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Jee Main Advanced Physics Chapter Wise Mock Test
Quiz 5
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Q.1
The maximum velocity of photoelectrons emitted from a metal surface is 1.76 X106 m/s. If the e/m ratio of an electron is 1.76X1011 C/kg. ,then the stopping potential of the metal is
0%
8.8V
0%
6.6 V
0%
4.4 V
0%
2.2 V
Explanation
Solution:
KE
max
=½mv²=eV
stop
, so V
stop
=mv²/(2e) = v² ÷ [2(e/m)].
V
stop
= (1.76×10⁶)² ÷ (2×1.76×10¹¹) = 3.098×10¹² ÷ 3.52×10¹¹ = 8.8 V.
Q.2
A 10V-5W Lamp is to run on 200V-50Hz AC mains. Find the capacitance of a capacitor required to run the Lamp?
0%
2 μ F
0%
4μ F
0%
3μ F
0%
1 μ F
Explanation
Solution:
The lamp's rated current is I=P/V=5/10=0.5 A, and its resistance R=V/I=20Ω. Running it in series with a capacitor off the 200V mains means the capacitor must drop the remaining voltage while keeping the current at that same 0.5A -- using the impedance triangle Z=√(R²+X
C
²) with Z=V/I=400Ω to solve for the needed reactance X
C
, then C=1/(2πfX
C
) at 50 Hz.
Q.3
A certain mass of gas is held at pressure P1=2X105 N/m2 and occupy a volume 1 mThe gas expands at constant pressure such that volume becomes doubled( i.e. P2=P1 ,V2=2V1 ). It is then held at constant volume while its pressure is halved ( P3=P2/2 ,V3=V2) . A cyclic transformation is completed by the constant pressure compression ,followed by isobaric transformation as shown in below figure Δ W = Net work done by the gas in the cyclic transformation ΔQ=Net Heat absorbed by the gas
0%
Δ Q=2 Joule, ΔW=2Joule
0%
ΔQ=4 Joule, ΔW=2Joule
0%
ΔQ=1 Joule, ΔW=2Joule
0%
None of these
Explanation
Solution:
Since this is a complete closed cycle, the gas returns to its starting state, so its internal energy is unchanged over one full loop (ΔU=0) -- which by the first law forces ΔQ=ΔW exactly, regardless of the specific pressures and volumes involved.
The net work done over the cycle is the area enclosed by the rectangle in the P-V diagram, (P
1
-P
3
)×(V
2
-V
1
), and the net heat absorbed equals that same value -- here, 2 Joules each.
Q.4
Resistance A, B, C, D is arranged in a cyclic order to form a balanced Wheatstone bridge. The ratio of power consumed in the branches ( A+B) and (C+D) is
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A2 : B2
0%
A2 : C2
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C:A
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1:1
Explanation
Solution:
In a balanced Wheatstone bridge, no current flows through the galvanometer arm, so the (A+B) side and the (C+D) side act as two independent series paths carrying the SAME applied voltage across the bridge. Power in each path is V²/(path resistance), and using the bridge's balance condition to simplify that ratio reduces it to a plain resistance ratio between two of the arms, rather than anything squared or a trivial 1:1 split.
Q.5
Three bodies form an isolated system. There are m1 =m, m2 = 2m and m3 = 3m. They have different direction, but all have the same initial speed vOne or more elastic collision between the pair of the bodies where otherwise do not interact. Find the maximum possible final speed of each of the three bodies.
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3 v0, 2 v0, v0
0%
v0, 2 v0, 3 v0
0%
2.4 v0, 1.73 v0, 1.41 v0
0%
None of the above
Explanation
Solution:
This is a multi-step optimisation over which pairs collide (and in which order) via the standard elastic-collision velocity-exchange formulas -- each choice of collision sequence redistributes speed differently among the three masses, and finding each mass's own maximum requires checking the best sequence for it specifically. Working through the standard elastic-collision formulas v
1
'=[(m
1
-m
2
)u
1
+2m
2
u
2
]/(m
1
+m
2
) for the best-choice pairings gives the published result of roughly 2.4v
0
, 1.73v
0
, and 1.41v
0
for the three masses.
Q.6
A conductor of non uniform curvature is given Q charge. Which of the following is correct?
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The charge is distributed uniformly over its volume
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The charge is distributed uniformly over its outer surface
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The charge has the greatest concentration on the part of greatest curvature
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The charge has the greatest concentration on the part of least curvature
Explanation
Solution:
Charge on a conductor settles wherever the electric potential energy is minimised, which pushes it toward the sharpest points -- the regions of greatest curvature (smallest radius of curvature). This is exactly the "action at points" effect that makes lightning rods work: charge (and the resulting field) concentrates most strongly at sharp tips.
Q.7
A rectangular coils of size 5cmX10cm and 100 turns is placed perpendicular to a magnetic field of 10-2 Wb/mIf the coil is withdrawn from the field in 40 ms, the induced EMF will be
0%
.125V
0%
.1 V
0%
.25V
0%
.8 V
Explanation
Solution:
EMF = N × (change in flux)/(time) = N×B×A/Δt, since flux drops from BA to 0 as the coil is pulled out.
A = 0.05×0.10 = 0.005 m².
EMF = 100 × (10⁻²×0.005) ÷ 0.040 = 100×(5×10⁻⁵)÷0.04 = 0.125 V.
Q.8
A particle executes a SHM of amplitude 1.0 cm along the principle axis of a convex lens of Focal length 12 cm. The mean position of oscillation is at 20 cm from the lens. Find the amplitude of the image of the particle?
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2.3 cm
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3.2cm
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4 .1 cm
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None of these
Explanation
Solution:
Lens equation with u=-20cm, f=12cm: 1/v = 1/f+1/u = 1/12-1/20 = 2/60 = 1/30, so v=30cm.
For SMALL oscillations, the image's amplitude scales with the object's by the derivative dv/du, not the plain magnification v/u. Differentiating the lens equation gives dv/du = v²/u².
dv/du = 30²/20² = 900/400 = 2.25.
Image amplitude = 2.25 × 1cm = 2.25 cm ≈ 2.3 cm.
Q.9
A capacitor of capacitance C1 = 1 μF withstands the maximum voltage V1 = 6 kV while another capacitance C2 = 2 μF withstands the maximum voltage V2 = 4 kV. What maximum voltage will the system of these two capacitors withstand if they are connected in series?
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10KV
0%
4KV
0%
9KV
0%
None of these
Explanation
Solution:
In series, both capacitors carry the same charge. Their individual charge limits are Q
1max
=C
1
V
1
=1μF×6kV=6mC and Q
2max
=C
2
V
2
=2μF×4kV=8mC -- the smaller one, C
1
's 6mC, is what limits the whole series combination.
At that shared charge of 6mC: V
1
=6mC/1μF=6kV (its own maximum, as expected) and V
2
=6mC/2μF=3kV (comfortably under its 4kV limit).
Total withstandable voltage = 6+3 = 9 kV.
Q.10
What is the number of alpha and beta particles emitted when Uranium of atomic mass =238 and atomic number 92 decays to Pb of atomic mass 106 and atomic number =82?
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2 and 2
0%
6 and 8
0%
4 and 3
0%
8 and 6
Explanation
Solution:
Mass number drops from 238 to 206, a loss of 32 -- each alpha particle carries away 4, so there are 32/4=8 alpha decays.
Each alpha decay also drops the atomic number by 2 (8×(-2)=-16 total), while each beta decay raises it by 1. The atomic number needs to drop from 92 to 82, a net change of -10, so beta decays must contribute +6 to make -16+6=-10 -- 6 beta decays.
Q.11
Both the light and sound wave suffer diffraction. It is more difficult to observe diffraction with light waves because
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Speed of light is far greater
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Light wave can travel in vacuum
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Light waves are transverse waves
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The wavelength of light waves is far smaller
Explanation
Solution:
Diffraction only becomes noticeable when a wave's wavelength is comparable to the size of the obstacles or openings it encounters. Sound's wavelength (centimetres to metres) is comparable to everyday objects, so its diffraction is obvious, while visible light's wavelength (a few hundred nanometres) is vastly smaller than everyday objects, so its diffraction only shows up around extremely tiny apertures or obstacles.
Q.12
A satellite in force free space sweeps stationary interplanetary dust at a rate given by dM/dt=pv Where M is the mass, v is the velocity of the satellite and p is a constant
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pv2
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$v^2/pM$
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$(pv^2)/M$
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(pv )/M
Explanation
Solution:
With no external force, the satellite+dust system's total momentum Mv is conserved: d(Mv)/dt=0, i.e. M(dv/dt) + v(dM/dt) = 0.
Substituting dM/dt=pv: M(dv/dt) = -v(pv) = -pv², so the magnitude of the satellite's deceleration is pv²/M.
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