Category: Uncategorized

1. c    2. a    3. c    4. d    5. a

1. Buddhist
2. Sangam
3. temples
4. Krishna, Arjuna
5. wood

1. T    2. F    3. T    4. T    5. T

1. c    2. a    3. d    4. b    5. e

1. Medical science in the ancient period was quite developed. Justify the statement.
Medical science was highly developed in ancient India. Doctors knew about surgery, medicines and treatment of diseases. Sushruta performed surgeries, and Charaka wrote about medicines and herbs.
2. What is the story of the epic Mahabharata?
The Mahabharata tells the story of a great war between the Pandavas and the Kauravas. It teaches moral values and duties. The Bhagavad Gita is a part of this epic.
3. What were the different styles of temple building in north and south India?
In North India, temples had tall and curved towers called Shikharas. In South India, temples had pyramidal towers called Vimanas.
4. What are Puranas?
Puranas are ancient religious texts that contain stories about gods, kings and creation of the world.
5. What is special about the Iron Pillar?
The Iron Pillar at Mehrauli is famous because it has not rusted even after many centuries. It shows the advanced knowledge of metallurgy in ancient India.

1. Write a few lines on the Iron Pillar located at Mehrauli.
The Iron Pillar is located in Delhi near the Qutub Minar. It was built during the Gupta period. It is about 7 metres high and is made of pure iron. It has not rusted over time, which shows the advanced scientific knowledge of that period.
2. What are Stupas? Mention some of their striking features.
Stupas are dome-shaped structures built over the relics of Buddha. They have a hemispherical dome, a central pillar and gateways. The Great Stupa at Sanchi is a famous example.
3. Write a note on paintings in the Gupta Period.
The Gupta period is known for beautiful wall paintings. The Ajanta caves contain fine paintings that show scenes from the life of Buddha and daily life. The paintings are colourful and realistic.
4. How were the stupas and temples built?
Stupas were built using bricks and stones. Temples were first made of wood and later of stone. Skilled craftsmen and architects designed them carefully.
5. State the developments made by Aryabhatta in the field of science.
Aryabhatta was a great mathematician and astronomer. He explained that the Earth rotates on its axis. He calculated the value of pi and studied eclipses scientifically.

1. Panchatantra and the Jataka Tales
Panchatantra – Collection of moral stories written in Sanskrit.
Jataka Tales – Stories about the previous births of Buddha.
2. Aryabhatiyam and Panchsiddhantika
Aryabhatiyam – Book written by Aryabhatta on mathematics and astronomy.
Panchsiddhantika – Book written by Varahamihira on astronomy.

1. Secular literature is an important source of history.
Secular literature gives information about social life, culture, economy and daily activities of people. It helps historians understand the past clearly.
2. The epics like Mahabharata and Ramayana are so popular even today.
These epics teach moral values, duties and devotion. Their stories are interesting and continue to influence people even today.

1. a    2. d    3. a    4. a    5. b

1. Samudragupta, Napoleon
2. Allahabad
3. chandalas
4. Kalidasa
5. Kanauj

1. d    2. e    3. c    4. a    5. b

1. F    2. T    3. F    4. T    5. F

1. How does Samudragupta’s prashasti describe him?
Samudragupta’s prashasti describes him as a brave warrior, a powerful conqueror and a generous ruler. It praises his military victories and compares him to great conquerors. He is also shown as a lover of music and learning.
2. Write a note on Chandragupta II.
Chandragupta II, also called Vikramaditya, was a powerful Gupta ruler. He defeated the Shakas and expanded the empire. His period is known for peace, prosperity, trade and cultural development. Great scholars like Kalidasa were in his court.
3. What factors led to the decline of the Gupta Empire?
The decline was caused by weak successors, internal conflicts, invasions by the Hunas and decentralisation of administration. These factors weakened central authority.
4. Who controlled the local assemblies under Pallavas?
The local assemblies were controlled by village elders and important landowners.
5. What were the main sources of revenue? How was this money spent?
The main sources of revenue were land tax, trade tax and tribute from subordinate rulers. The money was spent on administration, army, building temples, public works and welfare activities.

1. What do you know about Harshavardhana? Describe the main events of his career.
Harshavardhana became king in 606 CE. He made Kanauj his capital and expanded his empire in northern India. He was a capable administrator and maintained friendly relations with other rulers. He organised religious assemblies and supported education and culture.
2. How is Hiuen-Tsang’s account insightful?
Hiuen-Tsang visited India during Harsha’s reign. He described the administration, religious practices, education system and social life. His writings help historians understand this period clearly.
3. How did decentralisation weaken the Guptas and Harsha?
Local governors and feudatories were given more power. Over time they became independent and reduced the control of the central ruler, leading to decline.
4. The Chalukyas and the Pallavas were great architects. Give examples.
They built beautiful temples and rock-cut caves. The Pallavas built temples at Mahabalipuram. The Chalukyas built caves at Badami and temples at Aihole and Pattadakal.
5. How do we know that Harsha was a patron of art?
Harsha wrote plays and supported scholars and poets. He organised cultural gatherings and encouraged learning.

1. Bhakti and Vishayas
Bhakti – Devotion towards a personal god.
Vishayas – Administrative divisions of the empire.
2. Sabha and Ur
Sabha – Assembly of Brahmins in villages.
Ur – General assembly of the village people.
3. Ruler of Aryavarta and Ruler of Dakshinapatha
Ruler of Aryavarta – Controlled northern region.
Ruler of Dakshinapatha – Controlled southern region.

1. Direction of Incident Ray

In Class 10 CBSE problems, we assume that: The incident ray always travels from left to right. This direction is taken as the positive direction along the principal axis. Therefore: • Right side of Pole → Positive • Left side of Pole → Negative

---

2. Reference Point

All distances are measured from the Pole (P) of the mirror along the principal axis.

---

3. Horizontal Sign Rules (Along Principal Axis)

• Distance measured in the direction of incident light → Positive • Distance measured opposite to incident light → Negative Since light is assumed left → right: • Right side → Positive • Left side → Negative

---

4. Object Distance (u)

The object is placed in front of the mirror (left side). \[ u = \text{Always Negative} \]

---

5. Image Distance (v)

• Image formed in front of mirror → Negative • Image formed behind mirror → Positive Therefore: Real image → v negative Virtual image → v positive

---

6. Focal Length (f)

• Concave mirror → Focus in front → \( f \) negative • Convex mirror → Focus behind → \( f \) positive

---

7. Radius of Curvature (R)

\[ f = \frac{R}{2} \] • Concave mirror → \( R \) negative • Convex mirror → \( R \) positive

---

8. Vertical Sign Rules (Height)

• Height measured above principal axis → Positive • Height measured below principal axis → Negative So: Erect image → Image height positive Inverted image → Image height negative

---

9. Magnification Formula Connection

For mirrors: \[ m = -\frac{v}{u} \] • If \( m \) is positive → Image is erect • If \( m \) is negative → Image is inverted

---

10. Mirror Formula

\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Before substituting: 1. Identify mirror type 2. Assign sign of focal length 3. Object distance is always negative 4. Decide image nature 5. Assign correct sign to image distance 6. Substitute carefully

---

11. Special Case – Plane Mirror

• Object distance → Negative • Image distance → Positive • Focal length → Infinite • Radius → Infinite • Magnification → +1 Plane mirror always forms virtual, erect, same-size image.

---

12. Quick Summary Table

Quantity	Concave Mirror	Convex Mirror	Plane Mirror
u	Negative	Negative	Negative
v (Real)	Negative	Not formed	Not formed
v (Virtual)	Positive	Positive	Positive
f	Negative	Positive	Infinite
m (Real)	Negative	—	—
m (Virtual)	Positive	Positive	+1

---

13. Golden Memory Rules

Left side = Negative Right side = Positive Concave mirror → f negative Convex mirror → f positive Object distance → Always negative Real image → v negative Virtual image → v positive Above principal axis → Positive Below principal axis → Negative

1. A concave mirror forms a real image three times the size of the object at a distance of 36 cm from the mirror. Find: (a) Image distance (b) Object distance (c) Focal length (d) Nature of image
   Hint: Real image → image distance negative. Inverted image → magnification negative.


2. An object is placed at 15 cm from a concave mirror and produces a virtual image at 30 cm. Find focal length and magnification.
   Hint: Virtual image → image distance positive. Concave mirror → focal length negative.


3. A convex mirror forms an image half the size of the object at 20 cm behind the mirror. Find object distance and focal length.
   Hint: Convex mirror → focal length positive. Image behind mirror → positive image distance.


4. An object placed in front of a concave mirror produces an image at the same distance as the object. Find the position of object in terms of focal length.
   Hint: Use mirror formula and consider real image case.


5. A concave mirror of focal length 25 cm produces an image at 50 cm. Find object distance and magnification.
   Hint: Decide sign based on image type before substituting.


6. A convex mirror has radius of curvature 40 cm. An object is placed 30 cm in front of it. Find image distance and magnification.
   Hint: f = R/2. Convex mirror → positive focal length.


7. A concave mirror produces a magnification of –4. If image is formed at 40 cm in front of mirror, find object distance and focal length.


8. An object of height 5 cm is placed 20 cm in front of a convex mirror of focal length 15 cm. Find image height and nature of image.


9. A concave mirror of radius 60 cm forms an image 20 cm from mirror. Find object distance and magnification.


10. An object is placed between pole and focus of a concave mirror of focal length 12 cm. If object distance is 8 cm, find image distance and magnification.

1. An object is placed 24 cm in front of a concave mirror of focal length 12 cm. Find image distance and magnification.
   Sign Hint: Object in front → \(u\) negative. Concave mirror → \(f\) negative. Real image (if formed in front) → \(v\) negative.


2. A convex mirror of focal length 20 cm forms an image when object is at 30 cm. Find image distance.
   Sign Hint: Convex mirror → \(f\) positive. Object in front → \(u\) negative. Convex mirror forms virtual image → \(v\) positive.


3. An object is placed at 10 cm from a plane mirror. Find image position.
   Sign Hint: Object in front → \(u\) negative. Image behind mirror → \(v\) positive.


4. A concave mirror forms a real image at 36 cm when object is placed at 18 cm. Find focal length.
   Sign Hint: Real image in front → \(v\) negative. Object in front → \(u\) negative. Concave mirror → \(f\) negative.


5. A convex mirror forms image 15 cm behind mirror. If focal length is 25 cm, find object distance.
   Sign Hint: Image behind mirror → \(v\) positive. Convex mirror → \(f\) positive.

1. An object is placed at 30 cm in front of concave mirror of focal length 15 cm. Find image position.
   Sign Hint: Object in front → negative. Concave mirror → \(f\) negative.


2. A convex mirror of radius 40 cm forms image at 10 cm. Find object distance.
   Sign Hint: Radius of convex mirror → positive. \(f = R/2\). Image behind mirror → positive.


3. A concave mirror produces magnification –2. If image distance is 20 cm, find object distance.
   Sign Hint: Negative magnification → inverted → real image → \(v\) negative.


4. An object is placed 12 cm from plane mirror. Where is image formed?
   Sign Hint: Plane mirror image distance equals object distance but positive.


5. A concave mirror forms virtual image at 25 cm. Find object distance if focal length is 20 cm.
   Sign Hint: Virtual image → behind mirror → \(v\) positive. Concave mirror → \(f\) negative.

1. A concave mirror of focal length 18 cm forms image at 9 cm. Find object distance.
   Sign Hint: Concave mirror → \(f\) negative. Decide sign of \(v\) based on image nature.


2. A convex mirror forms image half the size of object at 16 cm. Find focal length.
   Sign Hint: Convex mirror → \(f\) positive. Image always virtual → \(v\) positive.


3. An object placed at focus of concave mirror. Where is image formed?
   Sign Hint: Object at focus → \(u = f\) (both negative).


4. A concave mirror produces image equal in size to object. Find object position.
   Sign Hint: Equal size real image → object at centre of curvature.


5. Object placed at 20 cm in front of convex mirror of focal length 15 cm. Find image position.
   Sign Hint: Object negative, focal length positive.

1. A concave mirror produces real image 3 times magnified at 30 cm. Find object distance.
   Sign Hint: Real image → inverted → magnification negative.


2. A convex mirror forms image 12 cm behind mirror when object is at 24 cm. Find focal length.
   Sign Hint: Both distances follow sign rule before substitution.


3. Plane mirror produces image 8 cm behind mirror. Find object position.
   Sign Hint: Image positive → object equal magnitude but negative.


4. Concave mirror radius is 50 cm. Find focal length and its sign.
   Sign Hint: Concave mirror → focal length negative.


5. A concave mirror produces virtual image 15 cm behind mirror. Find magnification.
   Sign Hint: Virtual image → positive \(v\). Magnification positive for erect image.

1. A concave mirror of focal length 10 cm forms image at 20 cm. Find object distance.
   Sign Hint: Check if image is real or virtual before assigning sign.


2. A convex mirror of focal length 30 cm forms image at 10 cm. Find object distance.
   Sign Hint: Convex → \(f\) positive. Image behind mirror → positive.


3. An object 15 cm from plane mirror. Find image distance and magnification.
   Sign Hint: Plane mirror → \(m = +1\).


4. A concave mirror forms image at 40 cm for object at 20 cm. Find focal length.
   Sign Hint: Both object and real image in front → negative values.


5. A convex mirror produces image of height 2 cm for object 8 cm placed at 32 cm. Find focal length.
   Sign Hint: Magnification positive for convex mirror.

1. Real image three times the size at 36 cm Real image → \( v = -36\,cm \) Magnification \( m = -3 \) \[ m = -\frac{v}{u} \] \[ -3 = -\frac{-36}{u} \] \[ -3 = \frac{36}{u} \] \[ u = -12\,cm \] Mirror formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] \[ \frac{1}{f} = \frac{1}{-36} + \frac{1}{-12} \] \[ \frac{1}{f} = \frac{-1-3}{36} \] \[ f = -9\,cm \] Image is real and inverted.

---

2. Virtual image → \( v = +30\,cm \) Object → \( u = -15\,cm \) \[ \frac{1}{f} = \frac{1}{30} + \frac{1}{-15} \] \[ \frac{1}{f} = \frac{1-2}{30} \] \[ f = -30\,cm \] \[ m = -\frac{30}{-15} = +2 \]

---

3. Convex mirror \( v = +20\,cm \) \( m = 0.5 \) \[ 0.5 = -\frac{20}{u} \] \[ u = -40\,cm \] \[ \frac{1}{f} = \frac{1}{20} + \frac{1}{-40} \] \[ f = +40\,cm \]

---

4. Image at same distance as object For real image: \[ v = u \] From mirror formula: \[ \frac{1}{f} = \frac{2}{u} \] \[ u = 2f \] Object is at centre of curvature.

---

5. \( f = -25\,cm \) Image at 50 cm (real) → \( v = -50\,cm \) \[ \frac{1}{-25} = \frac{1}{-50} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-2+1}{50} \] \[ u = -50\,cm \] \[ m = -\frac{-50}{-50} = -1 \]

---

6. Convex mirror \( R = 40\,cm \) \[ f = \frac{R}{2} = +20\,cm \] \( u = -30\,cm \) \[ \frac{1}{20} = \frac{1}{v} + \frac{1}{-30} \] \[ \frac{1}{v} = \frac{1}{20} + \frac{1}{30} \] \[ \frac{1}{v} = \frac{3+2}{60} \] \[ v = +12\,cm \] \[ m = -\frac{12}{-30} = +0.4 \]

---

7. \( m = -4 \) \( v = -40\,cm \) \[ -4 = -\frac{-40}{u} \] \[ u = -10\,cm \] \[ \frac{1}{f} = \frac{1}{-40} + \frac{1}{-10} \] \[ \frac{1}{f} = \frac{-1-4}{40} \] \[ f = -8\,cm \]

---

8. Convex mirror \( u = -20\,cm \) \( f = +15\,cm \) \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-20} \] \[ \frac{1}{v} = \frac{1}{15} + \frac{1}{20} \] \[ v = +8.6\,cm \] \[ m = -\frac{8.6}{-20} = +0.43 \] \[ h_i = 0.43 \times 5 = 2.15\,cm \] Virtual and erect.

---

9. Concave mirror \( R = -60\,cm \) \[ f = -30\,cm \] \( v = -20\,cm \) \[ \frac{1}{-30} = \frac{1}{-20} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-2+3}{60} \] \[ u = +60\,cm \] \[ m = -\frac{-20}{60} = 0.33 \]

---

10. Object inside focus \( u = -8\,cm \) \( f = -12\,cm \) \[ \frac{1}{-12} = \frac{1}{v} + \frac{1}{-8} \] \[ \frac{1}{v} = \frac{-1+1.5}{12} \] \[ v = +24\,cm \] \[ m = -\frac{24}{-8} = +3 \] Virtual and magnified.

1. Given: \( u = -24\,cm \), \( f = -12\,cm \) \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] \[ \frac{1}{-12} = \frac{1}{v} + \frac{1}{-24} \] \[ \frac{1}{v} = \frac{1}{-12} – \frac{1}{-24} \] \[ \frac{1}{v} = \frac{-2+1}{24} = \frac{-1}{24} \] \[ v = -24\,cm \] \[ m = -\frac{v}{u} = -\frac{-24}{-24} = -1 \]

---

2. Given: \( u = -30\,cm \), \( f = +20\,cm \) \[ \frac{1}{20} = \frac{1}{v} + \frac{1}{-30} \] \[ \frac{1}{v} = \frac{1}{20} + \frac{1}{30} \] \[ \frac{1}{v} = \frac{3+2}{60} = \frac{5}{60} \] \[ v = 12\,cm \]

---

3. Plane mirror Image distance = +10 cm \[ m = +1 \]

---

4. Given: \( u = -18\,cm \), \( v = -36\,cm \) \[ \frac{1}{f} = \frac{1}{-36} + \frac{1}{-18} \] \[ \frac{1}{f} = \frac{-1-2}{36} \] \[ f = -12\,cm \]

---

5. Given: \( v = +15\,cm \), \( f = +25\,cm \) \[ \frac{1}{25} = \frac{1}{15} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{1}{25} – \frac{1}{15} \] \[ \frac{1}{u} = \frac{3-5}{75} = \frac{-2}{75} \] \[ u = -37.5\,cm \]

---

6. Given: \( u = -30\,cm \), \( f = -15\,cm \) \[ \frac{1}{-15} = \frac{1}{v} + \frac{1}{-30} \] \[ \frac{1}{v} = \frac{-2+1}{30} \] \[ v = -30\,cm \]

---

7. \( R = +40\,cm \) \[ f = \frac{R}{2} = +20\,cm \]

---

8. Given: \( m = -2 \), \( v = -20\,cm \) \[ -2 = -\frac{v}{u} \] \[ -2 = -\frac{-20}{u} \] \[ u = -10\,cm \]

---

9. Plane mirror Image distance = +12 cm \[ m = +1 \]

---

10. Given: \( v = +25\,cm \), \( f = -20\,cm \) \[ \frac{1}{-20} = \frac{1}{25} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-5-4}{100} \] \[ u = -11.1\,cm \]

---

11. Given: \( f = -18\,cm \), \( v = -9\,cm \) \[ \frac{1}{-18} = \frac{1}{-9} + \frac{1}{u} \] \[ \frac{1}{u} = \frac{-1+2}{18} \] \[ u = +18\,cm \]

---

12. Given: \( m = 0.5 \), \( v = +16\,cm \) \[ 0.5 = -\frac{16}{u} \] \[ u = -32\,cm \] \[ \frac{1}{f} = \frac{1}{16} + \frac{1}{-32} \] \[ f = +32\,cm \]

---

13. Object at focus \[ v = \infty \]

---

14. Equal size real image \[ u = 2f \]

---

15. Given: \( u = -20\,cm \), \( f = +15\,cm \) \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{-20} \] \[ v = 60\,cm \]

---

16. Given: \( m = -3 \), \( v = -30\,cm \) \[ -3 = -\frac{-30}{u} \] \[ u = -10\,cm \]

---

17. Given: \( u=-24\,cm \), \( v=+12\,cm \) \[ \frac{1}{f} = \frac{1}{12} + \frac{1}{-24} \] \[ f = +24\,cm \]

---

18. Plane mirror Object = -8 cm Image = +8 cm

---

19. \( f = -25\,cm \)

---

20. Given: \( v = +15\,cm \) \[ m = -\frac{15}{u} \] \[ m > 0 \]

---

21. Given: \( f=-10\,cm \), \( v=-20\,cm \) \[ \frac{1}{-10} = \frac{1}{-20} + \frac{1}{u} \] \[ u = -20\,cm \]

---

22. Given: \( f=+30\,cm \), \( v=+10\,cm \) \[ \frac{1}{30} = \frac{1}{10} + \frac{1}{u} \] \[ u = -15\,cm \]

---

23. Plane mirror Image = +15 cm \[ m = +1 \]

---

24. Given: \( u=-20\,cm \), \( v=-40\,cm \) \[ \frac{1}{f} = \frac{1}{-40} + \frac{1}{-20} \] \[ f = -13.3\,cm \]

---

25. Given: \( m = \frac{2}{8} = 0.25 \) \[ 0.25 = -\frac{v}{u} \] Solve to get: \[ f = +32\,cm \]

Given:
Focal length = 12 cm (Concave mirror)
Object distance = 8 cm (between pole and focus)
Step 1: Apply Sign Convention For concave mirror: \[ f = -12\,cm \] Object placed in front of mirror: \[ u = -8\,cm \]
Step 2: Use Mirror Formula \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Substitute values: \[ \frac{1}{-12} = \frac{1}{v} + \frac{1}{-8} \] \[ \frac{1}{-12} = \frac{1}{v} – \frac{1}{8} \]
Step 3: Solve for v \[ \frac{1}{v} = \frac{1}{-12} + \frac{1}{8} \] Take LCM = 24 \[ \frac{1}{v} = \frac{-2 + 3}{24} \] \[ \frac{1}{v} = \frac{1}{24} \] \[ v = +24\,cm \]
Step 4: Calculate Magnification \[ m = -\frac{v}{u} \] \[ m = -\frac{24}{-8} \] \[ m = +3 \]
Final Answer:
Image distance: \[ v = +24\,cm \] Magnification: \[ m = +3 \]
Nature of Image:
• Image is virtual (since v is positive).
• Image is erect (since m is positive).
• Image is magnified (three times).