Assertion (A): A ladder is more likely to slip when a person is near the top than when he is near the bottom.
Reason (R): The friction between the ladder and floor decreases as he climbs up.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is true; a ladder is more likely to slip when a person is higher up due to increased outward horizontal thrust. Reason (R) is false; the normal force at the base remains constant, so the maximum static friction available does not decrease.
Assertion (A): It is more difficult to open the door by applying the force near the hinge.
Reason (R): Torque is maximum at hinge.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is true: Torque \(\tau = rF\sin\theta\) requires a larger force \(F\) for a smaller lever arm \(r\) (near the hinge). Reason (R) is false: Torque is zero at the hinge (pivot point) as \(r=0\). Thus, (A) is true, (R) is false.
A solid cylinder of mass $2 \text{ kg}$ and radius $4 \text{ cm}$ is rotating about its axis at the rate of $3 \text{ rpm}$. The torque required to stop after $2\pi$ revolutions is
(2019)
1. $2 \times 10^{-6} \text{ N m}$
2. $2 \times 10^{-3} \text{ N m}$
3. $12 \times 10^{-4} \text{ N m}$
4. $2 \times 10^{6} \text{ N m}$
View Answer
Here $I = \frac{1}{2}MR^2 = 1.6 \times 10^{-3} \text{ kg m}^2$, $\omega_0 = 3 \times \frac{2\pi}{60} = \frac{\pi}{10} \text{ rad/s}$, and $\theta = 4\pi^2 \text{ rad}$. Using $\omega^2 = \omega_0^2 + 2\alpha\theta$, $\alpha = -\frac{1}{800} \text{ rad/s}^2$. The required torque magnitude is $\tau = I\alpha = 2 \times 10^{-6} \text{ N m}$.
A rope is wound around a hollow cylinder of mass $3 \text{ kg}$ and radius $40 \text{ cm}$. What is the angular acceleration of the cylinder if the rope is pulled with a force of $30 \text{ N}$?
(2017-Delhi)
1. $0.25 \text{ rad/s}^2$
2. $25 \text{ rad/s}^2$
3. $5 \text{ m/s}^2$
4. $25 \text{ m/s}^2$
View Answer
For a hollow cylinder, the moment of inertia is $I = MR^2$. The torque provided by the rope is $\tau = F \times R = I\alpha$. Substituting $I$, we get $F \times R = MR^2 \alpha$, which gives $\alpha = \frac{F}{MR} = \frac{30}{3 \times 0.4} = 25 \text{ rad/s}^2$.
An automobile moves on a road with a speed of $54 \text{ km/h}$. The radius of its wheels is $0.45 \text{ m}$ and the moment of inertia of the wheel about its axis of rotation is $3 \text{ kgm}^2$. If the vehicle is brought to rest in $15 \text{ s}$, the magnitude of average torque transmitted by its brakes to wheel is:
(2015 Re)
1. $2.86 \text{ kg m}^2/\text{s}^2$
2. $6.66 \text{ kg m}^2/\text{s}^2$
3. $8.58 \text{ kg m}^2/\text{s}^2$
4. $10.86 \text{ kg m}^2/\text{s}^2$
View Answer
Initial angular velocity $\omega_0 = \frac{v}{r} = \frac{15}{0.45} = \frac{100}{3} \text{ rad/s}$. The angular acceleration is $\alpha = \frac{\omega_0}{t} = \frac{100/3}{15} = \frac{20}{9} \text{ rad/s}^2$. Torque is $\tau = I\alpha = 3 \times (\frac{20}{9}) = 6.66 \text{ kg m}^2/\text{s}^2$.
The moment of inertia of a body about a given axis is $1.2 \text{ kgm}^2$. Initially, the body is at rest. In order to produce a rotational kinetic energy of $1500 \text{ joule}$, an angular acceleration of $25 \text{ rad/sec}^2$ must be applied about that axis for a duration of:
(1990)
1. $4 \text{ s}$
2. $2 \text{ s}$
3. $8 \text{ s}$
4. $10 \text{ s}$
View Answer
Using $K = \frac{1}{2} I \omega^2$, we substitute the values to get $1500 = \frac{1}{2}(1.2) \omega^2$, giving $\omega = 50 \text{ rad/s}$. Applying kinematics equation $\omega = \omega_0 + \alpha t$, we find $50 = 0 + 25t$, yielding $t = 2 \text{ s}$.
Find the torque about the origin when a force of $3 \hat{j} \text{ N}$ acts on a particle whose position vector is $2 \hat{k} \text{ m}$.
(2020)
1. $6 \hat{j} \text{ N m}$
2. $-6 \hat{i} \text{ N m}$
3. $6 \hat{k} \text{ N m}$
4. $6 \hat{i} \text{ N m}$
View Answer
Torque is given by the cross product $\vec{\tau} = \vec{r} \times \vec{F}$. Substituting the given vectors, $\vec{\tau} = (2\hat{k}) \times (3\hat{j}) = 6(\hat{k} \times \hat{j})$. Since $\hat{k} \times \hat{j} = -\hat{i}$, the torque is $-6\hat{i} \text{ N m}$.