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8 months ago

Class 11th Physics

The motion of a simple pendulum executing S.H.M. is represented by the following equation. $y = A s i n (π t + ?),$ where time is measured in second. The length of pendulum is

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Payal Gupta

Contributor-Level 10

According to question, we can write

$ω = π = \sqrt[]{\frac{g}{l}} \Rightarrow l = \frac{g}{π^{2}} = \frac{9.8}{{(3.14)}^{2}} = 0.99395 = 99.4 c m$

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New answer posted

8 months ago

Class 11th Physics Gravitation

The TV transmission tower at a particular station has a height of 125 m. For doubling the coverage of its range, the height of the tower should be increased by

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Payal Gupta

Contributor-Level 10

According to question, we can write

$d = \sqrt[]{2 h R} \Rightarrow \frac{d_{2}}{d_{1}} = \sqrt[]{\frac{h_{2}}{h_{1}}} \Rightarrow h_{2} = {(\frac{d_{2}}{d_{1}})}^{2} h_{1} = {(\frac{2}{1})}^{2} * 125 = 500 m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

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New answer posted

8 months ago

Class 11th Physics

The time period of a satellite revolving around earth in a given orbit is 7 hours. If the radius of orbit is increased to three its previous value, then approximate new time period of the satellite will be

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Payal Gupta

Contributor-Level 10

According to question, we can write

$d = \sqrt[]{2 h R} \Rightarrow \frac{d_{2}}{d_{1}} = \sqrt[]{\frac{h_{2}}{h_{1}}} \Rightarrow h_{2} = {(\frac{d_{2}}{d_{1}})}^{2} h_{1} = {(\frac{2}{1})}^{2} * 125 = 500 m$

Increment in height of tower = h₂ – h₁ = 500 – 125 = 375 m

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New answer posted

8 months ago

Class 11th Physics

Starting with the same initial conditions, an ideal gas expands form volume V₁ to V₂ in the three different ways. The work done by the gas in W₁ if the process is purely isothermal, W₂, if the process is purely adiabatic and W₃ if the process is purely isobaric. Then, choose the correct option

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Payal Gupta

Contributor-Level 10

Process-AB $\Rightarrow$ Isobaric,

Process-AC $\Rightarrow$ Isothermal, and

Process-AD? Adiabatic

W₂ < W₁ < W₃

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New answer posted

8 months ago

Class 11th Maths

Let S = (0, 2) – ${\frac{π}{2}, \frac{3 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}} .$ Let y = y(x), $x \in S,$ be the solution curve of the differential equation $\frac{d y}{d x} = \frac{1}{1 + s i n 2 x}, y (\frac{π}{4}) = \frac{1}{2} .$ If the sum of abscissas of all the points of intersection of the curve y = y(x) with the curve $y = \sqrt[]{2} s i n x i s \frac{k π}{12},$ then k is equal to

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Payal Gupta

Contributor-Level 10

$\frac{d y}{d x} = \frac{1}{1 + s i n 2 x}$

$d y = \frac{s e c^{2} x d x}{{(1 + t a n x)}^{2}}$

$\Rightarrow y = - \frac{1}{1 + t a n x} + c$

When

$x = \frac{π}{4}, y = \frac{1}{2}$ gives c = 1

$x + \frac{π}{4} = \frac{5 π}{6} o r \frac{13 π}{6} \Rightarrow x = \frac{7 π}{12} o r \frac{23 π}{12}$

sum of all solutions =

$π + \frac{7 π}{12} + \frac{23 π}{12} = \frac{42 π}{12}$

Hence k = 42

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New answer posted

8 months ago

Class 11th Maths

Let $A = \sum_{i = 1}^{10} \sum_{j = 1}^{10} m i n {i, j} a n d B = \sum_{i = j}^{10} \sum_{j = 1}^{10} m a x {i, j} .$ Then A + B is equal to

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Payal Gupta

Contributor-Level 10

Each element of ordered pair (i, j) is either present in A or in B.

So, A + B = Sum of all elements of all ordered pairs {i, j} for $1 \leq i \leq 10$ and $1 \leq j \leq 10$

= 20 (1 + 2 + 3 + … + 10) = 1100

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New answer posted

8 months ago

Class 11th Maths Statistics

The value of the integral $\frac{48}{π^{4}} \int_{0}^{π} (\frac{3 π x^{2}}{2} - x^{3}) \frac{s i n x}{1 + c o s^{2} x} d x$ is equal to

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Payal Gupta

Contributor-Level 10

$l = \frac{48}{π^{4}} \int_{0}^{π} [{(\frac{π}{2} - x)}^{3} - \frac{3 π^{2}}{4} (\frac{π}{2} - x) + \frac{π^{3}}{4}] \frac{s i n x d x}{1 + c o s^{2} x}$

Using

$\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$

we get

$l = \frac{48}{π^{4}} \int_{0}^{π} [- {(\frac{π}{2} - x)}^{3} + \frac{3 π^{2}}{4} (\frac{π}{2} - x) + \frac{π^{3}}{4}] \frac{s i n x d x}{1 + c o s^{2} x}$

Adding these two equations, we get

$\Rightarrow l = \frac{12}{π} {[- t a n^{- 1} (c o s x)]}_{0}^{π} = \frac{12}{π} . \frac{π}{2} = 6$

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8 months ago

Class 11th Maths

Let A = ${n \in N : H . C . F . (n, 45) = 1}$ and Let B = ${2 k : k \in {1, 2, . . . . . ., 100}} .$ Then the sum of all the elements of $A \cup B i s . . . . . . . . . .$

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Payal Gupta

Contributor-Level 10

Sum of all elements of $A \cap B = 2$ [Sum of natural number upto 100 which are neither divisible by 3 nor by 5]

$= 2 [\frac{100 * 101}{2} - 3 (\frac{33 * 34}{2}) - 5 (\frac{20 * 21}{2}) + 15 (\frac{6 * 7}{2})]$

= 10100 – 3366 – 2100 + 630

= 5264

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New answer posted

8 months ago

Class 11th Maths

If $s i n^{2} (1 0^{0}) s i n (2 0^{0}) s i n (4 0^{0}) s i n (5 0^{0}) s i n (7 0^{0}) = α - \frac{1}{16} s i n (1 0^{0}),$ then 16 + 1 is equal to

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Payal Gupta

Contributor-Level 10

$(s i n 10 ° . s i n 50 ° . s i n 70 °) . (s i n 10 ° . s i n 20 ° . s i n 40 °)$

$= (\frac{1}{4} s i n 30 °) . [\frac{1}{2} s i n 10 ° (c o s 20 ° - c o s 60 °)]$

$= \frac{1}{32} [s i n 30 ° - s i n 10 ° - s i n 10 °]$

$\frac{1}{64} - \frac{1}{16} s i n 10 °$

Clearly $α = \frac{1}{64}$

Hence 16 + a^-1= 80

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New answer posted

8 months ago

Class 11th Maths

Let the solution curve y = y(x) of the differential equation (4 + x²)dy – 2x(x² + 3y + 4)dx = 0 pass through the origin. They y(2) is equal to

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Payal Gupta

Contributor-Level 10

$(4 + x^{2}) d y - 2 x (x^{2} + 3 y + 4) d x = 0$

$\Rightarrow \frac{d y}{d x} = (\frac{6 x}{x^{2} + 4}) y + 2 x$

$e^{- 3 l n (x^{2} + 4)} = \frac{1}{{(x^{2} + 4)}^{3}}$

$\frac{y}{{(x^{2} + 4)}^{3}} = \int \frac{2 x}{{(x^{2} + 4)}^{3}} d x + c$

$\Rightarrow y = - \frac{1}{2} (x^{2} + 4) + c {(x^{2} + 4)}^{3}$

When x = 0, y = 0 gives

$c = \frac{1}{32}$

So, for x = 2, y = 12

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