1
JEE Main 2017 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is :

JEE Main 2017 (Online) 8th April Morning Slot Physics - Rotational Motion Question 167 English
A
$${{2\,\,mg} \over {2\,m + M}}$$
B
$${{2\,\,Mg} \over {2\,m + M}}$$
C
$${{2\,\,mg} \over {2\,M + m}}$$
D
$${{2\,\,Mg} \over {2\,M + M}}$$
2
JEE Main 2017 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current I = Io cos ($$\omega $$t). The emf induced in the smaller inner loop is nearly :
A
$${{\pi {\mu _o}{I_o}} \over 2}.{{{a^2}} \over b}\,\omega \sin \left( {\omega t} \right)$$
B
$${{\pi {\mu _o}{I_o}} \over 2}.{{{a^2}} \over b}\,\omega \cos \left( {\omega t} \right)$$
C
$$\pi {\mu _o}{I_o}\,{{{a^2}} \over b}\omega \sin \left( {\omega t} \right)$$
D
$${{\pi {\mu _o}{I_o}\,{b^2}} \over a}\omega \cos \left( {\omega t} \right)$$
3
JEE Main 2017 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
What is the conductivity of a semiconductor sale having electron concentration of $$5 \times {10^{18}}\,\,{m^{ - 3}},$$ hole concentration of $$5 \times {10^{19}}\,\,{m^{ - 3}},$$ electron mobility of 2.0 m2 V$$-$$1 s-1 and hole mobility of 0.01 m2 V$$-$$1 s$$-$$1 ?

(Take charge of electronas 1.6 $$ \times $$ 10 $$-$$19 c)
A
1.68 ($$\Omega $$-m)$$-$$1
B
1.83 ($$\Omega $$-m)$$-$$1
C
0.59 ($$\Omega $$-m)$$-$$1
D
1.20 ($$\Omega $$-m)$$-$$1
4
JEE Main 2017 (Online) 8th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Magnetic field in a plane electromagnetic wave is given by

$$\overrightarrow B $$ = B0 sin (k x + $$\omega $$t) $$\widehat j\,T$$

Expression for corresponding electric field will be :
Where c is speed of light.
A
$$\overrightarrow E $$ = B0 c sin (k x + $$\omega $$t) $$\widehat k$$ V/m
B
$$\overrightarrow E $$ = $${{{B_0}} \over c}$$ sin (k x + $$\omega $$t) $$\widehat k$$ V/m
C
$$\overrightarrow E $$ = $$-$$ B0 c sin (kx +$$\omega $$t) $$\widehat k$$ V/m
D
$$\overrightarrow E $$ = B0 c sin (kx $$-$$$$\omega $$t) $$\widehat k$$ V/m
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