datavorous/entrance-exam-dataset huggingface dataset:
A sound wave of frequency f propagating through the air at a speed c is reflected from a surface which is moving away from the source with a constant speed v . The frequency of the reflected wave as perceived by the observer at the position of the source will be

Options:
A f(c+2v)c+v
B f(c+v)c-v
C f(c-v)c+v
D f(c-v)c-2v

The correct answer is:
f(c-v)c+v

Solution:
In the first case observer (surface) is moving away, so the perceived frequency is
f ′ =f( c−v c )
In the second case source (surface) is moving away, so the perceived frequency is
f ″ = f ′ ( c c+v )=f( c−v c+v )

Topic: Physics, Waves and Sound, JEE Main
A sound wave of frequency f travelling horizontally to the right is reflected by a large vertical plane surface moving to left with a speed v. There is a detector (1) attached on the surface and another detector (2) fixed on ground. If the speed of sound in the medium is c, then the difference in the frequency received by detector 2 after being reflected from the surface and the frequency received by detector 1 is

Options:
A ∆f=fcc-vvc+v
B ∆f=fvc-vcc+v
C ∆f=fcc+vvc-v
D ∆f=fvc+vcc-v

The correct answer is:
∆f=fvc+vcc-v

Solution:
f1=fc+vc

f2=f1cc-v=fc+vc-v

∆f=fc+vc-v-c+vc

∆f=fcc+v-c-vc+vcc-v

∆f=fc2+cv-c2-v2cc-v

∆f=fvc+vcc-v

Topic: Physics, Wave Optics, JEE Main
A sound wave of frequency $\mathrm{v} \mathrm{Hz}$ initially travels a distance of $1 \mathrm{~km}$ in air. Then, it gets reflected into a water reservoir of depth $600 \mathrm{~m}$. The frequency of the wave at the bottom of the reservoir is
$$
\text { ( } V_{\text {air }}=340 \mathrm{~m} / \mathrm{s} V_{\text {water }}=1484 \mathrm{~m} / \mathrm{s} \text { ) }
$$

Options:
A $>v \mathrm{~Hz}$
B $ < v \mathrm{~Hz}$
C $v \mathrm{~Hz}$
D 0 (the sound wave gets attenuated by water completely)

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

Solution:
The frequency of sound wave remains constant

Topic: Physics, Waves and Sound, TS EAMCET, TS EAMCET 2017
A sound wave passing through the air at NTP produces a pressure of 0.001 dyne cm-2 during a compression. The corresponding change in temperature (given Î³ = 1.5 and assume gas to be ideal) is

Options:
A 8.97×10-4 K

B 9.87Ã10-6 K
C 8.97Ã10-8 K
D 9.87Ã10-4 K

The correct answer is:
8.97Ã10-8 K

Solution:
TγP1-γ= constant

Differentiating
γ T γ - 1 dT P 1 - γ + T γ 1 - γ P - γ dP = 0
or, dT = γ - 1 T γ P dP
or, dT = 1.5 - 1 1.5 2 7 3 7 6 × 13.6 × 9 8 1 × 0.001

⇒8.97×10-8 K

Topic: Physics, Thermodynamics, JEE Main
A source and a detector both start moving simultaneously from the position shown in the figure, one along x-axis and the other along y-axis with speeds 40 m s-1 and 30 m s-1 respectively. If n is the frequency of the source then the graph between the apparent frequency n' observed by detector and time t would be [assume that source and detector do not collide]

.

Options:
A
B
C
D

The correct answer is:

Solution:
When the source and detector start moving towards the origin then frequency detected by the detector is greater than the original frequency.
After crossing origin frequency detected by the detector is less than the original frequency.
We can see from the figure that in one second the source and the detector will be at the origin. After that when they continue their motion, their respective distances from the origin maintains a constant ratio, hence the line joining them remains at a constant angle. So whatever apparent frequency the detector registers, it will remain constant.

Topic: Physics, Wave Optics, JEE Main
A source and a detector move away from each other in absence of wind with a speed of 20 m s-1, with respect to the ground. If the detector detects a frequency of 1800 Hz of the sound coming from the source, then the original frequency of source considering the speed of sound in the air 340 m s-1 will be _____ Hz.

Options:

The correct answer is:
2025

Solution:
The formula of apparent frequency in terms of velocity of the source and observer is, f′=fv−v0v+vs, now put the values which are given in the question,

⇒1800=f340−20340+20⇒f=1800×360320 Hz⇒f=2025 Hz

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2021 (26 Aug Shift 1)
A source and an observer are situated on two perpendicular tracks as shown in the figure. The observer is at rest and the source is moving with a speed of 50 m s-1. The source emits a sound wave of frequency 75 Hz, which travel in the medium with a velocity of 200 m s-1. The frequency of the sound heard by the observer when the source crosses the origin is

Options:
A 70 Hz
B 80 Hz
C 75 Hz
D 85 Hz

The correct answer is:
80 Hz

Solution:
At the time of emission, if the source and observer line is making an angle Î¸ with the line of motion of source, then
fapp=fovvâvscosÎ¸

x=vst and x=lcosÎ¸

[As this sound is heard by observer when source is at O]

â´cosÎ¸=vsv
and fapp=fov2v2âvs2
Putting the values, fapp=80 Hz

Topic: Physics, Waves and Sound, JEE Main
A source and listener are both moving towards each other with speed $v / 10$, where $v$ is the speed of sound. If the frequency of the note emitted by the source is $f$, the frequency heard by the listener would be nearly

Options:
A 1.11 f
B 1.22 f
C f
D 1.27 f

The correct answer is:
1.22 f

Solution:
When source and listener both are moving towards each other then, the frequency heard
$n^{\prime}=n\left(\frac{v+v_0}{v-v_S}\right) \Rightarrow n^{\prime}=f\left(\frac{v+v / 10}{v-v / 10}\right)=1.22 f$.

Topic: Physics, Waves and Sound, JEE Main
A source of light is placed at a distance of $50 \mathrm{~cm}$ from a photo cells and the stopping potential is found to be $\mathrm{V}_{0^*}$. If the distance between the light source and photo cells is made $25 \mathrm{~cm}$, the new stopping potential will be:

Options:
A $V_0 / 2$
B $V_0$
C $4 V_0$
D $2 V_0$

The correct answer is:
$V_0$

Solution:
Stopping potential is independent of distance hence new stopping potential will remain unchanged in required.
Stopping potential $=V_0$

Topic: Physics, Dual Nature of Matter, NEET, NEET 2013 (Karnataka)
A source of light is placed in front of a screen. The intensity of light on the screen is I. Two Polaroids P1 and P2 are so placed in between the source of light and screen that the intensity of light on the screen is I2. Then the P2, should be rotated by an angle of (degrees) so that the intensity of light on the screen becomes 3I8.

Options:

The correct answer is:
30

Solution:
Recall the law of Malus I'=I2cos2θ

Here θ is the angle between two polaroids,

⇒3I8=I2cos2θ⇒θ=30°.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2021 (26 Aug Shift 2)
A source of sound A emitting waves of frequency $1800 \mathrm{~Hz}$ is falling towards ground with a terminal speed $\mathrm{v}$. The observer $\mathrm{B}$ on the ground directly beneath the source receives waves of frequency $2150 \mathrm{~Hz}$. The source A receives waves, reflected from ground of frequency nearly: (Speed of sound $=343 \mathrm{~m} / \mathrm{s}$ )

Options:
A
$2150 \mathrm{~Hz}$

B
$2500 \mathrm{~Hz}$

C
$1800 \mathrm{~Hz}$

D
$2400 \mathrm{~Hz}$

The correct answer is:
$2500 \mathrm{~Hz}$

Solution:
Given $\mathrm{f}_{\mathrm{A}}=1800 \mathrm{~Hz}$
$$
\begin{aligned}
&\mathrm{v}_{\mathrm{t}}=\mathrm{v} \\
&\mathrm{f}_{\mathrm{B}}=2150 \mathrm{~Hz}
\end{aligned}
$$
Reflected wave frequency received by $\mathrm{A}$, $\mathrm{f}_{\mathrm{A}}{ }^{\prime}=$ ?
Applying doppler's effect of sound, $\mathrm{f}^{\prime}=\frac{\mathrm{v}_{\mathrm{s}} \mathrm{f}}{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{t}}}$
Here, $\mathrm{v}_{\mathrm{t}}=\mathrm{v}_{\mathrm{s}}\left(1-\frac{\mathrm{f}_{\mathrm{A}}}{\mathrm{f}_{\mathrm{B}}}\right)=343\left(1-\frac{1800}{2150}\right)$ $\mathrm{v}_{\mathrm{t}}=55.8372 \mathrm{~m} / \mathrm{s}$
Now, for the reflected wave,
$$
\begin{aligned}
&\therefore \mathrm{f}_{\mathrm{A}}{ }^{\prime}=\left(\frac{\mathrm{v}_{\mathrm{s}}+\mathrm{v}_{\mathrm{t}}}{\mathrm{v}_{\mathrm{s}}-\mathrm{v}_{\mathrm{t}}}\right) \mathrm{f}_{\mathrm{A}} \\
&=\left(\frac{343+55.83}{343-55.83}\right) \times 1800 \\
&=2499.44 \approx 2500 \mathrm{~Hz}
\end{aligned}
$$

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2014 (12 Apr Online)
A source of sound emits sound waves at frequency f0 . It is moving towards an observer with fixed speed vs (vs<v) , where v is the speed of sound in air). If the observer were to move towards the source with speed v0 , one of the following two graphs (A and B) will give the correct variation of the frequency f heard by the observer as v0 is changed.
The variation of f with v0 is given correctly by:

Options:
A Graph A with slope =f0v-vs
B Graph A with slope =f0v+vs
C Graph B with slope =f0v-vs
D Graph B with slope =f0v+vs

The correct answer is:
Graph A with slope =f0v-vs

Solution:
fapp=v+v0v-vsf0
⇒f=f0v-vsv0+vf0v-vs
On comparing with y=mx+c
m=f0v-vs and c=vf0v-vs
⇒ Graph with +ve slope and +ve intercept.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2015 (11 Apr Online)
A source of sound having frequency 300 Hz and a receiver are located along the same line normal to the wall as shown in the figure. Both the source and the receiver are stationary and the wall recedes from the source with a velocity of 20 m s-1. If the beat frequency registered by the receiver is 240x Hz, then what is the value of x? [velocity of sound = 330 m s-1]

Options:

The correct answer is:
7

Solution:
As ; f1=f (for direct sound)
Now for reflected sound f2=V-20V+20f
If b is the beat frequency,
∴b=f1-f2
∴f-V-20V+20f=f×40V+20
=300×40350=2407 Hz

Topic: Physics, Waves and Sound, JEE Main
A source of sound is moving with a velocity of 50 m s-1 towards a stationary observer. The observer measures the frequency of sound as 500 Hz. The apparent frequency of sound as heard by the observer when source is moving away from him with the same speed is (Speed of sound at room temperature is 350 m s-1)

Options:
A 400 Hz
B 600 Hz
C 375 Hz
D 175.5 Hz

The correct answer is:
375 Hz

Solution:
Speed of sound, v=350 m s-1
Velocity of sound, v=50 m s-1
From Doppler's principle,
1st case, fâ²=vv-vsf ......(i)
500=350350-50f
Where, f= frequency of the source
2nd case, f''=vv+vsf
f''=350350+50f ....(ii)
From equation (i) and (ii) we get
f''500=350/400350/300=34
â´ f''=500Ã34=375 Hz

Topic: Physics, Waves and Sound, JEE Main
A source of sound is moving with constant velocity of $20 \mathrm{~m} / \mathrm{s}$ emitting a note of frequency 1000 $\mathrm{Hz}$. The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be (Speed of sound $v=340 \mathrm{~m} / \mathrm{s}$ )

Options:
A 9:8
B 8:9
C 1:1
D 9:10

The correct answer is:
9:8

Solution:
When source is approaching the observer, the frequency heard
$n_a=\left(\frac{v}{v-v_S}\right) \times n=\left(\frac{340}{340-20}\right) \times 1000=1063 \mathrm{~Hz}$
When source is receding, the frequency heard $n_r=\left(\frac{v}{v+v_S}\right) \times n=\frac{340}{340+20} \times 1000=944$ $\Rightarrow n_a: n_r=9: 8$
Short tricks: $\frac{n_a}{n_r}=\frac{v+v_S}{v-v_S}=\frac{340+20}{340-20}=\frac{9}{8}$.

Topic: Physics, Waves and Sound, JEE Main
A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 m s-1 at an angle of 60o with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer corresponding to the wave emitted by the source at the instant shown is (velocity of sound in air 330 m s-1 )
â

Options:
A 103âHz
B 106 Hz
C 97 Hz
D 100 Hz

The correct answer is:
103 Hz

Solution:
f=f0v-v0v-vs
f=100v-0v-+9.7
f=100vv1-9.7v
f=1001+9.7330=103 Hz

Topic: Physics, Waves and Sound, JEE Main
A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 m s-1 at an angle of 60o with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air 330 m s-1 ) is
â

Options:
A 103âHz
B 106 Hz
C 97 Hz
D 100 Hz

The correct answer is:
103 Hz

Solution:
f=f0v-v0v-vs
f=100v-0v-+9.7
f=100vv1-9.7v
f=1001+3.7330=103Hz

Topic: Physics, Waves and Sound, JEE Main
A source supplies heat to a system at the rate of 1000 W. If the system performs work at a rate of 200 W. The rate at which internal energy of the system increases is

Options:
A 600 W
B 800 W
C 500 W
D 1200 W

The correct answer is:
800 W

Solution:
According to the first law of thermodynamics, Q=W+U.

Upon differentiation, we get

dQdt=dWdt+dUdt⇒dUdt=dQdt-dWdt

Given: dQdt=1000 W

dWdt=200 W

Therefore, dUdt=1000 W-200 W=800 W

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2023 (06 Apr Shift 1)
A space station is at a height equal to the radius of the Earth. If ' \( V_{E} \) ' is the escape velocity on
the surface of the Earth, the same on the space station is __ times \( V_{E} \).

Options:
A \( \frac{1}{2} \)
B \( \frac{1}{4} \)
C \( \frac{1}{\sqrt{2}} \)
D \( \frac{1}{\sqrt{3}} \)

The correct answer is:
\( \frac{1}{\sqrt{2}} \)

Solution:
Given, a space station is at a height equal to radius of Earth, \( \frac{1}{2} M v_{s}^{2}=\frac{G M m}{(R+h)} \)
Here, \( h=R \) and \( v_{s} \) is velocity of space station
\[
\begin{array}{l}
\Rightarrow \frac{1}{2} M v_{s}^{2}=\frac{G M m}{2 R} \\
\Rightarrow v_{s}=\sqrt{\frac{G M}{R}}
\end{array}
\]
Now, we know that escape velocity is
\[
\begin{array}{l}
v_{e}=\sqrt{\frac{2 G M}{R}}=\sqrt{2} \sqrt{\frac{G M}{2}} \\
\Rightarrow v_{e}=\sqrt{2} v_{s} \\
\text { or } v_{s}=\frac{1}{\sqrt{2}} v_{e}
\end{array}
\]

Topic: Physics, Gravitation, JEE Main
A spacecraft of mass $100 \mathrm{~kg}$ breaks into two when its velocity is $10^4 \mathrm{~m} \mathrm{~s}^{-1}$. After the break, a mass of $10 \mathrm{~kg}$ of the spacecraft is left stationary. The velocity of the remaining part is

Options:
A $10^3 \mathrm{~m} \mathrm{~s}^{-1}$
B $11.11 \times 10^3 \mathrm{~m} \mathrm{~s}^{-1}$
C $11.11 \times 10^2 \mathrm{~m} \mathrm{~s}^{-1}$
D $10^4 \mathrm{~m} \mathrm{~s}^{-1}$

The correct answer is:
$11.11 \times 10^3 \mathrm{~m} \mathrm{~s}^{-1}$

Solution:
From conservation of momentum
$M v=m_1 v_1+m_2 v_2$
Here, $M=100 \mathrm{~kg}, v=10^4 \mathrm{~m} \mathrm{~s}^{-1}$
$\begin{array}{ll} & m_2=90 \mathrm{~kg}, v_2=? \\ \therefore & 100 \times 10^4=10 \times 0+90 \times v_2 \\ \text { or } & v_2=\frac{100 \times 10^4}{90}=11.11 \times 10^3 \mathrm{~m} \mathrm{~s}^{-1}\end{array}$

Topic: Physics, Center of Mass Momentum and Collision, AIIMS, AIIMS 2019 (25 May)
A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate dM(t)dt=bv2(t), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is:

Options:
A -bv3(t)
B -bv3M(t)
C -2bv3M(t)
D -bv32M(t)

The correct answer is:
-bv3M(t)

Solution:
dM(t)dt=-bv2
in free space no external force so there in only thrust force on rocket
fin =dMdtVrel
Ma=-bv2(t)v
a=-bv3M(t)

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2020 (05 Sep Shift 2)
A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in 24 hours around the planet?
[Given: Mass of planet =8×1022 kg ,
Radius of planet =2×106 m,
Gravitational constant G=6.67×10-11 Nm2/kg2 ]

Options:
A 17
B 9
C 13
D 11

The correct answer is:
11

Solution:
T=2π(R+h)3GM=2π 8.242×10186.67×10-11×8×1022
T≈7809 sec
Number of revolutions in 24 hours:
n=864007809=11.06
≈ 11 revolutions

Topic: Physics, Gravitation, JEE Main, JEE Main 2019 (10 Apr Shift 2)
A space-ship travelling in the interstellar medium is picking up dust at a rate dMdt=Î±v, where Î± is a positive constant, v is the instantaneous velocity and M is the total mass of the space-ship at any instant. The instantaneous acceleration of the space-ship is

Options:
A

−2α v 2 M

B

âÎ± v 2 M

C

−α v 2 2M

D

−α v 2

The correct answer is:

âÎ± v 2 M

Solution:
dMdt=Î±v

Fthrust=-vdMdt=Ma

â-vÎ±v=Ma

âa=âÎ±v2M

Topic: Physics, Center of Mass Momentum and Collision, JEE Main
A special dice with numbers $1,-1,2,-2,0$ and 3 is thrown thrice. What is the probability that the sum of the numbers occurring on the upper face is zero?

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

The correct answer is:
$25 / 216$

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

Topic: Mathematics, Probability, JEE Main
A special fair cubic die is rolled which has one blue side, two red sides and three yellow sides. The result is the colour of the top side after the die is rolled. If the die is rolled 8 times, then the probability of blue colour coming at least twice is

Options:
A 13×5768
B 68-13×5768
C 8×5768
D 1-8×5768

The correct answer is:
68-13×5768

Solution:
Let B is the event in which the top side of the die shows blue colour
PB=16=p (Let)
Probability of getting a blue colour at least twice
=1-C081-p8-C181-p7p1
=1-568-8Ã16Ã567
=68-13Ã5768

Topic: Mathematics, Probability, JEE Main
A specially designed Vernier calliper has the main scale least count of 1 mm. On the Vernier scale, there are 10 equal divisions and they match with 11 main scale divisions. Then, the least count of the Vernier calliper is

Options:
A 0.1 mm

B 0.909 mm

C 1.1 mm

D 0.09 mm

The correct answer is:
0.1 mm

Solution:
Here, 10 divisions of vernier scall
=11 main scale divisions So, 1 vernier scale division =1110 main scale divisions

Now, we use formula for least count, Least count =1 main scale division -1 vernier scale division.

⇒LC=1MSD−1VSD

=1−1110MSD

=−110MSD

=−110×1mm

=−01mm

So, magnitude of least count is 0.1 mm.

Topic: Physics, Experimental Physics, KVPY, KVPY 2019 (SA)
A species facing extremely high risk of extinction in the immediate future is called:

Options:
A Vulnerable
B Endemic
C Critically Endangered
D Extinct

The correct answer is:
Critically Endangered

Solution:
International Union for Conservation of Nature and Natural Resources
(IUCN), is now known as World Conservation Union (WCU) whose headquarters are at Morges, Switzerland looks after the species which need protection. They maintain Red Data Book, which enlists the different categories of protection needed by flora and fauna.

Critically endangered organisms face an extremely high risk of extinction and hence, require a greater degree of conservation and care. Examples of such organisms include various species of mammals (rhinos), amphibians (frogs), etc.

Topic: Biology, Biodiversity and its Conservation, NEET, NEET 2014
A species having carbon with sextet of electrons and can act as electrophile is called

Options:
A carbon free radical
B carbanion
C carbocation
D pentavalent carbon

The correct answer is:
carbocation

Solution:
A carbocation is a molecule in which a carbon atom has a positive charge and three bonds. We can say that, basically, they are carbon cations. Formerly, it was known as carbonium ion. Carbocation today is defined as any even-electron cation that possesses a significant positive charge on the carbon atom.

The carbon cations are very reactive and unstable due to an incomplete octet. In simple words, carbocations do not have eight electrons; therefore, they do not satisfy the octet rule.

In carbocation, the hybridisation of carbon will be sp2 and its shape is trigonal planar. There is also a vacant p- orbital which indicates its electron-deficient nature. The carbon has 6 electrons in its valence shell. Due to this, it is an electron-deficient species, also known as an electrophile.

Six electron species

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (31 Jan Shift 1)
A spectrometer gives the following reading when used to measure the angle of a prism.
Main scale reading: $58.5$ degree
Vernier scale reading : $09$ divisions
Given that $1$ division on main scale corresponds to $0.5$ degree. Total divisions on the vernier scale is $30$ and match with $29$ divisions of the main scale. The angle of the prism from the above data

Options:
A
$58.59^{\circ}$

B
$58.77^{\circ}$

C
$58.65^{\circ}$

D
$59^{\circ}$

The correct answer is:
$58.65^{\circ}$

Solution:
$\mathrm{L.C} =\frac{1}{60}$
Total Reading $=585+\frac{9}{60}=58.65$

Topic: Physics, Experimental Physics, JEE Main, JEE Main 2012 (Offline)
A spectrum is formed by a prism of dispersive power ' $\omega$ '. If the angle of deviation is ' $\delta$ ', then the angular dispersion is

Options:
A $\omega / \delta$
B $\delta / \omega$
C $1 / \omega \delta$
D $\omega \delta$

The correct answer is:
$\omega \delta$

Solution:
We know that $\frac{\delta_v-\delta_r}{\delta_{\text {mean }}}=\omega$
$\Rightarrow$ Angular dispersion $=\delta_v-\delta_r=\theta=\omega \delta$

Topic: Physics, Ray Optics, JEE Main
A sphere and a cube of same material and same total surface area are placed in the same evacuated space turn by turn after they are heated to the same temperature. Find the ratio of their initial rates of cooling in the enclosure.

Options:
A Ï 6 : 1
B π 3 : 1
C π 6 : 1
D π 3 : 1

The correct answer is:
Ï 6 : 1

Solution:
Rate of emission of energy = Ï T 4 S
Let m1 be the mass of sphere, C is specific heat and d Î¸ / dt , the rate of cooling.
For sphere
Ï T 4 S = m 1 C d Î¸ dt S ...(i)
Let m2 be the mass of cube, C its specific heat and d Î¸ / dt , the rate of cooling
For cube
Ï T 4 S = m 2 C d Î¸ dt C ...(ii)
From Eqs.(i) and (ii)
d Î¸ / dt s d Î¸ / dt c = m 2 m 1
= a 3 Ï 4 / 3 Ï r 2 Ï
where a is the side of cube and r is the radius of sphere, Ï is the density.
Required ratio R s R c = 3 a 3 4 Ï r 3
But since S (surface area) is the same,
6 a 2 = 4 Ï r 2
or a 2 = 2 / 3 Ï r 2
â´ R s R c = 3 2 Ï r 2 / 3 3 / 2 4 Ï r 3 = 2 Ï 2 Ï 3 4 Ï
= 2 Ï 1 2 = Ï 6

Topic: Physics, Thermal Properties of Matter, JEE Main
A sphere is rotating between two rough inclined walls as shown in figure. Coefficient of friction between each wall and the sphere is 13. If f1 and f2 be the frictional forces at P and Q. Then f1f2 is

Options:
A 4 3 + 1
B 1 3 + 2
C 1 2 + 3
D 1 + 2 3

The correct answer is:
4 3 + 1

Solution:
Let μ be the friction coefficient between sphere and each wall. Free body diagram of sphere is
Net force on the sphere in horizontal diarection is zero.
∴N1cos60ο+μN2cos60ο=N2cos30ο+μN1cos30ο
⇒N1+μN2=3N2+μN1
⇒N11-3μ=N23-μ
⇒N1N2=3-μ1-3μ
Substituting μ=13 we get,
⇒N1N2=3-131-33=33-13-3=1+43
⇒f1f2=μN1μN2=1+43

Topic: Physics, Rotational Motion, JEE Main