Exam Preparation | Class 11 NEB

DM

State the principle of homogeneity.

Solution

The principle of homogeneity states that, 'the dimensions of all term in an equation must be identical.i.e. dimension in L.H.S. of an equation is equal to the dimension in R.H.S. of an equation.'

DM

A student writes the formula for speed of longitudinal waves as \( v=\sqrt{\frac{\gamma P}{D}} \) where P is the pressure, D is density and \(\gamma\) is the ratio of specific heats which is dimensionless. Check the correctness of the formula.

Solution

The given equation is \[ v=\sqrt{\frac{\gamma P}{D}} \].

The dimensional formula of each of the quantity here are:
Dimensional formula of velocity i.e., Dimension of L.H.S. = \([LT^{-1}]\)
Dimensional formula of pressure = \(\frac{[MLT^{-2}]}{[L^2]}\) = \([ML^{-1}T^{-2}]\) \(\because P=\frac{F}{A}\)
Dimensional formula of density = \([ML^{-3}]\)
Checking whether principle of homogeneity follows or not,
\[Dimension \hspace{0.5cm} of \hspace{0.5cm} L.H.S. = Dimension \hspace{0.5cm} of \hspace{0.5cm} R.H.S.\] \[\begin{align} [LT^{-1}]&=(\frac{[ML^{-1}T^{-2}]}{[ML^{-3}]})^{\frac{1}{2}}\\ [LT^{-1}]&=[LT^{-1}]\\ \end{align}\]

As the principle of homogeneity holds true here, the given equation is dimensionally correct . However, the dimensional correctness is only necessary condition but not the sufficient condition for a physical equation to be valid. Thus, we cannot conclude if the formula is correct or not physically.

DM

Taking force, length and time to be fundamental quantities, find the dimensional formula of density.

Solution

We have, \[ \begin{align} \text{density} &= \frac{\text{mass}}{\text{volume}} \\ &= \frac{\text{Force}}{\text{acceleration} \times \text{volume}} \quad \left(\because F = ma \implies m = \frac{F}{a}\right) \\ &= \frac{[F]}{[LT^{-2}][L^3]} \\ &= [FL^{-4}T^2] \end{align} \]

Therefore, the dimensional formula of density is \([FL^{-4}T^2]\).

VA

State triangle law of vector addition. Use this law to derive an expression for the magnitude and direction of the resultant of two vectors.

Solution

Statement:
Triangle law of vector addition states that, "If two vectors are represented both in magnitude and direction by two adjacent sides of a triangle taken in the same order, then their resultant is completely represented both in magnitude and direction by the third side of the triangle taken in the opposite order."

Derivation:
Consider two vectors \(\vec{A}\) and \(\vec{B}\) inclined at an angle \(\theta\). Let these vectors be represented by sides \(OP\) and \(PQ\) of a triangle \(OPQ\) respectively. The third side \(OQ\) represents the resultant vector \(\vec{R}\), making an angle \(\phi\) with \(\vec{A}\).

\[\vec{R} = \vec{A} + \vec{B}\]

To find the magnitude \(R\), extend \(OP\) to \(S\) such that \(QS \perp OS\). In right-angled triangle \(\Delta OQS\):

\[ \begin{align} (OQ)^2 &= (OS)^2 + (QS)^2 \\ R^2 &= (OP + PS)^2 + (QS)^2 \\ R^2 &= (A + PS)^2 + (QS)^2 \end{align} \]

From right-angled triangle \(\Delta PQS\):

\[\sin\theta = \frac{QS}{PQ} \implies QS = B\sin\theta\] \[\cos\theta = \frac{PS}{PQ} \implies PS = B\cos\theta\]

Substituting the values of \(PS\) and \(QS\):

\[ \begin{align} R^2 &= (A + B\cos\theta)^2 + (B\sin\theta)^2 \\ R^2 &= A^2 + 2AB\cos\theta + B^2\cos^2\theta + B^2\sin^2\theta \\ R^2 &= A^2 + 2AB\cos\theta + B^2(\cos^2\theta + \sin^2\theta) \\ \therefore R &= \sqrt{A^2 + B^2 + 2AB\cos\theta} \end{align} \]

To find the direction \(\phi\) of \(\vec{R}\), from right-angled triangle \(\Delta OQS\):

\[ \begin{align} \tan\phi &= \frac{QS}{OS} = \frac{B\sin\theta}{A + B\cos\theta} \\ \therefore \phi &= \tan^{-1}\left(\frac{B\sin\theta}{A + B\cos\theta}\right) \end{align} \]

VA

Two vectors have equal magnitudes and their resultant also has the same magnitude. What is the angle between the vectors?

Solution

Let \(\vec{A}\) and \(\vec{B}\) be two vectors and \(\vec{R}\) be their resultant such that,

\[\vec{R}=\vec{A}+\vec{B}\] Given,
\[|\vec{A}|=|\vec{B}|=A (let)\] \[|\vec{R}|=A\] From the vector law of addition,
\[\begin{align} R&=\sqrt{A^2 + B^2 + 2AB cos\theta}\\ A^2&=2A^2 + 2A^2 cos\theta\\ cos\theta&=-\frac{1}{2}\\ \theta&=120^\circ\\ \end{align}\]

Thus, the angle between the given two vectors is \(120^\circ\).

VQ

What are necessary conditions for a physical quantity to be a vector quantity?

Solution

The necessary condition for a physical quantity to be a vector quantity are as follows:

It must have magnitude.

It must have direction.

It must obey the vector laws of addition/subtraction.

RE

Two forces of 30 N and 40 N are inclined to each other at an angle of \(60^\circ\). What is their resultant?

Solution

Given,

First force (\(F_1\)) = \(30\text{ N}\)
Second force (\(F_2\)) = \(40\text{ N}\)
Angle between them (\(\theta\)) = \(60^\circ\)
Resultant force (\(R\)) = ?

From the parallelogram law of vector addition, \[ \begin{align} R &= \sqrt{F_1^2 + F_2^2 + 2F_1 F_2 \cos\theta} \\ &= \sqrt{30^2 + 40^2 + 2 \times 30 \times 40 \times \cos 60^\circ} \\ &= \sqrt{900 + 1600 + 2400 \times \frac{1}{2}} \\ &= \sqrt{2500 + 1200} \\ &= \sqrt{3700} \\ &\approx 60.83\text{ N} \end{align} \]

Direction of the resultant:
If \(\phi\) is the angle made by the resultant \(R\) with the \(30\text{ N}\) force (\(F_1\)):

\[ \begin{align} \tan\phi &= \frac{F_2 \sin\theta}{F_1 + F_2 \cos\theta} \\ &= \frac{40 \sin 60^\circ}{30 + 40 \cos 60^\circ} \\ &= \frac{40 \times \frac{\sqrt{3}}{2}}{30 + 40 \times \frac{1}{2}} \\ &= \frac{20\sqrt{3}}{30 + 20} = \frac{34.64}{50} = 0.6928 \\ \phi &= \tan^{-1}(0.6928) \approx 34.72^\circ \end{align} \]

Therefore, the magnitude of the resultant force is \(60.83\text{ N}\) acting at an angle of \(34.72^\circ\) with the \(30\text{ N}\) force.

DM

Show that work done and kinetic energy have same dimensions.

Solution

1. Dimensional formula of Work Done:
We know that,

\[\text{Work Done} = \text{Force} \times \text{Displacement}\]

Substituting the dimensional formulas for force \([MLT^{-2}]\) and displacement \([L]\):

\[ \begin{align} [\text{Work Done}] &= [MLT^{-2}][L] \\ &= [M^1 L^2 T^{-2}] \quad \text{--- (i)} \end{align} \]

2. Dimensional formula of Kinetic Energy:
We know that,

\[\text{Kinetic Energy} = \frac{1}{2} m v^2\]

Since numerical constants (like \(\frac{1}{2}\)) are dimensionless, substituting the dimensions for mass \([M]\) and velocity \([LT^{-1}]\):

\[ \begin{align} [\text{Kinetic Energy}] &= [M][LT^{-1}]^2 \\ &= [M][L^2 T^{-2}] \\ &= [M^1 L^2 T^{-2}] \quad \text{--- (ii)} \end{align} \]

From equations (i) and (ii), the dimensional formula for both work done and kinetic energy is \([M^1 L^2 T^{-2}]\).

Hence, work done and kinetic energy have the same dimensions.

EC

How can you charge a body negatively by induction? Explain.

Solution

Charging Negatively by Induction

A neutral conducting sphere can be given a net negative charge by bringing a positively charged rod near it and grounding the sphere before removing the rod.

Four steps of charging negatively by induction

1. Step 1: Bring a positively charged glass rod close to an isolated neutral metal sphere (without touching). Free electrons in the sphere are attracted toward the near side, leaving the far side positively charged.
2. Step 2: Connect the far side of the sphere to the ground via a conducting wire. Free electrons flow up from the earth to neutralize the positive charges on the far side.
3. Step 3: Disconnect the ground wire while keeping the positively charged rod in position. The attracted negative charges remain bound on the near side.
4. Step 4: Remove the positively charged rod. The excess electrons redistribute uniformly over the entire surface of the sphere, giving it a net negative charge.

QN

Explain the meaning of the quantization of charge.

Solution

Quantization of charge means that electric charge is always composed of integral multiples of a certain minimum, elementary packet of a charge.

Any net charge \(q\) can be written as,

\[q=ne\] where, \(n=\pm1, \pm2,...\) and \(e\) is the elementary charge unit which is of value \(1.6 \times 10^{-19}\) C.

FR

Does charging by friction follow conservation of charge?

Solution

Yes, friction results in transfer of charge between the bodies which are rubbed against each other. The electrons lost by one body equals the electrons gained by the other and hence the conservation of charge is followed.

TM

Define temperature on the basis of Zeroth law of thermodynamics.

Solution

As per the Zeroth law, temperature is the unique thermal property of a system which determines whether or not the system is in thermal equilibrium with another system.

TD

If two bodies are in thermal equilibrium in one frame, will they be in the thermal equilibrium in all frames? Explain.

Solution

The zeroth law is defined independent of reference frame and holds true in every frame. The temperature of bodies remain equal between the bodies in any frame and hence they remain in thermal equilibrium.

TM

The steam point and ice point of a mercury thermometer are marked as 80⁰ and 20⁰ respectively. What will be the temperature in centigrade scale when this thermometer read 32⁰?

Solution

For an arbitrary thermometer,
upper fixed point (UFP) = 80 ⁰
lower fixed point (LFP) = 20 ⁰
reading (X) = 32 ⁰
For centigrade thermometer,
upper fixed point (UFP) = 100 ⁰ C
lower fixed point (LFP) = 0 ⁰ C
reading (C)=?
\[\begin{align} \frac{C-LFP}{UFP-LFP}&=\frac{X-LFP}{UFP-LFP}\\ \frac{C-0}{100-0}&=\frac{32-20}{80-20}\\ \therefore C&=20^\circ \hspace{0.1cm}C\\ \end{align}\] So, the centigrade thermometer reading is \(20^\circ\) C.

TM

Write short note about absolute scale of temperature.

Solution

The absolute scale of temperature is the thermometric scale where 0 K represents the absolute zero. Absolute zero is the theoretical state at which the molecular motion ceases (K.E. = 0) and a system possess minimum possible internal energy.

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