datavorous/entrance-exam-dataset huggingface dataset:
A stationary source emits sound waves of frequency 500 Hz . Two observers moving along a line passing through the source detect sound to be of frequencies 480 Hz and 530 Hz . Their respective speeds are, in m s-1 ,
(Given speed of sound =300 m/s )

Options:
A 16, 14
B 12, 16
C 12, 18
D 8, 18

The correct answer is:
12, 18

Solution:
Frequency observed by observer A, fA=500300-VA300
480=500(300-VA)300
VA=1500-14405
=12m/s
Frequency observed by observer B,  fB=500300+VB300
530=500300+VB300
VB=18m/s

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (10 Apr Shift 1)
A stationary source emits the sound of frequency f0=492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 m s-1. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz? (Given that the speed of sound in air is 330 m s-1 and the car reflects the sound at the frequency it has received).

Options:

The correct answer is:
6

Solution:
Frequency of sound as received by large car approaching the source.

f1=C+V0C f0=330+2330492 Hz.

This car now acts as source for reflected sound wave

â´ freflected=f1

Frequency of sound received by source,

f2=CC-V0 freflected=330330-2Ãf1=330328Ã332330Ã492Hz

â´ Beat frequency =f0-f2=332328-1Ã492 Hz

=6 Hz

Topic: Physics, Waves and Sound, JEE Main
A stationary source is emitted sound at a fixed frequency f0, which is reflected by two cars approaching the source. The difference between the is frequencies of sound reflected from the car is 1.2% of f0. What is the difference in the speed of the cars (in m s-1)? The cars are moving at constant speeds much smaller than the speed of sound which is 330 m s-1

Options:

The correct answer is:
1.98

Solution:
Firstly, car will be treated as an observer which is approaching the source. Then, it will be treated as a source, which is moving in the direction of sound.
Hence,
f1=fov+v1v-v1
f2=fov+v2v-v2
∴ f1-f2=1.2100fo
=fov+vv+v1-v+v2v-v2
Or
1.2100fo=2vv1-v2v-v1v-v2fo
as v1and v2 are very very less than v.
We can write, v-v1or v-v2≈v.
∴1.2100fo=2v1-v2vfo
Or v1-v2=v×1.2200
=300×1.2200=1.98m s-1

Topic: Physics, Waves and Sound, JEE Main
A stationary source is emitting sound at a fixed frequency f 0 , which is reflected by two cars approaching the source. The difference between the frequencies of sound reflected from the cars is 1.2% of f 0 . What is the difference in the speeds of the cars (in km per hour) to the nearest integer ? The cars are moving at constant speeds much smaller than the speed of sound which is 330  ms -1 .

Options:

The correct answer is:
7

Solution:
Let v1 and v2 be the velocities of sources, then apparent frequencies are
f1=c+v1c-v1f0 and f2=c+v2c-v2f0
Then difference between frequencies is,
âf=2âvCf0 âµâc>>v1â&âc>>v2
â´âv=7 kmph.

Topic: Physics, Waves and Sound, JEE Main
A stationary uniform rod of mass $m=1 \mathrm{~kg}$ and length $I=2 \mathrm{~m}$ leans against a smooth vertical wall making an angle $\theta=45^{\circ}$ with rough horizontal floor. Find the frictional force that is exerted by the floor on the rod.

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

The correct answer is:
$5 \mathrm{~N}$

Solution:
The correct option is A $5 \mathrm{~N}$
As the rod is stationary so the linear acceleration and angular acceleration of rod is zero.
$\begin{aligned}
& \text { i.e., } a_{\text {com }}=0 ; a=0 \\
& \mathrm{~N}_2=\mathrm{f} \\
& \left.\mathrm{N}_1=\mathrm{mg}\right\} \because \mathrm{a}_{\mathrm{com}}=0 \\
&
\end{aligned}$
Torque about any point of the rod should also be zero
$\begin{aligned}
& \therefore a=0 \\
& \tau_A=0 \Rightarrow m g \cos \theta \frac{1}{2}+f I \sin \theta=N_1 \cdot I \cos \theta \\
& \Rightarrow N_1 \cos \theta=f \sin \theta+\frac{m g \cos \theta}{2} \\
& \Rightarrow f=\frac{m g \cos }{2 \sin \theta} \frac{\theta \mathrm{mg} \cot \theta}{2} \\
& \Rightarrow f=\frac{1 \times 10 \times \cot 45^{\circ}}{2} \\
& \Rightarrow f=\frac{10}{2}=5 \mathrm{~N}
\end{aligned}$
So, option (a) is correct.

Topic: Physics, Rotational Motion, JEE Main
A stationary wave of amplitude A is generated between the two fixed ends x=0 and x=L. The particle at x=L3 is a node. There are only two particles between x=L6 and x=L3 which have maximum speed half of the maximum speed of the anti-node. Again there are only two particles between x=0 and x=L6 which have maximum speed half of that at the antinodes. The slope of the wave function at x=L3 changes with respect to time according to the graph shown. The symbols Î¼, Ï and A are having their usual meanings if used in calculations.

The time period of oscillations of a particle is

Options:
A T
B 2T
C T2
D 4T

The correct answer is:
2T

Solution:
x=0 isa Nodeâyâx=0 att=0 ây=Asinâ¡kx.sinâ¡Ït
âyâx= AkÂ·cosâ¡kxÂ·sinâ¡Ït
ÏÂ·2T=2Ï
T=ÏÏ
T'=2ÏÏ=2T

Topic: Physics, Waves and Sound, JEE Main
A statue of height 4 m stands on a tower of height 10 m. The angle subtended by the statue at the eyes of an observer of height 2 m, standing at a distance of 6 m from base of the tower is

Options:
A tan-1211
B tan-143
C tan-12
D tan-1713

The correct answer is:
tan-1211

Solution:
From the figure
tan⁡α=2, tan⁡β=43
tan⁡θ=2-431+83=211

Hence, angle subtended by the statue at the eyes of the observer is θ=tan-1211

Topic: Mathematics, Heights and Distances, JEE Main
A steady current flow in a long wire. It is bent into a circular loop of one turn and the magnetic field at the centre of the coil is B. If the same wire is bent into a circular loop of n turns, the magnetic field at the centre of the coil is

Options:
A Bn
B nB
C nB2
D n2B

The correct answer is:
n2B

Solution:
The magnetic field a circular coil (for n number of loops)
B=μ0ni2r
When, n1=1, n2=2
2πr1=n×2πr2
⇒ r1r2=n1
So B1B2=μ0n1i2r1μ0n2i2r2
BB2=n1n2.r1r2
BB2=1n.1n
⇒ B2=n2B

Topic: Physics, Magnetic Effects of Current, JEE Main
A steady current flows in a metallic conductor of non-uniform cross-section. The quantity/ quantities constant along the length of the conductor is/are

Options:
A Current, electric field and drift speed
B Drift speed only
C Current and drift speed
D Current only

The correct answer is:
Current only

Solution:
If E be electric field, then current density $j=$ $\sigma E$
Also we know that current density $j=\frac{i}{A}$
Hence $j$ is different for different area of crosssections. When $j$ is different, then $E$ is also different. Thus $E$ is not constant. The drift velocity $v_d$ is given by $v_d=\frac{j}{n e}=$ different for different $j$ values. Hence only current $i$ will be constant.

Topic: Physics, Current Electricity, JEE Main
A steady current flows in a metallic conductor of non-uniform cross-section. Which of these quantities is constant along the conductor?

Options:
A Electric field
B Drift velocity
C Current
D Current density

The correct answer is:
Current

Solution:
The current flowing through a conductor of non-uniform cross-section remain same in the whole of the conductor.

Topic: Physics, Current Electricity, AIIMS, AIIMS 2016
A steady current I is set up in a wire whose cross-sectional area decreases in the direction of the flow of the current. Then, as we examine the narrowing region,

Options:
A the current density decreases in value
B the magnitude of the electric field increases
C the current density remains constant
D the average speed of the moving charges remains constant

The correct answer is:
the magnitude of the electric field increases

Solution:
When current flows through a conductor of tapered cross-section, current flow through every section remains constant.

⇒ I1=I2

⇒ j1 A1=j2 A2

⇒ j1j2=A2A1<1

⇒ j1<j2

Current density increases in the narrow region. Also,
=nevd

⇒ nevd1<nevd2

⇒ vd1<vd2

Drift velocity increases in the narrow region.

and j=Eρ

where, ρ= resistivity of material.

⇒ E1ρ<E2ρ⇒E1<E2

Electric field magnitude increases in the narrow region.

Topic: Physics, Current Electricity, JEE Main
A steady current I is set up in a wire whose cross-sectional area decreases in the direction of the flow of the current. Then, as we examine the narrowing region,

Options:
A the current density decreases in value
B the magnitude of the electric field increases
C the current density remains constant
D the average speed of the moving charges remains constant

The correct answer is:
the magnitude of the electric field increases

Solution:
When current flows through a conductor of tapered cross-section, current flow through every section remains constant.

⇒ I1=I2

⇒ j1 A1=j2 A2

⇒ j1j2=A2A1<1

⇒ j1<j2

Current density increases in the narrow region. Also,
=nevd

⇒ nevd1<nevd2

⇒ vd1<vd2

Drift velocity increases in the narrow region.

and j=Eρ

where, ρ= resistivity of material.

⇒ E1ρ<E2ρ⇒E1<E2

Electric field magnitude increases in the narrow region.

Topic: Physics, Current Electricity, KVPY, KVPY 2018 (SA)
A steam at $100^{\circ} \mathrm{C}$ is passed into $1 \mathrm{~kg}$ of water contained in a calorimeter of water equivalent $0.2 \mathrm{~kg}$ at $9^{\circ} \mathrm{C}$ till the temperature of the calorimeter and water in it is increased to $90^{\circ} \mathrm{C}$. The mass of steam condensed in $\mathrm{kg}$ is nearly (specific heat of water $=1 \mathrm{cal} / \mathrm{g}{ }^{\circ} \mathrm{C}$, latent heat of vaporisation $=540 \mathrm{cal} / \mathrm{g}$ )

Options:
A $0.81$
B $0.18$
C $0.27$
D $0.54$

The correct answer is:
$0.18$

Solution:
By calorimetry,
Heat lost $=$ Heat gained
$$
\begin{aligned}
m \times L+ & m \times C \times\left(T_1-T_2\right) \\
& =\left(m_1+m_2\right) \times C \times\left(T_2-T_3\right) \\
m \times 540 & +m \times 1 \times(100-90) \\
& =(1000 \mathrm{~g}+200 \mathrm{~g}) \times 1 \times(90-9) \\
& =m \times 540+10 m=1200 \times 81 \\
550 m & =1200 \times 81 \\
m & =\frac{1200 \times 81}{550}=\frac{24 \times 81}{11} \mathrm{~g} \\
& =\frac{1944}{11} \mathrm{~g}=176 \mathrm{~g}=0.176 \mathrm{~kg}
\end{aligned}
$$
or $m=0.18 \mathrm{~kg}$

Topic: Physics, Thermal Properties of Matter, JEE Main
A steel and a brass wire, each of length $50 \mathrm{~cm}$ and cross-sectional area $0.005 \mathrm{~cm}^{2}$ hang from a ceiling and are $15 \mathrm{~cm}$ apart. Lower ends of the wires are attached to a light horizontal bar. A suitable downward load is applied to the bar so that each of the wires extends in length by $0.1 \mathrm{~cm}$. At what distance from the steel wire the load must be applied? [Young's modulus of steel is $2 \times 10^{12}$ dynes $/ \mathrm{cm}^{2}$ and that of brass is $1 \times 10^{12}$ dynes $\left./ \mathrm{cm}^{2}\right]$

Options:
A $7.5 \mathrm{~cm}$
B $5 \mathrm{~cm}$
C $10 \mathrm{~cm}$
D $3 \mathrm{~cm}$

The correct answer is:
$5 \mathrm{~cm}$

Solution:
Hint:

At equilibrium, Taking torque about point $\mathrm{O}$
$\begin{aligned}
T_{1} x=T_{2}(15-x) & \Rightarrow Y_{1} x=Y_{2}(15-x) \\
\left[Y=\frac{T}{A} \times \frac{L}{\ell} ; T \propto Y\right] & 1 \times 10^{12} x=2 \times 10^{12}(15-x) ; 3 x=15, \quad \text { So, } x=5 \mathrm{~cm}
\end{aligned}$

Topic: Physics, Rotational Motion, JEE Main
A steel ball of mass m1 = 1 kg moving with velocity 50 m s-1 collides with another ball of mass m2 = 200 g lying on the ground. Due to collision, the KE is lost and their internal energies change equally and T1 and T2 are the temperature changes of masses m1 and m2, respectively. If the specific heat of steel is unity and J = 4.18 J cal-1, then

Options:
A T1 = 7.1 °C and T2 = 1.47 °C

B T1 = 1.47 °C and T2 = 7.1 °C
C T1 = 3.4 °C and T2 = 17.0 °C
D T1 = 17.0 °C and T2 = 3.4 °C

The correct answer is:
T1 = 3.4 °C and T2 = 17.0 °C

Solution:
Half of KE is attained as heat by each ball
1 2 KE = m 1 s 1 T 2
⇒ 1 2 × 1 × 5 0 2 = 1 × 0.105 × 4 1 8 × 1 0 3 × 1 0 3 × T 1
T 1 = 5 0 × 5 0 2 × 0.105 × 4.18 × 1 0 3
= 2 5 2.1 × 4.18 = 2 5 8.778 ≅ 3.4 K
As m 2 = m 1 5 , so T2 = 5T1 = 17 K

Topic: Physics, Thermodynamics, JEE Main
A steel rail of length 5 m and area of cross section 40 cm2 is prevented from expanding along its length while the temperature rises by 10°C . If coefficient of linear expansion and Young's modulus of steel are 1.2×10-5 K-1 and 2×1011 N m-2 respectively, the force developed in the rail is approximately:

Options:
A 2×107 N
B 2×109 N
C 3×10-5 N
D 1×105 N

The correct answer is:
1×105 N

Solution:
F=YAαΔt
⇒F=2×1011×40×10-4×1.2×10-5×10
⇒F=9.6×104≈1×105 N.

Topic: Physics, Thermal Properties of Matter, JEE Main, JEE Main 2017 (09 Apr Online)
A steel ring of radius $r$ and cross-section area ' $A$ ' is fitted on to a wooden disc of radius $R(R\gtr)$. If Young's modulus be $E$, then the force with which the steel ring is expanded is

Options:
A $A E \frac{R}{r}$
B $A E\left(\frac{R-r}{r}\right)$
C $\frac{E}{A}\left(\frac{R-r}{A}\right)$
D $\frac{E r}{A R}$

The correct answer is:
$A E\left(\frac{R-r}{r}\right)$

Solution:
Initial length (circumference) of the ring $=2 \pi r$
Final length (circumference) of the ring $=2 \pi R$
Change in length $=2 \pi R-2 \pi r$.
strain $=\frac{\text { change in length }}{\text { original length }}=\frac{2 \pi(R-r)}{2 \pi r}=\frac{R-r}{r}$
Now Young's modulus
$E=\frac{F / A}{l / L}=\frac{F / A}{(R-r) / r}$
$\therefore F=A E\left(\frac{R-r}{r}\right)$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A steel rod has a radius of 20 mm and a length of 2.0 m. A force of 62.8 kN stretches it along its length. Young's modulus of steel is 2.0×1011N m-2. The longitudinal strain produced in the wire is _________×10-5.

Options:

The correct answer is:
25

Solution:
Young's modulus is given as

Y=StressStrain=FA∆LL

The given data is

Stress=62.8 kN

r=20×10-3 m

Y=2×1011 N m-2

Thus,

Strain=FA×Y⇒∆LL=62.8×103 Nπ×20×10-3 m2×2×1011 N m-2=25×10-5 N

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2023 (06 Apr Shift 1)
A steel rod is 4.000 cm in diameter at 30 °C . A brass ring has an interior diameter of 3.992 cm at 30 °C . In order that the ring just slides onto the steel rod, the common temperature of the two should be

αsteel=11×10-6/ °C and αbrass=19×10-6/ °C

Options:
A 200 °C
B 350 °C
C 280 °C
D 300 °C

The correct answer is:
280 °C

Solution:
41+11×10-6ΔT=3.992 1+19×10-6ΔT
⇒ 0.008=31.848×10-6∆T⇒∆T=250 °C

Topic: Physics, Thermal Properties of Matter, JEE Main
A steel rod is projecting out of a rigid wall. The shearing strength of steel is 345 MN m-2. The dimensions AB=5 cm, BC=BE=2 cm. The maximum load that can be put on the face ABCD is : (neglect bending of the rod) (g=10 m s-2)

Options:
A 3450 kg
B 1380 kg
C 13800 kg
D 345 kg

The correct answer is:
13800 kg

Solution:
F = shear strength x area on which shear stress acts
= 3 4 5 Ã 1 0 6 Ã 4 Ã 1 0 - 4
= 1 3 8 0 0 0 N
So the mass that can be placed =13800 kg

Topic: Physics, Mechanical Properties of Solids, JEE Main
A steel rod with y=2.0×1011 N m-2 and α=10-5 C∘-1 of length 4 m and area of cross-section 10 cm2 is heated from 0°C to 400°C without being allowed to extend. The tension produced in the rod is x×105 N where the value of x is _________.

Options:

The correct answer is:
8

Solution:
Thermal force F=AyαΔT

F=10×10-4×2×101110-5(400)
F=8×105 N
⇒x=8

Topic: Physics, Current Electricity, JEE Main, JEE Main 2021 (01 Sep Shift 2)
A steel tape measures the length of a copper rod as 90.0 cm when both are at 10 Â°C, the calibration temperature for the tape. What would be the tape reading for the length of the rod when both are at 30 Â°C. Given, Î± for steel 1.2Ã10-5 Â°C-1 and Î± for copper is 1.7Ã10-5 Â°C-1.

Options:
A 90.01 cm
B 89.90 cm
C 90.22 cm
D 89.80 cm

The correct answer is:
90.01 cm

Solution:
As the steel tape is calibrated at 10 Â°C, therefore, adjacent centimeter marks on the steel tape will be separated by a distance of
lt=l101+ Î±sâT=(1+Î±s20) cm
Length of copper rod at 30 Â°C
= 901+Î±c20cm
Therefore, number of centimeters read on the tape will be
= 901+Î±c2011+Î±s20= 901+1.7Ã10-5Ã2011+1.2Ã10-5Ã20
= 90Ã1.000341.00024=90.01 cm

Topic: Physics, Thermal Properties of Matter, JEE Main
A steel wire can sustain $100 \mathrm{~kg}$ weight without breaking. If the wire is cut into two equal parts, each part can sustain a weight of

Options:
A
$50 \mathrm{~kg}$

B
$400 \mathrm{~kg}$

C
$100 \mathrm{~kg}$

D
$200 \mathrm{~kg}$

The correct answer is:
$100 \mathrm{~kg}$

Solution:
Breaking force $\alpha$ area of cross section of wire
Load hold by wire is independent of length of the wire.

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2012 (19 May Online)
A steel wire has a length of $90 \mathrm{~cm}$ which is under a constant tension of $100 \mathrm{~N}$. The speed of the transverse waves that can be produced in the wire will be (take, the mass of the steel wire to be $6 \times 10^{-3} \mathrm{~kg}$ )

Options:
A $50 \mathrm{~m} / \mathrm{s}$
B $50 \mathrm{~cm} / \mathrm{s}$
C $1 / 3 \sqrt{6} \mathrm{~m} / \mathrm{s}$
D $50 \sqrt{6} \mathrm{~m} / \mathrm{s}$

The correct answer is:
$50 \sqrt{6} \mathrm{~m} / \mathrm{s}$

Solution:
Given, tension is the wire, $T=100 \mathrm{~N}$
Mass of steel wire, $m=6 \times 10^{-3} \mathrm{~kg}$
Iength, $l=90 \mathrm{~cm}=0.9 \mathrm{~m}$
Mass per unit length of steel wire,
$$
\begin{aligned}
\mu &=\frac{m}{l} \\
&=\frac{6 \times 10^{-3}}{0.9} \\
&=\frac{2}{3} \times 10^{-2} \mathrm{~kg}-\mathrm{m}^{-1}
\end{aligned}
$$
Speed of transverse wave,
$$
\begin{aligned}
v &=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{100}{\left(\begin{array}{c}
2 \\
3
\end{array}\right) \times 10^{-2}}} \\
&=\frac{300}{\sqrt{6}}=50 \sqrt{6} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main
A steel wire having a radius of 2.0 mm , carrying a load of 4 kg, is hanging from a ceiling. Given that g=3.1π m s-2, what will be the tensile stress that would be developed in the wire?

Options:
A 5.2×106 N m-2
B 6.2×106 N m-2
C 4.8×106 N m-2
D 3.1×106 N m-2

The correct answer is:
3.1×106 N m-2

Solution:
Stress =mgπr2=4×3.1×ππ×2×10-32
=3.1×106N/m2

Topic: Physics, Mechanical Properties of Solids, JEE Main, JEE Main 2019 (08 Apr Shift 1)
A steel wire having a radius of $2.0 \mathrm{~mm}$, carrying a load of $4 \mathrm{~kg}$, is hanging from a ceiling. Given that $\mathrm{g}=3.1 \mathrm{~mm} / \mathrm{s}^2$, what will be the tensile stress that would be developed in the wire?

Options:
A $4.8 \times 10^6 \mathrm{~N} / \mathrm{m}^2$
B $3.1 \times 10^6 \mathrm{~N} / \mathrm{m}^2$
C $6.2 \times 10^6 \mathrm{~N} / \mathrm{m}^2$
D $5.2 \times 10^6 \mathrm{~N} / \mathrm{m}^2$

The correct answer is:
$3.1 \times 10^6 \mathrm{~N} / \mathrm{m}^2$

Solution:
Tensile stress = Force/Area
Tensile stress $=(4)(3.1 \pi) / \pi\left(2 \times 10^{-3}\right)^2$
Tensile stress $=3.1 \times 10^6 \mathrm{Nm}^{-2}$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A steel wire of length $4.7 \mathrm{~m}$ and cross-sectional area $3.0 \times 10^{-5} \mathrm{~m}^{2}$ stretches by the same amount as a copper wire of length $3.5 \mathrm{~m}$ and cross-sectional area of $4.0 \times 10^{-5} \mathrm{~m}^{2}$ under a given load. What is the ratio of Young's modulus of steel to that of copper?

Options:
A $1.8: 1$
B $2.8: 1$
C $3.8: 1$
D $4.8: 1$

The correct answer is:
$1.8: 1$

Solution:
Given, for stoul wire,
Area, $A_{1}=3 \times 10^{-5} \mathrm{~m}^{\text {g }}$
Length, $L_{1}=4.7 \mathrm{~m}$
For copper wire,
$$
A_{2}=4 \times 10^{-5} \mathrm{~m}^{2}, L_{2}=3.5 \mathrm{~m}
$$
If $\gamma_{1}$ ynd $\gamma_{2}$ be the Young's modulus for the material steel wire und copper wire respectively, then
$$
\begin{aligned}
\frac{\gamma_{1}}{\gamma_{2}} &=\frac{\frac{K_{1} L_{1}}{A_{1} S}}{\frac{r_{2} L_{2}}{A_{2} A l}} \\
\Rightarrow \quad \frac{\gamma_{1}}{\gamma_{2}} &=\frac{L_{3} A_{2}}{L_{2} A_{1}} \quad\left[\because F_{1}=F_{2}\right] \\
&=\frac{4.7 \times 4 \times 10^{-5}}{35 \times 3 \times 10^{-5}}=1.8 \\
\Rightarrow \quad \gamma_{4}: \gamma_{2} &=1.8: 1
\end{aligned}
$$

Topic: Physics, Mechanical Properties of Solids, JEE Main
A steel wire with mass per unit length 7.0 × 10–3 kg m–1 is under tension of 70 N. The speed of transverse waves in the wire will be:

Options:
A 200π m s-1
B 100 m s-1
C 10 m s-1
D 50 m s-1

The correct answer is:
100 m s-1

Solution:
The formula to calculate the speed v of the transverse wave generated in a stretched string is given by

v= Tμ............................(1)

where, T is the tension in the string and μ is the mass density.

Therefore, required speed of the wave:

v=70 N7×10-3 kg m-1 =100 m/s

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (01 Feb Shift 1)
A step down transformer connected to an ac mains supply of 220 V is made to operate at 11 V, 44 W lamp. Ignoring power losses in the transformer, what is the current in the primary circuit?

Options:
A 2 A
B 0.4 A
C 0.2 A
D 4 A

The correct answer is:
0.2 A

Solution:
Power loss is zero. Hence, output Power is equal to input Power.

So, Input Power, P1=44 W

⇒V1i1=P1

⇒i1=P1V1=44220=0.2 A

Topic: Physics, Alternating Current, NEET, NEET 2021
A step down transformer is used to reduce the main supply from ' $V_1$ ' volt to ' $V_2$ ' volt. The primary coil draws a current ' $I_1$ ' A and the secondary coil draws ' $I_2$ ' A. $\left(I_1 < I_2\right)$. The ratio of input power to output power is

Options:
A $\frac{\mathrm{V}_1 \mathrm{~V}_2}{\mathrm{I}_1 \mathrm{I}_2}$
B $\frac{\mathrm{I}_1 \mathrm{~V}_1}{\mathrm{I}_2 \mathrm{~V}_2}$
C $\frac{\mathrm{I}_1 \mathrm{I}_2}{\mathrm{~V}_1 \mathrm{~V}_2}$
D $\frac{\mathrm{V}_1 \mathrm{I}_2}{\mathrm{~V}_2 \mathrm{I}_1}$

The correct answer is:
$\frac{\mathrm{I}_1 \mathrm{~V}_1}{\mathrm{I}_2 \mathrm{~V}_2}$

Solution:
$\begin{aligned} & \text { Input power }=\mathrm{I}_1 \mathrm{~V}_1 \\ & \text { Output power }=\mathrm{I}_2 \mathrm{~V}_2\end{aligned}$

Topic: Physics, Alternating Current, MHT CET, MHT CET 2021 (22 Sep Shift 2)
A step down transformer reduces \( 220 \mathrm{~V} \) to \( 110 \mathrm{~V} \). The primary draws \( 5 \mathrm{~A} \) of current and secondary supplies \( 9 \) A. The efficiency of the transformer is

Options:
A \( 20 \% \)
B \( 44 \% \)
C \( 90 \% \)
D \( 100 \% \)

The correct answer is:
\( 90 \% \)

Solution:
For a transformer, we all know that:

η=PoPi×100= EsisEpip×100.

This gives us:

% efficiency=PoutPin×100=110×9220×5×100
=90%

Topic: Physics, Alternating Current, JEE Main
A step up transformer is connected to an ac mains supply of 220 V to operate at $11000 \mathrm{~V}, 88$ watt. The current in the secondary circuit, ignoring the power loss in the transformer, is

Options:
A 8 mA
B 4 mA
C 0.4 A
D 4 A

The correct answer is:
8 mA

Solution:
In secondary circuit, $P=V i$
$\Rightarrow 88=11000 i$
$\Rightarrow \quad i=\frac{88}{11 \times 10^3}=8 \times 10^{-3} \mathrm{~A}$
$\Rightarrow \quad i=8 \mathrm{~mA}$

Topic: Physics, Alternating Current, NEET, NEET 2024 (Re-NEET)
A steroid hormone which regulate glucose metabolism is:

Options:
A cortisone
B cortisol
C corticosterone
D 11-deoxycorticosterone

The correct answer is:
cortisol

Solution:
Cortisol or hydrocortisone is the principal glucocorticoid hormone of many mammals including humans. It regulates the glucose metabolism and promotes gluconeogenesis, especially during starvation, and raises blood pressure. Cortisone is an inactive form of cortisol.
Related Theory
Cortisol is the main glucocorticoid released from the zona fasciculata layer of the adrenal cortex. The hypothalamus-pituitary-adrenal axis regulates both production and secretion of cortisol. Loss of regulation can lead to cortisol excess disorders, such as Cushing syndrome, or cortical insufficiency, such as Addison disease.

Topic: Biology, Chemical Coordination and Integration, NEET, NEET 2006