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Jee Main Advanced Physics Chapter Wise Mock Test
Quiz 1
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Q.1
A wheel starts from rest and spins with a constant angular acceleration. As time goes on the acceleration vector for a point on the rim
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increases in magnitude and becomes more nearly radial
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decreases in magnitude and becomes more nearly radial
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increases in magnitude and becomes more nearly tangent to the rim
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increases in magnitude but retains the same angle with the tangent to the rim
Explanation
Solution:
A point on the rim has two acceleration components: a tangential part a
t
=rα, which stays fixed since α is constant, and a radial (centripetal) part a
r
=ω²r, which keeps growing as ω increases from rest.
Since a
r
grows without bound while a
t
stays the same, the total acceleration vector's magnitude keeps increasing, and it points increasingly toward the radial direction (the fixed tangential part becomes a smaller and smaller fraction of the total).
Q.2
The following questions consists of two statements , Assertion and Reason. While answering these questions choose any of the following four responses Statement I : Two cylinder , one hollow (metal) and the other solid(wood) with the same mass and identical dimension are simultansouly allowed to roll without slipping down on a inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first Statement II: By the principle of conservation of energy ,the total kinetic energies of both the cylinders are identical when reach the bottom of the incline
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If both statement I and reason are true but Statement II is not a correct explanation of statement I.
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If statement I is false and Statement II is true
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If statement I is true and Statement II is false.
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If both statement I and Statement II are true and the Statement II is correct explanation of statement I.
Explanation
Solution:
For a solid cylinder, I=½MR²; for a hollow one, I=MR² -- twice as much. A larger moment of inertia means more of the available energy goes into spinning rather than moving forward, so the SOLID cylinder (lower I) actually accelerates faster and reaches the bottom first -- the hollow one does not, making Statement I false.
Statement II is true regardless: since both start from the same height with the same mass, conservation of energy guarantees both convert the same total PE into the same total KE (translational + rotational combined) by the time they reach the bottom, however that KE happens to be split between the two forms.
Q.3
A uniform sold sphere rolls on the horizontal surface at 20 m/s.it then rolls up the incline of 30°.If friction losses are negligible what will be the value of h where sphere stops on the incline
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28.6 m
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30 m
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28 m
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none of these
Explanation
Solution:
A rolling solid sphere carries both translational and rotational kinetic energy: KE = ½mv² + ½Iω² = ½mv² + ½×(2/5)mr²×(v/r)² = ½mv² + (1/5)mv² = (7/10)mv².
Rolling without slipping and without friction losses, this entire energy converts to gravitational PE at the point where it momentarily stops: (7/10)mv² = mgh, so h = 7v²/(10g) = 7(20²)/(10×9.8) = 2800/98 ≈ 28.6 m.
Q.4
A mass is moving with constant velocity along a line parallel to x-axis away from origin.its angular momentum with respect to origin is
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remains constant
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goes on decreasing
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is zero
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goes on increasing
Explanation
Solution:
Angular momentum about the origin is L = m(r×v), whose magnitude is mv times the perpendicular distance from the origin to the line of motion. For straight-line motion at constant velocity, that perpendicular distance never changes (the line doesn't bend), and neither do m or v -- so L stays constant throughout, even though the mass keeps moving further from the origin along the line itself.
Q.5
An ice skater spins with arms outstretch at 1.9 rev/s.Her moment of inertia at this time is 1.33 kgm2.She pulls her arms to increase her rate of spin.Her moment of inertia after she pulls her arm is .48kgm2.What is her new rate of spinning
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4.7 rev/s
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5.26 rev/s
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4.0 rev/s
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5.2 rev/s
Explanation
Solution:
Conservation of angular momentum: I
1
ω
1
= I
2
ω
2
(using rev/s consistently on both sides, since it's a ratio).
ω
2
= I
1
ω
1
/I
2
= (1.33 × 1.9) / 0.48 ≈ 5.26 rev/s.
Q.6
A wheel initially has an angular velocity of 18 rad/s. It has a constant angular acceleration of 2.0 rad/s2 and is slowing at first. What time elapses before its angular velocity is 18 rad/s in the direction opposite to its initial angular velocity?
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10 sec
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3 sec
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18 sec
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6 sec
Explanation
Solution:
Taking the initial spin direction as positive, ω
0
=18 rad/s and the final state is 18 rad/s in the OPPOSITE direction, i.e. ω=-18 rad/s.
Using ω=ω
0
+αt with α=-2.0 rad/s²: -18 = 18 + (-2)t ⇒ -36 = -2t ⇒ t = 18 s.
Q.7
A mass is whirled in a circular path with constant angular velocity and its angular momentum is L.If the string is now halve keeping the angular velocity same then angular momentum is
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L
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2L
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L/2
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L/4
Explanation
Solution:
For a mass whirled in a circle, L=Iω=(mr²)ω. Halving the radius while keeping ω the same scales L by (½)²=¼, since L depends on the SQUARE of the radius: L
new
= m(r/2)²ω = ¼(mr²ω) = L/4.
Q.8
A cylinder of Mass M and radius R rolls down a incline plane of inclination $\theta$.Find the linear accleration of the cylinder
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$\frac {2gcos \theta}{3}$
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$gsin \theta$
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$\frac {3gcos \theta}{2}$
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$\frac {2gsin \theta}{3}$
Explanation
Solution:
For an object rolling without slipping down an incline, the linear acceleration is a = g sinθ / (1 + I/(MR²)).
For a solid cylinder, I=½MR², so I/(MR²)=½, giving a = g sinθ / (1+½) = g sinθ / (3/2) = (2/3)g sinθ.
Q.9
For a wheel spinning with constant angular acceleration on an axis through its center, the ratio of the speed of a point on the rim to the speed of a point halfway between the center and the rim is:
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1/2
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4
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1/4
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2
Explanation
Solution:
Every point on a rotating wheel shares the same instantaneous angular speed ω, and linear speed is v=ωr -- proportional to distance from the centre. A point on the rim is at radius R, and a point halfway in is at R/2, so their speed ratio is R ÷ (R/2) = 2, regardless of what ω or the angular acceleration happen to be at that instant.
Q.10
A hoop of radius r and mass m rotating with an angular velocity $\omega _0$ is placed on a rough horizontal surface . The initial velocity of the center of the hoop is zero. What will be the velocity of the center of the hoop when it ceases to slip?
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$r \omega _0$
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$\frac {r \omega _0}{2}$
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$\frac {r \omega _0}{3}$
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$\frac {r \omega _0}{4}$
Explanation
Solution:
Friction decelerates the spin and accelerates the centre until rolling without slipping begins (v=ωr). With f as the friction force and I=mr² for a hoop:
m(dv/dt)=f, so v(t)=(f/m)t. And I(dω/dt)=-fr, so ω(t)=ω
0
-(f/(mr))t.
Setting v=ωr at the moment slipping stops: (f/m)t = rω
0
- (f/m)t, so 2(f/m)t = rω
0
, giving v = (f/m)t = rω
0
/2.
Q.11
Two wheels are identical but wheel B is spinning with twice the angular speed of wheel A. The ratio of the magnitude of the radical acceleration of a point on the rim of B to the magnitude of the radial acceleration of a point on the rim of A is:
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1/2
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1/4
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2
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4
Explanation
Solution:
Radial (centripetal) acceleration is a
r
=ω²r. For identical wheels (same r), doubling ω multiplies a
r
by 2²=4.
Q.12
Moment of inertia of a uniform rod of length L and mass M about an axis passing through L/4 from one end and perpendicular to its length
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7ML2/48
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11ML2/48
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ML2/12
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7ML2/36
Explanation
Solution:
The moment of inertia through the rod's centre is ML²/12. The point L/4 from one end is L/2 - L/4 = L/4 away from the centre.
By the parallel axis theorem: I = ML²/12 + M(L/4)² = ML²/12 + ML²/16.
Over a common denominator of 48: 4ML²/48 + 3ML²/48 = 7ML²/48.
Q.13
Four Solid spheres each of diameter $\sqrt 5$ cm and mass .5 kg are placed with their centers at the corners of a square of side 4 cm. The moment of inertia of the system about the diagonal of the sphere is $N \times 10^{-4} \ kg-m^2$, then N is
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9
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8
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4
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5
Explanation
Solution:
Set the square's corners at (0,0), (4,0), (4,4), (0,4) cm, with the diagonal axis running through (0,0) and (4,4). The two spheres at those corners sit exactly ON the axis; the other two, at (4,0) and (0,4), sit a perpendicular distance d=4/√2=2√2 cm from it.
Each sphere's own moment of inertia about a diameter through its own centre is (2/5)mr², with r=√5/2 cm. Using the parallel axis theorem for the two off-axis spheres:
I
total
= 4×(2/5)mr² + 2md² = 4(0.4)(0.5)(1.25×10⁻⁴) + 2(0.5)(8×10⁻⁴) = 1.0×10⁻⁴ + 8.0×10⁻⁴ = 9.0×10⁻⁴ kg·m², so N=9.
Q.14
A cylinder rolls up the incline plane reaches some height and then roll down without slipping through out this section.The direction of the frictional force acting on the cylinder are
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down the incline while ascending and up the incline while descending
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Up the incline while ascending and desending
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down the incline while ascending and desending
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Up the incline while ascending and down the incline while descending
Explanation
Solution:
For a cylinder rolling without slipping on an incline, friction has to supply exactly the torque needed to keep the spin rate matched to the translational speed at every instant. Working through the equations of motion for both the ascending (decelerating) and descending (accelerating) phases shows friction points UP the incline in both cases -- it's what keeps the rotation properly synced to the translation whichever way the cylinder is moving.
Q.15
One solid sphere X and another hollow sphere Y are of same mass and same outer radii. Their moment of inertia about their diameters are respectively Ix and Iy such that
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Ix > Iy
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Ix= Iy
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Ix/Iy=Dx/Dy Where Dx and Dy are their densities.
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Ix < Iy
Explanation
Solution:
For a solid sphere, I=(2/5)MR²; for a thin hollow shell of the same mass and radius, I=(2/3)MR² -- noticeably larger.
That's because in the hollow sphere all the mass sits out at the maximum radius R, as far from the axis as it can be, while in the solid sphere mass is spread all the way in toward the centre too -- closer to the axis on average, giving a smaller moment of inertia. So I
x
(solid) < I
y
(hollow).
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