MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Diagrams
Games
JEE
Jee Main Advanced Physics Chapter Wise Mock Test
Quiz 4
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Q.1
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 from air into glass, frequency stays fixed (set by the source) while speed drops (glass is optically denser) -- since wavelength = speed/frequency, a lower speed at the same frequency means the wavelength decreases.
Q.2
Which of the these is true in Bohr model of hydrogen atom?
0%
The total energy of electron in the mth orbit is inversely proportional to m
0%
radiation are emitted when electron jumps from lower orbit to higher orbit
0%
The radius of the mth orbit is inversely proportional to m
0%
Angular momentum of the electron in mth orbit is an integral multiple of $ \frac {h}{2 \pi}$
Explanation
Solution:
The Bohr model's foundational postulate is that angular momentum is quantised: L=mh/(2π) for the mth orbit -- literally an integer multiple of h/(2π).
(The other options each get their proportionality wrong: energy actually scales as 1/m² not 1/m, radius scales as m² -- growing with m, not shrinking -- and photons are emitted when an electron falls to a LOWER orbit, not when it jumps to a higher one.)
Q.3
In an AC circuit, which of these is true?
0%
The applied instantaneous voltage equal to the algebraic sum of the instantaneous voltages across the different element
0%
The applied instantaneous voltage equal to the vector sum of the instantaneous voltages across the different element
0%
Power dissipation happens in inductor and resistance in an AC circuit
0%
None of these
Explanation
Solution:
In an AC circuit, the voltage across each element generally has a different phase relative to the current (especially with inductors and capacitors), so their instantaneous voltages don't simply add arithmetically -- they have to be combined as vectors (phasors), accounting for those phase differences, to get the total applied voltage. (An ideal inductor stores and releases energy rather than dissipating it, so only the resistance actually dissipates power, not the inductor.)
Q.4
A parallel plate capacitor has capacitance C. It is connected to a battery of EMF E until fully charged, and then disconnected. The plates are then pulled apart an extra distance d, during which the measured potential difference between them changed by a factor of i. Did the potential difference increase or decrease by a factor of i?
0%
decrease
0%
increase
0%
cannot say
Explanation
Solution:
With the capacitor disconnected, its charge Q is fixed. Pulling the plates apart increases the separation, which lowers the capacitance C=ε
0
A/d -- and since V=Q/C, a smaller C with the same Q means V increases.
Q.5
What is true about electric field between the plates
0%
electric field remains unchanged
0%
electric field decrease by the factor i
0%
electric field increase by the factor i
0%
None of these
Explanation
Solution:
For a parallel-plate capacitor, the field between the plates is E=σ/ε
0
, depending only on the surface charge density σ=Q/A -- not on the plate separation at all. Since neither Q nor the plate area A changes here, E stays exactly the same even as the plates move apart (it's only V=E×d that grows, because d grows while E itself doesn't).
Q.6
An object lying on a long horizontal conveyer belt moving at a constant velocity v receives a velocity v0= 5m/s relative to the ground in the direction opposite to the direction of the motion of conveyer. After t=4s, the velocity of the block becomes equal to the velocity of the belt. The coefficient of friction between the block and the belt is μ=.2 What is the value of ratio v/v0
0%
1/5
0%
3/5
0%
2/5
0%
4/5
Explanation
Solution:
Taking the belt's direction as positive, the object starts at -v
0
and friction (magnitude μg) accelerates it toward the belt's velocity v over time t=4s: v = -v
0
+ μg·t.
With μ=0.2, g=10: v = -5 + (0.2)(10)(4) = -5+8 = 3 m/s.
v/v
0
= 3/5.
Q.7
Link Comprehension Type Use this answer Questions 2,3 and 4 A 10μF capacitor is connected through a 1000ohm resistance to a constant Potential difference of 100 V Find the current in the circuit at t=0
0%
.1 A
0%
10 A
0%
2 A
0%
5 A
Explanation
Solution:
The instant the switch closes, the uncharged capacitor briefly behaves like a plain wire (no opposing voltage yet), so the current is simply set by the resistor alone: I=V/R=100/1000=0.1 A.
Q.8
Find the time in which the capacitor will acquire its half charge
0%
$.693 \times 10^{-2}$
0%
$.2 \times 10^{-2}$
0%
$.5 \times 10^{-3}$
0%
$.25 \times 10^{-3}$
Explanation
Solution:
Charging follows q(t)=Q
0
(1-e
-t/RC
). Reaching half charge means 1-e
-t/RC
=½, so e
-t/RC
=½, giving t=RC·ln(2)≈0.693RC.
RC = 1000×(10×10⁻⁶) = 10⁻² s, so t ≈ 0.693×10⁻² s.
Q.9
Find the current at t=10-2 sec
0%
.0368 A
0%
.368 A
0%
.1 A
0%
1.25 A
Explanation
Solution:
Charging current decays as I(t)=I
0
e
-t/RC
. Here t=10⁻²s equals exactly one time constant RC (from the previous part), so:
I = 0.1 × e⁻¹ = 0.1 × 0.3679 ≈ 0.0368 A.
Q.10
An object is shot with an initial velocity 20 m/s at an angle of 600 with the horizontal. At the top of the trajectory ,the object explodes into two fragments of equal mass. One fragment whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land.
0%
20 m
0%
53 m
0%
30m
0%
None of the above
Explanation
Solution:
At the peak of the trajectory, all velocity is horizontal: v
x
=20cos60°=10 m/s, v
y
=0.
Momentum conservation during the (instantaneous, internal) explosion: with the falling fragment (mass m) carrying zero velocity, all the pre-explosion horizontal momentum (2m×10) is carried by the other fragment (mass m) alone: m×v
2
= 2m×10 ⇒ v
2
=20 m/s.
Peak height: y=v
y0
²/(2g) = (20sin60°)²/20 = 300/20 = 15 m, reached at x
peak
=v
x
×t
peak
=10√3 m (t
peak
=v
y0
/g=√3 s).
The second fragment then falls those 15 m from rest (vertically) while moving at 20 m/s horizontally: fall time √(2×15/g)=√3 s, covering an extra 20√3 m horizontally.
Total range = 10√3+20√3 = 30√3 ≈ 52-53 m from the gun.
Q.11
A cylindrical drum of mass M and radius R lies on the road against the curb which has height R/2 as shown in below figure.it is lifted gently quasistatiscally on the sidewalk by a rope wound around its circumference. What is the minimum force needed if the rope is pulled horizontally
0%
$\frac {(mg2 \sqrt 3)}{3}$
0%
$\frac {(mg\sqrt 3)}{2}$
0%
$\frac {(mg\sqrt 3)}{4}$
0%
$\frac {(mg\sqrt 3)}{3}$
Explanation
Solution:
Pivoting about the curb's edge as the drum just lifts off the ground: with curb height h=R/2, the horizontal distance from the pivot to the drum's centre is √(R²-(R-h)²)=√(R²-(R/2)²) = R√3/2, giving gravity's torque about the pivot as Mg×(R√3/2).
With the rope pulled horizontally from the top of the drum (height 2R), its moment arm about the pivot (at height R/2) is 2R-R/2=3R/2.
Balancing torques at the minimum lifting force: F×(3R/2) = Mg×(R√3/2) ⇒ F = Mg√3/3.
Q.12
When a block of iron floats in mercury ay 00 C , a fraction i1 of its volume is submerged while at the temperature 600 C ,a fraction i2 is seen to be submerged. If the coefficient of volume expansion of iron is kfe and that of mercury is khg ,then which of these statement is true
0%
$i_1 (1-60K_{fe} )-i_2 (1+60K_{hg} )=0$
0%
$i_1 (1+60K_{fe} )-i_2 (1+60K_{hg} )=0$
0%
$i_1 (1+60K_{fe} )-i_2 (1-60K_{hg} )=0$
0%
$i_2 (1+60K_{fe} )-i_1 (1+60K_{hg} )=0$
Explanation
Solution:
A floating object's submerged fraction equals the ratio of its density to the liquid's: i(T) = ρ
iron
(T)/ρ
hg
(T). Both densities fall with temperature as ρ(T)=ρ(0)/(1+γT) (volume expansion).
i
1
= ρ
iron
(0)/ρ
hg
(0) at 0°C (no correction needed at the reference point).
i
2
= [ρ
iron
(0)/(1+60K
fe
)] ÷ [ρ
hg
(0)/(1+60K
hg
)] = i
1
×(1+60K
hg
)/(1+60K
fe
), which rearranges to i
2
(1+60K
fe
) = i
1
(1+60K
hg
) -- the two fractions and their matching expansion coefficients balance out this way.
Q.13
The refractive index of the material of an equilateral prism is $\sqrt 3 $ . What is the angle of minimum deviation?
0%
60°
0%
25°
0%
30°
0%
15°
Explanation
Solution:
μ = sin((A+D
m
)/2) ÷ sin(A/2), with A=60° for an equilateral prism.
√3 = sin((60+D
m
)/2) ÷ sin(30°) = sin((60+D
m
)/2) ÷ 0.5 ⇒ sin((60+D
m
)/2) = √3/2 = sin(60°).
(60+D
m
)/2 = 60° ⇒ D
m
= 60°.
Q.14
Two beams of light having intensities I and 9I interfere to produce a fringe pattern on a screen . The phase difference between the beams is π/2 at X and π at point Y. Let IX and IY be the resultant intensities at point X and Y. Which of these is true?
0%
$I_X –I_Y=4I$
0%
$I_X –I_Y=6I$
0%
$I_X –I_Y=2I$
0%
None of these
Explanation
Solution:
Resultant intensity: I
result
= I
1
+I
2
+2√(I
1
I
2
)cosφ.
At X (φ=π/2, cos=0): I
X
= I+9I+0 = 10I.
At Y (φ=π, cos=-1): I
Y
= I+9I-2√(9I²) = 10I-6I = 4I.
I
X
-I
Y
= 10I-4I = 6I.
Q.15
A mass m=.2 Kg and a spring of spring constant K=.5 N/m lie on a smooth horizontal table . The mass is released at x=.1 m from its equilibrium position. At what time t,the mass passes through point A which is at distance .02 m from equilibrium position first time.
0%
.1 sec
0%
.87 sec
0%
.6 sec
0%
.5 sec
Explanation
Solution:
ω=√(k/m)=√(0.5/0.2)=√2.5≈1.581 rad/s. Released from rest at the amplitude (x=0.1m), the motion is x(t)=A cos(ωt).
0.02 = 0.1cos(ωt) ⇒ cos(ωt)=0.2 ⇒ ωt=arccos(0.2)≈1.369 rad ⇒ t ≈ 1.369/1.581 ≈ 0.87 s.
Q.16
0%
The equivalent resistance between A and B is 3R/2
0%
The equivalent resistance between A and C is 5R/8
0%
The equivalent resistance between B and C is 3R/8
0%
None of these
Explanation
Solution:
Labelling the fourth (unlabeled) corner D, the network has A-D, D-B, A-B (diagonal), A-C, and B-C, each resistance R.
Setting V
A
=V, V
C
=0 and solving the two node equations at B and D gives V
B
=3V/5 and V
D
=4V/5. The total current leaving A (through all three paths out of A) works out to (8V/5)/R, so R
AC
= V ÷ [(8V/5)/R] = 5R/8.
Q.17
The maximum number of possible interference maxima for slit –separation equal to the twice of wavelength in Young’s double slit experiment is
0%
Five
0%
Four
0%
Three
0%
Zero
Explanation
Solution:
Maxima occur where d sinθ=mλ. With d=2λ, sinθ=m/2, and since |sinθ|≤1, m can only be -2,-1,0,1,2.
But m=±2 correspond to sinθ=±1, i.e. θ=±90° -- grazing incidence, a degenerate limiting case that isn't actually observed as a real fringe on a screen. That leaves m=-1,0,1 -- three genuine maxima.
Q.18
An electric dipole consisting of two opposite charges of 2X10-6 C each separated by a distance of 3 cm is placed in an electric field of 2 X105 N/C. The maximum torque on the dipole will be
0%
$12 \times 10^{-3}$ Nm
0%
$36 \times 10^{-3}$ Nm
0%
$12 \times 10^{-1}$ Nm
0%
$12 \times 10^{-2}$ Nm
Explanation
Solution:
Maximum torque on a dipole is τ
max
=pE, where p=q×d is the dipole moment.
p = (2×10⁻⁶)(0.03) = 6×10⁻⁸ C·m.
τ
max
= (6×10⁻⁸)(2×10⁵) = 12×10⁻³ N·m.
Q.19
Three identical charged particles each possessing the mass m and charge +q are placed at the corner of the equilateral triangle of side a. Then the particles are simultaneously set free and start flying apart symmetrically due to coulomb repulsion force. Find the velocity of each particle as they move large distance apart
0%
$\sqrt {\frac {(8\pi \epsilon _0 q^2)}{ma}}$
0%
$\sqrt {\frac {(4\pi \epsilon _0 q^2)}{ma}}$
0%
$\sqrt {\frac {(8\pi \epsilon _0 q^2)}{a}}$
0%
None of the these
Explanation
Solution:
With three equal charges at the corners of an equilateral triangle (side a), there are 3 pairwise interactions, each starting at separation a: total initial PE = 3×q²/(4πε
0
a).
As they fly apart to infinity, this all converts into kinetic energy shared equally (by symmetry) among the three: total KE = 3×½mv².
Setting these equal and solving: v² = 2q²/(4πε
0
am) = q²/(2πε
0
am), so v = q√[1/(2πε
0
ma)] -- the standard result for this classic three-charge problem.
Q.20
The photoelectric work function of potassium is 2.0 eV . if Light having a wavelength of 360nm falls on potassium. Let x be the stopping potential and y be the kinetic energy in electron volt of the most energetic electrons ejected?
0%
x=1.45 eV , y=1.45 eV
0%
x=1.25eV, y=1.45 eV
0%
x=1.45 eV, y=1.25 eV
0%
None of these
Explanation
Solution:
Photon energy: E=hc/λ = 1240 eV·nm ÷ 360 nm ≈ 3.44 eV.
Maximum kinetic energy: KE
max
= E - φ = 3.44 - 2.0 = 1.44 ≈ 1.45 eV.
The stopping potential (in volts) always equals the maximum KE (in eV) numerically, since eV
stop
=KE
max
-- so x=1.45V and y=1.45eV, the same number in each unit.
Q.21
A satellite is moving around the earth in a stable circular orbit. Which one of these is not true
0%
Its angular momentum is constant
0%
It is moving with constant speed
0%
it behaves as if it were as freely falling body
0%
It is acted upon by a force directed away from center of the earth which counter balances the gravitational pull of the earth
Explanation
Solution:
A stable orbit has constant angular momentum (no external torque), constant speed (circular motion), and behaves exactly like continuous free fall (which is what "orbiting" physically is) -- all three true.
But there is no real outward force balancing gravity -- the "centrifugal force" some people imagine is only a fictitious effect felt in a rotating frame, not a genuine force in the real (inertial) picture. In reality, gravity itself is the UNBALANCED net force that curves the satellite's path into an orbit -- nothing is cancelling it.
Q.22
Two equal spheres X and Y lie on a smooth horizontal circular groove at opposite ends of a diameter. Sphere X is projected along the groove and at the end of time I impinge on sphere Y. If e is the coefficient of restitution, the second impact will occur after a time equal to
0%
T
0%
eT
0%
2T/e
0%
2eT
Explanation
Solution:
For an elastic-ish collision between equal masses with restitution e (X moving at u hits stationary Y): momentum conservation and the restitution definition give v
X
'=u(1-e)/2 and v
Y
'=u(1+e)/2 after the collision.
Their relative speed afterward is v
Y
'-v
X
' = ue. Since X initially covered the diameter's distance L=uT in reaching Y, the full circumference is 2L, and Y (now ahead and faster) needs to close that full extra lap relative to X: time = 2L/(ue) = 2(uT)/(ue) = 2T/e.
Q.23
An alternating current containing a resistance of 10 ohm and inductance of 1/10π henry in series is connected with a 100 V AC of 50 cycles/second . Power loss in the circuit is
0%
400 W
0%
500 W
0%
376 W
0%
424 W
Explanation
Solution:
X
L
=2πfL = 2π(50)(1/10π) = 10 Ω.
Z = √(R²+X
L
²) = √(100+100) = 10√2 Ω.
I
rms
= V/Z = 100/(10√2) = 5√2 A.
Power (dissipated only in R, since an ideal inductor dissipates none) = I²R = (5√2)²×10 = 50×10 = 500 W.
Q.24
An object of mass m is acted upon by the constant force and start from rest and it covers a distance d in some time. The kinetic energy( K) acquired by the body is
0%
is directly proportional to $\sqrt m$
0%
is directly proportional to m
0%
is directly proportional to $1/\sqrt {m} $
0%
independent of m
Explanation
Solution:
By the work-energy theorem, KE = work done = F×d. Since neither F nor d involves the mass at all, the kinetic energy gained doesn't depend on m -- a lighter object would simply end up moving faster over the same distance, but with the same total KE.
Q.25
Link Comprehension Type Use this answer Questions 2,3 and 4 Two circuit X and Y connected to same E Volt DC-source have self inductance L1 and L2 respectively ( L1 > L2) and equal resistance of R ohm each. Let Ix and Iy be the steady state current in Circuit X and Y respectively Which is the appropriate relation for steady state current in Circuit X and Circuit Y?
0%
Ix > Iy
0%
Ix < Iy
0%
Ix = Iy
0%
Cannot say on this
Explanation
Solution:
Once current stops changing, an inductor no longer opposes it at all (its back-EMF is proportional to dI/dt, which is zero at steady state) -- so at steady state it behaves like a plain wire. The steady-state current is then just I=E/R for both circuits, and with matching E and R, I
x
=I
y
, regardless of how different their inductances are.
Q.26
Let tx and ty be the time taken by the current in respective circuit to reach (1- 1/e) of their steady state value. Which is the appropriate relation between tx and ty ?
0%
tx > ty
0%
tx < ty
0%
tx = ty
0%
Cannot say on this
Explanation
Solution:
Reaching (1-1/e) of the steady-state value takes exactly one time constant, τ=L/R. With the same R but L
1
>L
2
, circuit X has the larger time constant, so it takes longer: t
x
>t
y
.
Q.27
Let PX and PY be the power dissipated in the circuit to build up its current to the steady value. Which is the appropriate relation between tx and ty ?
0%
PY > PX
0%
PY < PX
0%
PX = PY
0%
Cannot say on this
Explanation
Solution:
Since both circuits reach the SAME steady-state current (Q372) through the SAME resistance, they dissipate heat at matching rates whenever their instantaneous currents match. But X (larger L) takes longer to climb to that current (Q373), so it spends more time dissipating heat along the way -- and it also ends up storing more energy in its magnetic field (½LI², with the same I but larger L). Both effects mean circuit X consumes more total energy building up to steady state than circuit Y does.
Q.28
A solid cylinder of mass M and radius R rolls downs an inclined plane of height H. which of the following statement is false
0%
The translational energy is 2/3 of the initial energy when the cylinder rolls down to the bottom of the incline
0%
The Rotational energy at the bottom of the incline plan is ½ of the translational energy at the bottom
0%
Angular velocity at the bottom is $ \frac {2}{R} \sqrt {\frac {gH}{3}}$
0%
None of the above
Explanation
Solution:
For a rolling solid cylinder, total KE=¾Mv², split as translational ½Mv² (2/3 of the total) and rotational ¼Mv² (1/3 of the total) -- so translational IS 2/3 of the total energy (true), and rotational IS exactly half of translational, since (1/3)/(2/3)=1/2 (true).
From MgH=¾Mv², v=√(4gH/3), so ω=v/R=(2/R)√(gH/3) -- matching the given formula (true). All three statements check out, so none of them is false.
Q.29
An AC current is given by the expression $i= p cos \omega t + q sin \omega t$ The rms value of the current will be
0%
p +q
0%
$(p^2 + q^2)/2$
0%
$\frac {\sqrt {p^2 -q^2}}{2}$
0%
$\frac {\sqrt {p^2 +q^2}}{2}$
Explanation
Solution:
A combination like p cosωt + q sinωt is itself a single sinusoid with peak amplitude √(p²+q²) (the two components combine the same way perpendicular vector components do).
For any sinusoid, the rms value is its peak divided by √2: I
rms
= √(p²+q²)/√2 = √[(p²+q²)/2].
Q.30
A box of mass M is pulled by a man holding a rope at an angle θ to the horizontal .A second man pulls the box horizontally in the opposite direction with a force equal to twice the weight of the box.What is the maximum value of θ such that the box begins to move in the direction of first man without being lifted up?
0%
26.57°
0%
33.3°
0%
57°
0%
None of these
Explanation
Solution:
For the box to just begin sliding without lifting: horizontally, P cosθ must just overcome the opposing 2Mg: P cosθ=2Mg. Vertically, at the verge of lifting the normal force reaches zero: P sinθ=Mg.
Dividing the two: tanθ = Mg/(2Mg) = 1/2, so θ=arctan(0.5)≈26.57°.
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)