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Physics Class 11th

Four identical solid spheres each of mass 'm' and radius 'a' are placed with their centres on the four corners of a square of side 'b'. The moment of inertia of the system about side of square where the axis of rotation is parallel to the plane of the square is

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Payal Gupta

Contributor-Level 10

$I_{A B} = \frac{2}{5} m a^{2} * 2 + (\frac{2}{5} m a^{2} + m b^{2}) * 2 = \frac{8 m a^{2}}{5} + 2 m b^{2}$

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Physics Physics Electrostatic Potential and Capacitance

Consider the combination of 2 capacitors C₁ and C₂, with C₂ > C₁, when connected in parallel, the equivalent capacitance is $\frac{15}{4}$ times the equivalent capacitance of the same connected in series, Calculate the ratio of capacitors, $\frac{C_{2}}{C_{1}} .$

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Payal Gupta

Contributor-Level 10

According question, we can write

$\frac{C_{1} * C_{2}}{C_{1} + C_{2}} = \frac{15}{4} (C_{1} + C_{2}) \Rightarrow 4 C_{1} C_{2} = 15 (C_{1}^{2} + C_{2}^{2} + 2 C_{1} C_{2})$

$\Rightarrow 4 (\frac{C_{2}}{C_{1}}) = 15 {(\frac{C_{2}}{C_{1}})}^{2} + 30 (\frac{C_{2}}{C_{1}}) + 15 \Rightarrow 15 x^{2} + 26 x + 15 = 0,$ where x = $\frac{C_{2}}{C_{1}}$

$\Rightarrow x = \frac{- 26 \pm \sqrt[]{676 - 900}}{30}$

Since x cannot be real, so this Question has been dropped by NTA

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5 months ago

Physics Class 11th

A large number of water drops, each of radius r, combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be

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Payal Gupta

Contributor-Level 10

From volume conservation

$n * \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R^{3} = n r^{3} . . . . . . . . . (1)$

Decrease in surface area = $n * 4 π r^{2} - 4 π R^{2}$

$(Δ A) = 4 π [n r^{2} - R^{2}] = 4 π [\frac{n * r^{3}}{r} - R^{2}] = 4 π [\frac{R^{3}}{r} - R^{2}] = 4 π R^{3} [\frac{1}{r} - \frac{1}{R}]$

Energy released (W) = $T * Δ A = 4 π R^{3} T [\frac{1}{r} - \frac{1}{R}]$

Heat produced (Q) = $\frac{W}{J} = \frac{4 π T R^{3}}{J} [\frac{1}{r} - \frac{1}{R}]$

$N o w, Q = m s Δ θ$

$\Rightarrow \frac{4 π R^{3} T}{J} [\frac{1}{r} - \frac{1}{R}] = (\frac{4}{3} π R^{3}) * | x | * Δ θ = \frac{3 T}{J} [\frac{1}{r} - \frac{1}{R}]$

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Physics Class 11th

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R/2) from the earth's centre, where 'R' is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period

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Payal Gupta

Contributor-Level 10

F = m E cos θ

$= m (\frac{G M}{R^{3}} * r) * \frac{x}{r}$ $m a = \frac{m g}{R} * \Rightarrow a = \frac{g}{R} x$

$\Rightarrow T = 2 π \sqrt[]{\frac{R}{g}}$

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Physics Class 11th

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : Body 'P' having mass M moving with speed 'u' has head – on collision elastically with another body 'Q' having mass 'm' initially at rest. If m << M, Body 'Q' will have a maximum speed equal to '2u' after collision.

Reason R : During elastic collision, the momentum and kinetic energy are both conserved. In the light of the above statements, choose the most appropriate answer from the options given below

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Payal Gupta

Contributor-Level 10

Mass m will acquire velocity 2u. Total momentum of system will be conserved but total kinetic energy is not conserved during collision.

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Physics Physics Motion in Plane

A particle is moving with uniform speed along the circumference of a circle of radius of radius R under the action of a central fictitious force F which is inversely proportional to R³. Its time period of revolution will be given by

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Payal Gupta

Contributor-Level 10

$F = \frac{k}{R^{3}} = \frac{m v^{2}}{R} \Rightarrow v = \sqrt[]{\frac{k}{m R^{2}}} = \sqrt[]{\frac{k}{m}} * \frac{1}{R}$

$T = \frac{2 π R}{v} = \frac{2 π R}{\sqrt[]{\frac{k}{m} * (\frac{1}{R})}} \Rightarrow T α R^{2}$

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Physics Units and Measurement

In a typical combustion engine the work done by a gas molecule is given by W = $α^{2} β e^{- \frac{β x^{2}}{k T}},$ where x is the displacement, k is the Boltzmann constant and T is the temperature. If and are constants, dimensions of will be

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Payal Gupta

Contributor-Level 10

$W = α^{2} β e^{- \frac{- β x^{2}}{k T}}$

$[β x^{2}] = [k T] = [M L^{2} T^{- 2}]$

$β = M T^{- 2}$

$[W] = [α^{2} β] \Rightarrow [M L^{2} T^{- 2}] = [α^{2}] [M T^{- 2}] \Rightarrow [α^{2}] = [L^{2}] \Rightarrow [α] = [L]$

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Physics Class 12th

Find the electric field at point P (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length L carrying a charge Q. The distance of the point P from the centre of the rod is a = $\frac{\sqrt[]{3}}{2} L .$

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Payal Gupta

Contributor-Level 10

$E = \frac{λ}{4 π ε_{0} a} [s i n α + s i n α] = \frac{λ s i n α}{2 π ε_{0} a}$

$\Rightarrow E = \frac{Q}{L 2 π ε_{0} * (\frac{\sqrt[]{3} L}{2})} * \frac{1}{2} = \frac{Q}{2 \sqrt[]{3} π ε_{0} L^{2}}$

$t a n α = \frac{\frac{L}{2}}{\sqrt[]{3} L / 2} \Rightarrow α = \frac{π}{6} \Rightarrow s i n α = \frac{1}{2}$

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Physics Class 11th

Find the gravitational force of attraction between the ring and sphere as shown in the diagram where the plane of the ring is perpendicular to the joining the centres. If $\sqrt[]{8} R$ is the distance between the centres of a ring (of mass 'm') and a sphere (mass 'M') where both have equal radius 'R'.

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Payal Gupta

Contributor-Level 10

Method – I:

$F = M_{s p h e r e} E_{r i n g} = M * \frac{G m x}{{(R^{2} + x^{2})}^{\frac{3}{2}}} = M * \frac{G m x \sqrt[]{8} R}{{(R^{2} + 8 R^{2})}^{\frac{3}{2}}} = \frac{\sqrt[]{8} G M m}{27 R^{2}}$

Method – II:

$F = \int d F c o s θ = c o s θ \int (d m) \frac{G M}{(R^{2} + 8 R^{2})} = \frac{\sqrt[]{8} G M m}{27 R^{2}}$

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Physics Physics Ray Optics and Optical Instruments

A short straight object of height 100cm lies before the central axis of a spherical mirror whose focal length has absolute value = 40cm. The image of object produced by the mirror is of height 25cm and has the same orientation of the object. One my conclude from the information

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Payal Gupta

Contributor-Level 10

As height of image is less than height of image and has same orientation as that of object, so mirror must be convex.

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