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Jee Main Advanced Physics Chapter Wise Mock Test
Quiz 1
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Q.1
What will be the equivalent resistance between A and D?
0%
40 ohm
0%
30 ohm
0%
20 ohm
0%
10 ohm
Explanation
Solution:
With B and C left unconnected (open, dead-end branches), no current can flow through the resistors leading to them, so those two branches drop out of the calculation entirely, leaving a ladder network between A and D.
Solving the remaining four-node network (labelling the two surviving top junctions and two bottom junctions and writing current-balance equations at each) gives a total resistance of 3× the individual resistor value between A and D -- with each resistor at 10Ω, that works out to 30Ω.
Q.2
The radius vector describing the position of the particle A relative to origin r=t2i+(t-1)2j Find the rectangular components of the average velocity in the time interval between t and t+Δt
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(Δt+2t),(Δt+2t-2)
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(Δt+2t),(Δt+2t+2)
0%
(Δt-2t),(Δt+2t+2)
0%
(Δt+2t),(Δt-2t+2)
Explanation
Solution:
Average velocity component = (change in that coordinate)/Δt.
x: [(t+Δt)²-t²]/Δt = [2tΔt+Δt²]/Δt = 2t+Δt.
y: [(t+Δt-1)²-(t-1)²]/Δt, and with u=t-1 this is the same pattern: [2uΔt+Δt²]/Δt = 2u+Δt = 2(t-1)+Δt = 2t-2+Δt.
So the components are (Δt+2t, Δt+2t-2).
Q.3
A charged particle whose mass is $19.9 \times 10^{-27}$ kg and charge is $1.6 \times 10^{-19}$ C moves with a speed of $3 \times 10^5$ m/s at right angle to a magnetic field of .75 T. What is the force acting on the charge, centripetal acceleration and radius of the circle in which charged particle moves
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$3.6 \times 10^{-14}$ N , $1.9 \times 10^{12} m/s^2$, $49.9 \times 10^{-3}$ m
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$3.0 \times 10^{-14}$ N,$1.81 \times 10^{12} m/s^2$, $49.7 \times 10^{-3}$ m
0%
$3.0 \times 10^{-14}$ N,$1.81 \times 10^{12} m/s^2$, $40.7 \times 10^{-3}$ m
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$3.6 \times 10^{-14}$ N,$1.81 \times 10^{12} m/s^2$, $49.7 \times 10^{-3}$ m
Explanation
Solution:
Force: F=qvB = (1.6×10⁻¹⁹)(3×10⁵)(0.75) = 3.6×10⁻¹⁴ N.
Acceleration: a=F/m = 3.6×10⁻¹⁴ ÷ (19.9×10⁻²⁷) ≈ 1.81×10¹² m/s².
Radius: r=mv/(qB) = (19.9×10⁻²⁷×3×10⁵) ÷ (1.6×10⁻¹⁹×0.75) ≈ 49.7×10⁻³ m.
Q.4
Three particles A,B,C each of mass m are placed at the corner of the equilateral triangle of side a. Find the potential energy of the system
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$ \frac {-3Gm^2}{a}$
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$ \frac {-3Gm^2}{2a}$
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$ \frac {-Gm^2}{a}$
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None of these
Explanation
Solution:
Three equal masses at the corners of an equilateral triangle give 3 pairwise interactions, each at separation a. Total PE = 3 × (-Gm²/a) = -3Gm²/a.
Q.5
Two concentric co-planar circular loops made of wire with resistance per unit length 10-4 O/m have diameters .2 m and 2 m respectively. A time varying potential difference V=(4+2.5 t) V is applied to larger loop. Find the induced current in the small loop
0%
1.25 A
0%
2 A
0%
1 A
0%
1.5 A
Explanation
Solution:
The current in the large loop is I
L
(t)=V(t)/R
L
, producing a field at the common centre B(t)=μ
0
I
L
(t)/(2R
L
) -- and since the small loop sits well within this roughly-uniform central field, the flux through it is B(t)×πR
S
².
Differentiating and dividing by the small loop's own resistance (2πR
S
×resistance-per-length) works out algebraically to I
small
= (2.5μ
0
R
S
) / (8πR
L
²ρ²).
Plugging in R
S
=0.1m, R
L
=1m, ρ=10⁻⁴Ω/m, μ
0
=4π×10⁻⁷ gives I
small
≈ 1.25 A.
Q.6
The displacement time equation for a particle in linear motion is given as $x= \frac {a}{b} ( 1 - e^{-bt})$
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The velocity and acceleration of the particle at t=0 is a and -ab respectively
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The velocity will be decreasing as the time increasescorrect
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The displacement of the particle will fall between 0≤x≤a/bcorrect
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The maximum acceleration in the motion is -abcorrect
Explanation
Solution:
v=dx/dt = a·e⁻ᵇᴱ. At t=0: v=a. Acceleration=dv/dt=-ab·e⁻ᵇᴱ. At t=0: a=-ab -- confirming the first statement.
Since e⁻ᵇᴱ decays monotonically, |v| and |acceleration| both shrink as t grows -- velocity keeps decreasing, and the acceleration's largest magnitude, -ab, occurs right at t=0. And since x(t)=(a/b)(1-e⁻ᵇᴱ) climbs smoothly from x=0 at t=0 up toward a/b as t→∞, the displacement always stays within [0, a/b]. All four statements here are actually correct.
Q.7
A nearly mass-less rod is pivoted at one end so it can freely swing as a pendulum.Two masses 2 m and m are attached to it at distance a and 3a respectively from the pivot end. The rod is held horizontal and then released. Which of the following is correct?
0%
The angular acceleration at the instant it is released $ \frac {5g}{11a}$
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Moment of inertia of the system about pivot is $11ma^2$correct
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The angular acceleration at the instant when the rod makes an angle $\theta$ with horizontal $\frac {5 g cos \theta }{11a}$correct
0%
The net torque about the pivot end becomes zero when rod becomes vertical.correct
Explanation
Solution:
Moment of inertia about the pivot: I = 2m(a)²+m(3a)² = 2ma²+9ma² = 11ma².
Torque when horizontal: both weights pull the same way, τ=2mg(a)+mg(3a)=5mga, so α=τ/I=5mga/(11ma²)=5g/(11a).
At angle θ from horizontal, the effective lever arm scales by cosθ, giving α=5g cosθ/(11a) -- which drops to exactly zero once the rod swings all the way to vertical (cos90°=0). All four of these follow from the same torque analysis and are correct.
Q.8
A particle is constraint to move along x-axis acted by a force whose potential energy is given by $U=2x^3-15x^2+36x+100$ Choose the correct option
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The force acting on the particle is $6x^2-30x+36$
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If it is slightly displaced from x=3,it will oscillate
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The minimum K.E required at origin to reach at x=1 will be 23 J
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The minimum K.E required at origin to reach at x=2 will be 28 Jcorrect
Explanation
Solution:
F=-dU/dx=-(6x²-30x+36).
Setting dU/dx=0 gives equilibria at x=2 and x=3. Checking curvature (d²U/dx²=12x-30): at x=2 it's negative (an unstable peak), at x=3 it's positive (a stable valley) -- so a small displacement from x=3 genuinely does lead to oscillation.
U(x)=2x³-15x²+36x+100 rises monotonically from x=0 up through x=1 and x=2 (no dip in between, since the only critical points are at 2 and 3), so the minimum launch KE needed is simply the potential difference to that target point:
U(0)=100, U(1)=123, U(2)=128 ⇒ KE to reach x=1 is 123-100=23 J, and KE to reach x=2 is 128-100=28 J.
Q.9
Blocks A and B of mass 3 Kg and 2 Kg respectively ,placed on the horizontal smooth surface.They are connected by a mass-less spring. Block A is given a kicks which imparts a velocity of 10 m/s to its towards B
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The velocity of the Center of mass of the system remains constant through out the motion and its value is 6 m/s
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The velocity of block B in Center of mass frame at t=0 is -6 m/s
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The velocity of block A in Center of mass frame at t=0 is 4 m/s
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Total energy of the system is 150 Jcorrect
Explanation
Solution:
V
cm
= (m
A
v
A
+m
B
v
B
)/(m
A
+m
B
) = (3×10+2×0)/5 = 6 m/s, and since no external force acts on the spring-linked pair, this stays constant for the whole motion.
In the centre-of-mass frame, each block's velocity is just its lab velocity minus v
cm
: block B starts at 0-6=-6 m/s, block A starts at 10-6=4 m/s.
Total energy at t=0 (before the spring has stretched at all) is just the kinetic energy: ½(3)(10²)+½(2)(0²) = 150 J.
Q.10
Match the columns
Column-I
Column-II
A) A particle moving with constant velocity on a horizontal plane
p) P=constant
B) Motions of object under central forces
q) L=constant
C) A object fired from a gun explodes in air
r) K.E=constant
D) A object in Uniform circular motion in horizontal plane
s) PE is constant
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A(p,r,s),B(q),C(p,q),D(q,r,s)
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A(p,q,r,s),B(q,r),C(p),D(q,r,s)
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A(p,q,r,s),B(p),C(p),D(q,r,s)
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A(p,q,r,s),B(q),C(p),D(q,r,s)
Explanation
Solution:
Constant velocity in a horizontal plane: momentum, kinetic energy, and PE (height unchanged) all stay fixed -- p,q,r,s all apply.
Motion under a central force conserves angular momentum (the force always points through the centre, giving zero torque about it) specifically -- only q applies.
An object exploding in mid-air: momentum is conserved (explosion is purely an internal force) -- only p applies.
Uniform circular motion in a horizontal plane: angular momentum, kinetic energy (constant speed), and PE (constant height) all stay fixed -- q,r,s apply.
Q.11
Match the column
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A(s),B(p),C(r),D(q)
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A(p),B(s),C(q),D(r)
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A(s),B(q),C(p),D(r)
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A(s),B(p),C(q),D(r)
Explanation
Solution:
Ballistic pendulum (A): momentum conservation gives the block+bullet's speed as (0.005×300)/2=0.75 m/s, and v²=2gh gives h=0.75²/(2×9.8)≈0.0287 m (A→s).
Projectile down a 20° slope (B): solving where the parabolic path meets the line y=-x tan20° gives a landing point about 34.5 m down the slope (B→p).
Capillary rise (C): h=2T/(ρgr) with r=3.5×10⁻⁴m gives h≈0.017 m (C→q).
Skidding truck (D): a=F/m=5 m/s², and v²=2as gives s=900/10=90 m (D→r).
Q.12
A particle of mass m = 6 kg is moving with a uniform speed v = 3 m/sec in the XOY plane along the straight line Y = X - 4 . The magnitude of the angular momentum about origin is
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$36 \sqrt {2}$
0%
$18 \sqrt {2}$
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$64 \sqrt {2}$
0%
60
Explanation
Solution:
Angular momentum about the origin for a particle moving in a straight line is L=p×d, where d is the perpendicular distance from the origin to that line.
For y=x-4 (i.e. x-y-4=0), that distance is |0-0-4|/√(1²+1²) = 4/√2 = 2√2.
p = mv = 6×3 = 18 kg·m/s.
L = 18×2√2 = 36√2.
Q.13
Five Charges of equal amount (Q) are placed at five corners of a regular hexagon of side 15 cm. What will be the value of sixth charge placed at sixth corner of the hexagon so that the electric field at the centre of hexagon is zero
0%
Q/2
0%
Q/6
0%
Q
0%
none of these
Explanation
Solution:
If all SIX corners carried charge Q, the field at the centre would be exactly zero by the hexagon's perfect symmetry. The actual 5-charge setup is that same fully-symmetric configuration with the sixth charge removed -- so the field from the 5 charges alone must equal the field a lone -Q would produce at that sixth position.
To cancel that, the charge actually placed at the sixth corner needs to produce the opposite field, i.e. behave like +Q there -- so the sixth charge is simply Q.
Q.14
(N+n) number of identical resistors each of resistance R is combined to get the minimum and maximum resistances, what is the ratio of the maximum to minimum resistance
0%
$1:(N+n)^2$
0%
$(N+n)^2:1$
0%
$1:(N+n)$
0%
$(N+n):1$
Explanation
Solution:
Maximum resistance comes from putting all N+n resistors in series: (N+n)R. Minimum comes from putting them all in parallel: R/(N+n).
Ratio = (N+n)R ÷ [R/(N+n)] = (N+n)².
Q.15
A mass-less and in-extensible cord is wound round the circumference of a circular ring of mass M and radius R. The ring is free to rotate about an axis passing through its centre and perpendicular to its plane. A mass m is attached at the free end of the cord and is at rest. The angular speed of the ring when mass m has fallen through at height h is
0%
$ \sqrt {\frac {2gh}{R^2}}$
0%
$ \sqrt {\frac {2mgh}{MR^2}}$
0%
$ \sqrt {\frac {2mgh}{(M + 2m)R^2}}$
0%
$ \sqrt {\frac {2mgh}{(M + m)R^2}}$
Explanation
Solution:
Energy conservation as the mass falls height h: mgh = (rotational KE of ring) + (KE of falling mass).
Ring (I=MR²): KE=½MR²ω². Falling mass (v=ωR, matching the unwinding cord): KE=½m(ωR)²=½mR²ω².
mgh = ½R²ω²(M+m) ⇒ ω = √[2mgh/((M+m)R²)].
Q.16
A concrete sphere of radius R has a cavity of radius r which is packed with sawdust. The relative densities of concrete and sawdust are 2.4 and 0.3 respectively. For this sphere to float with its entire volume submerged under water, the ratio of the mass of concrete to the mass of sawdust will be
0%
4
0%
8
0%
3
0%
None of these
Explanation
Solution:
For the whole sphere to float fully submerged, its average density must equal water's (relative density 1):
[2.4(R³-r³) + 0.3r³] / R³ = 1
2.4R³ - 2.1r³ = R³ ⇒ 1.4R³ = 2.1r³ ⇒ R³/r³ = 1.5.
Mass ratio = [2.4(R³-r³)] / [0.3r³] = 2.4(1.5-1)/0.3 = 2.4×0.5/0.3 = 4.
Q.17
what is true for a elastic collision?
0%
Both Linear momentum and Energy are conserved
0%
Linear momentum is conserved but Energy is not conserved
0%
Energy is conserved but Linear momentum is not conserved
0%
Both Linear momentum and Energy are not conserved
Explanation
Solution:
By definition, an elastic collision is exactly the kind where both momentum and kinetic energy come out unchanged before and after.
Q.18
A gas behaves more closely as an ideal gas at
0%
low pressure and low temperature
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low pressure and high temperature
0%
high pressure & low temperature
0%
high pressure & high temperature
Explanation
Solution:
Real gases deviate from ideal behaviour because of intermolecular forces and the molecules' own finite size. Low pressure spreads molecules far enough apart that both effects become negligible, and high temperature gives molecules enough kinetic energy that intermolecular attractions barely matter -- together, these are exactly the conditions where real gases behave most like ideal ones.
Q.19
Two statements are made STATEMENT-I: Two bodies of mass m and M will reach ground in same time when fallen at same time from the tower of height h STATEMENT-II: Same Acceleration due to gravity is experienced by the both body in fall from tower
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Statement I is true ,statement II is true ,statement II is correct explanation for statement I
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Statement I is true ,statement II is true ,statement II is not a correct explanation for statement I
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Statement I is true,Statement II is false
0%
Statement I is False,Statement II is True
Explanation
Solution:
In free fall, every object experiences the exact same acceleration g, regardless of its mass -- heavier objects have more weight pulling them down but proportionally more inertia resisting that pull, and the two effects cancel exactly.
Since fall time (from t=√(2h/g)) depends only on h and g -- not mass at all -- both bodies falling from the same height reach the ground together, and it's precisely because they share the same g that this happens.
Q.20
A solid cylinder of Mass M and Radius R rolls down an inclined plane of height H Calculate the Angular velocity of the cylinder when it reaches bottom of the plane
0%
$ \sqrt { \frac {4gH}{3R^2}}$
0%
$ \sqrt { \frac {2gH}{3R^2}}$
0%
$ \sqrt { \frac {gH}{3R^2}}$
0%
$ \sqrt { \frac {4gH}{R^2}}$
Explanation
Solution:
MgH = ¾Mv² (total KE for a rolling solid cylinder) ⇒ v²=4gH/3 ⇒ v=√(4gH/3).
ω=v/R = √(4gH/3)/R = √(4gH/(3R²)).
0 h : 0 m : 1 s
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