Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
80 views
in Physics by (76.0k points)
closed by
Two blocks of masses `1 kg` and `2 kg` are connected by a metal wire goijng over a smooth pulley as shown in figure.
The breaking stress of the metal is `(40//3pi)xx10^(6)N//m^(2)`. If `g=10ms^(-12)`, then what should be the minimum radlus of the wire used if it is not to break?
image
A. `0.5mm`
B. `1mm`
C. `1.5mm`
D. `2mm`

1 Answer

0 votes
by (87.4k points)
selected by
 
Best answer
Correct Answer - B
`T=(2m_(1)m_(2))/(m_(1)+m_(2))g=(2xx1xx2)/(1+2)xx10N=40/3N`
If `r` is the minimum radius, then
Breaking stress `=(40/3)/(pir^(2))` or `40/(3pi)xx10^(6)=40/(3pir^(2))`
or `r^(2)=1/(10^(6))` or `r=1/(10^(3))m`
or `r=1/(10^(3))xx10^(3)m=1mm`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...