CLEAR
Toprankers-Info

+91 7676564400

Mobile App

Get it on Google Play
Discussions
Select Date
Tags:

Articles

Discussions
Submit Discussion
deepak mishra

· started a discussion

· 1 Months ago

wrong solution given (Question: 114053.0 )

1m^3=1000000cm^3
then how come 12.25cm^3=12.25*100?????
12.25cm^3=12.25/1000000=0.00001225m^3

Question:

100 persons can sleep in a room 25 m by 9.8 m. If each person requires 12.25 m3 of air then height of the room is:

Options:
A) 5 m
B) 6 m
C)  7 m
D) 8 m
Solution:
Ans(a)

We have,

Length of the room,\(l\)=25m

Breadth of the room,b=9.8m

Area of room,A= \(l\times b=25 \times 9.8 = 245^2\)

Amount of air required by 1 person =\(12.25m^3\)

Amount of air required by 100 persons = 12.25 \(\times 100=1225m^3\)

Volume of air inside room=1225 m\(^3\)

Let h = height of room= 1225

\(=l\times b\times h=1225\)

\(=A\times h=1225\)

\(=245h=1225 \)

\(h=\cfrac{1225}{245}=5m\)


हमारे पास है,

कमरे की लंबाई, l = 25 मी

कमरे की चौड़ाई, b = 9.8 मी

कमरे का क्षेत्रफल, A = l × b = 25 × 9.8 = 2452

1 व्यक्ति = 12.25m3 द्वारा आवश्यक हवा की मात्रा

100 व्यक्तियों द्वारा आवश्यक वायु की मात्रा = 12.25 × 100 = 1225 मी 3

कमरे के अंदर हवा की मात्रा = 1225 मी 3

माना h = कमरे की ऊंचाई = 1225

= l × b × h = 1225

= a × h = 1225

= 245h = 1225

h = 1225/245 = 5m

Knowledge Expert

· commented

· 1 Months ago

Dear Student,

We fixed the ambiguity of this question. it is m^3 in question.

Keep learning with us!!
Best wishes
Team Toprankers
-logo
We at aim to provide most comprehensive content & test for practice which is carefully divided into various chapters and topics to help you focus on your weak areas.Our Practice mock test gives you unmatched analytics to help you realize your strong areas,time management and improvement areas.Our Short & Crisp notes on each topic will help you to quickly revise the topic along with the tips & tricks of the exams.
More About
-Info
-Info

Mobile App

Get it on Google Play
About Us Campus Ambassador Careers Contact Us Refund Policy

All Rights Reserved Top Rankers

User Policy
Disclaimer
Terms & Conditions