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
167 views
in Physics by (86.4k points)
closed by
A steel rod of cross-sectional area `16 cm^(2)` and two brass rods each of cross-sectional area `10 cm^(2)` together support a load of `5000 kg ` as shown in the figure. ( Given, `Y_(steel) = 2xx10^(6) kg cm^(-2) and Y_(brass) = 10 ^(6) kg cm^(-2))`. Choose the correct option(s).
image
A. `120,161`
B. `161,120`
C. `120,140`
D. `141,120`

1 Answer

0 votes
by (85.7k points)
selected by
 
Best answer
Correct Answer - A
`A_(S) = 16cm^(2), A_(B) = 2xx10 = 20cm^(2)`
Load, `F = 5000kg`
`Y_(S) = 2.0xx10^(6)kg//cm^(2), Y_(B) = 1.0xx10^(6) kg//cm^(2)`
`l_(S) = 30cm` and `l_(B) = 20cm`
Let `sigma_(S)` = stress in steel and `sigma_(B)` = stress in brass decreases in length of steel rod = decreases in length of brass rod
`(sigma_(S))/(Y_(S)) xx l_(S) = (sigma_(B))/(Y_(B)) xx l_(B)`
`rArr sigma_(S) = (Y_(S))/(Y_(B)) xx (l_(B))/(l_(S)) xx sigma_(B).....(i)`
now using the relation `F = sigma_(S) A_(S) + sigma_(B) A_(B)`
`5000 = sigma_(S) xx 16 + sigma_(B) xx20.........(ii)`
solving equ.(i) and (ii) and we get
`sigma_(B) = 120.9kg//cm^(2)` and `sigma_(S) = 161.2kg//cm^(2)`

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

...