Grass-G/jee-main-questions huggingface dataset:
Subject: Physics — Newton's Laws of Motion — Constraint motion
Difficulty: Moderate

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

1) $|\overset{\rightarrow}{V_{C}} - \overset{\rightarrow}{V_{D}}| = 4|\overset{\rightarrow}{V_{B}}|$
2) $|\overset{\rightarrow}{V_{B}} - \overset{\rightarrow}{V_{C}}| = |\overset{\rightarrow}{V_{B}} - \overset{\rightarrow}{V_{D}}| = 2|\overset{\rightarrow}{V_{B}}|$
3) $|\overset{\rightarrow}{V_{C}}| - |\overset{\rightarrow}{V_{D}}| = 2|\overset{\rightarrow}{V_{B}}|$
4) All of the above

Correct option: 4

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, $\frac{V_{C} + V_{D}}{2} = V_{B}$
$\Rightarrow V_{D} = - 2m/s$
$\therefore V_{D} = 2m/s \uparrow$
$|\overset{\rightarrow}{V_{C}} - \overset{\rightarrow}{V_{D}}| = 6 + 2 = 8m/s$
$|\overset{\rightarrow}{V_{B}} - \overset{\rightarrow}{V_{C}}| = 4m/s$
$|\overset{\rightarrow}{V_{B}} - \overset{\rightarrow}{V_{D}}| = 4m/s$
$|\overset{\rightarrow}{V_{c}}| - |\overset{\rightarrow}{V_{D}}| = 6 - 2 = 4m/s$
Subject: Physics — Current Electricity — KVL
Difficulty: Moderate

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

1) The potential difference across C_1 is 5V
2) The potential difference between terminals of 15V battery is 9V
3) The potential difference V_a - V_b = - 4V
4) All of the above

Correct option: 4

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
$V_{a} - 7 + \frac{q}{5} = V_{b}$
$\Rightarrow V_{a} - V_{b} = 7 - \frac{q}{5}$
$= 7 - \frac{55}{5}$
$= - \frac{20}{5} = - 4V$
Subject: Physics — Sound wave — Doppler effect
Difficulty: Moderate

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,

1) the wavelength of sound reaching the hill is [IMAGE]
2) the wavelength of sound reaching the hill is [IMAGE]
3) the beat frequency observed by the observe [IMAGE]
4) the beat frequency observed by the observer is [IMAGE]

Correct option: 4

Solution:
[IMAGE] Beat frequency head by observer = f′′−f′′′
Subject: Physics — Gravitation — Kepler's laws
Difficulty: Moderate

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.


Answer: 4

Solution:
For satellite
Subject: Physics — Oscillation — Position of particle
Difficulty: Moderate

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

4) None of these

Correct option: 1

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

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

1) Zero
2) 1
3) 2
4) - 1

Correct option: 1

Solution:
Subject: Physics — Relative motion — Plane problem
Difficulty: Moderate

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


Answer: 6

Solution:
The angle θ = $\pi/6$
Subject: Physics — EMI — Mutual induction
Difficulty: Moderate

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).

1) $\frac{\pi\mu_{0}{Ir}^{2}}{2R}$
2) $\frac{\pi\mu_{0}Ir}{2}$
3) $\frac{\pi\mu_{0}{IR}^{2}}{2r}$
4) $\frac{\pi\mu_{0}IR}{2}$

Correct option: 1

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.
Subject: Physics —  — 
Difficulty: Tough

Question: Find the acceleration of each block.

3) zero

Correct option: 3

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
Subject: Physics — Centre of Mass — Conservation of linear momentum
Difficulty: Moderate

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?

1) 100 m/s upwards
2) 600 m/s upwards
3) 100 m/s downwards
4) 300 m/s upwards

Correct option: 1

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.
Subject: Physics — Ray Optics — Linear magnification
Difficulty: Easy

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)


Correct option: 1

Solution:
At focus, magnification is[IMAGE].
Subject: 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.


Answer: 04

Solution:
[IMAGE] F = [IMAGE] m = ρ × [IMAGE] πr3
F = Gρ2 × [IMAGE] × [IMAGE] ∝ r4
Subject: Physics — Thermodynamics — Carnot engine
Difficulty: Moderate

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


Answer: 6

Solution:
Subject: Physics — Electrostatics — Equivalent capacitance
Difficulty: Difficult

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.

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

Correct option: 2

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)$$
Subject: Physics — Nuclear physics — Radioactivity
Difficulty: Moderate

Question: [IMAGE] is maximum is

1) at no time this is possible
2) 2

Correct option: 2

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


Answer: 05

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
Subject: Physics —  — 

Question: has four choices

1) ,
2) ,
3) ,
4) out of which ONLY ONE is correct]


Solution:
Subject: 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


Answer: 1

Solution:
[IMAGE] TOPIC:Electromagnetic induction
SUB TOPIC:Faraday's law
LEVEL:Moderate
Subject: Physics — Mechanical Properties of Fluids — Bernoulli's equation
Difficulty: Moderate

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


Answer: 4

Solution:
Subject: Physics — Rigid Body Dynamics — Moment of inertia
Difficulty: Moderate

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


Answer: 7

Solution:
$I = \frac{{ma}^{2}}{12} + m\left( \frac{a}{\sqrt{2}} \right)^{2} = \frac{7ma^{2}}{12}$
Subject: Physics — Work Power & Energy — Conservation of mechanical energy
Difficulty: Moderate

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$k = 10N/m$)
($g = 10m/s^{2})$

1) $(2 + \sqrt{19})m$
2) 5m
3) 1m
4) None of these

Correct option: 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
Subject: Physics — EM wave — RMS value of oscillating magnetic field
Difficulty: Moderate

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

1) 1.6 × 10^-8T
2) 16 × 10^-9 T
3) 144 × 10^8T
4) 11.3 × 10^-8T

Correct option: 4

Solution:
$B_{0} = \frac{E_{0}}{c} = \frac{48}{3 \times 10^{8}} = 16 \times 10^{- 8}T$
$B_{rms} = \frac{B_{0}}{\sqrt{2}} = \frac{16 \times 10^{- 8}}{\sqrt{2}} = 8\sqrt{2} \times 10^{- 8}T$
Subject: Physics — Ray Optics — Optical instruments
Difficulty: Tough

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

1) 2 mm
2) 33 mm
3) 12 mm
4) 22 mm

Correct option: 4

Solution:
Case I
If final image is at least distance of clear vision
If final image is at infinity
Subject: Physics — Newton's Laws of Motion — Friction
Difficulty: Tough

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 $g = 10m/s^{2}$).

1) $\mu_{0}M + \frac{\mu_{0}M^{2}}{m}$
2) $\mu_{0}m + \frac{\mu_{0}M^{2}}{m}$
3) $\mu_{0}m + \frac{\mu_{0}m^{2}}{M}$
4) $\mu_{0}M + \frac{\mu_{0}m^{2}}{M}$

Correct option: 3

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);
$t_{0} = \mu_{0}m + \frac{\mu_{0}m^{2}}{M}$
Subject: Physics — Mechanical Properties of Fluids — Surface tension
Difficulty: Moderate

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


Answer: 2

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
Subject: Physics — Rigid Body Dynamics — Moment of inertia
Difficulty: Moderate

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?

1) $\frac{10}{3}mR^{2}$
2) $\frac{5}{3}mR^{2}$
3) $\frac{5}{6}mR^{2}$
4) $\frac{11}{3}mR^{2}$

Correct option: 1

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.
$I = \frac{mR^{2}}{4} + mR^{2} = \frac{5}{4}mR^{2}$
$I_{total} = \frac{5}{6}{mR}^{2} + 2 \times \left( \frac{5}{4}{mR}^{2} \right)$
$I_{total} = \frac{10}{3}{mR}^{2}$
Subject: Physics —  — 
Difficulty: Moderate

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° ( )


Answer: 60

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
Subject: Physics — Capacitance — Energy stored in capacitor
Difficulty: Moderate

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)________.


Answer: 6

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
Subject: Physics — Newton's Laws of Motion — Newton's second law
Difficulty: Moderate

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)


Correct option: 3

Solution:
(i) $a = \frac{2mg - mg}{3m} = \frac{g}{3}$ (ii)
$a = \frac{2mg - mg}{m} = g$
(iii) $a = \frac{2mg}{m} = 2g$ (iv) $a = \frac{2g}{3}$
Subject: Physics — Work Power & Energy — Work done by variable force
Difficulty: Moderate

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

1) 23 J
2) 32 J
3) zero
4) can't be obtained

Correct option: 2

Solution:
$W = \int\overline{F} \cdot d\overline{s} = \int(3t\ \widehat{i} + 5\widehat{j}).(4t\ dt\ \widehat{i})$
$= \int_{0}^{2}\mspace{2mu} 12t^{2}dt = \frac{12\left\lbrack t^{3} \right\rbrack_{0}^{2}}{3} = 32\ J$
Subject: Physics — Ray Optics — Lens Maker's formula
Difficulty: Moderate

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.


Answer: 6

Solution:
$\frac{1}{f} = (\mu - 1)(\frac{1}{R_{1}} - \frac{1}{R_{2}})$
$R_{1} = \infty$ $R_{2} = - 30cm$
$\frac{1}{f} = (1.5 - 1)(\frac{1}{\infty} - \frac{1}{- 30})$
$\frac{1}{f} = \frac{0.5}{30}$ $f = 60cm$
Subject: Physics — Projectile Motion — Projectile on an inclined plane
Difficulty: Moderate

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 $g = 10m/s^{2})$

1) 16
2) 18
3) 12
4) 15

Correct option: 1

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
$\sqrt{3}d = \frac{u^{2}sin2\theta}{g}$ $u = 2m/s$
Subject: Physics — Communication system — Modulation
Difficulty: Moderate

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.


Answer: 5

Solution:
Subject: Physics —  — 
Difficulty: Easy

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

4) None of these
 (F), (P) (T)

Correct option: 3

Solution:
x + y = 0 x - y = 1 - 2x - 2y + z = 0
x = [IMAGE] y = - [IMAGE] z = 0
TOPIC: UNIT & DIMENSION
SUB TOPIC: DIMENSIONS
Subject: Physics — Mechanical Properties of Fluids — Buoyant force
Difficulty: Moderate

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)

1) Tension in string will be$\left( 1 + \frac{a}{g} \right)$ times
2) Tension in string will be $\left( 1 - \frac{a}{g} \right)$ times
3) Upthrust force on block become 2$\left( 1 + \frac{a}{g} \right)$
times
4) Upthrust force on block becomes $\left( 1 - \frac{a}{g} \right)$
times

Correct option: 1

Solution:
Let σ is density of liquid &ρ is density at object
$B_{i} = \sigma vg\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ B_{f} = \sigma v(g + a) = B_{i}\left( \frac{g + a}{g} \right)$
$T_{i} = (\sigma - \rho)vg\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T_{f} = (\sigma - \rho)v(g + a) = T_{i}\left( \frac{g + a}{g} \right)$
Subject: Physics — E-M waves — Electric and magnetic field
Difficulty: Moderate

Question: A plane electromagnetic wave of wavelength $\\lambda\$ has an
intensity I. It is propagating along the positive Y-direction. The
allowed expressions for the electric and magnetic fields are given by

1) $\overset{\rightarrow}{E} = \sqrt{\frac{2I}{\varepsilon_{0}c}}cos\lbrack\frac{2\pi}{\lambda}(y - ct)\rbrack\;\ \overset{\rightarrow}{B} = + \frac{1}{c}E\overset{\hat{}}{i}$
2) $\overset{\rightarrow}{E} = \sqrt{\frac{I}{\varepsilon_{0}c}}cos\lbrack\frac{2\pi}{\lambda}(y - ct)\rbrack\overset{\hat{}}{k}\;\ \overset{\rightarrow}{B} = + \frac{1}{c}E\overset{\hat{}}{i}$
3) $\overset{\rightarrow}{E} = \sqrt{\frac{2I}{\varepsilon_{0}c}}cos\lbrack\frac{2\pi}{\lambda}(y - ct)\rbrack\overset{\hat{}}{k}\;\ \overset{\rightarrow}{B} = - \frac{1}{c}E\overset{\hat{}}{i}$
4) $\overset{\rightarrow}{E} = \sqrt{\frac{I}{\varepsilon_{0}c}}cos\lbrack\frac{2\pi}{\lambda}(y - ct)\rbrack\;\ \overset{\rightarrow}{B} = \frac{1}{c}E$

Correct option: 1

Solution:
If E_0 is magnitude of electric field then
$\frac{1}{2}\varepsilon_{0}E^{2} \times C = I$
$E_{0} = \sqrt{\frac{2I}{C\varepsilon_{0}}}$
$B_{0} = \frac{E_{0}}{C}$
Direction of $\overset{\rightarrow}{E} \times \overset{\rightarrow}{B}$
will be along$+ \overset{\hat{}}{j}$.
Subject: Physics —  — 
Difficulty: Tough

Question: A block of mass m is at rest on an incline having an angle of
inclination θ with the horizontal. The incline is now being rotated with
constant angular velocity ω about an axis passing through the
perpendicular side. If the surface of the incline is rough having
coefficient of friction as µ, then find the distance along the
incline so that block just remains stationary with respect to incline
and is on the verge of slipping.
 from the top of the incline as shown in the figure.]

4) [IMAGE] m θ 
 ω µ 
 
 )

Correct option: 3

Solution:
N + mrω2 sin θ = mg cos θ
mg sin θ + mrω2 cos θ = Fdown
µ (mg cos θ - mrω2 sin θ) = Fup
Fdown = Fup
r =[IMAGE] x cos θ = r
x =[IMAGE] =[IMAGE] =[IMAGE] TOPIC: CIRCULAR MOTION
SUB TOPIC: DYNAMICS
Subject: Physics —  — 
Difficulty: Moderate

Question: In new system of unit 1 unit mass is 5 kg 1 unit length is 20
meter and 1 unit time is 2 second then 1 unit force in new system is
how much newton ?

1) 100 N
2) 50 N
3) 30 N
4) 25 N
 1 5 kg 1 20 1
 2 1

Correct option: 4

Solution:
1 unit of mass = 5kg
1 unit of length = 20 m
1 unit of time = 2 sec
1 unit of F = [IMAGE] = [IMAGE] = 25 N
TOPIC: UNIT & DIMENSION
SUB TOPIC: DIMENSIONS
Subject: Physics — Ray Optics — Reflection (mirror)
Difficulty: Moderate

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

1) $\left( \frac{L}{4},\frac{\sqrt{2}L}{\pi} \right)$
2) $\left( \frac{L}{3},\frac{\sqrt{3}L}{2\pi} \right)$
3) $\left( \frac{3L}{4},\frac{\sqrt{2}L}{\pi} \right)$
4) $\left( \frac{2L}{3},\frac{\sqrt{3}L}{\pi} \right)$

Correct option: 4

Solution:
The slope of curve at such points is
Subject: Physics — Current Electricity — Potentiometer
Difficulty: Moderate

Question: In the potentiometer circuit shown, AB is a $10\Omega$ uniform
potentiometer wire of length 50 cm, E_1 is 2V accumulator of negligible
internal resistance. R_1 and R_2 are $15\Omega$ and 5$\Omega$
respectively. When K_1 and K_2 are both open, the galvanometer shows
no deflection when AJ is 31.25 cm. When K_1 and K_2 both are closed
the balance length AJ =5 cm. The emf of cell E_2is ____

1) 0.5 V
2) 1 V
3) 1.5 V
4) 2 V

Correct option: 1

Solution:
$V_{AB} = i10 = \frac{20}{25}$
$E_{2} = \frac{V_{AB}}{50} \times 31.25$
$= \frac{20}{25} \times \frac{31.25}{50}$
$= 0.5V$
Subject: Physics — Rigid Body Dynamics — Conservation of mechanical energy
Difficulty: Moderate

Question: One end of a straight uniform [IMAGE] long bar
is pivoted on horizontal table. It is released from rest when itmakes an
angle [IMAGE] from the horizontal (see figure). Its
angular speed when it hits the table is given as$\sqrt{3n}s^{- 1},$where
n is an integer. The value of n is.......


Answer: 5

Solution:
Subject: Physics — Wave Optics — YDSE
Difficulty: Moderate

Question: In the shown figure three slits, separated by
[IMAGE] and illuminated by monochromatic parallel beam of
light of wavelength[IMAGE]. P is a point on the line
perpendicular to the plane of slits through[IMAGE]. If
[IMAGE] is the intensity of the wave from each slit, the
intensity at p is


Correct option: 4

Solution:
Phase difference between
[IMAGE] &[IMAGE] Phase difference between [IMAGE] &[IMAGE] Now using phase diagram