The magnetic field in a plane electromagnetic wave is $${\mathrm{B}}_{\mathrm{y}}=\left(3.5\times 10^{-7}\right)\sin \left(1.5\times 10^{3}x+0.5\times 10^{11}t\right)\mathrm{T}$$. The corresponding electric field will be :

A plane EM wave is propagating along $$x$$ direction. It has a wavelength of $$4\mathrm{mm}$$. If electric field is in $$y$$ direction with the maximum magnitude of $$60{\mathrm{Vm}}^{-1}$$, the equation for magnetic field is :

The electric field in an electromagnetic wave is given by $$\overrightarrow{\mathrm{E}}=\hat{i}40\cos \omega (\mathrm{t}-z/\mathrm{c}){\mathrm{NC}}^{-1}$$. The magnetic field induction of this wave is (in SI unit) :

Arrange the following in the ascending order of wavelength: A. Gamma rays $$\left({\lambda }_1\right)$$ B. $$x$$ - rays $$\left({\lambda }_2\right)$$ C. Infrared waves $$\left({\lambda }_3\right)$$ D. Microwaves $$\left({\lambda }_4\right)$$ Choose the most appropriate answer from the options given below

Average force exerted on a non-reflecting surface at normal incidence is $$2.4\times 10^{-4}\mathrm{N}$$. If $$360\mathrm{W}/{\mathrm{cm}}^2$$ is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:

Match List I with List II : .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top} LIST I EM-Wave LIST II Wavelength Range A. Infra-red I. $$<10^{-3}$$ nm B. Ultraviolet II. 400 nm to 1 nm C. X-rays III. 1 mm to 700 nm D. Gamma rays IV. 1 nm to $$10^{-3}$$ nm Choose the correct answer from the options given below :

In the given electromagnetic wave $${\mathrm{E}}_{\mathrm{y}}=600\sin (\omega t-\mathrm{kx}){\mathrm{Vm}}^{-1}$$, intensity of the associated light beam is (in $$\mathrm{W}/{\mathrm{m}}^2$$ : (Given $${ϵ}_0=9\times 10^{-12}{\mathrm{C}}^2{\mathrm{N}}^{-1}{\mathrm{m}}^{-2}$$ )

Electromagnetic waves travel in a medium with speed of $$1.5\times 10^8\mathrm{m}{\mathrm{s}}^{-1}$$. The relative permeability of the medium is 2.0. The relative permittivity will be:

A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle $$(\theta )$$ of thread deflection in the extreme position will be :

A bullet is fired into a fixed target looses one third of its velocity after travelling $$4\mathrm{cm}$$. It penetrates further $$\mathrm{D}\times 10^{-3}\mathrm{m}$$ before coming to rest. The value of $$\mathrm{D}$$ is :

A body of mass $$2\mathrm{kg}$$ begins to move under the action of a time dependent force given by $$\vec{F}=\left(6t\hat{i}+6t^2\hat{j}\right)N$$. The power developed by the force at the time $$t$$ is given by:

A block of mass $$100\mathrm{kg}$$ slides over a distance of $$10\mathrm{m}$$ on a horizontal surface. If the co-efficient of friction between the surfaces is 0.4, then the work done against friction $$(\mathrm{in}J$$) is :

The potential energy function (in $$J$$ ) of a particle in a region of space is given as $$U=\left(2x^2+3y^3+2z\right)$$. Here $$x,y$$ and $$z$$ are in meter. The magnitude of $$x$$-component of force (in $$N$$ ) acting on the particle at point $$P(1,2,3)\mathrm{m}$$ is :

A bob of mass '$$m$$' is suspended by a light string of length '$$L$$'. It is imparted a minimum horizontal velocity at the lowest point $$A$$ such that it just completes half circle reaching the top most position B. The ratio of kinetic energies $$\frac{(K.E{)}_A}{(K.E{)}_B}$$ is :

A block of mass $$1\mathrm{kg}$$ is pushed up a surface inclined to horizontal at an angle of $$60^{\circ }$$ by a force of $$10\mathrm{N}$$ parallel to the inclined surface as shown in figure. When the block is pushed up by $$10\mathrm{m}$$ along inclined surface, the work done against frictional force is : $$\left[g=10\mathrm{m}/{\mathrm{s}}^2\right]$$

A particle is placed at the point $$A$$ of a frictionless track $$ABC$$ as shown in figure. It is gently pushed towards right. The speed of the particle when it reaches the point B is : (Take $$g=10\mathrm{m}/{\mathrm{s}}^2$$).

A particle of mass $$m$$ moves on a straight line with its velocity increasing with distance according to the equation $$v=\alpha \sqrt{x}$$, where $$\alpha$$ is a constant. The total work done by all the forces applied on the particle during its displacement from $$x=0$$ to $$x=\mathrm{d}$$, will be :

If a rubber ball falls from a height $$h$$ and rebounds upto the height of $$h/2$$. The percentage loss of total energy of the initial system as well as velocity ball before it strikes the ground, respectively, are :

A body of $$m\mathrm{kg}$$ slides from rest along the curve of vertical circle from point $$A$$ to $$B$$ in friction less path. The velocity of the body at $$B$$ is: (given, $$R=14\mathrm{m},g=10\mathrm{m}/{\mathrm{s}}^2$$ and $$\sqrt{2}=1.4$$)

A block is simply released from the top of an inclined plane as shown in the figure above. The maximum compression in the spring when the block hits the spring is :