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Three capacitors 2 pF, 3 pF and 6 pF are joined in series with eachother. The equivalent capacitance is
The correct answer is (d) 1/2 µF. For capacitors in series, the equivalent capacitance is given by: 1/Ceq = 1/C₁ + 1/C₂ + 1/C₃ Substituting C1 = 2 pF, C₂ = 3 pF and C₃ = 6 pF, we get Ceq = 1 pF = 1/2 µF. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-2/
The correct answer is (d) 1/2 µF.
For capacitors in series, the equivalent capacitance is given by:
1/Ceq = 1/C₁ + 1/C₂ + 1/C₃
Substituting C1 = 2 pF, C₂ = 3 pF and C₃ = 6 pF, we get Ceq = 1 pF = 1/2 µF.
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If a cos θ + b sin θ = m and a sin θ – b cos θ = n, then a² + b² =
We are given the equations: a cosθ + b sinθ = m — (1) a sinθ - b cosθ = n — (2) We need to find the value of a² + b². Step 1: Square both equations Square both sides of equation (1): (a cosθ + b sinθ)² = m² Expand the left-hand side: a²cos²θ + 2ab cosθ sinθ + b²sin²θ = m² — (3) Square both sides ofRead more
We are given the equations:
a cosθ + b sinθ = m — (1)
a sinθ – b cosθ = n — (2)
We need to find the value of a² + b².
Step 1: Square both equations
Square both sides of equation (1):
(a cosθ + b sinθ)² = m²
Expand the left-hand side:
a²cos²θ + 2ab cosθ sinθ + b²sin²θ = m² — (3)
Square both sides of equation (2):
(a sinθ – b cosθ)² = n²
Expand the left-hand side:
a²sin²θ – 2ab sinθ cosθ + b²cos²θ = n² — (4)
Step 2: Add equations (3) and (4)
Add the expanded forms of equations (3) and (4):
(a²cos²θ + 2ab cosθ sinθ + b²sin²θ) + (a²sin²θ – 2ab sinθ cosθ + b²cos²θ) = m² + n²
Simplify the terms:
– The terms involving 2ab cosθ sinθ cancel out.
– Combine the remaining terms:
a²(cos²θ + sin²θ) + b²(sin²θ + cos²θ) = m² + n²
Step 3: Use the Pythagorean identity
From the Pythagorean identity, we know:
cos²θ + sin²θ = 1.
Substitute this into the equation:
a²(1) + b²(1) = m² + n²
Simplify:
a² + b² = m² + n².
The question is based on Chapter 8 of the Class 10th NCERT Mathematics textbook, titled “Introduction to Trigonometry.” Provide your response in line with the concepts covered in this chapter.
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If sin θ + sin² θ = 1, then cos² θ + cos⁴ θ =
We are given the equation: sin θ + sin² θ = 1. We need to find the value of cos² θ + cos⁴ θ. Step 1: Express sin² θ in terms of sin θ Rearrange the given equation: sin² θ = 1 - sin θ. Step 2: Use the Pythagorean identity From the Pythagorean identity, we know: sin² θ + cos² θ = 1. Substitute sin² θRead more
We are given the equation:
sin θ + sin² θ = 1.
We need to find the value of cos² θ + cos⁴ θ.
Step 1: Express sin² θ in terms of sin θ
Rearrange the given equation:
sin² θ = 1 – sin θ.
Step 2: Use the Pythagorean identity
From the Pythagorean identity, we know:
sin² θ + cos² θ = 1.
Substitute sin² θ = 1 – sin θ into the identity:
(1 – sin θ) + cos² θ = 1.
Simplify:
cos² θ = sin θ.
Step 3: Express cos⁴ θ in terms of cos² θ
Since cos² θ = sin θ, we can write:
cos⁴ θ = (cos² θ)² = (sin θ)² = sin² θ.
Step 4: Substitute into cos² θ + cos⁴ θ
Now substitute cos² θ = sin θ and cos⁴ θ = sin² θ into the expression cos² θ + cos⁴ θ:
cos² θ + cos⁴ θ = sin θ + sin² θ.
From the given equation, we know:
sin θ + sin² θ = 1.
Thus:
cos² θ + cos⁴ θ = 1.
This question related to Chapter 8 Mathematics Class 10th NCERT. From the Chapter 8 Introduction to Trigonometry. Give answer according to your understanding.
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If a cot θ + b cosec θ = p and b cot θ + a cosec θ = q, then p² – q² =
We are given the equations: a cotθ + b cosecθ = p — (1) b cotθ + a cosecθ = q — (2) We need to find the value of p² - q². Step 1: Recall the identity for p² - q² The difference of squares formula states: p² - q² = (p + q)(p - q). Thus, we first calculate p + q and p - q. Step 2: Add equations (1) anRead more
We are given the equations:
a cotθ + b cosecθ = p — (1)
b cotθ + a cosecθ = q — (2)
We need to find the value of p² – q².
Step 1: Recall the identity for p² – q²
The difference of squares formula states:
p² – q² = (p + q)(p – q).
Thus, we first calculate p + q and p – q.
Step 2: Add equations (1) and (2)
Add the two equations:
(a cotθ + b cosecθ) + (b cotθ + a cosecθ) = p + q.
Simplify:
(a + b)cotθ + (a + b)cosecθ = p + q.
Factor out (a + b):
(a + b)(cotθ + cosecθ) = p + q. — (3)
Step 3: Subtract equation (2) from (1)
Subtract equation (2) from equation (1):
(a cotθ + b cosecθ) – (b cotθ + a cosecθ) = p – q.
Simplify:
(a – b)cotθ + (b – a)cosecθ = p – q.
Factor out (a – b):
(a – b)(cotθ – cosecθ) = p – q. — (4)
Step 4: Substitute into p² – q²
Using the identity p² – q² = (p + q)(p – q), substitute the expressions for p + q and p – q from equations (3) and (4):
p² – q² = [(a + b)(cotθ + cosecθ)] × [(a – b)(cotθ – cosecθ)].
Simplify:
p² – q² = (a + b)(a – b) × (cotθ + cosecθ)(cotθ – cosecθ).
Step 5: Simplify further using the difference of squares
The term (cotθ + cosecθ)(cotθ – cosecθ) is a difference of squares:
(cotθ + cosecθ)(cotθ – cosecθ) = cot²θ – cosec²θ.
From the trigonometric identity:
cosec²θ – cot²θ = 1,
we can write:
cot²θ – cosec²θ = -1.
Substitute this into the expression for p² – q²:
p² – q² = (a + b)(a – b) × (-1).
Simplify:
p² – q² = -(a + b)(a – b).
Expand (a + b)(a – b) using the difference of squares:
(a + b)(a – b) = a² – b².
Thus:
p² – q² = -(a² – b²).
This question is related to Chapter 8, “Introduction to Trigonometry,” from the Class 10th NCERT Mathematics textbook. Answer the question using your knowledge and understanding of the chapter.
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variable capacitor is connected to a 200 V battery. If its capacitance is changed from 2 µF to X µF, the decrease in energy of the capacitor is 2 X 10⁻² J. The value of X is
The correct answer is (a) 1 µF. The energy stored in a capacitor is given by U = 1/2CV². The decrease in energy is: ΔU = 1/2(2 × 10⁻⁶ - X) (200)² = 2 × 10⁻²J Solving for X, we get 1 µF. For more visit here: https://www.tiwariacademy.com/ncert-solutions/class-12/physics/chapter-2/
The correct answer is (a) 1 µF.
The energy stored in a capacitor is given by
U = 1/2CV². The decrease in energy is:
ΔU = 1/2(2 × 10⁻⁶ – X) (200)² = 2 × 10⁻²J
Solving for X, we get 1 µF.
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