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Jee Main Advanced Physics Chapter Wise Mock Test
Quiz 1
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Q.1
Consider the following two statement Statement A: If heat is added to the system,it temperature must increase Statement B: If dW >0 then dV >0 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:
Statement A is false: heat added to a system can just as easily do work or drive a phase change instead of raising temperature -- e.g. boiling water at constant temperature while absorbing heat, or an isothermal expansion.
Statement B is true: work done by a gas is dW=PdV, and since pressure P is always positive, dW and dV always share the same sign -- positive work directly implies positive volume change.
Q.2
A spring weighing machine inside a stationary lift read W N weight when a stone is kept on it.What will be the reading when the lift start moving downward with acceleration a
0%
$W(1- \frac {a}{g})$
0%
$W(1 + \frac {a}{g})$
0%
W
0%
none of these
Explanation
Solution:
With mg=W (the stationary reading), an elevator accelerating downward at a reduces the apparent weight to m(g-a) = W(1-a/g) -- a smaller reading than W, since the floor doesn't need to push up as hard when the whole system (scale, stone, and the floor under them) is accelerating downward together.
Q.3
Boyle's law is an example of a ?
0%
Latent Heat Process
0%
Isothermal Process
0%
Adiabatic Process
0%
Isochoric Process
Explanation
Solution:
Boyle's law (PV=constant) applies specifically to a gas at fixed temperature -- that condition is exactly what defines an isothermal process.
Q.4
A Alpha nucleus of energy $.5 mv^2$ bombards a heavy nuclear target of charge Ze. Then the distance of closest approach for the alpha nucleus will be proportional to
0%
$ \frac {1}{Ze}$
0%
$ \frac {1}{m}$
0%
$v^2$
0%
$ \frac {1}{v^4}$
Explanation
Solution:
At closest approach, all the kinetic energy ½mv² has converted into electrostatic PE: ½mv² = k(Ze)(2e)/d, giving d = 4kZe²/(mv²).
Of the four options, only "d ∝ 1/m" matches this correctly -- d actually grows with Ze (not shrinks, ruling out 1/(Ze)) and shrinks with v² (not grows, ruling out v² and matching 1/v² rather than 1/v⁴).
Q.5
A projectile has been fired with speed 10 m/s making an angle of 30° with the horizontal. Find the angle made by velocity vector of projectile with ground after 2 sec of its firing? Take g=10 m/s2
0%
45°
0%
60°
0%
Cant say anything
0%
None of the above
Explanation
Solution:
Time of flight for this launch: T = 2(u sinθ)/g = 2(10×sin30°)/10 = 2(5)/10 = 1 second.
Since the projectile lands after just 1 second, by t=2s it's no longer in flight at all -- the question describes a moment after the motion has already ended, so there's no meaningful velocity vector left to find an angle for.
Q.6
The ratio of the concentration of electrons to that of holes in a semiconductor is 7/5 and the ratio of currents is 7/Find the ratio of the drift velocities of electron and holes
0%
5/4
0%
4/5
0%
3/4
0%
4/7
Explanation
Solution:
Current for each carrier type is I ∝ (concentration)×(drift velocity), so I
n
/I
p
= (n/p)×(v
n
/v
p
).
7/4 = (7/5)×(v
n
/v
p
) ⇒ v
n
/v
p
= (7/4)÷(7/5) = 5/4.
Q.7
What is the time taken to heat 1 L of water from 10° C to 40° C by a heater of 836 W
0%
40 s
0%
80 s
0%
200 s
0%
150 s
Explanation
Solution:
Heat needed: Q=mcΔT = (1 kg)(4200 J/kg·K)(30 K) = 126000 J.
Time = Q/P = 126000/836 ≈ 150.7 s ≈ 150 s.
Q.8
The acceleration of a particle is increasing with time ,then the v-t curve would be like
0%
concave Up
0%
concave down
0%
straight line
0%
None of these
Explanation
Solution:
Acceleration is the slope of the v-t curve. If acceleration keeps growing over time, the curve's slope keeps getting steeper as time goes on -- exactly what "concave up" means (the same shape a curve like y=x² has, where the slope itself keeps increasing).
Q.9
A U tube containing a liquid is accelerate horizontally with constant acceleration a .The separation between the limb's is L. The Difference in the height of the liquid in the two arms would be
0%
$L\sqrt { \frac {a}{g}}$
0%
$ \frac {aL}{g}$
0%
$\frac {L}{2}$
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$ \frac {a^2L}{g^2}$
Explanation
Solution:
In the tube's accelerating frame, the liquid feels an effective backward pseudo-force, tilting its free surface at an angle θ where tanθ=a/g -- the same tilt effect used to find the level difference over the horizontal separation L: Δh = L tanθ = aL/g.
Q.10
An elastic spring of un-stretched length L and force constant k is stretched by a small length a .It is further stretched by another small length b.The work done in first stretching and second stretching is W1 and WWhich of the following is true
0%
$W_1=\frac {Ka^2}{2} ,W_2=\frac {Kb^2}{2}$
0%
$W_1=\frac {Ka^2}{2} ,W_2=\frac {Kb(2a+b)}{2}$
0%
$W_1=\frac {Ka^2}{2} ,W_2=\frac {K(a^2+b^2)}{2}$
0%
$W_1=\frac {Ka^2}{2}, W_2= \frac {K(a+b)^2}{2}$
Explanation
Solution:
Energy stored in a stretched spring at extension x is ½kx².
W
1
(stretching from 0 to a) = ½ka².
W
2
(the ADDITIONAL work stretching from a to a+b) is the difference between the energy at (a+b) and at a:
W
2
= ½k(a+b)² - ½ka² = ½k[(a²+2ab+b²)-a²] = ½k(2ab+b²) = ½kb(2a+b).
Q.11
Which of these is false for electric charge?
0%
Electric charge is additive
0%
e ($1.6 \times 10^{-19}$ C) is the elementary charge
0%
charge can not be destroyed or created
0%
1 Coulomb of charge contain $6.25 \times 10^{17}$ electron
0%
Coulomb law of electric force between the charges follows Newton's third law of motion
0%
The net force on any charge equals the vector sum of the forces exerted on it by all the charges
Explanation
Solution:
Charge being additive, e=1.6×10⁻¹⁹C being the elementary charge, charge conservation, Coulomb force pairs obeying Newton's third law, and the superposition principle are all standard, correct facts.
But the electron count in 1 Coulomb is Q/e = 1÷(1.6×10⁻¹⁹) = 6.25×10¹⁸ electrons -- a factor of 10 more than the "6.25×10¹&sup7;" claimed here, making that the false statement.
Q.12
Electromotive force represents
0%
Work
0%
Momentum
0%
Power
0%
Energy per unit change
0%
Force
Explanation
Solution:
EMF is defined as the energy supplied to charge carriers per unit charge that passes through the source -- energy per unit charge, measured in volts (joules per coulomb).
Q.13
Which of the following is False?
0%
Kirchhoff's first law of electric circuit is law of conservation charge
0%
Kirchhoff's second law of electric circuit is law of conservation of energy
0%
Kirchhoff's second law is based on law of conservation of momentum
0%
None of these
Explanation
Solution:
Kirchhoff's current law is charge conservation at a junction -- true. Kirchhoff's voltage law is energy conservation around a loop -- true. It has nothing to do with momentum conservation, which is an unrelated mechanical principle -- claiming the voltage law comes from momentum conservation is the false statement.
Q.14
Both electric forces and magnetic forces can deflect electron. Which of the following statement is false?
0%
Kinetic energy of the electron does not change in magnetic field
0%
Kinetic energy of the electron do change in electric field
0%
Kinetic energy neither changes in electric field nor in magnetic field
0%
None of these
Explanation
Solution:
A magnetic force is always perpendicular to velocity, so it can never do work -- kinetic energy stays unchanged in a magnetic field, true. An electric force generally does have a component along the motion, so it can speed up or slow down a charge -- kinetic energy does change in an electric field, also true. Since KE genuinely does change in an electric field, claiming it changes in "neither" field is the false statement.
Q.15
A particle is moving vertically downward.it passes through a magnetic field directed from south to north in a horizontal plane.which of the following is false?
0%
If the particle is positively charged ,it will be deflected towards east
0%
If the particle is negatively charged ,it will be deflected towards east
0%
If the particle has no charge ,it will not be deflected
0%
None of these
Explanation
Solution:
Taking east/north/up as x/y/z axes, the particle moves along -z (downward) through a field along +y (south to north). F=qv×B works out to point along +x (east) for a positive charge, and along -x (west) for a negative one -- so a negative charge here would actually be deflected toward the WEST, not the east. The claim that it deflects east is the false one; "no charge, no deflection" is correctly true, since F=qv×B is automatically zero when q=0.
Q.16
A charged particle at rest experiences no forces. At the location of the particle
0%
electric field may or may not be zero, Magnetic Field must be zero
0%
electric field may or may not be zero, Magnetic Field may or may not be zero
0%
electric field must be zero,Magnetic Field must be zero
0%
electric field must be zero, Magnetic Field may or may not be zero
Explanation
Solution:
Total force is F=qE+qv×B. Since the particle is at rest, v=0, so the magnetic term qv×B is automatically zero no matter what B is -- B is completely unconstrained by the "no force" condition.
For the remaining term qE to vanish (with q≠0), E itself must be zero -- so E must be zero, while B may or may not be.
Q.17
A cross-section of the long cylindrical conductor of radius R1 containing a long cylindrical hole of RThe axes of the cylinder and hole are parallel and are at distance d apart. A current 1 Ampere is uniformly distributed over the cylindrical conductor. The magnetic field at the center of the hole is
0%
$ \frac {(\mu _0 id)}{(2\pi (R_1^2-R_2^2))}$
0%
Zero
0%
$ \frac {(\mu _0 id)}{(2\pi (R_1^2+R_2^2))}$
0%
$ \frac {(\mu _0 id)}{(4\pi (R_1^2-R_2^2))}$
Explanation
Solution:
Treat this as a full solid cylinder (radius R
1
, current density J) PLUS a smaller cylinder at the hole's location carrying the exact opposite current density -J, so the two together cancel to leave zero current in the hole -- exactly matching the real setup.
At the hole's centre, the full cylinder alone (evaluated at distance d from ITS axis) contributes B=μ
0
Jd/2, while the cancelling cylinder contributes exactly zero at its OWN centre (r=0 there). With J=i/(π(R
1
²-R
2
²)), this gives B = μ
0
id/(2π(R
1
²-R
2
²)).
Q.18
The electron emitted in beta radiation originates from
0%
Photon escaping from the nucleus
0%
Inner orbits of atoms
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Free electrons existing in nuclei
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decay of a neutron in a nucleus
Explanation
Solution:
Beta decay is fundamentally a neutron inside the nucleus converting into a proton, an electron, and an antineutrino (n → p + e⁻ + ν̄) -- the emitted electron is that newly created particle, not one that was already sitting somewhere in the atom.
Q.19
A Small solid sphere of radius r can roll without slipping on the inner surface of a fixed large spherical shell of Radius R . Find the period of small oscillation of small sphere around its equilibrium position?
0%
$2 \pi \sqrt {((2(R-r))/5g)}$
0%
$2\pi \sqrt {((7(R+r))/5g)}$
0%
$2\pi \sqrt {((7(R-r))/5g)}$
0%
2\pi \sqrt {((7(R-r))/2g)}
Explanation
Solution:
This is the standard "ball rolling inside a spherical bowl" result: the effective pendulum length is (R-r) (the sphere's centre traces a circle of that radius), and the (7/5) factor comes from including the small sphere's own rotational inertia (I=₂ₛ₅mr²) in the restoring dynamics, the same way it enters a rolling cylinder's acceleration down an incline: T=2π√[7(R-r)/(5g)].
Q.20
What happens the to the capacitance ?
0%
Capacitance increases
0%
Capacitance decreases
0%
Capacitance remains unchanged
0%
insufficient information
Explanation
Solution:
Capacitance C=ε
0
A/d falls as the plate separation d increases -- pulling the plates apart directly lowers the capacitance.
Q.21
The displacement of particle is as per equation $y=at^2+bt+c$ Find the acceleration at time t=0
0%
2a
0%
b
0%
c
0%
2a+b
Explanation
Solution:
Velocity: dy/dt = 2at+b. Acceleration: d²y/dt² = 2a -- a constant, the same at every instant including t=0.
Q.22
A particle moves such that $x=a sin \omega t$ and $y =b cos \omega t$ what is the trajectory of the particle?
0%
Eclipse
0%
circle
0%
straight line
0%
parabola
Explanation
Solution:
(x/a)²+(y/b)² = sin²ωt+cos²ωt = 1 -- the standard equation of an ellipse with semi-axes a and b (a circle only in the special case a=b).
Q.23
A line source emits a cylindrical wave.if the medium absorbs no energy,the amplitude will vary with distance y from the source as proportional to
0%
$y^{-1}$
0%
$y^{-2}$
0%
$\frac {1}{ \sqrt {y}}$
0%
$ \sqrt {y}$
Explanation
Solution:
For a cylindrical wave from a line source, the wave's energy spreads out over a cylindrical surface whose area grows as 2πyL -- proportional to y itself (not y², the way a point source's spherical wavefront grows). With no absorption, intensity I ∝ 1/y, and since I ∝ amplitude², the amplitude ∝ 1/√y.
Q.24
The displacement of a particle having wave motion given by $ y=cos^2(t/4) sin(50t)$ This expression may be considered to be a result of the superposition of how many wave motions
0%
One
0%
Two
0%
Three
0%
Four
Explanation
Solution:
Using cos²θ=(1+cos2θ)/2: cos²(t/4) = (1+cos(t/2))/2.
y = [(1+cos(t/2))/2]×sin(50t) = ½sin(50t) + ½cos(t/2)sin(50t).
Using the product-to-sum identity on the second term: cos(t/2)sin(50t) = ½[sin(50.5t)+sin(49.5t)].
So y = ½sin(50t) + ¼sin(50.5t) + ¼sin(49.5t) -- a superposition of three distinct sinusoids, at angular frequencies 50, 50.5 and 49.5.
Q.25
There are two statement about Ideal gases Statement A: The $V_{rms}$ of gas molecules depends on the mass of the gas molecule and the temperature Statement B : The $V_{rms}$ is same for all the gases at the same temperature which one of the following is correct
0%
A & B both are correct
0%
A is correct only
0%
B is correct only
0%
A &B both are wrong
Explanation
Solution:
Statement A is true: v
rms
=√(3RT/M) depends on both the gas's molar mass M and the temperature T.
Statement B is false precisely because of that M dependence -- at the same temperature, a light gas (like hydrogen) has a much higher v
rms
than a heavy one (like oxygen); they're not the same.
Q.26
A metallic sphere has a cavity of diameter D at its center.If the sphere is heated,the diameter of the cavity will
0%
Decrease
0%
Increase
0%
Remain unchanged
0%
None of the above
Explanation
Solution:
Thermal expansion scales every linear dimension of the metal uniformly, including the boundary of the cavity -- a cavity in a heated solid expands exactly as if that empty space were filled with the same material, so the hole grows right along with the rest of the sphere.
Q.27
The horizontal and vertical displacement of the projectile at time t are $x=36t$ $y=48t-4.9t^2$ where x and y are in meters and t in second. Initial velocity of the projectile in m/s
0%
15
0%
30
0%
45
0%
60
Explanation
Solution:
The x and y equations give the velocity components directly: v
x
=dx/dt=36 m/s (constant), and v
y
=dy/dt=48-9.8t, which at t=0 is 48 m/s.
Initial speed = √(36²+48²) = √(1296+2304) = √3600 = 60 m/s.
Q.28
A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m/s at an angle of 30° with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the ground? take g = 10 m/s2
0%
5.20 m
0%
4.33 m
0%
2.60 m
0%
8.66 m
Explanation
Solution:
Since the ball is thrown from the roof (10 m up) and we want when it's back at 10 m from the ground, that's exactly the point where it returns to its LAUNCH height -- which is what the standard range formula describes.
R = u²sin(2θ)/g = 10²×sin(60°)/10 = 10×(√3/2) = 5√3 ≈ 8.66 m.
Q.29
A transverse wave in a medium is described by the equation $y=A sin^2(\omega t-kx)$. The magnitude of the maximum velocity of particles in the medium is equal to that of the wave velocity.if the value of A is
0%
$ \frac {\lambda}{2 \pi}$
0%
$ \frac {\lambda}{4 \pi}$
0%
$ \frac {\lambda}{ \pi}$
0%
$ \frac { 2 \lambda}{ \pi}$
Explanation
Solution:
y=A sin²(ωt-kx). Differentiating: dy/dt = A×2sin(ωt-kx)cos(ωt-kx)×ω = Aω·sin(2(ωt-kx)), whose maximum magnitude is Aω.
The wave's own phase velocity is v=ω/k = fλ = ωλ/(2π).
Setting the two equal: Aω = ωλ/(2π), so A = λ/(2π).
Q.30
A charge q is placed at the center of the line joining two equal charges Q.The system of three charges will be in equilibrium if q equal is?
0%
-Q/2
0%
Q/2
0%
Q/4
0%
-Q/4
Explanation
Solution:
For the two outer charges Q (a distance 2d apart, with q sitting exactly at their midpoint) to each be in equilibrium, the repulsion each Q feels from the other Q, kQ²/(2d)², must be exactly balanced by an attraction from the central charge q, kQ|q|/d².
Setting these equal: |q|/d² = Q/(4d²), so |q|=Q/4 -- and since this force needs to be attractive (pulling each Q back toward the centre, opposing their mutual repulsion), q must carry the opposite sign to Q, giving q=-Q/4.
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