Laws of Motion: NEET previous year questions

The box stays with the trolley only while static friction can supply the force needed to accelerate it. The maximum friction is $\mu mg$, so the greatest common acceleration is $a = \mu g = 0.12 \times 10 = 1.2$ m/s$^2$, option (4). The mass cancels out, and beyond this acceleration the box slips relative to the trolley.

The two perpendicular forces give a resultant $F = \sqrt{8^2 + 6^2} = 10\ \text{N}$, so $a = F/m = 10/5 = 2\ \text{m s}^{-2}$. The resultant makes an angle $\theta$ with the 8 N force where $\tan\theta = 6/8 = 3/4$, i.e. $\tan^{-1}(3/4)$ measured from the 8 N force (option (4)). Option (1) forgets to divide by the mass, and the others attach the angle to the wrong force or invert the ratio.

Minimum wear and tear is the exam's way of saying no sideways friction is needed, i.e. the car is at the optimum (design) speed of the banked track. Then the horizontal component of the normal reaction alone supplies the centripetal force, giving $\tan\theta = \frac{v^2}{rg} = \frac{10^2}{50 \times 10} = \frac{1}{5}$, so $\theta = \tan^{-1}(1/5)$. NCERT Class 11, Laws of Motion (motion on a banked road).

For a system of two masses $m_1$ and $m_2$ connected by a massless string over a frictionless pulley, the acceleration a is given by $a = \frac{(m_2 - m_1)g}{m_1 + m_2}$. Substituting $m_1 = 4\ kg$ and $m_2 = 6\ kg$, $a = \frac{(6 - 4)g}{4 + 6} = \frac{2g}{10} = \frac{g}{5}$. Chapter: Laws of Motion.

Why the other options are wrong:

- (1) $g/2$: this divides the net pull $(6 - 4)g = 2g$ by the 4 kg block alone, but both blocks accelerate together, so the full 10 kg must be used.
- (3) $g/10$: this takes the net pull as $g$, but the unbalanced weight is $(6 - 4)g = 2g$, which gives twice that.
- (4) $g$: this is free fall. The 4 kg block's weight pulls back against the 6 kg block and the net pull moves all 10 kg, so $a$ must be less than $g$.