Physics

1.6

In the adjoining figure ABCD is a square with sides AB = BC = CD = DA = 10 cm

Diagonals, AC = BD  = $\sqrt[]{{10}^2+{10}^2}$ = $10\sqrt[]{2}$cm

AO = OC = DO = BO = $5\sqrt[]{2}$ cm

At the centre of the square ABCD, O, a charge of 1 $\mu Cisplaced.$$\text{exception 25:}$

The force of repulsion between the charges placed at A and at O is equal in magnitude but opposite in direction between the charges placed at point C and centre O. Similarly ,the force of attraction between the charges placed at B & O and D & O will be equal in magnitude but opposite in direction. These charges will cancel each other.

Hence, the net charge at centre O will be zero.

Here is a deeper explanation. 

What you just saw is the use of the superposition principle. The total force on the 1 μC charge at the centre (O) is the vector sum of the forces from each of the four corner charges. 

Pairs can break down the analysis. 

• Forces from qA and qC: The force exerted by qA (+2 μC) on the central charge is repulsive. The force from qC (+2 μC) is also repulsive. Now, since qA and qC are equal and are at an equal distance from the centre, the forces they exert are equal in magnitude. But they point in opposite directions along the diagonal AC. So, they cancel each other out.

• Forces from qB and qD: Similarly, the force exerted by qB (–5 μC) is attractive, and the force from qD (–5 μC) is also attractive. Because these charges are equal and equidistant from the centre, the forces are equal in magnitude but opposite in direction along the diagonal BD. That causes them to cancel out as well.

As both pairs of forces nullify each other, the net electrostatic force on the central charge is zero. Read more about Coulomb's Law

3.27

(a) Distance travelled by the particle between t = 0 s and t = 10 s is the area of the triangle = (1/2) x base x height = (1/2) x 10 x 12 = 60 m

The average speed of the particle is 60/10 m/s = 6 m/s

(b) Distance travelled by the particle between t = 2 s and t = 6 s

Let S₁ be the distance travelled by the particle in time 2 to 5 s and S₂ be the distance travelled between 5 to 6s.

For the motion  from 0 to 5 sec, u = 0, t = 5, v = 12 m/s

From the equation v = u + at we get a = (v-u)/t = 12/5 = 2.4 m/s²

Distance covered from 2 to 5 sec, S₁ = distance covered in 5 s – distance covered in 2 s

From the equation s = ut + at² we get,

= (1/2) * 2.4 * 25 – (1/2) * 2.4 * 4 = 30 – 4.8 = 25.2 m

For the motion from 5 sec to 10 sec, u = 12 m/s and a = -2.4 m/s²

and t = 5 sec to t = 6 sec means n = 1 for this motion

Distance covered in the 6th sec is S2 = u + (1/2) a (2n-1) = 12 + (1/2) (-2.4) (2 * 1-1) = 10.8 m

Therefore the total distance covered from t = 2 to 6 s = 25.2 + 10.8 m = 36 m

Initial velocity, u₁ = 15m/s, acceleration, a = -g = -10 m/s

From the relation  s₁=s₀+u₁t+ (1/2)at² where

s₀ = cliff height, s₁ = total height of the fall of the first stone, we get

s₁ = 200 + 15t – 5t²  ………. (1)

When the stone hit the floor, s₁ = 0, so the equation (1) becomes

0 = 200 +15t - 5t² = t² -3t – 40 = (t-8) (t+5) = 0

So t = 8s or -5s

Since the stone was thrown at t=0, so t cannot be –ve. Hence t = 8s

For the second stone,

Initial velocity, u₁ = 30 m/s, acceleration, a = -g = -10 m/s

From the relation  s₂=s₀+u₁t+ (1/2)at² where

s₀ = cliff height, s₂= total height of the fall of the second stone, we get

s₂ = 200 + 30t – 5t²  ………. (2)

When the stone hit the floor, s₂ = 0, so the equation (2) becomes

0 = 200 +30t - 5t² = t² -6t – 40 = (t-10) (t+4) = 0

So t = 10s or -4s

Since the stone was thrown at t=0, so t cannot be –ve. Hence t = 10s

Subtracting equation (1) from equation (2), we get

s_{2 –} s₁ = (200 + 30t – 5t²  ) – (200 + 15t – 5t² )  = 15t ……. (3)

Equation (3) represents the linear trajectory of the two stones because to this linear relation between (s_{2 –} s₁₎ and t, the projection is straight line till 8s (from the graph)

The maximum distance between the two stones is at t = 8s. So

(s_{2 –} s₁) _max = 15 x 8 = 120 m. This value has been depicted correctly in the graph.

After 8s, only the second stone is in motion (t for first stone has been found out to be 8s), the trajectory of second stone is obtained in equation (2), = 200 + 30t – 5t²  , which is a curved path.

3.23 The distance covered by the 3 wheeler on a straight line in the nth second can be expressed as:

Sn = u + a (2n-1)/2 …… (1),

Where

a = acceleration

u = initial velocity

n = time = 1,2,3, ……., n

Given, u = 0, a = 1m/s², from equation (1) we get Sn = (2n-1)/2 ……. (2)

With various values of n, we get Sn

            n                      Sn

            1                      0.5

            2                      1.5

            3                      2.5

            4                      3.5

            5                      4.5

            6                      5.5

            7                      6.5

            8                      7.5

            9                      8.5

            10                    9.5