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Class 11th Physics

A block of mass 40kg slides over a surface, when a mass of 4kg is suspended through an inextensible massless string passing over frictionless pulley as shown below.

The coefficient of kinetic friction between the surface and block is 0.02. The acceleration of block is. (given g = 10 ms^-².)

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Payal Gupta

Contributor-Level 10

According to Newton's laws of motion, we can write

4a = 4g – T . (1), and

40a = T - fk = T $μ$ (40g) ……………… (2)

Adding equations (1) and (2), we can write

44a = 40 – 0.02 * (400) = 32

$\Rightarrow a = \frac{32}{44} = \frac{8}{11} m / s^{2}$

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4 months ago

Class 11th Physics

A vessel contains 16g of hydrogen and 128g of oxygen at standard temperature and pressure. The volume of the vessel in cm³ is

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Payal Gupta

Contributor-Level 10

According to question, we can write

Total moles of gas = n = n_Oxygen+ n_Oxygen = $\frac{16}{2} + \frac{128}{32} = 12 m o l e s$

Volume of gas = 12 * 22.4 litre = 268.8 litre = 2.688 * 10⁵cm3 $\approx 27 * 1 0^{4} c m^{3}$

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New answer posted

4 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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4 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

4 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

4 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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4 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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4 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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4 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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