Question:
The current $I_{1}$ (in A) flowing through $1\Omega$ resistor in
the following circuit is

Solution:
[IMAGE] Resistance of branch PQ = 2.5 [IMAGE] Apply Ohm's law:
I = = 0.4A \\
i = = 0.2A

Physics


Question:
A small spherical droplet of density d is floating exactly half
immerse in a liquid of density ρ and surface tension T. The radius of
the droplet is (take note that the surface tension applies an upward
force on the droplet)

Solution:
$\Rightarrow d \cdot \frac{4}{3}\pi r^{3}g = \rho \cdot \frac{2}{3}\pi r^{3}g + 2\pi rT$
$\Rightarrow \frac{2}{3}r^{2}g(2d - \rho) = 2T$
$\Rightarrow r = \sqrt{\frac{3T}{(2d - \rho)}}$

Physics


Question:
A container of a large uniform cross-sectional area A resting on
a horizontal surface holds two immiscible, non-viscous and
incompressible liquids of densities and each of height
(1/2)H as shown. The smaller density liquid is open to atmosphere. A
homogeneous solid cylinder of length
[IMAGE] cross-sectional area (1/5) A is immersed such that it floats with its
axis vertical to the liquid-liquid interface with length (1/4) L in
denser liquid. If D is the density of the solid cylinder then D=5d/n.
Find n.,d yEck,d leku vuqizLFkdkV
gSa,d leku ?kuRo
jgk gS fd bldh dqy yEckbZ dk ?kuRo okys nzo A;fn D csyu

Physics


Question:
One end of a uniform rod of mass m and length
[IMAGE] is clamped. The rod lies on a smooth
horizontalsurface and rotates on it about the clamped end at a uniform
angular velocity$\omega$. The force exerted bythe clamp on the rod has a
horizontal component

Physics


Question:
The vessel shown in the figure has two sections of areas of
cross-section [IMAGE] and
[IMAGE]. A liquid of density ρ fills both the
sections, up to a height h in each. Neglect atmospheric pressure

the pressure at the base of the vessel is $2\ h\ \rho\ g$

the force exerted by the liquid on the base of the vessel is
$\ 2\ h\rho gA_{2}$

the walls of the vessel at the level X exert a downward force
$h\rho g\left( A_{2} - A_{1} \right)$ on the liquid.$\ $

Solution:
$0 + \rho g(2h) =$Press at base
$F = A_{2}(\rho g2h)$

Physics


Question:
A uniform rod AB of mass m and length I at rest on a smooth
horizontal surface. An impulse P is applied to the end B. The time taken
by the rod to turn through a right angle is

Solution:
Impulse = change in momentum
$\therefore\ \ \ \ \ \ \ P \cdot \frac{\mathcal{l}}{2} = \frac{mc^{2}}{12} \cdot \omega$
(about centre of AB)
$\Rightarrow \omega = \frac{6P}{m\mathcal{l}}$
$\frac{\pi}{2} = \omega t \Rightarrow t = \frac{\pi}{2\omega} = \frac{\pi m\mathcal{l}}{2 \times 6p}$
$\Rightarrow t = \frac{\pi m\mathcal{l}}{12p}$

Physics


Question:
The figure shows the P-V plot of an ideal gas taken through a
cycle DCBAD then
dks pØ DCBAD

maximum temperature will occur during the process C → B → A.

negative work is done by the gas in the cycle DCBAD.

heat flows out of the gas during C → B → A.

Solution:
Since temperature at C and A are same so maximum temperature
occurs at a state between them during process CBA.
Since cycle in PV graph is anticlockwise so work done by gas in
negative.
During CBA work done by gas is negative and [IMAGE] so
[IMAGE]. So heat is released by gas.
D;ksafdC rFkk A ij rkieku leku gS rks eku bu izØe CBA
ds nkSjku gksxkA D;ksfd PV vkjs[k okekorZ

Physics


Question:
A boy sits in a chair connected to a rope that passes over a
frictionless pulley. He pulls the loose end of the string with such a
force that the spring scale reads 250 N. The boy's weight is 320 N, and
the chair weighs 160 N. The force that the boy exerts on the chair is
(Take g = 10 m/s2)

Solution:
[IMAGE] TOPIC: NLM
SUB TOPIC: EQUILIBRIUM

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
If $\rho$ is density of the material of a uniform rope of length
$\mathcal{l}$ and $\sigma$ is the breaking stress. It is rotated in
horizontal circular path then maximum angular velocity it can have for
circular path:
$\mathcal{l}$yEckbZ dh le:i jLlh ds inkFkZ dk ?kuRo $\rho$ gS rFkk
jLlh dh ruu lkeF;Z(breaking stress)$\sigma$gSA bl ry esa
o`Ùkh; iFk ij ?
osx Kkr djks

Solution: $\omega^{2} = \frac{2\sigma}{\mathcal{l}^{2}\rho} \Rightarrow \omega = \frac{1}{\mathcal{l}}\sqrt{\frac{2\sigma}{\rho}}$

Physics


Question:
A particle is executing SHM between points
-[IMAGE] and [IMAGE], as shown in
figure-I. The velocity V(t) of the particle is partially graphed and
shown in figure-II. Two points A and B corresponding to time
are marked on the V(t) curve

At time [IMAGE], its speed is increasing.

At time [IMAGE], its position lies in between
[IMAGE] and O

The phase difference ∆φ between points A and B must be expressed
as 90°
<∆φ< 180°.

Solution:
At time [IMAGE], velocity of the particle is
negative i.e. going towar. From the graph, at
time [IMAGE], its speed is decreasing. Therefore
particle lies in between [IMAGE] and

Physics


Question:
A body is given a velocity $\sqrt{2gr}$ at the highest point of
a half circular smooth track which is joined by arough horizontal track
whose co-efficient of friction is $\mu = 0.5$. Then the distance
travelled by body before it stops on horizontal track is: (r=1m)

Solution:
$W_{g} + W_{t} = 0 - \frac{1}{2}m(2gr)$
$mg2r - \mu mgx = mgr\ $ $x = \frac{r}{\mu} = \frac{1}{0.5} = 2m$.

Physics


Question:
The dimensions of$\frac{B^{2}}{2\mu_{0}}$, where B is magnetic
field and $\mu_{0}$ is the magnetic permeability of vacuum, is

Solution:
Energy density in magnetic field$= \frac{B^{2}}{2\mu_{0}}$
$= \frac{\ Force\ \times \ displacement\ }{(\ displacement\ )^{3}} = \frac{{MLT}^{- 2}L}{L^{3}} = {ML}^{- 1}T^{- 2}$

Physics


Question:
Consider two charged metallic spheres $S_{1}$ and $S_{2}$ of
radii$R_{1}$and $R_{2}$, respectively. The electric fields $E_{1}$(on
$S_{1}$) and $E_{2}$ (on $S_{2}$) on their surfaces are such that
$E_{1}/E_{2} = R_{1}/R_{2}$. Then the
ratio$V_{1}(onS_{1})/V_{2}(onS_{2})$ of the electrostatic potentials on
each sphere is

Solution:
$\frac{E_{1}}{E_{2}} = \frac{r_{1}}{r_{2}}$
$\frac{V_{1}}{V_{2}} = \frac{E_{1}r_{1}}{E_{2}r_{2}} = \frac{r_{1}}{r_{2}} \times \frac{r_{1}}{r_{2}} = {(\frac{r_{1}}{r_{2}})}^{2}$

Physics


Question:
A bead of mass m is kept at a distance from the axis of
rotation of a smooth massless tube inside it. The tube is rotating on a
smooth horizontal surface about a vertical axis perpendicular to the
length L of the tube and passing through one of its end with
constant angular velocity[IMAGE].

The speed of the bead relative to tube when it is at a distance r
from axis of rotation is2[IMAGE]

The speed of the bead relative to tube when it is at a distance
'r' from axis of rotation is[IMAGE]

The work done by the centrifugal force in the frame of tube when
the bead reaches at a distance the axis of rotation starting

The work done by the centrifugal force in the frame of the tube
when the bead reaches at a from the axis of rotation

Solution:
$\frac{v^{2}}{2} = \omega^{2}\frac{\left( r^{2} - a^{2} \right)}{2}$
$V = \omega\sqrt{r^{2} - a^{2}}$
Applying WET equation
$W_{(Centrifugal\ force\ )} = \frac{1}{2}{mV}^{2}$
$W = \frac{1}{2}m\omega^{2}\left( r^{2} - a^{2} \right)$

Physics

P-10 Physics paper 1 30 October Jee Main.docx


Question:
Consider a particle initially moving with a velocity 7 m/s
starts decelerating at a constant rate of$2m/s^{2}$.
The distance travelled in the fourth second is$\frac{x}{10}m$. Find
the value of x.

Solution:
Here particle is decelerating so velocity of particle becomes
 after t=3.5 sec. the particle has a turning point at t=3.5 sec.
$0 = 7 - 2t$
$t = 3.5\sec$
$x_{1} = x_{3} - \frac{1}{2} \times 2 \times 3^{2} = 12m$
$y_{2} = 7 \times 3.5 - \frac{1}{2} \times 2 \times (3.5)^{2} = \frac{49}{4}m$
$\Rightarrow y_{2} - x_{1} = \frac{49}{4} - 12 = \frac{1}{4}m$.
Due to symmetry displacement of the particle at t=3 and t=4sec. is same.
so total distance traveled in the fourth sec
is$= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}m$

Physics


Question:
Both the diodes used in the circuit shown are assumed to be
ideal and have negligible resistance whenthese are forward biased. built
in potential in each diode is 0.7 V. For the input voltages shown in
thefigure, the voltage (in volts) at point A is..........
ekusa rFkk vxzfnf'kd ¼QkjoMZ½ bu varjfufeZr foHkokUrj(built in potential), 0.7 VgSAfp=k
 (input) oksYVrk fcUnq A ij oksYVrk

Solution:
[IMAGE] Let [IMAGE] Right diode is reversed biased and left diode is forward biased
 + Ik'p

Physics

P-21 Physcis Paper 19 Dec Hindi Eng..docx


Question:
An electron gun is placed inside a long solenoid of radius R on
its axis. The solenoid has n turns/lengthand carries a current I. The
electron gun shoots an electron along the radius of the solenoid with
speedv. If the electron does not hit the surface of the solenoid,
maximum possible value of v is (all symbolshave their standard meaning)

Solution:
$R_{\max} = \frac{R}{2} = \frac{{mv}_{\max}}{e\mu_{0}in}$
$V_{\max} = \frac{{Re}{\mu_{0}in}}{2m}$

Physics


Question:
An elevator of height ascends with constant acceleration
 When it crosses a platform bolt drops from the top of the
elevator. If the time for the bolt to hit the floor of the elevator
is$\sqrt{\frac{\lambda h}{g + a}}$ then find $\lambda$.

Solution:
Choose the floor of the elevator as reference. The observer is
inside the elevator take the downwarddirection as positive.
Acceleration of bolt relative to elevators is
$a^{'} = g - ( - a) = g + a$
$h = \frac{1}{2}a^{'}t^{2} = \frac{1}{2}(g + a)t^{2}$
$t = \sqrt{\frac{2h}{g + a}}$

Physics


Question:
When photon of energy [IMAGE] strikes the surface
of a metal A, the ejected photoelectrons have maximum kinetic energy Ta
eV and de-Broglie wavelength[IMAGE]. The maximum kinetic
energy of photoelectrons liberated from another metal B by photon of
energy [IMAGE] is [IMAGE] If the de-Broglie
wavelength of these photoelectrons[IMAGE], then the work
function of metal B is

Solution: Relation between De-Broglie wavelength and K. E. is

Physics


Question:
A long, straight wire of radius a carries a current distributed
uniformly over its cross-section. The ratioof the magnetic fields due to
the wire at distance $\frac{a}{3}$ and 2a, respectively from the axis of
the wire is

Solution:
$B_{A} = \frac{\mu_{0}ir}{2\pi a^{2}} = \frac{\mu_{0}i\frac{a}{3}}{2\pi a^{2}} = \frac{\mu_{0}i}{\pi a^{2}}\frac{a}{6} = \frac{\mu_{0}i}{6\pi a}$
$B_{B} = \frac{\mu_{0}i}{2\pi(2a)}$
$\frac{B_{A}}{B_{B}} = \frac{4}{6} = \frac{2}{3}$

Physics


Question:
Pressure versus temperature graph of the ideal gas at constant
volume V of an ideal gas is shown by a straight line A. Now, pressure P
of the gas is doubled and the volume is halved: then the corresponding
pressure versus temperature graph will be shown by the line:-,d vkn"kZ xSLk ds fLFkj xSl dk ncko cuke rkieku xzkQ
rkieku xzkQ ykbu tk,xkA

Solution:
$P = \frac{nRT}{V} = \frac{mRT}{MV}$
(Where[IMAGE] number of moles)
So, at constant volume pressure versus temperature graph is a straight
line passing through origin with slope mR/MV. As volume is halved,
slope becomes two times. Therefore, pressure versus temperature graph
will be shown by the line B.
 eku ds fo:) fLFkj <ky lkFk mRifÙk
 <yku nks xquk gks A ncko
cuke rkieku xzkQ tk,xkA

Physics


Question:
A litre of hydrogen at [IMAGE] and
[IMAGE] pressure expands isothermally until its volume
becomes double and then expands adiabatically until it becomes four
times. [IMAGE]. Final temperature (in
27°C o 106 izlkj tc rd bl nqxquk gks rFkk fQj:)ks"e izØe }kjk
pkj xqus vk;ru rd izlkfjr A vfUre

Physics


Question:
A pebble is thrown horizontally from the top of a 20 m high tower
with an initial velocity of 10 m/s. The air drag is negligible. The
speed of the pebble when it is at the same distance from top as well as
base of the tower (g = 10 m/s2)

Physics

P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx


Question:
A beam of electromagnetic radiation of intensity
[IMAGE] is comprised of wavelength,
[IMAGE] It falls normally on a metal (work
function[IMAGE] ) of surface area of
[IMAGE]. If one in [IMAGE] photons
ejects an electron, total number of electrons ejected
[IMAGE] then x is....

Solution:
Energy of photon. [IMAGE] (so photoelectric
effect will take place)
[IMAGE] No. of photons falling per second [IMAGE] No. of photoelectron emitted per second [IMAGE]

Physics


Question:
In the figure shown switch S_1 remains connected and switch
S_2 remains open for a long time. Now S_2 is also closed. Assuming
$\varepsilon = 10V$ and L = 1H. Find the magnitude of rate of change of
current (in A/s) in inductor just after the switch S_2 is closed.

Solution:
$i_{1} + i_{2} = \varepsilon/R$
$\frac{\varepsilon - X}{R} + \frac{- 3\varepsilon - X}{2R} = \frac{\varepsilon}{R}$
$\Rightarrow X = - \varepsilon$
Using ${kv}^{\mathcal{l}}$
$3\varepsilon + \frac{Ldi}{dt} - 2\varepsilon = 0$
$\frac{di}{dt} = - 10A/s$

Physics


Question:
The potential energy for a force field is given by U (x, y) = sin
(x + y), the magnitude of force acting on the particle at (0, π/4) is

Solution:
[IMAGE] = - cos (x + y) [IMAGE] -
cos (x + y) [IMAGE] TOPIC: WPE
SUB TOPIC: POTENTIAL ENERGY

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
The position-time (x-t) graph of two particles moving along
x-axis are as shown
[IMAGE] The separation between them

First increases then decreases then increases then decreases then
increases

First decreases then increases then decreases

Solution:
Separation first increases then decreases then increases then
decreases then increases.

Physics


Question:
Two long conductors are arranged as shown above to form
overlapping cylinders, each of radius r,whose centers are separated by a
distance d. Current of density J flows into the plane of the page along
the shaded part of one conductor andan equal current flows out of the
plane of the page along the shaded portion of the other, as shown. What
are the magnitude and direction of themagnetic field at point A?

$\left( \frac{\mu_{0}}{2\pi} \right)\pi dJ,\ $in the +y- direction

$\left( \frac{\mu_{0}}{2\pi} \right)\frac{d^{2}}{r},$ in
the +y direction

$\left( \frac{\mu_{0}}{2\pi} \right)\frac{4d^{2}J}{r},$ in
the -y direction

$\left( \frac{\mu_{0}}{2\pi} \right)\frac{{Jr}^{2}}{d},$ in
the -y direction

Solution:
[IMAGE] Fill the over-lapping part with +J and -J current density
Using principle of super position
Magnetic field due to conductor
$B_{M}(2\pi OA) = \mu_{0}J\pi(OA)^{2}$
$B_{M} = \frac{\mu_{0}J}{2\pi}\pi(OA)$ (in y-direction)
And magnetic field due to conductor
$B_{N}\left( 2\pi O^{'}A \right) = \mu_{0}J\pi\left( O^{'}A \right)^{2}$
$B_{N} = \frac{\mu_{0}J}{2\pi}\pi\left( O^{'}A \right)$ in y-direction
So Net is given by
$B_{Net} = B_{m} + B_{N} = \frac{\mu_{0}J}{2\pi}\pi\left\lbrack OA + O^{'}A \right\rbrack$
$= \frac{\mu_{0}J\pi d}{2\pi}$ in y -direction

Physics


Question:
A polarizer - analyzer set is adjusted such that the intensity
of light coming out of the analyzer is just 10 of the original
intensity. Assuming that the polarizer-analyzer set does not absorb any
light, the angle by which the analyzer needs to be rotated further to
reduce the output intensity to be zero is

Solution:
[IMAGE] Which is 0.707
So [IMAGE] and [IMAGE] so only one
option is correct
i.e. 18.4º
Angle rotated should be [IMAGE]

Physics


Question:
Two balls A and B having masses m kg. and2 m kg. moving with
in opposite direction, collide head on.
After collision A moves with a speed of [IMAGE] in the
same direction. Then correct statementare-
 The velocity of B after collision is
[IMAGE] opposite to the direction before collision.
 The coefficient of restitution is 0.2.
 The loss of kinetic energy due to collision
 All of the above

Solution: By COLM

Physics

P-11 Physics paper-2  1Nov Jee Main.docx


Question:
A projectile of mass 3 m is projected from ground with velocity
$20\ \sqrt{2}$m/s at an angle of 45º. At highestpoint it explodes into
two pieces. One of mass 2 m and the other of mass m. Mass 2 m falls at a
distance of 100 m from point of projection towards right of point of
projection. If the distance(inm) of second mass from point of projection
where it strikes the ground is n × 10, then what is the value of n.(g=10

Solution:
Range of the projectile in the absence of explosion
The path of centre of mass of projectile will not change, i.e.,
x_COM is still 80 m, Now, from the definition of centre of mass
$x_{COM} = \frac{m_{1}x_{1} + m_{2}x_{2}}{m_{1} + m_{2}}$
or $80 = \frac{(m)\left( x_{1} \right) + (2m)(100)}{m + 2m}$
Solving this equation, we get
x_1 = 40m
Therefore, the mass m will fall at a distance x_1=40 cm from point of
projection.

Physics


Question:
In figure a wire perpendicular to a long straight wire is
moving parallel to it with a speed
v = 10 m/s in the direction of the current flowing in the wire. The
current is 10A. What is the magnitude of the potential difference (in
µV) between the ends of the moving wire?

Physics


Question:
Magnitude of Magnetic field at point due to given current
distribution is

Solution:
${\overset{\rightarrow}{B}}_{0} = \frac{\mu_{0}I}{4R}\overset{\hat{}}{j} + \frac{\mu_{0}I}{2\pi R}\overset{\hat{}}{k}$
$B = \sqrt{\left( \frac{\mu_{0}I}{4R} \right) + \left( \frac{\mu_{0}I}{4\pi R} \right)^{2}}$
$= \frac{\mu_{0}I}{4\pi R}\sqrt{\pi^{2} + 1}$

Physics

Magnetic field due to current carrying conductor


Question:
A thin rod having mass m and length 4[IMAGE] is
free to rotate about horizontal axis passing through a point distant
[IMAGE] from one of its end, as shown in figure. It is
released, from the horizontal position as shown:
[IMAGE] What will be acceleration of centre of mass at this instant?

Solution:
Torque equation
$\tau_{Hinge\ } = I_{Hinge\ } \propto$
$mg\mathcal{l =}\left( \frac{m(4\mathcal{l})^{2}}{12} + m\mathcal{l}^{2} \right) \propto$
$\frac{3g}{7\mathcal{l}} = \propto$
Tangential acceleration$= \propto \mathcal{l =}\frac{3g}{7\mathcal{l}}$
Radial acceleration$= \omega^{2}\mathcal{l} = 0$

Physics


Question:
The circuit shown below is working as a 8V dc regulated voltage
source. When 12V is used as input,the power dissipated (in mW) in each
diode is 8n. Find n. (considering both zener diodes are
identical)__________.

Solution:
$i = \frac{(12 - 8)}{(200 + 200)}A = \frac{4}{400} = 10^{- 2}A$
Power loss in each diode = (4)(10^-2 ) W = 40 mW

Physics


Question:
An impulsive force F acts horizontally on a solid sphere of
radius R placed on a horizontal surface. The line of action of the
impulsive force is at a height h above the centre of the sphere. If the
rotational and translational kinetic energies of the sphere just after
the impulse are equal, then h is equal to$\sqrt{\frac{2}{n}}R$. Find n.

Solution:
Transitional k.e.= rotational k.e.
$\frac{1}{2}{mv}^{2} = \frac{1}{2}I\omega^{2}$
$V^{2} = \frac{2}{5}\omega^{2}R^{2}$....(i)
Impulse equation
Fdt = mV....(ii)
Fhdt $= I\omega = \frac{2}{5}mR^{2}\omega$....(iii)
dividing (2) and (3)
$h = \frac{2}{5}\frac{R^{2}\omega}{v}$...(iv)
 Solving (1) and (4)
h$= \sqrt{\frac{2}{5}}R$

Physics

P-10 Physics paper 1 30 October Jee Main.docx


Question:
A steel wire is rigidly fixed at both ends. Its length, mass and
cross sectional area are 1m, 0.01 kg and
[IMAGE] respectively. Then the temperature of the wire is
lowered by[IMAGE]. If the transverse waves are setup by
plucking the wire at [IMAGE] from one end and assuming
that wire vibrates with minimum number of loops possible for such a
case. The frequency of vibration is 44n Hz. Find n. [Coefficient of
linear expansion of steel [IMAGE] and Young's

Solution:
[IMAGE] The mechanical strain [IMAGE] The tension in wire[IMAGE] speed of wave in wire
[IMAGE] Since the wire is plucked at [IMAGE] from one end
The wire shall oscillate in [IMAGE] overtone (for minimum
number of loops)
TOPIC:Waves in a string
SUB TOPIC: Wire fixed at both ends
LEVEL:Moderate

Physics


Question:
Two wires of the same length and same area of cross-section are
suspended as shown. Their Young's moduli are in the ratio 2: 3. Blocks
are in equilibrium in each arrangement.

At equilibrium strain will be in ratio 3:4

At equilibrium stress will be in ratio 2: 3

When stretched and released, they will oscillate with time periods
in the ratio 3: 2

When stretched and released, they will oscillate with time periods
in the ratio [IMAGE] leku yEckbZ,oa leku vuqizLFk dkV {kS=kQy ds nks rkj fp=k vuqlkj vkyfEcr gSA mu xq.kkadksa

Solution: [IMAGE] Ratio of time period vkorZdky

Physics


Question:
Two moles of hydrogen are mixed with n moles of helium. The root
mean square speed of gas molecules in the mixture is $\sqrt{2}$ times
the speed of sound in the mixture. Then n is.
gkbMªkstu ds nks eksyksa n eksyksa ds lkFk

Solution:
$V_{rms} = \sqrt{\frac{3RT}{M_{mix}}}$
$V_{sound\ } = \sqrt{\frac{\gamma RT}{M_{mix}}}$
$V_{rms} = \sqrt{2}V_{sound\ }$
$\sqrt{\frac{3RT}{M_{mix}}} = \sqrt{2}\sqrt{\frac{\gamma RT}{M_{mix}}}$
$\gamma_{mix} = \frac{3}{2}$
$\gamma_{mix} = \frac{n_{1}C_{P_{1}} + n_{2}C_{P_{2}}}{n_{1}C_{v_{1}} + n_{2}C_{v_{2}}}$
$\frac{3}{2} = \frac{2 \times \frac{7R}{2} + n \times \frac{5R}{2}}{2 \times \frac{5R}{2} + n \times \frac{3R}{2}} \Rightarrow \frac{3}{2} = \frac{14 + 5n}{10 + 3n} \Rightarrow 30 + 9n = 28 + 10n \Rightarrow n = 2$

Physics


Question:
An ideal gas may expands from V_0 to 2V_0 according to
following three processes. Molar specific heat for processes b will be
izlkfjr gks ldrh

positive if a is isothermal and c is adiabatic processes

negative if a is isothermal and c is adiabatic processes

positive if c is isothermal and a is adiabatic processes

negative if c is isothermal and a is adiabatic processes

Solution:
If a is isothermal and c is adiabatic processes then for b
$\Delta U$ will be negative and Q will be positive so specific heat will
be negative.
 rFkk c:)ks"e izØe gks rks bds fy, $\Delta U$_.kkRed
gksxk rFkk _.kkRed gksxhA

Physics


Question:
A composite bar of length
[IMAGE] is made up from a rod of material 1 and
of length [IMAGE] attached to a rod of material 2 and of
length [IMAGE] as shown. If
[IMAGE] and [IMAGE] ar their respective coefficients of linear
expansion, then equivalent coefficient of linear expansion for the
composite rod is:,d la;qDr NM+ dh yEckbZ L = L_1+ L_2gS tks inkFkZ 1 dh,d NM+ dh
yEckbZ L_1gS,oa inkFkZ 2 dh,d NM+ dh yEckbZ L_2gS tks fp=kkuqlkj
la;qDr NM+

Solution:
When it has been given a temperature change ∆T, the new length
of the composite bar is the sum of the individual final lengths (Let
 be required coefficient of linear expansion)
tc b kUrj ifjorZuΔT, la;qDr NM+ dh ubZ yEckbZ vfUre yEckbZ gksxk (ekuk 'α izlkj
xq.kkad gS)

Physics


Question:
A ball of mass 1 kg is at rest in position P by means of two light
strings OP and RP as shown in the figure. The string RP is now cut and
the ball swings upto a position Q. If θ = 60° and tension in the string
in position OP and OQ are T1 (Just before cutting) and T2 respectively,
then [IMAGE] is equal to

4
1 kg OP RP
 P RP Q

θ = 60° OP OQ T_1 ( )
 T_2, [IMAGE]

Solution:
T1 cos θ = mg........(i)
T2 = mg cos θ........(ii)
[IMAGE] = cos2 θ = 4
TOPIC: CIRCULAR MOTION
SUB TOPIC: DYNAMICS

Physics

P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx


Question:
Two infinite planes each with uniform surface charge density
$+ \sigma$ are kept in such a way that the anglebetween them is 30°. The
electric field in the region shown between them is given by

$\frac{\sigma}{2\epsilon_{0}}\left\lbrack (1 + \sqrt{3})\overset{\hat{}}{y} - \frac{\overset{\hat{}}{x}}{2} \right\rbrack$

$\frac{\sigma}{2\epsilon_{0}}\left\lbrack \left( 1 - \frac{\sqrt{3}}{2} \right)\overset{\hat{}}{y} - \frac{\overset{\hat{}}{x}}{2} \right\rbrack$

$\frac{\sigma}{2\epsilon_{0}}\left\lbrack (1 + \sqrt{3})\overset{\hat{}}{y} + \frac{\overset{\hat{}}{x}}{2} \right\rbrack$

$\frac{\sigma}{\epsilon_{0}}\left\lbrack \left( 1 + \frac{\sqrt{3}}{2} \right)\overset{\hat{}}{y} + \frac{\overset{\hat{}}{x}}{2} \right\rbrack$

Solution: $\overset{\rightarrow}{E} = \frac{\sigma}{2\varepsilon_{0}}\left\lbrack \left( 1 - \frac{\sqrt{3}}{2} \right)\overset{\hat{}}{y} - \frac{1}{2}\overset{\hat{}}{x} \right\rbrack$.

Physics

Electric field due to infinite sheet of charge


Question:
Two immiscible liquids are poured in a U-tube having densities
$\rho_{1} = 1.0 \times 10^{3}\ kg/m^{3}$ and
$\rho_{3} = 3.0 \times 10^{3}\ kg/m^{3}$. If the ratio of heights (of
the liquids above their interface) $\frac{h_{1}}{h_{2}}$ is x, then find
the value of x.

Solution:
$p_{0} + \rho_{1}{gh}_{1} = p_{0} + \rho_{2}{gh}_{2}$
$\Rightarrow \frac{h_{1}}{h_{2}} = \frac{\rho_{2}}{\rho_{1}} = 3$

Physics


Question:
A small 1 kg mass is launched from the top of a cliff with
speed V at an angle of 30° above the horizontal. When the mass reaches
the ground, its velocity is directed at 45° below the horizontal. Which
one of the following choices is the magnitude of the total impulse that
was imparted to the mass during its flight ? Ignore air resistance.

Solution:
V cos 30° = V1 cos 45°
V [IMAGE] = [IMAGE] V1 = [IMAGE] J = pf - pi
=

Physics


Question:
A conducting loop of area 10 cm^2 and resistance 1 W is kept
in a time varying magnetic field, perpendicular to its plane, which is
zero at t = 0. The current I induced in the loop as a function of time t
is given as I = (8t + 1) mA. The magnetic field B as a function of time
t is

Solution:
[IMAGE] = (8t + 1) ´ 10^-3 A
⇒ ε = R(8t + 1) × 10^-3
[IMAGE] = R (8t + 1) × 10^-3 and [IMAGE] ∴ [IMAGE] (8t + 1) ´ 10^-3 ⇒ [IMAGE] ⇒ [IMAGE] ⇒ [IMAGE] ⇒ [IMAGE] ⇒ [IMAGE]

Physics


Question:
The activity of a radioactive sample falls from
[IMAGE] in 30 minutes. Its half life is close to

Solution:
[IMAGE] (i) (ii)
$\Rightarrow \text{\quad}\frac{ln2}{ln(7/5)} = \frac{t_{\frac{1}{2}}}{(30\min)} \Rightarrow \text{\quad}(2.06004)30 = t_{\frac{1}{2}} = 61.8\text{min}$

Physics


Question:
The mass of an object is 0.074 kg and its volume is 290 cm^3.
Its density expressed to the correct significant figures (with rounding
off) will be

Solution:
B
= 0.2551724 (up to two significant figures)

Physics


Question:
Mass per unit area of a circular disc of radius a depends on the
distance r from its centre as [IMAGE] The moment of
inertia of the disc about the axis, perpendicular to the plane and
passing through its centre is

Physics


Question:
70 calories of heat is required to raise the temperature of 2
moles of an ideal gas at constant pressure from 30°C to 35°C (R= 2 cal
/mol-°C) The gas may be:
ls] bl gks tkrk
gs rks xSl gks ldrh

Solution: [IMAGE] So gas is

Physics


Question:
A point object in air is in front of the curved surface of a
plano-convex lens. The radius of curvature of thecurved surface is 30 cm
and the refractive index of the lens material is 1.5, then the focal
length of thelens (in cm) is_______.

Solution:
$\frac{1}{f} = (\mu - 1)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
$R_{1} = \infty$
$R_{2} = - 30cm$
$\frac{1}{f} = (1.5 - 1)\left( \frac{1}{\infty} - \frac{1}{- 30} \right)$
$\begin{aligned}
 & \frac{1}{f} = \frac{0.5}{30} \\
 & f = 60cm
\end{aligned}$

Physics


Question:
A wire of circular cross section has inner portion of radius R
made of material of resisitivity ρ and is surrounded by an outer portion
of thickness R made of a material of double resisitivity. Find the
resistance of length L of such a wire.

Solution: R = [IMAGE] = [IMAGE] =

Physics


Question:
The densities of wood and benzene at 0°C
are$\ 880{kgm}^{- 3}$and $\ 900\ {kgm}^{- 3}$, respectively. The
coefficients of volume expansion of wood is
$1.2 \times 10^{- 3^{{^\circ}}}C^{- 1}$ and that of benzene is
$1.5 \times 10^{- 3^{{^\circ}}}C^{- 1}$. Value of temperature (in °C) at
which a piece of this wood would just sink in benzene at the same
temperature is T. Then find value of [IMAGE] (where T is
temperature in °C).

Solution:
For just sink condition
$\rho_{ow\ }\left( 1 - \gamma_{w\ }\Delta T \right) = \rho_{ob\ }\left( 1 - \gamma_{b}\Delta T \right)$
$880\left( 1 - 1.2 \times 10^{3}(T - 0) \right) = 900\left( 1 - 1.5 \times 10^{- 3}(T - 0) \right)$
$T10^{- 3}(1.5 - 1.2) = 900 - 880$
$T \times 10^{- 3} \times 0.3 = 20$
$T = \frac{20 \times 10^{4}}{3}{^\circ}C$

Physics


Question: In the given circuit, value of Y is

Physics


Question:
A litre of dry air at STP expands adiabatically to a volume of 3
litres. If [IMAGE] the work done by air is:
 air to be an ideal gas]
1 yhVj gok tks fd ekud rki o nkcSTP ij gS]:)ks"e
çfØ çlkfjr gksdj 3 yhVjvk;ru

Physics


Question:
Temperature of 100 g of water in a thermoflaskremains fixed for
a pretty long time at 50ºC. Anequal mass of sand at 20ºC is poured in
the flaskand shaken for some time so that the temperatureof the mixture
is 40ºC. Now the experiment is repeated with 100 g of a liquid at 50ºC
and an equalamount of sand at 20ºC when the temperature ofthe mixture is
found to be 30ºC. The specific heat of the liquid (in
$kJ{kg}^{- 1}K^{- 1}$ ) is

Solution:
$m_{\omega}\ S_{\omega}\ \Delta T_{\omega} = m_{S}\ S_{S}\ \Delta T_{S}$
$100 \times 1 \times 10 = 100 \times S_{S} \times 20$
$\Rightarrow S_{S} = \frac{1}{2}cal/{gm}^{\circ}C$
$m_{S}\ S_{S}\ \Delta T_{S} = m_{\mathcal{l}}{\ S}_{\mathcal{l}}\ \Delta T_{\mathcal{l}}$
$100 \times \frac{1}{2} \times 10 = 100 \times {\ S}_{\mathcal{l}} \times 20$
$\Rightarrow S_{\mathcal{l}} = \frac{1}{4}cal/ºC\ gm$
$S_{\mathcal{l}} = \frac{4.2kJ}{4}/^{\circ}cgm = 1.05kJ/ºC\ gm$

Physics


Question:
A Comet is observed as it passes our solar system. Its closest
approach to the sun is r = 6.67 × 10^10 m,at that time it has velocity
v = 2 × 10^6 m/s. If G = 6.67 × 10^-11 N m^-2 kg^-2 and mass of
the sun is 2 × 10^30kg. Then which of the following is true in respect
with this comet?

The comet will never return to the solar system

Solution:
Total energy of the comet = U+K
$= \frac{- GM_{s}m}{r} + \frac{1}{2}m\left( 2 \times 10^{6} \right)^{2}$ where m mass
of the Comet
$= m\left\lbrack \left( - 2 \times 10^{9} \right) + \left( 2 \times 10^{12} \right) \right\rbrack$>0
$\therefore$Comets will never returns to the solar system because total
energy is +ve.

Physics


Question:
For a fluid which is flowing steadily, the level in the vertical
tubes is best represented by
LFkk;h:i ls izokfgr nzo

Solution:
From continuity equation, velocity at cross-section (1) is more
than that at cross-section (2).
lkarR; lehdj.k ls vuqizLFk dkV(1) ij osx ] vuqizLFk dkV(2) ij osx

Physics


Question:
A particle of mass m is dropped from a height h above the ground.
At the same time another particle of the same mass is thrown vertically
upwards from the ground with a speed of[IMAGE]. If they
collide head-on completely in elastically, the time taken for the
combined mass to reach the ground, in units of [IMAGE] is

Solution: [IMAGE] time for collision [IMAGE] After [IMAGE] And [IMAGE] at the time of collision

Physics

P-11 Physics paper-2  1Nov Jee Main.docx


Question:
The circuit shown below is working as a 8V dc regulated voltage
source. When 12V is used as input,the power dissipated (in mW) in each
diode is 8n. Find n. (considering both zener diodes are identical)
__________.
 A tc blesa 12V oksYVrk yxk;h tkrh gS rks + esa gksus okyh mW esa 8n gksxh (nksuks T+khuj
Mk;ksM,d leku gSa) __________.

Solution:
$i = \frac{(12 - 8)}{(200 + 200)}A = \frac{4}{400} = 10^{- 2}A$
Power loss in each diode gkfu = (4)(10^-2 ) W
= 40 mW

Physics

P-21 Physcis Paper 19 Dec Hindi Eng..docx


Question: The current i in the network is

Solution: Both diodes are in reverse biased

Physics


Question: In the given circuit, value of Y is

Solution:
$$
[IMAGE] = 0

Physics


Question:
A beam of electromagnetic radiation of intensity
[IMAGE] is comprised of wavelength,
[IMAGE] It falls normally on a metal (work function
[IMAGE] ) of surface area of [IMAGE]. If
one in [IMAGE] photons ejects an electron, total number
of electrons ejected in[IMAGE] is
[IMAGE] then x is....

Solution:
Energy of photon. [IMAGE] (so photoelectric
effect will take place)
[IMAGE] No. of photons falling per second
[IMAGE] No. of photoelectron emitted per second

Physics


Question:
An electron starting from rest and accelerated through V =
20,000 V in an x-ray tube loses half the energy it acquired as it
approaches an atom of the target material. The remaining KE is lost in
subsequent interactions with other target atoms. Find the wavelength of
the radiation emitted in the first interaction.

Solution: ∆E = 1 × 104 eV = [IMAGE] λ = [IMAGE] nm = 0.124 nm

Physics


Question:
A circular platform is free to rotate in a horizontal plane about
a vertical axis passing through its centre. A tortoise is sitting at the
edge of the platform. Now the platform is given an angular velocity
$\omega_{0}$. When the tortoise moves along a chord of the platform with
a constant velocity (with respect to the platform) the angular velocity
of the platform $\omega(t)$ will vary with time t as

Solution:
since, there is no external torque, angular momentum will
remain conserved. The moment of inertia will first decrease till the
tortoise moves from A to C and then increase as it moves from C and D.
The re fore $\omega$ will initially increase and then decrease. Let R be
the radius of platform m the mass of disc and M is the mass of platform.
Moment of inertia when the tortoise is at A
and moment of inertia when the tortoise is at B
$I_{2} = {mr}^{2} + \frac{{MR}^{2}}{2}$
here $r^{2} = a^{2} + {\lbrack\sqrt{R^{2} - a^{2}} - vt\rbrack}^{2}$
From conservation of angular momentum
$\omega_{0}I_{1} = \omega(t)I_{2}$
substituting the values we can see that variation of $\omega(t)$ is
nonlinear.

Physics


Question:
A vessel contains two immiscible liquids of densities
[IMAGE]. A solid block of volume
[IMAGE] and density [IMAGE] is tied to one end of a string and other
is tied to the bottom of the vessel as shown in figure. The block is
immersed with [IMAGE] of its volume in the liquid of higher density and
[IMAGE] in the liquid of lower density. The entire system is kept in an elevator
which is moving upwards with an acceleration of a = g/2. Find the
tension in the string. (Take g = 10 m
[IMAGE] ) (in N),d ik=k esa nks ρ_1 = 1000 kg
m^-3rFkk ρ_2 = 1500 kg m^-3 Hkjs gSaV = 10^-3 m^3vk;ru rFkk d
= 800 kg m^-3?kuRo dk,d Bksl CykWd jLlh ca
dk nqljk fljk fp=kkuqlkj ik=k ds ry ij ca
 $\frac{2^{th}}{5}$Hkkx mPp ?kuRo nzo esa rFkk$\frac{3^{th}}{5}$Hkkx
fuEu ?kuRo ds nzo esa Mqck gSA lEiw.kZ Roj.k

Solution:
[IMAGE] We analysed this problem From the reference frame of Elevator. Total
buoyant force On the block,
CykWd ij dqy mRIykou cy
[IMAGE] For equilibrium,

Physics


Question:
A particle of mass m is released from point A and it stops at B
find out the coefficient of friction

Solution:
$W_{g} + W_{f} = K_{f} + K_{i}$
$mg\left( \frac{3h}{4} \right) - \mu mg3h = 0 - 0 \Rightarrow \mu = \frac{1}{4}$

Physics


Question:
Four functions given below describe motion of a particle. (I) y
= sin ωt - cos ωt, (II) y = sin3 wt, (III) y = 5 cos
[IMAGE], (IV) y = 1 + ωt + ω2t2. Therefore, simple
harmonic motion is represented by

Solution:
I and III are showing S.H.M. on the basis of superposition
principle.

Physics


Question:
Find the moment of inertia of a uniform rectangular plate of
mass M, edges of length
 and about itsaxis passing through
centre and perpendicular to

Solution:
Using perpendicular axis theorem l_3 = I_1 + I_2
$I_{2} = \frac{M\mathcal{l}^{2}}{12}$
$I_{3} = \frac{M\left( \mathcal{l}^{2} + b^{2} \right)}{12}$
$I_{3} = \frac{M\left( \mathcal{l}^{2} + b^{2} \right)}{12}$

Physics


Question:
The radius of a coil of wire with 21 turns is 0.22 m, and i_2
current flows clockwise in the coil as shown. A long straight wire
carrying a current is toward the left is located 0.05 m from the edge of
the coil. Themagnetic field at the centre of the coil is zero tesla. The
ratio $\frac{i_{1}}{9i_{2}}$ is (Use$\pi = 22/7$)

Solution:
$\frac{\mu_{0} \cdot i_{1}}{2\pi(d + r)} = \frac{N \times \mu_{0}i_{2}}{2r}$
$\Rightarrow N = \frac{i_{1}}{i_{2}} \cdot \frac{r}{\pi(d + r)}$
$\frac{i_{1}}{i_{2}} = 81$

Physics

Magnetic field due to current carrying conducting


Question:
A uniform solid sphere of radius R and mass M purely rolls down
an inclined plane. The coefficient of friction between the sphere and
the inclined plane is[IMAGE]. If the maximum value of
angle of inclination for this to be possible is
[IMAGE] then n is: (Take[IMAGE] )

Solution:
[IMAGE] put in (1)
[IMAGE] Now [IMAGE] TOPIC:Rotaional dynamics
SUB TOPIC: Rolling down the incline
LEVEL: Moderate

Physics


Question:
A small sphere of mass 1 kg is moving with a velocity
$(6\widehat{i} + \widehat{j})ms^{- 1}$. It hits a fixed smooth wall and
rebound with velocity $(4\widehat{i} + \widehat{j})ms^{- 1}$. The
coefficient of restitution between the sphere and the wall is

Solution:
Impulse = Change in momentum
$= 1(4\widehat{i} + \widehat{j}) - 1(6\widehat{i} + \widehat{j}) = - 2\widehat{i}$

Which is perpendicular to the wall. Componentof initial velocity along
$\widehat{i} = 6\widehat{i}$
$\Rightarrow$ Speed of approach =6
Therefore $e = \frac{4}{6} = \frac{2}{3}$

Physics


Question:
The distance between two slits in a Young's double slit
experiment is 3 mm. The distance of the screenfrom the slits is 1 m.
Waves of wavelength 1 mm are incident on the plane of the slits
normally.The distance of the first maxima on the screen from the central
maxima will be: (in cm)

Solution:
At first maxima d sinθ = λ
$3sin\theta = 1 \Rightarrow sin\theta = \frac{1}{3}$
$\tan\theta = \frac{y}{D} \Rightarrow y = D \times \tan\theta = 1 \times \frac{1}{\sqrt{8}}m = \frac{100}{\sqrt{8}}cm = 35.4cm$

Physics


Question:
Two infinite planes each with uniform surface charge density
$+ \sigma$ are kept in such a way that the anglebetween them is 30°. The
electric field in the region shown between them is given by

$\frac{\sigma}{2\epsilon_{0}}\left\lbrack (1 + \sqrt{3})\overset{\hat{}}{y} - \frac{\overset{\hat{}}{x}}{2} \right\rbrack$

$\frac{\sigma}{2\epsilon_{0}}\left\lbrack \left( 1 - \frac{\sqrt{3}}{2} \right)\overset{\hat{}}{y} - \frac{\overset{\hat{}}{x}}{2} \right\rbrack$

$\frac{\sigma}{2\epsilon_{0}}\left\lbrack (1 + \sqrt{3})\overset{\hat{}}{y} + \frac{\overset{\hat{}}{x}}{2} \right\rbrack$

$\frac{\sigma}{\epsilon_{0}}\left\lbrack \left( 1 + \frac{\sqrt{3}}{2} \right)\overset{\hat{}}{y} + \frac{\overset{\hat{}}{x}}{2} \right\rbrack$

Solution: $\overset{\rightarrow}{E} = \frac{\sigma}{2\varepsilon_{0}}\left\lbrack \left( 1 - \frac{\sqrt{3}}{2} \right)\overset{\hat{}}{y} - \frac{1}{2}\overset{\hat{}}{x} \right\rbrack$.

Physics

Electric field due to infinite sheet of charge


Question:
The densities of wood and benzene at 0 ºC are 880 kg m^-3 and
900 kg m^-3, respectively. The coefficientsof volume expansion of wood
is 1.2 × 10^-3 ºC^-1.and that of benzene is
1.5 × 10^-3 ºC^-1. Value oftemperature (in ºC) at which a piece of
this wood would just sink in benzene at the same temperature isT. Then
find value of 3×10^-4T (where T is tempture in ºC).
ydM+h rFkk csathu dk ?kuRo 0ºC ij Øe'k 880 kg m^-3rFkk 900 kg
m^-3gSA ydM+h rFkk csathu izlkj xq.kkad Øe'k 1.2 × 10^-3
C^-1rFkk 1.5 × 10^-3 ºC^-1gSA ftl rkieku ij ydM+h csathu esa
iw.kZ:Ik ls Mqcrh gS rks og rkieku T (ºC esa) gSA rc 3×10^-4T

Solution:
For just sink condition
Mqcus

Physics


Question:
Two blocks of masses 2 kg and 4 kg are connected through a massless
inextensible string. The
co-efficient of friction between 2 kg block and ground is 0.4 and the
coefficient of friction between 4 kg block and ground is 0.6. Two forces
F1 = 10 N and F2 = 20 N are applied on the blocks as shown in the
figure. Calculate the frictional force ( in N )between 4 kg block and
ground (Assume initially the tension in the string was just zero before
forces F1 and F2 were applied)
[IMAGE] 2 kg 4 kg 2 kg
 0.4 4 kg
0.6 F1 = 10 N F2 = 20 N
4 kg (, F1 F2 )

Solution:
System will not move (Equilibrium)
For equilibrium of 2 kg block
[IMAGE] 10 = 8 + T
⇒ T = 2N
For equilibrium of 4 kg block
[IMAGE] T + f2 = 20
⇒ f2 = 18 N
TOPIC: NLM
SUB TOPIC: FRICTION

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
A ball is thrown horizontal from the top of a tower 40 m high. The
ball strikes the ground at a point 80 m from the bottom of the tower.
Find the angle that the velocity vector makes with the horizontal just
before the ball hits the ground.

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
If the length of steel rod is 5cm longer than copper rod at all
temperatures. ($\alpha$ for copper $= 1.7 \times 10^{- 5}ºC^{- 1}$ and
$\alpha$ for steel $\left. \ = 1.1 \times 10^{- 5}C^{- 1} \right)$

Length of steel rod is 14.17cm and length of copper rod is 9.17cm

Length of steel rod is 15.17cm and length of copper rod is 6.17cm

Length of steel rod is 14.17cm and length of copper rod is 10.17cm

Length of steel rod is 15.17cm and length of copper rod is 9.17cm

Solution:
$\mathcal{l}_{1}\left( 1 + \alpha_{1}\Delta T \right) - \mathcal{l}_{2}\left( 1 + \alpha_{2}\Delta T \right) = \mathcal{l}_{1} - \mathcal{l}_{2}$
$\mathcal{l}_{1}\alpha_{1}\Delta T = \mathcal{l}_{2}\alpha_{2}\Delta T$
$\mathcal{l}_{1}\alpha_{1} = \mathcal{l}_{2}\alpha_{2}$
$\Rightarrow \mathcal{l}_{1} \times 1.7 = \mathcal{l}_{2} \times 1.1$
$17\mathcal{l}_{1} = 11\mathcal{l}_{2}$
As $\left( \mathcal{l}_{2} - \mathcal{l}_{1} = 5 \right)$
$\Rightarrow 17\mathcal{l}_{2} - 11\mathcal{l}_{2} = 5 \times 17$
$6\mathcal{l}_{2} = 85$
$\Rightarrow \mathcal{l}_{2} = \frac{85}{6} = 14.17cm$
$\mathcal{l}_{1} = 9.17cm$

Physics


Question:
The surface tension of a liquid is 5 Newton per meter. If a film
of this liquid is held on a ring of area $0.02m^{2}$, if its surface
energy is about $\left( n \times 10^{- 1} \right)$ Joule then n is

Solution:
We know that surface energy
$U_{s} = T \times \ Area\ $
Here, as 2 films are formed because of ring.
So $U_{s} = T \times 2 \times (A)$
$= 5\frac{N}{m} \times 2 \times 0.02m_{2} = 0.2J$

Physics


Question:
32 g of [IMAGE] is contained in a cubical
container of side 1 m and maintained at a temperature of
[IMAGE]. The isothermal bulk modulus of elasticity of
the gas is [IMAGE]. Then the value of K is:[IMAGE] 32 s 1m Hkqtk ds ?kukdkj crZu esa 127°C A rks K

Solution: Bulk modulus of elasticity at constant temperature ij

Physics


Question:
If the dipole moment vector[IMAGE] of a dipole
and the electric field[IMAGE] (uniform) are oriented as
shown then which of the following statement is incorrect?

potential energy in I = potential energy in II

potential energy in I = negative of potential energy in II

Solution:
[IMAGE] where α is always the shorter angle
between[IMAGE] and [IMAGE] ∴ in both cases I and II, α = π − θ ⇒ (cos α)_I = (cos α)_II
∴ U_I = U_II ⇒ (B) is wrong.
(C) and (D) are correct as direction of [IMAGE]

Physics


Question:
The minimum time in which the potential energy of a particle
executing SHM changes from maximum to minimum is 5 s. Then the time
period of SHM is

Solution:
P.E. is maximum at extreme position and minimum at mean
position.
Time to go from extreme position to mean position is,
 where T is time period of SHM.

Physics


Question:
The speed of a projectile at the highest point is $v_{1}$ and at
the point half the maximum height is $v_{2}$. If
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{2}{5}}$, then the angle of projection
is π/n. Find n.

Solution:
$0 = {(v_{2}sin\beta)}^{2} - 2g\frac{H}{2}$
$v_{1}^{2} = v_{1}^{2} - gh$
Also $\frac{v_{1}}{v_{2}} = \sqrt{\frac{2}{5}}$
From above
$v_{1} = \sqrt{\frac{2gH}{3}}$
$v_{2} = \sqrt{\frac{5gH}{3}}$
$H = \frac{u^{2}\sin^{2}\theta}{2g} \Rightarrow sin2\theta = \frac{2gH\cos^{2}\theta}{v_{1}^{2}}$
$\tan^{2}\theta = \frac{2gH}{v_{1}^{2}}{or\ tan}^{2}\theta = \frac{2gH \times 3}{2gH}$
$tan\theta = \sqrt{3}$
$\theta = 60^{\circ}$

Physics


Question:
A block of mass 60 kg is released from rest when compression in
the spring is 2 m (natural length of spring is 8 m). Surface AB is
smooth while surface BC is rough. Block travels x distance before coming
to complete rest. Value of x is: $\lbrack g = 10m/s^{2}\rbrack$

Solution:
By Work Energy
Theorem$W_{net} = \Delta K.E = 0 \Rightarrow \frac{1}{2}k(x_{0}^{2} - x^{2}) - \mu mgx = 0$
$\Rightarrow \frac{1}{2} \times 200(2^{2} - x^{2}) = \frac{1}{2} \times 60 \times 10x$
$\Rightarrow x = 1m$
Also at this moment $f_{\max} > kx$
So, block will not move so total distance travelled =2+1=3 m.

Physics


Question:
The diagram below shows the lowest four energy levels for an
electron in a hypothetical atom. The electron is excited to the -1eV
level of the atom and transitions to the lowest energy state by emitting
only two photons. Which of the following energies could not belong to
either of the photons ?

Solution: 3eV to -7eV is not possible.

Physics


Question:
An electron (mass m) with initial velocity [IMAGE] is in an electric field [IMAGE] If [IMAGE] is initial de-Broglie wavelength of electron, its de-Broglie wave length
at time t is given by

Solution: Initially [IMAGE] Velocity as a function of time [IMAGE] so wavelength [IMAGE]

Physics


Question:
An impulsive force F acts horizontally on a solid sphere of
radius R placed on a horizontal surface. Theline of action of the
impulsive force is at a height h above the centre of the sphere. If the
rotational andtranslational kinetic energies of the sphere just after
the impulse are equal, then h is equal to

Solution:
Transitional k.e.= rotational k.e.
$\frac{1}{2}{mv}^{2} = \frac{1}{2}I\omega^{2}$
$V^{2} = \frac{2}{5}\omega^{2}R^{2}$...(1)
Impulse eq. Fdt = mV...(2)
$\ Fhdt\ = I\omega = \frac{2}{5}mR^{2}\omega$...(3)
Dividing (2) and (3)$h = \frac{2}{5}\frac{R^{2}\omega}{v}$...(4)
Solving (1) and (4)$h = \sqrt{\frac{2}{5}}R$

Physics


Question:
Two liquids of densities 1 are mixed in equal
volumes. A cube of external dimension 20 cm has an inner cubical portion
of side 10 cm whose density is twice that of the outer portion. If this
cube is just floating in the liquid mixture, find the density of the
inner portion.

Solution:
ρ = [IMAGE] = 2 mg = 2ρ1 × 103 + ρ1 × (203 - 103)
= 9ρ1 × 103 g = ρ × 8 × 103 g

Physics


Question:
Consider a mixture of n moles of helium gas and 2 n moles of
oxygen gas (molecules taken to be rigid) as an ideal gas. Its
[IMAGE] value will be:
ds 2n eksYl dh feJ.k xSl eku s rks bl feJ.k

Physics


Question:
Find out velocity of block B in the given figure: (all the
pulleys are ideal and string are inextensible massless)

Solution:
$\mathcal{l}_{1}^{'} + \mathcal{l}_{2}^{'} = 0$
$x = 4$`<!- ->`0
$\mathcal{l}_{3}^{'} + \mathcal{l}_{4}^{'} = 0$
$40 - y + 40 - 80 = 0$
$y = 0$

Physics


Question:
In the figure shown switch [IMAGE] remains
connected and switch S2 remains open for a long time. Now
[IMAGE] is also closed. Assuming
[IMAGE] and [IMAGE]. The magnitude of
rate of change of current is 2n A/s in inductor just after the switch S2
is closed. Find n

Solution:
[IMAGE] Using [IMAGE] TOPIC: Electromagnetic induction
SUB TOPIC: Self induction
LEVEL: Moderate

Physics


Question:
A uniform elastic rod of cross-section area A, natural length L
and Young's modulus Y is placed on asmooth horizontal surface. Now two
horizontal forces (of magnitude F and 3F) directed along the lengthof
rod and in opposite direction act at two of its ends as shown. After the
rod has acquired steady state,the extension of the rod will be.

Solution:
Tension in rod at a distance x from right edge is
$T = F\left( 3 - 2\frac{x}{L} \right)$
[IMAGE] $\therefore$net extension in
rod $= \int_{0}^{L}\mspace{2mu}\frac{T}{YA}dx = \frac{2F}{YA}L$

Physics

Elongation in rod due to unbalanced force


Question:
In the fig. shown a cart moves on a smooth horizontal surface
due to an external constant force of magnitude F. The initial mass of
the cart is $M_{0}$and velocity is zero. Sand falls on to the cart with
negligible velocity at constant rate$\mu\ kg/s$ and sticks to the cart.
The velocity of the cart at time t is

Solution:
[IMAGE] Formula $F = m\frac{dv}{dt} + (V - u)\frac{dm}{dt}$
Here u=velocity of sand =0
$m = M_{o} - \mu t = mass\ at\ time\ t$
and$\frac{dm}{dt} = \mu$
$\therefore F = (M_{0} + \mu t)\frac{dm}{dt} + v\mu$
$(F - \mu v)dt = (M_{o} + \mu t)dv$
$\int_{0}^{t}\frac{dt}{M_{0} + \mu t} = \int_{0}^{v}\frac{dv}{F - \mu v}$
${\lbrack log(M_{0} + \mu t)\rbrack}_{0}^{t} = \frac{1}{\mu}\lbrack log(F - \mu v)\rbrack_{0}^{v}$
$log\frac{(M_{0} + \mu t)}{M_{0}} = log(\frac{F}{F - \mu v})$
$F - \mu v = \frac{M_{o}F}{M_{0} + \mu t} \Rightarrow v = \frac{Ft}{M_{0} + \mu t}$

Physics


Question:
Power delivered to a particle varies with time as P = (3t2 - 2t)
watt, where is in seconds, then change in its kinetic energy
between time t = 1 to t = 3s.
 P = (3t2 - 2t),, t = 1 t = 3s

Solution:
[IMAGE] W =[IMAGE] = ∆ K.E.
TOPIC: WPE
SUB TOPIC: POWER

Physics

P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx


Question:
A simple open U-tube contains mercury. Now water is poured
slowly upto height 27.2 cm in the left arm. How high (in cm) does the
mercury rise in the right arm from its initial level in equilibrium
state?
(Take density of mercury $= 13600\ kg/m^{3}$ and that of water
$= 1000\ kg/m^{3}$and $g\ = 10\ m/s^{2}$)
 27.2cm rd Hkjk fLFkfr
Hkqtk es a blds çkjfEHkd Lrj ls ikjk fdruh (cm esa) rd tk;sxkA
ikjs dk ?kuRo $= 13600\frac{kg}{m^{3}}$ rFkk ikuh

Solution:
[IMAGE] Pressure at A that is at interface of water and mercury and B is same as
both are at same horizontal level in same liquid. Thus
Aij vFkkZr~ ikuh dh vkUrfjd lrg rFkk ikjs ij rFkkBij nkc leku gksxk
D;ksafd nksauks leku nzo esa leku kjk çkIr
$= \frac{x}{2} = 1\ cm$

Physics


Question:
The efficiency $\eta$ of the cyclic process
$\left( \gamma = \frac{5}{3} \right)$ is given by $\frac{50n}{23}$ then
find out the value of n.
pØh; izØe $\eta\frac{50n}{23}$gS rksn

Solution:
$\eta = \frac{W}{+ Q}$
Temperature at A is T_0 then$P_{0}V_{0} = {nRT}_{0}$
 T_0gS rks$P_{0}V_{0} = {nRT}_{0}$
at$T_{A} = T_{0},T_{B} = 2T_{0},T_{C} = 6T_{0},T_{D} = 3T_{0}$
$W = 2P_{0}V_{0}$
$+ Q = Q_{AB} + Q_{BC}$
$= n \cdot C_{v}dT + nC_{p}dT$
$= \frac{3}{2}{nRT}_{0} + \frac{5}{2}nR\left( 4T_{0} \right)$
$= \frac{23}{2}{nRT}_{0} = \frac{23}{2}P_{0}V_{0}$
$\eta = \frac{2P_{0}V_{0}}{\frac{23}{2}P_{0}V_{0}} = \frac{4}{23}$

Physics


Question:
32 g of O_2 is contained in a cubical container of side 1 m and
maintained at a temperature of 127ºC. The isothermal bulk modulus of
elasticity of the gas in terms of universal gas constant R is

Solution:
Bulk modulus of elasticity at constant temperature
$B = \frac{dP}{dV/V} - = - V\left( \frac{dP}{dV} \right)_{T} = + P = \frac{nRT}{v}$
$= \frac{1 \times R \times 400}{1^{3}} = 400R$

Physics


Question:
A fiber sheet with relative permittivity 5 is inserted between
two parallel metal plates 0.25 cm apart. A potential difference of 2500
v is applied between the plates. If dielectric strength of air is
$3 \times 10^{6}v/m$ than

Potential gradient strength in fiber sheet exceeds dielectric
strength of air

Solution:
As $V = V_{1} + V_{2}$
$2500 = 0.0002E_{1} + 0.0023E_{2}$
$2E_{1} + 23E_{2} = 25 \times 10^{6}$
Also $E_{1} = 5E_{2}$
Solving $E_{2} = 0757 \times 10^{8}V/m$
As $E_{1} > 3.6 \times 10^{6}$ Hence Air will breakdown

Physics


Question:
Two particle of mass m each are fixed to a massless rod of
length [IMAGE]. The rod is smoothly hinged at one end to
a ceiling. It performs oscillation of small angle in vertical plane. The
length of the equivalent simple pendulum is

Physics


Question:
A shot is fired at an angle θ to the horizontal such that it strikes
the hill while moving horizontally Find initial angle of projection θ.

Solution:
[IMAGE] sin θ =[IMAGE] 4 sin θ = 8 sin θ - 6 cos θ
4 sin θ = 6 cos θ ⇒ tan θ =
TOPIC:KINEMATIC
SUB TOPIC: 2 D

Physics

P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx


Question:
An electric field
$\overset{\rightarrow}{E} = 4 \times \overset{\hat{}}{i} - (y^{2} + 1)jN/C$
passes through the box shown in figure. The flux of the electric field
through surface ABCD and BCGF and marked as $\phi_{I}$ and $\phi_{\Pi}$
respectively. The difference between
$(\phi_{I} - \phi_{\Pi})is\ (in{Nm}^{2/C})$

Solution:
Flux via ABCD
$\phi_{1} = \int\overset{\rightarrow}{E} \cdot d\overset{\rightarrow}{A} = 0$
Flux via BCEF
$\phi_{2} = \int\overset{\rightarrow}{E} \cdot d\overset{\rightarrow}{A} = 0$
$\phi_{2} = \overset{\rightarrow}{E} \cdot \overset{\rightarrow}{A} = (4 \times \overset{\hat{}}{i} - (y^{2} + 1)\overset{\hat{}}{j}) \cdot 4\overset{\hat{}}{i}$
$= 16x,x = 3$
$\phi_{2} = 48\frac{N - m^{2}}{C};\phi_{1} - \phi_{2} = - 48\frac{N - m^{2}}{C}$

Physics


Question:
A point moves according to the law, x = at and y = at (1- αt),
where a and α are positive constants and t is time. Find the moment at
which angle between velocity vector and acceleration vector

Not possible
 x = at y = at (1- αt) a α
 t
 [IMAGE]

Solution:
x = at ⇒ vx = a ⇒ ax = 0
y = at - aαt2 ⇒ vy =[IMAGE] = a - 2αat
ay =[IMAGE] = - 2αa
[IMAGE] = av cos θ
= (2αa) [IMAGE] Solve to get, [IMAGE] TOPIC:KINEMATIC
SUB TOPIC: ACCERATION

Physics

P-04 Paper 2 (Physics) Eng.+ Hindi 24-8-2020.docx


Question:
The force exerted by a compression device is given by
$F(x) = kx(x - \lambda)$ for $0 \leq x \leq \lambda$, where $\lambda$
is the maximum possible compression, x is the compression and k is a
constant. The work required to compress the device by a distance d will
be maximum when

Solution:
For W to be maximum
i.e. $F(x) = 0 \Rightarrow x = \lambda,x = 0$
Clearly for $d = \lambda$,the work done is maximum.

Physics


Question:
A sinusoidal travelling wave in a string has a velocity of
propagation 300 m/sec. The time period of oscillations of the particles
of the string is 0.04 sec. The phase difference between the oscillations
of two points at distances 10 m and 16 m respectively from source of
oscillation is nπ. Find n.

Physics


Question:
Three bars having lengths [IMAGE] and
[IMAGE] and area of cross section A, A/2, 2A are joined
rigidly end toend. The compound rod is subjected to a stretching force
F. It young's modules of the material of eachrod is Y then the increase
in length of the rod is.
 yEckbZ dh o vuqizLFk dkV
{ks=kQyA,[IMAGE], 2Adh rhu NM+s n`<+:i ls fljs ls fljs
NM+ xq.kkadY

Solution:
$\ \Delta l = \Delta l_{1} + \Delta l_{2} + \Delta l_{3}$
$\frac{F}{\frac{YA}{\mathcal{l}}} + \frac{F}{\frac{YA}{4\mathcal{l}}} + \frac{F}{\frac{2YA}{3\mathcal{l}}} = \frac{F\mathcal{l}}{YA}\left( 1 + 4 + \frac{3}{2} \right) = \frac{13F\mathcal{l}}{2YA}$

Physics


Question:
A non-viscous ideal fluid is flowing vertically downwards in a
pipe. The area of cross-section at section is double that at
section Vertical distance between section A and section B is h
and the height of water column in tube (2) is more than that in tube

by $\frac{h}{2}$ distance. If the velocity(m/s) of the fluid at section
A is x, then find the value of x. (if h = 2.7 m)

Solution:
Applying Bernoulli's equation between section A and B
$P_{A} + \frac{1}{2}\rho v_{A}^{2} + \rho{gh}_{A} = P_{B} + \frac{1}{2}\rho v_{B}^{2} + \rho{gh}_{B}$
$P_{A} + \frac{1}{2}\rho v^{2} + \rho gh = P_{B} + \frac{1}{2}\rho(2v)^{2} + 0$
$P_{B} - P_{A} = \rho gh - \frac{3}{2}\rho v^{2}\ $
and $P_{B} - P_{A} = \rho g\frac{h}{2}$
$\rho g\frac{h}{2} = \rho gh - \frac{3}{2}\rho v^{2}$
$\Rightarrow v = \sqrt{\frac{gh}{3}} = 3$

Physics


Question:
In the arrangement shown in figure, wavelength of light used is
[IMAGE]. The distance between slits
[IMAGE] The distance between [IMAGE] and
[IMAGE] is [IMAGE] If the ratio of
maximum to minimum intensity received on screen P is K. Then K is

Solution:
If intensity of light incident on slits [IMAGE] and
The intensity through [IMAGE] will be
[IMAGE] The intensity through [IMAGE] will be
[IMAGE] TOPIC:Wave Optics
SUB TOPIC:YDSE
LEVEL:Moderate

Physics


Question: The emf of cell E_2 is

Solution:
$i = \frac{2}{15 + 10}$
$V_{AB} = i10 = \frac{20}{25}$
$E_{2} = \frac{V_{AB}}{50} \times 31.25$
$= \frac{20}{25} \times \frac{31.25}{50}$
=0.5V

Physics


Question:
The ratio of r.m.s. speed to the r.ms. angular speed of a
diatomic gas at certain temperature is:(assume m= mass of one molecule
mass I = moment of inertia of the molecules)
ewy ekfu, m =,d v.kq
vk.kfod nzO;eku I = v.kqvksa

Solution:
$\frac{1}{2}{mV}^{2} = \frac{3}{2}kT$
$\frac{1}{2}I\omega^{2} = \frac{2}{2}kT$
$\frac{V}{\omega} = \sqrt{\frac{3I}{2m}}$

Physics


Question:
In a fluorescent lamp choke (a small transformer) 100 V of
reverse voltage is produced when the choke current changes uniformly
from 0.25 A to 0 in a duration of 0.025 ms. The self-inductance of the
choke(in mH) is estimated to be 2.5n. Find n.

Solution:
$100 = \frac{L(0.25)}{0.025} \times 10^{3}$
$\therefore L = 100 \times 10^{- 4}H$
$= 10mH$

Physics


Question:
In the circuit, if the forward voltage drop for the diode is
[IMAGE] the current in the circuit will be (approximately)

Physics


Question:
The frequency of incident light falling on photo sensitive
substance is doubled. If E_1 is the kinetic energy of emitted electron
in first case and E_2 is the kinetic energy in second case, then

Solution:
. [IMAGE] Dividing both equations,
[IMAGE] Hence, E_2 > 2E_1

Physics


Question:
A train of mass M is moving on a circular track of radius ' R
' with constant speed V. The length of the train is half of the
perimeter of the track. The linear momentum of the train will be.

Solution:
If we treat the train as a ring of mass then its COM will
be at a distance$\frac{2R}{\pi}$from the centre of thecircle. Velocity
of centre of mass is:
$V_{CM} = R_{CM} \cdot \omega$
$= \frac{2R}{\pi} \cdot \omega$
$= \frac{2R}{\pi} \cdot \left( \frac{V}{R} \right)$
$\left( \because\omega = \frac{V}{R} \right)$
$\Rightarrow V_{CM} = \frac{2V}{\pi} \Rightarrow MV_{CM} = \frac{2MV}{\pi}$
As the linear momentum of any system = MV_CM
[IMAGE] The linear momentum of the train
$= \frac{2MV}{\pi}$.

Physics

P-10 Physics paper 1 30 October Jee Main.docx


Question:
In the circuit, if the forward voltage drop for the diode is
0.5V, the current in the circuit will be(approximately) -
ifjiFk fy, vxzfoHko iru 0.5V gS] rc ifjiFk

Physics

P-21 Physcis Paper 19 Dec Hindi Eng..docx


Question:
A rod of length L has non-uniform linear mass density given by
[IMAGE] where a and b are constants
and[IMAGE]. The value of x for the centre of mass of the
rod is at

Solution: $x_{cm} = \frac{1}{M}\int_{0}^{l}x \cdot dM$

Physics

P-11 Physics paper-2  1Nov Jee Main.docx


Question:
A non-viscous ideal fluid is flowing vertically downwards in a
pipe. The area of cross-section at section is double that at
section Vertical distance between section A and section B is h
and the height of water column in tube (2) is more than that in tube

by [IMAGE] distance. If the velocity(m/s) of the
fluid at section A is x, then find the value of x. (if h = 2.7 m),d nzo uyh esa uhps dh rjQ çokfgr gSAHkkx A ij
vuqçLFk dkV
chp nwjh h

Solution:
Applying Bernoulli's equation between section A and B
ArFkk B

Physics


Question:
Radiation, with wavelength falls on a metal surface to
produce photoelectrons. The electronsare made to enter a uniform
magnetic field of $3 \times 10^{- 4}$T. If the radius of the largest
circular pathfollowed by the electrons is 10 mm, the work function of
the metal is close to

Solution:
${KE}_{\max} = E - \phi$
$= \frac{12400}{\lambda(inÅ)} - \phi$ (in eV)
$r = \frac{\sqrt{2mKE}}{eB}$
$KE_{\max} = \frac{r^{2}e^{2}B^{2}}{2m}$ (in J)
$= \frac{r^{2}eB^{2}}{2m}$ (in eV)

Physics


Question:
Three particles each of mass m can slide on fixed frictionless
horizontal circular tracks in the same horizontal plane as shown in the
figure. The coefficient of restitution being e=0.5. Assuming that
m_2and m_3 are at rest initially and lie along a radial line before
impact and the spring is initially unstretched, then maximum extension
in spring in subsequent motion.

Solution:
Velocity just after collision
$V_{0} = V_{1} + V_{2}$.....(1)
$\frac{1}{2} = \frac{- V_{1} + V_{2}}{V_{0}}$
$- V_{1} + V_{2} = \frac{V_{0}}{2}$.....(2)
From (1) & (2)
$\Rightarrow V_{1} = \frac{V_{0}}{4}\& V_{2} = \frac{3V_{0}}{4}$
When maximum extension occurs then angular speed with rest to centre
for mass m_2& m_3 are same. Using angular momentum conservation
about centre of circle.
$m\frac{3}{4}V_{0}(2R) + 0 = {mR}^{2}\omega + m(2R)^{2}\omega$
$\Rightarrow \omega = \frac{3}{10}\frac{V_{0}}{R}$
So velocity of$m_{2} = \frac{3}{5}V_{0}$
& velocity of$m_{3} = \frac{3}{10}V_{0}$
When extension is maximum using energy conservation.
$\frac{1}{2}m{(\frac{3}{4}V_{0})}^{2} = \frac{1}{2}m{(\frac{3}{5}V_{0})}^{2} + \frac{1}{2}m{(\frac{3}{10}V_{0})}^{2} + \frac{1}{2}kx_{max.}^{2} \Rightarrow x_{\max} = \frac{3}{4}V_{0}\sqrt{\frac{m}{5k}}$

Physics


Question:
A container is filled with a liquid of density ρ. Container is
accelerating on a horizontal surface with acceleration $\frac{3g}{4}$
towards right hand side and liquid is at rest with respect to container
as shown in figure. If there is a point A in liquid then, which of the
following is correct: (assume atmospheric pressure is zero and AB is
vertical and AC is horizontal line)

Pressure at A is $\rho g\mathcal{l}_{1}$

Pressure at A is $\rho\frac{3g}{4}\mathcal{l}_{2}$

$\theta = \tan^{- 1}\left( \frac{3}{4} \right)$

Solution:
for point A from top surface
$P_{A} = \rho g \cdot \mathcal{l}_{1}$and from right wall
$P_{A} = \frac{\rho^{3}g}{4} \cdot \mathcal{l}_{2}$
$\mathcal{l}_{1} = \frac{3}{4}\mathcal{l}_{2}$ and

Physics


Question:
[IMAGE] gas follows the process,
[IMAGE] constant. If then the ratio of work done by gas
to change in itsinternal energy is [IMAGE] then n is:
vuqikr [IMAGE] gks rks n

Physics


Question:
A slightly tapering wire of length[IMAGE] and end
radii is subjected to tension F. Y is the
Young'sModulus. The extension produced in the wire is.,d FkksM+s Vsifjx rkj ftl rFkk fljks dh
f= ij F ruko
rkj

Physics

Extension in a wire of non-uniform cross-section


Question:
A small circular loop of conducting wire has radius a and
carries current I. It is placed in a uniform magnetic field B
perpendicular to its plane such that when rotated slightly about its
diameter and released, it starts performing simple harmonic motion of
time period T. If the mass of the loop is $\\mathrm{m}\$ then

Solution:
$\tau = MBsin\theta = I\alpha$
$\pi R^{2}IB\theta = \frac{{mR}^{2}}{2}\alpha$.
$\omega = \sqrt{\frac{2\pi IB}{m}} = \frac{2\pi}{T};T = \sqrt{\frac{2\pi m}{IB}}$

Physics

Current carrying conductor in magnetic field


Question:
In the fig. shown a cart moves on a smooth horizontal surface
due to an external constant force of magnitude F. The initial mass of
the cart is M_0and velocity is zero. Sand falls on to the cart with
negligible velocity at constant rate and
sticks to the cart. The velocity of the cart at time t is

Solution:
[IMAGE] Formula $F = m\frac{dv}{dt} + (V - u)\frac{dm}{dt}$
Here, u = velocity of s and = O
m = M_o[IMAGE] t = mass at time t
and [IMAGE] $\therefore$ $F = \left( M_{0} + \mu t \right)\frac{dv}{dt} + v\mu$
$(F - \mu v)dt = \left( M_{0} + \mu t \right)dv$
$\int_{0}^{t}\mspace{2mu}\frac{dt}{M_{0} + \mu t} = \int_{0}^{v}\mspace{2mu}\frac{dv}{F - \mu v}$
$\frac{1}{\mu}\left\lbrack \mathcal{l}n\left( M_{0} + \mu t \right) \right\rbrack_{0}^{t} = - \frac{1}{\mu}\mathcal{\lbrack l}n(F - \mu v)\rbrack_{0}^{v}$
$\mathcal{l}n\frac{\left( M_{0} + \mu t \right)}{M_{0}}\mathcal{= l}n\left\{ \frac{F}{F - \mu v} \right)$
$F - \mu v = \frac{M_{0}F}{M_{0} + \mu t} \Rightarrow v = \frac{Ft}{M_{0} + \mu t}$

Physics

P-10 Physics paper 1 30 October Jee Main.docx


Question:
Three particles each of mass m can slide on fixed frictionless
horizontal circular tracks in the same horizontal plane as shown in the
figure. The coefficient of restitution being e=0.5. Assuming that
m_2and m_3 are at rest initially and lie along a radial line before
impact and the spring is initially unstretched, then maximum extension
in spring in subsequent motion.

Solution:
Velocity just after collision
$V_{0} = V_{1} + V_{2}$.....(1)
$\frac{1}{2} = \frac{- V_{1} + V_{2}}{V_{0}}$
$- V_{1} + V_{2} = \frac{V_{0}}{2}$.....(2)
From (1) & (2)
$\Rightarrow V_{1} = \frac{V_{0}}{4}\& V_{2} = \frac{3V_{0}}{4}$
When maximum extension occurs then angular speed with rest to centre
for mass m_2& m_3 are same. Using angular momentum conservation
about centre of circle.
$m\frac{3}{4}V_{0}(2R) + 0 = {mR}^{2}\omega + m(2R)^{2}\omega$
$\omega = \frac{3}{10}\frac{V_{0}}{R}$
So velocity of$m_{2} = \frac{3}{5}V_{0}$ & velocity
of$m_{3} = \frac{3}{10}V_{0}$
When extension is maximum using energy conservation.
$\frac{1}{2}m{(\frac{3}{4}V_{0})}^{2} = \frac{1}{2}m{(\frac{3}{5}V_{0})}^{2} + \frac{1}{2}m{(\frac{3}{10}V_{0})}^{2} + \frac{1}{2}kx_{max.}^{2} \Rightarrow x_{\max} = \frac{3}{4}V_{0}\sqrt{\frac{m}{5k}}$

Physics


Question:
If the projectile hits the inclined plane perpendicularly when
thrown horizontally with $v_{0}$, from a tower of height H as shown then
the value of $\frac{4gH}{{(v_{0})}^{2}}$ is 2k. Find k.

Solution:
$a_{y} = gcos30^{\circ}$
$u_{x} = V_{0}cos30^{\circ}$
$u_{y} = V_{0}sin30^{\circ}$
$V_{x} = u_{x} + a_{x}t$
$0 = \frac{V_{0}\sqrt{3}}{2} - \frac{g}{2}t$
$t = \frac{v_{0}\sqrt{3}}{g}$....(i)
$S_{y} = u_{y}t + a_{y}t^{2}$
$(Hcos30^{\circ} - 0) = V_{0}sin30^{\circ}t + \frac{1}{2}gcos30^{\circ}t^{2}$....(ii)
From (i) (ii)
$\Rightarrow \frac{4gH}{{(v_{0})}^{2}} = 10$

Physics


Question:
An ideal gas obey the law [IMAGE] constant. x is
value for which it has zero molar specific heat at normal temperature.
Then find 5 x. ([IMAGE] for the gas is 2.5 R)
gS ftlds fy, ij bl j fof'k"B A rc 5x
dk eku Kkr djksa ¼xSl

Solution:
For a gas at normal temperature, ij xSl ds fy,[IMAGE] Specific heat for a polytropic process
i‚ayhVªksfid çØe ds fy, fof'k"B
[IMAGE] According to question iz"ukuqlkj C=0

Physics


Question:
A uniform disc of radius R lies in the x-y plane, with its
centre at origin. its moment of inertia about zaxis is equal to its
moment of inertia about line $y = x + c$. The value of c will be

Solution:
$I_{PQR} = \frac{1}{4}{MR}^{2} + M{(\frac{C}{\sqrt{2}})}^{2}$
But $I_{PQR} = \frac{1}{2}{MR}^{2}$
$\therefore C = \pm \frac{R}{\sqrt{2}}$
Hence (2) is correct.

Physics


Question:
When the temperature of copper coin is raised by 80ºC, its
diameter increases by 0.2.

Percentage rise in the area of a face is 0.4

Coefficient of linear expansion of copper is
$0.25 \times 10^{- 4}C^{- 1}$

Solution:
$100 \times \frac{\Delta D}{D} = \alpha\Delta T \times 100$
$100 \times \left( \frac{\Delta A}{A} \right) = \beta\Delta T \times 100$
$= 2\alpha\Delta T \times 100$ $= 2(0.2)$ $= 0.4\%$
$100 \times \frac{\Delta t}{t} = \alpha\Delta T \times 100$
$\alpha = \frac{\frac{\Delta d}{d}}{\Delta T} = \frac{0.2 \times 10^{- 2}}{80}\left( \frac{\Delta d}{d} \times 100 = 0.2 \right)$
$\alpha = 0.25 \times 10^{- 4}{}^{\circ}C^{- 1}$
$100 \times \frac{\Delta V}{V} = \gamma\Delta T \times 100$
$= 3(0.2) = 0.6\%$

Physics


Question:
The maximum peak to peak voltage of an AM wave is 24 mV and the
minimum peak to peak voltage is 8 mV. The modulation factor is

Physics


Question:
[IMAGE] As shown in the figure, a battery of emf$\varepsilon$ is connected to
an inductor L and resistance R in series. The switch is closed at t=0.
The total charge that flows from the battery, between t=0 and
$t = t_{c}$ ($t_{c}$is the time constant of the circuit) is

Solution:
$q = \int_{0}^{T_{C}}idt$
$= \frac{\varepsilon}{R}{\lbrack t - \frac{e^{- t/T_{C}}}{\frac{- 1}{T_{C}}}\rbrack}_{0}^{T_{C}}$;
$= \frac{\varepsilon}{R}\lbrack T_{C} + T_{C}e^{- 1} - T_{C}\rbrack$
$= \frac{\varepsilon}{R} \times \frac{1}{e} \times \frac{L}{R}$
$= \frac{\varepsilon L}{R^{2}e}$

Physics


Question:
The primary purpose of using a parabolic mirror in a reflecting
telescope is

To collect more light from a distant galaxy

Physics


Question:
An ideal gas at a pressure of 1 atmosphere andtemperature of 27°C
is compressed adiabaticallyuntil its pressure becomes 8 times the
initialpressure. Then the final temperature is [IMAGE]

Solution: Here, [IMAGE] As changes are adiabatic,

Physics


Question:
A steel wire is rigidly fixed at both ends. Its length, mass and
cross sectional area are 1m, 0.01 kg and
[IMAGE] respectively. Then the temperature of the
wire is lowered by[IMAGE]. If the transverse waves are setup by
plucking the wire at [IMAGE] from one end and assuming that wire
vibrates with minimum number of loops possible for such a case. The
frequency of vibration is 44n Hz. Find n. [Coefficient of linear
expansion of steel [IMAGE] and Young's

Solution:
[IMAGE] The mechanical strain [IMAGE] The tension in wire[IMAGE] speed of wave in wire
[IMAGE] Since the wire is plucked at
[IMAGE] from one end
The wire shall oscillate in
[IMAGE] overtone (for minimum number of loops)
TOPIC:Waves in a string
SUB TOPIC: Wire fixed at both ends
LEVEL:Moderate

Physics


Question:
A system is shown in the figure. Block A is moving with 1 m /s
towards left. Wedge is moving with 1 m /s towards right. Then speed of
the block B will be

Solution:
Velocity of block A w.r.t. wedge is 2 m So we have
${\overset{\rightarrow}{V}}_{BW} = {\overset{\rightarrow}{V}}_{B} - {\overset{\rightarrow}{V}}_{W} \Rightarrow {\overset{\rightarrow}{V}}_{B} = {\overset{\rightarrow}{V}}_{BW} + {\overset{\rightarrow}{V}}_{W}$
So
$V_{B} = \sqrt{{(V_{W})}^{2} + {(V_{BW})}^{2} + 2V_{W} \times V_{BW}(cos120^{\circ})}$
$= \sqrt{1^{2} + 2^{2} + 2 \times 1 \times 2 \times ( - 1/2)} = \sqrt{3}m/s$

Physics


Question:
[IMAGE] A parallel plate capacitor has plates of area A separated by distance
 between them. It is filled with adielectric which has a
dielectric constant that varies as k(x) = K(1 + $\alpha$x) where
is the distancemeasured from one of the plates. If ($\alpha$d) << 1,
the total capacitance of the system is best given by theexpression:
lekukUrj IysVksa ls cus,d la
muds chp dh nwjh A bu IysVksa ds chp,d ijkoS|qr inkFkZ Hkjk
gqvk gS ftldk ijk oS|qrkad k(x) = K(1+ αx)
mi;qDr eku gksxk

$\frac{AK \in_{0}}{d}\left( 1 + \frac{\alpha d}{2} \right)$

$\frac{A \in_{0}K}{d}\left( 1 + \left( \frac{\alpha d}{2} \right)^{2} \right)$

$\frac{AK\epsilon_{0}}{d}(1 + \alpha d)$

Solution:
Capacitance of element ?kVd Capacitance of element ?kVd
$\sum\frac{1}{C^{'}} = \int_{0}^{d}\mspace{2mu}\frac{dx}{K\varepsilon_{0}A(1 + \alpha x)}$
$\frac{1}{C} = \frac{1}{K\varepsilon_{0}A\alpha}\mathcal{l}n(1 + \alpha d)$
Given αd << 1
$\frac{1}{C} = \frac{1}{K\varepsilon_{0}A\alpha}\left( \alpha d - \frac{\alpha^{2}d^{2}}{2} \right)$
$\frac{1}{C} = \frac{d}{K\varepsilon_{0}A}\left( 1 - \frac{\alpha d}{2} \right)$
$C = \frac{K\varepsilon_{0}A}{d}\left( 1 + \frac{\alpha d}{2} \right)$

Physics

P-21 Physcis Paper 19 Dec Hindi Eng..docx


Question:
A ball is thrown vertically upwards from the ground. It crosses a
point at the height of 25 m. twice at an interval of 4 secs. The ball
was thrown with the velocity of

Solution:
Let initial velocity of the ball = u
$\therefore 25 = ut - \frac{1}{2}gt^{2} = ut - 5t^{2}$
$\Rightarrow 5t^{2} - ut + 25 = 0$
If t_1and t_2 are the two solutions,
[IMAGE] Then, $t_{2} - t_{1} = \sqrt{{(t_{2} + t_{1})}^{2} - 4t_{1}t_{2}}$
$\Rightarrow 4^{2} = \frac{u^{2}}{25} - 4 \times \frac{25}{5}$
$\Rightarrow 16 = \frac{u^{2}}{25} - 20 \Rightarrow u^{2} = 25 \times 36$
$\Rightarrow u = 30m/s$

Physics


Question:
A diatomic ideal gas is heated at constant volume until the
pressure is doubled and again heated at constant pressure until volume
is doubled. Find out the average molar heat capacity for whole process.
in K)
nkc nqxuk ugha gks tk;s
rd ml nqxuk ugha gks

Solution:
Let initial pressure, volume, temperature be P_0, V_0, T_0
indicated by state A in P-V diagram. The gas is then isochorically taken
to state B (2P_0, V_0, 2T_0) and then taken from state B to state C
(2P_0, V_0, 4T_0) isobarically
vxj bl:vkrh s P_0, V_0, T_0ekus tSlk
fd uhps P-V fp=k bls fQj levk;rfud izfØ B (2P_0, V_0,
2T_0) vkSj fQj Bls C (2P_0, V_0, 4T_0) (2P_0, V_0, 4T_0) rd
[IMAGE] Total heat absorbed by 1 mole of gas (by definition of molar heat
capacity)
dqy tks 1 eksy xSl }kjk xzg.k dh xbZ &.
$\Delta Q = C_{v}\left( 2T_{0} - T_{0} \right) + C_{p}\left( 4T_{0} - 2T_{0} \right)$
$= \frac{5}{2}RT_{0} + \frac{7}{2}R \times 2T_{0} = \frac{19}{2}{RT}_{0}$
Total change in temperature from state A to C is
 dqy ifjorZu AvoLFkk ls Crd gS &
$\Delta T = 3T_{0}$
Molar heat capacityeksyj m"ek

Physics


Question:
The figure gives experimentally measured B vs. H variation in a
ferromagnetic material. The retentivity, coercivity and saturation,
respectively, of the material are

Solution:
[IMAGE] x = retentivity
y = coercivity
z = saturation magnetization

Physics


Question:
When the temperature of copper coin is raised by 80oC, its
diameter increases by 0.2.

percentage rise in the area of a face is 0.4

coefficient of linear expansion of copper is
$0.25 \times 10^{- 4}C^{- 1}$

All of the above
tc rkacs ds flD eku 80ºC c<

Solution:
$100 \times \frac{\Delta D}{D} = \alpha\Delta T \times 100$
$100 \times \left( \frac{\Delta A}{A} \right) = \beta\Delta T \times 100$
$= 2\alpha\Delta T \times 100$
$= 2(0.2)$
$= 0.4\%$
$100 \times \frac{\Delta t}{t} = \alpha\Delta T \times 100$
$\alpha = \frac{\frac{\Delta d}{d}}{\Delta T} = \frac{0.2 \times 10^{- 2}}{80}\left( \frac{\Delta d}{d} \times 100 = 0.2 \right)$
$\alpha = 0.25 \times 10^{- 4}\ ^{\circ}C^{- 1}$
$100 \times \frac{\Delta V}{V} = \gamma\Delta T \times 100$
$= 3(0.2) = 0.6\%$

Physics


Question:
A simple seconds pendulum is constructed out of a very thin
string of thermal coefficient of linear expansion
[IMAGE] and a heavy particle attached to one end. The free end of the string is
suspended from the ceiling of an elevator at rest. The pendulum keeps
correct time at 0° C. When the temperature rises to 50° C, the elevator
operator of mass 60 kg being a student of Physics accelerates the
elevator vertically, to have the pendulum correct time. The apparent
weight (in N) of the operator when the pendulum keeps correct time at 50
°C is 330x. Find x. [IMAGE],d lSd.M ljy yksyd cgqr iryh NM+ ls cuk gS] ftldk çlkj
xq.kkad α= 20 x
0ºC ij lgh Zr tc rki 50 ºC c<+rk gS rks 60kg
 fy¶V pkyd tks HkkSfrd foKku dk Nk=k gS] yksyd lgh
 fy, fy¶V dks Rofjr tc yksyd 50 ºC ij lgh
le; ç gks rks fy¶V pkyd dk vkHkklh Hkkj330x ( )
gksrk

Solution:
[IMAGE] when the temperature is raised length changes to l(1 + α ∆T)
tc rki c<+ rc yEckbZ esa ifjorZu l(1 + α ΔT) gks A
when the lift acceleration upwards
tc fy¶V dh vksj Rofjr gSA
g_eff = g + a
new period of pendulum,
yksyd

Physics


Question:
A uniform solid cylinder of mass M = 2 kg and radius R = 10 cm
is connected about an axis through the centre of the cylinder to a
horizontal spring with spring constant 4 N/m. The cylinder is pulled
back, stretching the spring 1 m from equilibrium. When released, the
cylinder rolls without slipping. What is the speed of the center of the
cylinder when it returns to equilibrium ?

Solution: wsp = ∆k

Physics


Question:
When a capillary tube is dipped in a liquid, the liquid rises to
a height h in the tube. The free liquidsurface inside the tube is
hemispherical in shape. The tube is now pushed down so that the height
ofthe tube outside the liquid is less than h.

The liquid will come out of the tube like in a small fountain.

The liquid will come out of the tube slowly.

The liquid will fill the tube but not come out of its upper end.

The free liquid surface inside the tube will not be hemispherical.
tc,d s nzo uyh esa nzo h rd
uhps rkfd nzo ds ckgj uyh

Solution:
The angle of contact at the free liquid surface inside the
capillary tube will change such that the verticalcomponent of the
surface tension forces just balance the weight of the liquid column.
 vUnj nzo dh eqDr lrg ij lEidZ dks.k bl çdkj cny tk,xk fd
i`"B ruko cy dk ?kVd nzo LrEHk

Physics


Question:
The variation of internal energy vs density of a diatomic gas
is as shown in figure. The molar heat capacity of the gas is

Solution:
According to graph
TV = constant
or PV^2 = constant

Physics


Question:
For an adiabatic process graph between PV&V for a sample of
ideal gas will be

Solution:
$PV\ \alpha\ T$
 for adiabatic process,
${TV}^{\gamma - 1} = \ $constant

Physics


Question:
A thin uniform rod of mass is hinged at one
end. This rod is maintained in horizontal position by colliding very
tiny balls each of mass m/10 completely elastically 10 times per sec
striking at the opposite end as shown in figure. Find the speed of the
ball - (g = 10 m/sec^2)

Solution:
[IMAGE] Taking torque about hinge:
$MgL/2 = FL$
$mg\frac{L}{2} = 2(\frac{mV}{10})10L$
$V = 2.5m/sec$

Physics


Question:
A parallel beam of light of wavelength [IMAGE] is
incident on slit S. Another two slit
distance 2 m as shown in the figure. The minimum distance between
intensity will occur at O is.

Solution: [IMAGE], when n = 1

Physics


Question:
Mass per unit area of a circular disc of radius a depends on the
distance r from its centre as [IMAGE] The moment of
inertia of the disc about the axis, perpendicular to the plane and
passing through its centre is

Physics

P-11 Physics paper-2  1Nov Jee Main.docx


Question:
In the shown figure a ray of light enters the face XY of a right
angered prism at grazing incident. It emerges from the adjacent face. YZ
at an angle[IMAGE]. If x is the critical angle of the
prism held in water of refractive index [IMAGE], then the
angle [IMAGE] with normal is

Solution:
[IMAGE] As at XY
[IMAGE] At YZ
[IMAGE] On solving

Physics


Question:
System shown in figure is in equilibrium. The magnitude of change
in tension in the string just before and just after, when one of the
spring is cut. Mass of both the blocks is same and equal to m and
springconstant of both springs is k. (Neglect any effect of rotation)

Solution:
Before cutting the spring
After cutting the spring
[IMAGE] 2mg - mg = 2 ma
a = g/2
$T_{3} = mg/2$
$T_{2} - T_{3} = mg - mg/2 = mg/2$

Physics


Question:
A point object in air is in front of the curved surface of a
plano-convex lens. The radius of curvature of the curved surface is 30
cm and the refractive index of the lens material is 1.5, then the focal
length of the lens (in cm) is10n. Find n.

Physics


Question:
In the circuit, if the forward voltage drop for the diode is
[IMAGE] the current in the circuit will be
(approximately)

Physics


Question:
In the given A.C. circuit, if battery voltage
is[IMAGE], power delivered by battery will have a power
factor of[IMAGE]. Find n.

Solution:
[IMAGE] Phasor diagram
[IMAGE] TOPIC: Alternating current
SUB TOPIC: LCR circuit
LEVEL: Moderate

Physics


Question:
the frequency of the light used in an experiment on
photoelectric effect. If the work function is $\frac{x}{42}$ (in eV)
then value of x is. $\left( h = 6.4 \times 10^{- 34}Js \right)$

Solution:
$V = \frac{1.6}{4 \times 10^{14}}\left( v - 2 \times 10^{14} \right) = \frac{1.6}{4 \times 10^{14}}v - 0.8$
Compare it with $V = \frac{h}{e}v - \frac{\phi}{e}$
$\frac{\phi}{e} = 0.8$
$\phi = 0.8eV$

Physics


Question:
Take two infinite planes with same surface charge density. Place
them as shown it the figure. The electric field in the region between
them is given by

$\frac{\sigma}{2\epsilon_{0}}\lbrack(1 + \sqrt{3})\overset{\hat{}}{y} - \frac{\overset{\hat{}}{x}}{2}\rbrack$

$\frac{\sigma}{2\epsilon_{0}}\lbrack(1 - \frac{\sqrt{3}}{2})\overset{\hat{}}{y} - \frac{\overset{\hat{}}{x}}{2}\rbrack$

$\frac{\sigma}{2\epsilon_{0}}\lbrack(1 + \sqrt{3})\overset{\hat{}}{y} + \frac{\overset{\hat{}}{x}}{2}\rbrack$

$\frac{\sigma}{\epsilon_{0}}\lbrack(1 + \frac{\sqrt{3}}{2})\overset{\hat{}}{y} + \frac{\overset{\hat{}}{x}}{2}\rbrack$

Solution: $\overset{\rightarrow}{E} = \frac{\sigma}{2\varepsilon_{0}}\lbrack(1 - \frac{\sqrt{3}}{2})\overset{\hat{}}{y} - \frac{1}{2}\overset{\hat{}}{x}\rbrack$

Physics

Electric field due to infinite sheet of charge


Question:
In the figure $m_{A} = m_{B} = m_{C} = 60kg$. The co-efficient
of friction between C and ground is 0.5, B and ground is 0.3, A& B is
0.4. C is pulling the string with the maximum possible force without
moving. Then tension in the string connected to A will be

Solution:
Maximum frictional force between C and ground =300 Nt
Max. frictional force between B and ground =360 Nt
So man is unable to pull B Hence T=

Physics


Question:
One mole of an ideal gas is kept enclosed under a light piston
(area=[IMAGE] ) connected by acompressed spring (spring
constant 100 N / m). The volume of gas is [IMAGE] and
its temperature is 100K. The gas is heated so that it compresses the
spring further by 0.1 m. Then the work done by the gas in the process
is: (in Joule) (Take R =8.3 J / K -mole and suppose there is no
atmosphere).,d 100 N / m½
ls tqM+s gYds fiLVu ¼kjk
 ¼R =8.3 J ekusa fd ok;qe.My ugha

Solution:
Before heating let the pressure of gas be
[IMAGE] from the equilibrium piston.
 -
[IMAGE] Since during heating process, pwafd xeZ izfØ nkSjku,
The spring is compressed further by 0.1 m fLizax 0.1 m

Physics


Question:
A river is flowing with a speed of $1\ kmh^{- 1}$. A swimmer
wants to go to point c starting from A. He swims with a speed of
$5\ kmh^{- 1}$ at an angle $\theta$w.r.t. the river flow, at what angle
with the river bank should the swimmer swim?

Solution:
Path of swimmer relative to ground is a straight line at 45º
with bank, therefore, x and y components of swimmer's resultant
velocity must be equal.
Let velocity of swimmer$|{\overset{\rightarrow}{v}}_{m}| = v$
$V_{x} = V_{r} + V_{m,r}cos\theta$
$V_{y} = V_{m,r}sin\theta$
[IMAGE] Condition for reaching the point C,
$tan45^{\circ} = \frac{v_{y}}{v_{x}}$
$V_{y} = V_{x}$.
$(V_{r} + V_{m,r}cos\theta) = V_{m,r}sin\theta$
$1 + 5cos\theta = 5sin\theta$
On squaring$1 + 25\cos^{2}\theta + 10cos\theta = 25 - 25\cos^{2}\theta$
$50\cos^{2}\theta + 10cos\theta - 24 = 0$
On solving, we get$\theta = 53^{\circ}$

Physics


Question:
You have a ring balanced at the center of the table by three
forces ${\overset{\rightarrow}{F}}_{1},{\overset{\rightarrow}{F}}_{2}$
and ${\overset{\rightarrow}{F}}_{3}$. The forces
${\overset{\rightarrow}{F}}_{1}$ and ${\overset{\rightarrow}{F}}_{2}$
have
components.$F_{1x} = 10N,F_{1y} = - 50N,F_{2x} = - 40N,F_{2y} = 100N$
The force ${\overset{\rightarrow}{F}}_{3}$ is required to make the
ringstationary at the center of the table. Which one ofthe 4 vectors
in the diagram could represent${\overset{\rightarrow}{\ F}}_{3}$?

Solution:
${\overset{\rightarrow}{F}}_{1} + {\overset{\rightarrow}{F}}_{2} + {\overset{\rightarrow}{F}}_{3} = 0 \Rightarrow 10\overset{\hat{}}{i} - 50\overset{\hat{}}{j} - 40\overset{\hat{}}{j} + {\overset{\rightarrow}{F}}_{3} = 0$
$\Rightarrow {\overset{\rightarrow}{F}}_{3} = 30\overset{\hat{}}{i} - 50\overset{\hat{}}{j} \Rightarrow IVth\ quadrant.$

Physics


Question:
If the velocity of light denoted by acceleration due to
gravity are taken as fundamental
units, then the dimensions of length will be

Solution:
Let $L\alpha c^{a}g^{b}p^{c}$
$L = kc^{a}g^{b}p^{c}$
Putting dimensions on both sides:
$\left\lbrack M^{0}LT^{0} \right\rbrack = \left\lbrack LT^{- 1} \right\rbrack^{a}\left\lbrack LT^{- 2} \right\rbrack^{b}\left\lbrack ML^{- 1}T^{- 2} \right\rbrack^{c}$
$\left\lbrack M^{0}LT^{0} \right\rbrack = \left\lbrack M^{c}L^{a + b - c}T^{- a - 2b - 2c} \right\rbrack$
Comparing dimensions of both sides
$c = 0,\ \ \ \ \ a + b - c = 1$
$- a - 2b - 2c = 0$
solving we get $a = 2,b = - 1$
hence$\lbrack L\rbrack = \left\lbrack c^{2}/g \right\rbrack$

Physics


Question:
In the given circuit diagram, a wire is joining points B and D.
The current in this wire is

Solution:
$i = \frac{20}{2} = 10A$
$1 = \frac{41}{5} - \frac{3i}{5}$
$= + \frac{i}{5} = 2A$

Physics


Question:
Figure is the plot of the stopping potential versus the
frequency of the light used in an experiment on photoelectric effect. If
the work function is $\frac{x}{5}($ in
[IMAGE] then value of x is.

Physics


Question:
A uniform magnetic field B in positive z direction exists in a
circular region of radius R = 5 m. A loop of radius R = 5m lying in x -
y plane encloses the magnetic field at t = 0 and then pulled at uniform. The emf induced (in volts) is the
loop at t = 2 sec is 6V. Then magnitude of 4B is

Solution:
$= VB2\sqrt{R^{2} - \left( \frac{vt}{2} \right)^{2}} = VB\sqrt{4R^{2} - V^{2}t^{2}}$
$6V = 4 \times B\sqrt{4 \times 25 - 16 \times 4}$
$B = 0.25T$

Physics


Question:
An electron of mass m and magnitude of charge |e| initially at
rest gets accelerated by a constant electric field E. The rate of change
of de-Broglie wavelength of this electron at time t ignoring
relativistic effects is

Physics


Question:
A particle is projected up an inclined plane of inclination
$\beta$ at an elevation $\alpha$ to the horizontal. If the particle
strikes the inclined plane at right angles then $\tan\alpha =$

Solution:
$T = \frac{2usin(\alpha - \beta)}{gcos\beta}$......(1)
$O = ucos(\alpha - \beta) - (gsin\beta)T$
$\Rightarrow T = \frac{ucos(\alpha - \beta)}{gsin\beta}$......(2)
from (1) &(2)
$2tan(\alpha - \beta) = cot\beta$
$\Rightarrow tan\alpha = (cot\beta + 2tan\beta)$

Physics


Question:
A string of length L of variable density is clamped between two
rigid supports. Whose density variation is shown in figure, ρ = density
[IMAGE] Find the time after which the disturbance created at one end will reach
other end.
(A = area of cross section of string in [IMAGE], T =
tension in the string is constant in N)

Physics


Question:
For a particle moving in a straight line, the acceleration
versus displacement graph is as shown. The relation between which of the
following quantities cannot be a straight line?

Solution:
D
α = kx
or Force = ma = kmx
∴ (C) is a straight line.
Again [IMAGE] ∴ (A) is a straight line.
∴ (B) is a straight line.
or [IMAGE] (D) is a parabola.
∴

Physics


Question:
In an α -decay the Kinetic energy of α particle is
[IMAGE] and Q -value of the reaction
is[IMAGE]. If the mass number of the mother nucleus is
20 N. Find [IMAGE]: (Assume that daughter nucleus is in
ground state)

Physics


Question:
In the show circuit diagram all capacitors are initially
uncharged and cells are ideal. If C_1 = 1 $\mu F$,C_2 = 2 $\mu F$ and
potential of earth is taken zero. Choose the correct options.
 izkjEHk rFkk
 A;fn C_1 = 1 $\mu F$,C_2 = 2 $\mu F$ rFkk i`Foh

Potential of point A is (4/3) volts
fcUnq A

Potential of point B is (-13/3) volts
fcUnq B

Potential of point C is (10/3) volts
fcUnq C

Potential of point A is (-4/3) volts
fcUnq Adk foHko (-4/3) oksYV gSA

Solution:
$V_{C} = 5 - \frac{5}{3} = \frac{10}{3}$V
$\frac{10}{3} - V_{B} = 8$
$V_{B} = \frac{- 14}{3}$V
$\frac{- 14}{3} - V_{A} = \frac{10}{3}$
$V_{A} = \frac{- 24}{3} = - 8$V

Physics

Electrical potential in an electric circuit

P-21 Physcis Paper 19 Dec Hindi Eng..docx


Question:
If an object is placed at 20 cm in front of a half thin convex
lens of focal length 10 cm as shown in figure then taking object
position as shown the.

Solution:
$\frac{1}{v} + \frac{1}{- 20} = \frac{1}{10} \Rightarrow v = 20$
$m = - 1$

Physics


Question:
A force F is given by: F = at + bt^2, where t represents time.
What are dimensions of a and b?

Solution:
$\lbrack F\rbrack = \lbrack at\rbrack;\ \lbrack a\rbrack = \left\lbrack \frac{F}{t} \right\rbrack = \left\lbrack \frac{{MLT}^{- 2}}{T} \right\rbrack = \left\lbrack {MLT}^{- 3} \right\rbrack$
$$\lbrack F\rbrack = \left\lbrack \lbrack b\rbrack = \left\lbrack \frac{F}{t^{2}} \right\rbrack = \left\lbrack \frac{{MLT}^{- 2}}{T^{2}} \right\rbrack = \left\lbrack {MLT}^{- 4} \right\rbrack$$

Physics


Question:
Two identical wires A and B, each of length $\mathcal{l}$, carry
the same current I. Wire A is bent into acircle of radius R and wire B
is bent form a squareof side a. If B_Aand B_B are the values of
magnetic field at the centres of the circle and square respectively,
then the ratio B_A/ B_B is$\pi^{2}:n\sqrt{2}.\ $What is the value of
n.

Solution:
B at centre of a circle $= \frac{\mu_{0}I}{2R}$
B at centre of a
square $= 4 \times \frac{\mu I}{4\pi \cdot \frac{\mathcal{l}}{2}}\left( sin45^{\circ} + sin45^{\circ} \right)$
$= 4\sqrt{2}\frac{\mu_{0}I}{2\pi\mathcal{l}}$
 Now, $R = \frac{L}{2\pi}\ $and $\mathcal{l =}\frac{L}{4}(\ as\ L = 2\pi R = 4\mathcal{l)}$
 Where, L= length of wire
$\therefore{\ \ \ B}_{A} = \frac{\mu_{0}L}{2 \cdot \frac{L}{2\pi}} = \frac{\pi\mu_{0}I}{L} = \pi\left( \frac{\mu_{0}I}{L} \right)$
$B_{B} = 4\sqrt{2}\frac{\mu_{0}I}{2\pi\left( \frac{L}{4} \right)} = \frac{8\sqrt{2}\mu_{0}I}{\pi L} = \frac{8\sqrt{2}}{\pi}\left( \frac{\mu_{0}I}{L} \right)$
$\therefore{\ \ \ \ B}_{A}:B_{B} = \pi^{2}:8\sqrt{2}$

Physics

Magnetic field due to current carrying conductor


Question:
A very long wire ABDMNDC is shown in figure carrying current I.
AB and BC parts are straight, longand at right angle. At D wire forms a
circular turn DMND of radius R. AB, BC parts are tangential tocircular
turn at N and D. Magnetic field at the centre of circle is

Solution:
$\frac{\mu_{0}i}{4\pi R}\left\lbrack \sin 90^{\circ} - \sin 45^{\circ} \right\rbrack \otimes + \frac{\mu_{0}i}{2R} \odot + \frac{\mu_{0}i}{4\pi R}\left( \sin 45^{\circ} + \sin 90^{\circ} \right) \odot$
$= \frac{- \mu_{0}i}{4\pi R}\left\lbrack 1 - \frac{1}{\sqrt{2}} \right\rbrack + \frac{\mu_{0}j}{2R} + \frac{\mu_{0}i}{4\pi R}\left\lbrack \frac{1}{\sqrt{2}} + 1 \right\rbrack \odot$
$= \frac{\mu_{0}i}{4\pi R}\left\lbrack - 1 + \frac{1}{\sqrt{2}} + 2\pi + \frac{1}{\sqrt{2}} + 1 \right\rbrack \odot$
$= \frac{\mu_{0}i}{4\pi R}\lbrack\sqrt{2} + 2\pi\rbrack \odot$
$= \frac{\mu_{0}i}{2\pi R}\left\lbrack \frac{1}{\sqrt{2}} + \pi \right\rbrack \odot$

Physics

Magnetic field due to current carrying conductor


Question:
If the magnetic field in a plane electromagnetic wave is given
by
$\overset{\rightarrow}{B} = 3 \times 10^{- 8}sin\left( 1.6 \times 10^{3}x + 48 \times 10^{10}t \right)\overset{\hat{}}{j}\ T$,
then what will be expression for electric field?

$\overset{\rightarrow}{E} = \left( 3 \times 10^{- 8}sin\left( 1.6 \times 10^{3}x + 48 \times 10^{10}t \right)\overset{\hat{}}{j}V/m \right)$

$\overset{\rightarrow}{E} = \left( 3 \times 10^{- 8}sin\left( 1.6 \times 10^{3}x + 48 \times 10^{10}t \right)\overset{\hat{}}{i}V/m \right)$

$\overset{\rightarrow}{E} = \left( 60sin\left( 1.6 \times 10^{3}x + 48 \times 10^{10}t \right)\overset{\hat{}}{k}V/m \right)$

$\overset{\rightarrow}{E} = \left( 9sin\left( 1.6 \times 10^{3}x + 48 \times 10^{10}t \right)\overset{\hat{}}{k}V/m \right)$

Solution:
$\frac{E_{0}}{B_{0}} = C$(speed of light in vacuum)
$E_{0} = B_{0}C = 3 \times 10^{- 8} \times 3 \times 10^{8}$
= 9 N/C
So, $E = 9\sin\left( 1.6 \times 10^{3}x + 48 \times 10^{10}t \right)$

Physics


Question:
The potential energy of a particle varies with x according to
the relation $U(x) = x^{2} - 4x$. The point x=2 is a point of

Solution:
$U(x) = x^{2} - 4x$
$F = 0$
$\frac{dU(x)}{dx} = 0$
$2x - 4 = 0\ gives\ \ \ x = 2$
$\frac{d^{2}U}{{dx}^{2}} = 2 > 0$ i.e. U is minimum hence x = 2 is a
point of stable equilibrium.

Physics


Question:
Figure shows two identical capacitors each of capacity C in a
circuit with two ideal diodes. A 100 voltbattery in connected to the
input terminals. The output voltage across MN is.

Solution:
When terminal x is positive the diode D2 gets reverse biased
and D1 forward biased therefore circuit is
Tc fljk x rks rFkk D_1vxzck;l gks Hence vr V_MN = 50 V

Physics

P-21 Physcis Paper 19 Dec Hindi Eng..docx


Question:
A fiber sheet with relative permittivity 5 is inserted between
two parallel metal plates 0.25 cm apart. Apotential difference of 2500 v
is applied between the plates. If dielectric strength of air is 3 ×
10^6 v/m then.

Potential gradient strength in fiber sheet exceeds dielectric
strength of air

Solution:
As V = V_1 + V_2
2500 = 0.0002 E_1 + 0.0023 E_2
2E_1 + 23 E_2 = 25 x 10^6
Also; E_1 = 5E_2

E_1 = 3.788 x 10^6 V/m
As E_1> 3.6 x 10^6 Hence Air will breakdown

Physics


Question:
Under an adiabatic process, the volume of an ideal gas gets
doubled. Consequently the mean collisiontime between the gas molecules
changes from. If
[IMAGE] for this gas then a good estimate,d
bl j.k mlds v.kqvksa esa gksus okyh VDdjksa

Physics


Question:
A constant torque acting on a uniform circular wheel changes its
angular momentum from 10 J-sec to60 J-sec in 4 sec. The torque in Nm is
2.5n. Find the value of n.

Solution: $\tau = \frac{dL}{dt} = \frac{60 - 10}{4} = 12.5$ Nm

Physics


Question:
At a height 0.4 m from ground, the velocity of projectile in
vector form
is$(\overset{\rightarrow}{v} = 6\overset{\hat{}}{i} + 2\overset{\hat{}}{j})$.
The angle of projection is π/n. Find n.

Solution:
${(v_{y})}^{2} = u_{y}^{2} + 2a_{y}S$
$(2)^{2} = {(u_{y})}^{2} + 2 \times ( - 10) \times 0.4$
$4 + 8 = u_{y}^{2}$
$u_{y} = \sqrt{12}$
$tan\theta = \frac{u_{y}}{u_{x}}$
$= \frac{\sqrt{12}}{6}$
$= \frac{1}{\sqrt{3}}$
$\theta = 30^{\circ}$

Physics


Question:
A long, straight wire of radius a carries a current distributed
uniformly over its cross-section. The ratioof the magnetic fields due
to the wire at distance[IMAGE] and 2a, respectively from
the axis of the wire is n/3. Find n.

Solution:
Using Ampere's circuital law:
$B_{B} = \frac{\mu_{0}i}{2\pi(2a)}$
$\frac{B_{A}}{B_{B}} = \frac{4}{6} = \frac{2}{3}$

Physics


Question:
A galvanometer having a coil resistance $100\Omega$ gives a full
scale deflection when a current of 1 mA is passed through it. What is
the value of the resistance which can convert this galvanometer into a
voltmeter giving full scale deflection for a potential difference of 10
V?

Solution:
$V_{g} = i_{g}R_{g} = 0.1V$
$V = 10V$
$R = R_{g}(\frac{V}{V_{g}} - 1)$ $= 100 \times 99 = 9.9K\Omega$

Physics


Question:
A vessel contains two immiscible liquids of densities
$\rho_{1} = 1000\ {kgm}^{- 3}$ and $\rho_{2} = 1500\ {kgm}^{- 3}$. A
solid block of volume $V = 10^{- 3}m^{3}$ and density
$d = 800{kgm}^{- 3}$ is tied to one end of a string and other is tied to
the bottom of the vessel as shown in figure. The block is immersed with
$\frac{2^{th}}{5}$ of its volume in the liquid of higher density and
$\frac{3^{th}}{5}$ in the liquid of lower density. The entire system is
kept in an elevator which is moving upwards with an acceleration of a =
g/2. Find the tension in the string in Newton. (Take $g = 10ms^{- 2}$)
(in N)

Solution:
[IMAGE] We analysed this problem from the reference frame of Elevator. Total
buoyant force On the block,
$F_{B} = \left( \frac{2}{5}V\rho_{2} + \frac{3}{5}V\rho_{1} \right)(g + a)$
For equilibrium,
$F_{B} = T + Vd(g + a)$
or $T = F_{B} - Vd(g + a)$
$= (g + a)V\left\lbrack \frac{2}{5}\rho_{2} + \frac{3}{5}\rho_{1} - d \right\rbrack$
$T = \left( 10 + \frac{10}{2} \right) \times 10^{- 3}\left\lbrack \frac{2}{5} \times 1500 + \frac{3}{5} \times 1000 - 800\rbrack = 6\text{N} \right.\ $

Physics


Question:
The current I_1 (in A) flowing through 1$\Omega$ resistor in
the following circuit is

Solution: $i = \frac{I}{2} = 0.2A$

Physics


Question:
The balancing length for a cell is 560 cm in a potentiometer
experiment. When an external resistance of10[IMAGE] is
connected in parallel to the cell, the balancing length changes by 60
cm. If the internalresistance of the cell is [IMAGE],
where N is an integer then value of N/2 is ______.

Solution:
Let the emf of cell is $\varepsilon$ internal resistance is
 and potential gradient is x
Only cell connected:
$\epsilon = 560\ x$......(1)
After connecting the resistor
$\frac{\varepsilon \times 10}{10 + r} = 500x$......(2)
from (1) and (2)
$\frac{560 \times 10}{10 + r} = 500s$
$56 = 50 + 5r$
$r = \frac{6}{5} = 1.2\Omega$
n = 12

Physics


Question:
A man who can swim at the rate of $2km/hr$ (in still river)
crosses a river to a point exactly opposite on the other bank by
swimming at an angle of 60º perpendicular direction of river flow of the
water in the river. The velocity of the water current in $km/hr$ is

Solution:
Let speed of current be u. For net velocity ofman to be normal
to river flow.
$2sin60^{\circ} = u$
or$u = \sqrt{3}km/hr$

Physics


Question:
A particle at a height from the ground is projected with
an angle 30º fromthe horizontal, it strikes the ground making angle 45º
with horizontal. It is again projected from the same point with the same
speed but with an angle of 60º with horizontal. The tangent of the angle
it makes with the horizontal is $\sqrt{n}$when it strikes the ground

Solution:
Using $v = \sqrt{u^{2} + 2gh}$
$v = \sqrt{u^{2}\sin^{2}\theta + 2gh}$(Vertical comp. when striking)
$ucos\theta = \sqrt{u^{2}\sin^{2}\theta + 2gh}$
$u^{2}\cos^{2}\theta = u^{2}\sin^{2}\theta + 2gh$
$u^{2}\left( \frac{3}{4} - \frac{1}{4} \right) = 2gh$
$u^{2} = 4gh$
$u = 2\sqrt{gh}$

Physics

Projectile projected obliquely from a height


Question:
In a double-slit experiment, at a certain point on the screen
the path difference between the two interfering waves is
[IMAGE] of a wavelength. The ratio of the intensity of
light at that point to that at the centre of a bright fringe is

Physics


Question:
Consider two black bodies A and B. The surface temperature
of black body A is twice that of B. The wavelength corresponding to
maximum spectral radiancy for bodies A and B are λ_A and λ_B
respectively. When radiations of wavelength λ_A is incident on a
metallic surface the kinetic energy of fastest photoelectron is K_1
and when radiations of wavelength λ_B is incident on the same metallic
surface the kinetic energy of fastest photoelectron is K_2. The work
function of the metal is

Solution: D

Physics


Question:
Find the time when the displacement of the particle becomes zero.
Assume velocity and position at t = 0 to be 0 and -10 m respectively.

Physics

P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx


Question:
The pressure gauge reading in meter of water column shown in the
given figure will be

Solution:
$P_{0} + \rho_{w}g1 + \rho_{wg}0.2 - \rho_{Hg}g0.2 = P_{0}$
$P - P_{0} = \rho_{Hg}g0.2 - \rho_{w}g(1.2) = 14896$
$14896 = \rho_{wgh}$
h = 1.52

Physics


Question:
Consider a sphere of radius R which carries a uniform charge
density $\rho$. If a sphere of radius$\frac{R}{2}$iscarved out of it, as
shown, the
ratio$\left| \frac{{\overset{\rightarrow}{E}}_{A}}{{\overset{\rightarrow}{E}}_{B}} \right|$
of magnitude of electric field of magnitude of electric field
${\overset{\rightarrow}{E}}_{A}$and ${\overset{\rightarrow}{E}}_{B}$,
respectively, atpoints A and B due to the remaining portion is

Solution:
For a solid sphere
$E = \frac{\rho r}{3\varepsilon_{0}}$
$E_{A} = \frac{- \rho R}{2\left( 3\varepsilon_{0} \right)}$
$\left| E_{A} \right| = \frac{\rho R}{6\varepsilon_{0}}$
Electric field at point B = E_B = E_1A + E_2A
E_1A = Electric Field Due to solid sphere of radius R at point
$B = \frac{\rho R}{3\varepsilon_{0}}$
E_2A = Electric Field Due to solid sphere of radius R/2 (which having
charge density -$\rho$)

Physics

Electric field by uniformly charged sphere


Question:
An electron (mass m) with initial velocity
$\overset{\rightarrow}{v} = v_{0}\overset{\hat{}}{i} + v_{0}\overset{\hat{}}{j}$
is in an electric field
$\overset{\rightarrow}{E} = - E_{0}\overset{\hat{}}{k}$. If
[IMAGE] is initial de-Broglie wavelength of electron, its
de-Broglie wave length at time t is given by

Solution:
Initially m$(\sqrt{2}v_{0}) = \frac{h}{\lambda_{0}}$
Velocity as a function of
time$= v_{0}\overset{\hat{}}{i} + v_{0}\overset{\hat{}}{j} + \frac{{eE}_{0}}{m}t\overset{\hat{}}{k}$
so wavelength
$\lambda = \frac{h}{m\sqrt{2v_{0}^{2} + \frac{e^{2}E_{0}^{2}}{m^{2}}t^{2}}}$
$\lambda = \frac{\lambda_{0}}{\sqrt{1 + \frac{e^{2}E_{0}^{2}}{2m^{2}v_{0}^{2}}t^{2}}}$

Physics