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Jee Main Advanced Physics Chapter Wise Mock Test
Quiz 2
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Q.1
A spring balance is attached to the ceiling of a lift. A man hangs his bag on the spring and the spring reads 49 N, when the lift is stationary. If the lift moves downward with an acceleration of 5 m/s2, the reading of the spring balance will be
0%
24 N
0%
74 N
0%
15 N
0%
49 N
Explanation
Solution:
Stationary reading: mg=49N, so m=49/9.8=5 kg.
Accelerating downward at a=5 m/s², the apparent weight (and hence the scale reading) is m(g-a) = 5×(9.8-5) = 5×4.8 = 24 N.
Q.2
A block of mass M is pulled along a horizontal friction surface by a rope of mass m. If a force P is applied at the free end of the rope, the force exerted by the rope on the block is
0%
$\frac {Pm}{m+M}$
0%
P
0%
$ \frac {PM}{m+M}$
0%
$ \frac {Pm}{M-m}$
Explanation
Solution:
Treating rope+block together (total mass m+M), the whole system shares one common acceleration: a = P/(M+m).
Looking at the block alone (mass M), the only horizontal force on it is the tension the rope exerts on it, T -- so by Newton's second law, T = Ma = PM/(M+m).
(As a check: for a massless rope, m→0, this correctly gives T=P, the whole applied force passing straight through undiminished.)
Q.3
find the maximum attainable temperature of ideal gas when gas undergoes through the process $P=P_0-aV^2$
0%
$ \frac {2P_0}{3R} \sqrt { \frac {P_0}{3a}}$
0%
$ \frac {2P_0}{R} \sqrt { \frac {P_0}{3a}}$
0%
$ \frac {P_0}{R} \sqrt { \frac {P_0}{a}}$
0%
$ \frac {P_0}{R} \sqrt { \frac {P_0}{2a}}$
Explanation
Solution:
With PV=RT (one mole), T = PV/R = (P
0
-aV²)V/R = (P
0
V-aV³)/R.
Maximising over V: dT/dV = (P
0
-3aV²)/R = 0 ⇒ V²=P
0
/(3a), i.e. V=√(P
0
/(3a)).
At this V, P
0
-aV² = P
0
-P
0
/3 = 2P
0
/3, so T
max
= V×(2P
0
/3)/R = (2P
0
/3R)√(P
0
/3a).
Q.4
Two masses $m_1 = 6$ kg and $m_2 = 14$ kg tied to a string are hanging over a light friction-less pulley. What is the acceleration of the masses when lift free to move?(g = 10 m/s2)
0%
4 m/s2
0%
2 m/s2
0%
3 m/s2
0%
1 m/s2
Explanation
Solution:
For an Atwood machine, a = (m
2
-m
1
)g/(m
1
+m
2
) = (14-6)×10/(14+6) = 80/20 = 4 m/s².
Q.5
which of these is not true for adiabatic process ?
0%
dQ=0
0%
dU+dW=0
0%
dT=0
0%
Molar specific heat is zero
Explanation
Solution:
By definition dQ=0 in an adiabatic process -- true. The first law then gives dU+dW=0 directly -- true. Since dQ=0, the molar specific heat (C=dQ/(ndT)) works out to zero too -- true.
But temperature is NOT required to stay constant in an adiabatic process -- quite the opposite, adiabatic compression heats a gas up and adiabatic expansion cools it down. A process where dT=0 is an isothermal one, a different process entirely.
Q.6
which of these is inertial frame of reference?
0%
A observer in car slowing down
0%
A observer in train speeding up
0%
A observer in car in uniform circular motion
0%
A observer in cycle moving with constant velocity.
Explanation
Solution:
An inertial frame is one that isn't accelerating. A car slowing down, a train speeding up, and a car going around a curve (which constantly changes direction, and so is always accelerating even at constant speed) are all accelerating frames. Only the cyclist moving at a genuinely constant velocity -- unchanging speed AND direction -- counts as inertial.
Q.7
Find the number of electrons in 5 Coulomb?
0%
$31.25 \times 10^{18}$
0%
$31.5 \times 10^{18}$
0%
$31 \times 10^{18}$
0%
$32.25 \times 10^{18}$
Explanation
Solution:
Number of electrons = total charge ÷ charge per electron = 5 ÷ (1.6×10⁻¹ⁿ) = 3.125×10¹ⁿ = 31.25×10¹⁸ electrons.
Q.8
Consider the following two statement Statement A: Work done in a closed loop by electrical force is zero Statement B:The force on one charge due to another charge is not affected by the presence of other charges near by. Which of the below is correct option?
0%
A & B both are correct
0%
A is correct only
0%
B is correct only
0%
A & B both are wrong
Explanation
Solution:
Both are standard properties of the electrostatic force: it's conservative, so work done around any closed loop is zero -- true. And by the superposition principle, the force between any one pair of charges depends only on that pair, unaffected by any other charges nearby (their effects simply add on top, without altering this particular pairwise force) -- also true.
Q.9
What is not true of uniform circular motion ?
0%
Centripetal force=$\frac {mv^2}{R}$
0%
angular velocity=$ \frac {v}{r}$
0%
angular acceleration >0
0%
Speed is constant
Explanation
Solution:
Centripetal force mv²/R, angular velocity ω=v/r, and constant speed are all standard, correct properties of uniform circular motion. But "uniform" means the angular velocity itself doesn't change -- so angular acceleration is exactly zero throughout, not greater than zero.
Q.10
An circular coil of radius R consists of N turns of wire.The current in the wire is I.Let B is the magnetic field at the center of the coil Which of the following is true
0%
$B= \frac {\mu _0NI}{R}$
0%
$B= \frac {\mu _0I}{2R}$
0%
$B= \frac {\mu _0NI}{4R}$
0%
none of these
Explanation
Solution:
The standard formula for the field at the centre of a circular coil of N turns, radius R, carrying current I is B=μ
0
NI/(2R). None of the three specific formulas offered matches that (each is missing or has a wrong factor), so by elimination, none of them is correct.
Q.11
A hollow metal sphere of radius 10 cm is charged such that potential on its surface is 10 V Four statement are stated STATEMENT -1: The Potential at the center is 10 V STATEMENT -2: All the charged reside at the outer surface STATEMENT -3: Potential at a point 5 cm away from the surface is 10 V STATEMENT -4: Potential at a point 5 cm from the center of the sphere is 10 V which one of the following is correct
0%
All the statement are correct
0%
Statement 1,2 and 4 are correct
0%
Statement 1,2 and 3 are correct
0%
Statement 2,3 and 4 are correct
Explanation
Solution:
Since it's a conductor, the electric field inside is zero, which means the potential is exactly the same everywhere inside, all the way to the centre -- true for both statement 1 (centre) and statement 4 (5cm from centre, still inside).
Being hollow, all the charge sits on the outer surface -- true.
But 5cm OUTSIDE the surface (15cm from the centre) is outside the sphere, where the potential behaves like that of a point charge and keeps falling off with distance -- it is NOT still 10V out there, making statement 3 false.
Q.12
A train sounds its whistle as it approaches and leaves a station.An man at the station measures a frequency of 200 Hz as the train approaches and a frequency of 180 Hz as the train leave.Find the speed of the train. Speed of the sound is given as 340 m/s
0%
17.89 m/s
0%
19.89 m/s
0%
16.91 m/s
0%
17 m/s
Explanation
Solution:
Doppler shift for a moving source: approaching, f
1
=f
0
v/(v-v
s
); leaving, f
2
=f
0
v/(v+v
s
).
Dividing: f
1
/f
2
= (v+v
s
)/(v-v
s
) ⇒ 200/180 = (340+v
s
)/(340-v
s
).
10(340-v
s
) = 9(340+v
s
) ⇒ 3400-10v
s
= 3060+9v
s
⇒ 340 = 19v
s
⇒ v
s
≈ 17.89 m/s.
Q.13
An electron(charge = $1.6 \times 10^{-19}$ C) moves with a speed $2 \times 10^6$ m/s along the positive x- direction in a magnetic Field B=(i-3j-4k) tesla. Force experienced by the electron in Newton
0%
$1.6 \times 10^{-12}$ N
0%
$1.4 \times 10^{-12}$ N
0%
$1.6 \times 10^{-13}$ N
0%
$1.2 \times 10^{-12}$ N
Explanation
Solution:
F = q(v×B). With v=2×10⁶(1,0,0) and B=(1,-3,-4):
v×B = (0×(-4)-0×(-3), 2×10⁶×(-4)-0×1, 2×10⁶×(-3)-0×1) = (0, -8×10⁶, -6×10⁶).
|v×B| = √(8²+6²)×10⁶ = √100×10⁶ = 10⁷.
|F| = |q|×|v×B| = 1.6×10⁻¹ⁿ × 10⁷ = 1.6×10⁻¹² N.
Q.14
According to Kirchhoff's Ist law ,the algebraic sum of the currents flowing towards a branch point is
0%
greater then zero
0%
equal to zero
0%
greater or less than zero
0%
less than zero
Explanation
Solution:
Kirchhoff's current law is just charge conservation at a junction: whatever current flows in must equal whatever flows out, so the algebraic sum (counting outgoing as negative) always comes to exactly zero -- charge can't pile up or vanish at a point.
Q.15
What is true of energy stored in electric field?
0%
Energy remains unchanged
0%
Energy decrease by the factor i
0%
Energy increase by the factor i
0%
None of these
Explanation
Solution:
Energy stored is U=Q²/(2C). With Q fixed and C shrinking as the plates separate, U grows -- by the same factor the potential difference grows by, since U=½QV and Q is unchanged while V increases by that factor.
Q.16
If the change in the value of g at height H above the surface of the earth is the same as that at a depth D below it, then (both h and d are much smaller than the radius of the earth) the ratio $ \frac {H}{D} + \frac {D}{H}$ is
0%
1/2
0%
3/2
0%
2
0%
5/2
Explanation
Solution:
g at height H (H<<R): g
H
=g(1-2H/R), a drop of Δg
H
=2gH/R.
g at depth D: g
D
=g(1-D/R), a drop of Δg
D
=gD/R.
Setting these drops equal: 2gH/R = gD/R ⇒ D=2H, so H/D=1/2 on its own.
The question asks for H/D + D/H, though: with D=2H, that's 1/2 + 2 = 5/2.
Q.17
A body of 1 kg is acted by the following forces F1=2i+3j+4k F2=4i-3j-(24)1/2k F3=-2i+3j-4k Find the magnitude of the acceleration
0%
7
0%
4
0%
3
0%
10
Explanation
Solution:
Adding the three force vectors component-wise:
x: 2+4-2=4. y: 3-3+3=3. z: 4-√24-4=-√24.
|F
net
| = √(4²+3²+(√24)²) = √(16+9+24) = √49 = 7.
With mass 1 kg, acceleration = F
net
/m = 7 m/s².
Q.18
A particle is moving in the x-y plane with $x=at$ $y=bt^2$ where a and b are constant Find the trajectory of the particle?
0%
Parabola
0%
Circle
0%
Straight line
0%
none of these
Explanation
Solution:
From x=at, t=x/a. Substituting into y=bt²: y=b(x/a)² -- y proportional to x², the equation of a parabola.
Q.19
The radius of the Bohr's first orbit is awhat will be the angular momentum of the electron in the nth orbit ?
0%
$ \frac {nh}{2 \pi}$
0%
$n^2a_0$
0%
$\frac {h}{2n \pi}$
0%
$ \frac {2 \pi}{nh}$
Explanation
Solution:
Bohr's model quantizes orbital angular momentum directly: it's one of the theory's founding postulates that L=nh/(2π) for the nth orbit (n²a
0
is the RADIUS formula instead, a different quantity).
Q.20
A circular ring of Radius a with uniform positive charge density λ per unit length is located in the Y-Z plane with its center at the origin O. A particle of mass M0 and positive charge Q0 is projected from the point S (a√3,0,0) on the positive X axis directly towards O with initial speed vFind the smallest (non zero) value of the speed v0 such that the particle does not return to point S
0%
$\sqrt {\frac {Q_0}{2 \lambda M_0 \epsilon _0}}$
0%
$\sqrt {\frac {Q_0}{2 M_0 \epsilon _0}}$
0%
$\sqrt {\frac {Q_0}{2 a M_0 \epsilon _0}}$
0%
$\sqrt {\frac {Q_0 \lambda}{2 M_0 \epsilon _0}}$
Explanation
Solution:
The potential on the axis of a charged ring (radius a, total charge Q
ring
=2πaλ) at distance x from the centre is V(x)=Q
ring
/(4πε
0
√(x²+a²)) -- maximum at the centre (x=0) and falling off on either side, symmetric about the centre.
Since the particle is positive and repelled by the ring, it decelerates climbing from S (x=a√3) toward the centre; the critical speed is exactly enough to reach the centre with zero velocity -- any more and it sails through to the other side and never returns, any less and it slides back to S.
V(0)-V(a√3) = [Q
ring
/(4πε
0
a)] - [Q
ring
/(4πε
0
×2a)] = Q
ring
/(8πε
0
a) = λ/(4ε
0
) (substituting Q
ring
=2πaλ).
Energy conservation: ½M
0
v
0
² = Q
0
×λ/(4ε
0
), so v
0
= √(Q
0
λ/(2M
0
ε
0
)).
Q.21
A parallel plate capacitor contains one mica sheet of thickness $d_1=1.0 \times 10^{-3} m$ and one fibre sheet of thickness $d_2=.5 \times 10^{-3} m$ The dielectric constants of mica and fibre are 8 and 2.5 respectively . Fibre break down in an electric field of $ 6.4 \times 10^6$ V/m .What maximum voltage can be applied to the capacitor?
0%
6000 V
0%
5200 V
0%
5426 V
0%
None of these
Explanation
Solution:
In a series dielectric stack, the electric displacement D (and hence ε
0
KE, the product of dielectric constant and field) stays the same across the interface.
At fibre's breakdown field E
2
=6.4×10⁶ V/m: K
1
E
1
=K
2
E
2
⇒ E
1
= (2.5×6.4×10⁶)/8 = 2×10⁶ V/m.
Max voltage = E
1
d
1
+E
2
d
2
= (2×10⁶)(1.0×10⁻³) + (6.4×10⁶)(0.5×10⁻³) = 2000+3200 = 5200 V.
Q.22
A small coil of radius .002 m is placed on the axis of a magnet of magnetic moment $10^5 \; JT^{-1}$ and length .1 m at a distance of .15 m from the center of the magnet Find the net force on coil when a current of 2.0 A is passed through it?
0%
$4.4 \times 10^{-3}$ N
0%
$4.0 \times 10^{-2}$ N
0%
$5.4 \times 10^{-3}$ N
0%
$6.4 \times 10^{-3}$ N
Explanation
Solution:
Treating the bar magnet as two poles of strength q
m
=M/(2l) separated by its length (l=0.05 m half-length), the axial field at distance d=0.15 m from the centre is B(d)=(μ
0
/4π)q
m
[(d-l)⁻²-(d+l)⁻²], and its gradient works out to dB/dd ≈ -175 T/m at this distance.
The coil's own magnetic moment is m
coil
=Iπr²=2.0×π×(0.002)² ≈ 2.51×10⁻⁵ A·m².
Force on a small dipole in a field gradient: F=m
coil
×|dB/dd| ≈ 2.51×10⁻⁵×175 ≈ 4.4×10⁻³ N.
Q.23
A metal disk of radius s rotates with angular velocity ω about a vertical axis through a uniform field B pointing Up. A circuit is made by connecting one end of the resistor to the axle and other end to the the sliding constant which touches the outer edge of the disk.The resistance of the resistor is R. Find the current in the resistor?
0%
$\frac {\omega Bs^2}{2R}$
0%
$\frac {\omega Bs}{2R}$
0%
$\frac {\omega Bs^2}{2R^2}$
0%
None of these
Explanation
Solution:
This is a Faraday (homopolar) disc generator: each radial element of the spinning disc has a motional EMF dE = (v×B)·dr = Bωr dr, and integrating from the centre (r=0) to the rim (r=s) gives EMF = ∫Bωr dr = ½Bωs².
Current = EMF/R = ωBs²/(2R).
Q.24
A solid sphere of mass M and radius R rolls down an incline plane . which of the following statement is correct
0%
K.E is 2/5 rotational and 3/5 translational
0%
K.E is 2/9 rotational and 7/9 translational
0%
K.E is 1/7 rotational and 6/7 translational
0%
K.E is 2/7 rotational and 5/7 translational
Explanation
Solution:
For a rolling solid sphere, translational KE=½mv² and rotational KE=½Iω²=½(₂ₛ₅mr²)(v/r)²=₁ₛ₅mv², giving total KE=(½+₁ₛ₅)mv²=⁵ₛ₁₀mv².
Fraction rotational = (₁ₛ₅)/(⁵ₛ₁₀) = 2/7. Fraction translational = (½)/(⁵ₛ₁₀) = 5/7.
Q.25
The unit of farad-ohm is ?
0%
$sec^{-2}$
0%
$sec^{-1}$
0%
$sec^{2}$
0%
$sec$
Explanation
Solution:
Farad = coulomb/volt, and ohm = volt/ampere = volt·second/coulomb (since ampere = coulomb/second). Multiplying: (C/V)×(V·s/C) = s -- the coulombs and volts cancel, leaving plain seconds. (This matches the familiar fact that RC, resistance times capacitance, is always a time constant.)
Q.26
A bicycle generator creates 3.0 V when bicycle is travelling at a speed of 9.0 km/hr. How much emf is induced when the bicycle is travelling at 15 km/hr?
0%
3.2 V
0%
4 V
0%
5 V
0%
None of these
Explanation
Solution:
A simple generator's EMF scales directly with speed. Scaling from 9.0 km/hr to 15 km/hr: EMF
new
= 3.0V × (15/9) = 5.0 V.
Q.27
The mean lives of a radio-active substance are 1620 and 405 years for alpha and beta emission respectively. Find the time during which 3/4 of sample will decay if it is decaying both the alpha and beta emission simultaneously?
0%
449 years
0%
550 years
0%
446 years
0%
556 years
Explanation
Solution:
With two independent decay channels, the combined decay constant adds: λ
total
= 1/τ
α
+ 1/τ
β
= 1/1620 + 1/405 = 1/1620 + 4/1620 = 5/1620 = 1/324, giving an effective mean life of 324 years.
For 3/4 decayed (1/4 remaining): N/N
0
=1/4=e
-λt
⇒ t = ln(4)×324 ≈ 1.386×324 ≈ 449 years.
Q.28
The electric potential energy of an isolated metal sphere of radius R with total charge Q is
0%
$\frac {Q^2}{2 \pi \epsilon _0 R}$
0%
$\frac {Q^2}{3 \pi \epsilon _0 R}$
0%
$\frac {Q^2}{24 \pi \epsilon _0 R}$
0%
None of these
Explanation
Solution:
The self-energy of an isolated charged conducting sphere is U=½QV, where V=Q/(4πε
0
R) is its own surface potential:
U = ½Q×Q/(4πε
0
R) = Q²/(8πε
0
R) -- the standard, well-established result for a charged sphere's stored energy.
Q.29
Which of the following is best neutron moderator?
0%
Barium oxide
0%
graphite
0%
pure water
0%
Heavy water
Explanation
Solution:
A good moderator needs to slow down fast neutrons efficiently without absorbing too many of them. Heavy water (D
2
O) combines a very low neutron-capture rate with effective moderation, giving it the best moderating ratio of the options here -- which is exactly why reactors designed around heavy water (like CANDU reactors) can run on natural, unenriched uranium.
Q.30
n1 rows,each having n2 cells in series are connected in parallel. This battery is sending maximum current to 3 ohm resistor. The internal resistance of the each cell is .5 ohm then which of the following is true
0%
n1=2 and n2=12
0%
n1=12 and n2=2
0%
n1=4 and n2=6
0%
n1=6 and n2=4
Explanation
Solution:
For a mixed grouping of cells (n
1
rows of n
2
in series, all rows in parallel), current to an external resistance R is maximised when the battery's total internal resistance equals R:
(n
2
r)/n
1
= R ⇒ (n
2
×0.5)/n
1
= 3 ⇒ n
2
/n
1
= 6.
Of the given options, only n
1
=2, n
2
=12 satisfies that ratio (12/2=6).
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