Question:
There is a small source of light at some depth below the surface
of water (refractive index = 4/3) in a tank of large cross sectional
surface area. Neglecting any reflection from the bottom and absorption
by water, percentage of light that emerges out of surface is (nearly):
[Use the fact that surface area of a spherical cap of height h and
radius of curvature r is $2\pi rh$]

Solution:
Solid angle $d\Omega = 2\pi R^{2}(1 - cos\beta)$
Percentage of light
$= \frac{2\pi R^{2}(1 - cos\beta)}{4\pi R^{2}} \times 100$
$= \frac{1 - cos\beta}{2} \times 100 = (\frac{4 - \sqrt{7}}{8}) \times 100 \approx 17\%$

Physics


Question:
For two thermodynamic process temperature and volume diagram are
given. In first process, it is a straight line having initial and final
coordinates as (V_0, T_0) and (2V_0,2T_0), where as in second
process it is a rectangular hyperbola having initial and final
coordinates (V_0, T_0) and (2V_0, T_0/2). Then ratio of work done in
the two processes must be
nks izfØ fy, rkieku rFkk (V_0, T_0) rFkk (2V_0, 2T_0), gS] tgk¡ nwljs

Solution:
[IMAGE] First process is constant pressure izFke
Hence vr ] $W_{1} = nR\left( 2T_{0} - T_{0} \right) = nRT_{0}$
Equation of second process is$T = \frac{c}{V}$

Physics


Question:
Figure below shows two paths that may be taken by a gas to go
from a state A to a state C.
[IMAGE] In process AB, 400J of heat is added to the system and in process BC,
100 J of heat is added to the system. The heat absorbed by the system
in the process AC will be:;gk¡ nks iFk ftu
A ls voLFkk B rd ys tk;k tk ldrk gSA
 s AB] çØe esa 400 J rFkk çØe BC

Solution:
For a complete cycle
iw.kZ pfØ; çØe ds fy,
Q_cycle = W_cycle
$+ 400 + 100 + Q_{C \rightarrow A} = \frac{1}{2}\left( 2 \times 10^{- 3} \right)\left( 4 \times 10^{4} \right)$
⇒ Q~C→ A~ = - 460 J
⇒ Q_A→C = + 460 J

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:
Water flows in a horizontal tube (see figure). The pressure of
water changes by 700 Nm-2 between A and B where the area of cross
section are [IMAGE] and [IMAGE], respectively. Find the rate of flow of water
through the tube. (density of water [IMAGE] )

Solution:
using equation of continuity
$40V_{A} = 20V_{B}$
$\Rightarrow 2V_{A} = V_{B}$
Using Bernoulli's equation
$P_{A} + \frac{1}{2}\rho V_{A}^{2} = P_{B} + \frac{1}{2}\rho V_{B}^{2}$
$\Rightarrow P_{A} - P_{B} = \frac{1}{2}\rho\left( V_{B}^{2} - V_{A}^{2} \right)$
$\Rightarrow \Delta P = \frac{1}{2}1000\left( V_{B}^{2} - \frac{V_{B}^{2}}{4} \right)$
$\Rightarrow \Delta P = 500 \times \frac{3V_{B}^{2}}{4}$
$\Rightarrow V_{B} = \sqrt{\frac{(\Delta P) \times 4}{1500}} = \sqrt{\frac{(700) \times 4}{1500}m/s}$
Volume flow rate $= 20 \times 100 \times V_{B} = 2732{cm}^{3}/s$

Physics


Question:
A uniform rod of mass 200 gram and length $L = \frac{1}{3}m$ 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. Find the
angular speed (in rad/s) of rod when it becomes horizontal.

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

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:
In the given A.C. circuit, if battery voltage is
$V = 200\sqrt{2}sin(100\pi t)$, power delivered by battery will have a
power factor of

Solution:
$i_{1} = \frac{200\sqrt{2}}{\sqrt{30^{2} + 40^{2}}}sin(\omega t - \tan^{- 1}(\frac{40}{30}))$
$= 4\sqrt{2}sin(100\pi t - 53^{\circ})$
Phasor diagram
$\Rightarrow tan\theta = \frac{R_{y}}{R_{x}} = \frac{1}{3} \Rightarrow cos\theta = \frac{3}{\sqrt{10}}$

Physics


Question:
For an adiabatic process graph between PV & V for a sample of
ideal gas will be:
vkn'kZ xSl ds,d uewus dk:)ks"k çØe

Solution: [IMAGE] for adiabatic process,:)ks"k çØe

Physics


Question:
The positions of
 are shown on the
binding energy curve as shown in figure. The energy released in the
fusion reaction will be closest to [IMAGE] will be
closest to( in MeV )

Solution: Q = -2 × 1 - 7 × 5.5 + 8 × 7 = 56 - 2 - 38.5 = 15.5 ]

Physics


Question: Initially the block is at rest acceleration of the block is

Solution:
[IMAGE] $N + 24 - 100 = 0$ For vertical direction
$\therefore N = 76N$
Now, $0 \leq f_{s} \leq \mu_{s}N\ $
$0 \leq f_{s} \leq 76 \times 0.5$
$0 \leq f_{s} \leq 38N$
$\therefore 32 < 38\text{ Hence }f = 32$
$\therefore\ \ $Acceleration of block is zero.

Physics


Question:
In the shown figure three slits, separated by
$d = \sqrt{\frac{2\lambda D}{3}}$ and illuminated by monochromatic
parallel beam of light of wavelength $\lambda$. P is a point on the line
perpendicular to the plane of slits through S_3. If I_0 is the
intensity of the wave from each slit, the intensity at p is

Solution:
Phase difference between S_2 &
S_3$\ = \frac{2\pi}{\lambda}(\sqrt{D^{2} + d^{2}} - D)$
$= \frac{2\pi}{\lambda}(\frac{D \times d^{2}}{2D^{2}}) = \frac{2\pi}{3}$
Phase difference between S_1 &
S_3$\ = \frac{2\pi}{\lambda} = (\sqrt{D^{2} + 4d^{2}} - D)$
$= \frac{2\pi}{\lambda} \times D(\frac{4d^{2}}{2D}) = \frac{8\pi}{3}$
[IMAGE] Now using phase diagram
$A_{R} = A\sqrt{3}andIR = 3Io$

Physics


Question:
A stone is dropped from a certain height and can reach the ground
in 5s. But in its fall, the stone is stopped after, 3s of the fall for a
moment and is dropped again at once. Now the stone reaches the ground in
total time of seconds. Find the value of t?
 5s
 3s

 t

Solution:
h = [IMAGE] × 10 × 52 = 125 m
h1 = [IMAGE] × 10 × 32 = 45
80 = [IMAGE] × 10 × t2
t = 4 sec. Total time = 3 + 4 = 7 sec
TOPIC:KINEMATIC
SUB TOPIC: MOTION UNDER GRAVITY

Physics

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


Question:
In a meter bridge experiment S is a standard resistance. R is a
resistance wire. It is found thatbalancing length is $\mathcal{l}$ = 25
cm. If R is replaced by a wire of half length and half diameter that of
R ofsame material, then the balancing distance is$\mathcal{l}'$ (in cm).
What is the value of $\mathcal{l}^{'}/10$.

Solution:
$\frac{x}{R} = \frac{75}{25} = 3$
$R = \frac{\rho\mathcal{l}}{A} = \frac{4\rho\mathcal{l}}{\pi d^{2}}$
$R^{'} = \frac{4\rho\left( \frac{\mathcal{l}}{2} \right)}{\pi\left( \frac{d}{2} \right)^{2}} = 2R$
 then $\frac{X}{R^{'}} = \left( \frac{100 - \mathcal{l}}{\mathcal{l}} \right)$
$\frac{100 - \mathcal{l}}{\mathcal{l}} = \frac{X}{2R} = \frac{3}{2}$
$\mathcal{l =}40.00cm$

Physics


Question:
In a Young's double slit experiment, the separation between the
slits is 0.15 mm. In the experiment, asource of light of wavelength 589
mm is used and the interference pattern is observed on a screen kept1.5
m away. The separation between the successive bright fringes on the
screen is

Physics


Question:
A ring shaped tube contains two ideal gases with equal masses
and atomic mass numbers M_1=32 and M_2=28. The gases are separated by
one fixed partition X and another movable conducting partition Y which
can move freely without friction inside the ring. The angle is α, in
equilibrium as shown in thefigure (in degrees). Find the value of
[IMAGE],d vkdkj dh Vîqc esa nks vkn'kZ xSl ftudk æO;eku leku rFkk
ijek.kq æ M_1=32 rFkk M_2=28
foHkktd X ls vyx&vyx dh tkrh gS rFkk nqljk foHkktd,d xfreku pkydfoHkktd
X tks fcuk ?k"kZ.k xfr dj ldrk gSA fp=kuqlkj
dks.k α ¼fMxzh esa½ rks [IMAGE]

Solution:
At equilibrium, pressure and temperature are same. rFkk rki leku gSA
 Also, let cross-sectional area of tube be A and radius of ring be r.
ekuk nkc

Physics


Question:
The magnetic field is increasing at a constant rate$\alpha$
through a circular loop of radius aandthe axis of magnetic field
coincides with axis of loop as shown in figure. Theresistance per unit
length of loop wire is$\rho$. Choose correct option(s).

Current in loop PQRS is $\frac{a\alpha}{2\rho}$ anticlockwise.

Current in the loop PQRS is $\frac{a\alpha}{2\rho}$ clock wise.

Current in wire PR is $\frac{\pi a\alpha}{\rho}$.

Current in wire PR is$\frac{\pi a\alpha}{2\rho}$.

Solution:
Magnitude of Induced emf = $A\frac{dB}{dt}$
$\varepsilon = \pi a^{2}\alpha$
$I_{PQRS\ } = \frac{a\alpha}{2\rho}$
I_PR=0

Physics


Question:
A ball of mass m is suspended from a fixed point by an
inextensible string. A small particle of same mass m, moving downward at
an angle 37º with the vertical, collides head on with v_0 velocity.
Coefficient of restitution is 4/5.

Velocity of ball after impact is [IMAGE]

Velocity of ball after impact is
$\left( \frac{25}{34} \right)\text{v}_{\text{0}}$

Impulse produced by tension is m v_0[IMAGE]

Solution:
Say V_1 is velocity attained by ball after collision
and $\frac{4}{5} = \frac{\frac{3V_{1}}{5} + V}{V_{0}}$
$V_{1} = \left( \frac{27}{34} \right)v_{0}:V = \frac{11}{34}v_{0}$
Impulse produced by tension $= mv_{0}cos37^{\circ} + mvcos37^{\circ}$

Physics

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


Question:
In the figure, potential difference (in V) between A and B is
2n. Find n:
 fp=k

Solution:
Diode is in forward bias, so it will behave as simple wire so,
 voLFkk vr;g lk
[IMAGE] So vc $V_{ab} = \frac{30}{5 + 10} \times 5 = 10V$

Physics

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


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:
A reflecting surface is represented by the equation
[IMAGE] A ray travelling horizontally becomes vertical after reflection. The
co-ordinates of points where this ray incident is

Solution:
The slope of curve at such points is [IMAGE] or [IMAGE] TOPIC: Ray Optics
SUB TOPIC:Reflection by mirror
LEVEL:Moderate

Physics


Question:
In a meter bridge experiment S is a standard resistance. R is a
resistance wire. It is found that balancing length is
$\mathcal{l =}25cm$. If R is replaced by a wire of half length and half
diameter that of R of same material, then the balancing distance
$\mathcal{l}'$ (in cm) will now be_______.

Solution:
$\frac{X}{R} = \frac{75}{25} = 3$
$R = \frac{\rho\mathcal{l}}{A} = \frac{4\rho\mathcal{l}}{\pi d^{2}}$
$R^{'} = \frac{4\rho(\frac{\mathcal{l}}{2})}{\pi{(\frac{d}{2})}^{2}} = 2R$
Then $\frac{X}{R^{'}} = (\frac{100 - \mathcal{l}}{\mathcal{l}})$
$\frac{100 - \mathcal{l}}{\mathcal{l}} = \frac{X}{2R} = \frac{3}{2}$
$\mathcal{l =}40.00cm$

Physics


Question:
In the circuit shown, equivalent resistance between A and B is
nR,then value of n is

Physics


Question:
A particle of mass m is projected with a speed u from the ground
at an angle [IMAGE] horizontal (x-axis). When it has
reached its maximum height, it collides completely inelastically with
another particle of the same mass and velocity[IMAGE].
The horizontal distance covered by the combined mass before reaching the
ground is

Solution: [IMAGE] so horizontal range after collision[IMAGE]

Physics

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


Question:
The distance between two slits in a Young's double slit
experiment is 3 mm. The distance of the screen from the slits is 1 m.
Microwaves of wavelength 1 mm are incident on the plane of the slits
normally. The distance (in cm) of the first maxima on the screen from
the central maxima is $\frac{100}{\sqrt{n}}$. Find n.

Solution: At first maxima [IMAGE]

Physics


Question:
When a capillary tube is dipped in a liquid, the liquid rises to
a height h in the tube. The free liquid surface inside the tube is
hemispherical in shape. The tube is now pushed down so that the height
of the 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 be hemispherical.

Solution:
The angle of contact at the free liquid surface inside the
capillary tube will change such that the vertical component of the
surface tension forces just balance the weight of the liquid column.

Physics


Question:
A bullet of mass m hits the hanging block of mass M at its centre
completely inelastically with a velocity v and gets embedded in it as
shown, then choose the correct option

The tension in the string just before and after the collision changes
by$\frac{m^{2}v^{2}}{(m + M)^{\mathcal{l}}}$

The tension in the string just before and after the collision changes
by
$$\left\lbrack g + \right\rbrack m$$

Tension in the string just after the collision
becomes$m\left\lbrack g + \frac{mv^{2}}{(m + M)^{\mathcal{l}}} \right\rbrack$

Tension in the string will not change after the collision

Solution:
As T_1 = Mg
$T_{2} = (M + m)g + \frac{m^{2}v^{2}}{(M + m)^{\mathcal{l}}}$
$\therefore T_{2} - T_{1} = m\left\lbrack g + \frac{mv^{2}}{(m + M)^{\mathcal{l}}} \right\rbrack$

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: Select the correct alternative for an ideal gas

The change in internal energy in a constant pressure process from
temperature $T_{1}\ to\ T_{2}\ $is
equal to$\ nC_{v}\left( T_{2} - T_{1} \right)$ where $C_{v}\ $is the
molar specific heat at constant volume and n the number of moles of
the gas.

The change in internal energy of the gas and the work done by the
gas are equal in magnitude in an adiabatic process.

The internal energy does not change in an isothermal process and No
heat is added or removed in an adiabatic process.

Solution:
$\Delta U = n\frac{f}{2}R\left( T_{2} - T_{1} \right) = nC_{v}\left( T_{2} - T_{1} \right)$
 For adiabatic process $Q_{12} = 0$
$Q_{12} = \Delta U + W_{12}$
$W_{12} = - \Delta U$
For Isothermal process Temperature is constant.
So, change internal energy is zero.

Physics


Question:
A body A of mass m is moving in a circular orbit of radius R
about a planet. Another body B of mass [IMAGE] collides
with A with a velocity which is half [IMAGE] the
instantaneous velocity [IMAGE] of A. The collision
iscompletely inelastic. Then, the combined body

Escapes from the planet's Gravitational field

Falls vertically downwards towards the planet

Starts moving in a elliptical orbit around the planet

Solution:
Conserving momentum
[IMAGE] thus the combined mass will go on to an
elliptical path

Physics

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


Question:
Find the resultant focal length (in cm ) for following system
where the common radius of curvature is 15 cm. Glass has refractive
index of 1.5 and water has refractive index.

Physics


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/mthan.,d Qkbcj dh ifV~Vdk ftl fo|qr'khyrk 5 gS 0-25 lseh-
fo|qr foHko 2500 v o ok;q

Potential gradient strength in fiber sheet exceeds dielectric
strength of air
Qkbcj ifV~Vdk

All the above are correct
lHkh fodYi lgh gSA

Solution:
As D;ksafd V = V_1 + V_2
2500 = 0.0002 E_1 + 0.0023 E_2
2E_1 + 23 E_2 = 25 x 10^6
Also rFkk; E_1 = 5E_2
Solving gy E_2 = 0 757 x 10^6 V/m
E_1 = 3.788 x 10^6 V/m
As E_1> 3.6 x 10^6 Hence Air will breakdown

Physics

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


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

Solution:
For a solid sphere
$E = \frac{\rho r}{3\varepsilon_{0}}$
$E_{A} = \frac{- \rho R}{2(3\varepsilon_{0})}$
$|E_{A}| = \frac{\rho R}{6\varepsilon_{0}}$
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 $- \rho$)
$= - \frac{\rho R}{54\varepsilon_{0}}$
$E_{B} = E_{1A} + E_{2A} = \frac{\rho R}{3\varepsilon_{0}} - \frac{\rho R}{54\varepsilon_{0}} = \frac{17\rho R}{54\varepsilon_{0}}$
$\frac{|E_{A}|}{|E_{B}|} = \frac{9}{17}$

Physics

Electric field due to uniformly charged sphere


Question:
A bead can slide on a smooth straight wire and a particle of mass
m attached to the bead by a lightstring of length L. The particle is
held in contact with the wire and with the string taut and is then let
fall.
If the bead has mass 2 m then when the string makes an angle $\theta$
with the wire, the bead will haveslipped a distance

Solution:
Since the centre of mass will not move
2mx=my$\Rightarrow$y=2x
x + y = $L - L\cos\theta = L\left( 1 - \cos\theta \right)$
$3x = L(1 - cos\theta)$
⇒ $x = \frac{L(1 - cos\theta)}{3}$

Physics

Center of mass at rest in the horizontal direction


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

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:
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:
The rod is uniform & weigh 500 N. Find magnitude of tension so
that tension in both strings (S_1, S_2) aresame (strings ideal).

Solution:
[IMAGE] Taking torque about W.
- T_1 (0.4)+T_2 (0.3)+500(0.2)=0
T = 1000

Physics


Question:
A Young's double slit experiment is conducted in water
[IMAGE] as shown in the figure, and a glass plate of
thickness t and refractive index [IMAGE] is placed in the
path of[IMAGE]. The magnitude of the phase difference at
O is[IMAGE]. Find n. (Assume that
 is the wavelength of light in air). O is
symmetrical w.r.t. [IMAGE] and [IMAGE].

Solution:
Here path difference will be:
[IMAGE] TOPIC:Waves Optics
SUB TOPIC:Optical path difference
LEVEL:Moderate

Physics


Question:
For a fluid which is flowing steadily, the level in the vertical
tubes is best represented by

Solution:
From continuity equation, velocity at cross-section (1) is more
than that at cross-section (2).
Hence $P_{1} < P_{2}$
Hence option (1) is correct.

Physics


Question:
The P-V diagram of a cyclic thermodynamic process for an
ideal gas is as shown in figure. Its P-T curve will be

Solution:
A
Process A → B is isochoric process, therefore, options (C) and (D)
cannot be correct. Again B → C is an adiabatic process, therefore, B → C
should be described by equation [IMAGE] constant.
As γ is always greater than

Physics


Question:
$\ $Figure below shows two paths that may be taken by a gas to go
from a state A to a state C.
[IMAGE] In process AB, 400J of heat is added to the system and in process BC,
100J of heat is added to the system.The heat absorbed by the system in
the process AC will be

Solution:
For a complete cycle
$Q_{cycle\ } = W_{cycle\ }$
$+ 400 + 100 + Q_{C \rightarrow A} = \frac{1}{2}\left( 2 \times 10^{- 3} \right)\left( 4 \times 10^{4} \right)$
$\Rightarrow \ \ \ \ Q_{C \rightarrow A} = - 460J$
$\Rightarrow \ \ \ \ Q_{A \rightarrow C} = + 460J$

Physics


Question:
The length of a potentiometer wire is 120 cm and it carries a
current of 60 mA. For a cell of emf 5V and internal resistance of
$20\Omega$, the null point on it is found to be at 1000 cm. The
resistance of whole wire is

Solution:
[IMAGE] Potential gradient
$= \frac{5}{1000} = \frac{V_{p}}{1200}$
$V_{P} = 6V$
And $R_{P} = \frac{V_{P}}{I} = \frac{6}{60 \times 10^{- 3}} = 100\Omega$

Physics


Question:
In the figure shown cross-section area of holes are same and
very small in comparison to area of cross-section of water tank:
(tanks are the open from the top)
gq, gS)

Solution: For case-1: fLFkfr-1 ds fy, [IMAGE] For case-2: fLFkfr-2 ds fy, [IMAGE] For case-3: fLFkfr-3

Physics


Question:
The speed of water flowing out of orifice after the cylinder is
kept inside it

Solution:
[IMAGE] Let h′ be height of water column just after putting cylinder,
$\ \rho h'\left( A - \frac{A}{3} \right) = \left( \frac{H}{2} \right)A\rho$
$\ \Rightarrow h^{'} = \frac{3}{4}H$
$\ V' = \sqrt{2{gh}^{'}} = \sqrt{\frac{3}{2}gH}$

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 across C_2 is 2V.

the potential difference$\ V_{a} - V_{b} = - 4V$.

the potential difference between terminals of 15V battery is 9V.

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


Question:
A solid sphere of mass 10 kg is placed over two smooth inclined
planes as shown in figure. The normal reactions at 2 is 10x Newton. Find

Solution:
[IMAGE] Solving above equation
[IMAGE] =10 x
x = 5
TOPIC:Newton's first Law
SUB TOPIC:Free body diagram

Physics


Question:
During an experiment an ideal gas is found to obey condition
[IMAGE] constant. [IMAGE] density of gas).
The gas is initially at a temperature [IMAGE], pressure
[IMAGE] and density [IMAGE]. The gas
expands such that densitychanges to [IMAGE] then which of
following is/are correct?,d iz;ksx

The temperature of gas changes to [IMAGE].

The graph of above process on the P - T diagram is parabola.

The graph of above process on the P - T diagram is not rectangular
hyperbola.

Solution: PT = constant

Physics


Question:
When a person throws a meter stick it is found that the centre
of the stick is moving with speed 10 m/s and left end of stick with
speed 20m/s. Both points move vertically upwards at that moment. Then
angular speed of the stick is

Solution: Angular

Physics


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

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

Physics

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


Question:
The electric fields of two plane electromagnetic plane waves in
vacuum are given by:
${\overset{\rightarrow}{E}}_{1} = E_{0}\overset{\hat{}}{j}cos(\omega t - kx)\ and\ {\overset{\rightarrow}{E}}_{2} = E_{0}\overset{\hat{}}{k}cos(\omega t - ky)$
At t = 0, a particle of charge q is at origin with a velocity
j$\overset{\rightarrow}{v} = 0.8c\overset{\hat{}}{j}$(c is the speed
of the light in vacuum). The instantaneous force experienced by the
particle is

Solution:
Magnetic field vectors associated with this electromagnetic
wave are given by
${\overset{\rightarrow}{B}}_{1} = \frac{E_{0}}{c}\overset{\hat{}}{k}cos(kx - \omega t)\&{\overset{\rightarrow}{B}}_{2} = \frac{E_{0}}{c}\overset{\hat{}}{i}cos(ky - \omega t)$
$\overset{\rightarrow}{F} = q\overset{\rightarrow}{E} + q\left( \overset{\rightarrow}{V} \times \overset{\rightarrow}{B} \right)$
$= q\left( {\overset{\rightarrow}{E}}_{1} + {\overset{\rightarrow}{E}}_{2} \right) + q\left( \overset{\rightarrow}{V} \times \left( {\overset{\rightarrow}{B}}_{1} + {\overset{\rightarrow}{B}}_{2} \right) \right)$
By putting the value of
${\overset{\rightarrow}{E}}_{1},{\overset{\rightarrow}{E}}_{2},{\overset{\rightarrow}{B}}_{1}\&{\overset{\rightarrow}{B}}_{2}$
The net Lorentz force on the charged particle is
$\overset{\rightarrow}{F} = {qE}_{0}\lbrack 0.8cos(kx - \omega t)\overset{\hat{}}{i} + cos(kx - \omega t)\overset{\hat{}}{j} + 0.2cos(ky - \omega t)\overset{\hat{}}{k}\rbrack$
At t = 0 and at x = y = 0
$\overset{\rightarrow}{F} = qE_{0}\lbrack 0.8\overset{\hat{}}{i} + \overset{\hat{}}{j} + 0.2\overset{\hat{}}{k}\rbrack$

Physics

Lorentz force on a charged particle due to EM wave.


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.
${\overset{\rightarrow}{v}}_{A} = 3\overset{\hat{}}{j}m/s$and
${\overset{\rightarrow}{v}}_{B} = 4\overset{\hat{}}{i}m/s$. The shortest
distance between them will be

Solution:
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
$4\ sin53^{\circ} = 16/\mathbf{5}\mathbf{m}$

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] mass of electron).

Solution: [IMAGE] From (1) and (2)

Physics


Question:
The dependence of acceleration due to gravity g on the distance r
from the centre of the earthassumed to be a sphere of radius R of
uniformdensity is as shown in figure below:-
[IMAGE] The correct figure is

Solution:
The acceleration due to gravity at a depth d below the surface
of earth is
Where R-d=r= distance of location from thecentre of the earth. When
[IMAGE] From [IMAGE] till R=r, for which
[IMAGE] For [IMAGE] or [IMAGE] Here, R+h=r Therefore, the variation of g with distance r from centre
of earth will be as shown in figure(4). Thus, option (4) is correct

Physics

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


Question:
A block of mass 1 kg initially at rest is dropped from a height
h on the a spring of force constant k=50N/m. The maximum compression in
the spring is x=1 m then height hof the block is

Solution:
Change in gravitational potential energy
= Elastic potential energy stored in-compressed spring
$\Rightarrow mg(h + x) = \frac{1}{2}kx^{2}$
$\Rightarrow 1 \times 10(h + 1) = \frac{1}{2} \times 50 \times 1^{2}$
$\Rightarrow h = \frac{3}{2}$

Physics


Question:
A block of mass 10 kg is to be pulled on a horizontal rough
surface (µ =0.75) the minimum magnitude of applied force is

Solution:
$Fcos\theta - \mu N = 0$
$N = mg - {Fsin}\theta\ldots\ldots(2)$
From (1) and (2)
${Fcos}\theta = \mu\left( mg - {Fsin}\theta \right)$
$K = \frac{\mu mg}{cos\theta + \mu sin\theta}$
$F_{\min} = \frac{\mu mg}{\sqrt{1 + \mu^{2}}}\left( \because y = Asin\theta + Bcos\theta;ymax = \sqrt{A^{2} + B^{2}} \right)$
$F_{\min} = \frac{0.75 \times 10 \times 10}{\sqrt{1 + (0.75)^{2}}}$

Physics


Question: The internal energy of a system in adiabatic expansion

Solution:
Adiabatic expansion is adiabatic cooling.
$\because\ $temp. Decreases $\Rightarrow$ internal energy decreases

Physics


Question:
Position of a particle moving on x-axis as a function of time t
is given by
$x = \left( t^{3} - 6t^{2} + 3t + 4 \right)$meter.
What is the velocity of the particle when its acceleration is zero?

Solution:
$v = 3t^{2} - 12t + 3$
$a = 6t - 12 = 0$
$\Rightarrow t = 2\sec$
$v(t = 2sec) = - 9m/s$

Physics


Question:
A body of mass 6 kg is acted upon by a force which causes a
displacement in it given by $x = \frac{t^{2}}{4}$ metre where t is the
time in second. The work done by the force is 2 seconds is

Solution:
The velocity of the body a time t is given by
$= \frac{dx}{dt} = \frac{d}{dt}(\frac{t^{2}}{4}) = \frac{t}{2}$
At $t = 0, = u = 0\ and\ t = 2s, = 1{ms}^{- 1}$, Now, work done =
increase in KE
$= \frac{1}{2}m - \frac{1}{2}mu^{2} = \frac{1}{2}m - 0$
$= \frac{1}{2}m = \frac{1}{2} \times 6 \times (1)^{2} = 3J$

Physics


Question:
A Carnot engine operates between two reservoirs of temperatures
900 K and 300 K. The engineperforms 1200 J of work per cycle. The heat
energy (in J) delivered by the engine to the lowtemperature reservoir,
in a cycle, is 100n. Find n,d dkuksZ bUtu dks 900 K vkSj 300 K ds nks HkaMkjks a ds chp pyk;k
fuEu rki okys HkaMkj esa izfr pØ 100n (J esa) NksM+rk gSFind
n

Physics


Question:
A particle moving with initial velocity 3 m/s along x-axis
from origin. Its acceleration is varying with x as
[IMAGE] as shown in figure. At
[IMAGE] tangent to the graph makes an angle 60° with
positive x-axis as shown in diagram. Then at [IMAGE]

Solution:
TOPIC: Rectilinear motion
SUB TOPIC: Acceleration-displacement graph
LEVEL: Moderate

Physics


Question:
You are shown a photo of a car driven on the inside wall of a
huge vertical cylinder of radius 50 m. The coefficient of static
friction between the car tyres and the cylinder is $\mu_{s} = 0.8$. The
minimum speed in (m/s), at which the car can be driven like that is x.
Find x.

Solution:
F. B. D. Of car:-
[IMAGE] $\mu N = mg$....(1)
$N = \frac{mv^{2}}{R}$....(2)
From (1) and (2)
$V = 25m/s$

Physics


Question:
Sinusoidal waves 5.00 cm in amplitude are to be transmitted along
a string having a linear mass density [IMAGE]. If the
source can deliver an average power of 90 W and the string is under a
tension of 100 N, then the frequency at which the source can operate is
(take [IMAGE] )

Solution: [IMAGE] using data f = 30 Hz

Physics


Question:
A frictionless track ABCDE ends in a circular loop of radius R. A
body slides down the track from point A which is at height $h = 82cm$.
The maximum value of R for a body to complete the loop successfully is

Solution:
Using energy conservation:
$\frac{1}{2}mv_{B}^{2} = mgh$
$V_{B} = \sqrt{\frac{2mgh}{m}}$
$v_{B} = \sqrt{2hg}\ \ldots(1)$
Also to complete vertical circle $v_{B} = \sqrt{5gR}$...(2)
$\therefore R = \frac{2}{5}h = 32.8cm$

Physics


Question:
A ring shaped tube contains two ideal gases with equal masses
and atomic mass numbers M_1=32 and M_2=28. The gases are separated by
one fixed partition X and another movable conducting partition Y which
can move freely without friction inside the ring. The angle is $\alpha$,
in equilibrium as shown in thefigure (in degrees). Find the value
of$\frac{\alpha}{48}$.

Solution:
At equilibrium, pressure and temperature are same.
$\therefore P_{1} = P_{2}$ and $T_{1} = T_{2}$
Also, let cross-sectional area of tube be A and radius of ring be r.
$\therefore\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}$
$\Rightarrow \frac{360 - \alpha}{n_{1}} = \frac{\alpha}{n_{2}}$
$\left( \therefore V_{1} = (360 - \alpha)rA \right.\ $ and $V_{2} = \alpha rA$
$\Rightarrow M_{1}(360 - \alpha) = M_{2}\alpha\lbrack\therefore n_{1} = \frac{m}{M_{1}}andn_{2} = \frac{m}{M_{2}}\rbrack$
$\alpha = \frac{360M_{1}}{M_{1} + M_{2}} = 192^{\circ}$

Physics


Question:
A particle is moved under a force
$\overset{\rightarrow}{F} = x\overset{\hat{}}{j}N$(x in meter) from O to
A by two paths: path 1 is OBA andpath 2 is straight line OA. If w_1 and
w_2 are work done by the force $\overset{\rightarrow}{F}$ along the two
paths respectivelythe value of $\frac{w_{1}}{w_{2}}$is

Solution:
work done in path 2
$W_{2} = \int\overset{\rightarrow}{F} \cdot (dx\overset{\hat{}}{i} + dy\overset{\hat{}}{j}) = \int x\overset{\hat{}}{j} \cdot (dx\overset{\hat{}}{i} + dy\overset{\hat{}}{j})$as
the line: y = x
$= \int_{0}^{1}{xdx} = \frac{1}{2}J$
work done in path 1
$w_{OB} = 0$ (force is zero)
$W_{1} = \int\overset{\rightarrow}{F} \cdot dx\overset{\hat{}}{i} + \int\overset{\rightarrow}{F} \cdot dy\overset{\hat{}}{j}$
$= 0 + \int_{y = 0}^{1}{(1\overset{\hat{}}{j}) \cdot dy\overset{\hat{}}{j} = 1J}$
$\frac{w_{1}}{w_{2}} = 2$

Physics


Question:
In an [IMAGE] the Kinetic energy of
[IMAGE] is 48 MeV and Q-value of the reaction is 50 MeV.
If the mass number of the mother nucleus is 100N. Find N: (Assume that
daughter nucleus is in ground state)

Solution:
We have
[IMAGE] TOPIC:Modern physics
SUB TOPIC:Alpha decay
LEVEL: Moderate

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:
The particle displacement (in cm) in a stationary wave is given
by [IMAGE] Find the distance (in cm) between a node and
the next antinode.

Solution:
[IMAGE] compare with [IMAGE] λ = 75 cm
$\frac{\lambda}{4} = 18.75cm$

Physics


Question:
Four resistances of [IMAGE] and
[IMAGE] respectively in cyclic order to form Wheat
stone's network. The resistance (in ohms) that is to be connected in
parallel with the resistance of [IMAGE] to balance the
network is2n. Find n.

Solution: on solving

Physics


Question:
A non-isotropic solid metal cube has coefficients of linear
expansion as:$5 \times \ 10^{–5}\ /{^\circ}C$ along the x-axis and
$5 \times 10^{–6}\ /{^\circ}C$ along the y and the z-axis. If the
coefficient of volume expansion of the solid is
$C \times 10^{–5}\ /{^\circ}C$ then the value of C is.........
izdkj $5 \times \ 10^{–5}\ /{^\circ}C$ x- rFkk
izlkj xq.kkad $C \times 10^{–5}\ /{^\circ}C$ gks] rks C

Solution:
$V = 2\alpha_{2} + \alpha_{1}$
$= 10 \times 10^{- 6} + 5 \times 10^{- 5}$
$= 60 \times 10^{- 6}/\ ^{\circ}C$

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 6n. Find

Solution:
Right diode is reversed biased and left diode is forward biased
$\therefore V_{E} = 12.7 - 0.7 = 12Volt$

Physics


Question: Which of the following gives a reversible operation ?

Solution:
A logic gate is reversible if we can recover input data from
the outputeg. NOT gate,d rdZ gksxk ladsr
izkIr fd;k tk ldsA
eg. NOT }kj

Physics

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


Question:
The speed of water flowing out of orifice after the cylinder is
kept inside it
csyu dks VSd ckn fNnz

Solution:
[IMAGE] Let h' be height of water column just after putting cylinder,
ekuk csyu

Physics


Question:
An engine has an efficiency of 0.25 when temperature of sink is
reduced by58ºC, if ts efficiency is doubled, then the temperature of the
source in °C is 29n. Find n.

Solution:
Here,$\eta_{1} = 1 - \frac{T_{2}}{T_{1}}$
or $\ 0.25 = 1 - \frac{T_{2}}{T_{1}} \Rightarrow \frac{1}{4} = 1 - \frac{T_{2}}{T_{1}}$
$\Rightarrow \frac{T_{2}}{T_{1}} = 1 - \frac{1}{4} = \frac{3}{4}$.....(i)
According to question,
$\eta_{2} = 2\eta_{1},$ and $T_{2} = T_{2} - 58^{\circ}C$
$\therefore 2 \times \frac{1}{4} = 1 - \frac{\left( T_{2} - 58^{\circ}C \right)}{T_{1}}$
$\Rightarrow 1 - \frac{1}{2} = \frac{T_{2} - 58^{\circ}C}{T_{1}}$
$\frac{1}{2} = \frac{T_{2}}{T_{1}} - \frac{58^{\circ}}{T_{1}} \Rightarrow \frac{3}{4} - \frac{1}{2} = \frac{58}{T_{1}}$
$\Rightarrow T_{1} = 232^{\circ}C$

Physics


Question:
A uniform sphere of weight w and radius 3 m is being held by a
string of length 2 m. attached to a frictionless wall as shown in the
figure. The tension in the string is $\frac{5w}{n}$ then find the value
of n.

Solution:
[IMAGE] From geometry:
$\cos\theta = \frac{3}{5}$
$\sin\theta = \frac{4}{5}$
As sphere is at equilibrium
$T\sin\theta = w$
$T\left( \frac{4}{5} \right) = w$
$T = \frac{5w}{4}$

Physics


Question:
A conducting circular loop is pulled with a constant velocity v
towards a region of uniform magnetic field of induction B as shown in
figure then current induced in the loop is (d > 2r).

Solution: By lenz's law induced current will oppose change in flux.

Physics


Question:
The particle of mass 1 kg is acted upon by a
force$\overset{\rightarrow}{F} = F_{x}\overset{\hat{}}{i} + F_{y}\overset{\hat{}}{j}$
from t = 0 to t = 4 sec. Find the velocityof the particle at t=4 sec.,
if its initial velocity is
$\overset{\rightarrow}{u} = - \overset{\hat{}}{i} + \overset{\hat{}}{j}$and
the variation of applied force with time is as shown in the graphs

$\overset{\rightarrow}{v} = - \overset{\hat{}}{i} + \overset{\hat{}}{j}$

$\overset{\rightarrow}{v} = 2\overset{\hat{}}{i} + 10\overset{\hat{}}{j}$

Solution:
$\int_{0}^{4}{F_{x}dt = \Delta P_{x} = m(v_{x} - u_{x})}$
$\Rightarrow 0 = m(v_{x} - u_{x})$.
$\Rightarrow V_{x} = u_{x} = - 1$
$\int_{0}^{4}{F_{y}dt = \Delta P_{y} = m(v_{y} - u_{y})}$
$\Rightarrow \frac{1}{2} \times 4 \times 5 = 1(v_{y} - 1) \Rightarrow v_{y} = 11$
$\overset{\rightarrow}{v} = - \overset{\hat{}}{i} + 11\overset{\hat{}}{j}$

Physics


Question:
A projectile is projected at an angle of 37° to the horizontal. When
it is projected at 53°, its time of flight on level ground would have
been 1 sec. more. At what angle should it be projected so that its range
is 10 m less than either of the situations ?

Solution:
[IMAGE] T1 =[IMAGE] (sin 53° - sin 37°) = 1
R =[IMAGE] = 60 m
R' = 50 =[IMAGE] = sin 2θ
2θ = 53°
TOPIC:KINEMATIC
SUB TOPIC: 2 D

Physics

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


Question:
Light of wavelength 600 mm is incident upon a single slit with
width[IMAGE]. The figure shows the pattern observed on a
screen positioned 2 m from the slits. Determine the distance s (in cm).

Physics


Question:
Effective capacitance of parallel combination of two capacitors
C1 and C2 is 10$\mu F$. When these capacitorsare individually connected
to a voltage source of 1V, the energy stored in the capacitor C_2 is 4
times thatof C_1. If these capacitors are connected in series, their
effective capacitance will be

Solution:
Given C_1 + C_2$= 10\mu F$....(i)
$4\left( \frac{1}{2}C_{1}V^{2} \right) = \frac{1}{2}C_{2}V^{2}$
$\Rightarrow$ 4C_1=C_2....(ii)
from equation (i) and (ii)
$C_{1} = 2\mu F$
$C_{2} = 8\mu F$
If they are in series
$C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}} = 1.6\mu F$

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

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

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

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

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

Physics


Question:
There are two vectors
$\overset{\rightarrow}{OA}\&\overset{\rightarrow}{OB}$. Vector
$\overset{\rightarrow}{OA}$ is inxy plane and of magnitude 5 units and
vector $\overset{\rightarrow}{OB}$ isx-z plane and of magnitude
$\sqrt{2}$units as shown. Theangle between the vectors
$\overset{\rightarrow}{OA}\&\overset{\rightarrow}{OB}$is$\theta = \cos^{- 1}(\frac{3}{n\sqrt{2}})$.
Find n.

Solution:
$\overset{\rightarrow}{OA} = 5cos53^{\circ}\overset{\hat{}}{i} + 5sin53^{\circ}\overset{\hat{}}{j} = 3\overset{\hat{}}{i} + 4\overset{\hat{}}{j}$
$\overset{\rightarrow}{OB} = \sqrt{2}cos45^{\circ}\overset{\hat{}}{i} + \sqrt{2}sin45^{\circ}\overset{\hat{}}{k} = 1\overset{\hat{}}{i} + 1\overset{\hat{}}{k}$
$cos\theta = \frac{\overset{\rightarrow}{OA} \cdot \overset{\rightarrow}{OB}}{|\overset{\rightarrow}{OA}||\overset{\rightarrow}{OB}|} = \frac{3 + 0}{5 \cdot \sqrt{2}} = \frac{3}{5\sqrt{2}}$
$\Rightarrow \theta = \cos^{- 1}(\frac{3}{5\sqrt{2}})$

Physics


Question:
An object A is dropped from rest from the top of a 30 m high
building and at the same moment another object B is projected vertically
upwards with an initial speed of 15 m/s from the base of the building.
Mass of the object A is 2 kg while mass of the object B is 4kg. The
maximum height above the ground level attained by the centre of mass of
the A and B system is (take$g = 10m/s^{2}$)

Solution:
Using
$\frac{m_{1}{\overset{\rightarrow}{x}}_{1} + m_{2}{\overset{\rightarrow}{x}}_{2}}{m_{1} + m_{2}} = X_{cm}$
By putting values $X_{cm} = 10\ m$from base.
[IMAGE] Initial velocity of
COM$\Rightarrow \frac{m_{1}v_{1} + m_{2}v_{2}}{m_{1} + m_{2}} = 10m/supward$
Also, acceleration of COM is g thus using
$v^{2} = u^{2} + 2$as for COM
$0 = 10^{2} + 2 \times ( - 10) \times h_{\max}$
$h_{\max} = 5m$from the initial position of COM
Hence total height from base$= 10 + 5m = 15m$

Physics


Question:
In the given circuit, value of Y is:
 ifjiFk

will not execute ifjiFk dk;kZ fuor ugha gksxk

Physics

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


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 -$\rho$)$= - \frac{\rho R}{54\varepsilon_{0}}$

Physics

Electric field by uniformly charged sphere


Question:
A uniform sphere of mass 500 g rolls without slipping on a plane
horizontal surface with its centre moving at a speed
of[IMAGE]. Its kinetic energy is

Solution:
K.E. of the sphere = Translational K.E + Rotational K.E.
[IMAGE] K = Radius of gyration

Physics

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


Question:
A small ball of mass m is released from rest from a height h
above the liquid surface. The density of ball is $\frac{\rho}{3}$:
(assume no splash)

If $h \geq H$, then the ball will be able to enter the liquid of
higher density

If $h \geq \frac{7H}{2}$, then the ball will be able to strike the
bottom of the vessel

If h = 2H, then the ball will cross the interface of two liquids
with kinetic energy = mgH and If h = 0 then ball will float on the
surface.

Solution:
$mg\left( h + \frac{H}{2} \right) = \lbrack v\rho g\rbrack\frac{H}{2}$
h= H
$mg\left( h + \frac{H}{2} + \frac{H}{2} \right) = \lbrack v\rho g\rbrack\frac{H}{2} + \lbrack v2\rho g\rbrack\frac{H}{2}$
$mg\left( \frac{5H}{2} \right) - \frac{3}{2}mgH = K.E$
K.E. = mgH.

Physics


Question:
For the situation shown in figure below, there is no friction
between pulley and string. For which of the following values of mass of
block A, the system will stay at rest $(g = 10m/s^{2})$

Solution:
Tension in string =100 N in equilibrium F.B.D. of A
$F_{s} = 80 \leq \mu N$
$80 \leq \frac{1}{2}(mg - 60)$
$mg \geq 220$
$m \geq 22$

Physics


Question:
When a metallic surface is illuminated with monochromatic light
of wavelength [IMAGE] the stopping potential is
[IMAGE] When the same surface is illuminated with light
of wavelength [IMAGE] the stopping potential is
[IMAGE] Then the work function of the metallic surface
is $\frac{hc}{x\lambda},$ where value of x

Physics


Question:
The pressure on a square plate is measured by measuring the
force on the plate and the lengthof the sides of plate. If the maximum
error in themeasurement of force and length are respectively 4 and 2 then what is the maximum percentage error in themeasurement of pressure?

Solution: $P = \frac{F}{A};\ \ \lbrack P\rbrack = \left\lbrack \frac{F}{A} \right\rbrack = \left\lbrack \frac{{MLT}^{- 2}}{L^{2}} \right\rbrack = \left\lbrack {ML}^{- 1}T^{- 2} \right\rbrack$

Physics


Question:
The first member of the Balmer series of hydrogen atom has a
wavelength of The wavelength of the second member of the Balmer
series (in nm) is 81n. Find n.

Solution:
$\frac{1}{\lambda} = RZ^{2}(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})$
$\frac{1}{\lambda_{1}} = R(1)^{2}(\frac{1}{2^{2}} - \frac{1}{3^{2}}) = \frac{5R}{36}$
$\frac{1}{\lambda_{2}} = R(1)^{2}(\frac{1}{2^{2}} - \frac{1}{4^{2}}) = \frac{3R}{16}$
$\frac{\lambda_{2}}{\lambda_{1}} = \frac{20}{27}$
$\lambda_{2} = \frac{20}{27} \times 6561A = 4860Å$
$= 486nm$

Physics


Question:
V - T diagram for a process of a given mass of ideal gas is as
shown in the figure. During the process pressure of gas.

Solution:
$V = KT + C$
$P = \frac{nRT}{V}$
$\Rightarrow P = \frac{nRT}{KT + C}$
$\frac{dP}{dT} = \frac{nRC}{(KT + C)^{2}}$
As C < 0 by diagram
$\Rightarrow \frac{dP}{dT} < 0$ for all T
$\Rightarrow$ P continuously decreases

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 -$\rho$)
$= - \frac{{KQ}^{'} \times 4}{{OR}^{2}} = - \frac{\rho R}{54\varepsilon_{0}}$

Physics

Electric field by uniformly charged sphere


Question:
A uniformly thick wheel with moment of inertial I and radius R is
free to rotate about its centre of mass (see figure). A massless string
is wrapped over its rim and two blocks of masses [IMAGE] and [IMAGE] are attached to the ends of the string. The
system is released from rest. The angular speed of the wheel
when[IMAGE] descents by a distance his

Physics

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


Question:
The following figure depicts a wave travelling in a medium.
Which pair of particles are in phase.

Solution:
Point is moving down though
they are at the same distance from equilibrium position at the instant
shown. Therefore B and F are not in phase.
By same argument, A and G are in phase.
TOPIC:Waves in a string
SUB TOPIC: Particles in phase

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. The magnetic field at the centre of the coil is zero tesla.
The ratio $\frac{i_{1}}{4i_{2}}$ 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 railway flat car has an artillery gun installed on it and
contains shells for fring The combined system excluding shell has a mass
M and moves with a velocity $v_{0}$. The barrel of the gun makes an
angle $\alpha$ with the horizontal. A shell of mass m leaves the barrel
at a speed relative to initial state of barrel in the forward
direction. The speed of the flat car so that it may stop after the
firing is: (neglect frinstion)

Solution:
u - velocity of shell with respect to gun.
In horizontal direction ${\overset{\rightarrow}{F}}_{ext} = 0$, so
centre of mass move with constant velocity in horizontal plane.
$(M + m)v_{0} = m(v_{0} + ucos\alpha)$
$V_{0} = \frac{mucos\alpha}{M}$

Physics


Question:
Sound of frequency 1000 Hz from a stationary source is reflected
from an object approaching the source at 30 m/s, back to a stationary
observer located at the source. The speed of sound in air is 330 m/s.
The frequency of the sound heard by the observer is

Physics


Question:
A solid sphere of mass m and radius r is gently placed on a
conveyer belt moving with constant velocityV. If the coefficient of
friction between the belt and sphere is$\frac{2}{7}$, the distance
travelled by the centre of the sphere before it starts pure rolling is.

Solution:
Initial situation w.r.t. the conveyer belt:
$\mu mgR = \frac{2}{5}{mR}^{2} \cdot \alpha$ $\Rightarrow$
$\alpha = \frac{5}{7} \cdot \frac{g}{R}$
At pure rolling:
$v^{'} = v - (\mu g)t$and $\omega^{'} = 0 + \left( \frac{5}{7} \cdot \frac{g}{R} \right)t$
(Since at pure rolling)
$\Rightarrow$ $v - \mu gt = \frac{5}{7} \cdot g \cdot t$
$\Rightarrow$ $t = \frac{v}{g}\left( \because\mu = \frac{2}{7} \right)$
For centre of the sphere:$S = v\left( \frac{v}{g} \right) - \frac{1}{2} \cdot \left( \frac{2}{7} \cdot g \right)\left( \frac{v^{2}}{g^{2}} \right)$
$S = \frac{v^{2}}{g} - \frac{v^{2}}{7g}$ (w.r.t. belt)
In this time, the belt with speed will move a distance
of$s_{b} = vt = v\left( \frac{v}{g} \right) = \frac{v^{2}}{g}$
[IMAGE] Distance travelled w.r.t. ground will be:$S_{g} = \left| S - S_{b} \right| = \left| \left( \frac{v^{2}}{g} - \frac{v^{2}}{7g} \right) - \left( \frac{v^{2}}{g} \right) \right| = \frac{v^{2}}{7g}$

Physics


Question:
A leak proof cylinder of length 1m, made of a metal which has
very low coefficient of expansion is floating vertically in water at 0ºC
such that its height above the water surface is 20 cm. When the
temperature of water is increased to 4ºC, the height of the cylinder
above the water surface becomes 21 cm. The density of water at T = 4ºC,
relative to the density at T = 0ºC is close to:,d yhd iwQ 1m yEck csyukdkj crZu,d,slh cuk gqvk gS ftldk
4ºC rd c<+k rks blds ckgj jgus okys Hkkx

Physics


Question:
In a double-slit experiment, at a certain point on the screen
the path difference between the two interfering waves is
$\frac{1}{8}$thof a wavelength. The ratio of the intensity of light at
that point to that at the centre of a bright fringe is

Solution:
$I = I_{0}\cos^{2}(\frac{\Delta\phi}{2})$
$\frac{I}{I_{0}} = \cos^{2}\lbrack\frac{\frac{2\pi}{\lambda} \times \Delta x}{2}\rbrack = \cos^{2}(\frac{\pi}{8})$;
$\frac{I}{I_{0}} = 0.853$

Physics


Question:
The coercivity of a small magnet where the ferromagnet gets
demagnetized is 3 × 10^3 Am^-1. The current required to be passed in
a solenoid of length 10 cm and number of turns 100, so that the magnet
gets demagnetized when inside the solenoid, is

Solution:
For solenoid
$\frac{B}{\mu_{0}} = H$ or $B = \mu_{0}nI$
$H = nI$
$\Rightarrow 3 \times 10^{3} = \frac{100}{0.1} \times I$
Ι = 3A

Physics


Question:
Figure shows the variation of internal energy "U" with the
density "$\rho$ " of one mole of ideal diatomic gas. Process BA is a
part of rectangular hyperbola. If the work done by gas in the process BA
is 7n joules. Find n ?
 &ijek.kqd xSl ds,d fLFkfrt
 ?kuRo [IMAGE]

Solution: XY = constant constant (isobaric process) fu;rkad¼lenkch

Physics


Question:
V-T diagram for a process of a given mass of ideal gas is as
shown in the figure. During the process pressure of gas.
çfØ;k

Solution:
[IMAGE] As C<0 by diagram
tSlk fd fp= }kjkC<0
[IMAGE] P continuously decreasesyxkrkj ?kVrk gSA

Physics


Question:
Which one of the following choices is closest to the order of
the angular momentum of the Earth (expressed in base MKS units)
associated with its rotation about its axis ?

Solution:
m = 6 × 1024 kg
R = 6.4 × 106 m
I = [IMAGE] mR2
L = Iω = [IMAGE] × 6 × 1024 × (6.4 × 106)2
×[IMAGE] = [IMAGE] × 0.39 × 1034

Physics


Question:
A non-viscous fluid of density [IMAGE] 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)
Solving we get,
[IMAGE] (3) Work done by gravitation force per unit volume = negative of the
increase in gravitational potential energy per unit volume =
$- (\rho gh)$
(4) Work done by elastic force per unit volume = decrease in
pressure = [IMAGE] = $2\rho gh$
TOPIC:Fluid dynamics
SUB TOPIC:Bernoulli's theorem

Physics


Question:
The stress versus strain graphs for wires of two materials A and
B are as shown in the figure. If $Y_{A}$ and $Y_{B}$ are the Young's
modulii of the materials, then
izfrcy rFkk fod`fr xzkQksa dks nks inkFkksZ A rFkk B

Solution: $\frac{\ Stress\ }{\ Strain\ } = Y = \ slope\ of\ graph\ $

Physics


Question: The current (in A) flowing through battery is2n. Find the value

Solution:
The diode D_1is reverse-biased and acts as an open switch as
shown in Fig. So, there is no currentthrough D_1and the
However, D_2is forward biased and acts like a short-current or closed
switch. The current drawn isI = 12/(2 + 4) = 2A.

Physics


Question:
A particle of mass m and charge q is released from rest in a
uniform electric field. If there is no other force on the particle, the
dependence of its speed v on the distance x travelled by it is correctly
given by (graphs are schematic and not drawn to scale)

Solution:
$V^{2} = \frac{2qE}{m}x$
$$

Physics

Motion of charged particle in uniform electric field


Question: What would be the magnetic moment of given current distribution

Solution:
$M = \frac{I}{4}\left\{ \pi(2R)^{2} - \pi(R)^{2} \right\}$ =
$\frac{3\pi{IR}^{2}}{4}$

Physics


Question:
Water flows in a horizontal tube (see figure). The pressure of
water changes by [IMAGE] between Aand B where the area of
cross section are [IMAGE] respectively. Find the rate of
flow ofwater through the tube.
(density of water [IMAGE],d {kSfrt uyh esa ikuh cg jgk gS ¼fp= ns[ksa½A bl uyh esa A ls B ds
chp ikuh ds ncko vUrj
dh vuqçLFk dkV A uyh esa ikuh

Solution: Using equation of continuity lehdj.k Using Bernoullies equation cjuksyh lehdj.k

Physics


Question:
A particle is revolving in a circle of radius r and centre at
'O' with uniform angular velocity [IMAGE]. Let angular
velocity at A about O, B, and C is[IMAGE],
the correct option

Solution:
[IMAGE] TOPIC:Rigid body dynamics
SUB TOPIC: Angular velocity

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:
Two men of equal masses stand at opposite ends of the diameter of
a turntable disc of a certain mass, moving with some angular velocity.
The two men start moving towards the center of the turntable atequal
rates. Then choose the correct option

Kinetic energy of the system will increase.

Angular momentum of the system will increase.

Angular momentum of the system will decrease.

Solution:
$I_{1}\omega_{1} = I_{2}\omega_{2}$
Since, men move towards middle of turn table Ι decreases hence
$\omega_{2}$increases.
$\therefore$ $KE = \frac{L^{2}}{2I};L =$constant and I decreases so KE
increases

Physics


Question:
The expansion of unit mass of an ideal gas atconstant pressure
is show in the diagram. Here

Solution:
$PV = nRT(K)$
$V = \frac{nR}{P}T(K)$
$V = \frac{nR}{P}\lbrack T_{c}^{o} + 273\rbrack$
$V = \frac{nR}{P}T_{c}^{o} + 273\frac{nR}{P}$
$\Rightarrow T_{c}^{o} = \frac{P}{nR}V - 273$
$y = mx - c$

Physics


Question:
A observer standing in a cart which has acceleration g on
horizontal road is observing vertical circular motion of a pendulum. One
end of the pendulum string is at rest at O with respect to the cart.
Choose the possible position of minimum speed for the pendulum with
respect to cart during complete vertical circular motion from the figure
given below

Solution:
Opposite to geffect.
TOPIC: CIRCULAR MOTION
SUB TOPIC: DYNAMICS

Physics

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


Question:
If N Joules of heat is to be transferred to nitrogen in the
isobaric heating process for that gas to perform the work A =2.0 J then
the value of N is
[IMAGE] xSl dks nh x;h dk eku N twy gksrk gS] rks
N

Solution: W = 2J

Physics


Question:
Two liquids of densities $\rho_{1}\ $and
$\rho_{2}\left( \rho_{2} = 2\rho_{1} \right)$ are filled up behind a
square wall of side 10 m as shown in figure. Each liquid has a height of
5 m. The ratio of the forces due to these liquids exerted on upper part
MN to that at the lower part NO is (Assume that the liquids are not
mixing)

Physics


Question:
A balloon rises from rest on the ground with constant
acceleration $\left( \frac{g}{8} \right)$. A stone is dropped when the
balloon has risen to a height H metre. Then, the time taken by the stone
to reach the ground is: (t=0 is the moment when stone is dropped)

Solution:
$v^{2} = 0 + 2 \times \frac{g}{8} \times H \Rightarrow v = \frac{\sqrt{gH}}{2}$
$- H = \frac{\sqrt{gH}}{2}t - \frac{1}{2}{gt}^{2} \Rightarrow t = \frac{\sqrt{gH} \pm 3\sqrt{gH}}{2g}$
$\Rightarrow t = 2\sqrt{\frac{H}{g}}$ (taking positive value)

Physics


Question:
An open rectangular tank of height 3m and length and width
respectively 5m and 3m contains water up to height 2m. It is accelerated
horizontally along the longer side. Find the acceleration of the tank so
that 10 of water spills

Solution:
$\ \Rightarrow y = \frac{12}{5}$
$\ tan\theta = \frac{y}{5} = \frac{a}{g}$
$\ \Rightarrow a = \frac{24}{5}$

Physics


Question:
Two blocks of mass 10 kg and 4 kg are connected by a spring of
negligible mass and are placed on a frictionless horizontal surface. An
impulse gives a speed of $14{ms}^{- 1}$ to the heavier block in the
direction of the lighter block. Then, the velocity of the centre of mass
is

Solution: $= \frac{10 \times 14 + 4 \times 0}{10 + 4} = 10m/s$.

Physics


Question:
An asteroid is moving directly towards the centre of the earth.
when at a distance of 10R (R is theradius of the earth) from the earths
centre, it has a speed of 12 km/s. Neglecting the effect of
earth'satmosphere, what will be the speed of the asteroid when it hits
the surface of the earth (escape velocityfrom the earth is 11.2

Solution:
KE_i + PE_i = KE_f + PE_i
$\frac{1}{2}mu_{0}^{2} + \left( - \frac{GMm}{10R} \right) = \frac{1}{2}mv^{2} + \left( - \frac{GMm}{R} \right)$
$v^{2} = u_{0}^{2} + \frac{2GM}{R}\left\lbrack 1 - \frac{1}{10} \right\rbrack$
$v = \sqrt{u_{0}^{2} + \frac{9}{5}\frac{GM}{R}}$
$= \sqrt{12^{2} + \left( \frac{9}{5} \right)\frac{(11.2)^{2}}{2}}$
$= \sqrt{144 + 0.9(11.2)^{2}} = \sqrt{256.896}$
= 16.028 km/s $\simeq 16$

Physics


Question:
A body is acted upon by a force which is inversely proportional to
the distance covered. The work done will be proportional to

Solution:
[IMAGE] TOPIC: WPE
SUB TOPIC: WORK DONE BY VARIABLE FORCE

Physics

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


Question:
Two seconds after projection, a projectile is travelling in a
direction inclined at 30º to the horizontal after one more second, it is
travelling horizontally. Then the speed of projection(u) and angle of
projection ($\theta$) are given by

Solution:
$T = 6sec = \frac{2usin\theta}{g} \Rightarrow usin\theta = 30$
$\tan 30^{\circ} = \frac{{usin}\theta - gt}{{ucos}\theta} = \frac{30 - 20}{{ucos}\theta}$which
gives $ucos\theta = 10\sqrt{3}$
$$tan\theta =

Physics


Question:
A body of mass m and radius R rolling horizontally without
slipping at a speed v climbs a ramp to a height[IMAGE].
The rolling body can be.

Solution:
From energy
$K = \frac{R}{\sqrt{2}}$
[IMAGE] Body is a disc

Physics

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


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:
An ideal gas whose internal energy U is at absolute zero
temperature is equal to zero undergoes areversible adiabatic
compression. If U, P, V, T represents the internal energy, pressure,
volume and absolute temperature respectively for ideal gas
then $\left( \gamma = \frac{C_{P}}{C_{v}} \right)$.

Solution:
$U\ \alpha\ T$
For adiabatic process ${TV}^{\gamma - 1} =$ constant
$\therefore{UV}^{\gamma - 1} =$constant
$\Rightarrow {VU}^{\frac{1}{\gamma - 1}} =$constant

Physics


Question:
The current (in A) flowing through battery in the given circuit
 ifjiFk

Solution:
The diode D_1is reverse-biased and acts as an open switch as
shown in Fig. So, there is no currentthrough D_1and the
/kkjk izokfgr ugha gksxhA
However, D_2is forward biased and acts like a short-current or closed
switch. The current drawn isI = 12/(2 + 4) = 2A.;)fi, D_2 vxz

Physics

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


Question:
A small ball of mass m is released from rest from a height h
above the liquid surface. The density of ballis $\frac{\rho}{3}$:
(assume no splash)
 xsan dks fLFkjkoLFkk

) If h ≥ H, then the ball will be able to enter the liquid of
higher density
 ?kuRo

If h ≥ $\frac{7H}{2}$, then the ball will be able to strike the
bottom of the vessel;fn h ≥ $\frac{7H}{2}$, gks rks xsan ik=k

If h = 2H, then the ball will cross the interface of two liquids
with kinetic energy = mgH and if h = 0 then ball will float on the
surface;fn h = 2H, gks rks xsan nksuks nzoks lrg ij xfrt = mgH h = 0 gks rks xasn lrg ij rSjxhA

Solution: [IMAGE] K.E. = mgH.

Physics


Question:
In the system shown in figure mA = 4m, mB = 3m and mC = 8m.
Friction is absent everywhere. String is light and inextensible. If the
system is released from rest find the acceleration of block B

Solution:
$T = m_{B}b$[Newtons II law for B in horizontal direction]
$T = \left\lbrack m_{A} + m_{e} \right\rbrack c$ I law for A
and C in horizontal direction]
$m_{A}g - T = m_{A}a$ [Newton's II law for A in vertical direction].
$\frac{m_{A}g - T}{m_{A}} = \frac{T}{m_{B}} + \frac{T}{m_{A} + m_{C}}$
$\frac{4mg - T}{4m} = \frac{T}{3m} + \frac{T}{12m} \Rightarrow T = \frac{3mg}{2} \Rightarrow b = \frac{g}{2}$

Physics


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: $\frac{1}{\text{v}} - \frac{1}{- 20} = \frac{1}{10} \Rightarrow \text{v} = 20$

Physics


Question:
Figure shows the variation of internal energy "U" with the
density "$\rho$ " of one mole of ideal diatomic gas. Process BA is a
part of rectangular hyperbola. If the work done by gas in the process BA
is 7n joules. Find n ?

Solution:
For rectangular hyperbola
XY = constant
$U\rho =$ constant
$nC_{V}T\rho =$ constant
$n\frac{5}{2}RT\frac{M}{V} =$constant
$\frac{5}{2}P\frac{VM}{V} =$constant $\Rightarrow$ P = constant
(isobaric process)
$\Delta U_{B \rightarrow A} = 37 - 2 = 35J = \frac{5}{2}nR\Delta T$
$14J = nR\Delta T$
$\therefore$ $W = 14J$

Physics


Question:
A particle moving with kinetic energy E has de Broglie
wavelength[IMAGE]. If energy [IMAGE] is
added to its energy, the wavelength becomes[IMAGE]. Value
of[IMAGE], is

Physics


Question:
A particle is released from a certain height H = 400 m. Due to the
wind the particle gathers the horizontal velocity component vx = ay
where [IMAGE] and y is the vertical displacement of the
particle from point of release, then the horizontal drift of the
particle when it strikes the ground is

Solution:
Vy = [IMAGE]... (1)
Vx =[IMAGE]... (2)
[IMAGE] Dividing (1) by (2)
[IMAGE] or [IMAGE] dy = 2dx
Integrating
[IMAGE] ∴ x = 2.67 km
TOPIC:KINEMATIC
SUB TOPIC: 2 D

Physics

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


Question:
A horizontal conductor is oriented north south and carries some
current. A positively charged particle located vertically above it and
having a velocity directed northward experiences an upward force. What
is the direction of the force if this charged particle were located to
the east of the conductor and had a velocity directed towards the
conductor.

Solution:
12 in first case according to [IMAGE] Magnetic field
will be towards west at position of charge particle for this current in
conductor should be towards south in second case magnetic field at
position of particular is upward so according to
[IMAGE] force will be towards north.

Physics


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

Physics


Question:
A particle is projected at an angle tan-1[IMAGE].
Find the height at which velocity of the particle is
[IMAGE] (m/sec). (g = 10 ms-2)

Solution:
tan θ
[IMAGE] Uy [IMAGE] 9 = 27 + 2(-10)Sy
Sy = 0.9

Physics

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


Question: [IMAGE] = [IMAGE] X

Solution:
L.H.S. is dimensionless. So
TOPIC: UNIT & DIMENSION
SUB TOPIC: DIMENSIONS

Physics

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


Question:
The velocity of an object moving rectilinearly is given as a
function of time by $v = 4t - 3t^{2}$, where v is in m/s and t is in
seconds. The average velocity of particle between t=0 to t=2 seconds is

Solution: $< v > = \frac{\int_{0}^{2}\mspace{2mu} vdt}{\int_{0}^{2}\mspace{2mu} dt} = \frac{\int_{0}^{2}\mspace{2mu}\left( 4t - 3t^{2} \right)dt}{2} = 0$

Physics


Question:
A small block of mass m is released from rest from point A inside
a smooth hemisphere bowl of radius R, which is fixed on ground such that
OA is horizontal. The ratio (x) of magnitude of centripetal force and
normal reaction on the block at any point B varies with $\theta$ as

Solution:
$\frac{mv^{2}}{R} = N - mg\ sin\theta$
By energy conservation,
$mgR\ sin\theta = \frac{1}{2}{mv}^{2}$
$\frac{{mv}^{2}}{R} = 2mg\ sin\theta$
$N = 3mg\ sin\theta$
$Ratio = \frac{{mv}^{2}}{RN} = \frac{2}{3}$ (Constant)
$x = \frac{2}{3}$

Physics


Question:
Projectiles are simultaneously projected from the same point
with speeds u and 2 u, at angles $\theta$ and $2\theta$ respectively
with horizontal. The locus of second projectile with respect to the
first is

$y = (\frac{2cos2\theta - cos\theta}{2sin2\theta - sin\theta})x$

$y = {(\frac{2sin2\theta - sin\theta}{2cos2\theta - cos\theta})}^{x}$

$y = (\frac{2cos2\theta + cos\theta}{2sin2\theta + sin\theta})x$

$y = (\frac{2sin2\theta + sin\theta}{2cos2\theta + cos\theta})x$

Solution:
For first projectile$y_{1} = usin\theta t - \frac{1}{2}gt^{2}$
$y_{2} = 2usin2\theta t - \frac{1}{2}{gt}^{2}$
$x_{1} = ucos\theta t$ $x_{2} = 2ucos2\theta t$
$y = y_{2} - y_{1} = t\lbrack 2usin2\theta - usin\theta\rbrack$
$x = x_{2} - x_{1} = t\lbrack 2ucos2\theta - ucos\theta\rbrack$
$\frac{y}{x} = \frac{2usin2\theta - usin\theta}{2ucos2\theta - ucos\theta}$
$y = x\lbrack\frac{2sin2\theta - sin\theta}{2cos2\theta - cos\theta}\rbrack$

Physics


Question:
In the figure shown a parallel plate capacitor has a dielectric
of width d/2 and dielectric constant K = 2.The other dimensions of the
dielectric are same as that of the plates. The plates P_1 and P_2 of
thecapacitor have area each. The energy of the capacitor is:
fp=k lekUrj iê la/kkfj=k esa ijkoS|qr dh pkSM+kbZ d/2
rFkk ijkoS|qrkad K = 2 gSAijkoS|qr

Solution:
$U = \frac{1}{2}C_{eq}v^{2}$
$C_{1} = \frac{k\varepsilon_{0}A}{d/2} = \frac{2\varepsilon_{0}A}{(d/2)}$
and $C_{2} = \frac{\varepsilon_{0}A}{d/2}$
$C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}};C_{eq} = \frac{\left( 2\frac{\varepsilon_{0}A}{d/2} \right) \times \frac{\varepsilon_{0}}{d/2}}{3\frac{\varepsilon_{0}}{d/2}}$
$= \frac{4}{3}\frac{\varepsilon_{0}A}{d}\ or\ C_{eq} = \frac{\varepsilon_{0}A}{d - \frac{d}{2} + \frac{d}{4}} = \frac{4\varepsilon_{0}A}{3d}$
$U = \frac{1}{2}\left( \frac{4}{3}\frac{\varepsilon_{0}A}{d} \right)V^{2} = \frac{2}{3}\left( \frac{\varepsilon_{0}}{d} \right){AV}^{2}$

Physics

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


Question:
In the figure shown a parallel plate capacitor has a dielectric
of width d/2 and dielectric constant K = 2.The other dimensions of the
dielectric are same as that of the plates. The plates P_1 and P_2 of
thecapacitor have area each. The energy of the capacitor is

Solution:
$U = \frac{1}{2}C_{eq}v^{2}$
$C_{1} = \frac{k\varepsilon_{0}A}{d/2} = \frac{2\varepsilon_{0}A}{(d/2)}$
and $C_{2} = \frac{\varepsilon_{0}A}{d/2}$
$C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}};$
$C_{eq} = \frac{\left( 2\frac{\varepsilon_{0}A}{d/2} \right) \times \frac{\varepsilon_{0}}{d/2}}{3\frac{\varepsilon_{0}}{d/2}}$
$= \frac{4}{3}\frac{\varepsilon_{0}A}{d}\ or\ C_{eq}$
$= \frac{\varepsilon_{0}A}{d - \frac{d}{2} + \frac{d}{4}} = \frac{4\varepsilon_{0}A}{3d}$
$U = \frac{1}{2}\left( \frac{4}{3}\frac{\varepsilon_{0}A}{d} \right)V^{2}$
$= \frac{2}{3}\left( \frac{\varepsilon_{0}}{d} \right){AV}^{2}$

Physics


Question:
A sphere of mass m is given some angular velocity about a
horizontal axis through its centre, andgently placed on a plank of mass
m. The coefficient of friction between the two is $\mu$. The plank rests
on a smooth horizontal surface. The initial acceleration of the sphere
relative to the plank will be $x(\mu g)$. Find the value of x.

Solution:
$a_{sphere\ \ } = \frac{\mu mg}{m} = \mu g( + \widehat{i})$
$a_{plank\ \ } = \frac{\mu mg}{m} = \mu g\left( - \widehat{i} \right)$
${\ a}_{relative}\ \ = 2\mu g$

Physics


Question: has four choices

Physics

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


Question:
A container is filled with a liquid of density ρ. Container is
accelerating on a horizontal surface with acceleration
[IMAGE] 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),d ik=$\rho$?kuRo ds nzo ls Hkjk gSA ik=k {kSfrt lrg
ij$\frac{3g}{4}$Roj.k vksj Rofjr gS rFkk nzo fp=kkuqlkj ik=k ds
fodYi gS P_atm = 0 o AB

Pressure at A is[IMAGE] Aij nkc[IMAGE] gksxkA

Pressure at A is [IMAGE] Aij nkc[IMAGE] gksxkA

Solution:
For point A from top surface PA=[IMAGE] and from
right wall P_A=[IMAGE] lrg ls fcUnq A ds fy, PA = [IMAGE] rFkk nka;h nhokj
lsP_A=[IMAGE] and rFkk

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:Fluid
SUB TOPIC:Bouyancy
LEVEL: Moderate

Physics


Question:
A disc has mass 30 kg and radius 1m. The moment of inertia in
kgm^2about a tangent to its rim in itsplane is 7.5n. Find the value on
n.

Solution:
Moment of inertia of a disc about its diameter is
$I_{d} = \frac{1}{4}{MR}^{2}$
Now, according to perpendicular axis theorem moment of inertia of disc
about a tangent passingthrough rim and in the plane of disc is
$$= = 37.5$$

Physics


Question:
The potential energy of a particle of mass 1 kg in a
conservative field is given as $U = (3x^{2}y^{2} + 6x)J$, where x and y
are measured in meter. Initially particle is at (1,1) & at rest then

Initial acceleration of particle is $6\sqrt{5}{ms}^{2}.$

If particle is left free it does not move in a straight line.

Work done to slowly bring the particle to origin is - 9 J.

Solution:
$\overset{\rightarrow}{F} = - \frac{\partial u}{\partial x}\overset{\hat{}}{i} - \frac{\partial u}{\partial y}\overset{\hat{}}{j}$
$\overset{\rightarrow}{F} = - (6xy^{2} + 6)\overset{\hat{}}{i} - (6x^{2}y)\overset{\hat{}}{j}$
At (1,1)
$\overset{\rightarrow}{F} = - 12\overset{\hat{}}{i} - 6\overset{\hat{}}{j}$
$|\overset{\rightarrow}{a}| = \frac{|\overset{\rightarrow}{F}|}{m}$ =
$\sqrt{12^{2} + 6^{2}} = 6\sqrt{5}ms^{- 2}$
Also,
$w_{ext} = \Delta u = u_{f} - u_{i} = 0 - \left( 3.1^{2}.1^{2} + 6.(1) \right) = - 9J$

Physics


Question:
A monoatomic ideal gas follows the process:
TV^3/2 = constant
The molar specific heat for this process is: (R = gas constant).

Solution:
TV^3/2 = constant
$\frac{(P \vee )V^{3/2}}{nR} =$constant
${PV}^{\frac{5}{2}} =$constant
${PV}^{x} =$constant
$x = \frac{5}{2}$
$C = C_{v} + \frac{R}{1 - x}$
For polytropic process,
$C = \frac{3}{2}R + \frac{R}{1 - \frac{5}{2}} = \frac{5R}{6}$

Physics


Question:
A thin rod of length 1 m is kept vertical and is hinged from the
lower end Its linear mass density varies with the distance (x)
from the end $H^{'}$ as $\lambda = (12x)kg/m$, where x is distance along
rod from point H. The potential energy (in Joules) of the rod is 5 N.
Find N (at H the potential energy is zero)
$\left( g = 10m/s^{2} \right)$.

Solution:
Lets take an element of dx width at a distance x from the lowest
end.
$dU = (dm)gx,U = 12g\int_{x = 0}^{x - 1}\mspace{2mu} x^{2}dx = 40J$

Physics


Question:
A ball is given velocity
[IMAGE] as shown. If the ratio of centripetal acceleration totangential
acceleration at the point where the ball leaves the circular path is
[IMAGE] then n is: [Neglect size of the ball].

Solution:
and [IMAGE].....(2)
[IMAGE] Now TOPIC: Circular motion
SUB TOPIC: Vertical circle
LEVEL: Moderate

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

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

Physics


Question:
A vertical capillary tube with inner radius 0.5 mm, is submerged
into water so that the length of its part above the water surface is h =
25 mm. Radius of curvature of meniscuses formed is
[IMAGE]. Here α is an integer. Find α. (surface
tension of water = 0.075 N/m. Density of water
[IMAGE], angle of contact = 0°)
0.5 mm vkUrfjd f=,d ikuh esa bl çdkj Mwch
gqbZ gS fd ikuh dh lrg ds blds Hkkx dh yEckbZ h = 25 mm gSA;gk¡α,d iw.kkZad gSAα dk eku Kkr dhft,A (ikuh dk i`"B ruko = 0.075
N/m, ty

Physics


Question:
If the block in the shown arrangement is acted upon by a constant
force F on a frictionless surface for t ≥ 0, its maximum speed will be

Solution:
For vmax, [IMAGE] ⇒ Fnet = 0
⇒ Kx = F
x =[IMAGE] W = ∆KE
[IMAGE] + Fx = [IMAGE] mv2 - 0
vmax =[IMAGE] TOPIC: WPE
SUB TOPIC: WORK ENERGY THEOREM

Physics

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


Question:
In the arrangement shown in figure, wavelength of light used
is[IMAGE]. The distance between slits
[IMAGE] is [IMAGE] The distance between
[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] is [IMAGE] The intensity through [IMAGE] will be

Physics


Question:
Two moles of an ideal gas with [IMAGE] are mixed
with 3 moles of another ideal gas with [IMAGE] Thevalue
of [IMAGE] for the mixture is
eksy=

Solution: [IMAGE] On rearranging we getiqu ij

Physics


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

Physics


Question:
A positively charged particle is moved in a straight line to
the right from position x = -4 m to position x = -2 m by an external
agent. The electric potential at these positions in space are V(x =
-4m) = -4 volts and V(x = -2m) = -2 volts. Which statement is true
about the work done by the external agent moving the charge between
these positions and the electric field component parallel to the
direction of motion that the moving charged particle experiences? Assume
the particle is moved at constant speed.

The external agent does negative work and the electric field is
directed to the right.

The external agent does positive work and the electric field is
directed to the right.

The external agent does negative work and the electric field is
directed to the left.

The external agent does positive work and the electric field is
directed to the left.

Solution:
wext = q (vs - vi) = q (-4 - (-2)) = -2q = -ve
E goes from higher potential to lower potential.

Physics


Question:
A rocket is fired vertically from the surface of the earthwith a
speed v. How far from the earth does the rocketgo before returning to
the earth (where [IMAGE] is the radius of the earth and
g is acceleration due to gravity)

Solution:
Let the rocket reached a height h from thesurface of earth.
Total energy at the surface of the earth is [IMAGE] Where m and [IMAGE] are the masses of rocket andearth
respectively.
At highest point, the velocity of the rocketbecomes zero.
Total energy at the highest point is
[IMAGE] According to law of conservation of energy,

Physics

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


Question:
One unit of power of a motor is 150 Watts in some system of units.
If the unit of force is doubled and unit of velocity tripled, then one
new unit of power of motor is ( in W )
 1 150,

Solution:
P = f v
nu = constant
n = no. attached, u = size of unit
P' =[IMAGE] so, size of power will become 6 times
TOPIC: UNIT & DIMENSION
SUB TOPIC: MAGNITUDE

Physics

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


Question:
The first diffraction minima due to a single slit diffraction is
at [IMAGE] for a light of
wavelength[IMAGE]. The width of the slit is

Physics


Question:
Two tuning forks are producing 6 beats/s. The frequency of
first tuning fork is 256 Hz. Now, first tuning fork is loaded with wax.
As a result the beat frequency becomes 4 beats/s. Now some portion of
the wax is removed from first tuning fork. As a result the beat
frequency becomes 0 beats/s. Then the frequency of first tuning fork
after waxing (the condition when beat frequency is 4 beats/s) will be

Solution:
Since beat frequency is 6 Hz, therefore, frequency of second
fork will be either 262 Hz or 250 Hz. Since on loading with wax beat
frequency decreases, frequency of second fork should be 250 Hz. So after
waxing, the frequency of first tuning fork should be either 252 or 246
Hz. Since after removing a portion of wax from first fork beat frequency
becomes zero. Therefore frequency of first fork after waxing should be
246 Hz.

Physics


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),d csyu&fiLVu mifLFkr gS rFkk;g P_1 nkc ij gSA blesa,d lkcqu dk cqycqyk ftldh mifLFkr gSA;fn
lkcqu ds cqycqys dh f= djuh gks rks] csyu
eku r csyu

Solution: Isothermal process. ¼lerkih; çØe

Physics


Question:
Slider block A moves to the right with constant velocity 6 m/s
in the arrangement as shown in the figure. C and D are points on the
string. If, and are the velocities of B, C and D respectively then,
choose

$|\overset{\rightarrow}{V_{C}} - \overset{\rightarrow}{V_{D}}| = 4|\overset{\rightarrow}{V_{B}}|$

$|\overset{\rightarrow}{V_{B}} - \overset{\rightarrow}{V_{C}}| = |\overset{\rightarrow}{V_{B}} - \overset{\rightarrow}{V_{D}}| = 2|\overset{\rightarrow}{V_{B}}|$

$|\overset{\rightarrow}{V_{C}}| - |\overset{\rightarrow}{V_{D}}| = 2|\overset{\rightarrow}{V_{B}}|$

Solution:
Clearly $v_{C} = 6m/s \downarrow$
By string constraint,
$–6 + V_{B} + V_{B} + V_{B} = 0 \Rightarrow V_{B} = 2m/s \downarrow$
Also, $\frac{V_{C} + V_{D}}{2} = V_{B}$
$\Rightarrow V_{D} = - 2m/s$
$\therefore V_{D} = 2m/s \uparrow$
$|\overset{\rightarrow}{V_{C}} - \overset{\rightarrow}{V_{D}}| = 6 + 2 = 8m/s$
$|\overset{\rightarrow}{V_{B}} - \overset{\rightarrow}{V_{C}}| = 4m/s$
$|\overset{\rightarrow}{V_{B}} - \overset{\rightarrow}{V_{D}}| = 4m/s$
$|\overset{\rightarrow}{V_{c}}| - |\overset{\rightarrow}{V_{D}}| = 6 - 2 = 4m/s$

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


Question:
A car moves towards a hill with speed c v. It blows a horn of
frequency f which is heard by an observer following the car with speed
ofv. The speed of sound in air is v. then,

the wavelength of sound reaching the hill is [IMAGE]

the beat frequency observed by the observe [IMAGE]

the beat frequency observed by the observer is [IMAGE]

Solution: [IMAGE] Beat frequency head by observer = f′′−f′′′

Physics


Question:
The time period of a satellite of earth is 5 hours.If the
separation between the centre of earth and the satellite is increased to
4 times the previous value, the new time period in hours will become
10n. Find the value of n.

Solution: For satellite

Physics

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


Question:
A particle is performing small oscillation about X & Y axis as
[IMAGE] then path of particle will look like

Solution: $x = \sin wt;y = \sin 3wt;w = \frac{\pi}{6}$

Physics


Question:
If equation[IMAGE] = [IMAGE] is correct.
Find the value of X.

Physics

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


Question:
A plane is flying with an air speed 10 m/s toward north but
suddenly encounters a wind of 10 m/ s at 30º north of east. If angle
made by new direction of velocity of plane with respect to ground from
north direction is $\frac{\pi}{n}$ then find the value of n

Solution: The angle θ = $\pi/6$

Physics


Question:
Two concentric rings of radius r and R are in same plane such
that R >> r. Current Ι is maintained in thering of radius r. Flux
associated to the ring of radius R due to this current is (use proper
approximation).

Solution:
If current I is maintained in ring of radius R the flux
associated with ring of radius r will be:$\frac{\pi\mu_{0}{Ir}^{2}}{2R}$
[IMAGE] Since M_12 = M_21 so same flux would be associated to the ring of
radius R if I is maintained in ring ofradius r.

Physics


Question: Find the acceleration of each block.

Solution:
Tendency of 2 kg block is of coming down and of 1 kg is of going
up.
[IMAGE] Where F1 is friction between 1 kg and 2 kg block and F2 is friction
between 2 kg and inclined surface.
For 2 kg block 2 × 10 ×[IMAGE] - T - F1 - F2 = 2a
For 1 kg block T - 1 × 10 ×[IMAGE] × F1 = a
⇒ Add both 10 - 5 - 2F1 - F2 = 3a
5 - 2F1 - F2 = 3a
The max. value F1 = 0.1 × 1 × 10[IMAGE] F2max = 3 × 10 × 0.2 ×[IMAGE] Max (2F1 + F2) =[IMAGE] ⇒ Static friction will act. ⇒ a = 0
TOPIC: NLM
SUB TOPIC: FRICTION

Physics

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


Question:
A particle of mass 1 kg is thrown vertically upwards with speed
100 m/s. After 5s it explodes into twoparts. One part of mass 400g comes
back with speed 25 m/s, what is the speed of other part just
afterexplosion?

Solution:
Velocity of particle after 5 s
v = u - gt
v = 100 - 10 × 5
= 100 - 50 = 50 m/s (upwards)
Conservation of linear momentum gives
Mv = m1v1 + m2v2......(i)
Taking upward direction positive, the velocity v_1 will be negative.
$\therefore v_{1} = - 25m/s,v = 50m/s$
$\ $ Also M=1kg,m_1=400g=0.4kg
and m_2=(M - m_1)=1 - 0.4=0.6kg
Thus,Eq.(i) becomes,
$1*50 = 0.4 \times ( - 25) + 0.6v_{2}$
or$\ 5$`<!- ->`0= - 10+0.6v_2
or 0.6v_2=60
or$v_{2} = \frac{60}{0.6} = 100m/s$
As v_2 is positive, therefore the other part will move upwards with a
velocity 100 m/s.

Physics


Question:
An object is gradually moving away from the focal point of a
concave mirror along the axis of the mirror. The graphical
representation of the magnitude of linear magnification (m) versus
distance of the object from the mirror (x) is correctly given by (Graphs
are drawn schematically and are not to scale)

Solution: At focus, magnification is[IMAGE].

Physics


Question:
Gravitational force between two identical uniform solid gold
spheres of radius r each in contact is proportional to n power of r.
Then value of n is.

Solution:
[IMAGE] F = [IMAGE] m = ρ × [IMAGE] πr3
F = Gρ2 × [IMAGE] × [IMAGE] ∝ r4

Physics


Question:
A Carnot engine operates between two reservoirs of temperatures
900 K and 300 K. The engineperforms 1200 J of work per cycle. The heat
energy (in J) delivered by the engine to the lowtemperature reservoir,
in a cycle, is 100n. Find n,d dkuksZ bUtu dks 900 K vkSj 300 K ds nks HkaMkjks a ds chp pyk;k
fuEu rki okys HkaMkj

Physics


Question:
[IMAGE] A and d are the plate area and distance between plates of a parallel
plate capacitor repectively. The dielectric constant of dielectric
filled inside a parallel plate capacitor is variable and given as
$k(x) = K(1 + \gamma x$) where is the distancemeasured from the
left plate as shown in the figure. If$(\gamma d) < < 1$, then find the
capacitance of capacitor with dielectric.

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

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

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

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

Solution:
Capacitance of element$= \frac{k\varepsilon_{0}A}{dx}$
$\sum\frac{1}{C^{'}} = \int_{0}^{d}\mspace{2mu}\frac{dx}{K\varepsilon_{0}A(1 + \gamma x)}$
$\frac{1}{C} = \frac{1}{K\varepsilon_{0}A\alpha}ln(1 + \gamma d)$
$${\ Given\ \ \gamma d < < 1
$$C = \right)$$

Physics


Question: [IMAGE] is maximum is

Solution: $\frac{dN_{2}}{dt} = 0\text{\quad} \Rightarrow \text{\quad}\lambda N_{1} = 2\lambda N_{2}\text{\quad}or\frac{N_{1}}{N_{2}} = 2$

Physics


Question:
Photons are emitted at the rate of 1.6 × 10^16 per second by a
laser source operating at the wavelength of 632.8 nm. The power of
source is ( in mW )

Solution:
Energy of each photon [IMAGE] = 3.14 ´
10^-19 J
Total energy emitted in one second = 3.14 × 10^-19 × 1.6 × 10^16 = 5
mW
Power = Energy emitted per second = 5 mW

Physics


Question: has four choices

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
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:
Two liquids of densities [IMAGE] are filled up
behind a square wall of side 10m as shown in figure. Each liquid has a
height of 5 m. The ratio of the forces due to these liquids exerted on
upper part MN to that at the lower part NO is 1/n. Find n. (Assume that
the liquids are not mixing):
fHkUu ?kuRoksa [IMAGE] rFkk nks
æo 10m yEckbZ dh,d oxkZdkj nhokj ds ihNs Hkjs gq, gSa ¼fp= ns[ksa½A
çR;sd æo dh 5 m gSA rc bu æoksa }kjk nhokj ds Hkkx MN rFkk
fupys Hkkx NO ij yxkus okys cyksa dk vuqikr 1/n gksxkrks n dk eku Kkr
djksaA æo fefJr ugha gksrs gSa½

Physics


Question:
The moment of inertia of a thin uniform square sheet of mass M
and side a about the axis AB which is in the plane of sheet is
$\frac{nMa^{2}}{12}$. Find n

Solution: $I = \frac{{ma}^{2}}{12} + m\left( \frac{a}{\sqrt{2}} \right)^{2} = \frac{7ma^{2}}{12}$

Physics

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


Question:
Initially spring is in natural length and system is at rest.
System is released when a particle of mass m collides vertically to
block B with velocity v and sticks to it. Find maximum extension in the
spring. (Pulley and string are ideal). (Given that
$m = 1kg,v = 10\sqrt{2}m/sec$ and spring constant$k = 10N/m$)
($g = 10m/s^{2})$

Solution:
Just after collision velocity of (block B+
particle)$= \frac{v}{2}$
At the time of maximum extension both A and B block will have same
velocity (say v')
Conserving angular momentum about centre of pulley of system of (A+B+
particle)
$v^{'} = \frac{v}{4}$
By energy conservation,
$\frac{1}{2} \times 2m \times {(\frac{v}{2})}^{2} = 2 \times \frac{1}{2} \times 2m \times {(\frac{v}{4})}^{2} + \frac{1}{2} \times 10 \times A^{2} - 2mgA$
On solving,
A=5m

Physics


Question:
In an electromagnetic wave, the electric field oscillates
sinusoidal with amplitude 48 Vm^-1, the RMSvalue of oscillating
magnetic field will be nearly equal to

Solution:
$B_{0} = \frac{E_{0}}{c} = \frac{48}{3 \times 10^{8}} = 16 \times 10^{- 8}T$
$B_{rms} = \frac{B_{0}}{\sqrt{2}} = \frac{16 \times 10^{- 8}}{\sqrt{2}} = 8\sqrt{2} \times 10^{- 8}T$

Physics


Question:
If we need a magnification of 375 from a compound microscope of
tube length 150 mm and an objective of focal length 5 mm, the focal
length of the eye-piece, should be close to

Solution:
Case I
If final image is at least distance of clear vision
If final image is at infinity

Physics


Question:
A plank having mass M is placed on smooth horizontal surface.
Block of mass m is placed on it coefficient of friction between block
and plank is $\mu_{0} + kx$,where k is constant and x is relative
displacement of block w.r.t. plank. A force F is applied on block where
F=at, where a=10; t is in second. Find$t_{0}$ to when relative motion
will occur between block and plank (use $g = 10m/s^{2}$).

Solution:
Let at time t_o relative motion will occur.
[IMAGE] $\mu_{0}mg = Ma$..... (1)
[IMAGE] $10t_{0} - \mu_{0}mg = ma$..... (2)
From (1) and (2);
$t_{0} = \mu_{0}m + \frac{\mu_{0}m^{2}}{M}$

Physics


Question:
The surface tension of a liquid is 5 Newton per metre. If a film
of this liquid is held on a ring of area 0.02 m^2, if its surface
energy is about ([IMAGE] ) Joule then n is:
esa bl æo dh fQYe mifLFkr gSrks bldh i`"Bh;
([IMAGE] ) gks rks n

Solution:
We know that surface energy
[IMAGE] Here, as 2 films are formed because of ring.
[IMAGE] ge tkurs gS fd i`"B;gk¡ oy

Physics


Question:
A thin walled hollow cylinder is closed from both the ends. The
mass of the curved part is m and the mass of each circular part is also
m. The radius of the cylinder is R and its length is 2R. What will be
the moment of inertia of this hollow cylinder about the axis,
perpendicular to its length and passing through its centre of mass?

Solution:
First find I of the hollow tube
$I_{1} = \int\frac{{dmR}^{2}}{2} + \int_{x = - \frac{\mathcal{l}}{2}}^{x = + \frac{\mathcal{l}}{2}}{(\frac{m}{\mathcal{l}}dx)}x^{2}$
$I_{1} = \frac{{mR}^{2}}{2} + \frac{m^{\mathcal{l}^{2}}}{12}$ where
$\mathcal{l} = 2R$
So $I_{1} = \frac{5}{6}{mR}^{2}$
Now moment of inertias of each circular plate about that axis.
$I = \frac{mR^{2}}{4} + mR^{2} = \frac{5}{4}mR^{2}$
$I_{total} = \frac{5}{6}{mR}^{2} + 2 \times \left( \frac{5}{4}{mR}^{2} \right)$
$I_{total} = \frac{10}{3}{mR}^{2}$

Physics


Question:
At t = 2 sec, a projectile is moving upwards. Its speed is 20 m/s
and it is making at angle 30° with horizontal. What is angle of
projection of the projectile from ground ( in degree ) ?
t = 2 sec 20 m/sec
 30° ( )

Solution:
[IMAGE] u cos θ = 20 cos 30°
vy = uy + ayt
20 sin 30° = u sin θ - g × 2
u sin θ = 30 m/s
tan θ = [IMAGE] TOPIC:KINEMATIC
SUB TOPIC: 2 D

Physics

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


Question:
A 60 pF capacitor is fully charged by a 20V supply. It is then
disconnected from the supply and isconnected to another uncharged 60 pF
capacitor in parallel. The electrostatic energy that is lost in
thisprocess by the time the charge is redistributed between them is (in
nJ)________.

Solution:
[IMAGE] V_0 = 20 V
Heat loss = Ui - Uf
$= \frac{1}{2}CV_{0}^{2} - 2\left\lbrack \frac{1}{2}C\left( \frac{V_{0}}{2} \right)^{2} \right\rbrack$
$= \frac{CV_{0}^{2}}{4}$
$= \frac{\left( 60 \times 10^{- 12} \right)(20)^{2}}{4}J$
= 6 × 10^-9 J = 6 nJ

Physics


Question:
In which of the following cases the magnitude of acceleration of
the block A will be maximum (Neglect friction and mass of pulley and
string)

Solution:
(i) $a = \frac{2mg - mg}{3m} = \frac{g}{3}$ (ii)
$a = \frac{2mg - mg}{m} = g$
(iii) $a = \frac{2mg}{m} = 2g$ (iv) $a = \frac{2g}{3}$

Physics


Question:
A force
$\overrightarrow{F} = \left( 3t\widehat{i} + 5\widehat{j} \right)N$ acts
on a body due to which its position varies as
$\overrightarrow{s} = \left( 2t^{2}\widehat{i} - 5\widehat{j} \right)$.
Work done by this force in first two seconds is

Solution:
$W = \int\overline{F} \cdot d\overline{s} = \int(3t\ \widehat{i} + 5\widehat{j}).(4t\ dt\ \widehat{i})$
$= \int_{0}^{2}\mspace{2mu} 12t^{2}dt = \frac{12\left\lbrack t^{3} \right\rbrack_{0}^{2}}{3} = 32\ J$

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.

Solution:
$\frac{1}{f} = (\mu - 1)(\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$R_{1} = \infty$ $R_{2} = - 30cm$
$\frac{1}{f} = (1.5 - 1)(\frac{1}{\infty} - \frac{1}{- 30})$
$\frac{1}{f} = \frac{0.5}{30}$ $f = 60cm$

Physics


Question:
Two inclined planes OA and OB of inclinations to the horizontal
$\alpha\ and\ \beta$, each equal to 30º are placed as shown in the
figure. A particle is projected at an angle of 90º with plane OA from
point A and its strikes the plane OB at point B normally. If the speed
of projection in m/s is x. Fill the value of 8 x. (given that OA=OB=20
cm and $g = 10m/s^{2})$

Solution:
Since $\alpha = \beta$
$AO = BO = 20cm = d$
$cos120^{\circ} = \frac{d^{2} + d^{2} - (AB)^{2}}{2d^{2}}$
$AB = \sqrt{3}d$
Now ABis the range of projectile from A to B
$\sqrt{3}d = \frac{u^{2}sin2\theta}{g}$ $u = 2m/s$

Physics


Question:
The maximum peak to peak voltage of an AM wave is
[IMAGE] and the minimum peak to peak voltage
is[IMAGE]. The modulation factor is 10x. Find x.

Physics


Question:
If in a system force (F), pressure (P) and time (T) is
fundamental unit, then dimensional formulae for length will be

Solution:
x + y = 0 x - y = 1 - 2x - 2y + z = 0
x = [IMAGE] y = - [IMAGE] z = 0
TOPIC: UNIT & DIMENSION
SUB TOPIC: DIMENSIONS

Physics

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


Question:
A string is holding a solid block below the surface of the liquid
as shown in figure. If the system is given an upward acceleration a,
then as compared to previous state, choose the correct option(s)

Tension in string will be$\left( 1 + \frac{a}{g} \right)$ times

Tension in string will be $\left( 1 - \frac{a}{g} \right)$ times

Upthrust force on block become 2$\left( 1 + \frac{a}{g} \right)$
times

Upthrust force on block becomes $\left( 1 - \frac{a}{g} \right)$
times

Solution:
Let σ is density of liquid &ρ is density at object
$B_{i} = \sigma vg\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B_{f} = \sigma v(g + a) = B_{i}\left( \frac{g + a}{g} \right)$
$T_{i} = (\sigma - \rho)vg\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{f} = (\sigma - \rho)v(g + a) = T_{i}\left( \frac{g + a}{g} \right)$

Physics