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10 months ago

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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New answer posted

10 months ago

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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New answer posted

11 months ago

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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New answer posted

11 months ago

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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New answer posted

11 months ago

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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New answer posted

11 months ago

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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New answer posted

11 months ago

Physics NCERT Solutions

Which is the best conducting material under room temperature?

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Piyush Vimal

Beginner-Level 5

The best conductor of electricity is Silver (Ag) at room temperature. It is not used in normal wiring even though being the best conductor because of its high cost.

Electrical Conductivity ( $\sigma$ ): $6.3 \times 10^ {7} \ \text {S/m}$
Electrical Resistivity ( $\rho$ ): $1.59 \times 10^ {-8} \ \Omega \cdot \text {m}$ (the lowest among all metals)

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11 months ago

Physics Work, Energy and Power

Two solids A and B of mass 1kg and 2kg respectively are moving with equal linear momentum. The ratio of their kinetic energies will be so the value of A ${(K . E)}_{A} : {(K . E)}_{B}$ will be $\frac{A}{1},$ …………………….

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Vishal Baghel

Contributor-Level 10

A -> m_A = 1 kg, P_A = P

B -> m_B = 2 kg, P_B = P

$? k E = \frac{P^{2}}{2 m}$

$∴ \frac{k E_{A}}{k E_{B}} = \frac{(\frac{P^{2}}{2 m A})}{(\frac{P^{2}}{2 m_{B}})} = \frac{m_{B}}{m_{A}} = \frac{2}{1}$

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New answer posted

11 months ago

Physics Class 11th

Two cars are approaching each other at an equal speed of 7.2 km/hr. When they see each other, both blow horns of 676Hz. The beat frequency heard by each driver will be…………………. Hz.

[Velocity of sound in air is 340 m/s]

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Vishal Baghel

Contributor-Level 10

$7.2 * \frac{5}{18} = 2 m / s$ [V_A = V_B = 2m/s]

$f'_{B} = f_{A} (\frac{v + v_{B}}{v - v_{A}})$

$↓$

App freq heard by car B

$f'_{B} = 676 [\frac{340 + 2}{340 - 2}] = 676 * \frac{342}{338} = 684 H z$

Similarly, $f'_{A} = f_{B} (\frac{v + v_{B}}{v - v_{B}}) = 684 H z$

$∴$ Beat frequency heard by both = 684 – 676 = 8Hz

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New answer posted

11 months ago

Physics Class 12th

A cylindrical wire of radius 0.5mm and conductivity 5 * 10⁷ S/m is subjected to an electric field of 10m V/m. The expected value of current in the wire will be x³πmA. The value of x is………………….

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Vishal Baghel

Contributor-Level 10

radius r = 0.5 mm

$σ = 5 * 1 0^{7} s / m$

$\vec{E} = 10 * 1 0^{- 3} V / m = 1 0^{- 2} V / m$

As we know

$J = σ E$

$J = \frac{i}{A} = σ E$

$i = σ E A = 5 * 1 0^{7} * 1 0^{- 2} * π * {(0.5 * 1 0^{- 3})}^{2} = 1.25 π * 1 0^{- 1} A = 125 π m A = x^{3} π m A$

$∴$ x = 5

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