Physics MCQs for NEET — Practice Questions with Answers

When the sparrow sits on the wire, it causes a deflection in the wire, creating a vertical component of tension. The tension in the wire must counteract not only the weight of the sparrow but also the additional vertical components due to the deflection. Therefore, the tension T in the wire is greater than the weight W of the sparrow: $T > W$.

Fair = m ( g - a)

$ F = { \triangle P \over \triangle t } = { 25 \over 0.04 } = 625 N $

Newton's third law of motion states that for every action, there is an equal and opposite reaction. This principle is fundamental to the conservation of momentum. When two bodies interact, the forces they exert on each other are equal in magnitude and opposite in direction, leading to the total momentum of the system remaining constant if no external forces are acting on it.

$ \nu_o = \sqrt { 2 g h_1 } and \nu = \sqrt {2 g h_1 }$ $ madt = m ( \nu_2 + \nu_1 ) $ $ \therefore a = { \nu_2 + \nu_1 \over dt }$

To solve this, we use the formula for force, which is given by Newton's second law: \( F = ma \). Here, \( m \) is the mass of the vehicle (100 kg) and \( a \) is the acceleration. First, we need to find the acceleration, which is given by: \( a = \frac{\Delta v}{\Delta t} \). The change in velocity (\( \Delta v \)) is \( -5 \ \text{m/s} \) (since the vehicle is stopping), and the time (\( \Delta t \)) is \( \frac{1}{10} \ \text{s} \). Therefore, \( a = \frac{-5}{1/10} = -50 \ \text{m/s}^2 \). Now, applying Newton's second law: \( F = 100 \ \text{kg} \times -50 \ \text{m/s}^2 = -5000 \ \text{N} \). The negative sign indicates direction, but the magnitude of the force is 5000 N.

$ F = { dp \over dt } = 0 + 2 bt $ $ \therefore F \alpha t $

A body with constant speed and zero acceleration means it is moving in a straight line with uniform motion. In uniform motion, both speed and velocity are constant, and the acceleration is zero. Therefore, the correct statement is: “A body has a constant speed and zero acceleration.”

$ F = \sqrt { F_1^2 + F_2^2 } $ $ and a = { F/M } $

The frictional force can be found using the formula: \( f = \mu N \), where \( \mu \) is the coefficient of friction and \( N \) is the normal force. For a horizontal surface, the normal force \( N \) is equal to the weight of the block, \( mg \). Here, \( \mu = 0.6 \), \( m = 1 \ \text{kg} \), and \( g = 9.8 \ \text{m/s}^2 \). Thus, \( N = 1 \ \text{kg} \times 9.8 \ \text{m/s}^2 = 9.8 \ \text{N} \). The frictional force is then \( f = 0.6 \times 9.8 \ \text{N} = 5.88 \ \text{N} \). However, the maximum frictional force is \( \mu mg \), and if the truck is accelerating at \( 5 \ \text{m/s}^2 \), the frictional force required to prevent slipping is \( f = ma = 1 \ \text{kg} \times 5 \ \text{m/s}^2 = 5 \ \text{N} \). Therefore, the correct answer is 5 N.