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Jee Main Advanced Physics Chapter Wise Mock Test
Quiz 3
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Q.1
A parachutist after bailing out falls 50 m with out friction. When parachute opens ,it decelerates at 2 m/sHe reaches the ground with a speed of 3 m/s. At what height ,did he bail out?
0%
293 m
0%
213 m
0%
200 m
0%
111 m
Explanation
Solution:
Free-fall phase (50 m, no friction): v
1
²=2g(50)=2(9.8)(50)=980, so v
1
=√980 m/s.
Deceleration phase (2 m/s², ending at 3 m/s): v
1
²-2(2)d
2
= 3² ⇒ d
2
= (980-9)/4 = 242.75 m.
Total height = 50 + 242.75 ≈ 293 m.
Q.2
The time period of a earth satellite in circular orbit is dependent of
0%
Radius of the orbit
0%
Mass of the satellite
0%
Both the mass and radius
0%
Neither mass nor radius
Explanation
Solution:
Orbital period comes from GMm/r²=mv²/r, and the satellite's own mass m cancels out of both sides -- the period T=2π√(r³/GM) ends up depending only on the orbital radius r (and Earth's mass GM, which is fixed), never on the satellite's own mass.
Q.3
A surface S enclosed a electric charge q. Let F be the electric flux passing through it.The flux will be maximum if the surface S is like
0%
Cube
0%
Spherical
0%
cylindrical
0%
Same in all the three
Explanation
Solution:
Gauss's law makes the flux through any closed surface depend only on the total charge it encloses, q/ε
0
-- entirely independent of the surface's shape. A cube, a sphere, or a cylinder enclosing the same charge q all carry exactly the same flux.
Q.4
If suddenly the gravitational force of attraction between earth and a satellite revolving around it becomes zero,then the satellite will?
0%
Move towards earth
0%
Becomes stationary in the orbit
0%
continue to move in the orbit with same velocity
0%
Move tangentially to the original orbit with same velocity
Explanation
Solution:
Gravity is what was supplying the centripetal force bending the satellite's path into a circle. Remove that force entirely, and by Newton's first law the satellite simply continues in a straight line at whatever velocity it had the instant gravity vanished -- which, for circular motion, is tangent to the original orbit.
Q.5
M capacitors of same capacitance C are connected in parallel to the voltage source V.The total energy stored in the capacitor will be?
0%
CV
0%
$CV^2$
0%
$.5MCV^2$
0%
$ \frac {.5CV^2}{M}$
Explanation
Solution:
M identical capacitors in parallel combine to a total capacitance of MC. Total stored energy = ½(MC)V² = 0.5MCV².
Q.6
A thin ring is rotating about its own axis .If the density of the ring is 11300 Kg/m3,then the maximum permissible linear speed of the ring is
0%
60 m/s
0%
40 m/s
0%
44.6 m/s
0%
42 m/s
Explanation
Solution:
A spinning ring develops tensile (hoop) stress σ=ρv² from its own circular motion, where v is its rim speed. The ring holds together only as long as this stress stays below the material's breaking stress, so the maximum permissible speed is v
max
=√(σ
break
/ρ).
Working backward from the given answer with ρ=11300 kg/m³ (this is lead's density) shows the breaking stress used here is about 20×10⁶ N/m²: v
max
=√(20×10⁶/11300) ≈ 42 m/s.
Q.7
Gases begin to conduct electricity at low pressure because
0%
atoms break up into electrons and proton's
0%
at low pressure gases turn into plasma
0%
the electrons in atoms can move freely at low pressure
0%
colliding electrons can acquire higher kinetic energy due to increased mean free path leading to ionization of atoms
Explanation
Solution:
At low pressure, gas molecules are spread further apart, so free electrons travel a much longer mean free path between collisions -- picking up much more kinetic energy along the way. When they finally do collide with a gas atom, they're moving fast enough to knock electrons loose from it, ionising it and enabling conduction.
Q.8
An LCR series circuit is connected to a source of AC voltage.At resonance ,the phase difference between the applied voltage and current in the circuit is
0%
180°
0%
45°
0%
90°
0%
0°
Explanation
Solution:
At resonance, an LCR series circuit's inductive and capacitive reactances exactly cancel, leaving the circuit behaving purely resistively -- and for a pure resistor, voltage and current are always perfectly in phase, a difference of 0°.
Q.9
Two parallel plate air capacitors have their plate areas 100 cm2 and 500 cm2 respectively.If they have the same charge and potential and the distance between the plates of the first capacitor is 0.5 mm, what is the distance between the plates of the second capacitor?
0%
.25 cm
0%
.1 cm
0%
.60 cm
0%
1 cm
Explanation
Solution:
Equal charge Q and equal potential V on both capacitors means equal capacitance too, since C=Q/V is the same ratio for both.
With C=ε
0
A/d, equal C means A
1
/d
1
=A
2
/d
2
, so d
2
= d
1
×(A
2
/A
1
) = 0.5mm×(500/100) = 2.5mm = 0.25 cm.
Q.10
Which of these is false for movement of alpha and beta particles in electric and magnetic field
0%
Both the article move in circular path in magnetic field
0%
curvature of path of beta particle is more than the curvature of alpha particle in magnetic field
0%
Deflection of alpha particle is opposite to that beta particle in electric field
0%
None of these
Explanation
Solution:
Both alpha and beta particles, being charged, curve into circles in a magnetic field -- true. A beta particle's much smaller mass gives it a much tighter (more curved) path than an alpha particle's for comparable speeds -- true. And carrying opposite-sign charges, alpha (positive) and beta (negative) particles are deflected in opposite directions by an electric field -- also true. None of the three statements here is actually false.
Q.11
If the distance between the earth and sun were half its present value,the number of days in the years would have been
0%
129
0%
730
0%
181
0%
64.5
Explanation
Solution:
By Kepler's third law, T² ∝ r³. Halving the orbital radius scales the period by (½)^(3/2) = 1/(2√2).
T' = 365 ÷ (2√2) ≈ 365 ÷ 2.828 ≈ 129 days.
Q.12
A body is moving in straight line with a initial velocity u0 and constant acceleration a0 . It covers a distance 40 m in the 4th sec and distance of 60 m in the 6th sec. Which of them are true?
0%
$u_0 =5 m/s , a_0=10 m/s^2$
0%
$u_0 =5 m/s , a_0=15 m/s^2$
0%
$u_0 =15 m/s , a_0=5 m/s^2$
0%
$u_0 =10 m/s , a_0=5 m/s^2$
Explanation
Solution:
Distance in the nth second: s
n
= u + a(2n-1)/2.
4th second: u + 3.5a = 40. 6th second: u + 5.5a = 60.
Subtracting: 2a = 20, so a=10 m/s². Substituting back: u + 3.5(10) = 40, so u = 5 m/s.
Q.13
Which all is conserved in Nuclear reactions
0%
Energy only
0%
Mass only
0%
Momentum only
0%
Mass,energy and momentum
Explanation
Solution:
In any nuclear reaction, momentum is always conserved (no external force), and mass and energy are conserved together (via E=mc²) -- any apparent "missing" mass shows up as released energy, and vice versa, so at this level all three -- mass, energy, and momentum -- are treated as conserved quantities.
Q.14
A elevator of Mass M moves upward with acceleration .3 g pulled by the cable. What is the normal force exerted by the elevator floor on the person of mass m standing on the elevator?
0%
Mg
0%
.7mg
0%
mg
0%
1.3mg
Explanation
Solution:
For the person (mass m) accelerating upward with the elevator at a=0.3g, Newton's second law gives N-mg=ma, so N = m(g+a) = m(g+0.3g) = 1.3mg -- the floor has to push up harder than just their weight to also accelerate them upward.
Q.15
When a ray of light enters a glass slab from air
0%
Wavelength decreases
0%
Frequency increase
0%
Neither wavelength nor frequency changes
0%
None of these
Explanation
Solution:
Crossing into glass, a light ray's frequency stays fixed (frequency is set by the source and doesn't change with medium), but its speed drops (glass is optically denser than air). Since wavelength = speed/frequency, a lower speed at the same frequency means a shorter wavelength.
Q.16
A pool ball of mass m and radius r is given initial sliding velocity v0 ( no rotation) on a horizontal pool table.how long will it take for ball to start pure rolling if the coefficient of friction between ball and table is k?
0%
$ \frac {2v_0}{7kg}$
0%
$ \frac {7v_0}{2kg}$
0%
$ \frac {v_0}{kg}$
0%
None of the above
Explanation
Solution:
Sliding without rotating, kinetic friction f=kmg acts backward, decelerating the centre (a=-kg) while its torque about the centre (fr) spins the ball up: angular acceleration α=fr/I=kmgr/(₂ₛ₅mr²) = 5kg/(2r).
Pure rolling begins once v=ωr: v
0
-kgt = [5kg/(2r)]t×r = (5kg/2)t.
v
0
= kgt(1+5/2) = (7/2)kgt, so t = 2v
0
/(7kg).
Q.17
Two person X and Y,each carrying a source of sound of frequency k are standing a few meters apart in a quiet field. X starts moving towards Y with the velocity a .If b is the speed of sound,how many beats will be heard per sec by A?
0%
$ka(b-a)$
0%
$ \frac {2ka}{a+b}$
0%
$ \frac {2ka}{b}$
0%
$ \frac {ka}{b}$
Explanation
Solution:
X still hears their own source at its true frequency k (a source never Doppler-shifts relative to itself). But moving toward Y's stationary source at speed a, X hears Y's frequency shifted up to k(b+a)/b.
Beat frequency = difference between these two: k(b+a)/b - k = k[(b+a-b)/b] = ka/b.
Q.18
A monochromatic point source of light is at a distance of .2 m from a photo electric cell,the saturation current and cutoff voltage are 18 mA and .6 Volt respectively.If the same source is placed at .6 m away from the photo electric cell,then
0%
stopping voltage=.6 V,Saturation current =2 mA
0%
stopping voltage=.6 V,Saturation current = mA
0%
stopping voltage=.6 V,Saturation current =3 mA
0%
stopping voltage=.16 V,Saturation current =2 mA
Explanation
Solution:
Stopping voltage depends only on the light's frequency (via eV
stop
=hf-φ), not on how far away or how intense the source is -- moving the source doesn't change its colour, so it stays 0.6 V.
Saturation current tracks intensity, which falls off as 1/r². Moving from 0.2 m to 0.6 m is 3× the distance, so intensity (and hence current) drops by a factor of 3²=9: 18mA ÷ 9 = 2 mA.
Q.19
A proton is released from rest in a region of steady and uniform Electric Field E and Magnetic Field B which are parallel to each other. The proton trajectory will be
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Circle
0%
Straight line
0%
helix
0%
cycloid
Explanation
Solution:
Starting from rest, the proton has zero velocity, so the magnetic force qv×B is zero at that first instant -- only the electric force qE acts, accelerating it along E's direction, which is the same direction as B.
Since it only ever picks up velocity along that shared direction, its velocity stays parallel to B at every later instant too, keeping v×B at zero throughout -- the magnetic field never gets a chance to act, and the proton just travels in a straight line.
Q.20
Three capacitors have capacitance's of C F, 2C F and 3C F respectively. They are first connected to have maximum capacitance and then connected to have minimum capacitance. Find the ratio of maximum capacitance to minimum capacitance.
0%
6
0%
5
0%
11
0%
None of these
Explanation
Solution:
Maximum capacitance comes from connecting all three in parallel: C+2C+3C=6C.
Minimum comes from all three in series: 1/C
min
= 1/C+1/2C+1/3C = 11/(6C), so C
min
=6C/11.
Ratio = 6C ÷ (6C/11) = 11.
Q.21
A flat bad truck carries a box. The coefficient of friction between the box and truck is .3 What is the maximum acceleration the driver can have so that box does not slide? Take g=10 m/s2
0%
Data not sufficient
0%
.3 m/s2
0%
3 m/s2
0%
None of these
Explanation
Solution:
The box can only be pushed forward by friction from the truck bed beneath it, and the most that friction can provide is μmg. Setting this equal to ma (the force needed to accelerate the box along with the truck) gives the maximum shared acceleration: a
max
=μg=0.3×10=3 m/s².
Q.22
Two electron are released towards each other with equal velocities of 106 m/s. What is the closest distance between them?
0%
$3.56 \times 10^{-10}$ m
0%
$2.56 \times 10^{-7}$ m
0%
$1.6 \times 10^{-10}$ m
0%
$2.56 \times 10^{-10}$ m
Explanation
Solution:
By symmetry, two identical electrons approaching head-on with equal speed both come to rest simultaneously at their closest approach, converting all their kinetic energy into electrostatic PE:
mv² = ke²/d ⇒ d = ke²/(mv²)
= (9×10⁹)(1.6×10⁻¹⁹)² ÷ [(9.1×10⁻³¹)(10⁶)²] ≈ 2.56×10⁻¹⁰ m.
Q.23
A heater boils a certain quantity of liquid in time x. Another heater boils the same quantity of liquid in time y. If the both the heater are connected in parallel,the combination will boil the liquid in time
0%
$ \frac {1}{x}+ \frac {1}{y}$
0%
$x +y $
0%
$ \frac {x+y}{2}$
0%
$ \sqrt {xy}$
Explanation
Solution:
In parallel, the two heaters' powers simply add: P
combined
=P
1
+P
2
. Since each heater's own power is (same heat needed)/(its own time), P
1
=Q/x and P
2
=Q/y -- so the combined heating RATE adds as 1/x + 1/y (the reciprocal-time relationship the combined power follows, the same way parallel resistors' conductances add).
Q.24
The effective capacitance of a number of capacitor connected in series is p F ,When one of the capacitor is removed,the The effective capacitance is q F.The capacitance of the capacitor which is removed
0%
$q-p$
0%
$ \frac {pq}{p+q}$
0%
$ \frac {pq}{q-p}$
0%
None of these
Explanation
Solution:
For capacitors in series, 1/C
total
is the sum of each 1/C. Removing one capacitor drops exactly its own 1/C term from that sum:
1/p = 1/q + 1/C
removed
⇒ 1/C
removed
= 1/p - 1/q = (q-p)/(pq) ⇒ C
removed
= pq/(q-p).
Q.25
When 30 J of work was done on a gas,50 J of heat energy was released. The final internal energy of the is 90 J.The initial internal energy of the gas is?
0%
110 J
0%
70 J
0%
10 J
0%
170 J
Explanation
Solution:
First law: ΔU = Q + W (W = work done ON the gas). Releasing 50J of heat means Q=-50J, and 30J done on the gas means W=+30J:
ΔU = -50+30 = -20 J.
Since ΔU = U
final
-U
initial
= 90-U
initial
= -20, U
initial
= 90+20 = 110 J.
Q.26
A Proton moving with constant velocity passes through a region without any change in its velocity. If E and B denote electric and magnetic field,this region cannot have
0%
E > 0, B >0
0%
E=0,B=0
0%
E > 0,B=0
0%
E=0,B > 0
Explanation
Solution:
For velocity to stay unchanged, the net force qE+qv×B must be zero. With E=0 and B=0, there's nothing to unbalance -- fine. With E=0 and B>0, the magnetic force vanishes too whenever v is parallel to B -- also possible. With both E>0 and B>0, a velocity-selector-style balance between qE and qv×B is possible for the right geometry.
But with E>0 and B=0, there's nothing at all to cancel the electric force qE -- the velocity would necessarily change, so this combination cannot be what's happening here.
Q.27
A bar magnet falls vertical down through a coil placed in horizontal plane? What will be the acceleration of bar magnet?
0%
g
0%
greater than g
0%
less than g
0%
zero
Explanation
Solution:
As the magnet falls through the coil, the changing flux induces a current, and by Lenz's law that induced current opposes the change -- creating a magnetic force that resists the magnet's fall. This retarding force works against gravity, so the magnet's acceleration ends up less than g (though generally still greater than zero, unless it reaches a full terminal velocity).
Q.28
which of these are true?
0%
A conducting rod XY moves parallel to the x-axis in a uniform magnetic field pointing in +z direction.The end X is towards -y direction.The end X will be positively charged
0%
A square coil of metal wire is stationary in a non-uniform magnetic field,An Emf is induced in the coil
0%
The self inductance of the straight conductor is very large
0%
Lenz's law is a consequence of the law of conservation of charge
Explanation
Solution:
With the rod moving along +x through B along +z, the magnetic force on its positive charges, qv×B, points along -y -- pushing them toward the -y end of the rod, which is end X, making X positively charged.
A coil sitting still in a field that's non-uniform in SPACE but constant in TIME sees no change in flux, so no EMF is induced -- that claim is false. A straight wire has very LOW self-inductance, not large -- also false. And Lenz's law follows from conservation of ENERGY, not charge -- also false.
Q.29
An inductor stores energy in its
0%
Magnetic field
0%
Electric field
0%
electric and magnetic field both
0%
None of these
Explanation
Solution:
An inductor's energy is stored in the magnetic field building up around its current-carrying coil (½LI²) -- a capacitor is the device that stores energy in an electric field instead.
Q.30
which of these is not true for radioactive decay?
0%
The nucleus of the atom emits the radio active radiation
0%
The penetrating power of Gamma rays is highest among the radioactive radiation
0%
alpha and beta particles are deflected by electric and magnetic field but gamma rays are not
0%
None of these
Explanation
Solution:
The nucleus really is the source of radioactive emissions -- true. Gamma rays, being the most energetic and having no charge to interact via, do penetrate matter more than alpha or beta -- true. And being charged, alpha and beta particles are deflected by electric and magnetic fields, while uncharged gamma photons are not -- also true. None of these three statements is actually false.
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