Class 11th

2.3 

1 Calorie = 4.2 J = 4.2 kgm²s^-2

Standard formula for conversion

Given unit / New unit = (M1/M2)^x (L1/L2)^y (T1/T2)^z

Formula for energy = M¹L²T^-2

Here x = 1, y = 2, z = -2

In this problem

M1 = 1 kg, L1 = 1 m, T1 = 1 s and

M2 =? kg, L2 =? m, T2 =? s

So 1 Calorie = 4.2 (1/? )¹ (1/? )² (1/? )^-2

= 4.2? ^-1? ^-2? ²

Ans.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.

8.12

Mass of the Sun, $M_s$ = $2*{10}^{30}$ kg, Mass of the Earth, $M_e$ = $6*{10}^{24}$ kg

Orbital radius, r = $1.5*{10}^{11}$ m

Let the mass of the rocket be, m

Let x be the distance from the centre of the Earth where the gravitational force acting on satellite P becomes zero.

From Newton's law of gravitation, we can equate gravitational forces acting on the satellite P under the influence of the Sun and the Earth as:

$\frac{G_m{M}_s}{({r-x)}^2}$ = $G_m\frac{M_e}{x^2}$ or( ${\frac{r-x}{x})}^2$ 

= $\frac{M_s}{M_e}$( ${\frac{r-x}{x})}^2$ = $\frac{{2*10}^{30}}{{6*10}^{24}}$ ,

 $\frac{r-x}{x}$ = 577.35 , r = 578.35x ,

x = $\frac{1.5*{10}^{11}}{578.35}$ = 2.59 $*{10}^8$ m

Ans.8.1

(a) No. Unlike electrical forces, gravitational force is independent of the status of the objects.

(b) Yes, the size of the space station is large enough and the astronaut will detect the change in Earth's gravity.

(c) Tidal effect depends inversely upon the cube of the distance while gravitational force depends inversely on the square of the distance. Since the distance between the Moon and the Earth is smaller than the distance between the Sun and the Earth, the tidal effect of the Moon's pull is greater than the tidal effect of the Sun's pull.

The differences between old and modern periodic table are as follows: