Physics Test-37

Given,

Acceleration, \(a=2 m / s ^{2}\)

\(u=20 m / s\)

\(v=30 m / s\)

From equation of motion,

\(v^{2}-u^{2}=2 a s\)

\(\Rightarrow s=\frac{\left(v^{2}-u^{2}\right)}{(2 \times a)}\)

\(\Rightarrow s=\frac{\left(30^{2}-20^{2}\right)}{(2 \times 2)}\)

\(\Rightarrow s=\frac{500}{4}\)

\(\Rightarrow s=125 m\)

When compact disk is illuminated by a source of white light, The small ripples on the compact disc split white light into the constituent colours. This is due to diffraction.

A compact disc contains many fine circular lines on it so it acts as diffraction grating. When white light falls upon it, it is diffracted as a result of which different colours are diffracted at different angles. We see different colours when we look at it through different angles.

As we know that the momentum of a body is given as,

P = m × v...(1)

Where m = mass, and v = velocity of the body

By equation (1) it is clear that if the momentum of a body is constant, then its velocity will also be constant and the acceleration will be zero.

An object is in translational equilibrium if the velocity of its translational motion is constant. To be in translational equilibrium, the net force acting on the object must be zero.

Therefore we can say that if the momentum of a body is constant, then it will be in translational equilibrium.

Isothermal dissipation of work and adiabatic dissipation of workexhibit external mechanical Irreversibility.

In the isothermal compression of a gas there is work done on the system to decrease the volume and increase the pressure.The work done in adiabatic process derivation can be derived from the first law of thermodynamics relating to the change in internal energy \(dU\) to the work \(dW\) done by the system and the heat \(d Q\) added to it. The work done \(d W\) for the change in volume \(V\) by \(dV\) is given as \(PdV\).

Velocity \(\mathrm{V}=\frac{\text { distance }}{\text { time }}=\frac{100}{10}=10 \mathrm{~m} / \mathrm{s}\)

Assuming his mass to be \(6 \mathrm{0kg}\), his kinetic energy is\(\mathrm{K} . \mathrm{E}=\frac{1}{2} \mathrm{mv}^{2}\)\(=\frac{1}{2} \times 60 \times 100\)

\(=60 \times 50\)\(=3000 \mathrm{~J}\)

For a dark fringe to form,

\(\frac{d y}{D}=\lambda\)

\(y=\frac{D \lambda}{d}\)

\(=\frac{2 \times 600 \times 10^{-9}}{10^{-3}}\)

\(=1.2\) mm

Distance between the first dark fringes on either side of central bright fringe \(=2 \times y=2.4\) mm

Given: Initially force per unit length on the wires \(= f _{ ab }= f _{ ba }= f =10^{-3}\) N and \(d =2\) m

If two current carrying long wires \(A\) and \(B\) are separated by a very small distance and placed parallel to each other. Then the force per unit length on wire \(A\) and wire \(B\) is given as,

\(f_{a b}=f_{b a}=f=\frac{\mu_{o} I_{a} I_{b}}{2 \pi d}\)

Where, \(I _{ a }=\) current in the wire \(A , I _{ b }=\) current in the wire \(B\), and \(d =\) distance between the wires

If the current in both the wires is doubled,

\(I_{a}^{\prime}=2 I_{a}\)

\(\Rightarrow I_{b}^{\prime}=2 I_{b}\)

And the distance between the wires is halved, so,

\(d^{\prime}=\frac{d}{2}\)

So, when the current in both the wires is doubled and the distance between the wires is halved, the new force per unit length on wire A and wire B will be,

\(f_{a b}^{\prime}=f_{b a}^{\prime}=f^{\prime}=\frac{\mu_{o} I_{a}^{\prime} I_{b}^{\prime}}{2 \pi d^{\prime}}\)

\(\Rightarrow f^{\prime}=2 \times \frac{\mu_{o} \times 2 I_{a} \times 2 I_{b}}{2 \pi d}\)

\(\Rightarrow f^{\prime}=8 \times \frac{\mu_{o} I_{a} I_{b}}{2 \pi d}\)

\(\Rightarrow f ^{\prime}=8 f\)

\(\Rightarrow f ^{\prime}=8 \times 10^{-3}\) N

Given,

Radius, \(r=1000 m\)

Banking angle, \(\theta=45^{\circ}\)

Coefficient of friction, \(\mu=0.5\)

Acceleration due to gravity, \(g=9.8 m/s^2\)

As we know,

Maximum safe velocity is calculated by,

\(v=\sqrt{\frac{r g(\tan \theta+\mu)}{1-\mu \tan \theta}} \)

\(\therefore v=\sqrt{\frac{1000 \times 9.8\left(\tan 45^{\circ}+0.5\right)}{1-0.5 \times \tan 45^{\circ}}} \)

\(\Rightarrow v=\sqrt{\frac{1000 \times 9.8 (1+0.5)}{1-0.5 \times 1}}\)

\(\Rightarrow v=\sqrt{\frac{1000 \times 9.8 \times 1.5}{1-0.5}}\)

\(\Rightarrow v=\sqrt{\frac{1000 \times 9.8 \times 1.5}{0.5}}\)

\(\Rightarrow v=\sqrt{29400}\)

\(\therefore v =172 m / s\)

Thermal conductivity is measured in watts per meter kelvin (W/(m⋅K)) and J/m °K sec.

The thermal conductivity of a material is a measure of its ability to conduct heat. Heat transfer occurs at a lower rate in materials of low thermal conductivity than in materials of high thermal conductivity.

Given: \(m_{2}=\frac{1}{4} m_{1}\)

The time period of the simple pendulum is given as,

\(\Rightarrow T=2 \pi \sqrt{\frac{l}{g}}\) .....(i)

Where, T = Time period of oscillation, l = length of the pendulum, and g = gravitational acceleration

Since the time period of the simple pendulum does not depend on the mass of the bob, so the change in mass of the bob will not affect the time period of the pendulum.