Physics

4.4

(a) The addition of two scalar quantities is meaningful – if they both represent the same physical quantity.

(b) The addition of a vector quantity with a scalar quantity is not meaningful.

(c) Multiplying a scalar with a vector is meaningful.

(d) The multiplication of two scalar quantities is meaningful.

(e) The addition of two vectors is meaningful.

(f) Adding a component of a vector to the same vector is meaningful.

4.3 Impulse is the only vector quantity from the above list. Impulse is the product of force and time. Force is a vector quantity and time is a scalar quantity. The product of 1 vector and 1 scalar is always a vector quantity.

4.2 Work & Current are scalar quantities. Work is the product of Force and displacement. Since the product of two vectors quantities are always scalar. Current is independent of direction, it is only magnitude, hence scalar.

4.1 A scalar quantity depends only on the magnitude – volume, mass, speeds, density, number of moles, angular frequency are all scalar quantities.

A vector quantity depends on magnitude and direction – velocity, acceleration, displacement and angular velocity are all vector quantities.

As per the NCERT Textbooks, Thomson proposed a Atomic Structure of Atom that tells" An atom consists of a positively charged sphere in which the electrons are embedded like the seeds are embedded in watermelon. This model is often compared to a pudding or watermelon with electrons distributed like raisins or plums, also known as “plum pudding model.”

9.5 Actual depth of the bulb in water, $d_1$ = 80 cm = 0.8 m

Refractive index of water, $\mu$ = 1.33

In the given figure, i = angle of incidence, r = angle of refraction = 90 $°$

The light source, bulb (B) is placed at the bottom of the tank.

Since the bulb is a point source, the emergent light can be considered as a circle of radius, with radius R = $\frac{AC}{2}$ = OA = OB

Using Snell's law, we can write the refractive index of water as:

$\mu =\frac{\sin ?r}{\sin ?i}$ = $\frac{\sin ?90°}{\sin ?i}$ = 1.33

i = 48.75 $°$

In Δ OBC, $\tan ?i$ = $\frac{OC}{OB}$

$\tan ?48.75°$ = $\frac{R}{80}$

R = 91.23 cm

Hence, the area of the water surface = $\pi R^2$ = 2.61 $*{10}^4$ ${cm}^2$ = 2.61 $m^2$

The presesnt day theory for structure of Atom is developed through many discoveries and hypothesises. In class 12 Physics, Atom chapter includes development of the sturucture of atom, and theories in the path of development of present theory. Students can check the ordered points below;

1. Thomson's Model of the Atom
2. Rutherford's Nuclear Model
3. Bohr's Model of the Hydrogen Atom
4. De Broglie's Hypothesis
5. Energy Emission Spectrum

These throries have been used to introduced the current theory of structure of atom.

Students who are preparing for the class 12 board exams needs additional practice questions after completing the NCERT Exercises. We have provided practice questions with accurate and atep-by-step solutions for students to better prepare for the board exams. Students can check the below provided link to access our additional practice questions along with previous year questions.

Chapter 12 Atoms NCERT Solutions 

9.3 Actual depth of the needle in water, $h_1$ = 12.5 cm

Apparent depth of the needle in water, $h_2$ = 9.4 cm

Refractive index of water =, it can be calculated as $\mu$ = $\frac{h_1}{h_2}$

So $\mu$ = $\frac{12.5}{9.4}$ = 1.33

So the refractive index of water = 1.33

The water is replaced by a liquid with refractive index, $\mu \text{'}$ = 1.63

From the relation of, ${\mu }^{\text{'}}$ = $\frac{h_1}{h_2^{\text{'}}}$ , where $h_2^{\text{'}}$ is the new apparent depth by microscope, we get

$h_2^{\text{'}}=$ $\frac{h_1}{{\mu }^{\text{'}}}$ = $\frac{12.5}{1.63}$ = 7.67 cm

So to focus again, microscope needs to be moved up by 9.4 – 7.67 cm = 1.73 cm