Physics MCQs for NEET — Practice Questions with Answers

According to the NCERT text, 'Thus, the stopping distance is proportional to the square of the initial velocity. Doubling the initial velocity increases the stopping distance by a factor of 4 (for the same deceleration).' This is derived from the formula $d_s = -v_0^2 / (2a)$. If $v_0$ becomes $2v_0$, then $d_s$ becomes $(2v_0)^2 / (2a) = 4v_0^2 / (2a)$, which is 4 times the original stopping distance.

The NCERT text states, 'Reaction time depends on complexity of the situation and on an individual.' Factors like fatigue, distraction, or intoxication (e.g., alcohol) would impair an individual's ability to observe, think, and act quickly, thereby increasing reaction time. The other options describe conditions that would likely lead to a decreased or normal reaction time.

The NCERT example states, 'The ruler drops under free fall. Therefore, $v_0 = 0$, and $a = -g = -9.8 \text{ m s}^{-2}$. The distance travelled $d$ and the reaction time $t_r$ are related by $d = \frac{1}{2} g t_r^2$.' This is derived from the equation of motion for constant acceleration, $s = ut + \frac{1}{2}at^2$, where initial velocity $u=0$, acceleration $a=g$, and distance $s=d$.

The NCERT text provides data: 'the braking distance was found to be 10 m, 20 m, 34 m and 50 m corresponding to velocities of 11, 15, 20 and 25 m/s'. Therefore, for a velocity of $25 \text{ m/s}$, the braking distance is $50 \text{ m}$ based on the given empirical data.

The NCERT text explicitly states, 'Stopping distance is an important factor considered in setting speed limits, for example, in school zones.' The primary reason for lower speed limits in areas like school zones is to allow drivers more time and distance to stop if unexpected situations (like a child running onto the road) arise, thereby enhancing safety.

The process described involves two phases: first, the time taken to react (reaction time) during which the vehicle continues to move (reaction distance), and second, the time taken for the vehicle to stop after the brakes are applied (braking distance or stopping distance). The NCERT mentions 'Reaction time is the time a person takes to observe, think and act. For example, if a person is driving and suddenly a boy appears on the road, then the time elapsed before he slams the brakes of the car is the reaction time.' This implies motion during reaction time. 'stopping distance...is an important factor for road safety and depends on the initial velocity (v0) and the braking capacity'.

From the formula $d_s = -v_0^2 / (2a)$, we can see that stopping distance ($d_s$) is inversely proportional to the magnitude of deceleration ($|a|$). A larger deceleration means the vehicle can slow down and stop more quickly, hence covering a shorter distance. The negative sign for 'a' implies deceleration, so we consider its magnitude.

As per the NCERT example, the reaction time ($t_r$) is calculated using the formula $d = \frac{1}{2} g t_r^2$. Rearranging for $t_r$: $t_r = \sqrt{\frac{2d}{g}}$. Given $d = 21.0 \text{ cm} = 0.21 \text{ m}$ and $g = 9.8 \text{ m/s}^2$. So, $t_r = \sqrt{\frac{2 \times 0.21}{9.8}} = \sqrt{\frac{0.42}{9.8}} = \sqrt{0.0428...} \approx 0.207 \text{ s}$. Rounding to two significant figures, $t_r \approx 0.21 \text{ s}$.

The NCERT definition of reaction time states: 'Reaction time is the time a person takes to observe, think and act.' It also mentions that 'Reaction time depends on complexity of the situation and on an individual,' disproving the first two options. The third option defines only the 'act' component, not the full process.

According to the provided text, the total magnetic field B is written as $B = \mu_0 (H + M)$. This equation defines the relationship between these three fundamental magnetic quantities.