0.6 V. If the current gain factor (β) is 24, then the base current is ____μA. (Round off to the Nearest Integer)

Options:

The correct answer is:
25

Solution:
β=ICIB=24; RC=1000

ΔV=0.6

IC=0.61000

IC=6×10-4
IB=ICβ=6×10-424=25 μA

Topic: Physics, Semiconductors, JEE Main, JEE Main 2021 (27 Jul Shift 1)
A transistor is operated in common-emitter configuration at $\mathrm{V}_{\mathrm{C}}=2 \mathrm{~V}$ such that a change in the base current from $100 \mu \mathrm{A}$ to $200 \mu \mathrm{A}$ produces a change in the collector current from $5 \mathrm{~mA}$ to $10 \mathrm{~mA}$. The current gain is :

Options:
A 50
B 75
C 100
D 150

The correct answer is:
50

Solution:
$$
\begin{aligned}
\Delta \mathrm{I}_{\mathrm{B}} & =100 \mu \mathrm{A} \\
\Delta \mathrm{I}_{\mathrm{C}} & =5 \mathrm{~mA} \\
\beta & =\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{5 \times 10^{-3}}{100 \times 10^{-6}}=50
\end{aligned}
$$

Topic: Physics, Semiconductors, NEET, NEET 2009 (Mains)
A transistor is used as a common emitter amplifier with a load resistance of \( 2 \mathrm{k} \Omega \). The input resistance is \( 150 \Omega \). Base
current is changed by \( 20 \mu \mathrm{A} \) which results in a change in collector current by \( 1.5 \mathrm{~mA} \). The voltage gain of the amplifier is

Options:
A \( 900 \)
B \( 1000 \)
C \( 1100 \)
D \( 1200 \)

The correct answer is:
\( 1000 \)

Solution:
Voltage gain =V0Vi=R0×∆ICRi×∆IB
= 2000×1.5× 10 −3 150×20× 10 −6 = 3 3000× 10 −6
= 1 ( 1000 ) −1 =1000

Topic: Physics, Semiconductors, JEE Main
A transistor is used in an amplifier circuit in common emitter mode. If the base current changes by 100 μA, it brings a change of 10 mA in collector current. If the load resistance is 2 kΩ and input resistance is 1 kΩ, the value of power gain is x×104. The value of x is _____ .

Options:

The correct answer is:
2

Solution:
βAC=ΔiCΔiB=10×10-3100×10-6=100

Power gain, AP=∆iC2Rout∆iB2Rin=βAC2Rout Rin=1002×2×103103=2×104

Therefore, x=2

Topic: Physics, Semiconductors, JEE Main, JEE Main 2022 (29 Jun Shift 1)
A transistor is used in common emitter mode as an amplifier. Then

Options:
A The base-emitter junction is forward biased
B The base-emitter junction is reverse biased
C The input signal is connected in series with the voltage applied to the base-emitter junction
D The input signal is connected in series with the voltage applied to bias the base collector junction

The correct answers are:
The base-emitter junction is forward biased, The input signal is connected in series with the voltage applied to the base-emitter junction

Solution:
The circuit of a CE amplifier is as Shown below.
This has been shown a NPN transistor. Therefore base emitter are forward, biased and input signal is connected between base and emitter.

Topic: Physics, Semiconductors, JEE Main
A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10mV is added to the base-emitter voltage, the base current changes by 10μA and the collector current changes by 1.5 mA. The load resistance is 5kΩ. The voltage gain of the transistor will be _____ .

Options:

The correct answer is:
750

Solution:
First, use the formula of current gain in terms of change in base current, and collector current,

β=ΔIcΔIB

given,

VBE=10×10-3 V, ∆ib=10×10-6 A,

∆ic=1.5×10-3 A and RL=5×103 Ω

So the formula of voltage gain,

βV=∆VCE∆VBE=∆ic×RL∆VBE

⇒βV=1.5×10-3×500010×10-3=750

Topic: Physics, Semiconductors, JEE Main, JEE Main 2022 (24 Jun Shift 1)
A transistor oscillator is
(i) An amplifier with positive feedback
(ii) An amplifier with reduced gain
(iii) The one in which DC supply energy is converted into AC output energy. Then

Options:
A All (i), (ii) and (iii) are correct
B (i) and (ii) are correct
C (i) and (iii) are correct
D (ii) and (iii) are correct

The correct answer is:
(i) and (iii) are correct

Solution:
A transistor oscillator is an amplifier with positive feedback.
A transistor oscillator is the one in which DC supply energy is converted into AC output energy.

Topic: Physics, Semiconductors, JEE Main
A transition metal $M$ forms a volatile chloride which has a vapour density of $94.8$. If it contains $74.75 \%$ of chlorine the formula of the metal chloride will be

Options:
A
$\mathrm{MCl}_3$

B
$\mathrm{MCl}_2$

C
$\mathrm{MCl}_4$

D
$\mathrm{MCl}_5$

The correct answer is:
$\mathrm{MCl}_4$

Solution:
$74.75 \%$ of chlorine means $74.75 \mathrm{~g}$ chlorine is present in $100 \mathrm{~g}$ of metal chloride.
Weight of metal $=100 \mathrm{~g}-74.75 \mathrm{~g}$ $=25.25 \mathrm{~g}$
Equivalent weight
$$
\begin{aligned}
& =\frac{\text { weight of metal }}{\text { weight of chlorine }} \times 35.5 \\
& =\frac{25.25}{74.75} \times 35.5=12
\end{aligned}
$$
Valency of metal
$$
\begin{array}{r}
=\frac{2 \times \text { V.D. }}{\text { Equivalent wt. of metal }+35.5} \\
=\frac{2 \times 94.8}{12+35.5}=4
\end{array}
$$
$\therefore$ Formula of compound $=\mathrm{MCl}_4$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2012 (26 May Online)
A transition metal ' \(\mathrm{M}\) ' among \(\mathrm{Sc}, \mathrm{Ti}, \mathrm{V}, \mathrm{Cr}, \mathrm{Mn}\) and \(\mathrm{Fe}\) has the highest second ionisation enthalpy. The spin-only magnetic moment value of \(\mathrm{M}^{+}\)ion is _______. \(\mathrm{BM}\) (Near integer)
(Given atomic number \(\mathrm{Sc}: 21, \mathrm{Ti}: 22, \mathrm{~V}: 23, \mathrm{Cr}: 24, \mathrm{Mn}: 25, \mathrm{Fe}: 26\))

Options:

The correct answer is:
6

Solution:
Among given metals, $\mathrm{Cr}$ has maximum $\mathrm{IE}_2$
because Second electron is removed from stable configuration $3 d^5$
$\mathrm{Cr}^{+}:[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^0$
$\therefore$ No of unpaired $\mathrm{e}^{-}$in $\mathrm{Cr}^{+}$is $5, \mathrm{n}=5$
$\begin{aligned}
\text { So, Magnetic moment } & =\sqrt{\mathrm{n}(\mathrm{n}+2)} \text { B.M } \\
& =\sqrt{5(5+2)}=5.92 \quad \mathrm{BM} \approx 6
\end{aligned}$

Topic: Chemistry, d and f Block Elements, JEE Main, JEE Main 2024 (09 Apr Shift 2)
A translucent white waxy solid 'A' reacts with excess of chlorine to give a yellowish white powder 'B'. 'B' reacts with organic compounds containing -OH group converting them into chloro derivatives. 'B' on hydrolysis gives 'C' and is finally converted to phosphoric acid. 'A', 'B' and 'C' are

Options:
A P4,PCl3,H3PO4
B P4,PCl5,H3PO3
C P4,PCl5,POCl3
D P4,PCl3,POCl3

The correct answer is:
P4,PCl5,POCl3

Solution:
P4A+10Cl2→4PCl5B
PCl5+C2H5OH→C2H5Cl+POCl3+HCl
PCl5+H2O→POCl3C+2HCl
POCl3+3H2O→H3PO4+3HCl

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna is 50 m. What is the maximum distance between them for satisfactory communication in line of sight (LOS) mode?

Options:
A 45.5 km
B 54.5 km
C 55.4 km
D 455 km

The correct answer is:
45.5 km

Solution:
Given,

Height of transmitting antenna is, hT=50 m

Height of receiving antenna is, hR=32 m

Thus, maximum distance for satisfactory communication will be,

dM=2RhT+2RhR where R is radius of the Earth,

dM=2RhT+2RhRdM=2×64×105×50+2×64×105×32dM=800010+640010dM=1440010 mdM=45536.7 mdM=45.5 km

Topic: Physics, Communication System, JEE Main, JEE Main 2014 (09 Apr Online)
A transmitting antenna at top of a tower has a height of 50 m, and the height of receiving antenna is, 80 m. What is the range of communication for the line of sight (LOS) mode?
[use radius of the earth=6400 km]

Options:
A 80.2 km
B 144.1 km
C 57.28 km
D 45.5 km

The correct answer is:
57.28 km

Solution:
The formula of range of the antenna is given by the formula,=2Rh1+2Rh2., here R is the radius of the earth then, putting the values, the range =57.28 km.

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (26 Aug Shift 2)
A transmitting antenna has a height of 320 m and that of receiving antenna is 2000 m. The maximum distance between them for satisfactory communication in line of sight mode is d. The value of d is ________ km.

Options:

The correct answer is:
224

Solution:
Recall the formula of maximum distance between transmitting antenna and receiving antenna for satisfactory communication in terms of the radius of the earth and height of antennas,

d=2Rh1+2Rh2⇒d=2×6400×0.32+2×6400×2≈224 km

Topic: Physics, Communication System, JEE Main, JEE Main 2021 (27 Aug Shift 1)
A transmitting antenna is kept on the surface of the earth. The minimum height of receiving antenna required to receive the signal in line of sight at 4 km distance from it is x×10-2 m. The value of x is _____.
(Let, radius of earth R=6400 km)

Options:
A 125
B 1250

C 12.5
D 1.25

The correct answer is:
125

Solution:
The formula to calculate the range of the antenna is given by

γ=2Rh ...1

Substitute the values of the known parameters into equation (1) and solve to obtain the required height of the antenna.

4×103=2×6400000×h⇒16×106=2×64×105×h⇒h=1602×64m=108m=10008×10-2 m=125×10-2 m

Topic: Physics, Communication System, JEE Main, JEE Main 2023 (11 Apr Shift 1)
A transmitting antenna of height \( h \) and the receiving antenna of height \( 45 \mathrm{~m} \) are separated by a distance of \( 40 \mathrm{~km} \) of satisfactory communication in line-of-sight mode. Then, the value of \( h \) is (given, radius of earth is \( 6400 \mathrm{~km} \) ),

Options:
A \( 15 \mathrm{~m} \)
B \( 20 \mathrm{~m} \)
C \( 30 \mathrm{~m} \)
D \( 25 \mathrm{~m} \)

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

Solution:
Given: Height of transmitting antenna, ht=h,

height of receiving antenna, hr=45 m,

distance between the two antennas, D=40 km,

radius of earth, R=6400 km.

For line-of-sight mode communication, above terms are related as,

D=12Rht+12Rhr⇒12Rht=D-12Rhr⇒12Rht=D-12Rhr2⇒ht=12RD-12Rhr2⇒ht=12×6.4×106×4×104-12×6.4×106×452=20 m

Topic: Physics, Communication System, JEE Main
A transmitting station releases waves of wavelength 960 m. A capacitor of 2.56 μF is used in the resonant circuit. The self-inductance of coil necessary for resonance is x×10-8 H. find x

Options:

The correct answer is:
10

Solution:
λ=960 m

C=2.56 μF=2.56×10-6 F

c=3×108 m s-1

L=?

Now at resonance, ω0=1LC [Resonant frequency]

2πf0=1LC

On substituting f0=cλ, we have 2πcλ=1LC

On substituting f0=cλ, we have 2πcλ=1LC

Squaring both sides: 4π2c2λ2=1LC

=4×10×3×10829602=1 L×2.56×10-6

⇒1 L=4×10×9×1016×2.56×10-6960×960

⇒L=10×10-8 H

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (25 Feb Shift 1)
A transparent cube of side d, made of a material of refractive index Î¼2, is immersed in a liquid of refractive index Î¼1 ( Î¼1<Î¼2). A ray is incident on the face AB at an angle Î¸ (shown in the figure). Total internal reflection takes place at the point E on the face BC.
Then, Î¸ must satisfy

Options:
A Î¸<sin-1Î¼22Î¼12-1
B θ>sin-1μ1μ2
C θ<sin-1μ1μ2
D θ>sin-1μ22μ12-1

The correct answer is:
θ<sin-1μ22μ12-1

Solution:
At surface (1)
Î¼1sinÎ¸=Î¼2sinr â¦1
At surface (2)
Î¼1=Î¼2sin90-r
Î¼1=Î¼2cosr ...(2)
From equation (1) and (2)
âsinâ¡Î¸=sinâ¡rcosâ¡r
âsinÎ¸=Î¼22-Î¼12Î¼2Î¼1Î¼2
âsinÎ¸=Î¼22Î¼12-1
Î¸ must satisfy,
âÎ¸<sin-1Î¼22Î¼12-1

Topic: Physics, Ray Optics, JEE Main, JEE Main 2019 (12 Apr Shift 2)
A transparent cube of side d, made of a material of refractive index Î¼2, is immersed in a liquid of refractive index Î¼1 ( Î¼1<Î¼2). A ray is incident on the face AB at an angle Î¸ (shown in the figure). Total internal reflection takes place at the point E on the face BC.

Then, Î¸ must satisfy

Options:
A Î¸<sin-1Î¼22Î¼12-1
B θ>sin-1μ1μ2
C θ<sin-1μ1μ2
D θ>sin-1μ22μ12-1

The correct answer is:
θ<sin-1μ22μ12-1

Solution:
At surface (1)
Î¼1sinÎ¸=Î¼2sinr â¦1
At surface (2)
Î¼1=Î¼2sin90-r
Î¼1=Î¼2cosr ...(2)
From equation (1) and (2)
âsinâ¡Î¸=sinâ¡rcosâ¡r
âsinÎ¸=Î¼22-Î¼12Î¼2Î¼1Î¼2
âsinÎ¸=Î¼22Î¼12-1
Î¸ must satisfy,
âÎ¸<sin-1Î¼22Î¼12-1

Topic: Physics, Ray Optics, JEE Main
A transparent solid cylindrical rod has a refractive index of $\frac{2}{\sqrt{3}}$. It is surrounded by air. A light ray is incident at the mid point of one end of the rod as shown in the figure.
The incident angle $\theta$ for which the light ray grazes along the wall of the rod is

Options:
A
$\sin ^{-1}\left(\frac{1}{2}\right)$

B
$\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

C
$\sin ^{-1}\left(\frac{2}{\sqrt{3}}\right)$

D
$\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

The correct answer is:
$\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)$

Solution:
$$
\begin{aligned}
& \sin c=\frac{\sqrt{3}}{2} \\
& \sin r=\sin (90-c)=\cos c=\frac{1}{2} \\
& \frac{\sin \theta}{\sin r}=\frac{\mu_2}{\mu_1} \\
& \sin \theta=\frac{2}{\sqrt{3}} \times \frac{1}{2} \\
& \theta=\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)
\end{aligned}
$$

Topic: Physics, Ray Optics, JEE Main, JEE Main 2009
A transverse harmonic wave on a string is given by yx, t=5 sin6t+0.003x where x and y are in cm and t in sec. The wave velocity is _________ m s-1.

Options:

The correct answer is:
20

Solution:
The general equation for a wave can be written as y=Asinωt+kx. Comparing it with the given equation, y=5 sin6t+0.003x,

ω=6 rad s-1, k=0.003 cm-1

Using the equation, v=ωk

v=60.003×102=20 m s-1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (10 Apr Shift 1)
A transverse wave is propagating on a stretched string of mass per unit length 32 g m-1. The tension on the string is 80 N. The speed of the wave over the string is

Options:
A 52 m s-1
B 52 m s-1
C 25 m s-1
D 50 m s-1

The correct answer is:
50 m s-1

Solution:
Speed of wave in a string,
v=Tμ
Given, T=80 N
μ=32 g m-1
=32×10-3 kg m-1
v=8032×10-3=8×10432
=100×12
=50 m s-1

Topic: Physics, Waves and Sound, JEE Main
A transverse wave is represented by : y = 1 0 π sin 2 π T t - 2 π λ x For what value of the wavelength the wave velocity is twice the maximum particle velocity?

Options:
A 40 cm
B 60 cm
C 10 cm
D 20 cm

The correct answer is:
40 cm

Solution:
wave velocity=nÎ»
(Vp)max=aÏ
nÎ»=2a2Ïn
Î»=2(a)2Ï
=2.10Ï.2Ï
= 40 cm

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2014 (09 Apr Online)
A transverse wave propagating along the x-axis is represented by y(x, t) = 8.0 sin(0.5 Ïx â 4Ït â Ï4) where x is in metres and t is in seconds. What is the speed (in m s-1) of the wave?

Options:

The correct answer is:
8

Solution:
y x t = 8 · 0 sin 0 · 5 π x - 4 π t - π 4
Compare with a standard wave equation,
y = a sin 2 π x λ - 2 π t T⁡ + ϕ
we get 2 π λ = 0 · 5 π or, λ = 2 π 0 · 5 π = 4 m
2πT⁡=4πor,T⁡=2π4π=12 s
υ = 1 / T = 2 Hz
Wave velocity, ν=λυ=4×2=8 m s-1

Topic: Physics, Waves and Sound, JEE Main
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, JEE Main
A transverse wave propagating on a stretched string of linear density 3×10-4 kgm-1 is represented by the equation

y=0.2sin(1.5x+60t)

where, x 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=0.2sin[1.5x+60t]

Comparing with standard equation, we get

y=Asin[kx+ωt]

k=1.5=2πλ

and ω=60=2πT

∴ Velocity of wave, v=ωk

=601.5=40 m s-1

Velocity of wave in stretched string, v=Tm

where, m is the linear density, T is tension in the string.

So, T=v2m=(40)2×3×10-4

=0.48 N

Topic: Physics, Waves and Sound, JEE Main
A transverse wave travels on a taut steel wire with a velocity of v when tension in it is 2.06×104N. When the tension is changed to T, the velocity changed to v2. The value of T is close to:

Options:
A 2.50×104N
B 5.15×103 N
C 30.5×104 N
D 10.2×102 N

The correct answer is:
5.15×103 N

Solution:
vâT
⇒v1v2=T1T2⇒vv2=2.06Ã104T
⇒T=2.06Ã1044N=0.515Ã104N

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2020 (08 Jan Shift 2)
A trapezium is formed by the pair of tangents of parabola P:y=x24+1 drawn from the centre of the ellipse E:x24+y2=14, tangent at the vertex of P and the tangent at end point of the minor axis of E. The area (in sq. units) of the trapezium is

Options:
A 34
B 32
C 316
D 38

The correct answer is:
34

Solution:
The equation of the tangents from the centre of the ellipse to the parabola are y=Â±x

The equation of the tangent at the vertex P of the parabola is y=1

The equation of the tangent at the end point of the minor axis of the ellipse is y=12

Hence, the area of trapezium ABCD = 1 2 â 1 2 1+2 = 3 4 sq. units

Topic: Mathematics, Ellipse, JEE Main
A trapezium is such that three of its sides have lengths as 6 cm, then the length (in cm) of the fourth side such that the area of trapezium is maximum, is

Options:

The correct answer is:
12

Solution:
From the figure above,
Area of the trapezium =6Ã6sinâ¡Î¸+2Ã12Ã6cosâ¡Î¸Ã6sinâ¡Î¸
A=36sinâ¡Î¸+36sinâ¡Î¸cosâ¡Î¸
A=36sinâ¡Î¸+18sinâ¡2Î¸
dAdÎ¸=0â36cosâ¡Î¸+36cosâ¡2Î¸=0
2cos2Î¸+cosâ¡Î¸-1=0
cosâ¡Î¸=12 or cosâ¡Î¸=-1
for Î¸=Ï3, d2AdÎ¸2<0
So, area is maximum when Î¸=Ï3
Length of fourth side =6cosÏ3Ã2+6=12 cm

Topic: Mathematics, Application of Derivatives, JEE Main
A travelling harmonic wave is represented by the equation yx, t=10-3sin50t+2x, where x and y are in meter and t is in seconds. Which of the following is a correct statement about the wave?

Options:
A The wave is propagating along the positive x -axis with speed 25 m s-1
B The wave is propagating along the positive x -axis with speed 100 m s-1
C The wave is propagating along the negative x -axis with speed 25 m s-1
D The wave is propagating along the negative x -axis with speed 100 m s-1

The correct answer is:
The wave is propagating along the negative x -axis with speed 25 m s-1

Solution:
Given wave equation, yx, t=10-3sin50t+2x

This equation is similar to the equation of wave travelling along the negative x-axis which is given by

yx,t=Asinωt+kx

After comparing, Angular frequency ω=50 s-1

Wave number k=2 m-1

The wave velocity, v=ωk=502=25 m s-1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (12 Jan Shift 1)
A travelling microscope has 20 divisions per cm on the main scale while its Vernier scale has total 50 divisions and 25 Vernier scale divisions are equal to 24 main scale divisions, what is the least count of the travelling microscope ?

Options:
A 0.001 cm
B 0.002 mm
C 0.002 cm
D 0.005 cm

The correct answer is:
0.002 cm

Solution:
One main scale division MSD=120 cm

One vernier scale division VSD=2425MSD=2425×120 cm

∴ Least count LC=MSD-VSD=1201-2425cm

=120×125=1500=0.002 cm

=0.002 cm

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2022 (29 Jul Shift 1)