Uncategorized - Class 9 & Class 10 Resources Center

Pause and Ponder

1. A weightlifter lifts a barbell. List two forces that are acting on the barbell. Are these forces balanced if the weightlifter keeps the barbell steady?

Answer: The two primary forces acting on the barbell are:

The downward gravitational force (weight of the barbell) exerted by the Earth.
The upward muscular force (or normal force) applied by the weightlifter’s hands.

Yes, if the weightlifter keeps the barbell steady (at rest), its acceleration is zero. Therefore, the net force acting on it is zero, meaning these two forces are perfectly balanced.

2. Two players R and S are participating in an arm-wrestling match. At the instant, when the arms tilt to the front direction, are the forces exerted by the players balanced? If not, which player exerted the larger force?

Answer: No, the forces are not balanced at the instant the arms tilt. Because there is motion (an acceleration causing the arms to move from a steady state), a net force is present. The player pushing in the direction of the tilt exerted a force of a larger magnitude.

3. An object is moving with a constant velocity. Is there a net force acting upon it?

Answer: No. According to Newton’s first law of motion, an object moving with a constant velocity has zero acceleration. Therefore, the net force acting upon it must be zero.

4. Suppose, no net force is acting on an object. Which of the following situations are possible?

Object remains at rest if at rest.
Object keeps moving with a constant velocity if already moving.
Object is moving with a constant acceleration.

Answer: Situations (i) and (ii) are possible. Both describe states of zero acceleration, which directly correspond to a condition where no net force acts on the object, as per Newton’s first law of motion.

5. In the real world, it is difficult to find a situation where no forces are acting on an object. But by applying additional forces, a condition can be achieved where the net force on the object is zero. Explain with the help of an example.

Answer: Consider pushing a heavy wooden block across a horizontal floor at a constant velocity. In the real world, a frictional force acts on the block in the direction opposite to its motion. By applying a continuous muscular push that is exactly equal in magnitude and opposite in direction to the frictional force, the two forces cancel each other out. Thus, the net force becomes zero, allowing the block to maintain its constant velocity.

6. A toy car of mass 100 g is moving with a constant velocity of 0.5 m s^-1. What is the net force acting on the toy car?

Answer: The net force is 0 N. Since the toy car is moving with a constant velocity, its acceleration is zero ($a = 0$). According to Newton’s second law ($F = ma$), if acceleration is zero, the net force must also be zero.

7. Two children of different masses are sitting on identical swings. To impart identical initial acceleration, for which child would you require to apply a larger force? Explain why.

Answer: A larger force must be applied to the child with the larger mass. According to Newton’s second law of motion, force is directly proportional to mass for a given acceleration ($F = ma$). Therefore, the heavier child requires a greater force to achieve the same initial acceleration.

8. How are glass items packed for transportation using a bubble wrap or hay protected from damage?

Answer: Bubble wrap or hay acts as a compressible cushion. In the event of sudden jolts or collisions during transportation, these materials compress and increase the time duration over which the glass item’s velocity comes to zero. As per Newton’s second law, a longer time of impact reduces the acceleration, which in turn significantly reduces the impact force exerted on the glass, thereby preventing it from breaking.

9. Why does a fireperson sometimes struggle when holding the pipe issuing water?

Answer: Water rushes out of the hose pipe at a very high velocity, carrying substantial momentum. According to Newton’s third law of motion, the forward force of the issuing water results in an equal and opposite reaction force (recoil) acting backwards on the pipe. The fireperson struggles because they must continuously apply a large muscular force to counteract this strong backward reaction force.

10. Suppose a spacecraft is moving in a region of space where the gravitational force acting upon it is negligible. Suggest how it can change its velocity.

Answer: The spacecraft can change its velocity by firing its thrusters. When the spacecraft expels exhaust gases at high speed in one direction, the gases exert an equal and opposite reaction force on the spacecraft (as per Newton’s third law of motion). This net force accelerates the spacecraft in the opposite direction, thereby changing its velocity.

Revise, Reflect, Refine

1. Using a horizontal force F, a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table?

Answer: The frictional force is precisely F, acting in the direction opposite to the applied force. Because the table moves at a constant velocity, the net force must be zero. Therefore, the applied force perfectly balances the frictional force.

2. For a ball moving on a smooth frictionless surface, choose the appropriate option that will make the following statements physically correct.

If no net force is applied on the ball, the velocity of the ball will (remain the same / increase / decrease).
If a net force is applied on the ball in the direction of its motion, the magnitude of the velocity of the ball will (remain the same / increase / decrease).
If a net force is applied on the ball in a direction opposite to the direction of its motion, the magnitude of the velocity of the ball will (remain the same / increase / decrease).

Answers:

(i) remain the same
(ii) increase
(iii) decrease

3. Two blocks P and Q on a smooth horizontal surface are shown. Two forces of magnitudes 4 N and 5 N are acting in opposite directions on block P, while block Q is moving with a constant velocity. Which of the following statements is correct?

P experiences a net force and Q does not experience a net force.
P does not experience a net force and Q experiences a net force.
Both P and Q experience a net force.
Neither P nor Q experiences a net force.

Answer: (i) P experiences a net force and Q does not experience a net force.

Reasoning: Block P has opposing forces of 5 N and 4 N, resulting in a net force of 1 N. Block Q is moving with a constant velocity, meaning its acceleration is zero, and thus the net force on it is zero.

4. While practising for the snake boat race, 100 oarsmen are rowing a boat together. Out of these, 95 row backwards to propel the boat forward. But by mistake, 5 oarsmen row in the opposite direction. If each oarsman applies a horizontal force of 200 N, what is the net force on the snake boat?

Answer:

Force applied by 95 oarsmen propelling the boat forward = $95 \times 200\text{ N} = 19,000\text{ N}$ (acting forwards).

Force applied by 5 oarsmen propelling the boat backwards = $5 \times 200\text{ N} = 1,000\text{ N}$ (acting backwards).

Net force on the boat = $19,000\text{ N} – 1,000\text{ N} = \mathbf{18,000\text{ N}}$ in the forward direction.

5. When a net force acts on an object, we observe that the object accelerates:

Answer: (iv) in the direction of force, with acceleration proportional to the force acting on the object. (In accordance with Newton’s second law, $\vec{F} = m\vec{a}$).

6. The position-time graph for four objects A, B, C and D moving along a straight line are given. A net force acts on:

Answer: (iii) Object C

Reasoning: Graphs A and D are straight inclined lines, indicating constant non-zero velocity. Graph B is a horizontal line, indicating zero velocity (rest). For A, B, and D, acceleration is zero, so the net force is zero. Graph C is a curve (parabolic), indicating that the velocity is changing over time. Therefore, Object C is accelerating, which means a net force is acting upon it.

7. A sailor jumps out from a small boat to the shore. As the sailor jumps forward, will the boat move? If yes, in which direction and why.

Answer: Yes, the boat will move backwards (away from the shore). When the sailor jumps forward, they apply a backward force on the boat with their feet. According to Newton’s third law of motion, the boat applies an equal and opposite forward force on the sailor. The backward reaction force on the unsecured boat causes it to move backwards.

8. During a high jump event, a landing mat or sand bed is placed for the athlete to fall upon. Explain the reason behind it.

Answer: A landing mat or sand bed is soft and yields under the athlete’s weight. This increases the time duration over which the athlete’s downward velocity reduces to zero. As the time of impact increases, the rate of change of momentum decreases. According to Newton’s second law ($F = \frac{\Delta p}{\Delta t}$), this significantly reduces the impact force exerted on the athlete’s body, preventing severe injuries.

9. A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision:

Answer: (iv) the loaded cart and the empty cart, both exert an equal magnitude of force on each other.

Reasoning: According to Newton’s third law of motion, forces always occur in equal and opposite pairs, regardless of the masses or velocities of the colliding objects.

10. The acceleration-mass graph for the acceleration produced by a force on objects of different masses is plotted. Plot the force-mass graph for this case.

Answer: From the given graph, let us calculate the force at various points using $F = ma$:

At $m = 1\text{ kg}, a = 10\text{ m s}^{-2} \implies F = 1 \times 10 = 10\text{ N}$
At $m = 2\text{ kg}, a = 5\text{ m s}^{-2} \implies F = 2 \times 5 = 10\text{ N}$
At $m = 4\text{ kg}, a = 2.5\text{ m s}^{-2} \implies F = 4 \times 2.5 = 10\text{ N}$
At $m = 5\text{ kg}, a = 2\text{ m s}^{-2} \implies F = 5 \times 2 = 10\text{ N}$

Since the force is constantly 10 N regardless of the mass, the force-mass graph will be a horizontal straight line parallel to the mass-axis (X-axis) intersecting the force-axis (Y-axis) at 10 N.

11. The velocity-time graph of an object of mass 10 kg moving along a straight line is shown. Calculate the force acting on the object by using the graph.

Answer:

From the graph, we can extract the initial and final velocities to find the acceleration.

At $t = 0\text{ s}$, initial velocity $u = 10\text{ m s}^{-1}$

At $t = 8\text{ s}$, final velocity $v = 30\text{ m s}^{-1}$

Acceleration ($a$) = $\frac{v – u}{t} = \frac{30 – 10}{8} = \frac{20}{8} = 2.5\text{ m s}^{-2}$

Mass of the object ($m$) = 10 kg

Force ($F$) = $ma = 10\text{ kg} \times 2.5\text{ m s}^{-2} = \mathbf{25\text{ N}}$

12. A bullet of mass 50 g moving with a speed of 100 m s^-1 enters a heavy stationary wooden block and stops after penetrating a distance of 50 cm. Estimate the stopping force acting on the bullet (assume that the bullet undergoes constant acceleration within the block).

Answer:

Given:

Mass of bullet ($m$) = 50 g = 0.05 kg
Initial velocity ($u$) = 100 m s^-1
Final velocity ($v$) = 0 m s^-1 (since it stops)
Distance penetrated ($s$) = 50 cm = 0.5 m

Calculation:

First, calculate the acceleration using the third equation of motion:

$v^2 – u^2 = 2as$

$(0)^2 – (100)^2 = 2 \times a \times 0.5$

$-10000 = 1 \times a \implies a = -10000\text{ m s}^{-2}$

Next, calculate the force using Newton’s second law:

$F = ma$

$F = 0.05\text{ kg} \times (-10000\text{ m s}^{-2}) = -500\text{ N}$

The negative sign indicates that the force is acting in the direction opposite to the motion. Therefore, the stopping force acting on the bullet is 500 N.

13. An ace footballer converted a penalty shot by kicking the football with a speed of 108 km h^-1. The estimated force they imparted was 800 N. The mass of the football was 0.4 kg. Calculate the time of contact between their foot and the ball.

Answer:

Given:

Initial velocity ($u$) = 0 m s^-1 (ball was initially at rest)
Final velocity ($v$) = 108 km h^-1 = $108 \times \frac{5}{18}\text{ m s}^{-1} = 30\text{ m s}^{-1}$
Force ($F$) = 800 N
Mass ($m$) = 0.4 kg

Calculation:

According to Newton’s second law of motion, $F = m \left( \frac{v – u}{t} \right)$.

Rearranging to solve for time ($t$):

$t = \frac{m(v – u)}{F}$

$t = \frac{0.4 \times (30 – 0)}{800}$

$t = \frac{12}{800}$

$t = 0.015\text{ s}$

The time of contact between the foot and the ball was 0.015 seconds.

14. An object of mass 2 kg moving with a constant velocity of 10 m s^-1 encounters a rough patch where the force of friction on the object is 7 N. At the same time, an additional constant force of 3 N opposing the motion is applied on the object. After entering the rough patch, how much distance does the object travel before coming to rest?

Answer:

Given:

Mass ($m$) = 2 kg
Initial velocity ($u$) = 10 m s^-1
Final velocity ($v$) = 0 m s^-1
Frictional force = 7 N (opposing motion)
Additional force = 3 N (opposing motion)

Calculation:

Total opposing force ($F$) = $-7\text{ N} – 3\text{ N} = -10\text{ N}$ (negative sign indicates it opposes motion).

Calculate acceleration ($a$):

$a = \frac{F}{m} = \frac{-10\text{ N}}{2\text{ kg}} = -5\text{ m s}^{-2}$

Calculate distance ($s$) using the third equation of motion:

$v^2 – u^2 = 2as$

$(0)^2 – (10)^2 = 2 \times (-5) \times s$

$-100 = -10s$

$s = \frac{-100}{-10} = 10\text{ m}$

The object travels a distance of 10 metres before coming to rest.

15. A tractor pulls a harrow of mass m₁ with a net force F resulting in an acceleration of a₁. The same tractor pulls a trolley of mass m₂ with a force F producing an acceleration of a₂. If the tractor now pulls the trolley with the harrow placed on it (with the same force F), then obtain an expression for the resulting acceleration in terms of a₁ and a₂. Ignore friction.

Answer:

From Newton’s second law, $m = \frac{F}{a}$.

For the harrow: $m_1 = \frac{F}{a_1}$

For the trolley: $m_2 = \frac{F}{a_2}$

When pulling both together, the total mass $M = m_1 + m_2$.

Substitute the expressions for mass:

$M = \frac{F}{a_1} + \frac{F}{a_2} = F \left( \frac{1}{a_1} + \frac{1}{a_2} \right) = F \left( \frac{a_1 + a_2}{a_1 a_2} \right)$

The new resulting acceleration ($a$) when the same force $F$ is applied to the combined mass $M$ is:

$a = \frac{F}{M}$

Substituting $M$:

$a = \frac{F}{F \left( \frac{a_1 + a_2}{a_1 a_2} \right)}$

$a = \frac{a_1 a_2}{a_1 + a_2}$

Therefore, the resulting acceleration is $\mathbf{\frac{a_1 a_2}{a_1 + a_2}}$.

16. When the pole of a bar magnet is brought close to a magnetic compass, the bar magnet and the compass needle exert a magnetic force on each other. As per Newton’s third law of motion, both the forces are equal in magnitude and opposite in direction. However, the compass needle moves, whereas the bar magnet does not move. Explain why.

Answer: While the magnetic forces acting on both the bar magnet and the compass needle are indeed equal in magnitude (due to Newton’s third law), their masses are vastly different. The bar magnet has a significantly larger mass compared to the extremely lightweight compass needle.

According to Newton’s second law ($a = \frac{F}{m}$), for the same magnitude of force, acceleration is inversely proportional to mass. Therefore, the acceleration produced in the heavy bar magnet is so minuscule that it is imperceptible, and frictional forces keep it at rest. Conversely, the small mass of the compass needle allows it to experience a prominent acceleration, causing it to move noticeably.

Exercise Set 1.1

(i) From the figure, $D_1$ lies on the x-axis at $x = 8$.

Distance from left wall (y-axis) = 8 ft

Distance from x-axis = 0 ft

(ii)

$$D_1 = (8, 0)$$

(iii)

$$\text{Width of door} = 11.5 – 8 = 3.5 \text{ ft}$$

A standard room door is about 3 ft wide, so 3.5 ft is comfortable. A wheelchair user can enter easily.

(iv)

$$\text{Bathroom door width} = 4 – 1.5 = 2.5 \text{ ft}$$

Since $2.5 < 3.5$, the bathroom door is narrower than the room door.

Exercise Set 1.2

Question 1 — Study table with feet at $(8,9)$, $(11,9)$, $(11,7)$

(i) The table is a rectangle. The fourth foot completes the rectangle:

$$\text{Fourth foot} = (8, 7)$$

(ii) The point $(8,7)$ is in the bedroom, clear of the wardrobe. It is a reasonable spot.

(iii)

$$\text{Width} = 11 – 8 = 3 \text{ ft}$$

$$\text{Length} = 9 – 7 = 2 \text{ ft}$$

Height cannot be found from a 2D floor map.

Question 2 — Bathroom door

The door is 2.5 ft wide. When it swings 90 degrees into the bedroom, the tip reaches up to $x = 2.5$. The wardrobe left edge is at $x = 3$. Since $2.5 < 3$, the door will not hit the wardrobe. If the door is made wider, a sliding door or outward opening door should be used.

Question 3 — Reiaan’s bathroom

(i) Four corners of the bathroom:

$$O = (0, 0), \quad F = (0, 10), \quad R = (-6, 10), \quad P = (-6, 0)$$

(ii) The showering area SHWR is a rectangle. Coordinates:

$$S = (-6, 6), \quad H = (-6, 9), \quad W = (-3, 9), \quad R = (-3, 6)$$

(iii) Washbasin $3 \times 2$ ft, one possible placement:

$$(-3, 8), \quad (-3, 10), \quad (0, 10), \quad (0, 8)$$

Toilet $2 \times 3$ ft, one possible placement:

$$(-6, 0), \quad (-6, 3), \quad (-4, 3), \quad (-4, 0)$$

Question 4 — Other rooms

(i) Dining room corners:

$$P(-6, 0), \quad A(12, 0), \quad (12, -15), \quad (-6, -15)$$

(ii) Centre of dining room:

$$x_c = \frac{-6 + 12}{2} = 3$$

$$y_c = \frac{0 + (-15)}{2} = -7.5$$

Dining table $5 \times 3$ ft centred at $(3, -7.5)$, feet at:

$$(0.5, -6), \quad (5.5, -6), \quad (5.5, -9), \quad (0.5, -9)$$

Section 1.4 — Distance Formula Worked Examples

For $A(3,4)$, $D(7,1)$, $M(9,6)$:

$$AD = \sqrt{(7-3)^2 + (1-4)^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \text{ units}$$

$$DM = \sqrt{(9-7)^2 + (6-1)^2} = \sqrt{4 + 25} = \sqrt{29} \text{ units}$$

$$MA = \sqrt{(3-9)^2 + (4-6)^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10} \text{ units}$$

End of Chapter Exercises

Question 1

The axes intersect at the origin:

$$x\text{-coordinate} = 0, \quad y\text{-coordinate} = 0$$

Question 2

Point $H$ on the vertical line through $W(-5, y)$:

$$H = (-5, y) \text{ for any value of } y$$

If $y > 0$, H is in Quadrant II. If $y < 0$, H is in Quadrant III. If $y = 0$, H is on the x-axis.

Question 3

Points $R(3,0)$, $A(0,-2)$, $M(-5,-2)$, $P(-5,2)$

(i) Side $AM$ is horizontal. Side $MP$ is vertical. Therefore:

$$AM \perp MP$$

(ii) Side $AM$ from $(0,-2)$ to $(-5,-2)$ is parallel to the x-axis.

(iii) $M(-5,-2)$ and $P(-5,2)$ have the same x-coordinate and opposite y-values. They are mirror images in the x-axis.

Question 4

Triangle with $I(0,0)$, $N(5,0)$, $Z(5,-6)$:

$$IZ = \sqrt{(5-0)^2 + (-6-0)^2} = \sqrt{25 + 36} = \sqrt{61} \text{ units}$$

$$ZN = \sqrt{(5-5)^2 + (0+6)^2} = \sqrt{0 + 36} = 6 \text{ units}$$

$$IN = \sqrt{(5-0)^2 + (0-0)^2} = \sqrt{25} = 5 \text{ units}$$

Verification using Pythagoras:

$$IN^2 + ZN^2 = 25 + 36 = 61 = IZ^2$$

Question 5

Without negative numbers, only Quadrant I (where $x \geq 0$ and $y \geq 0$) can be used. Points in Quadrants II, III and IV cannot be located. So the system cannot locate all points on a 2D plane.

Question 6

Check if $M(-3,-4)$, $A(0,0)$, $G(6,8)$ are collinear by comparing slopes:

$$\text{Slope of } MA = \frac{0-(-4)}{0-(-3)} = \frac{4}{3}$$

$$\text{Slope of } AG = \frac{8-0}{6-0} = \frac{8}{6} = \frac{4}{3}$$

Both slopes are equal. The three points are collinear.

Question 7

Check if $R(-5,-1)$, $B(-2,-5)$, $C(4,-12)$ are collinear:

$$\text{Slope of } RB = \frac{-5-(-1)}{-2-(-5)} = \frac{-4}{3}$$

$$\text{Slope of } BC = \frac{-12-(-5)}{4-(-2)} = \frac{-7}{6}$$

$$\frac{-4}{3} \neq \frac{-7}{6}$$

The points are not collinear.

Question 8

(i) Right-angled isosceles triangle using $O(0,0)$, $A(4,0)$, $B(0,4)$:

$$OA = 4, \quad OB = 4, \quad AB = \sqrt{4^2 + 4^2} = 4\sqrt{2}$$

Right angle is at $O$ and $OA = OB$, so this is a right-angled isosceles triangle.

(ii) Using $O(0,0)$, $C(-3,-4)$ in Quadrant III, $D(3,-4)$ in Quadrant IV:

$$OC = \sqrt{9 + 16} = 5, \quad OD = \sqrt{9 + 16} = 5$$

Since $OC = OD$, triangle $OCD$ is isosceles.

Question 9 — Midpoint Table

For $S(-3,0)$, $M(0,0)$, $T(3,0)$:

$$\frac{-3+3}{2} = 0 \quad \text{and} \quad \frac{0+0}{2} = 0$$

M is the midpoint. Yes.

For $S(2,3)$, $M(3,4)$, $T(4,5)$:

$$\frac{2+4}{2} = 3 \quad \text{and} \quad \frac{3+5}{2} = 4$$

M is the midpoint. Yes.

For $S(0,0)$, $M(0,5)$, $T(0,-10)$:

$$\frac{0+(-10)}{2} = -5 \neq 5$$

M is not the midpoint. No.

For $S(-8,7)$, $M(0,-2)$, $T(6,-3)$:

$$\frac{-8+6}{2} = -1 \neq 0$$

M is not the midpoint. No.

The midpoint formula is:

$$M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$

Question 10

$M(-7,1)$ is midpoint of $A(3,-4)$ and $B(x,y)$:

$$\frac{3 + x}{2} = -7 \implies 3 + x = -14 \implies x = -17$$

$$\frac{-4 + y}{2} = 1 \implies -4 + y = 2 \implies y = 6$$

$$B = (-17, 6)$$

Question 11

$A(4,7)$ and $B(16,-2)$. Point $P$ divides $AB$ in ratio $1:2$ from $A$:

$$P = \left(\frac{2 \times 4 + 1 \times 16}{3}, \frac{2 \times 7 + 1 \times (-2)}{3}\right) = \left(\frac{24}{3}, \frac{12}{3}\right) = (8, 4)$$

Point $Q$ divides $AB$ in ratio $2:1$ from $A$:

$$Q = \left(\frac{1 \times 4 + 2 \times 16}{3}, \frac{1 \times 7 + 2 \times (-2)}{3}\right) = \left(\frac{36}{3}, \frac{3}{3}\right) = (12, 1)$$

$$P = (8, 4), \quad Q = (12, 1)$$

Question 12

(i) Distance from origin to each point:

$$OA = \sqrt{1^2 + (-8)^2} = \sqrt{1 + 64} = \sqrt{65}$$

$$OB = \sqrt{(-4)^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}$$

$$OC = \sqrt{(-7)^2 + (-4)^2} = \sqrt{49 + 16} = \sqrt{65}$$

All three equal $\sqrt{65}$, so A, B, C lie on circle $K$ with centre $O$ and:

$$\text{Radius} = \sqrt{65} \approx 8.06 \text{ units}$$

(ii)

$$OD = \sqrt{(-5)^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61}$$

Since $\sqrt{61} < \sqrt{65}$, point $D$ lies inside circle $K$.

$$OE = \sqrt{0^2 + 9^2} = 9$$

Since $9 > \sqrt{65} \approx 8.06$, point $E$ lies outside circle $K$.

Question 13

Midpoints $D(5,1)$, $E(6,5)$, $F(0,3)$ of sides $BC$, $CA$, $AB$:

$$A = E + F – D = (6+0-5, 5+3-1) = (1, 7)$$

$$B = D + F – E = (5+0-6, 1+3-5) = (-1, -1)$$

$$C = D + E – F = (5+6-0, 1+5-3) = (11, 3)$$

$$A = (1, 7), \quad B = (-1, -1), \quad C = (11, 3)$$

Question 14

(i) Draw a $10 \times 10$ grid with scale $1 \text{ cm} = 200 \text{ m}$. The two main roads are the axes. Streets are numbered 1 to 5 on each side in both directions.

(ii)

(a) Street intersection $(4,3)$ — exactly 1 such intersection.

(b) Street intersection $(3,4)$ — exactly 1 such intersection.

Note: $(4,3)$ and $(3,4)$ are different intersections, just as coordinates are ordered pairs.

Question 15

(i) Check Circle A, centre $A(100,150)$, radius $= 80$:

$$100 – 80 = 20 > 0, \quad 100 + 80 = 180 < 800$$

$$150 – 80 = 70 > 0, \quad 150 + 80 = 230 < 600$$

Circle A lies entirely inside the screen.

Check Circle B, centre $B(250,230)$, radius $= 100$:

$$250 – 100 = 150 > 0, \quad 250 + 100 = 350 < 800$$

$$230 – 100 = 130 > 0, \quad 230 + 100 = 330 < 600$$

Circle B lies entirely inside the screen. Neither circle lies outside.

(ii) Distance between centres:

$$AB = \sqrt{(250-100)^2 + (230-150)^2} = \sqrt{150^2 + 80^2} = \sqrt{22500 + 6400} = \sqrt{28900} = 170 \text{ px}$$

Sum of radii $= 80 + 100 = 180$ px

Difference of radii $= 100 – 80 = 20$ px

Since $20 < 170 < 180$, that is $|r_1 – r_2| < AB < r_1 + r_2$, the two circles intersect at two points.

Question 16

Points $A(2,1)$, $B(-1,2)$, $C(-2,-1)$, $D(1,-2)$

Step 1 — Side lengths:

$$AB = \sqrt{(-1-2)^2 + (2-1)^2} = \sqrt{9+1} = \sqrt{10}$$

$$BC = \sqrt{(-2+1)^2 + (-1-2)^2} = \sqrt{1+9} = \sqrt{10}$$

$$CD = \sqrt{(1+2)^2 + (-2+1)^2} = \sqrt{9+1} = \sqrt{10}$$

$$DA = \sqrt{(2-1)^2 + (1+2)^2} = \sqrt{1+9} = \sqrt{10}$$

All four sides are equal.

Step 2 — Diagonals:

$$AC = \sqrt{(-2-2)^2 + (-1-1)^2} = \sqrt{16+4} = \sqrt{20}$$

$$BD = \sqrt{(1+1)^2 + (-2-2)^2} = \sqrt{4+16} = \sqrt{20}$$

Both diagonals are equal.

Step 3 — Midpoints of diagonals:

$$\text{Midpoint of } AC = \left(\frac{2+(-2)}{2}, \frac{1+(-1)}{2}\right) = (0, 0)$$

$$\text{Midpoint of } BD = \left(\frac{-1+1}{2}, \frac{2+(-2)}{2}\right) = (0, 0)$$

The diagonals bisect each other at the origin. Since all sides are equal, diagonals are equal, and they bisect each other, ABCD is a square.

$$\text{Area} = (\sqrt{10})^2 = 10 \text{ square units}$$

Property	Solution	Suspension	Colloid
Nature	Homogeneous	Heterogeneous	Heterogeneous (appears uniform)
Particle Size	< 1 nm	> 1000 nm	1–1000 nm
Visibility	Not visible	Visible	Not visible
Settling	No	Yes	No
Tyndall Effect	No	Yes	Yes

Class 9 & Class 10 Resources Center

Cell Biology Exercises

Chapter 6: How Forces Affect Motion

Pause and Ponder

Revise, Reflect, Refine

Class 10 Question Bank

Empower Free Education

Category: Uncategorized

Pause and Ponder

Revise, Reflect, Refine

Class 10 Question Bank

Exercise Set 1.1

Exercise Set 1.2

Section 1.4 — Distance Formula Worked Examples

End of Chapter Exercises

Chapter 5: Exploring Mixtures and Their Separation

5.1 Classification of Mixtures

Key Activity (Laser Test)

5.2 Solutions

5.2.1 Concentration of Solution

5.2.2 Methods to Express Concentration

1. Mass by Mass (% m/m)

2. Mass by Volume (% m/v)

3. Volume by Volume (% v/v)

5.2.3 Solubility

5.3 Separation of Homogeneous Mixtures

1. Crystallization

2. Distillation

Fractional Distillation

3. Paper Chromatography

5.4 Separation of Heterogeneous Mixtures

1. Separating Funnel

2. Sublimation

3. Suspensions

4. Centrifugation

5. Coagulation

5.4.4 Colloids

5.5 Tyndall Effect

Comparison Table

Important Points