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1

JEE Main 2021 (Online) 27th August Morning Shift

Numerical
A uniform conducting wire of length is 24a, and resistance R is wound up as a current carrying coil in the shape of an equilateral triangle of side 'a' and then in the form of a square of side 'a'. The coil is connected to a voltage source V0. The ratio of magnetic moment of the coils in case of equilateral triangle to that for square is 1 : $$\sqrt y $$ where y is ................. .
Your Input ________

Answer

Correct Answer is 3

Explanation

In triangle shape $${N_t} = {{24a} \over {3a}} = 8$$

In square $${N_s} = {{24a} \over {4a}} = 6$$

$${{{M_t}} \over {{M_3}}} = {{{N_t}I{A_t}} \over {{N_s}I{A_s}}}$$ [I will be same in both]

$$ = {{8 \times {{\sqrt 3 } \over 4} \times {a^2}} \over {6 \times {a^2}}}$$

$${{{M_t}} \over {{M_s}}} = {1 \over {\sqrt 3 }}$$

y = 3
2

JEE Main 2021 (Online) 26th August Evening Shift

Numerical
A coil in the shape of an equilateral triangle of side 10 cm lies in a vertical plane between the pole pieces of permanent magnet producing a horizontal magnetic field 20 mT. The torque acting on the coil when a current of 0.2 A is passed through it and its plane becomes parallel to the magnetic field will be $$\sqrt x $$ $$\times$$ 10$$-$$5 Nm. The value of x is .................
Your Input ________

Answer

Correct Answer is 3

Explanation



$$\overrightarrow \tau = \overrightarrow M \times \overrightarrow B = MB\sin 90^\circ $$

$$ = MB = {{i\sqrt 3 {l^2}} \over 4}B$$

$$ = \sqrt 3 \times {10^{ - 5}}$$ N $$-$$ m
3

JEE Main 2021 (Online) 26th August Evening Shift

Numerical
A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s$$-$$1 in a uniform horizontal magnetic field of 3.0 $$\times$$ 10$$-$$2 T. The maximum emf induced the coil will be ................. $$\times$$ 10$$-$$2 volt (rounded off to the nearest integer)
Your Input ________

Answer

Correct Answer is 60

Explanation

Maximum emf $$\varepsilon = N\,\omega AB$$

N = 20, $$\omega$$ = 50, B = 3 $$\times$$ 10$$-$$2 T

$$\varepsilon $$ = 20 $$\times$$ 50 $$\times$$ $$\pi$$ $$\times$$ (0.08)2 $$\times$$ 3 $$\times$$ 10$$-$$2 = 60.28 $$\times$$ 10$$-$$2

Rounded off to nearest integer = 60
4

JEE Main 2021 (Online) 26th August Morning Shift

Numerical
Two short magnetic dipoles m1 and m2 each having magnetic moment of 1 Am2 are placed at point O and P respectively. The distance between OP is 1 meter. The torque experienced by the magnetic dipole m2 due to the presence of m1 is ........... $$\times$$ 10$$-$$7 Nm.

Your Input ________

Answer

Correct Answer is 1

Explanation

$$\overrightarrow \tau = \overrightarrow {{M_2}} \times \overrightarrow {{B_1}} $$

$$\tau = {M_2}{B_1}\sin 90^\circ $$

$$ = 1 \times {{{\mu _0}} \over {4\pi }}{{{M_1}} \over {{{(1)}^3}}}1$$

= 10$$-$$7 N.m

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