JEE Physics Circular Motion MCQs Set 12

Answer: (d) 7 rpm

Explanation:
1. \( T=m\omega^2r\Rightarrow\omega\propto\sqrt T \).
2. \( \omega'=5\sqrt2\approx7.07 \) rpm.

Teacher's Note:
1. At fixed radius the angular speed varies as the square root of the tension.
2. Doubling the tension multiplies omega by root 2.

Answer: (a) 5 cm/s and 5 cm/s²

Explanation:
1. Angular velocity stays the same.
2. \( v=\omega r \) halves to 5 cm/s, and \( a=\omega^2r \) also halves to 5 cm/s².

Teacher's Note:
1. Both v and a are linear in r when omega is fixed.
2. Do not confuse this with a fixed-speed situation, where a would scale as 1/r.

Answer: (c) \( T_1>T_2 \)

Explanation:
1. The tension at a section provides the centripetal force for the part of the rod beyond it.
2. The part beyond L/4 is longer than the part beyond 3L/4, so it needs more force.

Teacher's Note:
1. Tension in a rotating rod decreases from the axis to the free end.
2. The direction of rotation does not matter since omega appears squared.

Answer: (d) \( \frac{u^2}{g\cos\theta} \)

Explanation:
1. Component of g normal to the velocity at launch: \( g\cos\theta \).
2. \( R=\frac{u^2}{g\cos\theta} \).

Teacher's Note:
1. Angle of the velocity with the horizontal is theta, so the normal component of gravity is g cos theta.
2. At the top this becomes u^2 cos^2(theta)/g.

Answer: (b) decreases

Explanation:
1. \( mg\cos\theta-N=\frac{mv^2}{R} \).
2. \( N=mg\cos\theta-\frac{mv^2}{R} \); as the car descends the angle from the vertical grows, so cos falls and N decreases.

Teacher's Note:
1. The radial component of weight is the only quantity that changes with position at constant speed.
2. At the top N is largest (mg - mv^2/R), and it drops as the slope steepens.

Answer: (d) \( T_1>T_2 \)

Explanation:
1. \( T=\frac{mu^2}{l}+mg(3\cos\theta-2) \).
2. Cosine decreases as the angle increases, so the tension is larger at 30 degrees.

Teacher's Note:
1. Tension in a vertical circle falls steadily as the body rises from the lowest point.
2. Compare cos 30 with cos 60.

Answer: (a) \( \text{not greater than }\frac{mv^2}{r} \)

Explanation:
1. At the top: \( N+mg=\frac{mv^2}{r} \) with \( N\ge0 \).
2. \( \text{so } mg\le\frac{mv^2}{r} \).

Teacher's Note:
1. The bucket pushes the water down with N (a non-negative force), adding to its weight.
2. So the required centripetal force is at least equal to the weight.

Answer: (c) \( \sqrt{\frac{g}{R}} \) rad/s

Explanation:
1. \( mg-N=m\omega^2R \) and N = 0 give \( \omega=\sqrt{g/R} \).

Teacher's Note:
1. Zero apparent weight means gravity just provides the centripetal force.
2. This gives a day of about 84 minutes.

Answer: (d) 0.6 m/s

Explanation:
1. Position along the wall \( x=d\tan\theta \).
2. \( v=\frac{dx}{dt}=d\sec^2\theta\,\omega=3\times2\times0.1=0.6 \) m/s.

Teacher's Note:
1. Differentiate the position of the spot with respect to time using the beam angle.
2. sec squared 45 degrees equals 2.

Answer: (a) 1 m

Explanation:
1. The component of acceleration perpendicular to the velocity is the centripetal part: \( a_n=4 \) m/s².
2. \( R=\frac{v^2}{a_n}=\frac{4}{4}=1 \) m.

Teacher's Note:
1. Only the component of acceleration normal to the velocity bends the path.
2. The component along v (2 m/s²) is tangential.

Answer: (c) 6 : 5 : 3

Explanation:
1. All turn with the same \( \omega \). Outermost section: \( T_{BC}=m\omega^2(3r)=3m\omega^2r \).
2. \( T_{AB}-T_{BC}=m\omega^2(2r)\Rightarrow T_{AB}=5m\omega^2r \).
3. \( T_{OA}-T_{AB}=m\omega^2r\Rightarrow T_{OA}=6m\omega^2r \).

Teacher's Note:
1. Work from the outermost particle inward, since each thread supplies the net inward force on the particles beyond it.
2. The tension in each section is the sum of the centripetal forces of all particles outside it.

Answer: (d) \( T'=\frac{\sqrt3}{2}T \)

Explanation:
1. Radius 2a: extension a, \( ka=m\omega^2(2a)\Rightarrow\frac{k}{m}=2\omega^2=\frac{8\pi^2}{T^2} \).
2. Radius 3a: extension 2a, \( k(2a)=m\omega'^2(3a)\Rightarrow\omega'^2=\frac{2k}{3m}=\frac{16\pi^2}{3T^2} \).
3. \( T'=\frac{2\pi}{\omega'}=\frac{\sqrt3}{2}T \).

Teacher's Note:
1. The spring tension supplies the centripetal force and depends on the extension beyond the natural length.
2. Extensions are a and 2a, not the full radii.

Answer: (a) 10 m/s

Explanation:
1. Let the top speed be v. \( T_{min}=\frac{mv^2}{L}-mg \), and \( T_{max}=\frac{mv^2}{L}+5mg \) (energy gives \( v_B^2=v^2+4gL \)).
2. Ratio 4: \( \frac{mv^2}{L}+5mg=4\frac{mv^2}{L}-4mg\Rightarrow v^2=3gL \).
3. \( v^2=3\times10\times\frac{10}{3}=100 \), so v = 10 m/s.

Teacher's Note:
1. Write the maximum (bottom) and minimum (top) tensions with the same unknown speed, using energy conservation to link them.
2. Their difference is always 6 mg.

Answer: (c) \( v^3 \)

Explanation:
1. Horizontal velocity stays \( v_0 \), so the normal component of g is \( a_n=g\frac{v_0}{v} \).
2. \( R=\frac{v^2}{a_n}=\frac{v^3}{gv_0}\propto v^3 \).

Teacher's Note:
1. Resolve g along the normal to the velocity using the constant horizontal component.
2. In general R is proportional to v cubed for a projectile.

Answer: (d) \( \frac{b}{\sqrt{b^2+\alpha^2t^4}} \)

Explanation:
1. \( a_t=\alpha,\ a_r=\frac{(\alpha t)^2}{b} \).
2. \( \cos\phi=\frac{a_t}{\sqrt{a_t^2+a_r^2}}=\frac{\alpha}{\sqrt{\alpha^2+\alpha^4t^4/b^2}}=\frac{b}{\sqrt{b^2+\alpha^2t^4}} \).

Teacher's Note:
1. The velocity is along the tangent, so the angle is the one between the total acceleration and its tangential component.
2. At t = 0 the angle is zero and it grows with time.

Answer: (b) 2 m/s²

Explanation:
1. Normal reaction \( N=\frac{mv^2}{R}=\frac{4}{2\sqrt2}=\sqrt2 \) N.
2. Friction (kinetic) \( f=\mu N=\sqrt2 \) N gives tangential deceleration \( a_t=\sqrt2 \) m/s². Radial acceleration \( a_r=\frac{v^2}{R}=\sqrt2 \) m/s².
3. \( a=\sqrt{2+2}=2 \) m/s².

Teacher's Note:
1. The wall's normal force provides the centripetal force, and friction slows the block along the tangent.
2. The floor is smooth, so no other horizontal force acts.

Answer: (c) 5 m

Explanation:
1. Radius of curvature \( R=\frac{(1+y'^2)^{3/2}}{|y''|} \).
2. At x = 0: \( y'=0,\ y''=\frac15 \), so R = 5 m.

Teacher's Note:
1. Use the curvature formula for a plane curve.
2. At the vertex the slope is zero.

Answer: (d) \( 90^\circ \)

Explanation:
1. At angle \( \theta \): \( v^2=3gl-2gl(1-\cos\theta)=gl(1+2\cos\theta) \).
2. Centripetal acceleration \( a_c=\frac{v^2}{l}=g(1+2\cos\theta) \) (towards the centre) and tangential acceleration \( a_t=g\sin\theta \).
3. Equal components give \( \sin\theta=1+2\cos\theta \), whose valid solution is \( \theta=90^\circ \).

Teacher's Note:
1. A 45 degree angle between the total acceleration and the string means the tangential and radial accelerations are equal.
2. The other algebraic root would need an impossible negative speed squared.

Answer: (a) 75 N

Explanation:
1. Take a small element subtending angle \( d\theta \): \( 2T\sin\frac{d\theta}{2}=dm\,\omega^2R \) so \( T=\frac{m\omega^2R}{2\pi} \).
2. With \( R=\frac{l}{2\pi} \) and \( \omega=2\pi n \): \( T=\frac{m(2\pi n)^2 l}{4\pi^2}=mn^2l=3\times25\times1=75 \) N.

Teacher's Note:
1. Balance the outward pull of rotation on a small element against the two tensions pulling inward.
2. The result T = m n^2 l follows after substituting the radius from the length.

Answer: (b) \( 45^\circ \)

Explanation:
1. \( v=10 \) m/s. Horizontal: \( T\sin\theta=\frac{mv^2}{r}=10m \); vertical: \( T\cos\theta=mg=10m \).
2. \( \tan\theta=1\Rightarrow\theta=45^\circ \).

Teacher's Note:
1. The bob needs the same centripetal acceleration as the car.
2. Compare the horizontal acceleration v^2/r with g.