Q.4.6 Establish the following vector inequalities geometrically or otherwise:

(a) |a+b| < |a| + |b|

(b) |a+b| > ||a| −|b||

(c) |a−−b| < |a| + |b|

(d) |a−−b| > ||a| − |b||

velocity of a freely falling body is v= $\sqrt[]{2gh}$

And $?=\frac{h}{mv}=\frac{h}{m\sqrt[]{2gh}}$

$?=h$ -1

 The wavelength of a photon needed to remove a proton from a nucleus which is bound to the nucleus with 1 MeV energy is nearly
a) 1.2 nm (b) 1.2 x 10-3 nm
(c) 1.2 x 10-6  nm (d). 1.2 x 10 nm

8.12

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

Orbital radius, r = $1.5\times {10}^{11}$ 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( ${\frac{r-x}{x})}^2$ 

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

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

x = $\frac{1.5\times {10}^{11}}{578.35}$ = 2.59 $\times {10}^8$ 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.