Physics Test - 24

Diamagnetic materials are feebly repelled by magnets. \(\mathrm{X}\) is negative and very small. Diamagnetic materials are those that some people generally think of as non-magnetic, and include water, wood, most organic compounds such as petroleum and some plastics, and many metals including copper, particularly the heavy ones with many core electrons, such as mercury, gold and bismuth.

Plane Mirror: 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.
• The laws of reflection are true for both plane mirrors as well as spherical mirrors.
• The plane mirror always forms the virtual and erect images.

We know that:

Time period of SHM of small vertical oscillations in a liquid is given by:

\(T=2 \pi \sqrt{\frac{l}{g}}\), where \(l\) is the length of cube/cylinder dipped in the water.

So according to law of floatation,

weight of the cube \(=\) weight of the water displaced

\(a b c \times d \times g=b c l \times l \times g\)

\(\Rightarrow l=d a\)

\(\Rightarrow T=2 \pi \sqrt{\frac{d a}{g}}\)

Potential applied across the heating element, \(V=230 \mathrm{~V}\) Initial current, \(I_{1}=3.2 \mathrm{~A}\)

After few seconds, steady current attained is, \(I_{2}=2.8 \mathrm{~A}\) Using ohm's law,

\(\mathrm{R}_{1}=\frac{230}{3.2}=71.87 \Omega\)

and, \(\mathrm{R}_{2}=\frac{230}{2.8}=82.14 \Omega\)

\(a=1.7 \times 10^{-4 \circ} \mathrm{C}^{-1} \)

\(\mathrm{t}_{1}=27^{\circ} \mathrm{C}\)

Using the relation, \(\mathrm{R}_{2}=\mathrm{R}_{1}\left[1+\alpha\left(\mathrm{t}_{2}-\mathrm{t}_{1}\right)\right] \)

\(\text { Therefore, } \mathrm{t}_{2}=\frac{\mathrm{R}_{2}-\mathrm{R}_{1}}{\mathrm{R}_{1} \cdot a}+\mathrm{t}_{1} \)

\(\Rightarrow \mathrm{t}_{2}=\frac{82.14-71.87}{71.87 \times 1.7 \times 10^{-4}}+27 \)

\(=840.56+27 \)

\(=867.56^{\circ} \mathrm{C} \)

\(=867^{\circ} \mathrm{C}\)

Let the north side is positive \(y\)-axis \((\hat{j})\) and the east side is positive \(x\)-axis \((\hat{i})\)

So, initial velocity \(v_{i}=5 \hat{i},\) final velocity \(v_{f}=5 \hat{j}, time~t =10\) s

We know that average acceleration \(\vec{a}\) is given as:

\(\vec{a}=\frac{\overrightarrow{\Delta v}}{\Delta t}\)

Where, \(\Delta v\) is change in velocity and \(\Delta t\) is change in time.

\(=\frac{\overrightarrow{v_{f}}-\overrightarrow{v_{i}}}{\Delta t}\)

\(=\frac{5 \hat{j}-5 \hat{i}}{10}\)

\(=\frac{1}{2}(\hat{j}-\hat{i})\)

\(=\frac{1}{\sqrt{2}}\) m/s\(^{2}\)

From its components, it is evident that the direction is north-west making 135 degrees with respect to the east.

As we know that the value of the horizontal component is always zero irrespective of the inclination of the body.

So, the Horizontal component will be,

\(T_{1} \cos 30^{\circ}=100 \cos 60^{\circ} \)

\(\therefore T_{1}=\frac{100 }{ \sqrt{3}} N\)

Vertical Component will be,

\(T_{1} \sin 30^{\circ}+100 \sin 60^{\circ}= W\)

By putting the value of \(T_{1}\) from above we get

\(\frac{100}{\sqrt{3}} \times \frac{1}{2}+100 \times \frac{\sqrt{3}}{2}= W \)

\( \therefore W =\frac{200 }{ \sqrt{3}} N\)

The magnitude of the difference between the individual measurement and true value of the quantity is called absolute error.

The absolute error is calculated by taking the absolute value of the difference between the measured value and the true value. Mathematically, it is expressed as:

Absolute Error = Measured Value − True Value

This ensures that the error is always positive (or zero), regardless of whether the measured value is greater or less than the true value.

The process in which an electron escapes from the metal surface is known as electron emission. This occurs due to various mechanisms such as heating (thermionic emission), exposure to light (photoelectric emission), or application of a strong electric field (field emission). Once freed, these electrons contribute to electrical conductivity or can be manipulated for various technological applications like vacuum tubes or semiconductor devices. 

Given a cup of water

We need to find the charge present in that cup

Let us assume that the cup is of \(250 \mathrm{~g}\) or \(250 \mathrm{~mL}\)

We know that the molecular mass of water is \(18 \mathrm{~g} / \mathrm{mol}\)

Each molecule of water contains \(2+8=10\) electrons and protons.

So, let the number of molecules present in a cup of water \(=\mathrm{n}\)

\(=\frac{6.022 \times 10^{23} \times 250}{18}=8.36 \times 10^{24}\)

Therefore,

Total number of electrons or protons present in a cup of water be \(\mathrm{N}\).

\(=n \times 10=8.36 \times 10^{25}\)

Total negative or positive charge is given by,

\(q=N e\)

\(=8.36 \times 10^{25} \times 1.6 \times 10^{-19} \mathrm{C}\)

\(=1.33 \times 10^{7} \mathrm{C}\)

So, the number of charges present in a \(250 \mathrm{~g}\) cup \(=1.33 \times 10^{7} \mathrm{C}\).

The corpuscular theory was largely developed by Sir Isaac Newton.

It stated that light is made up of small discrete particles called corpuscles which travel in a straight line with a finite velocity and possess impetus.
Newton’s corpuscular theory is based on the following points:

• Light consists of very thin particles called corpuscles.
• These corpuscles after emission travel in a straight line and at a very high velocity.
• When these particles enter the eyes they cause a sensation of vision.
• Corpuscles of different colors have different sizes.