Class 11th

A railway track (made of iron) is laid in winter when the average temperature is $18 ° C .$ The track consists of sections of 12.0m placed on after the other. How much gap (in cm) should be left between two such sections so that there is no compression during summer when the maximum temperature goes to $48 ° C ?$ Coefficient of linear expansion of iron = 11 * $1 0^{- 6} / ° C$ .

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Contributor-Level 10

$Δ L = L_{0} α Δ T$

= 12 * 11 * 10^-6 * 30

= 3960 * 10^-6 m

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2 months ago

Three rods of equal mass having length L, L and √2L are joined to form a triangular frame as shown in the figure. The system is rotating about an axis passing through the vertex of the triangle and perpendicular to the plane of the triangle with an angular velocity ω. Find the angular momentum of the system.

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Class 11th Physics Mechanical Properties of Fluids

Contributor-Level 10

L = Iω
I = (mL²/3) + (mL²/3) + [m (√2L)²/12] + m (L/√2)² = mL² [2/3 + 1/6 + 1/2] = (4mL²/3); L = (4/3)mL²ω

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2 months ago

A plastic circular disc of radius 10cm is placed on a thin oil film of thickness 2mm, spread over a flat horizontal surface. The torque (N-m) required to spin the disc about its central vertical axis with a constant angular velocity 8 rad/sec is (coefficient of viscosity of oil is 1.0 kg/m-s) (take p = 3.14)

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Contributor-Level 10

$τ = \int F . r = \int_{0}^{R} η (2 π r d r) (\frac{r w}{t}) . r$

$= \frac{2 π η W R^{4}}{4 t} = \frac{2 * 3.14 * 1 * 8 * 1 0^{- 4}}{4 * 2 * 1 0^{- 3}}$ f

$= 0.625 N - m ? 0.63 N - m$

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2 months ago

Class 11th Thermodynamics

The molar specific heat C_v of a diatomic gas is (5/2)R when we treat its molecules to be rigid having no vibrations. But this assumption is not always valid especially at high temperatures. In that case, molecules are treated as non-rigid having vibrational energy also. C_v for diatomic gas with non-rigid gas molecules will be:

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Contributor-Level 10

Vibrational energy of a non-rigid gas molecule is K? T so, total energy = (5/2)K? T + K? T = (7/2)K? T
∴ C? = (7/2)R

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2 months ago

The minimum and maximum distance of a satellite from the centre of the earth are 2R and 4R, respectively, where R is the radius of earth and M is the mass of earth. The radius of curvature at the point of maximum distance is

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Contributor-Level 10

$\frac{1}{2} m V_{a}^{2} - \frac{G M m}{2 x}$

= $\frac{1}{2} m \frac{V_{a}^{2}}{4} - \frac{G M m}{4 x}, V_{a} = \sqrt[]{\frac{2}{3} \frac{G M}{x}}$

ROC at point – B

$R O C = \frac{V_{b}^{2}}{a ⊥} = \frac{\frac{2 G M}{3 R} * \frac{1}{4}}{\frac{G M}{{(4 R)}^{2}}} = \frac{8 R}{3}$

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2 months ago

A satellite is moving around a planet of radius R in a circular orbit of radius r. The speed of the satellite is v. Find the acceleration due to gravity on the surface of the planet.

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Contributor-Level 10

v = √ (GM/r) (orbital speed)
⇒ GM = v²r ⇒ (GM/R²) = (v²r/R²) ⇒ g = (v²r/R²)

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2 months ago

Class 11th Physics Waves

The absolute temperature of air in a region linearly increases from $T_{1}$ to $T_{2}$ in a space of width d. The time taken by the sound wave to travel in this region is (The velocity of sound at 273 k is v ):

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Raj Pandey

Contributor-Level 9

$V = \sqrt[]{\frac{γ R T}{M}},$ where, $T = T_{1} + \frac{T_{2} - T_{1}}{d} x$ $_{} \frac{_{}_{}}{}$

Also, $v = \sqrt[]{\frac{γ R 273}{M}} \Rightarrow \sqrt[]{\frac{γ R}{M}} = \frac{v}{\sqrt[]{273}}$ $\frac{}{} \frac{}{} \frac{}{}$

Also, $V = \frac{d x}{d t} = \sqrt[]{\frac{γ R T}{M}}$ $\Rightarrow \int_{0}^{d} ? \frac{d x}{\sqrt[]{T_{1} + \frac{T_{2} - T_{1}}{d} x}} = \sqrt[]{\frac{γ R}{M}} \int_{0}^{t} ? d t$ $\frac{}{} \frac{}{}$

$\Rightarrow t = \frac{\sqrt[]{273}}{v} \cdot \frac{2 d}{T_{2} - T_{1}} (\sqrt[]{T_{2}} - \sqrt[]{T_{1}}) ∴ t = \frac{2 d}{v} \frac{\sqrt[]{273}}{\sqrt[]{T_{1}} + \sqrt[]{T_{2}}}$

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2 months ago

An inverted cone is rotating about its vertical axis. A particle is kept on the inner surface of cone and it is at rest relative to the cone at a height of 0.4 m above its vertex. The coefficient of friction between the surface of cone and the particle is 0.6 and the apex angle of cone is 90°. The maximum angular velocity of revolution of the cone can be: (take g = 10/s²)

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Class 11th Physics Oscillations

Contributor-Level 10

N = (mg/√2) + (mω²h/√2)
(mω²h/√2) = (mg/√2) + f?
(mω²h/√2) = (mg/√2) + μ (mg/√2) + (mω²h/√2)
ω = √ (g/h) (1+μ)/ (1-μ) = 10rad/s

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2 months ago

A solid cylinder attached to horizontal massless spring can roll without slipping along horizontal surface. Find time period of oscillation.

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Class 11th Motion in a Plane

Contributor-Level 10

$τ_{A} = I α$

$K x R = \frac{3}{2} M R^{2} α$

$K R θ R = \frac{3}{2} M R^{2} α$

$α = \frac{2 K}{3 M}$

$∴ T = 2 π \sqrt[]{\frac{3 M}{2 K}}$

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2 months ago

A particle is moving in a circle of radius 1m with angular velocity (ω) given as a function of angular displacement ( θ ) by the relation ω = θ² + 2θ. The total acceleration of the particle when θ = 1rad is:

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