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The electric power consumed by a 220 V-100 W bulb when operated at 110 V is:
The resistance of the bulb is R =220²/100= 484Ω . When operated at 110V, power consumed is P ′ = 110²/484 = 25W. thus, the Correct answer: (a) 25 W. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
The resistance of the bulb is
R =220²/100= 484Ω
. When operated at 110V, power consumed is
P ′ = 110²/484 = 25W.
thus, the Correct answer: (a) 25 W.
For more visit here:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-3/
What is the importance of Class 10 Mathematics Chapter 9 MCQ?
Class 10 Mathematics Chapter 9 MCQ assesses understanding of trigonometric applications, enhances problem-solving skills, and prepares students for real-life scenarios involving heights, distances, and angles, crucial for exams and advanced studies. For Practice MCQ visit here: https://www.tiwariacaRead more
Class 10 Mathematics Chapter 9 MCQ assesses understanding of trigonometric applications, enhances problem-solving skills, and prepares students for real-life scenarios involving heights, distances, and angles, crucial for exams and advanced studies.
For Practice MCQ visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-9/
What is the importance of Class 10 Mathematics Chapter 8 MCQ?
Class 10 Mathematics Chapter 8 MCQ evaluates knowledge of trigonometric ratios and identities, strengthens foundational concepts, and prepares students for advanced math applications, problem-solving, and scoring well in exams. For Practice MCQ visit here: https://www.tiwariacademy.in/ncert-solutionRead more
Class 10 Mathematics Chapter 8 MCQ evaluates knowledge of trigonometric ratios and identities, strengthens foundational concepts, and prepares students for advanced math applications, problem-solving, and scoring well in exams.
For Practice MCQ visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions-class-10-maths-chapter-8/
If the circumference of a circle is diminished by 10 %, then its area is diminished by
Given: - Circumference decreased by 10% ⇒ New radius = 0.9r. New area: A' = π(0.9r)² = 0.81πr² = 0.81A. Decrease in area: A - A' = A - 0.81A = 0.19A. Percentage decrease: (0.19A / A) × 100 = 19%. This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 Area related to CiRead more
Given:
– Circumference decreased by 10% ⇒ New radius = 0.9r.
New area:
A’ = π(0.9r)² = 0.81πr² = 0.81A.
Decrease in area:
A – A’ = A – 0.81A = 0.19A.
Percentage decrease:
(0.19A / A) × 100 = 19%.
This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 Area related to Circle. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
The area of incircle of an equilateral triangle is 154 cm². The perimeter of the triangle is
Given: - Area of incircle = 154 cm². Step 1: Find radius of incircle. πr² = 154 ⇒ (22/7)r² = 154 ⇒ r² = 49 ⇒ r = 7 cm. Step 2: Relate radius to side of equilateral triangle. r = (a√3)/6 ⇒ 7 = (a√3)/6 ⇒ a = 42/√3 = 14√3 cm. Step 3: Find perimeter. Perimeter = 3a = 3 × 14√3 ≈ 3 × 14 × 1.732 ≈ 72.7 cm.Read more
Given:
– Area of incircle = 154 cm².
Step 1: Find radius of incircle.
πr² = 154 ⇒ (22/7)r² = 154 ⇒ r² = 49 ⇒ r = 7 cm.
Step 2: Relate radius to side of equilateral triangle.
r = (a√3)/6 ⇒ 7 = (a√3)/6 ⇒ a = 42/√3 = 14√3 cm.
Step 3: Find perimeter.
Perimeter = 3a = 3 × 14√3 ≈ 3 × 14 × 1.732 ≈ 72.7 cm.
This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 Area related to Circle. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/