Physics Test - 10

Mirror Formula: The following formula is known as the mirror formula, which provided a relationship between focal length \((f)\), image distance \((v)\), and objects distant \((u)\) as shown below:

\( \frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)

The above formula is valid for spherical mirrors only.

The least count of the Screw gauge is equal to \(0.01 \mathrm{~mm}\).

The least count of the screw gauge \(=\frac{\text { Pitch }}{\text { Number of divisions circular scale }}\)

The number of divisions on a circular scale \(=50\)

\(0.01 \mathrm {~mm} =\frac{\text { Pitch }}{50}\)

Pitch \(=0.5 \mathrm{~mm}\)

Let \(E_{0}\) be the potential difference applied across the total length \(l(=10 {~cm})\) of potentiometer wire,

Potential gradient in the first case \(=\frac{E_{0}}{l}\)

As per question, \(E=\frac{l}{3}\left(\frac{E_{0}}{l}\right)=\frac{E_{0}}{3} \ldots .\) (i)

Potential gradient in second case \(=\frac{E_{0}}{\frac{3 l}{ 2}}=\frac{2 E_{0}}{3 l}\)

If \(x\) is the desired length of potentiometer to balance the emf \(E\) of the cell, then

\(E=x \times \frac{2 E_{0}}{3 l} \ldots \ldots\) (ii)

From (i) and (ii), we have

\(\frac{E_{0}}{3}=x \times \frac{2 E_{0}}{3 l}\)

or \(x=\frac{l}{2}=\frac{10}{2}=5 {~cm}\)

The magnetic field B inside the toroid is constant in magnitude for the ideal toroid of closely wound turns. The magnetic field in the open space inside (point P) and exterior to the toroid (point Q) is zero.

The magnetic field \(B\) inside the toroid is given as,

\(B=\frac{\mu_{o} N I}{2 \pi r}\)

Where, \(N =\) number of turns, \(I =\) current, and \(r =\) average radius of the toroid, If \(N\) and \(I\) are constant, then,

\(B \propto \frac{1}{r}\)

\(\Rightarrow B \propto r^{-1}\)

Here, \(F=?\)

Charge at \(P\),

\(q=1 \mu C=10^{-6} C\)

Charge at \(A\),

\(q_{1}=+2 \times 10^{-8} C\)

Charge at \(B\),

\(q_{2}=-10^{-8} C\)

\(A P=10 cm =0.1 m\)

\(B P=5 cm =0.05 m\)

\(\angle A P B=90^{\circ}\)

Force at \(P\) due to \(q_{1}\) charge at \(A\),

\(F_{1}=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q}{(A P)^{2}}\), along AP produced,

\(=\frac{9 \times 10^{9} \times 2 \times 10^{-8} \times 10^{-6}}{(0.1)^{2}}\)

\(=18 \times 10^{-3} N\)

Force at \(P\) due to \(q_{2}\) charge at \(B\),

\(F_{2}=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{2} q}{(B P)^{2}}\), along \(B P\)

\(=\frac{9 \times 10^{9} \times 10^{-8} \times 10^{-6}}{(0.05)^{2}}\)

\(=36 \times 10^{-3} N\)

As angle between \(\vec{F}_{1}\) and \(\vec{F}_{2}\) is \(90^{\circ}\),

\(\therefore\) Resultant force \(F=\sqrt{F_{1}^{2}+F_{2}^{2}}\)

\(F =\sqrt{\left(18 \times 10^{-3}\right)^{2}+\left(36 \times 10^{-3}\right)^{2}}\)

\(=18 \times 10^{-3} \times 2.236\)

\(F =4.024 \times 10^{-2} N\)

A plane mirror is a mirror with a flat (planar) reflective surface.

The characteristics of an image formed in a plane mirror:

• The image formed by the plane mirror is virtual and erect i.e. image cannot be projected or focused on a screen.
• The distance of the image ‘behind’ the mirror is the same as the distance of the object in front of the mirror.
• The size of the image formed is the same as the size of the object.
• The image is laterally inverted, i.e. left hand appears to be the right hand when seen from the plane mirror.
• If the object moves towards (or away from) the mirror at a certain rate, the image also moves towards (or away from) the mirror at the same rate.

From the above, it is clear that the image we see in the plane mirror is virtual and is laterally inverted.

Heating effect produced by current is due to the collision of electrons.

• When a potential difference is applied across the ends of a conductor, its free electrons get accelerated in the opposite direction of the applied field.
• But the speed of the electrons does not increase beyond a constant drift speed. This is because, during the course of their motion, the electrons collide frequently with the positive metal ions.
• The kinetic energy gained by the electrons during the intervals of free acceleration between collisions is transferred to the metal ions at the time of the collision.
• The metal ions begin to vibrate about their mean positions more and more violently.

Potential gradient (k):Potential difference (or fall in potential) per unit length of wire is called potential gradient i.e.,

\(k=\frac{V}{l}\)

Where, \(V=\) Potential difference and \(l =\) length of the wire

Potential gradient directly depends upon:

• The resistance per unit length (R/L) of potentiometer wire.
• The radius of potentiometer wire ( i.e., Area of crosssection)
• The specific resistance of the material of potentiometer wire (i.e., ρ)
• The current flowing through potentiometer wire (I).

Let us assume that the threshold frequency of the sphere is \(\lambda_{o} .\) Let the stopping potential is \(V'\) of the surface when the light of wavelength \(\lambda_{3}\) is used. Thus according to Einstein photoelectric equation.

\(\frac{h c}{\lambda_{1}}=\frac{h c}{\lambda_{o}}+e V\).....(i)

\(\frac{h c}{\lambda_{2}}=\frac{h c}{\lambda_{o}}+3 e V\).....(ii)

\(\frac{h c}{\lambda_{3}}=\frac{h c}{\lambda_{o}}+e V^{\prime}\).....(iii)

From equation (i) and (ii)

\(\frac{h c}{\lambda_{2}}=\frac{h c}{\lambda_{0}}+3\left(\frac{h c}{\lambda_{1}}-\frac{h c}{\lambda_{0}}\right)\)

\(\Rightarrow \frac{h c}{\lambda_{2}}=\frac{h c}{\lambda_{0}}+\frac{3 h c}{\lambda_{1}}-\frac{3 h c}{\lambda_{0}}\)

\(\Rightarrow \frac{h c}{\lambda_{2}}-\frac{3 h c}{\lambda_{1}}=-\frac{2 h c}{\lambda_{0}}\)\(\ldots(i v)\)

From equation (iii) and (iv)

\(\frac{h c}{\lambda_{2}}-\frac{3 h c}{\lambda_{1}}=-2\left(\frac{h c}{\lambda_{3}}-e V^{\prime}\right)\)

\(\Rightarrow \frac{h c}{\lambda_{2}}-\frac{3 h c}{\lambda_{1}}=-\frac{2 h c}{\lambda_{3}}+2 e V^{\prime}\)

\(\Rightarrow h c\left(\frac{1}{\lambda_{2}}+\frac{2}{\lambda_{3}}-\frac{3}{\lambda_{1}}\right)=2 e V^{\prime}\)

\(\Rightarrow \frac{h c}{e}\left(\frac{1}{ \lambda_{3}}+\frac{1}{2\lambda_{2}}-\frac{3}{2 \lambda_{1}}\right)=V^{\prime}\)

Let us assume that the charge \(q=\pm 10 \mu {C}=10^{-5} {C}\) is placed at a distance of \(5 {~cm}\) from the square \({ABCD}\) of each side \(10 {~cm}\). The square \({ABCD}\) can be considered as one of the six faces of a cubic Gaussian surface of each side \(10 {~cm}\).

Now, the total electric flux through the faces of the cube as per Gaussian theorem:

\(\phi=\frac{q}{\epsilon_{0}}\)

Therefore, the total electric flux through the square ABCD will be:

\(\phi_{{E}}=\frac{1}{6} \times \phi\)

\(=\frac{1}{6} \times \frac{{q}}{\epsilon_{0}}\)

\(=\frac{1}{6} \times \frac{10^{-5}}{8.854 \times 10^{-12}} \quad\) \((\because \epsilon_0 =8.854 \times 10^{-12})\)

\(=1.88 \times 10^{5} {Nm}^{2} {C}^{-1}\)