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10 months ago

Class 11th Work, Energy and Power

A block moving horizontally on a smooth surface with a speed of 40 m/s splits into parts with masses in the ratio of 1:2. If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is:-

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Vishal Baghel

Contributor-Level 10

Using conservation of linear momentum:

=> 40 * 3m = 60 * m + v * 2m

=> v = 30 m/s

$K E_{i} = \frac{1}{2} * 3 m * {(40)}^{2}$

$K E_{f} = \frac{1}{2} * m * {(60)}^{2} + \frac{1}{2} * 2 m * {(30)}^{2}$

$\Rightarrow \frac{K E_{f}}{K E_{i}} = \frac{54}{48}$

Fractional change in kinetic energy = 1 $- \frac{K E_{f}}{K E_{i}} = \frac{1}{8}$

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10 months ago

Class 11th NCERT Class 11 Maths

Thew length of the latus rectum of a parabola, whose vertex and focus are on the positive x-axis at a distance R and S (>R) respectively from the origin, is:

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Raj Pandey

Contributor-Level 9

Length of latus rectum = 4 (distance between vertex and focus) = 4 (S – R)

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Class 11th Physics

Statement-I: If three forces ${\vec{F}}_{1}, {\vec{F}}_{2} a n d {\vec{F}}_{3}$ are represented by three sides of triangle and, ${\vec{F}}_{1} + {\vec{F}}_{2} = - {\vec{F}}_{3},$ then these three forces are concurrent forces and satisfy the condition for equilibrium.

Statement-II: A triangle made up of three forces ${\vec{F}}_{1}, {\vec{F}}_{2} a n d {\vec{F}}_{3}$ as its sides taken in the same order, satisfy the condition for translator equilibrium.

In the light of the above statements, choose the most appropriate answer from the options given below:

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Vishal Baghel

Contributor-Level 10

For equilibrium net force acting on the system should be zero.

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10 months ago

Class 11th Thermodynamics

A mixture of hydrogen and oxygen has volume 500 cm³, temperature 300K, pressure 400 kPa and mass 0.76g. The ratio of masses of oxygen to hydrogen will be:-

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Vishal Baghel

Contributor-Level 10

Using Ideal gas Equation :

PV = nRT

$\Rightarrow n = \frac{P V}{R T} = \frac{400 * 1 0^{3} * 500 * 1 0^{- 6}}{8.3 * 100} = 0.008$

=> $m_{1} = 0.12, m_{2} = 0.64$

=> $\frac{m_{2}}{m_{1}} = \frac{16}{3}$

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10 months ago

Class 11th Physics

Four particle each of mass M, move along a circle of radius R under the action of their mutual gravitational attraction as shown in figure. The speed of each particle is:

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alok kumar singh

Contributor-Level 10

For circular motion F_{net

$= \frac{M V^{2}}{R}$}

_{$\sqrt[]{2 F} + F' = \frac{M V^{2}}{R}$}

$\sqrt[]{2} \frac{G M M}{{(\sqrt[]{2} R)}^{2}} + \frac{G M M}{{(2 R)}^{2}} = \frac{M V^{2}}{R}$

$\frac{G M}{R} [\frac{1}{\sqrt[]{2}} + \frac{1}{4}] = V^{2}$

$V = \frac{1}{2} \sqrt[]{\frac{G M}{R} (2 \sqrt[]{2} + 1)}$

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Class 11th Physics

If velocity [V], time [T] and force [F] are chosen as the base quantities, the dimensions of the mass will be:

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Vishal Baghel

Contributor-Level 10

Using F = MA = $m \frac{V}{T}$

=> m = FTV^-1

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10 months ago

Class 11th Physics

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50 * 10³ kg. The inner and outer radii of each column are 50cm and 100cm respectively. Assuming uniform local distribution, calculate the compression strain of each column.

[Use Y = 2.0 * 10¹¹ Pa, g = 9.8 m/s²]

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Vishal Baghel

Contributor-Level 10

Load of mass will be equally distributed among the four colours so force on each columns will be 125 * 10³N.

Cross section area of the column = $π [{(1)}^{2} - {(0.5)}^{2}] = 2.355 m^{2}$

Using young's modulus : $ε = \frac{σ}{Y} = \frac{F}{A Y} = \frac{125 * 1 0^{3}}{2.355 * 2 * 1 0^{11}} = 2.65 * 1 0^{- 7}$

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10 months ago

Class 11th Physics

For a body executing S.H.M. :

(a) Potential energy is always equal to its K.E.

(b) Average potential and kinetic energy over any given time interval are always equal.

(c) Sum of the kinetic and potential energy at any point of time is constant.

(d) Average K.E. in one time period is equal to average potential energy in one time period.

Choose the most appropriate option from the options given below:

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Vishal Baghel

Contributor-Level 10

In SHM sum of kinetic and potential energy will be constant and average kinetic energy & average potential energy in one time will be remains same.

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10 months ago

Class 11th NCERT Class 11 Maths

If p and q are the lengths of the perpendiculars from the origin on the lines.

x cosec a - y sec a = k cot 2a and x sin a + y cos a = k sin 2a respectively, then k² is equal to:

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Raj Pandey

Contributor-Level 9

Length of perpendicular from origin to xcosec a - y sec a = k cot2 a and x sin a + ycos a = k sin 2a are

$p = k c o t α s i n α c o s α = \frac{k}{2} . \frac{c o s 2 α}{s i n 2 α} s i n 2 α = \frac{k}{2} c o s 2 α \Rightarrow \frac{2 p}{k} = c o s 2 α$

and $q = k s i n 2 α \Rightarrow \frac{q}{k} = s i n 2 α$

on solving these two we get 4p² + q² = k²

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10 months ago

Class 11th Physics

Consider two separate ideal gases of electrons and protons having same number of particles. The temperature of both the gases are same. The ratio of the uncertainty in determining the position of an electron to that of a proton is proportional to:-

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Vishal Baghel

Contributor-Level 10

Heisenberg uncertainty principle :

$\begin{array}{l} \Rightarrow Δ x Δ p \geq ? \\ \Rightarrow Δ x (m Δ v) \geq ? \end{array}$

$\Rightarrow Δ x (m \sqrt[]{\frac{3 R T}{m}}) \geq ?$

$\Rightarrow Δ x \propto \frac{1}{\sqrt[]{m}}$

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