Class 11 Physics, Chapter 6 (NCERT). 15 questions by concept, 3 with pictures. Open “Show answer” for the answer and why.
How to teach this chapter · Common misconceptions & worked visual for this chapter
A wheel rolls without slipping on a stationary road. Its centre moves right at 3 m/s. What is the ground-frame velocity of the material point instantaneously touching the road?
(a) 0 m/s
At the bottom, rotational velocity relative to the centre is leftward with magnitude equal to centre speed. It cancels the rightward translation at that instant.
For two masses in regions with different gravitational acceleration, must the centre of gravity coincide with the centre of mass?
(c) No; unequal gravitational weighting can separate the two centres.
Centre of mass weights positions by mass alone. The gravitational moment weights masses by the local gravitational field; uniform gravity guarantees coincidence, but a varying field need not.
The shown uniform disc has the indicated circular hole removed. With the original centre at x=0 and positive x to the right, where is the remaining centre of mass?
(c) 2 cm
The removed area is one quarter of the original disc, so its mass is M/4. Subtract its first moment: X=−(M/4)(signed R/2)/(3M/4)=−signed R/6.
Where is the centre of mass of a hollow spherical shell with uniform mass density?
(a) At the geometric centre, even though that point contains no material.
Symmetrically placed equal mass elements have cancelling first moments about the geometric centre. The centre of mass is a weighted coordinate, not necessarily a material particle.
A system experiences a constant net external force of 6 N along x for 3 s. What is the change in total x momentum?
(a) 18 kg m/s
Total momentum changes by the external impulse. For constant force, ΔPx=Fx Δt, retaining the force sign.
A body rotates about the z axis with angular speed 3 rad/s. A particle is at (6,0,8) m. What is its linear speed?
(c) 18 m/s
The perpendicular distance from the z axis is |x| because y=0. The coordinate along the axis does not enter omega cross r.
For a right-handed set of mutually perpendicular unit vectors, what is k × j?
(c) −i
Reverse the order of the corresponding cyclic unit-vector product. Antisymmetry changes the sign of the vector.
A particle has position (-3,2,0) m and momentum (1,4,0) kg m/s. What is its z component of angular momentum about the origin?
(b) -14 kg m²/s
Use the z component of r cross p: Lz=x py−y px. The scalar product and reversed component order describe different operations.
A stationary car has mass 1200 kg and axle separation 2.0 m. Its centre of gravity is 1.2 m behind the front axle. Each axle has two equally loaded wheels. With g=9.8 m/s², what is the load on each front wheel?
(a) 2352 N
Take moments about the rear axle: total front reaction times axle spacing equals Mg times the CG distance ahead of the rear axle. Divide that front-axle reaction by two for the two front wheels; an off-centre CG does not give Mg/4 at every wheel.
The two equal opposite forces shown act on one rigid body. What is their total torque magnitude about any origin in the plane?
(c) 60 N m
The forces cancel as vectors but their separated lines produce moments in the same rotational sense. A couple has magnitude F times the full perpendicular separation d.
The force and distances in the diagram are given. What is the magnitude of the force’s torque about O?
(b) 48 N m
The moment arm is the perpendicular distance from O to the force line of action. For this vertical force it is the horizontal distance 4 m, not OP or the vertical coordinate.
A uniform solid cylinder of mass 20 kg and radius 0.25 m rotates about its symmetry axis at 100 rad/s. Which pair gives rotational energy and axial angular momentum?
(b) 3125 J; 62.5 kg m²/s
For a uniform solid cylinder I=MR²/2, here 0.625 kg m². Use K=I omega²/2 and L=I omega; substituting the thin-ring I doubles both results.
A wheel has angular velocity omega=6t² rad/s for 0≤t≤3 s, with t in seconds. What angle does it turn through during this interval?
(c) 54 rad
Angular displacement is the integral of angular velocity. Integrating b t² gives b T³/3; the endpoint mean is invalid because angular acceleration is not constant.
A rotor with unchanged moment of inertia initially has rotational energy 20 J. Its angular speed is multiplied by 2. What is its new rotational energy?
(a) 80 J
For fixed I, Krot=I omega²/2. The energy scales by the square of the angular-speed factor.
A rotating person reduces moment of inertia to 1/2 of its initial value with negligible external torque. What is final rotational kinetic energy divided by initial rotational kinetic energy?
(c) 2
At fixed angular momentum, K=L²/(2I). Reducing I raises the rotational kinetic energy inversely; internal work supplies the increase.
Wrong
The same push turns the door equally well anywhere.
Why it feels right. I focus on how hard I push.
Right
Turning effect also depends on the perpendicular lever arm.
Worked example
