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(i) In a triangle, exterior angle and adjacent internal angle make a linear pair.

x + 110° = 180°
⇒ x = 180° – 110° = 70°

In a triangle, sum of all angles is 180°.

∴ x + y + 60° = 180°
⇒ 70° + y + 60° = 180°
⇒ y + 130° = 180°
⇒ y = 180° – 130° = 50°
∴ x = 70°, y = 50°

(ii) In a triangle, exterior angle and adjacent internal angle make a linear pair.

x = 45° + 30° = 75°

∵ In a triangle, sum of all angles is 180°.

∴ y + 45° + 30° = 180°
⇒ y + 75° = 180°
∴ y = 180° – 75° = 105°
∴ x = 75°,y= 105°

(iii) In a triangle, exterior angle and adjacent internal angle make a linear pair.

∴ y + 80° = 180°
⇒ y = 180° – 80° = 100°

In a triangle, sum of all angles is 180°

∴ x + x + y = 180°
⇒ 2x + 100° = 180°
⇒ 2x = 180° – 100°
⇒ 2x = 80°
∴ x = 80°/2 = 40°
∴ x = 40°, y = 100°

(iv) Here x = 70° (vertically opposite angle)

∵ Sum of three angles of a triangle = 180°

∴ x + y + 40° = 180°
⇒ 70° + y + 40° = 180°
⇒ 110° + y = 180°
∴ y = 180° – 110° = 70°
∴ x = 70°,y = 70°

In any triangle exterior angle is always equal to the sum of two interior angles.

(i) x = 80° + 50° = 130°

(ii) x + 35° = 110°
⇒ x = 110° – 35° = 75°

(iii) x = 30° + 35° = 65°

(iv) x + 70° = 120°

⇒ x = 120° – 70° = 50°

(v) x + 35° = 65°

⇒ 65° – 35° = 30°

(vi) x = 55° + 40° = 95°

According to the Pythagoras theorem, we have

(Hypotenuse)² = (Base)² + (Height)²

Let c be hypotenuse and a and b be other two legs of right angled triangle

Then we have

c² = a² + b²

c² = 3² + 4²

c² = 9 + 16 = 25

c = 5 cm

(i) The greatest side is hypotenuse = 6.5 cm

(2.5)² + (6)² = (6.5)²

(According to pythagoras property)

6.25 + 36 = 42.25

42.25 = 42.25

∴ This is a right angled triangle. The given length can be the sides of a right angled triangle. The right angle between the lengths of 2.5 cm and 6 cm.

(ii) The greatest side is 5 cm

(2)² + (2)² = (5)² (According to pythagoras property)

4 + 4 ≠ 25

∴ It is not the lengths of right angled triangle.

(iii) The greatest side is 2.5 cms.

According to pythagoras property

(2.5)² = (1.5)² + (2)²

6.25 = 2.25 + 4

6.25 = 6.25

It is a right angled trianges, the given lengths are sides of the right angled triangles.

∴ The right angle between the lengths of 1.5 cms and 2 cms.

(i) ∠A + ∠B + ∠C = 180°

x° + 50° + 60° = 180° 

(By Angle sum property of a triangle)

∠x + 110° = 180°

∠x = 180° – 110°

∠x = 70°

(ii) ∠P + ∠Q + ∠R = 180°

90° + 30° + ∠x = 180° 

(By Angle sum property of a triangle)

120° + ∠x = 180°

∠x = 180° – 120° = 60°

(iii) ∠X + ∠Y + ∠Z = 180°

30° + 110° + ∠x = 180° 

(By Angle sum property of a triangle)

140° + ∠x = 180°

∴ ∠x = 180° – 140° = 40°

(iv) 50° + x + x = 180°

50° + 2x = 180°

(By Angle sum property of a triangle)

2x = 180° – 50° = 130°

x = \(\frac{180^o}{2}\)

∴ ∠x = 60°

(v) x + x + x = 180°

(By Angle sum property of a triangle)

3x = 180°

x = \(\frac{180^o}{2}\)

∴ x = 60°

(vi) x + 2x + 90° = 180° 

(ByAngle sum property of a triangle)

3x = 180°

x = \(\frac{180^o}{3} = 60^o\)

According to the Pythagoras theorem, we have

(Hypotenuse)² = (Base)² + (Height)²

Given that the two legs of a right triangle are equal and the square of the hypotenuse, which is 50

Let the length of each leg of the given triangle be x units.

Using the Pythagoras theorem, we get

x² + x² = (Hypotenuse)²

x² + x² = 50

2x² = 50

x² = 25

x = 5

Hence, the length of each leg is 5 units.

(i) In △APB, AP < AB + BP because the sum of any two sides of a triangle is greater than the third side.

(ii) In △APC, AP < AC + PC because the sum of any two sides of a triangle is greater than the third side.

(iii) AP < ½ (AB + AC + BC)

In △ABP and △ACP, we can write as

AP < AB + BP… (i) (Because the sum of any two sides of a triangle is greater than the third side)

AP < AC + PC … (ii) (Because the sum of any two sides of a triangle is greater than the third side)

On adding (i) and (ii), we have:

AP + AP < AB + BP + AC + PC

2AP < AB + AC + BC (BC = BP + PC)

AP < (AB – AC + BC)

(i) The condition for Pythagorean triplet is the square of the largest side is equal to the sum of the squares of the other two sides.

37² =1369

12² + 35² = 144 + 1225 = 1369

12² + 35² = 37²

Yes, they represent a Pythagorean triplet.

(ii) The condition for Pythagorean triplet is the square of the largest side is equal to the sum of the squares of the other two sides.

25² = 625

7² + 24² = 49 + 576 = 625

7² + 24² = 25²

Yes, they represent a Pythagorean triplet.

(iii) The condition for Pythagorean triplet is the square of the largest side is equal to the sum of the squares of the other two sides.

45² = 2025

27² + 36² = 729 + 1296 = 2025

27² + 36² = 45²

Yes, they represent a Pythagorean triplet.

(iv) The condition for Pythagorean triplet is the square of the largest side is equal to the sum of the squares of the other two sides.

39² = 1521

15² + 36² = 225 + 1296 = 1521

15² + 36² = 39²

Yes, they represent a Pythagorean triplet.

We know that the sum of all angles of a triangle is 180°

Therefore, for the given △FCB, we have

∠FCB + ∠CBF + ∠BFC = 180°

50° + ∠CBF + 90° = 180°

∠CBF = 180° – 50° – 90° = 40°

Using the above steps for △ABD, we can say that:

∠ABD + ∠BDA + ∠BAD = 180°

∠BAD = 180° – 90° – 40° = 50°

(i) False

Explanation:

We know that the sum of any two sides of a triangle is greater than the third side, it is not true for the given triangle.

(ii) True

Explanation:

We know that the sum of any two sides of a triangle is greater than the third side, it is true for the given triangle.

(iii) False

Explanation:

We know that the sum of any two sides of a triangle is greater than the third side, it is not true for the given triangle.

(i) No, because if there are two right angles in a triangle, then the third angle of the triangle must be zero, which is not possible.

(ii) No, because as we know that the sum of all three angles of a triangle is always 180°. If there are two obtuse angles, then their sum will be more than 180°, which is not possible in case of a triangle.

(iii) Yes, in right triangles and acute triangles, it is possible to have two acute angles.

(iv) No, because if each angle is less than 60°, then the sum of all three angles will be less than 180°, which is not possible in case of a triangle.

(b) 78^o

Because,

Consider the ΔABC,

We know that the exterior angle of a triangle is equal to the sum of its interior opposite angles.

∴ ∠ABC + ∠BAC = ∠ACD

= ∠ABC + 54^o = 132^o

= ∠ABC = 132^o – 54^o

= ∠ABC = 78^o

We know that the sum of any two sides of a triangle is always greater than the third side, in △OAB, we have,

OA + OB > AB ….. (i)

OB + OC > BC …… (ii)

OA + OC > CA ….. (iii)

On adding equations (i), (ii) and (iii) we get:

OA + OB + OB + OC + OA + OC > AB + BC + CA

2(OA + OB + OC) > AB + BC + CA

OA + OB + OC > (AB + BC + CA)/2

Or

OA + OB + OC > ½ (AB + BC +CA)

Hence the proof.

We know that the sum of the angles of a triangle is 180^o.

∴ ∠A + ∠B + ∠C = 180^o

= 72^o+ 63^o+ ∠C = 180^o

= 135^o + ∠C = 180^o

= ∠C = 180^o– 135⁰

= ∠C = 45^o

Hence, the measures of ∠C is 45^o.

In the given triangle, the angles are in the ratio 3: 2: 1.

Let the angles of the triangle be 3x, 2x and x.

We know that sum of angles in a triangle is 180°

3x + 2x + x = 180°

6x = 180°

x = 30°

Also, ∠ACB + ∠ACE + ∠ECD = 180°

x + 90° + ∠ECD = 180° (∠ACE = 90°)

We know that x = 30°

Therefore

∠ECD = 60°

The Pythagoras Theorem:

In a right triangle, the square of the hypotenuse is always equal to the sum of the squares of the other two sides.

Converse of the Pythagoras Theorem:

If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle, with the angle opposite to the first side as right angle.

A) Both ‘A’ and ‘R’ are correct and ‘A’ is supported by ‘R’.

We know that the smallest side is always opposite to the smallest angle, which in this case is 30°, it is AC.

Also, because the largest side is always opposite to the largest angle, which in this case is 100°, it is BC.

Speed of the water = 20 km/h = 20000 meter/60 min

\(\therefore\) Length of water covered in 20 min

= speed x time

= \(\frac{20000}{60}\times20=\frac{20000}3\) meter

Radius of base of the pipe is r = 3/2 meter

\(\therefore\) Volume of water flow = Area of cross section x Length of water flow

\(=\pi\times(3/2)^2\times\frac{20000}3\) = 15000 π m³

Standing under depth = 16 cm = \(\frac{16}{100}\) m

\(\therefore\) Irrigation area = \(\frac{Volume\,of\,water\,flow}{Standing\,water\,depth}\)

\(=\cfrac{15000\pi}{\frac{16}{100}}\)\(=\cfrac{15000\times3.14}{\frac{4}{25}}\) 

 = 294375 m²

= 29.43 hactares

i. In thermonuclear fusion experiments, scientists come across extremely hot gases or plasma where the temperature is of the order of millions of degree celsius.

ii. At such high temperatures, molecules of glowing gas are moving away and towards the observer with high speeds.

iii. Due to Doppler effect, the wavelength λ of a particular spectral line is apparently changed.

iv. One edge of the line now corresponds to an apparently increased wavelength λ due to molecules moving directly towards the observer and the other edge to an apparently decreased wavelength λ₂ due to molecules moving directly away from the observer.

v. The line is thus observed to be broadened. The breadth of the line can be measured by using a diffraction grating.

vi. Since ‘λ’ and ‘c’ are known, the velocity ‘v’ can be calculated using the formula, v = \(\sqrt{\cfrac{3RT}M}\), where ‘R’ is the molar gas constant, ‘T’ is absolute temperature and M is the mass of one mole.

Note: Doppler effect in light is symmetric, i.e., it depends only on the relative velocity of the source and the observer. The difference occurs because light does not require a medium for propagation and the speed of light is same for any observer whether the observer and/or the source is moving.

i. Doppler effect of light is used to determine the radial velocities of distant galaxies.

ii. It is used to measure the speed of rotation of the sun.

a. The east and west edges of the sun are photographed. Each contains absorption lines due to elements such as iron vaporised in the sun and also some absorption lines due to oxygen in the earth’s atmosphere.

b. When the two photographs are put together so that the oxygen lines coincide, the iron lines in the two photographs are displaced relative to each other.

c. In one case, the edge of the sun approaches the earth and in the other, the opposite edge recedes from the earth. Measurements show a rotational speed of nearly 2 km/s.

Correct option is (B) moving towards earth

विभव का शिखर मान V₀ = V_rms √2 = 200√2 वोल्ट.

धारा का शिखर मान i₀ = i_rms √2 = 8√2 ऐम्पियर

(iii) 4√2 ऐम्पियर

F = qvB sin θ = qvB sin 30°

F = qvB/2  [∵ sin 30° = 1/2]

दिया है, B = 3000 G = 0.3 T, a = 0.1 m, b = 0.05 m, i = 12 A 

कुंडली का क्षेत्रफल A = ab = 0.1 m x 0.05 m = 5 x 10^-3 m (a), (b), (c), (d) प्रत्येक दशा में कुंडली के तल पर अभिलम्ब, चुम्बकीय क्षेत्र के लम्बवत् है; अतः प्रत्येक दशा में बल-युग्म का आघूर्ण τ = iAB sin 90° = 12 x 5 x 10^-3 x 0.3 = 1.8 x 10^-2 N-m 

प्रत्येक दशा में बल शून्य है, क्योंकि एकसमान चुम्बकीय क्षेत्र में रखे धारालूप पर बल-युग्म कार्य करता है परन्तु बल नहीं। 

(a) τ = 1.8 x 10^-2 N-m ऋण y-अक्ष की दिशा में तथा बल शून्य है। 

(b) τ = 1.8 x 10^-2 N-m ऋण y-अक्ष की दिशा में तथा बल शून्य है। 

(c) τ = 1.8 x 10^-2 N-m ऋण x-अक्ष की दिशा में तथा बल शून्य है। 

(d) τ = 1.8 x 10^-2 N-m तथा बल शून्य है। 

(e) तथा (f) दोनों स्थितियों में कुंडली के तल पर अभिलम्ब चुम्बकीय क्षेत्र के अनुदिश है; अत: t = iAB sin 0° = 0 

अत: इन दोनों दशाओं में बल-आघूर्ण व बल दोनों शून्य हैं। यह स्थितियाँ सन्तुलन की स्थायी अवस्था में दर्शाती हैं।

बल-युग्म के आघूर्ण का परिमाण τ = NIAB sin θ 

यहाँ फेरों की संख्या N = 20; वर्गाकार कुण्डली के तल को क्षेत्रफल 

A = भुजा² = (0.10 मी)² = 0.01 मी 

कुण्डली में धारा I = 12 A; चुम्बकीय क्षेत्र B = 0.80 T तथा θ = 30° 

τ = 20 x 12 x 0.01 x 0.80 x sin 30° न्यूटन मीटर

= 240 x 0.008 x 1/(2) न्यूटन मीटर

= 0.960 न्यूटन मीटर।

परिनालिका के अन्दर उसकी अक्ष पर चुम्बकीय क्षेत्र B = 0.27 T (जिसकी दिशा अक्ष के अनुदिश ही होती है)। धारावाही तार अक्ष के लम्बवत् है, 

अतः θ = 90°; तार की लम्बाई L = 3.0 सेमी = 3.0 x 10^-2 मी; तार में धारा I = 10 A; अतः तार पर लगने वाला चुम्बकीय बल 

F = ILB sin θ न्यूटन 

= 10 x (3.0 x 10^-2) (0.27) x sin 90° न्यूटन 

= 81 x 10^-2 x 1 न्यूटन 

= 8.1 x 10^-2 न्यूटन

(i) p. = q x 2l = 3.2 x 10^-19 कूलॉम x 2.4 x 10^-10 मीटर = 7.68 x 10^-29 कूलॉम-मीटर 

(ii) वैद्युत-द्विध्रुव को साम्यावस्था अर्थात् वैद्युत-क्षेत्र E की दिशा से θ कोण घुमाने में आवश्यक कार्य W = pE (1 – cos θ) यहाँ θ = 180° एवं p = 7.68 x 10^-29 कूलॉम-मीटर 

E = 4.0 x 10⁵ वोल्ट/मीटर। 

W = (7.68 x 10^-29) x (4.0 x 10⁵) x (1 – cos 180°) 

= (7.68 x 10^-29) x (4.0 x 10⁵) x 2 (∵ cos 180° = – 1) 

= 6.144 x 10^-28 जूल 

(iii) वैद्युत-क्षेत्र E में द्विध्रुव को साम्यावस्था से से कोण पर रखे होने पर इसकी स्थितिज ऊर्जा U = – pE cos θ 

जहाँ θ = द्विध्रुव की अक्ष तथा इसकी साम्यावस्था अर्थात् वैद्युत-क्षेत्र की दिशा के बीच स्थित कोण साम्यावस्था में θ = 0°, 

अत: U₀ = – pE (cos 0° = 1) 

यहाँ p = 7.68 x 10^-29 कूलॉम-मीटर 

E = 4.0 x 10⁵ वोल्ट/मीटर 

U₀ = – (7.68 x 10^-29) x (4.0 x 10⁵) जूल = -3.072 x 10^-23 जूल

1. Non contact force.

2. metre/second.

3. Swinging of a pendulum.

1. Linear

2. Non Contact Force 

3. rotatory 

4. uniform

1. c 

2. d 

3. b 

4. a 

5. e

(a) 20^o

Given, a = 40^o

Then, 2a = 2 × 40 = 80^o

From the figure, angel formed on the straight line are equal to 180^o,

Then, 5b + 2a = 180^o

5b + 80^o = 180^o

5b= 180^o – 80^o

5b = 100^o

b = 100/5

b = 20^o

c. To and fro movement of a vibrating string

Given that AB = AC; ∠BAC = 36°, 

∠ADB = 45°, ∠AEC = 40°

From the figure 

∠ABD + ∠ABC = 180° 

∠ABD = 180° – 72° = 1086 

In ΔABD 

∠DAB + ∠ABD + ∠D = 180° 

∠DAB + 108° + 45° = 180° 

∠DAB = 180° – 153° = 27°

1. Forces 

2. rest

3. vibrations

4. periodic motion 

5. Robots 

6. robota 

7. Robotics

(b) revolution of the moon around the earth

Rotation of the earth is a periodic motion because it takes equal interval of time for all rotations.

There are two types: 

1. Uniform motion, 

2. Non-uniform motion.

(b) Non- Periodic motion

Motion :

Change of position of an object with respect to time is known as motion.

1. Based on Path : 1. Linear motion. – Ex.: Parade of the soldiers. 

2. Curvilinear motion. – Ex.: Paper flight moving. 

3. Circular motion. – Ex.: Swirling stone tied to the rope. 

4. Rotatory motion. – Ex.: Rotating top. 

5. Oscillatory motion. – Ex.: Clock pendulum. 

6. Zigzag (irregular) motion. – Ex.: Motion of a bee.

2. Based on Duration : 

1. Periodic motion. – Ex.: Motion of a bob of simple Pendulum.

2. Non periodic motion. – Ex.: Swaying of the branches of a tree.

3. Based on Speed: 

1. Uniform motion. – Ex.: Elour hand of a clock. 

2. Non – uniform motion. – Ex.: Motion of a train, as it leaves a station.

1. Linear motion; 

2. Curvilinear; 

3. Circular motion; 

4. Rotatory motion; 

5. Oscillatory motion 

6. Irregular motion.

Answer is (A) π/4

(a) Non-periodic Motion 

(b) Non-periodic Motion 

(c) Non-uniform Motion

(d) Periodic Motion 

(e) Uniform Motion

We know that:

θ (radian) = \(\frac { arc\; length }{ radius }\)

We have, radius = 5 cm

and arc length = 12 cm

θ = \(\frac { 12 }{ 5 }\)

Answer is (D) π radian

Then   \(\begin{bmatrix}1&-2&3\\[0.3em]1&2&1\\[0.3em]x&2&-3\end{bmatrix}\) = 0

⇒ 1 (- 6 – 2) + 2 (-3 – x) + 3(2 – 2x) = 0

⇒ -8 – 6 – 2x + 6 – 6x = 0

⇒ -8 = 8

⇒ x = 8/-8 = -1

Hence, x = -1

Down’s syndrome is a human genetic disorder caused due to trisomy of chromosome no. 21. Such individuals are aneuploid and have 47 chromosomes (2n + 1). The symptoms include mental retardation, growth abnormalities, constantly open mouth, dwarfness etc. The reason for the disorder is the non-disjunction (failure to separate) of homologous chromosome of pair 21 during meiotic division in the ovum. The chances of having a child with Down’s syndrome increase with the age of the mother (+40) because ova are present in females. Since their birth and therefore older cells are more prone to chromosomal non-disjunction because of various physico-chemical exposures during the mother’s life-time.