Grass-G/jee-main-questions huggingface dataset:
Subject: Physics —  — 
Difficulty: Moderate

Question: The displacement x of a body of mass 1 kg on horizontal smooth
surface as function of time t is given by x =[IMAGE].
The work done in the first one second is

4) [IMAGE] t 1 kg 
x x =[IMAGE]

Correct option: 2

Solution:
dW = F.dx
x =[IMAGE] v =[IMAGE] = t3 ⇒ dx = t3dt
[IMAGE] = 3t2
[IMAGE] t3 dt
TOPIC: WPE
SUB TOPIC: WORK DONE BY VARIABLE FORCE
Subject: Physics —  — 
Difficulty: Easy

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


Correct option: 4

Solution:
Phase difference between [IMAGE] Since D>>d then apply binomial approximation: [IMAGE] Phase difference between[IMAGE] Now using phase diagram
SUB TOPIC:YDSE
Subject: Physics —  — 

Question: has four choices

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

Correct option: 4

Solution:
Subject: Physics — Current Electricity — KVL
Difficulty: Moderate

Question: The series combination of two batteries, both of the same emf
10V, but different internal resistance of20$\Omega$ and 5$\Omega$, is
connected to the parallel combination of two resistors 30$\Omega$ and
R$\ \Omega$. The voltagedifference across the battery of internal
resistance 20$\Omega$ is zero, the value of R (in$\ \Omega$) is ___
(1)
(2)
(3)
(4)


Answer: 30

Solution:
V_1 = $\varepsilon_{1}$  - i. r_1
0 = 10 - i × 20
i = 0.5A
V_2 = $\varepsilon_{2}$ - ir_2
= 10 - 0.5 × 5
V_2 = 7.5V
$0.5 = 0.25 + \frac{7.5}{x}$
$0.5 = \frac{7.5}{30} + \frac{7.5}{x}$
$0.5 = 0.25 + \frac{7.5}{x}$
$\frac{7.5}{x} = 0.25;$
$x = \frac{7.5}{0.25} = 30$ohms
Subject: Physics — Simple Harmonic Motion — Block spring system
Difficulty: Tough

Question: One end of a spring is fixed to the ceiling and other end is
attached to a block. The block is released when spring is relaxed. The
product of time period and amplitude is 8 S.I. units. If spring is cut
in two equal parts and the two springs are attached to the block as
shown in figure. The block is released when both springs are relaxed. If
the product of time period and amplitude in S.I. units is x, then find
the value of x.


Answer: 1

Solution:
In first case,
[IMAGE] In second case,
Subject: Physics — Mechanical Properties of Fluids — Equation of continuity
Difficulty: Moderate

Question: An ideal fluid flows (laminar flow) through a pipe of
non-uniform diameter. The maximum and minimum diameters of the pipes are
6.4 cm and 4.8 cm, respectively. The ratio of the minimum and the
maximum velocities of fluid in this pipe is n/16. Find n:,d vkn'kZ nzo cnyrs gq,,d ikbi izokg esa cg jgk gSA
ikbi o Øe'k 6.4 cm
cgus okys nzo xfr


Answer: 9

Solution:
Using equation of continuity
lkrR;rk lehdj.k yxkus ij
$A_{1}V_{1} = A_{2}V_{2}$
$\frac{V_{1}}{V_{2}} = \frac{A_{2}}{A_{1}} = \left( \frac{4.8}{6.4} \right)^{2} = \frac{9}{16}$
Subject: Physics —  — 

Question: Two
produce stationary wave. The minimum distance of antinode from the
origin along positive x-axis in terms of the wavelength λ is
(y_1 and y_2 are in meter)



Solution:
B
y = y_1 + y_2 = [IMAGE] Amplitude is maximum at antinode
[IMAGE] General solution
For minimum distance of antinode from origin put n = 1
SECTION - B [INTEGER ANSWER section contains
Subject: Physics —  — 
Difficulty: Medium

Question: A ball was thrown at velocity of 10 m/s at 37° to the horizontal.
What is the rate of change of its speed when its radius of curvature is
51.2 m.

4) 37° 10 m/s 
 51.2 m ?

Correct option: 2

Solution:
[IMAGE] aN = g cos θ =[IMAGE] v cos θ = 10 cos 37°
v =[IMAGE] g cos3 θ =[IMAGE] cos3 θ =[IMAGE] cos θ =[IMAGE] θ = 60°
at =[IMAGE] = g sin θ = 2
TOPIC: CIRCULAR MOTION
SUB TOPIC: Radius of Curvature
Subject: Physics — Thermodynamics — P-V diagram and first law of thermodynamics
Difficulty: Moderate

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

1) $2P_{0}V_{0}$
2) $5P_{0}V_{0}$
3) $6P_{0}V_{0}$
4) $7{PoV}_{0}$

Correct option: 4

Solution:
$\Delta Q = \Delta U + \Delta W$
$W_{ABCA} = \frac{1}{2}\left( 2P_{0} \right)\left( 2V_{0} \right) = 2P_{0}V_{0}$
$W_{ABC} = 2P_{0}V_{0} - W = 2P_{0}V_{0} - \left( - 2P_{0}V_{0} \right) = 4P_{0}V_{0}$
$\Delta U = \frac{nf}{2}R\Delta T = \frac{f}{2}P\Delta V = \frac{3}{2}P_{0}\left( 2V_{0} \right) = 3P_{0}V_{0}$
$\therefore\Delta Q = 7P_{0}V_{0}$
Subject: Physics — Mechanical Properties of Fluids — Viscosity
Difficulty: Moderate

Question: A thin horizontal movable plate is separated from two fixed
horizontal plates [IMAGE] and
[IMAGE] by two highly viscous liquids of coefficient of viscosity
[IMAGE] and [IMAGE] as shown, where $\eta_{2} = 4\eta_{1}$.
Area of contact of movable plate with each fluid is same. If the
distance between two fixed plates is h, then the distance
[IMAGE] of movable plate from upper fixed plate such that the movable plate can
be moved with a constant velocity by applying a minimum constant
horizontal force F on movable plate is h/n (assume velocity gradient to
be uniform in each liquid). Find n.

1) 2
2) 3
3) 4
4) 5

Correct option: 2

Solution:
Let v be the velocity of the movable plate and F is equal to
viscous force
$F = \left\lbrack n_{1}\frac{v}{h_{1}} + n_{2}\frac{v}{h - h_{1}} \right\rbrack A$
$\Rightarrow \frac{dF}{{dh}_{1}} = 0\therefore h_{1} = \frac{h}{3}$
Subject: Physics — Atomic physics — Lyman series
Difficulty: Moderate

Question: The ratio of the longest and shortest wavelengths of the Lyman
series is.


Correct option: 1

Solution:
[IMAGE] Equation dividing
Subject: Physics — Thermodynamic — Thermodynamic processes
Difficulty: Moderate

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

4) None of these

Correct option: 2

Solution:
[IMAGE] (isothermal process) ¼lerkih
Subject: Physics — Kinetic Theory of Gases — Gases in equilibrium
Difficulty: Tough

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$\frac{\alpha}{48}$.,d vkdkj dh Vîqc esa nks vkn'kZ xSl ftu vyx&vyx dh tkrh gS rFkk nqljk foektd,d
xfreku pkydfoHkktd Y tks fcuk ?k"kZ.k xfr dj ldrk gSA
 fp=kuqlkj dks.k α ¼fMxzh esa½ rks [IMAGE]


Answer: 4

Solution:
At equilibrium, pressure and temperature are same.
 rFkk rki leku
 Also, let cross-sectional area of tube be A and radius of ring be r.
ekuk nkc
Subject: Physics — Current Electricity — Potentiometer
Difficulty: Moderate

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

1) $600\Omega$
2) $120\Omega$
3) $100\Omega$
4) $80\Omega$

Correct option: 3

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

Question: Two uniform solid spheres of equal radii R, butmass M and 4M have
a centre to centreseparation 6R, as shown in figure. A projectile of
mass m in projected from the surface of thesphere of mass M directly
towards the centre of the second sphere. The minimum speed ofthe
projectile so that it reaches the surface of the second sphere
is[IMAGE]. What is n?

1) 4
2) 5
3) 3
4) 1

Answer: 3

Solution:
[IMAGE] The projectile need to reach a point on the line joining the centres
of sphere where net electric force on it is zero.
Let the point N has no net electric field due to spheres.
Therefore, $\frac{GM}{x^{2}} = \frac{G.4M}{{(6R - x)}^{2}}$
- x = 2R
Apply conservation of mechanical energy at projectile initial position
and particle at N
E_i = E_f
SUB TOPIC: Conservation of mechanical energy
Subject: Physics —  — 
Difficulty: Moderate

Question: A force given by the relation F = 8t, acts on a body of mass 2 kg,
initially at rest. Find the work done by this force on the body during
first 2 seconds of its motion ( in joule )
m = 2 kg F = 8t 
 2 sec. ( )


Answer: 64

Solution:
F = 8t
m[IMAGE] = 8t
[IMAGE] mv = 4t2
or m [IMAGE] = 4t2 ⇒ dx =[IMAGE] W =[IMAGE] =[IMAGE] = 64 J
TOPIC: WPE
SUB TOPIC: WORK DONE BY VARIABLE FORCE
Subject: Physics — Center of mass — Center of mass of extended body
Difficulty: Easy

Question: The " y ' co-ordinate of the centre of mass of the system of
three rods of length as shown in
figure is: (Assume all rods to be of uniform density)

1) $\frac{9a}{8\sqrt{3}}$
2) $\frac{9a}{16\sqrt{3}}$
3) zero
4) $\frac{8a}{\sqrt{3}}$

Correct option: 2

Solution:
The y-coordinate of centre of mass is
$= \frac{2m(\frac{\sqrt{3}}{2}a) + 2m(0) + 2m(0) + m(\frac{\sqrt{3}}{4}a) + m(\frac{\sqrt{3}}{4}a)}{8m}$
$= \frac{9a}{16\sqrt{3}}$
Subject: Physics — Unit and dimension — Dimensional analysis
Difficulty: Moderate

Question: Dimensions of gravitational constant are


Correct option: 4

Solution:
$G = \frac{Fr^{2}}{m_{1}m_{2}};\ \lbrack G\rbrack = \left\lbrack \frac{Fr^{2}}{m_{1}m_{2}} \right\rbrack = \left\lbrack \frac{MLT^{- 2}L^{2}}{M^{2}} \right\rbrack = \left\lbrack M^{- 1}L^{3}T^{- 2} \right\rbrack$
Subject: Physics — System of particles — Linear momentum of center of mass
Difficulty: Moderate

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

1) Zero
2) $\frac{2M\ V}{\pi}$
3) MVR
4) MV

Correct option: 2

Solution:
If we treat the train as a ring of mass then its COM will
be at a distance[IMAGE] from the centre of the circle.
Velocity of centre of mass is:
$V_{CM} = R_{CM} \cdot \omega$
= $\frac{2R}{\pi} \cdot \omega$ =
$\frac{2R}{\pi} \cdot \left( \frac{v}{R} \right)$
$\left( \because\omega = \frac{v}{R} \right)$
$\Rightarrow$ $V_{CM} = \frac{2V}{\pi}$ $\Rightarrow$
${MV}_{CM} = \frac{2MV}{\pi}$
As the linear momentum of any system = MV_CM
[IMAGE] The linear momentum of the train
$= \frac{2MV}{\pi}$
Subject: Physics — Modern physics — De Broglie wavelength
Difficulty: Moderate

Question: A particle moving with kinetic energy E has de Broglie
wavelength$\lambda$. If energy $\Delta E$ is added to its energy, the
wavelength become $\lambda/2$. Value of$\Delta E$, is

1) 4E
2) 3E
3) 2E
4) E

Correct option: 2

Solution:
$\lambda = \frac{h}{\sqrt{2(KE)m}} = \lambda \propto \frac{1}{\sqrt{KE}}$
$\frac{\lambda}{\lambda/2} = \sqrt{\frac{{KE}_{f}}{{KE}_{i}}}$
$4{KE}_{i} = {KE}_{f}$
$\Rightarrow \Delta E = 4KE_{i} - KE_{i} = 3KE = 3E$
Subject: Physics —  — 

Question: A smooth ball of mass m moving with the velocity v_o makes an
elastic impact on
another smooth ball of mass M at rest. After impact the velocity
components v_1 and v_2 are

1) v_o cos α, 0
2) v_o sin a, 0
3) v_o sin a, v_o cos α
4) none of these


Solution:
B
Velocity components tangential to the direction
of impact remains unchanged before and after the collision.
∴ (B)
Subject: Physics — Rigid Body Dynamics — Radius of gyration
Difficulty: Moderate

Question: The radius of gyration of a uniform rod of length
[IMAGE], about an axis passing through a point
[IMAGE] away fromthe centre of the rod, and perpendicular
to it, is


Correct option: 1

Solution:
Subject: Physics — Mechanical Properties of Fluids — Fluid in motion
Difficulty: Moderate

Question: A U tube filled with a liquid is accelerating horizontally with
an acceleration a. The acceleration (ms^-2) of the tube is


Answer: 5

Solution:
For free surface $tan\theta = a/g$
or $a = gtan\theta = 10/2 = 5$
Subject: Physics — Ray Optics — Refraction (Lens)
Difficulty: Moderate

Question: Final image of point object formed by the combination is
located

1) On plane surface
2) At a distance 10 cm from plane surface
3) At a distance 30 cm from plane surface
4) At a distance 20 cm from curved surface.

Correct option: 1

Solution:
1st refraction at curved surface
[IMAGE] Reflection of mirror,
[IMAGE] 2nd refraction at curved surface
[IMAGE] So, final image from on plane surface.
Subject: Physics — Electronic device — Gates
Difficulty: Easy

Question: Which of the following gives a reversible operation?


Correct option: 4

Solution:
A logic gate is reversible if we can recover input data from
the output eg. NOT gate
Subject: Physics —  — 
Difficulty: Tough

Question: A thin circular ring of mass per unit length ρ and radius r is
rotating at an angular speed ω as shown in figure. The tension in the
ring is

1) ρω2r2
4) None
ρ r 
 ω

Correct option: 1

Solution:
2T sin[IMAGE] = dmRω2
T = ρR2ω2
TOPIC: CIRCULAR MOTION
SUB TOPIC: DYNAMICS
Subject: Physics — Center of mass — Center of mass of rod of non-uniformlinear mass density
Difficulty: Moderate

Question: The centre of mass of a non uniform rod of length L whose mass
per unit length $\lambda$varies as
$\lambda = \frac{k \cdot x^{2}}{L}$where k is a constant &x is the
distance of any point on rod from its one end, is (from the same end)

1) $\frac{3}{4}L$
2) $\frac{1}{4}L$
3) $\frac{k}{L}$
4) $\frac{3k}{L}$

Correct option: 1

Solution:
Subject: Physics —  — 
Difficulty: Moderate

Question: The acceleration of the blocks (A) and (B) respectively in
situation shown in the figure is: (pulleys & strings are massless)

4) none of these

Correct option: 1

Solution:
aA = 2aB...(1)
mAg - T = mAaA...(2)
2T - mBg = mBaB...(3)
Solving, we get aA =[IMAGE] & aB
TOPIC: NLM
SUB TOPIC: NEWTON'S SECOND LAW
Subject: Physics —  — 

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


Answer: 9

Solution:
If intensity of light incident on slits [IMAGE] and
The intensity through [IMAGE] will be
[IMAGE] The intensity through [IMAGE] will be
[IMAGE] TOPIC:Wave Optics
SUB TOPIC:YDSE
LEVEL:Moderate
Subject: Physics — Ray Optics — Circle of illuminance
Difficulty: Moderate

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

1) 50
2) 34
3) 17
4) 21

Correct option: 3

Solution:
[IMAGE] Solid angle [IMAGE] Percentage of light [IMAGE]
Subject: Physics — Oscillation — Damped oscillation
Difficulty: Moderate

Question: A simple pendulum of length 50cm and bob's mass 100g is
oscillating in the air. A damping force f = 0.0693V acts due to air,
where V is speed of bob. At any time it has a amplitude A. How much time
it will take so that amplitude becomes half of initial value.

1) 1 sec
2) 3 sec
3) 2 sec
4) none of these

Correct option: 3

Solution:
$A = A_{0^{e}}\frac{- bt}{2m}$
$\frac{\text{A}_{0}}{2} = \text{A}_{0^{\text{e}}}\frac{- \text{bt}}{2\text{m}} \Rightarrow \frac{\text{bt}}{2\text{m}} = \ln 2$
Subject: Physics — Newton's Laws of Motion — Newton's second law
Difficulty: Moderate

Question: Two men of masses m and m/2 starts climbing up on two massless
strings fixed at the ceiling with acceleration g and g/2 respectively.
The ratio of tensions in the two strings are 2x: y then find the value
of x + y.


Answer: 7

Solution:
FBD of man of mass (m)
$T_{1} = 2mg$
FBD of man of mass (m/2)
$T_{2} = \frac{mg}{2}\left\lbrack \frac{3}{2} \right\rbrack = \frac{3mg}{4}$
$\Rightarrow T_{1}:T_{2}::8:3$
Subject: 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

1) 2
2) 3
3) 5
4) 8

Correct option: 4

Solution:
and [IMAGE].....(2)
[IMAGE] Now [IMAGE] and [IMAGE] TOPIC: Circular motion
SUB TOPIC: Vertical circle
LEVEL: Moderate
Subject: Physics — Wave Optics — Diffraction
Difficulty: Moderate

Question: Visible light of wavelength [IMAGE] falls normally
on single slit and produces a diffractionpattern. It is found that the
second diffraction minimum is at 60º C from the central maximum. If the
first minimum is produced at[IMAGE], then
[IMAGE] is close to

1) 25º
2) 30º
3) 20º
4) 45º

Correct option: 2

Solution:
For 2^ndminima
So for 1^st minima
[IMAGE] (From equation (i)
[IMAGE] (From sin table)
Subject: Physics —  — 

Question: A thin uniform rod of mass is hinged at one
end. This rod is maintained in horizontal position by colliding very
tiny balls each of mass m/10 completely elastically 10 times per sec
striking at the opposite end as shown in figure. The speed of the balls
just before colliding rod is [IMAGE].Find n.


Answer: 5

Solution:
[IMAGE] Taking torque about hinge:
[IMAGE] TOPIC:Rotaional dynamics
SUB TOPIC: Torque
LEVEL: Moderate
Subject: Physics — Behaviour of perfect gases — Work done
Difficulty: Moderate

Question: One mole of an ideal gas is kept enclosed under a light piston
(area=[IMAGE] ) connected by acompressed spring (spring
constant 100 N / m). The volume of gas is
[IMAGE] and its temperature is 100K. The gas is
heated so that it compresses the spring further by 0.1 m. Then the work
done (in Joule)by the gas in the process is 1.5n. Find n: (Take R =8.3 J
/ K -mole and suppose there is no atmosphere).,d 100 N / m½
ls tqM+s gYds fiLVu ¼{ks=Qy
 rFkk bl 100K
xeZ


Answer: 1

Solution:
Before heating let the pressure of gas be
[IMAGE] from the equilibrium piston.
[IMAGE] Since during heating process, pwafd xeZ izfØ nkSjku,
The spring is compressed further by 0.1 m fLizax 0.1 m 
∴[IMAGE] Work done by gas
dk;Z[IMAGE] = 1.50 = 1.5 J
Subject: Physics — Rigid Body Dynamics — Conservation of angular momentum
Difficulty: Moderate

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).

1) 1000 N
2) 1500 N
3) 2000 N
4) 500 N

Correct option: 2

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

Question: Three polaroids are kept coaxially. Angle between the first and
third polaroid is 90°. Angle between the first and second polaroid is
60°. If unpolarized light energy incident on the first polaroid is I0.
Light energy that emerges from the system is

1) zero

Correct option: 2

Solution:
According to Malus law, I = I0 cos2 θ
After 2nd Polaroid, I = [IMAGE] cos2 60° =
[IMAGE] After 3rd Polaroid, I = [IMAGE] cos2 30° =
Subject: Physics — Rigid Body Dynamics — Pure rolling
Difficulty: Moderate

Question: A uniform disc of mass M and radius R, is resting on a table on
its rim. The coefficient of frictionbetween disc and table is
[IMAGE] Now the disc is pulled with aforce F as shown in
the figure. What is the maximum value of F for which the disc rolls
without slipping?

1) If there is no slipping, angular acceleration ofthe disc,
2) [IMAGE] Now torque of the disc (Using 2)
Substituting this in Eq. (i), we get
 Since there is no slipping,

Answer: 3

Solution:
Let a be the acceleration of the centre of massof disc. Then
If there is no slipping, angular acceleration ofthe disc,
[IMAGE] Now torque of the disc (Using 2)
Substituting this in Eq. (i), we get
 Since there is no slipping,
Subject: Physics — Mechanical Properties of Fluids — Velocity of efflux
Difficulty: Easy

Question: The speed of water flowing out of the orifice before the
cylinder kept inside the tank

1) $\sqrt{gH}$
2) $1.414\sqrt{gH}$
3) $\frac{\sqrt{gh}}{2}$
4) $\sqrt{\frac{gh}{2}}$

Correct option: 1

Solution:
Subject: Physics — Projectile Motion — Horizontal projectile from a height
Difficulty: Moderate

Question: A man throws a packet from the top of the tower directly aiming
at his friend who is standing on the ground at a certain distance from
the base of the tower and this distance is same as the height of the
tower. If packet is thrown with a speed of 4 m /s and it hits the ground
midway between the tower base and his friend then height of the tower
(in m) is 0.4k. Find k. $(g = 10m/s^{2})$

1) &$h = (\frac{4}{\sqrt{2}})t + \frac{1}{2}{gt}^{2}$....
2) $\frac{h}{2} = \frac{1}{2}gt^{2}$
from

Answer: 8

Solution:
$\frac{4}{\sqrt{2}} \cdot t = \frac{h}{2}$....(1)
&$h = (\frac{4}{\sqrt{2}})t + \frac{1}{2}{gt}^{2}$....(2)
$\frac{h}{2} = \frac{1}{2}gt^{2}$
from (1) & (2)
Subject: Physics — Wave on a string — SHM of a particle
Difficulty: Moderate

Question: A transverse sinusoidal wave move along a string in the positive
x-direction at a speed 5cm/s. The wavelength of the wave 0.2m and its
amplitude is 5cm. At given instant velocity of P is
[IMAGE] when its displacement is 2.5 cm. Which of the
following is correct?

2) $\overset{\rightarrow}{\text{v}} \cdot \overset{\rightarrow}{\text{a}} = \frac{2\sqrt{3}}{32}\pi^{3}\text{cm}^{2}/\text{s}^{3}$
3) $\overset{\rightarrow}{\text{v}} \cdot \overset{\rightarrow}{\text{a}} = \frac{25\sqrt{3}}{3}\pi^{3}\text{cm}^{2}/\text{s}^{3}$
4) $\overset{\rightarrow}{\text{v}} \cdot \overset{\rightarrow}{\text{a}} = \frac{25\sqrt{5}}{32}\pi^{3}\text{cm}^{2}/\text{s}^{3}$

Correct option: 1

Solution:
Subject: Physics — Moving charges and — Magnetic field due to current carrying coil
Difficulty: Moderate

Question: Two uniform circular coils x and y having equal number of turns
and carrying equal current in same sense, subtend same solid angle at
point O. if smaller coil x is midway between O and y, then, if we
represent magnetic field due to coil y as B_y and that due to coil x as
B_x then $(\frac{B_{y}}{B_{x}}) = K$ then 20 K is

1) 10
2) 30
3) 20
4) 40

Correct option: 1

Solution:
If radius of x is r, then radius of y is 2r
$B_{y} = \frac{\mu_{0}}{4\pi}\frac{2\pi I(2r)^{2}}{{\lbrack(2r)^{2} + (2d)^{2}\rbrack}^{3/2}}$
$B_{x} = \frac{\mu_{0}}{4\pi}\frac{2\pi{Ir}^{2}}{{\lbrack r^{2} + d^{2}\rbrack}^{3/2}}$
Subject: Physics —  — 

Question: The sphere at P is given a downward velocity
[IMAGE] and swings in a vertical plane at the end of a
rope of [IMAGE] attached to a support at O. The rope
breaks at angle 30° from horizontal, knowing that it can withstand a
maximum tension equal to three times the weight of the sphere. The value
of [IMAGE] is [IMAGE]. Find


Answer: 2

Solution:
[IMAGE] TOPIC:Vertical circle
SUB TOPIC: Simple pendulum
LEVEL: Moderate
Subject: Physics — Elasticity — Potential energy stored in elongated rubber cord
Difficulty: Moderate

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


Answer: 5

Solution:
v = 10^3 × 10^-2
v = 10.00 m/s
Subject: Physics — Mechanical properties of liquid — Buoyancy
Difficulty: Moderate

Question: A block is partially immersed in a liquid and the vessel is
accelerating upwards with acceleration "a". The block is observed by two
observers $O_{1}$ and $O_{2}$one at rest and the other accelerating with
an acceleration "a" upward. The total buoyant force on the block is:,d CykWd nzo:i ls Mqck gqvk gS rFkk ik= Roj.k "a" ls,d izs{kd fojke ij gS rFkk nwljk Roj.k"a" ls dh vksj Rofjr
gSA CykWd ij dk;Zjr~ dqy mRIykou cy gksxkA

1) same for $O_{1}$ and $O_{2}$
2) greater for $O_{1}$ than $O_{2}$
3) greater for $O_{2}$ than $O_{1}$
4) data is not sufficient

Correct option: 1

Solution:
Buoyant force mRIykou
cy$= F_{b} = V_{sub} \cdot \rho_{\mathcal{l}} \cdot g$
Where,$V_{sub},\rho_{\mathcal{l}}$and g all are same w.r.t.
$O_{1}$and$O_{2}$
Hence vr (1)