\(Radius\ of\ curvature=40cm\\ focal\quad length=?\\ R=2f\\ 40=2\times f\\ f=\frac { 40 }{ 2 } =20cm\\ Focal\ length=20cm\)

Generations of Computer:

(i) In atoms, electrons revolve around the nucleus in certain special or permissible orbits known as discrete orbits or shells or energy levels.
(ii) While revolving in these discrete orbits the electrons do not radiate energy.
(iii) The circular orbits are numbered as 1, 2, 3, 4, ... or designated as K, L, M, N, shells. These numbers are referred to as principal quantum numbers (n)
(iv) K shell (n = 1) is closest to the nucleus and is associated with lowest energy. L, M, N, .... etc are the next higher energy levels. As the distance from the nucleus increases the energy of the shells also increases.
(v) The energy of each orbit or shell is a fixed quantity and the energy is quantized.
(vi) As the distance from the nucleus increases, the size of the orbits also increases.
(vii) Maximum number of electrons that can be accommodated in an energy level is given by 2n² where n is the principal quantum number of the orbit.
(viii) When an electron absorbs energy, it jumps from lower energy level to higher energy level.
(ix) When an electron returns from higher energy level to lower energy level, it gives off energy.

Speed of the signal = Speed of the light = 3 x 10⁸ ms^-1
Time taken by the signal to reach the ground = 5 seconds
Speed = \(\frac { Distance }{ time\ taken } \)
3 x 10⁸ ms^-1 =\(\frac { Distance }{ 5s } \)
Distance = 3 x 10⁸ x 5
= 15 x 10⁸ m
Distance of the spaceship from the ground station = 15 x 10⁸ m.

Materials required: Glass trough, sand, flower pot, plugged at the bottom, pea or bean seeds and water.
Procedure:
1. A glass trough is taken and is filled with sand. A flower pot containing water, plugged at the bottom is kept at the centre of the glass trough.
2. Soaked pea or bean seeds are placed around the pot in the sand, what do we observe after 6 or 7 days?
Observation: It will be observed that radicle has grown towards the pot and moisture instead of growing vertically downward.
Inference: It proves that primary root is positively hydrotropic.

Position of image
Object distance u = − 15 cm (to the left of mirror )
Image distance v = ?
Focal length f = − 10 cm (concave mirror) Using mirror formula.
\(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f } \)
\(\frac { 1 }{ v } +\frac { 1 }{ -15 } =\frac { 1 }{ -10 } \)
\(\frac { 1 }{ v } -\frac { 1 }{ 15 } =\frac { -1 }{ 10 } \)
\(\frac { 1 }{ v } =\frac { -1 }{ 10 } +\frac { 1 }{ 15 } \)
\(\frac { 1 }{ v } =\frac { -1 }{ 30 } \)
∴ Image distance v = -30cm (negative sign indicates that the image is on the left side of the mirror)
∴ Position of image is 30 cm in front of the mirror
2.Nature of image: Since the image is in front of the mirror it is real and inverted.
3.Size of image: To find the size of the image, we have to calculate the magnification
\(m=\frac { -v }{ u } \)
Object distance u = − 15cm
Image distance v = − 30 cm 
\(m=\frac { -(-30) }{ (-15) } \)
m = − 2
We know that,\(m=\frac { { h }_{ 2 } }{ { h }_{ 1 } } \)
Here, height of the object h₁ = 1 cm
-2 =\(\frac { { h }_{ 2 } }{ 1 } \)
h₂ = -2\(\times\) 1
= -2 cm
The height of image is 2 cm (negative sign shows that the image is formed below the principal axis).

9.8 = 1 kg mass
98 N = ?
\(\text{Mass of the object }=\frac{98\times1}{9.8}=10\ kg\)