Class 11th
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New answer posted
7 months agoContributor-Level 10
The least count (LC) of the screw gauge is the pitch divided by the number of divisions on the circular scale.
LC = Pitch / N = 0.1 cm / 50 = 0.002 cm.
A measurement taken with this instrument must be a multiple of its least count. We check the options:
(A) 2.121 / 0.002 = 1060.5 (Not a multiple)
(B) 2.123 / 0.002 = 1061.5 (Not a multiple)
(C) 2.124 / 0.002 = 1062 (Is a multiple)
(D) 2.125 / 0.002 = 1062.5 (Not a multiple)
The correctly recorded measurement is 2.124 cm.
New answer posted
7 months agoContributor-Level 10
∴ σ² ≤ 1/4 (M - m)²
Where M and m are upper and lower bounds of values of any random variable.
∴ σ² < 1/4 (10 - 0)
⇒ 0 < < 5
∴ σ ≠ 6
New answer posted
7 months agoContributor-Level 10
Let P = (3t², 6t); N = (3t²,0)
M = (3t², 3t)
Equation of MQ: y = 3t
∴ Q = (3/4 t², 3t)
Equation of NQ
y = ( 3t / (3/4 t² - 3t²) ) (x - 3t²)
y - intercept of NQ = 4t = 4/3 ⇒ t = 1/3
∴ MQ = 9/4 t² = 1/4
PN = 6t = 2
New answer posted
7 months agoContributor-Level 10
A: D ≥ 0
⇒ (m + 1)² - 4 (m + 4) ≥ 0
⇒ m² + 2m + 1 - 4m - 16 ≥ 0
⇒ m² - 2m - 15 ≥ 0
⇒ (m - 5) (m + 3) ≥ 0
⇒ m ∈ (-∞, -3] U [5, ∞)
∴ A = (-∞, -3] U [5, ∞)
B = [-3,5)
A − B = (-∞, −3) U [5, ∞)
A ∩ B = {-3}
B - A = (-3,5)
A U B = R
New answer posted
7 months agoContributor-Level 10
Ellipse: x²/4 + y²/3 = 1
eccentricity = √ (1 - 3/4) = 1/2
∴ foci = (±1,0)
For hyperbola, given 2a = √2 ⇒ a = 1/√2
∴ hyperbola will be x²/ (1/2) - y²/b² = 1
eccentricity = √ (1 + 2b²)
∴ foci = (±√ ( (1+2b²)/2 ), 0)
∴ Ellipse and hyperbola have same foci
√ ( (1+2b²)/2 ) = 1
⇒ b² = 1/2
∴ Equation of hyperbola: x²/ (1/2) - y²/ (1/2) = 1
⇒ x² - y² = 1/2
Clearly, (√3/2, 1/2) does not lie on it.
New answer posted
7 months agoContributor-Level 9
Burning of fossil fuels (which contain sulphur and nitrogenous matter) such as coal and oil in power stations and furnaces produce sulphur dioxide and nitrogen oxides which causes acid rain.
New answer posted
7 months agoContributor-Level 10
α, β are roots of x² + px + 2 = 0
⇒ α² + pα + 2 = 0 and β² + pβ + 2 = 0
⇒ 1/α, 1/β are roots of 2x² + px + 1 = 0
But 1/α, 1/β are roots of 2x² + 2qx + 1 = 0
⇒ p = 2q
Also α + β = -p, αβ = 2
(α - 1/α) (β - 1/β) (α + 1/β) (β + 1/α)
= ( (α²-1)/α ) ( (β²-1)/β ) ( (αβ+1)/β ) ( (αβ+1)/α )
= ( (-pα-3) (-pβ-3) (αβ+1)² ) / ( (αβ)² )
= 9/4 (p²αβ + 3p (α + β) + 9)
= 9/4 (9 - p²) = 9/4 (9 - 4q²)
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