Physics MCQs for NEET — Practice Questions with Answers

$ \vec V_0 = V_0cosθ \hat i + V_0 sinθ \hat j$ $ \vec V = V_0 cosθ \hat i + (V_0sm \theta – gt) \hat j$ $ \vec V_0 . \vec V = 0 $ $ \therefore t = { V_0 \over gsin \theta } $ Now find x

At time t $ V_ x = V_0 = V_0 cos \theta $ $ V_y = Visinθ – gt$
$ tan \alpha = { Vy \over Vx } = {V_0sin \theta – gt \over V_0cos \theta } $

For a projectile motion where the initial and final points are on the same horizontal line, the average velocity is the horizontal component of the initial velocity, which is \( V_0 \cos \theta \).

When a body travels in a circle at constant speed, it is undergoing circular motion. In circular motion, even though the speed is constant, the direction of the velocity is constantly changing. This change in direction means there is an acceleration, called centripetal acceleration, which acts towards the center of the circle. Therefore, the body has an inward radial acceleration.

Centrifugal acceleration $ a_r = { v^2 \over r } = \left( { 2 \pi r \over T } \right) ^2 { 1 \over r } = 4 \pi^2 r f^2 $

In projectile motion, at the highest point, the vertical component of the velocity is zero but the horizontal component of the velocity still exists. Thus, the total velocity is not zero. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.

$ A = \int _0^R y dx $ $ take \, y = x tan \theta - { gx^2 \over 2V_0 x cos \theta } $

$ PV = RT \Rightarrow PV = { M \over M_o } RT $ $ \therefore {PMo \over RT } ={ M \over V} = \rho $
$ \rho = { PMo \over RT} = {P \times m \times N_A \over RT} = { Pm \over \left( { R \over N_A} \right)} T = { Pm \over k_B T } $