JEE Physics System Of Particles and Rotational Motion MCQs Set 22

Answer: (a) 110

Explanation:
1. Parallel axis theorem: \( I=\frac25mR^2+md^2=m\left(\frac25\times25+100\right)=110\,m \) (in cm²).
2. \( I=mk^2\Rightarrow k^2=110 \), so x = 110.

Teacher's Note:
1. Use I about the centre of mass plus m d².
2. Radius of gyration k is defined by I = m k².

Answer: (a) 32

Explanation:
1. Mass is proportional to volume: \( \frac{M'}{M}=\frac{(R/2)^2(L/2)}{R^2L}=\frac18 \).
2. Moments of inertia: \( I_1=\frac12MR^2,\ I_2=\frac12\cdot\frac M8\cdot\frac{R^2}4=\frac{MR^2}{64} \).
3. \( \frac{I_1}{I_2}=\frac{1/2}{1/64}=32 \).

Teacher's Note:
1. The carved-out cylinder has one eighth of the mass.
2. The radius appears squared in I, which gives another factor of 4.

Answer: (a) 17

Explanation:
1. \( I_{CM}=\frac12MR^2 \).
2. Parallel axis theorem: \( I_{AB}=\frac12MR^2+M\left(\frac23R\right)^2=MR^2\left(\frac12+\frac49\right)=\frac{17}{18}MR^2 \).
3. \( \frac{I_{AB}}{I_{CM}}=\frac{17/18}{1/2}=\frac{17}9 \), so x = 17.

Teacher's Note:
1. Add M d² to the moment of inertia about the centre.
2. 1/2 + 4/9 = 17/18.

Answer: (a) 5

Explanation:
1. Disc about a tangent in its plane: \( I_d=\frac14M_dR^2+M_dR^2=\frac54\times4R^2=5R^2 \).
2. Sphere about a tangent: \( I_s=\frac25M_sR^2+M_sR^2=\frac75\times5R^2=7R^2 \).
3. \( \frac{I_d}{I_s}=\frac57 \), so x = 5.

Teacher's Note:
1. A diameter of the disc in its plane has I = MR²/4.
2. Use the parallel axis theorem to shift to the tangent.

Answer: (a) 40 m/s

Explanation:
1. Total kinetic energy of a rolling solid sphere \( =\frac12mv^2+\frac12\cdot\frac25mr^2\omega^2=\frac7{10}mv^2 \).
2. \( 2240=\frac7{10}\times2\times v^2\Rightarrow v^2=1600 \), so v = 40 m/s.

Teacher's Note:
1. Rolling without slipping adds a rotational part, 2/5 of the translational one, for a solid sphere.
2. The total is 7/5 of the translational kinetic energy.

Answer: (a) 800

Explanation:
1. Components: \( u_x=u_y=10\sqrt2 \). At t = 2 s: \( x=20\sqrt2,\ y=20\sqrt2-20,\ v_x=10\sqrt2,\ v_y=10\sqrt2-20 \).
2. \( L=m(xv_y-yv_x)=0.1\left[20\sqrt2(10\sqrt2-20)-(20\sqrt2-20)10\sqrt2\right]=0.1\times(-200\sqrt2) \).
3. \( |L|=20\sqrt2=\sqrt{800} \), so K = 800.

Teacher's Note:
1. Angular momentum equals the torque of gravity integrated over time: τ = m g x.
2. Also L = m u cosθ x (g t²/2) = 0.1 x 10√2 x 20.

Answer: (a) 3

Explanation:
1. Mass \( m=Ad\times2=2Ad \) and \( I=\frac{ml^2}{12}=\frac{2Ad\times4}{12}=\frac{2Ad}3 \).
2. \( E=\frac12I\omega^2\Rightarrow\omega=\sqrt{\frac{2E}{I}}=\sqrt{\frac{3E}{Ad}} \), so \( \alpha=3 \).

Teacher's Note:
1. Moment of inertia of a rod about its centre is ml²/12.
2. Mass = density x volume = d x A x length.

Answer: (a) 10 cm/s

Explanation:
1. \( KE=\frac7{10}mv^2\Rightarrow7\times10^{-3}=\frac7{10}\times1\times v^2 \).
2. \( v^2=10^{-2}\Rightarrow v=0.1\ \text{m/s}=10\ \text{cm/s} \).

Teacher's Note:
1. For a solid sphere rolling without slipping, KE = 7mv²/10.
2. Convert metres per second into centimetres per second at the end.

Answer: (a) 54 J

Explanation:
1. Friction decelerates the centre of mass: \( a=\mu g=3\ \text{m/s}^2 \), so after 2 s \( v=18-3\times2=12\ \text{m/s} \).
2. For pure rolling \( \omega=\frac vr \) (check: \( \alpha=\frac{\mu mgr}{\frac12mr^2}=\frac{6}{r} \), so \( \omega=\frac{12}r \) at t = 2 s, consistent).
3. \( KE=\frac12mv^2+\frac12\left(\frac12mr^2\right)\omega^2=\frac34mv^2=\frac34\times0.5\times144=54\ \text{J} \).

Teacher's Note:
1. Rolling disc: kinetic energy = (1/2)mv² + (1/4)mv² = (3/4)mv².
2. Kinetic friction acts until v = ωr.

Answer: (a) 5

Explanation:
1. Mass \( m=\rho\pi R^2t\Rightarrow R^2\propto\frac{1}{\rho t} \) for equal masses.
2. \( \frac{R_1^2}{R_2^2}=\frac{\rho_2t_2}{\rho_1t_1}=\frac{5\times0.5}{3\times1}=\frac56 \).
3. About a diameter \( I=\frac14mR^2 \), so \( \frac{I_1}{I_2}=\frac{R_1^2}{R_2^2}=\frac56 \) and x = 5.

Teacher's Note:
1. For equal masses the moment of inertia is proportional to R².
2. The area of a disc is πR², so R² is proportional to mass/(ρ t).

Answer: (a) 2 m/s

Explanation:
1. Height of the incline \( h=0.6\sin30^\circ=0.3\ \text{m} \).
2. Energy conservation: \( mgh=\frac12mv^2+\frac14mv^2=\frac34mv^2\Rightarrow v^2=\frac43gh=\frac43\times10\times0.3=4 \).
3. \( v=2\ \text{m/s} \).

Teacher's Note:
1. A rolling solid cylinder has kinetic energy 3mv²/4.
2. Alternatively a = (2/3) g sinθ = 10/3 m/s² and v² = 2as.

Answer: (a) 3

Explanation:
1. Perpendicular axis theorem: \( I_z=I_x+I_y=\frac{MR^2}4+\frac{MR^2}4=\frac{MR^2}2 \) about the centre.
2. Parallel axis theorem: \( I_{edge}=\frac12MR^2+MR^2=\frac32MR^2 \), so x = 3.

Teacher's Note:
1. First find I about the axis normal to the disc through the centre.
2. Then shift the axis by R to the edge.

Answer: (d) 360 m/s

Explanation:
1. Time of fall \( t=\sqrt{\frac{2\times20}{10}}=2\ \text{s} \) for both.
2. Speeds after the collision: ball \( \frac{30}{2}=15\ \text{m/s} \), bullet \( \frac{120}{2}=60\ \text{m/s} \).
3. Momentum conservation: \( 0.01\,u=0.2\times15+0.01\times60=3.6\Rightarrow u=360\ \text{m/s} \).

Teacher's Note:
1. The horizontal speeds after the collision follow from distance divided by the common time of fall.
2. Momentum is conserved horizontally.

Answer: (d) 0.6 m/s

Explanation:
1. Rate of firing \( =\frac{180}{60}=3 \) bullets per second, so momentum given to the bullets per second \( =3\times0.02\times100=6\ \text{kg m/s} \).
2. Taking this as the momentum balanced by the gun: \( 10\,v=6\Rightarrow v=0.6\ \text{m/s} \).

Teacher's Note:
1. By conservation of momentum the gun gains the momentum that the bullets carry away.
2. This treats the bullets fired in one second as a single group.

Answer: (b) Fig (b)

Explanation:
1. Impulse equals the area under the force-time graph.
2. \( (a)=\frac12\times1\times0.5=0.25,\ (b)=0.5\times2=1.0,\ (c)=\frac12\times1\times0.75=0.375,\ (d)=\frac12\times2\times0.5=0.5 \) N s.
3. The largest area is in figure (b).

Teacher's Note:
1. Impulse is the integral of F dt.
2. A rectangle has twice the area of a triangle with the same base and height.