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## Kumar Ketan is teaching live on Unacademy Plus

Kumar Ketan
#Specialisation in Electronics #6 years teaching experience #Physics Faculty in Kota #"magical education" (211k+ subs) #Writer

U
sir, please complete whole part of transistor. i am waiting
DEEPAK KUMAR
7 months ago
Complete Analog Electronics will be completed by the End of this month.
Anil Nagar
7 months ago
Thanks sir
sir ek hee question (1st wala ) hua last wala question kaafi achaw tha
Kumar Ketan
8 months ago
Don't worry, keep practicing and try to solve more and more this type of questions
only one correct . but understood everything .
all correct

1. unacademy 62.9k views 4.8 Kumar Ketan Follow me on the Unacademy #15 Educator in IIT JEE #6 years teaching experience #IIT JEE Physics #YouTuber 'magical education" (170k+ subs) #Unacademy educator (60k+ lifetin eviews) 3k 2 37 Followers Following Courses Get updates about new courses . Watch all my lessons . Download slides and watch offline Message Lists (1) Kumar Ketan HINDI IT JEE Physics by Kumar Ketan for IT JEE 52 saves Kumar Ketan

2. When a dc voltage of 200 V is applied to a coil of self inductance (2-/3/ ) H, a current of 1 A flows through it. But by replacing dc source with ac source of 200 V, the current in the coil is reduced to 0.5 A. Then the frequency of ac supply is (a) 100 Hz (c) 60 Hz (b) 75 Hz (d) 50 Hz

3. Resistance ofcoil, R =-00 = 2 1 A With ac source, I = R2X2 200 R2 + XL or 0.5 2 or R (2TTUL)2 (400)2 2 2V3 = (400)--(200) 200 600 or 4V3u = 2V3 100 or u = 50 Hz.

4. Half-life of a radioactive substance is 20 minute. The time between 20% and 80% decay will be (a) 20 min (c) 40 min (b) 30 min (d) 25 min

5. According to radioactive decay N Noe t where, Number of radioactive nuclei present in the sample at t 0 No N Number of radioactive nuclei left undecayed after time t decay constant For 20% decay 80N 0 100

6. For 80% decay 20N 100 0 Dividing equation (i) by (ii), we get 4 eAti 2) Taking natural logarithms of both sides, we geu In 4 = (t2-ti) In 2 1/2 t2-ti 2 T1/2 = 2 20 min = 40 min

7. A juggler throws balls into air. He throws one whenever the previous one is at its highest point. If he throws n balls each second, the height to which each ball will rise is 2g (d 2 (a) & (b)g (c) 28 (d) & 4n 2n2 n2

8. Time taken by each ball to reach highest point, As the juggler throws the second ball, when the first ball is at its highest point, so v 0 Using, v u at, we get or ( 0 u + (-g)