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Questions : 4, 5, and 6
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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

Unacademy user
great work sir.. keep it up..and sir please make videos on ncert books matter for all require subject. ...
Kumar Ketan
6 months ago
Get plus subscription for daily live sessions and you can get a lot of live sessions from me daily. You can give your number I'll text you and explain about the discounts
2nd one wrong or else correct . please make more of such courses . awesome course.
Kumar Ketan
9 months ago
OK, to beta aapke jo question glt Huye aap us pr at least 8-10 question practice kriye... Aur sir ise Fir se solve kijiye.... As I said, in next 4 days you'll get 80 lessons back to back. Every hour lessons are publishing
Aswini Kumar Sahoo
9 months ago
sir can u please provide Ur no. for contact ??
Kumar Ketan
9 months ago
we can't provide our personal details.. but you can find me if you want. just try and use your mind
sir iss Baar toh teno Shi gye thnku sir
Kumar Ketan
9 months ago
That's great

  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. Que. Water from a tap emerges vertically downwards with an initial speed of 1 m s The cross-sectional area of the tap is 10 4 m2 Assuming pressure to be constant through out the stream of water and flow to be steady, the cross-sectional area 0.15 m below the tap is (a) 10-4 m2 (c) 0.5 x 10-4 m2 (b) 105 m2 (d) 0.2 x 10 4 m2

  3. By equation of continuity, Alvi = A2V2 or V,- ! 2 1/ 2 10-4 ><! or A2 2 VI+2 10 0.15 +2gh = 0.5 10-4 m2

  4. One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant k. A mass m hangs freely from the free end of the spring. The area of cross-section and the Young's modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period T equal to Que. m(YA + kL) YAk (a) 2 (b) 2 mYA N kL (d) 2rt, ml (c) 2 VYA

  5. For two springs in series, the equivalent force constant is kea-2. Here, ki-k and k1 + k2 YA 2 kYA kYA k =-L- eg YA kL YA m(YA kL) V YAk .. . T=2

  6. Que. The speed of light (c), acceleration due to gravity (g) and pressure (P) are taken as fundamental units, the dimensions of gravitational constant (G) are (a) Ic gP-3] (c) [cogP-] (b) c2gp-2] (d) c2gp-2]

  7. Let G = kcxopz -1r 21z Equating both sides, we get, On solving, x = 0 y = 2, z =-1. Thus, [G] = [cog2P-1]