JEE Main 2018 (Online) 15th April Evening Slot | Alternating Current and Electromagnetic Induction Question 179 | Physics

JEE Main 2018 (Online) 15th April Evening Slot

MCQ (Single Correct Answer)

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JEE Main 2018 (Online) 15th April Evening Slot Physics - Alternating Current and Electromagnetic Induction Question 186 English

A copper rod of mass m slides under gravity on two smooth parallel rails, with separation l and set at an angle of $$\theta $$ with the horizontal. At the bottom rails are joined by a resistance R. There is a uniform magnetic field B normal to the plane of the rails, as shown in the igure. The terminal speed of the copper rod is :

$${{mg\,R\,\tan \,\theta } \over {{B^2}\,{l^2}}}$$

$${{mg\,R\,\cot \,\theta } \over {{B^2}\,{l^2}}}$$

$${{mg\,R\,\sin \,\theta } \over {{B^2}\,{l^2}}}$$

$${{mg\,R\,\cos \,\theta } \over {{B^2}\,{l^2}}}$$

JEE Main 2018 (Online) 15th April Evening Slot

MCQ (Single Correct Answer)

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At the center of a fixed large circular coil of radius R, a much smaller circular coil of radius r is placed. The two coils are concentric and are in the same plane. The larger coil carries a current I. The smaller coil is set to rotate with a constant angular velocity $$\omega $$ about an axis along their common diameter. Calculate the emf induced in their smaller coil after a time t of its start of rotation.

$${{{\mu _o}{\rm I}} \over {2\,R}}$$ $$\omega $$ $$\pi $$ r² sin$$\omega $$ t

$${{{\mu _o}{\rm I}} \over {4\,R}}$$ $$\omega $$ $$\pi $$ r² sin$$\omega $$ t

$${{{\mu _o}{\rm I}} \over {4\,R}}$$ $$\omega $$ r² sin$$\omega $$ t

$${{{\mu _o}{\rm I}} \over {2\,R}}$$ $$\omega $$ r² sin$$\omega $$ t

JEE Main 2018 (Online) 15th April Evening Slot

MCQ (Single Correct Answer)

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A plane polarized monochromatic EM wave is traveling in vacuum along z direction such that at t = t₁ it is found that the electric field is zero at a spatial point z₁. The next zero that occurs in its neighbourhood is at z₂. The frequency of the electroagnetic wave is :

$${{3 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$$

$${{1.5 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$$

$${{6 \times {{10}^8}} \over {\left| {{z_2} - {z_1}} \right|}}$$

$${1 \over {{t_1} + {{\left| {{z_2} - {z_1}} \right|} \over {3 \times {{10}^8}}}}}$$

JEE Main 2018 (Online) 15th April Morning Slot

MCQ (Single Correct Answer)

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A monochromatic beam of light has a frequency $$v = {3 \over {2\pi }} \times {10^{12}}Hz$$ and is propagating along the direction $${{\widehat i + \widehat j} \over {\sqrt 2 }}.$$
It is polarized along the $$\widehat k$$ direction. The acceptable form for the magnetic field is :