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 2
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Q.1
Find the linear velocity of the cylinder when it reaches at the bottom
0%
$ \sqrt { \frac {4gH}{3}}$
0%
$ \sqrt { \frac {2gH}{3}}$
0%
$ \sqrt { \frac {gH}{3}}$
0%
none of these
Explanation
Solution:
From MgH=¾Mv²: v²=4gH/3, so v=√(4gH/3).
Q.2
Rotational energy of the cylinder when it reaches the bottom of the plane
0%
MgH/4
0%
2MgH/3
0%
MgH
0%
MgH/3
Explanation
Solution:
For a rolling solid cylinder, rotational KE is exactly 1/3 of the total kinetic energy (translational carries the other 2/3), and the total equals the full PE lost, MgH -- so rotational energy at the bottom is MgH/3.
Q.3
Two metallic sphere A and B of Radii R1 and R2 are far apart.They are connected by a thin wire. Q1 is the charge on metallic sphere A Q2 is the charge on metallic sphere B And it is given Q1 +Q1 =Q Find the value of Q1
0%
$ \frac {R_1Q}{R_1+R_2}$
0%
$ \frac {R_2Q}{R_1+R_2}$
0%
$ \frac {R_1Q}{R_1-R_2}$
0%
None of these
Explanation
Solution:
Connected by a wire, the two spheres must share the same potential: kQ
1
/R
1
= kQ
2
/R
2
, so Q
1
/R
1
=Q
2
/R
2
, i.e. Q
1
=Q
2
(R
1
/R
2
).
Combined with Q
1
+Q
2
=Q: Q
2
(R
1
/R
2
+1)=Q ⇒ Q
2
=QR
2
/(R
1
+R
2
), and so Q
1
=Q-Q
2
=QR
1
/(R
1
+R
2
).
Q.4
Find the value of Q2
0%
$ \frac {R_1Q}{R_1+R_2}$
0%
$ \frac {R_2Q}{R_1+R_2}$
0%
$ \frac {R_2Q}{R_1-R_2}$
0%
none of these
Explanation
Solution:
Following the same shared-potential argument as Q
1
: Q
2
= QR
2
/(R
1
+R
2
).
Q.5
Find the potential of Metallic sphere A
0%
$ \frac {Q}{4 \pi \epsilon _0(R_1+R_2)}$
0%
$ \frac {Q}{4 \pi \epsilon _0R_1}$
0%
$ \frac {Q}{4 \pi \epsilon _0R_2}$
0%
None of these
Explanation
Solution:
Potential of A = kQ
1
/R
1
= k×[QR
1
/(R
1
+R
2
)]/R
1
= kQ/(R
1
+R
2
) = Q/(4πε
0
(R
1
+R
2
)).
Q.6
Find the potential of Metallic sphere B
0%
$ \frac {Q}{4 \pi \epsilon _0(R_1+R_2)}$
0%
$ \frac {Q}{4 \pi \epsilon _0R_2}$
0%
$ \frac {Q}{4 \pi \epsilon _0R_1}$
0%
None of these
Explanation
Solution:
Potential of B = kQ
2
/R
2
= k×[QR
2
/(R
1
+R
2
)]/R
2
= kQ/(R
1
+R
2
) = Q/(4πε
0
(R
1
+R
2
)) -- the same as sphere A's, exactly as it should be for two points joined by a conducting wire.
Q.7
A point charge +Q is at the origin of a coordinates system . It is surrounded by a concentric uniform distribution of charge on a spherical shell at r=R for which the total charge is -2Q Which of the following is correct?
0%
Flux crossing spherical surface for r < R is Q/ε0
0%
Flux crossing spherical surface for r > R is -Q/ε0correct
0%
Flux crossing spherical surface for r > R is -2Q/ε0
0%
Flux crossing spherical surface for r < R is -Q/ε0
Explanation
Solution:
For r<R, only the central +Q is enclosed: flux = Q/ε
0
.
For r>R, both +Q and the shell's -2Q are enclosed: total enclosed charge = Q + (-2Q) = -Q, so flux = -Q/ε
0
. Both of these hold (the r<R and r>R=-Q/ε
0
results are both correct; the -2Q/ε
0
and second "r<R" options don't match either enclosed-charge calculation).
Q.8
Let V, Vrms and Vp respectively denotes the mean speed, root mean square speed and most probable speed of the molecule in ideal mono-atomic gas at absolute temperature T. The mass of the molecule is m then
0%
No molecule can have a speed greater the (2)1/2Vrms
0%
No molecule can have a speed less the Vp / (2)1/2
0%
Vp < V < Vrms
0%
The average kinetic charge of a molecule (3/4 )mVp2correct
Explanation
Solution:
For the Maxwell-Boltzmann speed distribution, the three standard speeds order as V
p
< V < V
rms
(most probable is smallest, rms is largest, with the mean sitting in between) -- a well-known result of the distribution's shape.
Also true: average KE = (3/2)kT, and since V
p
²=2kT/m, (3/4)mV
p
² = (3/4)(2kT) = (3/2)kT -- exactly matching, so that statement holds too.
Q.9
A thick slab extending from z=-b to z=+b carries a uniform current density K and direction of current is positive x axis. Which of the following is true
0%
Magnetic field points in negative y axis for z > 0
0%
Magnetic field points in negative y axis for z < 0
0%
Magnetic field points in positive y axis for z > 0
0%
Magnetic field points in positive y axis for z < 0correct
Explanation
Solution:
For an infinite current sheet/slab carrying surface current in +x̂, the field on either side runs along ±ŷ (by the right-hand rule), pointing oppositely on the two sides. Outside a THICK slab, Ampere's law gives exactly the same field as an equivalent thin sheet would -- here, that means the field points in -ŷ on the z>0 side and +ŷ on the z<0 side, and both of those hold.
Q.10
Match the columns
Column (A)
( it tells us about the location)
Column (B)
( it tells us about the gravitational force experienced)
a)R
1
< d < R
2
p) $ \frac {G ( M _1 + M_2 + M_3 ) m }{d^2}$
b) d < R
1
q) $ \frac {G ( M_1 + M_2 ) m}{ d^2}$
c) R
1
< R
2
< d < R
3
r) $ \frac {GM_1}{ d^2}$
d) R
1
< R
2
< R
3
< d
s) 0
0%
A(p),B(q),C(r),D(s)
0%
A(r),B(s),C(q),D(p)
0%
A(p),B(r),C(q),D(s)
0%
A(s),B(q),C(r),D(p)
Explanation
Solution:
Between R
1
and R
2
, only the innermost shell's mass M
1
is enclosed (outer shells contribute nothing by the shell theorem): force=GM
1
m/d² (a→r).
Inside R
1
, no mass at all is enclosed: force=0 (b→s).
Between R
2
and R
3
, M
1
+M
2
is enclosed: force=G(M
1
+M
2
)m/d² (c→q).
Beyond R
3
, all three masses are enclosed: force=G(M
1
+M
2
+M
3
)m/d² (d→p).
Q.11
Match the columns
Column (A)
Column (B)
a)which coil will produce maximum Magnetic induction at the center
p) A
b) which coil will produce minimum Magnetic induction at the center
q) B
c) Which coil has maximum magnetic moment per turn
r) C
d)Which coil has minimum magnetic moment per turn
s) F
0%
A(q),B(r),C(q),D(r)
0%
A(p),B(s),C(p),D(s)
0%
A(q),B(r),C(r),D(p)
0%
A(p),B(s),C(s),D(p)
Explanation
Solution:
A coil's field at its centre (B=μ
0
NI/(2R)) grows with more turns and current and shrinks with larger radius, while its magnetic moment per turn (m=IπR²) grows with the SQUARE of the radius -- so the coil with the smallest radius (for the same current) gives the strongest central field but the weakest moment per turn, and vice versa for the largest-radius coil. Matching each specific coil (A, B, C, F) to these extremes depends on their individual radius/turn values from the accompanying figure.
Q.12
In Millikan's oil drop experiment ,a charged oil drop of mass 3.2 × 10-14 kg is held stationary between two parallel plates 6 mm apart by applying a potential difference of 1200 V between them .How much excess electrons does the oil drop carry? Take g=10 m/s2
0%
7
0%
8
0%
9
0%
10
Explanation
Solution:
At equilibrium, the electric force balances gravity: qE=mg, where E=V/d.
E = 1200 ÷ 0.006 = 2×10⁵ V/m.
q = mg/E = (3.2×10⁻¹⁴×10) ÷ (2×10⁵) = 1.6×10⁻¹⁸ C.
Number of excess electrons = q/e = 1.6×10⁻¹⁸ ÷ 1.6×10⁻¹⁹ = 10.
Q.13
A liquid is kept in a cylindrical vessel .When the vessel is rotated about its axis,the liquid rises at its sides.If the radius of the vessel is .05 m and speed of the rotation is 2 rev/sec,The difference in the heights of the liquid at the center and at the sides of the vessel will be (take g=10 m/s2 and π2=10
0%
2 cm
0%
8 cm
0%
4 cm
0%
1 cm
Explanation
Solution:
For a liquid spinning in a cylindrical vessel, the surface forms a paraboloid, and the height difference between the rim (radius r) and the very centre is h=ω²r²/(2g).
ω=2π(2 rev/s)=4π rad/s, so ω²=16π²=16×10=160 (using the given π²=10).
h = 160×(0.05)²/(2×10) = 160×0.0025/20 = 0.02 m = 2 cm.
Q.14
Two circular loops X and Y of radii Rx and Ry respectively are made from a uniform wire .The ratio of their moments of inertia about axis passing through their centers and perpendicular to their planes is IY : IX=8:1,Then RY : RX
0%
2
0%
8
0%
4
0%
6
Explanation
Solution:
For a loop made of uniform wire, its mass is proportional to its radius (mass=linear density × 2πR), and its moment of inertia about the central axis is I=mR² -- so I is proportional to R³ overall.
I
Y
/I
X
= (R
Y
/R
X
)³ = 8, so R
Y
/R
X
= 2.
Q.15
Two cylindrical rods of length L1 and L2,radii R1 and R2 have thermal conductivity K1 and K2 The ends of the rods are maintained at the same temperature difference .if L1 = 2L2 and R2 = 2R1, the rates of the heat flow will be same in the two rods provided K 1/ K2
0%
8
0%
2
0%
1
0%
4
Explanation
Solution:
Heat flow rate = KAΔT/L. Setting the two rods' rates equal (same ΔT): K
1
πR
1
²/L
1
= K
2
πR
2
²/L
2
.
With R
2
=2R
1
(so R
2
²=4R
1
²) and L
1
=2L
2
: K
1
/L
1
= 4K
2
/L
2
⇒ K
1
/K
2
= 4L
1
/L
2
= 4×2 = 8.
Q.16
The Potential Energy U in Joule of a particle of mass m=2 kg moving in the x-y plane is given by the formula $U=6x + 8y+5$ Here x and y are coordinates of the particle in meter The particle is at (4, 6) at t=0 and v0=0 i and j are the unit vectors across x and y axis respectively Find the acceleration of the particle
0%
9
0%
3
0%
4
0%
5
Explanation
Solution:
Force is the negative gradient of potential energy: F = -(∂U/∂x, ∂U/∂y) = -(6,8) N.
|F| = √(6²+8²) = √100 = 10 N.
a = F/m = 10/2 = 5 m/s².
Q.17
A ball is dropped from a height of 12 m. It loses 25% of its K.E. on striking the ground. Calculate the height to which it would bounce?
0%
6 m
0%
8 m
0%
9 m
0%
1 m
Explanation
Solution:
Energy at impact equals the full drop height's worth: mg(12).
Retaining 75% after the bounce: KE
after
= 0.75×12mg = 9mg, which converts entirely into height on the way back up: mgh
bounce
=9mg ⇒ h
bounce
=9 m.
Q.18
What is true for gravitational force
0%
work done between the two points does not depend on the path travel
0%
work done in a closed loop is zerocorrect
0%
W = - ΔUcorrect
0%
it is a conservative forcecorrect
Explanation
Solution:
Gravity is a conservative force, and that one fact is really what underlies all of these: work done between two points is the same regardless of path, work done around any closed loop is exactly zero, and the work-energy relation W=-ΔU holds. All four statements describe the same conservative-force property from different angles, and all are true.
Q.19
A object is moving in a plane with velocity vector given by u=u0i+(ωacosωt)j Where i and j are unit vectors along x and y axes respectively . It is given at t=0, r=0 Where r is radius vector from origin The trajectory of the particle is
0%
$y=xa sin ( \frac {\omega}{u_0})$
0%
$y=x sin ( \frac {\omega a}{u_0})$
0%
$y=a sin ( \frac {\omega x}{u_0})$
0%
$y=a cos ( \frac {\omega x}{u_0})$
Explanation
Solution:
x(t) = ∫u
0
dt = u
0
t.
y(t) = ∫ωa cos(ωt) dt = a sin(ωt) + C, and y=0 at t=0 gives C=0, so y(t)=a sin(ωt).
Eliminating t using t=x/u
0
: y = a sin(ωx/u
0
).
Q.20
Find the radius vector of the object at t=3π/2ω
0%
$-a\mathbf{i} + (\frac {2\pi u_0}{2 \omega})\mathbf{j}$
0%
$ (\frac {2\pi u_0}{2 \omega})\mathbf{j}$
0%
$a\mathbf{i} + (\frac {2\pi u_0}{2 \omega})\mathbf{j}$
0%
$a\mathbf{i} - (\frac {2\pi u_0}{2 \omega})\mathbf{j}$
Explanation
Solution:
At t=3π/(2ω): x=u
0
t = 3πu
0
/(2ω).
y=a sin(ωt) = a sin(3π/2) = -a.
So the radius vector is r = [3πu
0
/(2ω)]i - aj -- following directly from the same x(t) and y(t) used to derive the trajectory equation in the previous part.
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
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)