Physics Test-11

Given,

\(x(t)=\frac{v}{A}\left(1-e^{-A t}\right)\)

As we know that \( (1- {e}^{\text {-At }}) \) is a constant value and will have no dimension.

Thus, the dimension of \(\frac{v}{A}\) will be equal to the dimension of \(x\).

Dimension of position, \(x=\left[M^{0} L^{1} T^{0}\right]\)

Thedimension of velocity,\(v=\left[\mathrm{M}^{0} \mathrm{~L}^{1} \mathrm{~T}^{-1}\right]\)

\(\Rightarrow x=\frac{v}{A}\)

\(\Rightarrow\left[M^{0} L^{1} T^{0}\right]=\frac{\left[M^{0} L^{1} T^{-1}\right]}{A}\)

\(\Rightarrow A=\left[T^{-1}\right]\)

α particle comprises of 2 protons and 2 neutrons, which have 4 times higher mass than neutrons, and thus low penetration power.

Beta particles are much smaller and more penetrating than alpha particles, but their range in tissue is still limited.

Due to their large mass and being chargeless in nature, neutrons have high kinetic energy and thus, high penetration power.

The value of the Prandtl number for air is about 0.7

Prandtl number (Pr) - is a dimensionless number approximating the ratio of momentum diffusivity (kinematic viscosity) to thermal diffusivity and is often used in heat transfer and free and forced convection calculations.

A bar magnet is a magnetic dipole. It is analogous to an electric dipole which is a system of two equal and opposite equal charges. Isolated magnetic charges or magnetic monopoles do not exist in nature. Even if a bar magnet is broken down to the atomic level, the north and south poles can never be separated.

Use the formula of force between two bar magnets.

\(F=\frac{\mu_{0}}{4 \pi} \frac{6 M_{1} M_{2}}{r^{4}}\)

Here,

\(\mathrm{F}\) is force between two bar magnets.

M₁ and M₂ are magnetic moments.

\(r\) is the distance between bar magnets.

\(\mu_{0}\) is the permeability of free space.

If the length of the bar magnets is negligible compared then \(r=d\)

Thus, the force acting between them will be proportion to \(\frac{1}{d^{4}}\).

In rotational motion, Power = Torque × Angular velocity ⇒ P = τω

Bernoulli's principle is based on the principle of conservation of energy.

Bernoulli's principle is based on the principle of conservation of energy. This states that in a steady flow the sum of all forms of mechanical energy in a fluid along a streamline is the same at all points on that streamline. This requires that the sum of kinetic energy, potential energy and pressure energy remain constant. So at each point, the net energy is conserved in the fluid.

We know, energy stored in the capacitor, \((E_c)=\frac{1}{2} C V^2\)

And energy stored in the inductor, \((E_L)=\frac{1}{2} L I^2\)

Where, \(C\) is the value of capacitance, \(V\) is the potential difference across plates ofcapacitor (its maximum value is given \(400 V ), L\) is the value of inductance \((1 H)\) and \(I\) is the current flowingthrough the inductor \((1 A)\).

Energy stored in the capacitor \(=\) Energy stored in the Inductor

\(\frac{1}{2} C V^2=\frac{1}{2} L I^2\)

\(\Rightarrow C V^2=L I^2\)

The value of capacitance \((C)=\frac{L I^2}{V^2}\)

Now substituting the values, we get

\(C=\frac{1 \times 2^2}{400^2}\)

\(\Rightarrow C=\frac{4}{16000}\)

\(\Rightarrow C=2.5 \times 10^{-5}\)

\(\Rightarrow C=25 \times 10^{-6} F\)

\(\Rightarrow C=25 \mu F\)

Impure semiconductors in which the charge carriers are produced due to impurity atoms are called extrinsic semiconductors. They are obtained by doping an intrinsic semiconductor with impurity atoms. Electric current flows through "free" electrons and "holes", also called charge carriers. Doping is done by adding elements such as phosphorus or boron to a semiconductor such as silicon, which substantially increases the amount of free electrons or holes available in the semiconductor.

According to the given question, let us consider:

\(\Rightarrow\) After time ' \(t\) ' the man caches the bus

\(\Rightarrow\) So after time ' \(t\) ' the distance covered by the bus is \(\frac {1}{2} t^{2}= \frac {t^{2}}{2}\)

\(\Rightarrow\) The total distance is equal to \(48+ \frac {t^{2}} {2}\) \((\) Initial velocity of the bus = \(0)\)

\(\Rightarrow\) We know that, \(v t=s\)

\(\Rightarrow 10 t=48+ \frac {t^{2}}{2}\)

\(\Rightarrow 20 t=96+t^{2}\)

\(\Rightarrow t^{2}-20 t+96=0\)

\(\Rightarrow (t-8)(t-12)=0\)

Implies that,

\(\Rightarrow t=8\) sec, \(t=12\) sec

Since 12 is not present in the option, so 8 will be the correct answer.

As we know,

\(X_C=\frac{1}{2 \pi f C} \)

\(\therefore C=\frac{1}{2 \pi f X_C}\)

\(\therefore C=\frac{1}{2 \pi \times \frac{400}{\pi} \times 25}\)

\(=50 \times 10^{-6}\)

\(=50 \mu F\)