Laws Of Motion Ques 40
- A block of mass $m _1=1 kg$ another mass $m _2=2 kg$ are placed together (see figure) on an inclined plane with angle of inclination $\theta$. Various values of $\theta$ are given in List $I$. The coefficient of friction between the block $m _1$ and the plane is always zero. The coefficient of static and dynamic friction between the block $m _2$ and the plane are equal to $\mu=0.3$.
(2014 Adv.)
In List II expressions for the friction on the block $m _2$ are given. Match the correct expression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration due to gravity is denoted by $g$.
[Useful information $\tan \left(5.5^{\circ}\right) \approx 0.1$;
$\left.\tan \left(11.5^{\circ}\right) \approx 0.209 ; \tan \left(16.5^{\circ}\right) \approx 0.300\right]$
| List I | List II | ||
|---|---|---|---|
| P. | $\theta=5^{\circ}$ | 1. | $m _2 g \sin \theta$ |
| Q. | $\theta=10^{\circ}$ | 2. | $\left(m _1+m _2\right) g \sin \theta$ |
| R. | $\theta=15^{\circ}$ | 3. | $\mu m_2 g \cos \theta$ |
| S. | $\theta=20^{\circ}$ | 4. | $\mu\left(m _1+m _2\right) g \cos \theta$ |
Codes
(a) P-1, Q-1, R-1, S-3
(b) P-2, Q-2, R-2, S-3
(c) P-2, Q-2, R-2, S-4
(d) P-2, Q-2, R-3, S-3
Show Answer
Answer:
Correct Answer: 40.(d)
Solution:
Formula:
- Block will not slip if the static friction is greater than or equal to the applied force
$\left(m _1+m _2\right) g \sin \theta \leq \mu m _2 g \cos \theta$
$$ \begin{array}{cc} \Rightarrow & 3 \sin \theta \leq \frac{3}{10}(2) \cos \theta \\ \Rightarrow \quad \tan \theta \leq 1 / 5 \\ \quad \theta \leq 11.5^{\circ} \end{array} $$
(P) $\theta=5^{\circ}$ friction is static friction
$$ f=\left(m _1+m _2\right) g \sin \theta $$
(Q) $\theta=10^{\circ}$ friction is kinetic
$f=\left(m_1+m_2\right) g \sin \theta$
(R) $\theta=15^{\circ}$ friction is static
$f=\mu m_2 g \cos \theta$
(S) $\theta=20^{\circ}$ friction is static
$\Rightarrow f=\mu m_2 g \cos \theta$
BETA
