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2 months ago

Class 11th Work, Energy and Power

An elevator in a building can carry a maximum of 10 persons, with the average mass of each person being $68 k g$ . The mass of the elevator itself is $920 k g$ and it moves with a constant speed of $3 m / s$ . the frictional force opposing the motion is $6000 N$ . If the elevator is moving up with its full capacity, the power delivered by the motor to the elevator $(g = 10 m / s^{2})$ must be at least:

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alok kumar singh

Contributor-Level 10

Net force on motor will be

$F_{min} = [920 + 68 (10)] g + 6000$

$= 22000 N$

So, required power for motor

$\begin{array}{r} P_{min} & = {\overset{?}{F}}_{min} \cdot \overset{?}{v} \\ = 22000 * 3 \\ = 66000 watt \end{array}$

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2 months ago

Class 11th Thermodynamics

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from $τ_{1}$ to $τ_{2}$ . If $\frac{C_{p}}{C_{v}} = γ$ for the gas then a good estimate for $\frac{τ_{2}}{τ_{1}}$ is given by

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alok kumar singh

Contributor-Level 10

Relaxation time $(τ) \propto \frac{V}{\sqrt[]{T}}$

and $T \propto \frac{1}{V^{γ - 1}}$

$\begin{array}{r} τ \propto V^{1 + \frac{γ - 1}{2}} τ \propto V^{\frac{1 + γ}{2}} \\ \frac{τ_{f}}{τ_{i}} = {(\frac{2 V}{V})}^{\frac{1 + γ}{2}} \frac{τ_{f}}{τ_{i}} = (2)^{\frac{1 + γ}{2}} \end{array}$

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Class 11th System of Particles and Rotational Motion

As shown in the figure, $a$ bob of mass $m$ is tied by $a$ massless string whose other end portion is wound on a fly wheel (disc) of radius $r$ and mass $m$ . When released from rest the bob starts falling vertically. When it has covered a distance of $h$ , the angular speed of the wheel will be

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Vishal Baghel

Contributor-Level 10

$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} l ω^{2}$

$v = ω R$ (no slipping)

$m g h = \frac{1}{2} m ω^{2} R^{2} + \frac{1}{2} \frac{m R}{2} ω^{2}$

$m g h = \frac{3}{4} m ω^{2} R^{2}$

$ω = \sqrt[]{\frac{4 g h}{3 R^{2}}} = \frac{1}{R} \sqrt[]{\frac{4 g h}{3}}$

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Class 11th Physics

The dimension of stopping potential $V_{0}$ $_{}$ in photoelectric effect in units of Planck's constant ' $h$ ', speed of light ' $c$ ' and Gravitational constant ' $G$ ' and ampere $A$ is:

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Raj Pandey

Contributor-Level 9

Bonus $V = K (h)^{a} (I)^{b} (G)^{c} (C)^{d} (V$

is voltage $)$

We know, $\begin{array}{r} [h] = {M L}^{2} T^{- 1} \\ [l] = A \\ [G] = M^{- 1} L^{3} T^{- 2} \\ [C] = L T^{- 1} \\ [V] = M L^{2} T^{- 3} A^{- 1} \end{array}$

$M L^{2} T^{- 3} A^{- 1} = {(M L^{2} T^{- 1})}^{a} (A)^{b} {(M^{- 1} L^{3} T^{- 2})}^{c} {(L T^{- 1})}^{d}$

$M L^{2} T^{- 3} A^{- 1} = M^{a - c} L^{2 a + 3 c + d} T^{- a - 2 c - d} A^{b}$

$a - c = 1$

$2 a + 3 c + d = 2$

$- a - 2 c - d = - 3$

$b = - 1$

On solving,

$c = - 1$ $a = 0$

$d = 5, b = - 1$

$V = K (h)^{?} (I)^{- 1} (G)^{- 1} (C)^{5}$

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Class 11th System of Particles and Rotational Motion

Three point particles of masses $10 k g, 1.5 k g$ and $2.5 k g$ are placed at there corners of a right angle triangle of sides $4.0 c m, 3.0; c m$ and $5.0 c m$ as shown in the figure. The centre of mass of the system is at a point:

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Vishal Baghel

Contributor-Level 10

Take $1 k g$ mass at origin

$X_{c m} = \frac{1 * 0 + 1.5 * 3 + 2.5 * 0}{5} = 0.9 c m$

$Y_{c m} = \frac{1 * 0 + 1.5 * 0 + 2.5 * 4}{5} = 2 c m$

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Class 11th Physics System of Particles and Rotational Motion

The radius of gyration of a uniform rod of length $l$ , about an axis passing through a point $\frac{l}{4}$ $\frac{}{}$ away from the centre of the rod, an perpendicular to it is:

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Vishal Baghel

Contributor-Level 10

$\frac{M l^{2}}{12} + M (\frac{l}{4}) = {M K}^{2}$

$\frac{l^{2}}{12} + \frac{l^{2}}{16} = K^{2}$

$K = \sqrt[]{\frac{7}{48}} l$

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2 months ago

Class 11th Physics

A mass of $10 k g$ is suspended by a rope of length $4 m$ , from the ceiling. A force $F$ is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of $45^{?}$ with the vertical. Then $F$ equals: (Take $g = 10 {m s}^{- 2}$ and the rope to be massless)

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alok kumar singh

Contributor-Level 10

$\frac{T}{\sqrt[]{2}} = 100 \frac{T}{\sqrt[]{2}} = F$

$F = 100 N$

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2 months ago

Class 11th Physics Mechanical Properties of Fluids

An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are $6.4 c m$ and $4.8 c m$ , respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is

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alok kumar singh

Contributor-Level 10

Using equation of continuity

$\begin{array}{r} A_{1} V_{1} = A_{2} V_{2} \\ \frac{V_{1}}{V_{2}} = \frac{A_{2}}{A_{1}} = {(\frac{d_{2}}{d_{1}})}^{2} = {(\frac{4.8}{6.4})}^{2} = \frac{9}{16} \end{array}$

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2 months ago

Class 11th Physics

The plot that depicts the behavior of the mean free time $τ$ (time between two successive collisions) for the molecules of an ideal gas, as a function of temperature $(T)$ , qualitatively, is: (Graphs are schematic and not drawn to scale)