Physics Test - 12

The following figure shows the cyclic process of gas. If an object returns to its initial position after one or more processes it went through.

\(A B C D A\) is the cycle. The pressure at the points both \(D\) and \(C\) remain the same, which is \(2 P\). The volume at the point \(D\) is \(V\) and at the point \(C\) is \(3 V\). So, the work done by the gas from point \(D\) to point \(C, W_{D C}=2 P(3 V-V)=4 P V\)

The pressures at the points \(C\) and \(B\) are \(2 P\) and \(P\) respectively. The volume at the points both \(C\) and \(B\) remain the same, which is \(3 V\). So, the work done by the gas from point \(C\) to point \(B, \)

\(W_{C B}=P(3 V-3 V)=0\)

The pressure at the points both \(B\) and \(A\) remain the same, which is \(P\). The volume at the point \(B\) is \(3 V\) and at the point \(A\) is \(V\),

So, the work done by the gas from point \(B\) to point \(A, W_{B A}=P(V-3 V)=-2 P V\)

The pressures at the points \(A\) and \(D\) are \(P\) and \(2 P\) respectively. The volume at the points both \(A\) and \(D\) remain the same, which is \(V\).

So, the work done by the gas from point \(A\) to the point \(D, W_{A D}=P(V-V)=0\)

Hence the total work done in the whole cycle,

\(W=4 P V-2 P V=2 P V\)

We know the heat rejected from the cycle is equal to the amount of total work done by the gas, so \(Q=W\)

\( Q=2 P V\)

Given:

Potential difference (V) = 200 V

Current (I) = 0.5 A

Electric power can be written as,

​⇒ P = VI

⇒ P = 200 × 0.5 = 100 W

Given,

Force, \(F=90 \mathrm{~N}\)

Acceleration,\(a=2.6\) \(\mathrm{m} / \mathrm{s}^{2}\)

The force required to accelerate a body of mass \(m\) with acceleration \(a\) is given by,

\(\mathrm{F}={ma}\)

\(90=\mathrm{m} \times 2.6\)

\(\mathrm{~m}=\frac{90} {2.6}\)

\(\mathrm{~m}=34.6 \mathrm{~kg}\)

Let \(\mathrm{E}_{1}=\) energy required to remove \(\mathrm{e}^{-}\) from singly ionised Helium atom

\(=\frac{+13 \cdot 6 z^{2}}{n^{2}}\)

\(=\frac{(13.6)(2)^{2}}{(1)^{2}}\)

\(=54.4 \mathrm{eV}\)

\(E_{2}=\) energy required to remove an \(\mathrm{e}^{-}\)from \(\mathrm{He}\)-atom

given \(\mathrm{E}_{1}=2.2 \mathrm{E}_{2}\)

\(\therefore \mathrm{E}_{2}=\frac{\mathrm{E}_{1}}{2.2}=\frac{54.4}{2.2}=24.7 \mathrm{eV}\)

\(\therefore\) Energy required to remove both \(\mathrm{e}^{-1} \mathrm{~s}\) from He-atom \(=24.7+54.4\)

\(=79.1 \mathrm{eV}\)

When the metal heated, the length, surface area, volume of the metal also increased. The increase in temperature which results in metal expands and this expansion is termed as the thermal expansion of metal i.e. expansion in metal due to the heating effect.

So, the iron blade has a ring in which the wooden handle is fixed. The ring is slightly smaller in size than a wooden handle. When the ring is heated which is made up of metal expands, after the ring cools, it tightly fits in the wooden handle.

Kinematics pairs are those which have two elements thatpermit relative motion.

Kinematics pair can be classified according to:

• Nature of contact
• Nature of relative motion
• Nature of mechanical constraint

Photoelectric effect

\({K} {E}_{\max }={h} {v}-{W}_{{o}}\)

Where, \(W_{o}\) is the work function of the metal.

Clearly \({K} {E}\) varies linearly with the the frequency of the photons.

Given:

\(\mathrm{R}=0.8 \mathrm{~A} / \mathrm{W}\) and \(P=10 \mu \mathrm{W}\)

In a photodiode, the responsivity is given by:

\(R=\frac{I}{P}\)

\(I =\) photocurrent generated

\(P=\) Optical power incident

\(R=\) Responsitivity

\(0.8=\frac{I}{10 \mu}\)

\(I=0.8 \times 10 \mu \mathrm{A}=8 \mu \mathrm{A}\)

An eye that suffers from myopia as well as from hypermetropia is said to suffer from presbyopia. A person with this defect cannot see objects distinctly placed at any distance from him. To correct this defect, a person is prescribed bifocal lens that has both types of lenses convex and concave.

In a spherical shell, the internal energy per unit volume is given by \(\frac{U}{V} \alpha T^{4}\)

\(\Rightarrow U=C V T^{4}\)

Here \(C\) is a constant

Next the value of \(P\),

\(=\frac{1}{3}\left(\frac{U}{V}\right)=\frac{1}{3}\left(\frac{C V T^{4}}{V}\right)\)

From adiabatic expansion, \(d Q=0\) and \(d U=-d W\)

\(d(C V T)^{4}=-P d V\)

We get,

\(\Rightarrow 4 V d T=-\frac{4}{3} T d V \)

\(\Rightarrow \frac{d T}{T}=\frac{d V}{3 V}\)

On integrating,

\(\Rightarrow T V^{ \frac{1}{3}}=C^{1} \)

\(\Rightarrow T\left(\frac{4}{3} \pi R^{3}\right)^{\frac{1}{3}}=C^{1}\)

\(T R=\) Constant

\(\Rightarrow T \alpha \frac{1}{R}\)