Physics MCQs for NEET — Practice Questions with Answers

The emission spectra of atoms, which consist of discrete lines, provided experimental evidence for the quantization of energy in atoms, as proposed by Niels Bohr's atomic model.

When a particle moving in a circle with a constant angular speed.
Its velocity vector is tangent to the circle and the acceleration vector is along the radius of the circle. The velocity and acceleration vectors are perpendicular to each other.

If the speed of body is increasing then acceleration is in the direction of velocity.It may be positive or negative. If acceleration is in negative direction then acceleration is increasing but in negative side, so it will be called as decreasing.

The magnitude of average speed and average velocity can only be equal if object moves in a straight line without turning back. In that condition distance will be equal to displacement.

x = 10t – 2t²

v = 0, at the time of coming to rest, so
10 – 4t = 0
t = 2.5 s

The given equation represents a simple harmonic motion, where x = A sin(ωt) is the displacement. The acceleration (a) is the second derivative of displacement with respect to time, which is a = -ω^2 x. Hence, the correct option is (2) a = -ω^2 x.

Body starts from rest and moves with a constant acceleration, then the distance travelled in equal time intervals
will be in the ratio of odd number. (Galileo's law of odd number)
x1 : x2 ⇒ 1 : 3
x : x2 ⇒ 1 : 3

x = 3x

$$\begin{gathered} \frac{\mathrm{Heat}\left(\mathrm{Q}\right)}{\mathrm{Time}\left(\mathrm{t}\right)} = \frac{\mathrm{K} \times \mathrm{A}\left(\mathrm{Area}\right) \times ∆\mathrm{T}\left(\mathrm{Temperature} \mathrm{difference}\right)}{\mathrm{l}\left(\mathrm{Length}\right)} \\ Where K is the thermal conductivity \\ \mathrm{So}, \mathrm{expression} for \mathrm{K} \mathrm{in} \mathrm{terms} \mathrm{of} \mathrm{other} va\mathrm{riables}, \\ \mathrm{K} = \frac{\mathrm{Q} \times \mathrm{l}}{\mathrm{t} \times \mathrm{A} \times ∆\mathrm{T}} \\ \mathrm{Putting} \mathrm{dimension} \mathrm{of} \mathrm{individual} \mathrm{quantities} \\ \mathrm{K} = \frac{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right] \times \left[\mathrm{L}\right]}{\left[\mathrm{T}\right] \times \left[{\mathrm{L}}^2\right] \times \left[\mathrm{K}\right]} = \left[{\mathrm{MLT}}^{-3}{\mathrm{K}}^{-1}\right] \end{gathered}$$

$$\begin{gathered} Q=mL \\ L=\frac{Q}{m}=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[M\right]}=\left[M^0{L}^2{T}^{-2}\right] \end{gathered}$$