#Class11 #Physics #NCERT #Problem #Solutions #JEEMAINS #CBSE #infinityvision #JEEADVANCE
A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of 45o with the vertical. Then F equals.
(Take g = 10 ms-2 and the rope to be mass less)
(a) 90 N (b) 75 N (c) 70 N (d) 100 N
[JEE (Main)-7th Jan 2020-Shift-1]
IITJEE PHYSICS QUESTION BANK PLAYLIST LINK :
[ Ссылка ]
---------------------------------------------------------------------------------------------------------------------------------------------------------------Our services which are FREE for all | Only serious students apply:
• Take FAST & avail up to 100% Scholarship | Register by calling at 8010996622
• Apply for TEST SERIES for PRAYAAS Crash Course JEE Main | KVPY | NTSE and CBSE (Class 8 to 10) by calling at 8010996622
For any sort of assistance or support, please call our Student Care number: 8010996622 or enquire at: www.infinityvision.in
---------------------------------------------------------------------------------------------------------------------------------------------------------------
Infinity Vision is a premier organization for competitive exam preparation. With us, students prepare for the toughest of examinations such as IIT – JEE, NEET, KVPY, NTSE, IMO, etc. & pave the way towards a successful career. We offer intensive Foundation Courses for Class 6th to 10th which is ideal for the preparation of NTSE, KVPY, IMO, NSEJS, INO, RIMC, and for different school level competitions and Olympiads. For Class 11th & 12th, we provide an extensive online/classroom program to prepare for IIT – JEE/NEET, KVPY & Board Exams.
Like us on Facebook: [ Ссылка ]
Follow us on Instagram: [ Ссылка ]
Subscribe to our YouTube Channel for regular updates : [ Ссылка ]
Website: www.infinityvision.in
Ещё видео!