Question:
A person with a normal near point (25 cm) is using a compound
microscope in normal adjustment with objective focal length
[IMAGE] and an eye-piece of focal length
[IMAGE] can bring an object placed
[IMAGE] from the objective in sharp focus. The
magnifying power of the microscope is 11n. Find n

Solution:
Angular magnification of the eye-piece for image at
[IMAGE] TOPIC: Optical instruments
SUB TOPIC: Microscope
LEVEL: Moderate

Physics


Question:
A uniform metre stick is held vertically with one end on the
floor and is allowed to fall. The speed of theother end when it hits the
floor assuming that the end at the floor does not slip

Solution:
using energy conservation
$mg\frac{\mathcal{l}}{2} = \frac{1}{2}I\omega^{2}$
$mg\frac{\mathcal{l}}{2} = \frac{1}{2} \cdot \frac{m\mathcal{l}^{2}}{3}\omega^{2}$
$= 1m$ $\omega = \sqrt{\frac{3g}{}}$
$V_{A} = \omega = \sqrt{3g} = (\sqrt{3g})$

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.

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:
Planet P has mass M_P and radius R_P. Planet Q has mass M_P/2.
If the escape velocities from the Planets P and Q are equal, then what
is the radius of planet Q?

Physics


Question:
Error in the measurement of radius of a sphere is1.What is the
percentage error in the calculated value of its volume?

Solution:
$V = \frac{4}{3}\pi r^{3}$
Take natural log on both side
$ln(V) = ln\left( \frac{4}{3}\pi \right) + 3ln(r)$
Differentiate both side
$\frac{\Delta V}{V} = 3\frac{\Delta r}{r}$
$\%\frac{\Delta V}{V} = 3 \times 1 = 3\%$

Physics


Question:
The van der Waals' equation of state for somegases can be
expressed as:
$$\left( P + \right)(V - b) = RT$$
Where P is the pressure, V the molar volume andT is the absolute
temperature of the given sampleof gas, a, b, and R constants. The
dimensions is

Solution: $\lbrack P\rbrack = \left\lbrack \frac{a}{V^{2}} \right\rbrack\lbrack a\rbrack = \lbrack P\rbrack\left\lbrack V^{2} \right\rbrack = \left\lbrack M^{1}L^{- 1}T^{- 2} \right\rbrack\left\lbrack L^{6} \right\rbrack = \left\lbrack M^{1}L^{5}T^{- 2} \right\rbrack$

Physics


Question:
In the fig. shown a cart moves on a smooth horizontal surface
due to an external constant force ofmagnitude F. The initial mass of the
cart is M_0and velocity is zero. Sand falls on to the cart
withnegligible 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 sand = 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


Question:
A bob of mass m is connected to a spring of spring constant k and a
string which is fixed to the ceiling as shown in figure. Original length
of the spring is L and the bob rotates in a horizontal circle centred at
the fixed rod. Assuming the spring remains horizontal, then elongation
in the spring is

Solution:
[IMAGE] T cos θ = mg... (1)
T sin θ + kx = mw2 (L + x)... (2)
from (1) and (2)
x =[IMAGE] TOPIC: CIRCULAR MOTION
SUB TOPIC: DYNAMICS

Physics

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


Question:
A disc of mass is attached to a spring of
stiffness During its motion, the disc rolls on the ground. When
released from some stretched position, the centre of the disc will
execute simple harmonic motion with a time period of

Solution: Torque = Iα about the point of contact

Physics


Question:
The vertical height between two points A and B is 30 m. A particle
is projected downward from A and at the same time another particle is
projected upward from B with the same speed. Both particles reach the
ground simultaneously. The velocity of the projection of the particle is
(B is on ground)

Solution:
S = 0 = ut -[IMAGE] gt2
[IMAGE] 30 = ut + [IMAGE] gt2 =[IMAGE] by adding
30 = 2ut =[IMAGE] u2 =[IMAGE] ⇒[IMAGE] = u
TOPIC:KINEMATIC
SUB TOPIC: MOTION UNDER GRAVITY

Physics

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


Question:
On heating water, bubbles beings formed at the bottom of the
vessel detach and rise. Take the bubbles to be spheres of radius R and
making a circular contact of radius r with the bottom of the vessel. If
r << R, and the surface tension of water is T, value of r just before
bubbles detach is: (density of water is
ikuh dks xeZ
gksdj dh vksj mBrs gSaA cqycqyksa dks f= xksyk eku ysa
ikuh dk i`"B ruko T gSa, rc cqycyksa ds cl foyXu gksus ls tjk igys r

Solution:
[IMAGE] The bubble will detach if -
Buoyant force ≥ Surface tension force
[IMAGE] Solving [IMAGE] cqycqyk vyx gksxk;fn -
mRIykou cy≥i`"B ruko

Physics


Question:
The pressure gauge reading in metre of water column shown in the
given figure will be:
 fp= esa ty LrEHk

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

Physics


Question:
A uniform disc of mass M and radius R is lifted using a string
as shown in the figure. Then,

Its rate of change of angular momentum is MgR

Solution:
$\frac{3Mg}{2} + \frac{Mg}{2} - Mg = Ma$
$a = g \uparrow$
$\tau = I\alpha$
$\frac{3Mg}{2}R - \frac{Mg}{2}R = \frac{{MR}^{2}}{2} \cdot \alpha$
$\alpha = \frac{2g}{R}$
$\frac{dL}{dt} = \tau = MgR$

Physics

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


Question:
40 m
80 m

Solution:
With standard kinematics, we get
[IMAGE] y = [IMAGE] gt2, t = [IMAGE] = 2.86 s
x = vxt, vx =[IMAGE] = 28.0 m/s
vy2 - v02 = 2a(y - y0)
vy = -[IMAGE] = -[IMAGE] = 28.0 m/s
θ = A tan[IMAGE] = 315°
TOPIC:KINEMATIC
SUB TOPIC: 2 D

Physics

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


Question:
A solid cylinder and a solid sphere, both having the same mass
and radius, are released from an inclined plane having angle of
inclination[IMAGE] one by one. They roll on the incline
without slipping. The statement that holds good in this motion is

The force of friction that acts on the two is the same

The force of friction is greater in case of a sphere than for a
cylinder.

The force of friction is greater in case of cylinder than for sphere.

The force of friction will depend on the nature of the surface of
the body that is moving and that of the inclined surface, and is
independent of the shape and size of the moving body.

Solution:
[IMAGE] Equation of motion
mg sin [IMAGE]....(i)
$fR = I\alpha = I\frac{a}{R}$
$\Rightarrow f = I\frac{a}{R^{2}}$....(ii)
From (i) and (ii)
$mgsin\theta - f = m\frac{{fR}^{2}}{I} \Rightarrow mgsin\theta = f\left\{ 1 + \frac{mR^{2}}{I} \right)$
$\Rightarrow$ $\ \ f = \frac{mgsin\theta}{1 + \frac{mR^{2}}{I}}$
Moment of inertia of solid cylinder is greater, so friction force f will
be greater.

Physics

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


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

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:
The graph which depicts the results of Rutherford gold foil
experiment with α -particles is:
α -particles is:
Y = Number of scattered alpha -particles detected
(Plots are schematic and not to scale)

Physics


Question:
The following graph shows two isotherms for a fixed mass of an
ideal gas. The ratio of r.m.s speed of the molecules at temperatures
T_1 and T_2 is

Solution:
$PV = \frac{1}{3}M\left( v_{rms} \right)_{1}^{2} = RT$
On lower curve, we have V=1m^3, corresponding to P=1×10^5Pa = 1
atmosphere. At T_1, we have
de oØ tks P=1×10^5Pa = 1 okrkoj.k ds
vuq:i gSA T_1 ij] gekjs ikl
$1 \times 1 = \frac{1}{3}M\left( v_{rms} \right)_{1}^{2} = {RT}_{1}$
or$\left( v_{rms} \right)_{1}^{2} = \frac{3}{M}$
Similarly, for upper curve we have
blh rjg] oØ ds fy, gekjs ikl gS
V=1m^3, P = 2 atmosphere
$\therefore 2 \times 2 = \frac{1}{3}M\left( v_{ans} \right)_{2}^{2} = {RT}_{2}$
$\therefore\left( v_{mas} \right)_{2}^{2} = \frac{4 \times 3}{M} = \frac{12}{M}$
$\therefore\ Required\ ratio\ = \sqrt{\frac{12/M}{3/M}} = 2$

Physics


Question:
A hollow conducting sphere of inner radius R and outer radius
2R is given a charge Q as shown in the figure, then the

Solution:
[IMAGE] charge is on the outer surface hence [IMAGE] remains
constant.

Physics


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:
A 100 turns closely wound current carrying coilof radius 10 cm
is placed in a vertical plane. Thecoil carries a current of 3.2 A and
free to rotateabout a horizontal axis which coincides with itsdiameter.
A uniform horizontal magnetic field of2 Tesla exists in the region. Due
to magnetic torquecoil starts rotating. The angular velocity in
rad/secof the coilwhen it has rotated by $\frac{\pi}{2}$ is 10n. Find n.

Solution:
$\tau = MBsin\theta$
$I\alpha = MBsin\theta$
$\alpha = \frac{MB}{I}\sin\theta$
$\frac{d\omega}{d\theta}*\frac{d\theta}{dt} = \frac{MB}{I}sin\theta$
$\int_{0}^{\omega}\mspace{2mu}\omega d\omega = \int_{0}^{\frac{\pi}{2}}\mspace{2mu}\frac{MB}{I}\sin{\theta d\theta}$
$\left( \frac{\omega^{2}}{2} \right)_{0}^{\omega} = \frac{MB}{I}\lbrack - cos\theta\rbrack_{0}^{\pi/2}$
$\omega = \sqrt{\frac{2MB}{I}} = \sqrt{\frac{2NIAB}{I_{rot}}}\ $
$= \sqrt{\frac{2 \times 100 \times 3.2 \times \pi \times 100 \times 10^{- 4} \times 2}{0.1}}$
= 20 rad/sec.

Physics

Torque on current carrying loop in magnetic


Question: has four choices

Physics


Question:
A capacitor is made of two square plates each of side
making a very small angle $\alpha$ between them, as shown in figure. The
capacitance will be close to

$\frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{2d})$

$\frac{\varepsilon_{0}a^{2}}{d}(1 + \frac{\alpha a}{2d})$

$\frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{4d})$

Solution:
$\Rightarrow c = \frac{\varepsilon_{0}a}{\alpha}\lbrack ln(d + \alpha x)\rbrack_{0}^{a}$
$= \frac{\varepsilon_{0}a}{\alpha}In(1 + \frac{\alpha a}{d}) \approx \frac{\varepsilon_{0}a^{2}}{d}(1 - \frac{\alpha a}{2d})$

Physics


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.
V=U+at
$0 = 7 - 2t$
$t = 3.5\sec$
$y_{2} = 7 \times 3.5 - \frac{1}{2} \times 2 \times (3.5)^{2} = \frac{49}{4}m$
$y_{2} - x_{1} = \frac{49}{4} - 12 = \frac{1}{4}m$.
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:
A ball of mass M collides elastically with another
identical ball at rest as shown in figure. Initially velocity of
ball After collision

Solution:
[IMAGE] Velocity along line of impact will get exchanged because both bodies are
identical. $v_{B} = ucos\theta$
$V_{A} = usin\theta$

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. [IMAGE]

Physics


Question:
In a cyclic waste power plant for thermal decomposition of waste
hot gases are supplied from a "Kentuki Chamber" where gases are heated
under kentuki process [IMAGE] where P, V and T are
pressure, volume and temperature of gases in chamber and k is positive
constant. Find work done by 5 mole of hot gases when temperature is
raised from 500 K to 550 K:,d pØ IykUV fo?kVu
"dsUVqdh psEcj" ls rIr xSals çnku dh tkrh
çfØ xeZ tgk¡ P, V o T psEcj
eksyksa 500 K ls 550 K rd o`f)

Physics


Question:
A uniform cylinder floats with its axis vertical with
[IMAGE] parts of its length in the upper liquid and
[IMAGE] part of its length in lower liquid. If the
densities of upper and lower liquids are [IMAGE] then
the density of the material of the cylinder is

Solution:
Let [IMAGE] be the be the length of the cylinder
then
 is the density of the cylinder.
TOPIC: Hydrostatics
SUB TOPIC:Bouyancy
LEVEL: Moderate

Physics


Question:
P-V diagram of ideal monoatomic gas is as shown. Find net heat
given to the gas in the process ABC.

Physics

P-V diagram and first law of thermodynamics


Question:
n a meter bridge, the wire of length 1m has a non-uniform
cross-section such that, the variation$\frac{dR}{d^{\mathcal{l}}}$ of
its resistance R with
length$\mathcal{l}$ is $\frac{dR}{d\mathcal{l}} \propto \frac{1}{\sqrt{\mathcal{l}}}$.
Two equal resistances are connected as shown in thefigure. The
galvanometer has zero deflection when the jockey is at point P. What is
the length AP ?

Solution:
According to Question
$\int_{0}^{\mathcal{l}}\mspace{2mu} c\frac{d\mathcal{l}}{\sqrt{\mathcal{l}}} = \int_{\mathcal{l}}^{1}\mspace{2mu} c\frac{d\mathcal{l}}{\sqrt{\mathcal{l}}}$
$(2\sqrt{\mathcal{l}})_{0}^{\mathcal{l}} = (2\sqrt{\mathcal{l}})_{t}^{1}$
$2\sqrt{\mathcal{l}} = 2 - 2\sqrt{\mathcal{l}}$
$2\sqrt{\mathcal{l}} = 2 - 2\sqrt{\mathcal{l}}$
$4\sqrt{\mathcal{l}} = 2$
$\mathcal{l =}\frac{1}{4} = 0.25m$.

Physics


Question:
A block of weight 10 N produces an extension of 9 cm when it is
hung by an elastic spring of length 60 cm and is in equilibrium. The
spring is cut into two parts, one of length 40 cm and the other of
length 20 cm. Another weight of 20 N hangs in equilibrium supported by
both parts as shown in the figure. Find the total extension of the two
parts of the spring (in cm) now.
10N 60 cm 9cm, 40
cm 20 cm 20N
 (cm )

Solution:
10 = k ×[IMAGE] k =[IMAGE] k1 × 4 = k × 60 =[IMAGE] × 60
k1 =[IMAGE] k2 × 20 = k × 60 ⇒ k2 =[IMAGE] kq = k1 + k2 = 500
10 = 500 x ⇒ x =[IMAGE] m = 2 cm
TOPIC: NLM
SUB TOPIC: SPRING

Physics

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


Question:
The maximum peak to peak voltage of an AM wave is 24 mV and the
minimum peak to peak voltage is8 mV. The modulation factor is.

Solution:
Here ¼;gk¡½, $V_{\max} = \frac{24}{2}$=12mV and¼rFkk½
$V_{\min} = \frac{8}{2} = 4mV$

Physics

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


Question:
A particle moving on a circular path travels first one third
part of circumference in 2 sec & nextone third part in 1 sec. Average
angular velocity of the particle is $\frac{n\pi}{9}$ (in rad/sec). Then
valueof n.

Solution: $\omega_{avg} = < \omega > = \frac{\ total\ \theta}{\ total\ \ time\ }$

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
vuur yEckbZ vkSj pkSM+kbZ okys nks leryksa

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

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

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

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

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

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


Question:
A coal based thermal power plant producing electricity operates
between the temperatures 27ºC and227ºC. The plant works at 80 of its
maximum theoretical efficiency. Complete burring of 1 kg of coalyields
3600 KJ of heat. A house needs 10 units of electricity each day. Coal
used for supplying theamount of energy for the house in one year is,d 1 kg iw.kZ tyus ij 3600 KJ mRiUu gksrh,d o"kZ rd nsus

Solution:
Maximum efficiency of engine
$= \left( 1 - \frac{T_{2}}{T_{1}} \right) \times 100 = 40\%$
$= \left( 1 - \frac{T_{2}}{T_{1}} \right) \times 100 = 40\%$
$\therefore$overall efficiency = 0.4 x 0.8 = 0.32
$\therefore$dqy n{krk = 0.4 x 0.8 = 0.32
Requirement of house in one year,d lky esa ?kj
= 10 × 365 units = 3650 kwh = 3650 × 3600 kJ
$\therefore$ in put =$\frac{3650 \times 3600}{0.32}$kJ
$\therefore$ coalrequired =$\frac{\ In\ put\ }{36000}$=1141kg

Physics


Question:
In a Young's double slit experiment sources of equal
intensities are used. Distance between slits is d and wavelength of
light used is λ(λ << d). Angular separation of the nearest points on
either side of central maximum where intensities becomes half of the
maximum value is

Solution:
[IMAGE] I + I + 2 [IMAGE] cos φ
2I = 2I + 2I cos φ
cos φ = 0
φ = [IMAGE] ∆x = [IMAGE] x = [IMAGE] θ =[IMAGE]

Physics


Question:
A thin metal rod of length L_0 is shaped into a ring with a
small gap x as shown. On heating the system

Data insufficient to arrive at a conclusion
L_0 yackbZ dh,d iryh NM+ dks vYi varjky x

Solution:
On uniform heating of an object which is free to expand,
distance between any two points increases.
oLrq tks çlkfjr gksus ds fy, LorU= gS]

Physics


Question:
A uniform disc of mass m and radius R is resting on a smooth
horizontal surface and it is free to rotateabout its own axis which is
vertical. A small particle of mass m collides its periphery with a
velocity of Vin tangential direction. If the collision is perfectly in
elastic and the particle sticks to the periphery, theangular velocity of
the system just after the collision will be

Solution:
Applying angular momentum conservation about the rotating axis
L_i = L_f
O + mv R= $\left( \frac{mR^{2}}{2} + mR^{2} \right)\omega_{f}$
$\omega = \frac{2V}{3R}$

Physics


Question:
Two particle A and B are located in x-y plane at points (0,0)
and (0,4 m). They simultaneously start moving with velocities.
${\overset{\rightarrow}{v}}_{A} = 2\overset{\hat{}}{j}m/s$and${\overset{\rightarrow}{v}}_{B} = 2\overset{\hat{}}{i}m/s$.
Select the correct alternative(s).

Time after which they are at minimum distance is 1 s

The distance between them first decreases and then increases

The shortest distance between them is $2\sqrt{2}m$

Solution:
${\overset{\rightarrow}{v}}_{AB} = {\overset{\rightarrow}{v}}_{A} - {\overset{\rightarrow}{v}}_{B} = 2(j - \overset{\hat{}}{i})m/s\ \ \ or\ |{\overset{\rightarrow}{v}}_{AB}| = 2\sqrt{2}m/s$
Assuming B to be at rest, A will move with velocity
${\overset{\rightarrow}{v}}_{AB}$ in the direction shown in figure.
The distance between them will first decrease from A to Cand then
increase beyond C.
[IMAGE] Minimum distance between them is BC which is equal to
$\frac{4}{\sqrt{2}}$ or $2\sqrt{2}$ and the time after which they are
at closest distance is:
$t = \frac{AC}{|{\overset{\rightarrow}{v}}_{AB}|} = \frac{2\sqrt{2}}{2\sqrt{2}} = 1s$

Physics


Question:
A certain quantity of oxygen [IMAGE] is
compressed isothermally until its pressure is doubled
[IMAGE]. The gas is then allowed to expand adiabatically
until its original volume is restored. Then the final pressure(P) in
terms of initial pressure [IMAGE] is:
vkWDlhtu,d izkjfEHkd nkc

Solution: [IMAGE] (isothermal process) ¼lerkih

Physics


Question:
Radiation, with wavelength [IMAGE] falls on a
metal surface to produce photoelectrons. The electronsare made to enter
a uniform magnetic field of[IMAGE]. If the radius of the
largest circular pathfollowed by the electrons is 10 mm, the work
function of the metal is closest to what integral value in eV

Solution:
[IMAGE] (in eV)
[IMAGE] (in eV)
[IMAGE] TOPIC:Dual nature of particle
SUB TOPIC:Photoelectric effect

Physics


Question:
A sound wave of frequency n travels horizontally to right. It is
reflected from a large vertical wall moving to the left with a speed V.
The speed of sound in medium is C. Then

the frequency of reflected wave for stationary listener is

the number of beats heard by stationary listener to the left of
reflecting wall is [IMAGE]

Solution:
[IMAGE] (number of waves striking the wall per sec)
[IMAGE] (frequency observed by stationary listener)
TOPIC:Sound waves
SUB TOPIC:Doppler effect
LEVEL:Moderate
2. In YDSE arrangement, white light is used to illuminate the slits. At
point on the screen directly in front of slits, which of these
wavelength is missing?
([IMAGE] = wavelength of light used)
(d = distance between slits)
(D = distance between slits and screen)
(given d << D)

Physics


Question:
A rubber cord has a cross-section area 1 cm^2 and natural
length of [IMAGE] It is stretched by
[IMAGE] to fire a small object of
mass[IMAGE]. If the Young's modulus (Y) is
[IMAGE] Speed acquired by the object in (m/s) is 2n. Find
n. Assume Hook's law is valid and neglect gravity.,d jcj dh Mksjh dk vuqçLFk dkV kjk çkIr
osx (m/s) esa½2n.nKkr djsaA ekfu,sa fd gqd

Solution:
v = 10^3 × 10^-2
v = 10.00 m/s

Physics

Potential energy stored in elongated rubber cord


Question:
A small ball of mass 1 kg strikes a wedge of mass 4kg
horizontally with a velocity of 10 m/s. Friction is absent everywhere
and collision is elastic. Select the correct answers

The linear momentum of the ball along common normal and common
tangent will not change during the collision.

The linear momentum of the ball along common normal and common
tangent will change during the collision.

The linear momentum of the ball along common normal will not
change but remain same along common tangent during the collision.

The linear momentum of the ball along common normal will change
but remain same along common tangent during the collision.

Solution:
Since impulse acts along common normal the linear momentum of the
ball along common normal will change. There is no impulse along common
tangent therefore linear momentum of the ballwill not change during the
collision.
TOPIC: Collision
SUB TOPIC:Elastic collision
LEVEL:Moderate

Physics


Question:
A particle is projected vertically upwards. It will be at
$\left( \frac{3}{4} \right)$^th of its greatest height at time
$t = t_{1}$and $t = t_{2}$. $\left( t_{1} < t_{2} \right)$. Then the
ratio $\left( t_{1}:t_{2} \right)$ is

Solution:
$H = \frac{u^{2}}{2g}$;$\frac{3}{4}\left( \frac{u^{2}}{2g} \right) = ut - \frac{1}{2}gt^{2} \Rightarrow t = \frac{u}{g} \pm \frac{u}{2g}$
$\Rightarrow t_{1} = \frac{u}{2g}$and$t_{2} = \frac{3u}{2g} \Rightarrow t_{1}:t_{2} = 1:3$

Physics


Question:
For D1 forward bias resistance is 20 Ohm and reverse bias
resistance is 600 Ohm. For D2 forward bias resistance is 30 Ohm and
reverse bias resistance is 200 Ohm. Find the equivalent resistance
between A and B if (i) A is at a higher potential (ii) B is at a higher
potential

Physics


Question:
A charged particle of mass moving under
the influence of uniform electric field $E\overset{\rightarrow}{i}$and a
uniform magnetic field $B\overset{\rightarrow}{k}$follows a trajectory
from point P to Q as shown in figure. The velocitiesat P and Q are
respectively,
$v\overset{\rightarrow}{i}$and$- 2v\overset{\rightarrow}{j}$. Then which
of the following statements (A, B, C, D) are thecorrect ? (Trajectory
shown is schematic and not to scale)

Rate of work done by the electric field at P
is$\frac{3}{4}\left( \frac{{mv}^{3}}{a} \right)$and at Q is zero.

The difference between the magnitude of angular momentum of the
particle at P and Q is 2mav

Solution:
(A) by work energy theorem
$0 + {qE}_{0}2a = \frac{3}{2}{mv}^{2}$
$$E =
(B) Rate of work done at A= power of electric force
$= {qE}_{0}V$
$= \frac{3}{4}\frac{{mv}^{3}}{a}$
(C) at$Q,\frac{dW}{dt} = 0$for both forces
(D)
$\Delta\overset{\rightarrow}{L} = ( - m2v2a\overset{\hat{}}{k}) - ( - mva\overset{\hat{}}{k})$
$|\Delta\overset{\rightarrow}{L}| = 3mva$

Physics


Question:
A uniform rod of mass 200 gram and length
[IMAGE] is initially at rest in vertical position. The rod
is hinged at centre such that it can rotate freely without friction
about a fixed horizontal axis passing through its centre. Two particles
of mass m = 100 gram each having horizontal velocity of equal magnitude
u = 17 m/s strikes the rod at top and bottom simultaneously as shown and
stick to the rod.
The angular speed (in rad/s) of rod when it becomes horizontal is 8.5n.
Find n

Solution:
From conservation of angular momentum.
$mu\frac{L}{2} + mu\frac{L}{2} = \left\lbrack 2m\frac{L^{2}}{12} + m\left\{ \left( \frac{L}{2} \right)^{2} + \left( \frac{L}{2} \right)^{2} \right\} \right\rbrack\omega$
$muL = \left\lbrack \frac{{mL}^{2}}{6} + \frac{{mL}^{2}}{4} + \frac{{mL}^{2}}{4} \right\rbrack\omega$
$= \frac{2{mL}^{2}}{3}\omega$
or $\omega = \frac{3u}{2L}$
$= 76.50$

Physics

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


Question:
A charged particle of mass moving under
the influence of uniform electric field $E\overset{\rightarrow}{i}$and a
uniform magnetic field $B\overset{\rightarrow}{k}$follows a trajectory
from point P to Q as shown in figure. The velocitiesat P and Q are
respectively,
$v\overset{\rightarrow}{i}$and$- 2v\overset{\rightarrow}{j}$. Then which
of the following statements (A, B, C, D) are the correct ? (Trajectory
shown is schematic and not to scale)

Rate of work done by the electric field at P
is$\frac{3}{4}\left( \frac{{mv}^{3}}{a} \right)$and at Q is zero.

The difference between the magnitude of angular momentum of the
particle at P and Q is 2mav

Physics


Question:
A LCR circuit behaves like a damped harmonic oscillator.
Comparing it with a physical spring-mass damped oscillator having
damping constant the correct equivalence would be

$L \leftrightarrow \frac{1}{b},C \leftrightarrow \frac{1}{m},R \leftrightarrow \frac{1}{k}$

$L \leftrightarrow k,C \leftrightarrow b,R \leftrightarrow m$

$L \leftrightarrow m,C \leftrightarrow \frac{1}{k},R \leftrightarrow b$

$L \leftrightarrow m,C \leftrightarrow k,R \leftrightarrow b$

Solution:
In damped oscillation
$ma + bv + kx = 0$
[IMAGE] $m\frac{d^{2}x}{{dt}^{2}} + b\frac{dx}{dt} + kx = 0$....(i)
In the circuit
$- iR - L\frac{di}{dt} - \frac{q}{c} = 0$
$L\frac{d^{2}q}{{dt}^{2}} + R\frac{dq}{dt} + \frac{1}{c} \cdot q = 0$....(ii)
Comparing equation (i) and (ii)
$m = L,b = R,k = \frac{1}{c}$

Physics


Question:
A non-viscous fluid of density ρ is flowing in a tube as shown
in figure. Area of section

Work done by gravitational force per unit volume from section

Work done by elastic forces (pressure) per unit volume from
section

Solution:
Applying Bernoulli is equation from section-(1) and (2)
[k.M-(1).M-(2) esa cjukWyh lehdj.k yxkus ij
[IMAGE] andrFkkP_1- P_2= $\rho$g(2h)
Solving we get, gy djus ij çkIr gksxk
$V = \sqrt{\frac{2gh}{3}}$
(C) Work done by gravitation force per unit value
[IMAGE] decrease in gravitational
PT_value = $\rho{gh}_{1} - \rho{gh}_{2}$

PT_value = $\rho{gh}_{1}\ - \rho{gh}_{2}$
$W_{gr} = 0 - \rho gh$
(D) Work done by elastic force volume, çR;kLFk cyksa }kjk çfr bdkbZ
W_e=decrease in elastic P.E._vol = decrease in pressure
W_e=çR;kLFk fLFkfrt P.E._vol= nkc

Physics


Question:
A small smooth disc of mass m and radius r moving with an initial
velocity along the positive x-axis collided with a big disc of mass
2 m and radius 2 r which was initially at rest with its centre at origin
as show in figure.
[IMAGE] If the coefficient of restitution is 0 then velocity of larger disc
after collision is

Solution:
[IMAGE] The larger sphere will move along line of impact.
e=0, magnitude of velocity of larger sphere
$v^{'} = \frac{mvcos\theta}{m + 2m} = \frac{vcos\theta}{3}$
Cos$\theta$= $2\sqrt{2}/3$
Sin $\theta$= 1/3
Velocity of larger sphere
$= v^{'}cos\theta\overset{\hat{}}{i} - v^{'}sin\theta\overset{\hat{}}{j}$
$= \frac{v}{3}\cos^{2}\theta\overset{\hat{}}{i} - \frac{v}{3}sin\theta cos\theta\overset{\hat{}}{j} = \frac{8}{27}v\overset{\hat{}}{i} - \frac{2\sqrt{2}}{27}v\overset{\hat{}}{j}$

Physics


Question:
Initially spring is in its natural length and blocks A and B are
at rest. Find maximum value of constant force F that can be applied on B
such that block A remains at rest. (g = 10 ms-2).

Solution:
Value of $\mu mg$ for A and B are 2N and 2N.
Work energy theorem on B gives
Fx - 2x - (k/2) x^2 = 0
F = 2 + kx/2....................(i)
For equilibrium of A, kx = $\mu mg = 2$
So, F = 2 + 1 = 3N

Physics


Question:
In the system shown in the figure the friction and mass of rope is
negligible, then acceleration of 2 m is

Solution:
2mg - 2T = 2ma
[IMAGE] 2 [T - mg = m (2a)]
0 = 6 ma ∴ a = 0
TOPIC: NLM
SUB TOPIC: EQUILIBRIUM

Physics

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


Question:
In a cylinder-piston arrangement, air is under a pressure P_1.
A soap bubble of radius r lies inside the cylinder, soap bubble has
surface tension T. The radius of soap bubble is to be reduced to half,
The new pressure P_2 to which air should be compressed isothermally.
(Assume r is very small as compared to height of cylinder)

Solution:
Isothermal process.
$\left( P_{1} + \frac{4T}{r} \right)\left( \frac{4}{3}\pi r^{3} \right) = \left( P_{2} + \frac{4T}{r/2} \right)\left( \frac{4}{3}\pi(r/2)^{3} \right)$
$P_{2} = 8P_{1} + \frac{24T}{r}$

Physics


Question:
A man sitting inside a train moving with uniform acceleration
[IMAGE]. Man throws a ball from point O (on the floor of
train) with relative velocity [IMAGE] (where
[IMAGE] = velocity of ball with respect to train
[IMAGE] Ball falls on the train at point A (where OA = R) after time T.
Choose correct option

Solution:
[IMAGE] Time of flight = T = $\frac{2\ (20\sqrt{}2)sin45{^\circ}}{10} = 4s$
Solving we get[IMAGE] TOPIC:Projectile motion
SUB TOPIC:Relative motion
LEVEL:Moderate

Physics


Question:
Suppose the potential energy between electron and proton at a
distance r is given by [IMAGE] where k is constant, e is
charge of proton and r is radius of orbit. Application of Bohr's theory
to hydrogen atom in this case shows that

energy of electron in [IMAGE] orbit is
proportional to [IMAGE] and [IMAGE] (m =
mass of electron).

Solution:
From (1) and (2)
[IMAGE] TOPIC:Structure of Atom
SUB TOPIC:Bohr's model of hydrogen
LEVEL:Moderate

Physics


Question:
A uniform rope of linear mass density $\lambda$ and length 1 is
coiled on a smooth horizontal surface. One end is pulled up with
constant velocity v. Then the average power applied by the external
agent in pulling the entire rope just off the ground is

$\frac{1}{2}\lambda v^{3} + \frac{\lambda\mathcal{l}vg}{2}$

$\lambda\mathcal{l}gv + \frac{1}{2}\lambda v^{3}$

Solution:
Initial K.E. = 0
Initial P.E. = 0
When the rope is just pulled off the table.
Final K.E. = $\frac{1}{2}(\lambda l)v^{2}$
Final P.E$= \frac{\left( \lambda\mathcal{l} \right)g\mathcal{l}}{2}$
Time taken $t = 1/v$
Average power = $\frac{\text{ net change in energy }}{\text{ time }}$
$= \frac{\frac{1}{2}\lambda\mathcal{l}v^{2} + \lambda\mathcal{l}g\mathcal{l/}2}{\mathcal{l/}v} = \frac{1}{2}\lambda v^{3} + \frac{\lambda\mathcal{l}vg}{2}$

Physics


Question:
A particle moves along x-axis with initial position x=0. Its
velocity varies with x-coordinate as shown in graph. The acceleration
'a" of this particle varies with x as

Solution:
The linear relationship between V and x is $V = - mx + C$
where m and C are positive constants.
Acceleration
$a = v\frac{dV}{dx} = - m( - mx + C)$
Hence the graph relating a tox is.

Physics


Question:
A particle is projected vertically upward from the surface of
the earth with a speed of [IMAGE], R being the radius
of the earth and g is the acceleration due to gravity on the surface of
the earth. Then the maximum height ascended is nR,then value of n
is(neglect cosmic dust resistance)

Solution:
[IMAGE] and g = [IMAGE] Therefore [IMAGE] ×[IMAGE] × gR - Rg =
- [IMAGE] We get h = 3R

Physics


Question:
On a smooth pulley of radius R hangs a homogeneous flexible rope of
mass / length λ and length  (Fig.). Find the maximum tension in the
rope.

Solution:
[IMAGE] Tmax - T0 = λgR [1 - cos θ]
Tmax = λgR + λg[IMAGE] = λ
TOPIC: NLM
SUB TOPIC: EQUILIBRIUM

Physics

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


Question:
The minimum time in which the potential energy of a particle
executing SHM changes from maximum to minimum is 5 s. The time period of
SHM in seconds is 4T. Find T.

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:
Three metallic plates out of which middle is given a charge Q,
as shown in the figure. The outer plates can be earthed with switches S1
and S2. The area of plates is same.The charge appearing on outer surface
of external plates when S_1 and S_2 are open and external platesare
uncharged.
Ckgjh ifV~Vdkvksa dh ckgjh lrgks ij S_1 rFkk S_2

Solution:
[IMAGE] (conductor pkyd)
[IMAGE] So vr -x = x + Q + y - y
2x = - Q
$x = \frac{- Q}{2}$
$- x = \frac{Q}{2}$

Physics

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


Question:
System is shown in the figure. Assume that cylinder remains in
contact with the two wedges. The velocity of cylinder is

Solution:
[IMAGE] As cylinder will remains in contact with wedge A
Vx = 2u
As it also remain in contact with wedge B
u sin 30° = Vy cos30° - Vxsin30°
Vy = Vx$\frac{Sin30{^\circ}}{Cos30{^\circ}}$ +
$u\frac{Sin30{^\circ}}{Cos30{^\circ}}$
Vy = Vx tan30° + u tan 30°
Vy = 3utan30° = $\sqrt{3}$u
V = $\sqrt{7}$u

Physics


Question:
A thin metal rod of length L_0 is shaped into a ring with a
small gap x as shown. On heating the system

Data insufficient to arrive at a conclusion
L_0 yackbZ dh,d iryh NM+ dks vYi varjky x

Solution:
On uniform heating of an object which is free to expand,
distance between any two points increases.
oLrq tks çlkfjr gksus ds fy, LorU= gS]

Physics


Question:
Which of the following is an equivalent cyclic process
corresponding to the thermodynamic cyclic givenin the figure ? where, 1
→ 2 is adiabatic. (Graphs are schematic and are not to scale)

Solution:
For process A - B izØe ds fy,; Volume is constant
PV = nRT T increases cM+rk gS
For process B - C izØe
$\Rightarrow {TV}^{\gamma - 1}$ = Constant
For process C - A izØe ds fy,; pressure is constant

Physics


Question:
Two stones are in the same horizontal line and are$50\sqrt{5}m$
away. They are projected with a velocity of 10
respectively and with an angle of 37º and 53º with horizontal
respectively. The minimum distance between them in meters is 5n. Find s.
Suppose they don't strike the ground during the motion.
(take$g = 10m/s^{2}$)

Solution: $d_{\min} = 50\sqrt{5}\ \ \ \ \ \ \ \ sin\theta = 50\sqrt{5} \times \frac{2}{5\sqrt{5}} = 20m$

Physics


Question:
A leak proof cylinder of length 1 m, made of a metal which has
very low coefficient of expansion is floating vertically in water at 0°C
such that its height above the water surface is 20 cm. When the
temperature of water is increased to 4°C, the height of the cylinder
above the water surface becomes 21 cm. The density of water at T = 4°C,
relative to the density at T = 0°C is close to

Solution:
$mg = A(79)\rho_{4^{\circ}C}g$
$\frac{\rho_{4^{\circ}C}}{\rho_{0^{\circ}C}} = \frac{80}{79} = 1.01$

Physics


Question:
A particle of mass $m = \frac{5}{42}kg$ is tied to an end of an
inextensible string and is whirled in a vertical circle as shown below.
Then, the difference in tension of string at A and B i.e.
$T_{A} - T_{B}$ in Newton is: (particle completes the circle)

Solution:
$\Delta T = \frac{3mg\Delta h}{I}$
$h_{B} - h_{A} = I(\frac{3}{5} + \frac{4}{5}) = \frac{7I}{5}$
$T_{A} - B_{B} = \frac{3mg}{I}(h_{B} - h_{A})$
$= \frac{3mg}{I} \times \frac{7I}{5}$
$= \frac{21mg}{5} = \frac{21}{5} \times \frac{5}{42} \times 10$
$= 5N$

Physics


Question:
An ideal fluid flows (laminar flow) through a pipe of
non-uniform diameter. The maximum and minimum diameters of the pipes are
6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the
maximum velocities of fluid in this pipe is

Solution:
Using equation of continuity
$\frac{V_{1}}{V_{2}} = \frac{A_{2}}{A_{2}} = \left( \frac{4.8}{6.4} \right)^{2} = \frac{9}{16}$

Physics


Question:
Consider a uniform rod of mass [IMAGE] and length
[IMAGE] pivoted about its centre. A mass m moving
withvelocity v making angle [IMAGE] to the rod's long
axis collides with one end of the rod and sticks to it. The angular
speed of the rod-mass system just after the collision is

Physics

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


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
velocity[IMAGE]. The emf induced (in volts) is the loop
at t = 2 sec is 6V. Then magnitude of 4B is

Solution:
[IMAGE] TOPIC:Electromagnetic induction
SUB TOPIC:Faraday's law
LEVEL: Moderate

Physics


Question:
In the figure shown $C_{1} = 11\mu F$ and $C_{2} = 5\mu F$ then
at steady state

The potential difference across C_1 is 5V

The potential difference between terminals of 15V battery is 9V

The potential difference V_a - V_b = - 4V

Solution:
[IMAGE] at steady state
$I(3) + 1(2) = 15$
I = 3
$kvL$
$C \rightarrow D \rightarrow E \rightarrow a \rightarrow b \rightarrow C$
$V_{c} - I(3) + \frac{q}{11} - 7 + \frac{q}{5} = V_{c}$
$\frac{q}{11} + \frac{q}{5} = 7 + 3 \times 3 = 16$
$q = 55\mu C$
Now KVL for a$\rightarrow$b
$V_{a} - 7 + \frac{q}{5} = V_{b}$
$\Rightarrow V_{a} - V_{b} = 7 - \frac{q}{5}$
$= 7 - \frac{55}{5}$ $= - \frac{20}{5} = - 4V$

Physics