datavorous/entrance-exam-dataset huggingface dataset:
A tower subtends an angle of 60o at a point on the same level as the foot of the tower and at a second point just 10 meters above the first point the angle of depression of the foot of the tower is 15o. The height of the tower is (in meters)

Options:
A 1032-3
B 1032-3
C 1032+3
D 1032+3

The correct answer is:
1032+3

Solution:
In the figure AB is the vertical tower and C,D are the two points
Now, in right angled ΔABD
ABBD=tan⁡60o
∴hx=tan⁡60o⇒h=3x … 1
In right angled ΔBCE, BEEC=tan⁡15o
∴10x=tan⁡15o ⇒x=10tan⁡15o … 2
From 1 and 2, we get
h=10tan⁡15o×3=1032+3 meters

Topic: Mathematics, Heights and Distances, JEE Main
A tower subtends angles Î±, 2Î± and 3Î±, respectively, at points A,B and C all lying on a horizontal line through the foot of the tower. If ABBC=1+pcos(pÎ±), then the value of p is

Options:

The correct answer is:
2

Solution:
Let PM be the tower.

Given, â BAP=Î±,â CBP=2Î±âandââ MCP=3Î±

â´â APB=â PBCââ BAP=2Î±âÎ±=Î±

Similarly, â BPC=Î±

âBP=AB

By sine formula in triangle BCP, we have

BCsin(â BPC)=BPsin(â BCP)

âBCsinÎ±=ABsin(180oâ3Î±)

âABBC=sin3Î±sinÎ±=3sinÎ±â4sin3Î±sinÎ±

=3â4sin2Î±

=3â2(1âcos2Î±)

=1+2cos2Î±

Topic: Mathematics, Heights and Distances, JEE Main
A tower subtends angles Î±, 2Î± and 3Î± respectively at points, A,B and C (all points lying on the same side on a horizontal line through the foot of the tower), then the value of ABBC is equal to

Options:
A 1+2cos⁡2α
B 1-2cos⁡2α
C 1+3cos⁡2α
D 1-3cos⁡2α

The correct answer is:
1+2cos⁡2α

Solution:
Let, the height of the tower is PQ=h
Where, Q is the foot of the tower and P is the top of the tower, then
QA=hcotâ¡Î±
QB=hcotâ¡2Î±
QC=hcotâ¡3Î±

ABBC=QA-QBQB-QC=cotâ¡Î±-cotâ¡2Î±cotâ¡2Î±-cotâ¡3Î±
=cosâ¡Î±sinâ¡Î±-cosâ¡2Î±sinâ¡2Î±cosâ¡2Î±sinâ¡2Î±-cosâ¡3Î±sinâ¡3Î±=sinâ¡2Î±cosâ¡Î±-cosâ¡2Î±sinâ¡Î±sinâ¡3Î±cosâ¡2Î±-cosâ¡3Î±sinâ¡2Î±Ãsinâ¡3Î±sinâ¡Î±
=sinâ¡3Î±sinâ¡Î±=3-4sin2Î±=1+2cosâ¡2Î±

Topic: Mathematics, Heights and Distances, JEE Main
A tower $T_1$ of height $60 \mathrm{~m}$ is located exactly opposite to a tower $T_2$ of height $80 \mathrm{~m}$ on a straight road. From the top of $T_1$, if the angle of depression of the foot of $T_2$ is twice the angle of elevation of the top of $T_2$, then the width (in $m$ ) of the road between the feet of the towers $T_1$ and $T_2$ is

Options:
A
$20 \sqrt{2}$

B
$10 \sqrt{2}$

C
$10 \sqrt{3}$

D
$20 \sqrt{3}$

The correct answer is:
$20 \sqrt{3}$

Solution:
$$
\text { Let the distance between } T_1 \text { and } T_2 \text { be } x
$$
From the figure
$E A=60 \mathrm{~m}\left(T_1\right)$ and $D B=80 \mathrm{~m}\left(T_2\right)$ $\angle D E C=\theta$ and $\angle B E C=2 \theta$
Now in $\triangle D E C$,
$$
\tan \theta=\frac{D C}{A B}=\frac{20}{x}
$$
and in $\triangle B E C$,
$$
\tan 2 \theta=\frac{B C}{C E}=\frac{60}{x}
$$
We know that
$$
\tan 2 \theta=\frac{2 \tan \theta}{1-(\tan \theta)^2}
$$
$$
\begin{aligned}
&\Rightarrow \frac{60}{x}=\frac{2\left(\frac{20}{x}\right)}{1-\left(\frac{20}{x}\right)^2} \\
&\Rightarrow x^2=1200 \Rightarrow x=20 \sqrt{3}
\end{aligned}
$$

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
A tower T1 of height 60 m is located exactly opposite to a tower T2 of height 80 m on a straight road. From the top of T1, if the angle of depression of the foot of T2 is twice the angle of elevation of the top of T2, then the width (in m) of the road between the feet of the towers T1 and T2 is

Options:
A 20Â 2
B 10Â 2
C 10Â 3
D 20 3

The correct answer is:
20 3

Solution:
Let, the width of the road the road is d.
If the angle of the elevation is Î¸, then
tanÎ¸=20d, here 20 is the diï¬erence between the heights of T1 and T2.
Given that, the angle of depression is twice of the angle of elevation.
tanâ¡2Î¸=60d
We know, tanâ¡2Î¸=2tanâ¡Î¸1-tan2â¡Î¸
â´60d=40/d1-400/d2
â400d2=13
d=20 3

Topic: Mathematics, Heights and Distances, JEE Main
A toy-car, blowing its horn, is moving with a steady speed of 5 m s-1, away from a wall. An observer, towards whom the toy car is moving, is able to hear 5 beats per second. If the velocity of sound in air is 340 m s-1 the frequency of the horn of the toy car is close to

Options:
A 680 Hz
B 510 Hz
C 340 Hz
D 170 Hz

The correct answer is:
170 Hz

Solution:
Frequency heard by the observer directly from source f340340-5=f1

Frequency heard by the observer coming after relection from wall
f340340+5 = f 2
Beats frequency=f1-f2
5=f340340-5-340340+5
⇒ f=170 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2016 (10 Apr Online)
A train accelerating uniformly from rest attains a maximum speed of $40 \mathrm{~ms}^{-1}$ in $20 \mathrm{~s}$. It travels at the speed for $20 \mathrm{~s}$ and is brought to rest with uniform retardation in further $40 \mathrm{~s}$. What is the average velocity during the period ?

Options:
A $80 \mathrm{~m} / \mathrm{s}$
B $25 \mathrm{~m} / \mathrm{s}$
C $40 \mathrm{~m} / \mathrm{s}$
D $30 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$25 \mathrm{~m} / \mathrm{s}$

Solution:
(i) $\mathrm{v}=\mathrm{u}+$ at $_{1}$

$$

\begin{array}{l}

40=0+a \times 20 \\

a=2 m / s^{2} \\

v^{2}-u^{2}=2 a s \\

40^{2}-0=2 \times 2 s_{1} \\

\text { (ii) } s_{1}=400 \mathrm{~m} \\

\text { (iii) } v=u+a t \\

\quad 0=40+a \times 40 \\

a=-1 \mathrm{~m} / \mathrm{s}^{2} \\

0^{2}-40^{2}=2(-1) \mathrm{s}_{3} \\

\mathrm{~s}_{3}=800 \mathrm{~m} \\

\text { Total distance travelled }=\mathrm{s}_{1}+\mathrm{s}_{2}+\mathrm{s}_{3} \\

=400+800+800=2000 \mathrm{~m} \\

\text { Total time taken }=20+20+40=80 \mathrm{~s} \\

\text { Average velocity }=\frac{2000}{80}=25 \mathrm{~m} / \mathrm{s}

\end{array}

$$

Topic: Physics, Motion In One Dimension, JEE Main
A train approaching a hill at a speed 40 km h-1 sounds a whistle of frequency 580 Hz when it is a distance of 1 km from the hill. A wind with speed 40 km h-1 is blowing in the direction of motion of the train. (Velocity of sound in air = 1200 km h-1)
The frequency of echo is

Options:
A 599 Hz

B 590 Hz
C 610 Hz
D 620 Hz

The correct answer is:
620 Hz

Solution:
fhill=fv+vm+v0v-vm-vs=5801200+40+01200+40-40=580×12401200

fecho=fhillv-vm+v0v-vm
Here fhill is the frequency that incidents on the hill
=580×124012001200-40+401200-40
=580×12401200×12001160=620 Hz

Topic: Physics, Waves and Sound, JEE Main
A train blowing a whistle of frequency 320 Hz approaches an observer standing on the platform at a speed of 66 m s-1. The frequency observed by the observer will be (given speed of sound =330 m s-1) ______ Hz.

Options:

The correct answer is:
400

Solution:
Using Doppler's effect, the frequency observed by the observer when the train moves towards the stationary observer is

fapp=fvv-vs

Given here, vs=330 m s-1 and v=66 m s-1

Putting the values,

fapp=320330330-66

⇒fapp=400 Hz.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (25 Jan Shift 2)
A train is moving on a straight track with speed 20 m s-1 . It is blowing its whistle at the frequency of 1000 Hz. The percentage change in the frequency heard by a person standing near the track as the train passes him is (speed of sound =320 m s-1 ) close to:

Options:
A 24%
B 6%
C 12%
D 18%

The correct answer is:
12%

Solution:
Before train passes
f′=vv-vs×f0
=320320-20×f0
=320300×f0
=16f015
After train passes
f′′=vv+vs×f0
f′′=320320+20f0
=320340f0
=1617f0
Change in frequency
=1615-1617f0
∴ Percentage change in frequency
=1615-1617f0f0×100 ≈12% nearly

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2015 (04 Apr)
A train is moving towards east and a car is along north, both with same speed. The observed direction of car to the passenger in the train is

Options:
A East-north direction
B West-north direction
C South-east direction
D None of these

The correct answer is:
West-north direction

Solution:
Let $\vec{v}_c$ be the velocity of car and $\vec{v}_t$ be the velocity of train.
We know that
$\vec{v}_{c t}=\vec{v}_c-\vec{v}_t$
Hence, from the figure, velocity of car w.r.t train $\left(\mathrm{v}_{\mathrm{ct}}\right)$ is towards West - North.

Topic: Physics, Motion In Two Dimensions, JEE Main
A train moves towards a stationary observer with speed 34 m s-1. The train sounds a whistle and its frequency registered by the observer is f1. If the train's speed is reduced to 17 m s-1, the frequency registered is f2. If the speed of sound is 340 m s-1, then the ratio f 1 f 2

Options:
A 1819
B 12
C 2
D 1918

The correct answer is:
1918

Solution:
Here, vs1=34 m s-1,
v=340 m s-1
f1=vÃnv-vs1=340Ãn340-34=340306n
f2=vÃnv-vs2=340Ãn(340-17)=340n323
f1f2=323306=1918

Topic: Physics, Waves and Sound, JEE Main
A train moves towards a stationary observer with speed 34 m s-1. The train sounds a whistle and its frequency registered by the observer is v1. If the train's speed is reduced to17 m s-1, the frequency registered is v2. If the speed of sound is 340 m s-1, then the ratio v1v2 is

Options:
A 2
B 12
C 1819
D 1918

The correct answer is:
1918

Solution:
Perceived frequency by observer in Ist case
v1=vvv-vs
∴ v1=v340340-34=340v306
Perceived frequency by observer in IInd case
v2=v340340-17=340v323
Therefore,
v1v2=340v306×323340v=323306=1918

Topic: Physics, Waves and Sound, JEE Main
A train moves towards a stationary observer with speed 34 m/s . The train sound a whistle and its frequency registered by the observer is f1 . If the train' s speed is reduced to 17 m/s , the frequency registered is f2 . If the speed of sound is 340 m/s , then the ratio of f1f2 is

Options:
A 18/19
B 1/2
C 2
D 19/18

The correct answer is:
19/18

Solution:
f1=vv-vsf⁡=340340-34f⁡=340306f &
f2=340340-17f⁡=340323f
f1f2=1918

Topic: Physics, Waves and Sound, JEE Main
A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its frequency registered by the observer is f1. If the speed of the train is reduced to 17 m/s, the frequency registered is f2. If speed of sound is 340 m/s, then the ratio f1/f2 is:

Options:
A 21/20
B 20/19
C 19/18
D 18/17

The correct answer is:
19/18

Solution:
For stationary observers, f=vv-u×original frequency

f1=340340-34f
f2=340340-17f
f1f2=340-17340-34=1918

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2019 (10 Jan Shift 1)
A train moving at a speed of 20 m s-1 towards a stationary object emits a sound of frequency 930 Hz. Some part of the wave gets reflected from the object, back to the train as an echo. The frequency of this echo, as detected by the driver of the train is [speed of sound in air is 330 m s-1]

Options:
A 1400 Hz
B 1750 Hz
C 1050 Hz
D 1550 Hz

The correct answer is:
1050 Hz

Solution:
The speed of the train, vT=20 m s-1
The speed of sound in air, v=330 m s-1
The frequency of the echo detected by the driver of the train is
v'=vv+vTv-vT

v'=930330+20330-20=930×350310 Hz

⇒v'=1050 Hz

Topic: Physics, Waves and Sound, JEE Main
A train moving at a speed of $220 \mathrm{~ms}^{-1}$ towards a stationary object, emits a sound of frequency $1000 \mathrm{~Hz}$. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by the driver of the train is (speed of sound in air is $330 \mathrm{~ms}^{-1}$ )

Options:
A $3500 \mathrm{~Hz}$
B $4000 \mathrm{~Hz}$
C $5000 \mathrm{~Hz}$
D $3000 \mathrm{~Hz}$

The correct answer is:
$5000 \mathrm{~Hz}$

Solution:
From Doppler's shift, we know for this case
$$
\begin{aligned}
n^{\prime} & =n\left(\frac{v+v_s}{v-v_s}\right) \\
& =1000\left(\frac{330+220}{330-220}\right) \\
& =1000\left(\frac{550}{110}\right)=5000 \mathrm{~Hz}
\end{aligned}
$$

Topic: Physics, Waves and Sound, NEET, NEET 2012 (Mains)
A train of 150 m length is going towards north direction at a speed of 10 m s-1. A parrot flies at a speed of 5 m s-1 towards south direction parallel to the railway track. The time taken by the parrot to cross the train is equal to

Options:
A 12 s
B 8 s
C 15 s
D 10 s

The correct answer is:
10 s

Solution:
The relative velocity of the bird with respect to the train is :
⇒V→BT=V→B−V→T
VBT=5−−10=15 m s-1
[because they are going in the opposite direction]

∴ time taken by the bird to cross the train = 1 5 0 15 = 1 0 s

Topic: Physics, Motion In One Dimension, JEE Main
A train of 150 meter length is going towards north direction at a speed of $10 \mathrm{~m} / \mathrm{sec}$. A parrot flies at the speed of $5 \mathrm{~m} / \mathrm{sec}$ towards south direction parallel to the railway track. The time taken by the parrot to cross the train is

Options:
A 12 sec
B 8 sec
C 15 sec
D 10 sec

The correct answer is:
10 sec

Solution:
Relative velocity
$\begin{aligned} & =10+5=15 \mathrm{~m} / \mathrm{sec} \\ & \therefore t=\frac{150}{15}=10 \mathrm{sec}\end{aligned}$

Topic: Physics, Motion In Two Dimensions, JEE Main
A train of length L move with a constant speed Vt. A person at the back of the train fires a bullet at time t=0 towards a target which is at a distance of D (at time t=0 ) from the front of the train (on the same direction of motion). Another person at the front of the train fires another bullet at time t=T towards the same target. Both bullets reach the target at the same time. Assuming the speed of the bullets Vb are same, the length of the train is

Options:
A TVb×2Vt
B TVb+Vt
C 2TVb+2Vt
D 2TVb-2Vt

The correct answer is:
TVb+Vt

Solution:
Bullets from both person reaches target at same instant, so we equate time to get,
L+D ( v t + v b ) = D ( v b + v t ) +T
L=( v b + v t )T

Topic: Physics, Motion In One Dimension, JEE Main
A train, standing at the outer signal of railway station blows a whistle of frequency $400 \mathrm{~Hz}$ in still air. (i) What is the frequency of the whistle for a platform observer when the train
(a) approaches the platform with a speed of $10 \mathrm{~m} / \mathrm{s}$
(b) recedes from the platform with a speed of $10 \mathrm{~m} / \mathrm{s}$ ?
(ii) What is the speed of sound in each case?
[Speed of sound in still air $=340 \mathrm{~m} / \mathrm{s}$ ]

Options:

None

Solution:
(i) Given, $v=400 \mathrm{~Hz}, v=340 \mathrm{~m} / \mathrm{s}$
(a) When the train approaches the platform $v_s=10 \mathrm{~m} / \mathrm{s}$ $v^{\prime}=\frac{v}{v-v_s} \times v=\frac{340 \times 400}{340-10}=412.12 \mathrm{~Hz}$
(b) When the train recedes $\quad v^{\prime}=\frac{v \times v}{v+v_s}$ $=\frac{340 \times 400}{340+10}=388.6 \mathrm{~Hz}$
(ii) The speed of sound remains same in both cases i.e. $340 \mathrm{~m} / \mathrm{s}$.

Topic: Physics, Waves and Sound
A train, standing in a station-yard, blows a whistle of frequency $400 \mathrm{~Hz}$ in still air. The wind starts blowing in the direction from the yard to the station with a speed of $10 \mathrm{~m} / \mathrm{s}$. What are the frequency, wavelength and speed of sound for an observer standing on the station's platform? Is the situation exactly identical to the case when the air is still and the observer runs towards the yard at a speed of $10 \mathrm{~ms}^{-1}$ ? The speed of sound in still air can be taken as $340 \mathrm{~ms}^{-1}$

Options:

None

Solution:
$v=400 \mathrm{~Hz}, v_{\mathrm{m}}=10 \mathrm{~m} / \mathrm{s}, v=340 \mathrm{~m} / \mathrm{s}$
When the wind is blowing in the direction of sound
$\therefore$ Effective speed of sound $=v+v_{\mathrm{m}}=340+10=350 \mathrm{~m} / \mathrm{s}$
As the source and listener both are at rest, therefore frequency remains unchanged, i.e. $400 \mathrm{~Hz}$.
Wavelength of sound for the stationary observer.
$$
\lambda^{\prime}=\frac{v+v_{\mathrm{m}}}{v}=\frac{350}{400}=0.875 \mathrm{~m}
$$
In case the air is still but the observer is running towards the yard.
When the observer moves towards the stationary engine (source) in still air,
$$
\begin{aligned}
v_{\mathrm{o}} &=-10 \mathrm{~m} / \mathrm{s}, v_{\mathrm{s}}=0 \\
v^{\prime} &=\frac{v-v_o}{v-v_s} \times v^{\prime}=\frac{340+10}{340-0} \times 400 \\
&=\frac{350}{340} \times 400=411.8 \mathrm{~Hz} .
\end{aligned}
$$
As wavelength of sound waves is not affected by motion of the observer, it remains unchanged.
Speed of sound relative to the observer,
$$
v^{\prime \prime}=340+10=350 \mathrm{~ms}^{-1} \text {. }
$$
Thus the situations in both the cases are not exactly identical.

Topic: Physics, Waves and Sound
A train starting from rest first accelerates uniformly up to a speed of $80 \mathrm{~km} / \mathrm{h}$ for time $t$, then it moves with a constant speed for time $3 t$. The average speed of the train for this duration of journey will be (in $\mathrm{km} / \mathrm{h}$ ) :

Options:
A 40
B 80
C 30
D 70

The correct answer is:
70

Solution:
$\begin{aligned} \text { Average speed } & =\frac{\text { total distance }}{\text { time taken }} \\ & =\frac{\frac{80 \times \mathrm{t}}{2}+80 \times 3 \mathrm{t}}{4 \mathrm{t}}=70 \mathrm{~km} / \mathrm{hr} .\end{aligned}$

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2024 (06 Apr Shift 1)
A transducer used at the transmitting end, serves the purpose of converting

Options:
A sound signal to electrical form
B sound signal to magnetic form
C electrical signal to sound form
D electrical signal to magnetic form

The correct answer is:
sound signal to electrical form

Solution:
In this case transducer takes the sound signal and convert them to electrical signal.
Tranducer acts as a conversion device. At the transmitting end it converts high frequency current into electromagnetic waves. At the receiving end it transforms electromagnetic waves into electrical signals that is fed into the input of the receiver. several types of antenna are used in communication.

Topic: Physics, Communication System, JEE Main
A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW . If the current in the secondary coil is 10 A , then the input voltage and current in the primary coil are:

Options:
A 440 V and 20 A
B 220 V and 20 A
C 440 V and 5 A
D 220 V and 10 A

The correct answer is:
440 V and 5 A

Solution:
Assuming 100% efficiency,

V2V1=i1i2=N2N1

i1i2=N2N1i110=150300i1=5 A

P=VI2200=5×V1V1=440 V

Topic: Physics, Alternating Current, JEE Main
A transformer consisting of 300 turns in the primary and 150 turns in the secondary gives output power of 2.2 kW . If the current in the secondary coil is 10 A , then the input voltage and current in the primary coil are:

Options:
A 440 V and 20 A
B 220 V and 20 A
C 440 V and 5 A
D 220 V and 10 A

The correct answer is:
440 V and 5 A

Solution:
Assuming 100% efficiency,

V2V1=i1i2=N2N1

i1i2=N2N1i110=150300i1=5 A

P=VI2200=5×V1V1=440 V

Topic: Physics, Alternating Current, JEE Main, JEE Main 2019 (10 Apr Shift 1)
A transformer has 100 turns in the primary coil and carries 8 A current. If the input power is 1 kW, the number of turns in secondary coil to have 500 V of output will be

Options:
A 100
B 200
C 400
D 300

The correct answer is:
400

Solution:
Given that,
Np=100, Ip=8 A
Vs=500 V, Pi=1 kW=100 W
Input power, Pi=VPIP
1000=VpÃ8
VP=125 V
Again, we know that
NsNP=VsVp
â Ns100=500125
Ns=500Ã100125
=400

Topic: Physics, Alternating Current, JEE Main
A transformer has an efficiency of 80% and works at 10 V and 4 kW. If the secondary voltage is 240 V, then the current in the secondary coil is:

Options:
A 1.59 A
B 13.33 A
C 1.33 A
D 15.1 A

The correct answer is:
13.33 A

Solution:
Efficiency η=PoutputPinput=ESISEPIP

⇒0.8=240IS4000

⇒IS=3200240=13.33 A

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (01 Feb Shift 2)
A transformer having an efficiency of 90% is working on 200 V, 3 kW power supply. If the current in the secondary coil is 6 A, then the voltage across the secondary coil and the current in the primary coil respectively, are

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

The correct answer is:
450 V, 15 A

Solution:
Power output =3 kW×90100=2.7 kW
Is=6 A
∴Vs=2.7 kW6 A=450 V
⇒Ip=3 kW200 V=15 A

Topic: Physics, Alternating Current, JEE Main
A transformer is used to light \( 100 \mathrm{~W}-110 \mathrm{~V} \) lamp from \( 220 \mathrm{~V} \) mains. If the main current is \( 0.5 \)
A, the efficiency of the transformer is

Options:
A \( 90 \% \)
B 95\%
C \( 96 \% \)
D \( 99 \% \)

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

Solution:
Given, transformer lights $100 \mathrm{~W}-100 \mathrm{~V}$ lamp; input voltage $=220 \mathrm{~V}$; input current $=0.5 \mathrm{~A}$
Therefore, efficiency $\eta=\frac{\text { Output power }}{\text { Input power }} \times 100$
Here, output power $=100 \mathrm{~W}$
Input power $=$ Input voltage $\times$ Input current $=220 \times 0.5=110 \mathrm{~W}$
Thus, $\eta=\frac{100}{110} \times 100=90 \%$

Topic: Physics, Current Electricity, KCET, KCET 2015
A transformer is used to light 140 W, 24 V lamp from 240 V AC mains. The current in the main cable is 0.7 A. The efficiency of the transformer is

Options:
A 63.8%
B 84%
C 83.3%
D 48%

The correct answer is:
83.3%

Solution:
Mains voltage is V=240 V and
mains current is I=0.7 A
input power will be Pinput=240×0.7=168 W
But output power is given as Poutput=140 W
Therefore efficiency =outputinput×100 % =83.33 %

Topic: Physics, Alternating Current, NEET
A transformer is used to light a 100 W and 110 V lamp from a 220 V mains. If the current is 0.5 A, the efficiency of the transformer is

Options:
A 86%
B 91%
C 96%
D 98%

The correct answer is:
91%

Solution:
Given: Output power, P=100 W
The voltage across the primary coil, Vp=220 V
Current in the primary coil, Ip=0.5 A
The efficiency of a transformer, Î· = output power input power Ã 1 0 0
=PVpIpÃ100=100220Ã0.5Ã100â91%

Topic: Physics, Alternating Current, JEE Main
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 $0.5 \mathrm{~A}$, the efficiency of the transformer is approximately

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

The correct answer is:
$90 \%$

Solution:
The efficiency of transformer $=\frac{\text { Energy obtained from the secondary coil }}{\text { Energy given to the primary coil }}$
Or $\eta=\frac{\text { Output power }}{\text { Input power }}$
or
$$
\eta=\frac{V_{s} I_{s}}{V_{p} I_{p}}
$$
Given, $\quad V_{s} I_{s}=100 \mathrm{~W}, V_{p}=220 \mathrm{~V}, I_{p}=0.5 \mathrm{~A}$
$$
\text { Hence, } \quad \eta=\frac{100}{220 \times 0.5}=0.90=90 \%
$$
Note The efficiency of an ideal transformer is 1 (or $100 \%$ ). But in practice due to loss in energy, the efficiency of transformer is always less than 1 (or less than $100 \%$ ).

Topic: Physics, Alternating Current, MHT CET, MHT CET 2008
A transformer is used to light a 100 W and 110 V lamp from a 220 V mains. If the main current is 0.5 A , the efficiency of the transformer is approximately

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

The correct answer is:
$90 \%$

Solution:
The efficiency of transformer
$=\frac{\text { Energy obtained from the secondary coil }}{\text { Energy given to the primary coil }}$
or
$\begin{array}{lrl}
\text { or } & \eta & =\frac{\text { Output power }}{\text { Input power }} \\
\text { or } & \eta & =\frac{V_s I_s}{V_p I_p} \\
\text { Given, } & V_s I_s & =100 \mathrm{~W}, V_p=220 \mathrm{~V}, I_p=0.5 \mathrm{~A} \\
\text { Hence, } & \eta & =\frac{100}{220 \times 0.5}=0.90=90 \%
\end{array}$

Topic: Physics, Alternating Current, JIPMER, JIPMER 2009
A transformer is used to light a 100 W and 110 V lamp from a 220 V mains. If the main current is 0.5 A. The efficiency of the transformer is approximately -

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

The correct answer is:
90%

Solution:
Efficiency of Transformer =PowerÂ outputPowerÂ input
% Î·=PoutPinÃ100=100220Ã0.5Ã100=90%

Topic: Physics, Alternating Current, JEE Main
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, JEE Main